TECHTIPS

Mathematical & Logical Puzzles -A7

noreply@blogger.com (Nisheeth Ranjan) — Wed, 18 Aug 2010 13:59:00 +0530

25. Another Escalator Puzzle

Let the number of visible steps be n and the little boy move up one step in unit time. Therefore n - 50 steps disappear in 50 units of time. As the bigger boy travels three times the speed of his little brother, n - 75 steps disappear in 75 ¸ 3 units of time, or 25 units of time.

Therefore: n - 50 = n - 75

50 25

From this equation, n is readily deduced as 100. Therefore the escalator had 100 steps.

26. The Ladder Against The Wall

Call AD = a and EC = b.

Since triangle ADF and triangle FEC are similar then:

a = 4 or ab = 16...............................................................................................(1)

4 b

AB = a + 4 BC = b + 4.

Therefore by Pythgoras’ Theorem:

AB² + BC² = AC² or (a + 4)² + (b + 4)² = 400.

Expanding and collecting terms together:

a² + b² + 8(a + b) + 32 = 400..............................................................................(2)

Now from (1) 32 = 2ab

Substituting in (2):

a² + b² + 8(a + b) + 2ab = 400...........................................................................(3)

Now: a² + 2ab + b² = (a + b)²

So substituting in (3) we get:

(a + b)² + 8(a + b) = 400

Add 16 to both sides:

(a + b)² + 8(a + b) + 16 = 416

Or: (a + b + 4)² = 416

So: a + b + 4 = Ö416..........................................................................................(4)

From (1) b = 16

Substituting in (4):

a + 16 + 4 = Ö416

By using the formula for solution of quadratic equations, a is found to be 15.36 metres or 1.04 metres. This makes AB = 19.36 metres or 5.04 metres.

27. A Coin Puzzle

The rearrangement from Figure 1 to Figure 2 can be effected in four moves.

1. Move C and D to the right leaving room for another 20c and another 10c between E and C.

2. Move A and B to the right so that A touches D.

3. Move D and A into the gap between E and C.

4. Move E and D into the gap between C and B.

28. The Dilemma Of A Racecourse Owner

The racecourse owner needs six sets of equipment set out as in the diagram. A race between any two sets will be a different multiple of 1000 metres.

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Mathematical & Logical Puzzles -Q7

noreply@blogger.com (Nisheeth Ranjan) — Wed, 18 Aug 2010 12:53:00 +0530

25. Another Escalator Puzzle

There are very long escalators in some New York subway stations. You don’t have to climb them since the moving steps will do the job for you. However, two brothers have to get to a baseball game and are in a hurry, and so they run up the moving steps, adding their speed to that of the escalator.

The taller boy climbs three times as quickly as his little brother and whilst he runs up he counts 75 steps. The little one counts only 50 steps. How many steps has the visible part of this New York escalator?

26. The Ladder Against The Wall

AB is a wall at right angles to the ground BC. DFEB is a square box 4 feet on a side. AFC is a ladder 20 feet long whose extreme ends touch the wall and the ground, and whose length rests on the corner of the box. How far up the wall does the ladder reach? (i.e. What is the length of AB?)

27. A Coin Puzzle

The problem is to change the arrangement of twenty-cent and ten-cent coins shown in Figure 1 to that shown in Figure 2, in the smallest number of moves.

The only move allowed consists of sliding a pair of touching coins, which must be one twenty-cent coin and one ten-cent coin, from one point on the imaginary base line (which of course is the same in both drawings) to another. The coins must remain in contact throughout the move and may not be rotated during the move to reverse the order of the coins moved. Gaps are allowed at the end of every move except the last.

28. The Dilemma Of A Racecourse Owner

A racecourse owner wished to install photoelectric timing devices on his racecourse, such that any distance between 1000 metres and 31000 metres (in multiples of 1000 metres) could be timed. The photoelectric devices were extremely expensive and so he wanted to purchase the fewest possible.

It must be possible, by choosing any two sets of these devices, to time one of the desired distances between them. What is least number that he requires, and how would they be set out? Assume for the purposes of this puzzle that the racecourse is a circular one.

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Mathematical & Logical Puzzles -A6

noreply@blogger.com (Nisheeth Ranjan) — Wed, 18 Aug 2010 12:36:00 +0530

21. The Knight’s Tour (1)

1	10	35	28	3	12
18	27	2	11	34	29
9	36	17	30	13	4
26	19	8	33	22	31
7	16	21	24	5	14
20	25	6	15	32	23

There are numerous solutions to this problem. The solution given here is just one of them to prove to those who gave up that it is possible. The numbers indicate the order of the moves starting at 1 and finishing at 36. Note that the solution given here is a circular tour where the knight finishes on a cell exactly a knight’s move from where it started.

22. The Knight’s Tour (2)

1	14	63	40	3	24	57	42
62	39	2	13	60	41	4	23
15	64	61	36	25	56	43	58
38	33	12	55	52	59	22	5
11	16	37	26	35	54	51	44
32	29	34	53	50	47	6	21
17	10	27	30	19	8	45	48
28	31	18	9	46	49	20	7

The numbers show the 63 moves in numerical order. There are many more solutions to this problem. The solution shown here is a circular tour where the knight finishes on a cell exactly a knight’s move away from where it started.

23. Alice In Wonderland

The Jub-Jub lives in a Tum-Tum tree, the Bandersnatch is an animal, and the Jabberwock has four wings and feeds on Toves. This problem yields readily to a little trial and error.

24. The Spider And The Fly

The easiest way to solve this puzzle is to imagine the room as a carboard box and to open it out flat as shown in the diagram above. The shortest walking path for the fly to be reached is exactly 40 metres. The spider crawls along the straight dotted line labelled 40m in the diagram of the unfolded room above.

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Mathematical & Logical Puzzles -Q6

noreply@blogger.com (Nisheeth Ranjan) — Wed, 18 Aug 2010 12:27:00 +0530

21. The Knight’s Tour (1)

Place a knight on a 6 by 6 chess board (i.e. an ordinary chess board which is 2 squares short each way). The knight may be placed on any square. The problem is to visit each square once and once only by taking ordinary knight’s chess moves.

22. The Knight’s Tour (2)

On an ordinary 8 by 8 chess board, a knight stands in the top left-hand corner square. The problem is to visit each square once and once only using the knight’s chess move.

23. Alice In Wonderland

“Oh, dear,” said Alice mournfully, “I wish I didn’t have to go into the Wabe, for that is where the Jub-Jub lives.”

“Nonsense!”, exclaimed the Mad Hatter. “It lives in a Tum-Tum tree.” And addressing the March Hare, “Doesn’t it?”

“Of course it does,” agreed the March Hare. “Everybody knows that, don’t you agree?” (Here he prodded the Dormouse who was asleep on his shoulder).

The Dormouse opened one sleepy eye and wearily muttered, “Yes, yes of course,” and promptly went back to sleep again.

“Talking of birds,” said the March Hare, “the Bandersnatch is exceptionally shy this year.”

“But I’m sure the Bandersnatch is an animal,” said Alice.

“My, you are ignorant,” said the Mad Hatter, “of course it’s a bird.”

The Dormouse, who appeared to be sleeping was consulted once more and he wearily announced, “I agree with Alice.” Then he returned to snoring once again.

“The Jabberwock,” announced the Mad Hatter, “is an interesting animal. It has four wings and feeds on Borogoves.”

“It has four wings,” agreed Alice, “but surely it feeds on Raths.”

“Poppycock!” snorted the March Hare, disturbing the poor Dormouse, who woke up suddenly and appeared to be at least half-listening for a change. “It has five wings and feeds on Toves.”

“Five wings and feeds on Raths,” corrected the Dormouse, who found staying awake much too much effort, especially when no-one seemed to know what he was talking about.

At this point in the proceedings, the Cheshire Cat appeared, sitting on the branch of a near-by Tum-Tum tree. He grinned and informed them that of the four zoological facts that each had volunteered, two each were true and two each were false. Whereupon he disappeared until only the grin remained, and this slowly faded, starting from the corners and ending in the middle. Which still leaves the following unanswered: Where does the Jub-Jub live? What is a Bandersnatch? And how many wings has a Jabberwock and what does it feed on?

24. The Spider And The Fly

In a rectangular room 30 metres by 12 metres by 12 metres, a spider is in the middle of one end wall and one metre from the ceiling.

A fly is in the middle of the opposite end wall and one metre from the floor. It is too paralysed with fear to move.

What is the shortest distance that the spider must crawl across the surfaces of the room in order to reach the fly?

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Mathematical & Logical Puzzles -Q5

noreply@blogger.com (Nisheeth Ranjan) — Tue, 27 Jul 2010 17:37:00 +0530

17. Missionaries and Cannibals

Three missionaries and three cannibals have to cross a river with the aid of a two-seater row-boat. Since the man-eating savages are ravenously hungry, it is imperative that on neither shore they are ever in the majority. The priests’ difficulties are further aggravated by the fact that, of the three savages only one knows how to row, and of the three priests only one knows how to row. How does this uneasy company get across the river?

18. Road Safety Addition

H A L T

A T

M A J O R

R O A D

A H E A D

The figure represents an addition sum where the digits have been replaced by letters. There are ten letters (A, D, E, H, J, L, M, O, R, T) each representing one of the digits 0 through 9 but not necessarily in that order. What values can you give the letters to make the sum total correctly?

19. The Lonesome Eight

x x 8 x x

x x x ) x x x x x x x x

x x x

x x x x

x x x

x x x x

The figure represents a long division, Xs representing unknown digits. Fill in these unknown digits making the division correct. There is only one possible solution.

Note: This problem is a lot easier that it at first appears, yielding readily to a few elementary insights.

20. The Farmer And The Hundred Dollars

A farmer went to market to buy some animals. Cows, sheep and pigs must all be represented in his purchase. Cows sold at $5 each, sheep at $1 each and pigs at $1 for 20. The farmer has $100 to spend and he wants to buy 100 animals. How many of each did he buy?

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Mathematical & Logical Puzzles -A5

noreply@blogger.com (Nisheeth Ranjan) — Tue, 27 Jul 2010 17:33:00 +0530

17. Missionaries and Cannibals

M = misionary C = cannibal Bracketed letters are the ones who can row.

Left bank	River	Right bank
MM(M)CC(C)	-	-
MMC(C)	(M)C ®	-
MMC(C)	¬ (M)	C
MM(M)	(C)C ®	C
MM(M)	¬ (C)	CC
M(C)	(M)M ®	CC
M(C)	¬ (M)C	MC
MC	(M)(C) ®	MC
MC	¬ (M)C	M(C)
CC	(M)M ®	M(C)
CC	¬ (C)	MM(M)
C	(C)C ®	MM(M)
C	¬ (C)	MM(M)C
-	(C)C ®	MM(M)C
-	-	MM(M)CC(C)

A total of thirteen crossings.
18. Road Safety Addition

There are seven basic solutions, values for D and H are the two remaining digits in each case. These two digits can be put in either order making 14 solutions altogether. If zero cannot be the first digit, #4 has only one, not two solutions (H not equal 0) making only 13 solutions in total.

	R	T	A	M	L	O	J	E
1	2	9	7	6	1	0	5	3
2	2	9	7	6	0	1	4	3
3	2	9	7	6	0	1	5	4
4	2	9	6	5	4	8	7	3
5	2	4	7	6	3	9	0	8
6	6	7	3	2	0	5	9	8
7	6	7	3	2	4	1	5	0

19. The Lonesome Eight

When two digits are brought down instead of one, there must be a zero in the quotient. This occurs twice, so we know at once that the quotient must be x080x.

When the divisor is multiplied by the quotient’s last digit, the produxt is a four-digit number. But we see that 8 times the divisor is a three-digit number. So the quotient’s last digit must be nine.

The divisor must be less than 125 because 8 ´ 125 = 1000, a four-digit number.

We can now deduce that the quotient’s first digit must be more than 7, for 7 times a divisor less than 125 would give a product which would be more than 100 less than 1000, the lowest four-digit number, and therefore would give a difference of more than 100 (a three digit number) when subtracted from the first four digits of the dividend, instead of the two-digit difference it does, in fact, give.

The first digit of the quotient cannot be nine because nine times the divisor is a four figure number, so it must be 8. This makes the full quotient 80809.

The divisor must be more than 123 because 123 ´ 80809 is a seven-digit number and our dividend has eight digits. Therefore it must be 124.

The final result is now easily found.

20. The Farmer And The Hundred Dollars

To buy 100 animals with $100, the number of anomals bought must equal the number of dollars spent. Since sheep sell at the price of one sheep for $1, they can be ignored until we have bought a number of cows and pigs such that the number of cows and pigs bought, equals the number of dollars paid for them. The rest of the $100 can be made up of sheep, which will not affect the balance of animals bought against dollars spent.

For every cow bought, we have spent $4 more than we need to, to make the number of cows bought equal to dollars spent. For every pig bought, we have spent 95 cents less than necessary to make the number of pigs bought equal to the number of dollars spent.

Thus if we buy cows and pigs in the inverse ratio of this excess and insufficiency, (i.e. 19 cows for every 80 pigs) the number of animals bought will be equal to the number of dollars spent.

But 80 + 19 = 99 so we can only buy one such balanced set for under $100. The other $1 can be used to buy a sheep. Therefore the farmer bought 19 cows, 1 sheep and 80 pigs.

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Mathematical & Logical Puzzles -Q4

noreply@blogger.com (Nisheeth Ranjan) — Fri, 23 Jul 2010 12:50:00 +0530

13. The Tribulations Of A Father Of Five

A father rowing a boat, has to transport his five sons across a river in a minimum number of one-way crossings, such that finally all the children have had an identical number of crossings.

The boat will hold the father (who does all the rowing) and not more than two children. No pair of children of immediately neighbouring ages can be left together in the absence of their father or they will start to fight. But children more separated in age will content themselves with more peaceful occupations. What is more, the father, whilst handling the oars, is not able to control his children at the same time.

How does he get the five children safely across the river, in the minimum number of one-way crossings, so that each child has the same number of such trips?

14. Three Wives & Their Jealous Husbands

Three jealous husbands with their wives, having to cross a river, find a boat without a boatman. But the boat is so small that it can contain no more than two of them at a time. How can these six persons cross the river so that none of the women shall be left in the company of any of the men unless her husband is present? (Women to row when absolutely necessary to prevent a violation of the above conditions).

15. Four Wives & Their Jealous Husbands

Four couples have to cross a river with the aid of one small boat with the capacity of only two people. Nowhere, either on land or in the boat, is any woman allowed to be left in the company of any of the men unless her husband is present. How can the transfer be executed if there is an island in the middle of the river where the boat may land and leave some of its cargo?

16. Getting The Nuggets Across

During the Alaska gold-rush, three prospectors who had struck it rich had to cross the Yukon. But they could find only a small row-boat with the capacity of either two men or one man and a bag of gold nuggets. Each of them owned a bag of the precious metal but the contents of each of the three bags of nuggets were not the same. Smith’s gold was worth $8000, Jones’ $5000 and Brown’s $3000.

None of them trusted the other two, and so after some argument they agreed that the passage should be arranged so that none of them, either on shore or in the boat should be in the presence of nuggets worth more than he owned. How did the men proceed to get themselves and their properties across the Yukon in the least number of crossings consistent with the restrictions?

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Mathematical & Logical Puzzles -A4

noreply@blogger.com (Nisheeth Ranjan) — Fri, 23 Jul 2010 12:48:00 +0530

13. The Tribulations Of A Father Of Five

Let A,B,C,D,E be the five children in descending order of age.

Left Bank	Crossing	Right Bank
ABCDE	-	-
ACE	BD ®	-
ACE	¬ B	D
BE	AC ®	D
BE	¬ AD	C
BD	AE ®	C
BD	¬ CE	A
BD	CE ®	A
BD	¬ -	ACE
-	BD ®	ACE
-	-	ABCDE

The minimum number of crossings is nine with three crossings per child. Note that if C and E have only one crossing each, the total number of crossings is reduced from nine to seven.

14. Three Wives & Their Jealous Husbands

Let the three husbands be A,B and C and their wives a,b and c respectively. The following moves are typical of a minimum solution with men rowing where possible.

Left Bank	Crossing	Right Bank
ABCabc	-	-
ACac	Bb ®	-
ACac	¬ B	b
ABC	ac ®	b
ABC	¬ a	bc
Aa	BC ®	bc
Aa	¬ Bb	Cc
ab	AB ®	Cc
ab	¬ c	ABC
b	ac ®	ABC
b	¬ B	ACac
-	Bb®	ACac
-	-	ABCabc

Note that only a, B, and c need be able to row.

15. Four Wives & Their Jealous Husbands

Let the husbands be ABCD and their wives abcd respectively. Then the following crossings are necessary:

Left bank	River	Island	River	Right bank
ABCDabcd	-	-	-	-
ABCDcd	ab ®	-	-	-
ABCDcd	¬ b	a	-	-
ABCDd	bc ®	a	-	-
ABCDd	¬ c	ab	-	-
CDcd	AB ®	ab	-	-
CDcd	-	AB	ab ®	-
CDcd	-	AB	¬ b	a
CDcd	-	b	AB ®	a
CDcd	-	b	¬ B	Aa
CDcd	¬ B	b	-	Aa
BCD	cd ®	b	-	Aa
BCD	¬ d	bc	-	Aa
Dd	BC ®	bc	-	Aa
Dd	-	bc	BC ®	Aa
Dd	-	bc	¬ a	ABC
Dd	-	c	ab ®	ABC
Dd	-	c	¬ C	ABab
Dd	¬ C	c	-	ABab
d	CD ®	c	-	ABab
d	-	c	CD ®	ABab
d	-	c	¬ b	ABCDa
d	-	-	bc ®	ABCDa
d	-	-	¬ c	ABCDab
d	¬ c	-	-	ABCDab
-	cd ®	-	-	ABCDab
-	-	-	cd ®	ABCDab
-	-	-	-	ABCDabcd

16. Getting The Nuggets Across

Let Smith, Jones and Brown be designated by the numbers 3, 2 and 1 respectively and their respective properties by (3), (2) and (1) . In any group on either bank the total of the numbers in parentheses must never exceed the total of the plain numbers. One solution is as follows:

Left bank	River	Right bank
123(1)(2)(3)	-	-
13(1)(3)	2(2)®	-
13(1)(3)	¬ 2	(2)
12(3)	3(1)®	(2)
12(3)	¬ 3	(1)(2)
3(3)	12 ®	(1)(2)
3(3)	¬ 2(2)	1(1)
2(2)	3(3)®	1(1)
2(2)	¬ 1(1)	3(3)
(1)(2)	12 ®	3(3)
(1)(2)	¬ 3	12(3)
(1)	3(2)®	12(3)
(1)	¬ 1	23(2)(3)
-	1(1)®	23(2)(3)
-	-	123(1)(2)(3)

A minimum of thirteen trips.

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Mathematical & Logical Puzzles -Q3

noreply@blogger.com (Nisheeth Ranjan) — Fri, 23 Jul 2010 12:42:00 +0530

9. The Weights And The Scales

A shopkeeper has a beam balance and requires to weigh any article of a whole number of grams from 1 to 40 inclusive. What is the least number of weights necessary to do this?

10. The Efficient Electrician

An electrician is faced with the following annoying dilemma. In the basement of a three-storey house he finds, bunched together in a hole in the wall, the exposed ends of eleven wires, but he has no way of knowing which end above belongs to which end below. His problem is to match ends.

To accomplish his task, he may do one of two things:

· Short circuit the wires at either end by twisting the ends together in any manner he chooses.

· Test for continuity of a closed circuit by means of a “continuity tester”, consisting of a battery and a bell. The bell rings when the instrument is connected to two ends of a continuous unbroken circuit.

Not wishing to exhaust himself with needless stair climbing and having a passionate interest in operations research, the electrician sat down on the top floor with a paper and pencil and soon devised the most efficient method of labelling the wires. What was his method?

11. How Old Are You?

Mr. Jones said to Mr. Smith: “When I was as old as you are now, you were twice as old as you were when I was three times as old as you.” How old is Mr. Jones if Mr. Smith is 28?

12. Ann And Mary

The present sum of the ages of the two girls is 44. Mary is twice as old as Ann was when Mary was half as old as Ann will be when she is three times as old as Mary was when she was three times as old as Ann. How old are Ann and Mary?

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Mathematical & Logical Puzzles -A3

noreply@blogger.com (Nisheeth Ranjan) — Fri, 23 Jul 2010 12:40:00 +0530

9. The Weights And The Scales

Four weights are quite sufficient for this purpose. 1 gram, 3 grams, 9 grams and 27 grams.

To weight two grams, the 3-gram weight is put on one pan and the 1-gram weight on the other, making a difference of two grams.

To weight five grams, the 9-gram weight is put on one pan and the and the 3-gram weight and the 1-gram weight are put on the other, making a difference of five grams.

By using this method of adding and subtracting weights, the four weights will weigh any number of whole grams from one to forty grams inclusive.

10. The Efficient Electrician

On the top floor, the electrician shorted five pairs of wires by twisting them together, leaving the last wire free. Then he walked to the basement and identified the lower ends of the shorted pairs by means of his continuity tester. He labelled the bottom ends of the first shorted pair A1 and A2 respectively, the ends of the second shorted pair B1 and B2 respectively, the ends of the third shorted pair C1 and C2 respectively, the ends of the fourth shorted pair D1 and D2 respectively, the ends of the fifth shorted pair E1 and E2 respectively and the unpaired end he labelled F. He shorted the bottom ends of the wires by leaving A1 free, connecting A2 to B1, B2 to C1, C2 to D1, D2 to E1, and E2 to F.

Back on the top floor, he removed all the shorts but left the pairs twisted together at insulated portions so that the pairs were still identifiable. He then checked for continuity between the free wire, which he knew to be the upper end of F and some other wire. When he found the other wire he was at once able to label it E2 and identify its mate as E1. He next tested for contintuity between E1 and another end which when found could be marked D2 and its mate D1. Continuing in this fashion, the remaining ends were easily identified.

11. How Old Are You?

If when Jones was 28, Smith was 2x years old, then he was x years old when Jones was 3x years old. Whatever their respective ages, Jones is always the same number of years older than Smith. Age difference when Smith is x and Jones is 3x is 2x years.

Age difference when Jones is 28 and Smith 2x is 28 - 2x.

Now the age differences are equal, therefore 2x = 28 - 2x or 4x = 28 or x = 7.

Age difference is 2x or 14 years. Therefore when Jones was 28 years old, Smith was 14 years old.

Smith is 28 so Jones is now 28 + 14 = 42 years old.

12. Ann And Mary

Work from the end backwards. Let Ann be x years old, then Mary was 3x when she was three times as old as Ann. Age difference is 2x. Ann will be three times 3x when she is 9x years old. Mary will be 4½x when she is half as old as 9x. Ann will be 4½x - 2x or 2½x when Mary is 4½x. Therefore Mary now being twice as old as 2½x is now 5x. Ann is 2x years younger or 3x years old. But we are told that 5x + 3x = 44. And from this x = 5½.

Mary is 5x or 27½ years old whilst Ann is 3x or 16½ years old. Easy isn’t it?

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Mathematical & Logical Puzzles -Q2

noreply@blogger.com (Nisheeth Ranjan) — Thu, 22 Jul 2010 18:05:00 +0530

5. The Professor On The Escalator

When Professor Stanislaw Slapernarski, the Polish mathematician walked very slowly down the down-moving escalator, he reached the bottom after taking 50 steps. As an experiment, he then ran up the same escalator one step at a time, reaching the top after taking 125 steps.

Assuming that the Professor went up five times as fast as he went down (i.e. he took five steps to every one before), and that he made each trip at a separate constant speed, how many steps would be visible if the escalator stopped running?

6. The Encounter On The Bridge

On a foggy night, a passenger car and a truck meet on a bridge which is so narrow that the two vehicles can neither pass nor turn. The car has proceeded twice as far onto the bridge as the truck, but the truck has required twice as much time as the car to reach this point. Each vehicle has only half its forward speed when in reverse. Which of the two vehicles should back up to allow both of them to cross the bridge in the minimum time?

7. The Early Commuter

A commuter is in the habit of arriving at his suburban station each evening at exactly five o’clock. His wife always meets the train and drives him home. One day he takes an earlier train and arrives at his station at four o’clock. The weather is pleasant so instead of telephoning home, he starts walking along the route always taken by his wife. They meet somewhere along the way.

He gets into the car and they drive home arriving ten minutes earlier than they usually do. Assuming that the wife always drives with a constant speed, and on this particular occasion she left just in time to meet the five o’clock train, can you determine how long the husband walked before he was picked up?

8. The Two Ferryboats

Two ferryboats start at the same instant from opposite sides of the river travelling across the water en route at right angles to the shores. Each travels at a constant speed but one is faster than the other. They pass at a point 720 metres from the nearest shore. Both boats remain at their slips for ten minutes before starting back. On the return trip, they pass 400 metres from the other shore. How wide is the river?

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Mathematical & Logical Puzzles -A2

noreply@blogger.com (Nisheeth Ranjan) — Thu, 22 Jul 2010 18:04:00 +0530

5. The Professor On The Escalator

Let n be the number of steps visible when the escalator is not moving, and let a unit of time be the time it takes the professor to walk down one step. If he walks down the down-moving escalator in 50 steps, then n - 50 steps have gone out of sight in 50 units of time. It takes him 125 steps to run up the same escalator taking five steps to every one before. In this time 125 - n steps have disappeared in 125 ¸ 5 units of time. Since the escalator can be assumed to run at a constant speed, we have the following linear equation that readily yields n = 100 steps:

n - 50 125 - n

50 25

6. The Encounter On The Bridge

If we call the length of the bridge 3d, the two vehicles will meet when the car is 2d over the bridge and the truck, d. To reach this point, the car has required t seconds and the truck 2t seconds. To back up the car would need 2t seconds for the distance 2d; the truck 4t seconds for the distance d. Now let us consider the possiblities:

Case 1 The car backs up. It requires 2t seconds, meanwhile the truck proceeds. It needs 4t seconds to finish the journey across the bridge so the car has to wait 2t seconds until it can start across again. It requires 1½t seconds to drive over the bridge. Thus 5½t seconds are required for the operation.

Case 2 The truck backs up. It requires 4t seconds. Simultaneously the car proceeds though only with one-eighth its speed because of the truck’s snail pace. They arrive at the end of the bridge together, then the truck drives over the bridge which takes 6t seconds. Thus in this case 10t seconds are need to complete the operation.

Therefore, Case 1 is the solution most favourable to both vehicles.

7. The Early Commuter

The commuter walked for 55 minutes before his wife picked him up.

Since they arrive home 10 minutes earlier than usual, this means that the wife had chopped 10 minutes off her usual travel time to and from the station. It follows that she must have met her husband five minutes before his usual pick-up time of five o’clock or at 4:55 pm. He started walking at 4pm and was therefore walking for 55 minutes.

8. The Two Ferry Boats

When the two ferry boats meet for the first time, the combined distance travelled by the boats is equal to the width of the river. When they reach the opposite shores, the combined distance travelled is twice the width of the river and when they meet for the second time, the combined distance is three times the width of the river.

Since both boats are moving at a constant speed for the same period of time, it follows that each boat has gone three times as far as when they first met (after they had travelled a combined distance of one river’s width). Since the white boat, (slow boat), had travelled 720 metres before the first meeting, its total distance at the time of the second meeting must be 3 x 720 or 2160 metres. This is 400 metres more than the width of the river, so we must subtract 400 from 2160 to obtain 1760 metres as the width of the river.

Go Back To The Questions

Mathematical & Logical Puzzles -A1

noreply@blogger.com (Nisheeth Ranjan) — Thu, 22 Jul 2010 11:12:00 +0530

1. The Flight Around The World

Three aeroplanes are sufficient to ensure the flight of one of them around the world. There are many ways in which this may be done, but the following seems to be the most efficient.

Planes A and B and C take off together. After going one-eighth of the way, C transfers ¼ tank to A and ¼ tank to B. This gives A and B full tanks and leaves C with a ¼ tank sufficient just to get back to base.

Planes A and B continue a further one-eighth of the way when B transfers ¼ tank to A. B now has ½ tank left, sufficient just to get back to base.

Plane A meanwhile, with a full tank continues until he runs out of fuel ¼ way from base (¾ way round the world). Here A is met by C which has been refuelled at base. C transfers ¼ tank to A and both planes head for home.

A and C run out of fuel one-eighth way from base where they are met by refuelled plane B. B transfers ¼ tank to A and ¼ tank to C and the three planes then have just enough fuel to reach base with empty tanks.

2. Who Is The Engineer?

· From 2 we see that Mr. Robinson lives in Los Angeles.

· From 3 and 6 we can deduce that the distinguished mathematical physicist lives in Omaha.

· From 4 we see that Mr. Jones cannot live in Omaha (6) or Los Angeles (2) so he must live in Chicago.

· From 5 we now know that the brakeman’s name is Jones.

From Smith cannot be the fireman; he cannot be the brakeman (5) so he must be the engineer.

3. The Coloured Hats

C asks himself, “Can my hat be green?” If so then A will know immediately that he has a red hat, for only a red hat on his head would cause B to lift his hand. A would have therefore left the room. B would reason in the same way and also leave. Since neither has left the room, C deduces that his own hat must be red.

4. The Coloured Hats Again

If Bill and Colin are wearing red hats, Arthur would have known his hat was blue, because there were only two red hats. Since Arthur didn’t know the correct answer, Colin concluded that there remained only two possiblities for himself and Bill. Either both had blue hats, or one had a blue hat and the other a red hat. If he himself had a red hat, Colin reasoned, Bill would have concluded that he, Bill, had a blue hat, because otherwise Arthur would have known that he, Arthur, must have a blue hat. So Bill, because he was not able to tell the colour of his own hat, involuntarily betrayed to Colin that his, Colin’s hat was not red. Therefore Colin concluded that his hat was blue.

Go Back To The Questions

Mathematical & Logical Puzzles -Q1

noreply@blogger.com (Nisheeth Ranjan) — Thu, 22 Jul 2010 11:10:00 +0530

1. The Flight Around The World

A group of aeroplanes is based on a small island. The tank of each plane holds just enough fuel to take it half-way around the world. Any desired amount of fuel may be transferred from the tank of one plane to the tank of another whilst the planes are in flight. The only source of fuel is on the island, and for the purposes of this problem it is assumed that there is no time lost in refuelling either in the air or on the ground.

What is the smallest number of planes that will ensure the flight, non-stop, of one plane around the world on a great circle assuming that each of the planes has the same constant ground speed and rate of fuel consumption, and that all planes return safely to their island base?

2. Who Is The Engineer?

Smith, Jones and Robinson are engineer, brakeman and fireman on a train but not necessarily in that order. Also riding the train are three passengers with the same three surnames, to be identified in the following premises with a “Mr.” before their names.

· Mr. Robinson lives in Los Angeles.

· The brakeman lives in Omaha.

· Mr. Jones long ago forgot all the algebra he ever learnt at High School.

· The passenger with the same name as the brakeman lives in Chicago.

· The brakeman and one of the passengers, a distinguished mathematical physicist, attend the same church.

· Smith beat the fireman at billiards.

What is the name of the engineer?

3. The Coloured Hats

Three men, Alan, Brian and Colin are blindfolded and are told that either a red hat or a green hat will be placed on each of their heads. After this is done, the blindfolds are removed and the men are asked to do the following:

· To raise a hand if they see a red hat.

· To leave the room as soon as they are sure of the colour of their own hat.

All the hats happen to be red, so all the men raise a hand. Several minutes elapse until Colin, who is more astute than the others, leaves the room. How did he deduce the colour of his hat?

4. The Coloured Hats Again

At a party, four people played a game. Three of them sat one behind the other, so that Arthur saw Bill and Colin, and Bill saw only Colin who sat in front and saw nobody.

Dave had five hats which were showed to his three friends. Three of the hats were blue and two were red. Now Dave placed a hat on the head of each of his three friends, putting aside the remaining two hats.

He now asked Arthur what colour his hat was. Arthur said he couldn’t tell. Bill, asked the colour of his own hat didn’t know for certain either. Colin, however, although he couldn’t see any hat at all, gave the correct answer when asked what the colour of his hat was! What was the colour of Colin’s hat and how did he deduce it?

Check Out The Answers Here

Mutating Table Errors

noreply@blogger.com (Nisheeth Ranjan) — Thu, 22 Jul 2010 10:03:00 +0530

Mutating Table Problems with DELETE CASCADE

A mutating table is a table that is currently being modified by an UPDATE, DELETE, or INSERT statement, or it is a table that might need to be updated by the effects of a declarative DELETE CASCADE referential integrity constraint. The restrictions on such a table apply only to the session that issued the statement in progress.

For all row triggers, that were fired as the result of a DELETE CASCADE, there are two important restrictions regarding mutating tables. These restrictions prevent a trigger from seeing an inconsistent set of data.

The SQL statements of a trigger cannot read from (query) or modifya
mutating table of the triggering statement.

Example 1

Select in a mutating table from a row trigger (Tested on Oracle 8.1.7)

We want to explain this situation on an example. We have two tables "A" and "B". "A" is the master table and "B" the detail table. We specified a foreign key between "B" and "A" with the CASCADE DELETE option.

Here are the CREATE statements

drop table B;
drop table A;

create table A (
ida number not null,
vala varchar2(10),
primary key(ida));

create table B (
idb number,
valb varchar2(10),
foreign key (idb) references A (ida) on delete cascade)
/

create or replace trigger b_br
after delete on B
for each row
declare
n integer;
begin
select count(*) into n from A;
dbms_output.put_line('there are ' || n || ' rows in A');
dbms_output.put_line('after statment on B');
dbms_output.new_line;
end;
/

insert into A values(1,'Table A');
insert into A values(2,'Table A');
insert into B values(1,'Table B');
insert into B values(1,'Table B');
commit;

set serveroutput on;
delete from A where idA = 1;

ERROR at line 1:
ORA-04091: table SCOTT.A is mutating, trigger/function may not see
ORA-06512: at "SCOTT.B_BR", line 4
ORA-04088: error during execution of trigger 'SCOTT.B_BR'

Notice that the SQL statement ( "select count(*) into n from A" ) is run for the first row of the table, and then the AFTER row trigger B_BR is fired. In turn, a statement in the AFTER row trigger body attempts to query the original table A. However, because the table A is mutating due to the CASCADE DELETE foreign key, this query is not allowed by Oracle. If attempted, a runtime error occurs, the effects of the trigger body and triggering statement are rolled back, and control is returned to the user or application.

Solution: Use statement trigger instead of row trigger

If you delete the line "FOR EACH ROW" from the trigger above, then the trigger becomes a statement trigger, the table is not mutatingwhen the trigger fires, and the trigger does output the correct data.

SQL> delete from A where idA = 1;

there are 1 rows in A
after statment on B

1 row deleted.

SQL> select count(*) from B;

COUNT(*)
----------
0

It is not always possible to change the row trigger to a statement trigger. In this case, the oracle manual proposes the following solution:

Example 2

Cascading Update with Oracle 8i

Before Oracle8i, there was a "constraining error" that prevented a row trigger from modifying a table when the parent statement implicitly read that table to enforce a foreign key constraint. As of Oracle8i, there is no constraining error. In addition, checking of the foreign key is deferred until at least the end of the parent statement. The mutating error still prevents the trigger from reading or modifying the table that the parent statement is modifying.

This allows most foreign key constraint actions to be implemented via their obvious after-row trigger, providing the constraint is not self-referential. Update cascade, update set null, update set default, delete set default, inserting a missing parent, and maintaining a count of children can all be implemented easily.

However there is one problem with a multirow update - let's demonstrate this on an example.

create table P (
p1 number not null,
primary key (p1));

create table F (
f1 number,
foreign key (f1) references P (p1) on delete cascade);

create trigger pu_ar
after update on p
for each row
begin
if (:new.p1 != :old.p1) then
    update f
       set f1 = :new.p1
     where f1 = :old.p1;
end if;
end;
/

insert into p values (1);
insert into p values (2);
insert into p values (3);
insert into f values (1);
insert into f values (2);
insert into f values (3);
commit;

update p set p1 = p1+1;
commit;

SQL> select * from p;

        P1
----------
         2 <===== OK
         3 <===== OK
         4 <===== OK

SQL> select * from f;

        F1
----------
         4   <===== Wrong !
         4   <===== Wrong !
         4   <===== Wrong !

The statement first updates (1) to (2) in p, and the trigger updates (1) to (2) in f, leaving two rows of value (2) in f. Then the statement updates (2) to (3) in p, and the trigger updates both rows of value (2) to (3) in f. Finally, the statement updates (3) to (4) in p, and the trigger updates all three rows in f from (3) to (4). The relationship of the data in p and f is lost.

To avoid this problem, you must forbid to update to an existing primary key, if this primary key have any childs, this can be accomplished with the following trigger.

create or replace trigger pu_ar
after update on P
for each row

declare
echildexists exception;

-- check if child table have child records with this new
-- key, this is not allowed.

cursor curs_exists is
select 'x'
from F
where f1 = :new.p1;

rtf curs_exists%rowtype;

begin
if(:new.p1 <> :old.p1) then

open curs_exists;
fetch curs_exists into rtf;

    if (curs_exists%found) then
      close curs_exists;
      raise echildexists;
    end if;

close curs_exists;

    update f
       set f1 = :new.p1
     where f1 = :old.p1;
end if;

exception
when echildexists then
    raise_application_error (
    -20102,'error: this primary key: '
    ||to_char(:new.p1)
    ||' exists and has child rows in f, this triggers
    again an update and so on ...');
end pu_ar;
/

Test the update

SQL> update p set p1 = p1+1;

ERROR at line 1:
ORA-20102: error: this primary key: 2 exists and has child
rows in f, this triggers again an update and so on ...
ORA-06512: at "SCOTT.PU_AR", line 34
ORA-04088: error during execution of trigger 'SCOTT.PU_AR'

Example 3

Using a temporary table

If you need to update a mutating table, then you could use a temporary table, a PL/SQL table, or a package variable to bypass these restrictions. For example, in place of a single AFTER row trigger that updates the original table, resulting in a mutating table error, you may be able to use two triggers - an AFTER row trigger that updates a temporary table, and an AFTER statement trigger that updates the original table with the values from the temporary table.

In the next example, "from the real world", we want to show this. The table CUG can only have records of the following types

o A: Type = 1

o B: Type = 2 (Leader for C or D)

o C: Type = 3 (Lead by B)

o D: Type = 4 (Lead by B)

Note, that the types C and D must be leaded by the type B.

Create table CUG

drop table CUG cascade constraints;
create table CUG (
id_cug      number(12) not null primary key,
id_B        number(12) not null,
type        number(1),
foreign key (id_B) references CUG (id_cug)
on delete cascade);

Next we create a temporary table to avoid the "Mutating Table Problem".

drop table CUGTMP;
create global temporary table CUGTMP (
id_B number(12),
type number(1))
on commit delete rows;

The following trigger checks new rows (Inserts) in CUG

create or replace trigger bi_r
before insert on CUG
for each row
declare
l_type     CUG.type%type;
begin
if (:new.type in (3,4)) then
    select type into l_type from CUG
     where id_cug = :new.id_B;
end if;
if (l_type != 2) then
     raise_application_error(-20002,
     'C and D CUGs must have a leading B');
end if;
end;
/

The following Trigger saves the new values for id_B in the temporary table.

create or replace trigger au_r
after update of id_B on CUG
for each row
begin
insert into CUGTMP (id_B,type)
values (:new.id_B,:new.type);
end;
/

The following Trigger finally checks, that C and D CUGs belong to a B CUG.

create or replace trigger au_s
after update of id_B on CUG
declare
l_id_B        number(12);
l_typeCD      number(1);
l_typeB       number(1);
cursor cur_cugtmp is
select id_B,type
   from CUGTMP;
begin
open cur_cugtmp;
loop
    fetch cur_cugtmp into l_id_B,l_typeCD;
    exit when cur_cugtmp%notfound;
     select type into l_typeB from CUG
      where id_cug = l_id_B;
    if (l_typeB != 2) then
       raise_application_error(-20002,
       'C and D CUGs must have a leading B');
    end if;
end loop;
close cur_cugtmp;
end;
/

Test insert and update

insert into CUG (id_cug,id_B,type)
values (0,0,0);
insert into CUG (id_cug,id_B,type)
values (1,0,2);
insert into CUG (id_cug,id_B,type)
values (2,0,2);
insert into CUG (id_cug,id_B,type)
values (3,1,3);
insert into CUG (id_cug,id_B,type)
values (4,2,3);
insert into CUG (id_cug,id_B,type)
values (5,1,4);
insert into CUG (id_cug,id_B,type)
values (6,2,4);
commit;

SQL> select * from CUG;

    ID_CUG       ID_B       TYPE
---------- ---------- ----------
         0          0          0
         1          0          2
         2          0          2
         3          1          3
         4          2          3
         5          1          4
         6          2          4

Now, we want that that the CUGs 3,4,5,6 changes the leadership to CUG 2

SQL> update CUG set id_B = 2 where id_cug in (3,4,5,6);

4 rows updated.

SQL> select * from cug;

    ID_CUG       ID_B       TYPE
---------- ---------- ----------
         0          0          0
         1          0          2
         2          0          2
         3          2          3
         4          2          3
         5          2          4
         6          2          4

Next we delete the "Leader" with ID_CUG = 2. All childs must be deleted automatically with the DELETE CASCADE.

SQL> select * from cug;

    ID_CUG       ID_B       TYPE
---------- ---------- ----------
         0          0          0
         1          0          2

Everything looks fine now.

Oxis International

noreply@blogger.com (Nisheeth Ranjan) — Tue, 20 Jul 2010 11:30:00 +0530

Oxis International,Inc. is the premiere company which provides the potent antioxidants, Ergothioneine and pure l-ergothioneine through thier patented synthetic manufacturing process, worldwide.
The products of the company include therapeutic nutraceutical products, cosmeceutical products and proprietary formulations and clinical products that are developed internally and/or out-licensed to biotech and pharmaceutical companies as drug candidates. The company is focused on naturally occurring protective substances since they are more likely to be both safe and efficacious.

The primary products of Oxis incorporate and emphasize the multifaceted “super antioxidant” compound, L- Ergothioneine (“ERGO”) , GLUTATHIONE PEROXIDASE (GPX) as a key component. the company owns several patents and pending applications related to ERGO that cover current and planned products relevant to our nutraceutical and cosmecuetical businesses. Our patents and patent applications address ERGO’s protective effects and activities and the ERGO manufacturing process. It produces the compounds which TREAts OXIDATIVE STRESS by controlling free radicles and hence it is very helpful as an anti aging compound as well.

Oxis is listed under the category of penny stocks which can be a great opportunity for inverstors.Investers can contact us through our website for more information.

To know more about us visit :http://www.oxis.com

Lateral Thinking Puzzles

noreply@blogger.com (Nisheeth Ranjan) — Mon, 19 Jul 2010 10:58:00 +0530

The Puzzles

1. The man in the Elevator

A man lives on the tenth floor of a building. Every day he takes the elevator to go down to the ground floor to go to work or to go shopping. When he returns he takes the elevator to the seventh floor and walks up the stairs to reach his apartment on the tenth floor. He hates walking so why does he do it?

This is probably the best known and most celebrated of all lateral thinking puzzles. It is a true classic. Although there are many possible solutions which fit the initial conditions, only the canonical answer is truly satisfying.

Answer – He’s a Dwarf

2. The Man in the Bar

A man walks into a bar and asks the barman for a glass of water. The barman pulls out a gun and points it at the man. The man says 'Thank you' and walks out.

This puzzle has claims to be the best of the genre. It is simple in its statement, absolutely baffling and yet with a completely satisfying solution. Most people struggle very hard to solve this one yet they like the answer when they hear it or have the satisfaction of figuring it out.

Answer – he Has Hiccups

3. The Man who Hanged Himself

Not far from Madrid, there is a large wooden barn. The barn is completely empty except for a dead man hanging from the middle of the central rafter. The rope around his neck is ten feet long and his feet are three feet off the ground. The nearest wall is 20 feet away from the man. It is not possible to climb up the walls or along the rafters. The man hanged himself. How did he do it?

Answer – He stands on a block of ice

4. Death in a Field

A man is lying dead in a field. Next to him there is an unopened package. There is no other creature in the field. How did he die?

Answer – His parachute didn’t open

5. Anthony and Cleopatra

Anthony and Cleopatra are lying dead on the floor of a villa in Egypt. Nearby is a broken bowl. There is no mark on either of their bodies and they were not poisoned. How did they die?

Answer – Their fish

6. The Coal, Carrot and Scarf

Five pieces of coal, a carrot and a scarf are lying on the lawn. Nobody put them on the lawn but there is a perfectly logical reason why they should be there. What is it?

Answer – A snowman

7. Trouble with Sons

A woman had two sons who were born on the same hour of the same day of the same year. But they were not twins. How could this be so?

Answer – Their two of triplets.

8. Push that Car

A man pushed his car. He stopped when he reached a hotel at which point he knew he was bankrupt. Why?

Answer – He’s playing monopoly

9. The Arm of the Postal Service

One day a man received a parcel in the post. Carefully packed inside was a human arm. He examined it, repacked it and then sent it on to another man. The second man also carefully examined the arm before taking it to the woods and burying it. Why did they do this? This one probably has more variations than any other. A great one to puzzle out. It requires plenty of good questions.

Answer –

The three men had been stranded on a desert island. Desperate for food, they had agreed to amputate their left arms in order to eat them. They swore an oath that each would have his left arm cut off. One of them was a doctor and he cut the arms off his two companions. They were then rescued. But his oath was still binding so he later had to have his arm amputated and sent to his colleagues.

This is often told with a further twist whereby a doctor pays a tramp a large sum in order to amputate the tramp's arm which the doctor then sends to another man who inspects it etc. This variation can make for a long night of questioning!

10. Heaven

A man died and went to Heaven. There were thousands of other people there. They were all naked and all looked as they did at the age of 21. He looked around to see if there was anyone he recognised. He saw a couple and he knew immediately that they were Adam and Eve. How did he know?

Answer – They don’t have belly buttons

11. Friday

A man rode into town on Friday. He stayed for three nights and then left on Friday. How come?

A silly one - but it is surprisingly popular.

Answer – The man’s horse is called friday

12. Manhole Covers

Why is it better to have round manhole covers than square ones?

This is logical rather than lateral, but it is a good puzzle which can be solved by lateral thinking techniques. It is supposedly used by a very well-known software company as an interview question for prospective employees.

Answer – A square manhole cover can be turned and dropped down the diagonal of the manhole. A round manhole cannot be dropped down the manhole. So for safety and practicality, all manhole covers should be round.

13. The Deadly Party

A man went to a party and drank some of the punch. He then left early. Everyone else at the party who drank the punch subsequently died of poisoning. Why did the man not die?

Answer – The poison in the punch came from the ice cubes. When the man drank the punch the ice was fully frozen. Gradually it melted, poisoning the punch.

14. The Deadly Dish

Two men went into a restaurant. They both ordered the same dish from the menu. After they tasted it, one of the men went outside the restaurant and shot himself. Why?

Answer – The dish that the two men ordered was albatross. They had been stranded many years earlier on a desert island. When the man tasted albatross he realized that he had never tasted it before. This meant that the meat he had been given on the island was not albatross as he had been told. He correctly deduced that he had eaten the flesh of his son who had died when they first reached the island.

15. The Realization

A man was walking downstairs in a building when he suddenly realized that his wife had just died. How?

Answer – The man was in a hospital building when the electricity failed. His wife was on a life support machine.

16. The Blind Beggar

A blind beggar had a brother who died. What relation was the blind beggar to the brother who died? (Brother is not the answer).

Answer – His Sister

17. The Broken Match

A man is found dead in a field. He is clutching a broken match. What happened?

Answer – He and a number of other passengers were making a balloon trip in a desperate attempt to flee a country. The balloon had to lose weight to stop it from crashing. He drew the short match and had to jump.

18. The Music Stopped

The music stopped. She died. Explain.

Answer – She was a circus tight-rope walker who walked blindfolded over a high wire. The band played as she crossed and when the music stopped it was the signal that she had reached the end of the walk and could safely alight. One day the conductor was taken ill and the stand-in conductor ended the piece of music too early. She stepped off to her death.

19. Swimmer in the Forest

Deep in the forest was found the body of a man who was wearing only swimming trunks, snorkel and facemask. The nearest lake was 8 miles away and the sea was 100 miles away. How had he died?

This is supposedly based on a true incident. Does this make it an urban legend? Many urban legends can be restated as lateral thinking puzzles. This is a very good one of this type.

Answer – He was swimming when a plane came by to pick up water to put out forest fires, he wsa picked up in the water and dropped over the forest.

20. The Elder Twin

One day Kerry celebrated her birthday. Two days later her older twin brother, Terry, celebrated his birthday. How come?

Answer – At the time she went into labor, the mother of the twins was travelling by boat. The older twin, Terry, was born first early on March 1st. The boat then crossed the International Date line (or any time zone line) and Kerry, the younger twin, was born on February the 28th. In a leap year the younger twin celebrates her birthday two days before her older brother.

Form Builder

noreply@blogger.com (Nisheeth Ranjan) — Mon, 19 Jul 2010 10:38:00 +0530

Developer Forms Tutorial

Form Builder

Developer Release 6 tools are rapid application development programs. Several parts make up the entire set. We will focus on the Forms component as it is used to generate client side graphical interface applications to work with Oracle servers. Access to Oracle is not a necessity but if you have data coming from Oracle servers then you must have access to it. A valid Oracle userid and password are needed.

Form builder is an application found in the program group for Developer. Locate it and start the application by double clicking on the icon or name of the application. You see the following screen on start.

Figure 1 Starting Screen

Choose the first option (Use the Data Block Wizard) if not selected. We will use the wizard to define our data requirements (the source and the details)for the application. Click OK. Click Next to move to the next window.

The window allows you to choose either a Table or View or a Stored Procedure to be created. A Stored Procedure is a program written in PL/SQL or some other language (JAVA) that may be compiled and stored in the database for use later. A Data Block is a logical space where columns from database tables and elsewhere may be defined. Fields need not always be from a stored table.

Select Table or View option, if not already chosen. Click Next button to move on to the next screen.

Figure 2 Select Type of Data Block

Once you get to the next screen you have to make several decisions. The first , of course, is to select the source table. We are assuming that the data is coming from a table on our server. If that is not the case then we could simply provide any name which would be used by the application. Because we want to use a base table (table on our server), we click the Browse button to browse through the tables and views we have on the server. The system responds with a request to log in to Oracle (see Figure 3).

Once you are connected to the server, you will see a listing of all tables and views in your account. You may also see other tables and views to which you have access privileges. In addition to viewing and selecting tables and views the listing gives a number of other choices. We will stick with tables we have created so far.

Select the TRACKS table from the list and click OK.

Figure 3 Getting the Tables

The following window is displayed. All of the available columns are displayed in the Available Columns window and highlighted. Any or all of these columns may be moved to the Database Items window by clicking on either > or >> respectively. Database items are what the data block wizard uses for defining the data use for the application. It is possible to include a subset of the available columns in our application. This is true if you are building a retrieval only application, or an update application of selected columns. For regular data entry (full rows) you need to specify all the columns.

Once the columns are selected click the Next button. A message screen informing that the data block wizard is completed successfully is displayed. Press Finish button to proceed to Layout design.

Figure 4 Selecting Columns for Use

Figure 5 Data Block Defined

Figure 6 Selecting Type of Canvas

Different types of canvases are available for our application. The pop-up list for Type, shown below, indicates the different kinds of canvases. The common type is the Content type which is a single window screen. Other type used often is the Tabbed canvas to tab several screen one behind the other each with a tab. Clicking on a tab will bring that screen to the foreground. Stacked canvas may also be used if the application workflow is such that screens are disposed off in a linear sequence.

Figure 7 Pull Down Menu of Canvas Types

Select the type of canvas and click Next. You now have a choice to include selected items or all items for display on the screen. Remember that in the Data Block Wizard we selected the database items to be defined for the block. Now we get to decide whether all of these are to be used for display as well. Any item not chosen for display will go onto the control page/block which only the developer can access!

Figure 8 Select Columns for Display

Clicking Next brings you to the following screen. Here you can change the Prompt text to read better. The width and height of the displayed item can be controlled as well. The values are given in pixels. A better place to adjust these would be on the layout.

Figure 9 Define Prompt Labels

The next screen asks you to select the type of layout desired. You may choose a Form layout or Tabular layout. If you plan to display one record at a time you should choose Form layout whereas if you plan to display multiple records at a time the Tabular layout is preferred.

In the example, we choose the Tabular layout even though Figure 10 shows otherwise.

Figure 10 Choose Type of Interface

Once the layout is decided, you get a new screen where you define the name of the display frame and number of records to be displayed. The number of records to be displayed should be chosen with a view of the screen real estate. By adding a scroll bar option you can view more records than will be displayed at any one time. If you want the records to be spaced you may also include a value for distance between records.

Figure 11 Settings for Tabular Display

Click the Next button and on the next screen press Finish. The system will now generate the layout and display that along with the Object Navigator window in the background. The layout window shows the lay of the columns we chose for display. It also shows the prompts we have chosen. You will notice that the prompts for the Tabular display are on top of the items whereas for Form display they will be to the left of the items.

Figure 12 Layout of the Application

Figure 13 Object Navigator Window

If there are no changes to be made to the layout then we can test the application by clicking on the Traffic Signal Icon. You should see the following window. You are now ready to work with the application.

To save the application design and the compiled version of the Form click Save and name the file. Choose a directory where you have easy access.

To run the application again, start Forms Runtime. Browse to the file that you named earlier. Enter connection details in the appropriate fields and click OK.

Figure 14 Completed Application

Cursors in SQL & PL/SQL

noreply@blogger.com (Nisheeth Ranjan) — Fri, 25 Jun 2010 15:53:00 +0530

SQL Cursor

A SQL cursor is a private Oracle SQL working area. There are two types of SQL cursor: implicit or explicit cursor. The implicit cursor is used by Oracle server to test and parse the SQL statements and the explicit cursors are declared by the programmers.
Using the implicit cursor, we can test the outcome of SQL statements in PL/SQL. For example,

SQL%ROWCOUNT, return the number of rows affected;
SQL%FOUND, a BOOLEAN attribute indicating whether the recent SQL statement matches to any row;
SQL%NOTFOUND, a BOOLEAN attribute indicating whether the recent SQL statement does not match to any row;
SQL%ISOPEN, a BOOLEAN attribute and always evaluated as FALSE immediately after the SQL statement is executed.

To write the explicit cursor, please refer to the following example. Note that a cursor definition can array a number of arguments.
For example,
DECLARE
CURSOR csr_ac (p_name VARCHAR2) IS
SELECT empno, name, sal
FROM employee
WHERE name LIKE '%p_name%';
BEGIN
FOR rec_ac IN csr_ac ('LE')
LOOP
DBMS_OUTPUT.PUT_LINE(rec_ac.empno || ' ' ||rec_ac.name || ' '||v_sal);
END LOOP ;
CLOSE csr_ac;
END;
/
Another way of writing the above code, is to use the basic loop and the SQL%NOTFOUND cursor, as shown in the following.

SQL> DECLARE

2 CURSOR csr_ac (p_name VARCHAR2) IS

3 SELECT empno, ename, sal

4 FROM emp

5 6 WHERE ename LIKE '%SMITH%';

7 8 v_a emp.empno%TYPE;

9 v_b emp.ename%TYPE;

10 v_c emp.sal%TYPE;

11 12 BEGIN

13 OPEN csr_ac('');

14 LOOP

15 FETCH csr_ac INTO v_a, v_b, v_c;

16 EXIT WHEN csr_ac%NOTFOUND;

17 18 DBMS_OUTPUT.PUT_LINE(v_a || ' ' || v_b || ' '||v_c);

19 20 END LOOP;

21 CLOSE csr_ac;

22 END;

23

24 25 /

OUTPUT

7369 SMITH 800

Cursors in PL/SQL

Every SQL statement executed by the RDBMS has a private SQL area that contains information about the SQL statement and the set of data returned. In PL/SQL, a cursor is a name assigned to a specific private SQL area for a specific SQL statement. There can be either static cursors, whose SQL statement is determined at compile time, or dynamic cursors, whose SQL statement is determined at runtime. Static cursors are covered in greater detail in this section. Dynamic cursors in PL/SQL are implemented via the built-in package DBMS_SQL. See the book Oracle Built-in Packages and the corresponding Oracle PL/SQL Built-ins Pocket Reference, both from O'Reilly & Associates, for full coverage on DBMS_SQL and the other built-in packages.

Explicit Cursors

Explicit cursors are SELECT statements that are DECLAREd explicitly in the declaration section of the current block or in a package specification. Use OPEN, FETCH, and CLOSE in the execution or exception sections of your programs.

1.1 Declaring explicit cursors

To use an explicit cursor, you must first declare it in the declaration section of a block or package. There are three types of explicit cursor declarations:

· A cursor without parameters, such as:

      CURSOR company_cur

IS

         SELECT company_id FROM company;

· A cursor that accepts arguments through a parameter list:

      CURSOR company_cur (id_in IN NUMBER) IS

      SELECT name FROM company

      WHERE  company_id = id_in;

· A cursor header that contains a RETURN clause in place of the SELECT statement:

      CURSOR company_cur (id_in IN NUMBER)

      RETURN company%ROWTYPE IS

      SELECT * FROM company;

1.2 Opening explicit cursors

To open a cursor, use the following syntax:

OPEN cursor_name [(argument [,argument ...])];

where cursor_name is the name of the cursor as declared in the declaration section. The arguments are required if the definition of the cursor contains a parameter list.

You must open an explicit cursor before you can fetch rows from that cursor. When the cursor is opened, the processing includes the PARSE, BIND, OPEN, and EXECUTE statements. This OPEN processing includes: determining an execution plan, associating host variables and cursor parameters with the placeholders in the SQL statement, determining the result set, and, finally, setting the current row pointer to the first row in the result set.

When using a cursor FOR loop, the OPEN is implicit in the FOR statement. If you try to open a cursor that is already open, PL/SQL will raise an "ORA-06511: PL/SQL: cursor already open" exception.

1.3 Fetching from explicit cursors

The FETCH statement places the contents of the current row into local variables. To retrieve all rows in a result set, each row needs to be fetched. The syntax for a FETCH statement is:

FETCH cursor_name INTO record_or_variable_list;

where cursor_name is the name of the cursor as declared and opened.

1.4 Closing explicit cursors

The syntax of the CLOSE statement is:

CLOSE cursor_name;

where cursor_name is the name of the cursor declared and opened.

After all rows have been fetched, a cursor needs to be closed. Closing a cursor releases the private SQL area used by the cursor, freeing the memory used by that cursor.

If you declare a cursor in a local anonymous, procedure, or function block, that cursor will automatically close when the block terminates. Package-based cursors must be closed explicitly, or they stay open for the duration of your session. Closing a cursor that is not open raises an INVALID CURSOR exception.

1.5 Explicit cursor attributes

There are four attributes associated with cursors: ISOPEN, FOUND, NOTFOUND, and ROWCOUNT. These attributes can be accessed with the % delimiter to obtain information about the state of the cursor. The syntax for a cursor attribute is:

cursor_name%attribute

where cursor_name is the name of the explicit cursor.

The behaviors of the explicit cursor attributes are described in the following table.

Attribute	Description
%ISOPEN	TRUE if cursor is open. FALSE if cursor is not open.
%FOUND	INVALID_CURSOR is raised if cursor has not been OPENed. NULL before the first fetch. TRUE if record was fetched successfully. FALSE if no row was returned. INVALID_CURSOR if cursor has been CLOSEd.
%NOTFOUND	INVALID_CURSOR is raised if cursor has not been OPENed. NULL before the first fetch. FALSE if record was fetched successfully. TRUE if no row was returned. INVALID_CURSOR if cursor has been CLOSEd.
%ROWCOUNT	INVALID_CURSOR is raised if cursor has not been OPENed. The number of rows fetched from the cursor. INVALID_CURSOR if cursor has been CLOSEd.

Frequently a cursor attribute is checked as part of a WHILE loop that fetches rows from a cursor:

DECLARE

   caller_rec caller_pkg.caller_cur%ROWTYPE;

BEGIN

   OPEN caller_pkg.caller_cur;

   LOOP

      FETCH caller_pkg.caller_cur into caller_rec;

      EXIT WHEN caller_pkg.caller_cur%NOTFOUND

OR

               caller_pkg.caller_cur%ROWCOUNT > 10;

      UPDATE call

         SET caller_id = caller_rec.caller_id

         WHERE call_timestamp < SYSDATE;

   END LOOP;

   CLOSE caller_pkg.caller_cur;

END;

2 Implicit Cursors

Whenever a SQL statement is directly in the execution or exception section of a PL/SQL block, you are working with implicit cursors. These statements include INSERT, UPDATE, DELETE, and SELECT INTO statements. Unlike explicit cursors, implicit cursors do not need to be declared, OPENed, FETCHed, or CLOSEd.

SELECT statements handle the %FOUND and %NOTFOUND attributes differently from explicit cursors. When an implicit SELECT statement does not return any rows, PL/SQL immediately raises the NO_DATA_FOUND exception and control passes to the exception section. When an implicit SELECT returns more than one row, PL/SQL immediately raises the TOO_MANY_ROWS exception and control passes to the exception section.

Implicit cursor attributes are referenced via the SQL cursor. For example:

BEGIN

   UPDATE activity SET last_accessed := SYSDATE

   WHERE UID = user_id;

   IF SQL%NOTFOUND THEN

      INSERT INTO activity_log (uid,last_accessed)

      VALUES (user_id,SYSDATE);

   END IF

END;

SQL Attributes	Description
%ISOPEN	Always FALSE since the cursor is opened implicitly and closed immediately after the statement is executed.
%FOUND	NULL before the statement. TRUE if one or more rows were inserted, updated, or deleted or if only one row was selected. FALSE if no row was selected, updated, inserted, or deleted.
%NOTFOUND	NULL before the statement. TRUE if no row was selected, updated, inserted, or deleted. FALSE if one or more rows were inserted, updated, or deleted.
%ROWCOUNT	The number of rows affected by the cursor.
%BULK_ROWCOUNT (Oracle8i)	A pseudo index-by table containing the numbers of rows affected by the statements executed in bulk bind operations. See the "Bulk Binds (Oracle8i)" section for more information on %BULK_ROWCOUNT.

Use the RETURNING clause in INSERT, UPDATE, and DELETE statements to obtain data modified by the associated DML statement. This clause allows you to avoid an additional SELECT statement to query the results of the DML statement. For example:

BEGIN

   UPDATE activity SET last_accessed := SYSDATE

   WHERE UID = user_id

   RETURNING last_accessed, cost_center

   INTO timestamp, chargeback_acct;

2.1 The SELECT FOR UPDATE clause

By default, the Oracle RDBMS locks rows as they are changed. To lock all rows in a result set, use the FOR UPDATE clause in your SELECT statement when you OPEN the cursor, instead of when you change the data. Using the FOR UPDATE clause does not require you to actually make changes to the data; it only locks the rows when opening the cursor. These locks are released on the next COMMIT or ROLLBACK. As always, these row locks do not affect other SELECT statements unless they, too, are FOR UPDATE. The FOR UPDATE clause is appended to the end of the SELECT statement and has the following syntax:

SELECT ...

  FROM ...

   FOR UPDATE [OF column_reference] [NOWAIT];

where column_reference is a comma-delimited list of columns that appear in the SELECT clause. The NOWAIT keyword tells the RDBMS to not wait for other blocking locks to be released. The default is to wait forever.

In the following example, only columns from the inventory (pet) table are referenced FOR UPDATE, so no rows in the dog_breeds (dog) table are locked when hounds_in_stock_cur is opened:

DECLARE

   CURSOR hounds_in_stock_cur IS

      SELECT pet.stock_no, pet.breeder, dog.size

        FROM dog_breeds dog ,inventory pet

       WHERE dog.breed = pet.breed

         AND dog.class = 'HOUND'

         FOR UPDATE OF pet.stock_no, pet.breeder;

BEGIN

2.2 The WHERE CURRENT OF clause

UPDATE and DELETE statements can use a WHERE CURRENT OF clause if they reference a cursor declared FOR UPDATE. This syntax indicates that the UPDATE or DELETE should modify the current row identified by the FOR UPDATE cursor. The syntax is:

[UPDATE | DELETE ] ...

   WHERE CURRENT OF cursor_name;

By using WHERE CURRENT OF, you do not have to repeat the WHERE clause in the SELECT statement. For example:

DECLARE

   CURSOR wip_cur IS

      SELECT acct_no, enter_date FROM wip

       WHERE enter_date < SYSDATE -7

         FOR UPDATE;

BEGIN

   FOR wip_rec IN wip_cur

   LOOP

      INSERT INTO acct_log (acct_no, order_date)

         VALUES (wip_rec.acct_no, wip_rec.enter_

            date);

      DELETE FROM wip

         WHERE CURRENT OF wip_cur;

   END LOOP;

END;

Cursor Variables

A cursor variable is a data structure that points to a cursor object, which in turn points to the cursor's result set. You can use cursor variables to more easily retrieve rows in a result set from client and server programs. You can also use cursor variables to hide minor variations in queries.

The syntax for a REF_CURSOR type is:

TYPE ref_cursor_name IS REF CURSOR

   [RETURN record_type];

If you do not include a RETURN clause, then you are declaring a weak REF CURSOR. Cursor variables declared from weak REF CURSORs can be associated with any query at runtime. A REF CURSOR declaration with a RETURN clause defines a "strong" REF CURSOR. A cursor variable based on a strong REF CURSOR can be associated with queries whose result sets match the number and datatype of the record structure after the RETURN at runtime.

To use cursor variables, you must first create a REF_CURSOR type, then declare a cursor variable based on that type.

The following example shows the use of both weak and strong REF CURSORs:

DECLARE

   -- Create a cursor type based on the companies

      table.

   TYPE company_curtype IS REF CURSOR

      RETURN companies%ROWTYPE;

   -- Create the variable based on the REF CURSOR.

   company_cur company_curtype;

   -- And now the weak, general approach.

   TYPE any_curtype IS REF CURSOR;

   generic_curvar any_curtype;

The syntax to OPEN a cursor variable is:

OPEN cursor_name FOR select_statement;

FETCH and CLOSE a cursor variable using the same syntax as for explicit cursors. There are a number of restrictions on cursor variables:

· Cursor variables cannot be declared in a package since they do not have a persistent state.

· You cannot use the FOR UPDATE clause with cursor variables.

· You cannot assign NULLs to a cursor variable nor use comparison operators to test for equality, inequality, or nullity.

· Neither database columns nor collections can store cursor variables.

· You cannot use RPCs to pass cursor variables from one server to another.

· Cursor variables cannot be used with the dynamic SQL built-in package DBMS_SQL

C Aptitute

noreply@blogger.com (Nisheeth Ranjan) — Fri, 25 Jun 2010 15:32:00 +0530

1.    void main()
{
            int const * p=5;
            printf("%d",++(*p));
}

Answer:
Compiler error: Cannot modify a constant value.
Explanation:
p is a pointer to a "constant integer". But we tried to change the value of the "constant integer".

2.    main()
{
            char s[ ]="man";
            int i;
            for(i=0;s[ i ];i++)
            printf("\n%c%c%c%c",s[ i ],*(s+i),*(i+s),i[s]);
}

Answer:
                        mmmm
                       aaaa
                       nnnn
Explanation:
s[i], *(i+s), *(s+i), i[s] are all different ways of expressing the same idea. Generally array name is the base address for that array. Here s is the base address. i is the index number/displacement from the base address. So, indirecting it with * is same as s[i]. i[s] may be surprising. But in the case of C it is same as s[i].

3.      main()
{
            float me = 1.1;
            double you = 1.1;
            if(me==you)
printf("I love U");
else
                        printf("I hate U");
}

Answer:
I hate U

Explanation:
For floating point numbers (float, double, long double) the values cannot be predicted exactly. Depending on the number of bytes, the precession with of the value represented varies. Float takes 4 bytes and long double takes 10 bytes. So float stores 0.9 with less precision than long double.
Rule of Thumb:
Never compare or at-least be cautious when using floating point numbers with relational operators (== , >, <, <=, >=,!= ) .

4.      main()
            {
            static int var = 5;
            printf("%d ",var--);
            if(var)
                        main();
            }

Answer:
5 4 3 2 1

Explanation:
When static storage class is given, it is initialized once. The change in the value of a static variable is retained even between the function calls. Main is also treated like any other ordinary function, which can be called recursively.

5.      main()
{
            int c[ ]={2.8,3.4,4,6.7,5};
            int j,*p=c,*q=c;
            for(j=0;j<5;j++) {
                        printf(" %d ",*c);
                        ++q;     }
            for(j=0;j<5;j++){
printf(" %d ",*p);
++p;     }
}

Answer:
                        2 2 2 2 2 2 3 4 6 5

Explanation:
Initially pointer c is assigned to both p and q. In the first loop, since only q is incremented and not c , the value 2 will be printed 5 times. In second loop p itself is incremented. So the values 2 3 4 6 5 will be printed.

6.      main()
{
            extern int i;
            i=20;
printf("%d",i);
}

Answer:
Linker Error : Undefined symbol '_i'
Explanation:
                        extern storage class in the following declaration,
                                    extern int i;
specifies to the compiler that the memory for i is allocated in some other program and that address will be given to the current program at the time of linking. But linker finds that no other variable of name i is available in any other program with memory space allocated for it. Hence a linker error has occurred .

7.      main()
{
            int i=-1,j=-1,k=0,l=2,m;
            m=i++&&j++&&k++||l++;
            printf("%d %d %d %d %d",i,j,k,l,m);
}

Answer:
                        0 0 1 3 1

Explanation :
Logical operations always give a result of 1 or 0 . And also the logical AND (&&) operator has higher priority over the logical OR (||) operator. So the expression ‘i++ && j++ && k++’ is executed first. The result of this expression is 0    (-1 && -1 && 0 = 0). Now the expression is 0 || 2 which evaluates to 1 (because OR operator always gives 1 except for ‘0 || 0’ combination- for which it gives 0). So the value of m is 1. The values of other variables are also incremented by 1.

8.      main()
{
            char *p;
            printf("%d %d ",sizeof(*p),sizeof(p));
}

Answer:
                        1 2

Explanation:
The sizeof() operator gives the number of bytes taken by its operand. P is a character pointer, which needs one byte for storing its value (a character). Hence sizeof(*p) gives a value of 1. Since it needs two bytes to store the address of the character pointer sizeof(p) gives 2.

9.      main()
{
            int i=3;
            switch(i)
            {
                default:printf("zero");
                case 1: printf("one");
                           break;
               case 2:printf("two");
                          break;
              case 3: printf("three");
                          break;
              }
}

Answer :
three

Explanation :
The default case can be placed anywhere inside the loop. It is executed only when all other cases doesn't match.

10. main()
{
printf("%x",-1<<4);
}

Answer:
fff0

Explanation :
-1 is internally represented as all 1's. When left shifted four times the least significant 4 bits are filled with 0's.The %x format specifier specifies that the integer value be printed as a hexadecimal value.

11.      main()
{
            char string[]="Hello World";
            display(string);
}
void display(char *string)
{
            printf("%s",string);
}

Answer:
Compiler Error : Type mismatch in redeclaration of function display

Explanation :
In third line, when the function display is encountered, the compiler doesn't know anything about the function display. It assumes the arguments and return types to be integers, (which is the default type). When it sees the actual function display, the arguments and type contradicts with what it has assumed previously. Hence a compile time error occurs.

12.      main()
{
            int c=- -2;
            printf("c=%d",c);
}

Answer:
                                    c=2;

Explanation:
Here unary minus (or negation) operator is used twice. Same maths rules applies, ie. minus * minus= plus.
Note:
However you cannot give like --2. Because -- operator can only be applied to variables as a decrement operator (eg., i--). 2 is a constant and not a variable.
13.      #define int char
main()
{
            int i=65;
            printf("sizeof(i)=%d",sizeof(i));
}

Answer:
                        sizeof(i)=1

Explanation:
Since the #define replaces the string int by the macro char
14.      main()
{
int i=10;
i=!i>14;
Printf ("i=%d",i);
}

Answer:
i=0

Explanation:
In the expression !i>14 , NOT (!) operator has more precedence than ‘ >’ symbol. ! is a unary logical operator. !i (!10) is 0 (not of true is false). 0>14 is false (zero).

15. #include
main()
{
char s[]={'a','b','c','\n','c','\0'};
char *p,*str,*str1;
p=&s[3];
str=p;
str1=s;
printf("%d",++*p + ++*str1-32);
}

Answer:
77

Explanation:
p is pointing to character '\n'. str1 is pointing to character 'a' ++*p. "p is pointing to '\n' and that is incremented by one." the ASCII value of '\n' is 10, which is then incremented to 11. The value of ++*p is 11. ++*str1, str1 is pointing to 'a' that is incremented by 1 and it becomes 'b'. ASCII value of 'b' is 98.
Now performing (11 + 98 – 32), we get 77("M");
So we get the output 77 :: "M" (Ascii is 77).

16. #include
main()
{
int a[2][2][2] = { {10,2,3,4}, {5,6,7,8} };
int *p,*q;
p=&a[2][2][2];
*q=***a;
printf("%d----%d",*p,*q);
}

Answer:
SomeGarbageValue---1

Explanation:
p=&a[2][2][2] you declare only two 2D arrays, but you are trying to access the third 2D(which you are not declared) it will print garbage values. *q=***a starting address of a is assigned integer pointer. Now q is pointing to starting address of a. If you print *q, it will print first element of 3D array.

17.      #include
main()
{
struct xx
{
      int x=3;
      char name[]="hello";
};
struct xx *s;
printf("%d",s->x);
printf("%s",s->name);
}

Answer:
Compiler Error

Explanation:
You should not initialize variables in declaration

18. #include
main()
{
struct xx
{
int x;
struct yy
{
char s;
struct xx *p;
};
struct yy *q;
};
}

Answer:
Compiler Error

Explanation:
The structure yy is nested within structure xx. Hence, the elements are of yy are to be accessed through the instance of structure xx, which needs an instance of yy to be known. If the instance is created after defining the structure the compiler will not know about the instance relative to xx. Hence for nested structure yy you have to declare member.

19. main()
{
printf("\nab");
printf("\bsi");
printf("\rha");
}

Answer:
hai

Explanation:
\n - newline
\b - backspace
\r - linefeed

20. main()
{
int i=5;
printf("%d%d%d%d%d%d",i++,i--,++i,--i,i);
}

Answer:
45545

Explanation:
The arguments in a function call are pushed into the stack from left to right. The evaluation is by popping out from the stack. and the evaluation is from right to left, hence the result.

21. #define square(x) x*x
main()
{
int i;
i = 64/square(4);
printf("%d",i);
}

Answer:
64

Explanation:
the macro call square(4) will substituted by 4*4 so the expression becomes i = 64/4*4 . Since / and * has equal priority the expression will be evaluated as (64/4)*4 i.e. 16*4 = 64

22.      main()
{
char *p="hai friends",*p1;
p1=p;
while(*p!='\0') ++*p++;
printf("%s   %s",p,p1);
}

Answer:
ibj!gsjfoet

Explanation:
                        ++*p++ will be parse in the given order
Ø *p that is value at the location currently pointed by p will be taken
Ø ++*p the retrieved value will be incremented
Ø when ; is encountered the location will be incremented that is p++ will be executed
Hence, in the while loop initial value pointed by p is ‘h’, which is changed to ‘i’ by executing ++*p and pointer moves to point, ‘a’ which is similarly changed to ‘b’ and so on. Similarly blank space is converted to ‘!’. Thus, we obtain value in p becomes “ibj!gsjfoet” and since p reaches ‘\0’ and p1 points to p thus p1doesnot print anything.

23. #include
#define a 10
main()
{
#define a 50
printf("%d",a);
}

Answer:
50

Explanation:
The preprocessor directives can be redefined anywhere in the program. So the most recently assigned value will be taken.

24.      #define clrscr() 100
main()
{
clrscr();
printf("%d\n",clrscr());
}

Answer:
100

Explanation:
Preprocessor executes as a seperate pass before the execution of the compiler. So textual replacement of clrscr() to 100 occurs.The input program to compiler looks like this :
                        main()
                        {
                             100;
                             printf("%d\n",100);
                        }
            Note:
100; is an executable statement but with no action. So it doesn't give any problem

25.   main()
{
printf("%p",main);
}

Answer:
                        Some address will be printed.

Explanation:
            Function names are just addresses (just like array names are addresses).
main() is also a function. So the address of function main will be printed. %p in printf specifies that the argument is an address. They are printed as hexadecimal numbers.

26. main()
{
clrscr();
}
clrscr();

Answer:
No output/error

Explanation:
The first clrscr() occurs inside a function. So it becomes a function call. In the second clrscr(); is a function declaration (because it is not inside any function).

27. enum colors {BLACK,BLUE,GREEN}
main()
{

printf("%d..%d..%d",BLACK,BLUE,GREEN);

return(1);
}

Answer:
0..1..2

Explanation:
enum assigns numbers starting from 0, if not explicitly defined.

28.       void main()
{
char far *farther,*farthest;

printf("%d..%d",sizeof(farther),sizeof(farthest));

}

Answer:
4..2

Explanation:
            the second pointer is of char type and not a far pointer

29. main()
{
int i=400,j=300;
printf("%d..%d");
}

Answer:
400..300

Explanation:
printf takes the values of the first two assignments of the program. Any number of printf's may be given. All of them take only the first two values. If more number of assignments given in the program, then printf will take garbage values.

30. main()
{
char *p;
p="Hello";
printf("%c\n",*&*p);
}

Answer:
H

Explanation:
* is a dereference operator & is a reference operator. They can be applied any number of times provided it is meaningful. Here p points to the first character in the string "Hello". *p dereferences it and so its value is H. Again & references it to an address and * dereferences it to the value H.

31.       main()
{
    int i=1;
    while (i<=5)
    {
       printf("%d",i);
       if (i>2)
              goto here;
       i++;
    }
}
fun()
{
   here:
     printf("PP");
}

Answer:
Compiler error: Undefined label 'here' in function main

Explanation:
Labels have functions scope, in other words The scope of the labels is limited to functions . The label 'here' is available in function fun() Hence it is not visible in function main.

32.       main()
{
   static char names[5][20]={"pascal","ada","cobol","fortran","perl"};
    int i;
    char *t;
    t=names[3];
    names[3]=names[4];
    names[4]=t;
    for (i=0;i<=4;i++)
            printf("%s",names[i]);
}

Answer:
Compiler error: Lvalue required in function main

Explanation:
Array names are pointer constants. So it cannot be modified.

33.     void main()
{
            int i=5;
            printf("%d",i++ + ++i);
}

Answer:
Output Cannot be predicted exactly.

Explanation:
Side effects are involved in the evaluation of   i

34.       void main()
{
            int i=5;
            printf("%d",i+++++i);
}

Answer:
Compiler Error

Explanation:
The expression i+++++i is parsed as i ++ ++ + i which is an illegal combination of operators.

35.       #include
main()
{
int i=1,j=2;
switch(i)
{
case 1: printf("GOOD");
                break;
case j: printf("BAD");
               break;
}
}

Answer:
Compiler Error: Constant expression required in function main.

Explanation:
The case statement can have only constant expressions (this implies that we cannot use variable names directly so an error).
            Note:
Enumerated types can be used in case statements.

36. main()
{
int i;
printf("%d",scanf("%d",&i)); // value 10 is given as input here
}

Answer:
1

Explanation:
Scanf returns number of items successfully read and not 1/0. Here 10 is given as input which should have been scanned successfully. So number of items read is 1.

37. #define f(g,g2) g##g2
main()
{
int var12=100;
printf("%d",f(var,12));
}

Answer:
100

38. main()
{
int i=0;

for(;i++;printf("%d",i)) ;
printf("%d",i);
}

Answer:
1

Explanation:
before entering into the for loop the checking condition is "evaluated". Here it evaluates to 0 (false) and comes out of the loop, and i is incremented (note the semicolon after the for loop).

39. #include
main()
{
char s[]={'a','b','c','\n','c','\0'};
char *p,*str,*str1;
p=&s[3];
str=p;
str1=s;
printf("%d",++*p + ++*str1-32);
}

Answer:
M

Explanation:
p is pointing to character '\n'.str1 is pointing to character 'a' ++*p meAnswer:"p is pointing to '\n' and that is incremented by one." the ASCII value of '\n' is 10. then it is incremented to 11. the value of ++*p is 11. ++*str1 meAnswer:"str1 is pointing to 'a' that is incremented by 1 and it becomes 'b'. ASCII value of 'b' is 98. both 11 and 98 is added and result is subtracted from 32.
i.e. (11+98-32)=77("M");

40.       #include
main()
{
struct xx
   {
      int x=3;
      char name[]="hello";
   };
struct xx *s=malloc(sizeof(struct xx));
printf("%d",s->x);
printf("%s",s->name);
}

Answer:
Compiler Error

Explanation:
Initialization should not be done for structure members inside the structure declaration

41.       #include
main()
{
struct xx
{
              int x;
              struct yy
               {
                 char s;
                 struct xx *p;
               };
                        struct yy *q;
           };
            }

Answer:
Compiler Error

Explanation:
in the end of nested structure yy a member have to be declared

42. main()
{
extern int i;
i=20;
printf("%d",sizeof(i));
}

Answer:
Linker error: undefined symbol '_i'.

Explanation:
extern declaration specifies that the variable i is defined somewhere else. The compiler passes the external variable to be resolved by the linker. So compiler doesn't find an error. During linking the linker searches for the definition of i. Since it is not found the linker flags an error.

43. main()
{
printf("%d", out);
}

int out=100;
Answer:
Compiler error: undefined symbol out in function main.

Explanation:
The rule is that a variable is available for use from the point of declaration. Even though a is a global variable, it is not available for main. Hence an error.

44.    main()
{
extern out;
printf("%d", out);
}
int out=100;

Answer:
100

            Explanation:
This is the correct way of writing the previous program.

45. main()
{
show();
}
void show()
{
printf("I'm the greatest");
}

Answer:
Compier error: Type mismatch in redeclaration of show.

Explanation:
When the compiler sees the function show it doesn't know anything about it. So the default return type (ie, int) is assumed. But when compiler sees the actual definition of show mismatch occurs since it is declared as void. Hence the error.
The solutions are as follows:
1. declare void show() in main() .
2. define show() before main().
3. declare extern void show() before the use of show().

46. main( )
{
int a[2][3][2] = {{{2,4},{7,8},{3,4}},{{2,2},{2,3},{3,4}}};
printf(“%u %u %u %d \n”,a,*a,**a,***a);
printf(“%u %u %u %d \n”,a+1,*a+1,**a+1,***a+1);
}

Answer:
100, 100, 100, 2
114, 104, 102, 3

47.   main( )
{
int a[ ] = {10,20,30,40,50},j,*p;
for(j=0; j<5; j++)
    {
printf(“%d” ,*a);
a++;
    }
    p = a;
   for(j=0; j<5; j++)
      {
printf(“%d ” ,*p);
p++;
      }
}

Answer:
Compiler error: lvalue required.

Explanation:
Error is in line with statement a++. The operand must be an lvalue and may be of any of scalar type for the any operator, array name only when subscripted is an lvalue. Simply array name is a non-modifiable lvalue.

48.       main( )
{
static int a[ ]   = {0,1,2,3,4};
int *p[ ] = {a,a+1,a+2,a+3,a+4};
int **ptr = p;
ptr++;
printf(“\n %d %d %d”, ptr-p, *ptr-a, **ptr);
*ptr++;
printf(“\n %d %d %d”, ptr-p, *ptr-a, **ptr);
*++ptr;
printf(“\n %d %d %d”, ptr-p, *ptr-a, **ptr);
++*ptr;
      printf(“\n %d %d %d”, ptr-p, *ptr-a, **ptr);
}

Answer:
            111
            222
            333
            344

49. main( )
{
void *vp;
char ch = ‘g’, *cp = “goofy”;
int j = 20;
vp = &ch;
printf(“%c”, *(char *)vp);
vp = &j;
printf(“%d”,*(int *)vp);
vp = cp;
printf(“%s”,(char *)vp + 3);
}

Answer:
g20fy

Explanation:
Since a void pointer is used it can be type casted to any other type pointer. vp = &ch stores address of char ch and the next statement prints the value stored in vp after type casting it to the proper data type pointer. the output is ‘g’. Similarly the output from second printf is ‘20’. The third printf statement type casts it to print the string from the 4th value hence the output is ‘fy’.

50. main ( )
{
static char *s[ ] = {“black”, “white”, “yellow”, “violet”};
char **ptr[ ] = {s+3, s+2, s+1, s}, ***p;
p = ptr;
**++p;
printf(“%s”,*--*++p + 3);
}

Answer:
ck

Explanation:
In this problem we have an array of char pointers pointing to start of 4 strings. Then we have ptr which is a pointer to a pointer of type char and a variable p which is a pointer to a pointer to a pointer of type char. p hold the initial value of ptr, i.e. p = s+3. The next statement increment value in p by 1 , thus now value of p = s+2. In the printf statement the expression is evaluated *++p causes gets value s+1 then the pre decrement is executed and we get s+1 – 1 = s . the indirection operator now gets the value from the array of s and adds 3 to the starting address. The string is printed starting from this position. Thus, the output is ‘ck’.

51.    main()
{
int i, n;
char *x = “girl”;
n = strlen(x);
*x = x[n];
for(i=0; i
   {
printf(“%s\n”,x);
x++;
   }
}

Answer:
(blank space)
irl
rl
l

Explanation:
Here a string (a pointer to char) is initialized with a value “girl”. The strlen function returns the length of the string, thus n has a value 4. The next statement assigns value at the nth location (‘\0’) to the first location. Now the string becomes “\0irl” . Now the printf statement prints the string after each iteration it increments it starting position. Loop starts from 0 to 4. The first time x[0] = ‘\0’ hence it prints nothing and pointer value is incremented. The second time it prints from x[1] i.e “irl” and the third time it prints “rl” and the last time it prints “l” and the loop terminates.

52.     int i,j;
            for(i=0;i<=10;i++)
            {
            j+=5;
            assert(i<5);
            }

Answer:
Runtime error: Abnormal program termination.
                                    assert failed (i<5), ,

Explanation:
asserts are used during debugging to make sure that certain conditions are satisfied. If assertion fails, the program will terminate reporting the same. After debugging use,
            #undef NDEBUG
and this will disable all the assertions from the source code. Assertion
is a good debugging tool to make use of.

53.       main()
            {
            int i=-1;
            +i;
            printf("i = %d, +i = %d \n",i,+i);
            }

Answer:
i = -1, +i = -1

Explanation:
Unary + is the only dummy operator in C. Where-ever it comes you can just ignore it just because it has no effect in the expressions (hence the name dummy operator).

54. What are the files which are automatically opened when a C file is executed?
Answer:
stdin, stdout, stderr (standard input,standard output,standard error).

55. what will be the position of the file marker?
            a: fseek(ptr,0,SEEK_SET);
            b: fseek(ptr,0,SEEK_CUR);

Answer :
            a: The SEEK_SET sets the file position marker to the starting of the file.
                        b: The SEEK_CUR sets the file position marker to the current position
            of the file.

56.       main()
            {
            char name[10],s[12];
            scanf(" \"%[^\"]\"",s);
            }
            How scanf will execute?
Answer:
First it checks for the leading white space and discards it.Then it matches with a quotation mark and then it reads all character upto another quotation mark.

57.       What is the problem with the following code segment?
            while ((fgets(receiving array,50,file_ptr)) != EOF)
                                    ;
Answer & Explanation:
fgets returns a pointer. So the correct end of file check is checking for != NULL.

58.   main()
            {
            main();
            }

Answer:
Runtime error : Stack overflow.

Explanation:
main function calls itself again and again. Each time the function is called its return address is stored in the call stack. Since there is no condition to terminate the function call, the call stack overflows at runtime. So it terminates the program and results in an error.

59.      main()
            {
            char *cptr,c;
            void *vptr,v;
            c=10; v=0;
            cptr=&c; vptr=&v;
            printf("%c%v",c,v);
            }

Answer:
Compiler error (at line number 4): size of v is Unknown.

Explanation:
You can create a variable of type void * but not of type void, since void is an empty type. In the second line you are creating variable vptr of type void * and v of type void hence an error.

60.       main()
            {
            char *str1="abcd";
            char str2[]="abcd";
            printf("%d %d %d",sizeof(str1),sizeof(str2),sizeof("abcd"));
            }

Answer:
2 5 5

Explanation:
In first sizeof, str1 is a character pointer so it gives you the size of the pointer variable. In second sizeof the name str2 indicates the name of the array whose size is 5 (including the '\0' termination character). The third sizeof is similar to the second one.

61.      main()
            {
            char not;
            not=!2;
            printf("%d",not);
            }

Answer:
0

Explanation:
! is a logical operator. In C the value 0 is considered to be the boolean value FALSE, and any non-zero value is considered to be the boolean value TRUE. Here 2 is a non-zero value so TRUE. !TRUE is FALSE (0) so it prints 0.

62.       #define FALSE -1
            #define TRUE   1
            #define NULL   0
            main() {
               if(NULL)
                        puts("NULL");
               else if(FALSE)
                        puts("TRUE");
               else
                        puts("FALSE");
               }

Answer:
TRUE

Explanation:
The input program to the compiler after processing by the preprocessor is,
            main(){
                        if(0)
                                    puts("NULL");
            else if(-1)
                                    puts("TRUE");
            else
                                    puts("FALSE");
                        }
Preprocessor doesn't replace the values given inside the double quotes. The check by if condition is boolean value false so it goes to else. In second if -1 is boolean value true hence "TRUE" is printed.

63.     main()
            {
            int k=1;
            printf("%d==1 is ""%s",k,k==1?"TRUE":"FALSE");
            }

Answer:
1==1 is TRUE

Explanation:
When two strings are placed together (or separated by white-space) they are concatenated (this is called as "stringization" operation). So the string is as if it is given as "%d==1 is %s". The conditional operator( ?: ) evaluates to "TRUE".

64.       main()
            {
            int y;
            scanf("%d",&y); // input given is 2000
            if( (y%4==0 && y%100 != 0) || y%100 == 0 )
                 printf("%d is a leap year");
            else
                 printf("%d is not a leap year");
            }

Answer:
2000 is a leap year

Explanation:
An ordinary program to check if leap year or not.

65.       #define max 5
            #define int arr1[max]
            main()
            {
            typedef char arr2[max];
            arr1 list={0,1,2,3,4};
            arr2 name="name";
            printf("%d %s",list[0],name);
            }

Answer:
Compiler error (in the line arr1 list = {0,1,2,3,4})

Explanation:
arr2 is declared of type array of size 5 of characters. So it can be used to declare the variable name of the type arr2. But it is not the case of arr1. Hence an error.
Rule of Thumb:
#defines are used for textual replacement whereas typedefs are used for declaring new types.

66.       int i=10;
            main()
            {
            extern int i;
              {
                 int i=20;
                        {
                        const volatile unsigned i=30;
                        printf("%d",i);
                        }
                  printf("%d",i);
               }
            printf("%d",i);
            }

Answer:
30,20,10

Explanation:
'{' introduces new block and thus new scope. In the innermost block i is declared as,
            const volatile unsigned
which is a valid declaration. i is assumed of type int. So printf prints 30. In the next block, i has value 20 and so printf prints 20. In the outermost block, i is declared as extern, so no storage space is allocated for it. After compilation is over the linker resolves it to global variable i (since it is the only variable visible there). So it prints i's value as 10.

67.       main()
            {
                int *j;
                {
                 int i=10;
                 j=&i;
                 }
                 printf("%d",*j);
}

Answer:
10

Explanation:
The variable i is a block level variable and the visibility is inside that block only. But the lifetime of i is lifetime of the function so it lives upto the exit of main function. Since the i is still allocated space, *j prints the value stored in i since j points i.

68.      main()
            {
            int i=-1;
            -i;
            printf("i = %d, -i = %d \n",i,-i);
            }

Answer:
i = -1, -i = 1

Explanation:
-i is executed and this execution doesn't affect the value of i. In printf first you just print the value of i. After that the value of the expression -i = -(-1) is printed.

69.       #include
main()
{
   const int i=4;
   float j;
   j = ++i;
   printf("%d %f", i,++j);
}

Answer:
Compiler error

Explanation:
i is a constant. you cannot change the value of constant

70. #include
main()
{
int a[2][2][2] = { {10,2,3,4}, {5,6,7,8} };
int *p,*q;
p=&a[2][2][2];
*q=***a;
printf("%d..%d",*p,*q);
}

Answer:
garbagevalue..1

Explanation:
p=&a[2][2][2] you declare only two 2D arrays. but you are trying to access the third 2D(which you are not declared) it will print garbage values. *q=***a starting address of a is assigned integer pointer. now q is pointing to starting address of a.if you print *q meAnswer:it will print first element of 3D array.

71.      #include
main()
{
    register i=5;
    char j[]= "hello";
     printf("%s %d",j,i);
}

Answer:
hello 5

Explanation:
if you declare i as register compiler will treat it as ordinary integer and it will take integer value. i value may be stored either in register or in memory.

72.      main()
{
              int i=5,j=6,z;
              printf("%d",i+++j);
            }

Answer:
11

Explanation:
the expression i+++j is treated as (i++ + j)

Java Interview Q & A -Part1

noreply@blogger.com (Nisheeth Ranjan) — Tue, 22 Jun 2010 12:35:00 +0530

1. How can you achieve Multiple Inheritance in Java?

Java's interface mechanism can be used to implement multiple inheritance, with one important difference

from c++ way of doing MI: the inherited interfaces must be abstract. This obviates the need to choose

between different implementations, as with interfaces there are no implementations.

2. Replacing Characters in a String

// Replace all occurrences of 'a' with 'o'

String newString = string.replace('a', 'o');

Replacing Substrings in a String

static String replace(String str,

String pattern, String replace) {

int s = 0;

int e = 0;

StringBuffer result = new StringBuffer();

while ((e = str.indexOf(pattern, s)) >= 0) {

result.append(str.substring(s, e));

result.append(replace);

s = e+pattern.length();

}

result.append(str.substring(s));

return result.toString();

}

Converting a String to Upper or Lower Case

// Convert to upper case

String upper = string.toUpperCase();

// Convert to lower case

String lower = string.toLowerCase();

Converting a String to a Number

int i = Integer.parseInt("123");

long l = Long.parseLong("123");

float f = Float.parseFloat("123.4");

double d = Double.parseDouble("123.4e10");

Breaking a String into Words

String aString = "word1 word2 word3";

StringTokenizer parser =

new StringTokenizer(aString);

while (parser.hasMoreTokens()) {

processWord(parser.nextToken());

3. Searching a String

String string = "aString";

// First occurrence.

int index = string.indexOf('S'); // 1

// Last occurrence.

index = string.lastIndexOf('i'); // 4

// Not found.

index = string.lastIndexOf('z'); // -1

4. Connecting to a Database and Strings Handling

Constructing a String

If you are constructing a string with several appends, it may be more efficient to construct it using a

StringBuffer and then convert it to an immutable String object.

StringBuffer buf = new StringBuffer("Initial Text");

// Modify

int index = 1;

buf.insert(index, "abc");

buf.append("def");

// Convert to string

String s = buf.toString();

Getting a Substring from a String

int start = 1;

int end = 4;

String substr = "aString".substring(start, end); // Str

5. What is a transient variable?

A transient variable is a variable that may not be serialized. If you don't want some field not to be serialized,

you can mark that field transient or static.

6. What is the difference between Serializalble and Externalizable interface?

When you use Serializable interface, your class is serialized automatically by default. But you can override writeObject() and readObject()two methods to control more complex object serailization process. When you use Externalizable interface, you have a complete control over your class's serialization process.

7. How many methods in the Externalizable interface?

There are two methods in the Externalizable interface. You have to implement these two methods in order to make your class externalizable. These two methods are readExternal() and writeExternal().

8. How many methods in the Serializable interface?

There is no method in the Serializable interface. The Serializable interface acts as a marker, telling the object serialization tools that your class is serializable.

9. How to make a class or a bean serializable?

By implementing either the java.io.Serializable interface, or the java.io.Externalizable interface. As long as one class in a class's inheritance hierarchy implements Serializable or Externalizable, that class is serializable.

10. What is the serialization?

The serialization is a kind of mechanism that makes a class or a bean persistence by having its properties or fields and state information saved and restored to and from storage.

11. What are synchronized methods and synchronized statements?

Synchronized methods are methods that are used to control access to an object. A thread only executes a synchronized method after it has acquired the lock for the method's object or class. Synchronized statements are similar to synchronized methods. A synchronized statement can only be executed after a thread has acquired the lock for the object or class referenced in the synchronized statement.

12. What is synchronization and why is it important?

With respect to multithreading, synchronization is the capability to control the access of multiple threads to shared resources. Without synchronization, it is possible for one thread to modify a shared object while another thread is in the process of using or updating that object's value. This often causes dirty data and leads to significant errors.

13. What is the purpose of finalization?

The purpose of finalization is to give an unreachable object the opportunity to perform any cleanup processing before the object is garbage collected.

14. What classes of exceptions may be caught by a catch clause?

A catch clause can catch any exception that may be assigned to the Throwable type. This includes the Error and Exception types.

15. What is the difference between the Reader/Writer class hierarchy and the InputStream/OutputStream class hierarchy?

The Reader/Writer class hierarchy is character-oriented, and the InputStream/OutputStream class hierarchy is byte-oriented.

16. What happens when a thread cannot acquire a lock on an object?

If a thread attempts to execute a synchronized method or synchronized statement and is unable to acquire an object's lock, it enters the waiting state until the lock becomes available.

17. What restrictions are placed on method overriding?

Overridden methods must have the same name, argument list, and return type. The overriding method may not limit the access of the method it overrides. The overriding method may not throw any exceptions that may not be thrown by the overridden method.

18. What restrictions are placed on method overloading?

Two methods may not have the same name and argument list but different return types.

19. How does multithreading take place on a computer with a single CPU?

The operating system's task scheduler allocates execution time to multiple tasks. By quickly switching between executing tasks, it creates the impression that tasks execute sequentially.

20. How is it possible for two String objects with identical values not to be equal under the == operator?

The == operator compares two objects to determine if they are the same object in memory. It is possible for two String objects to have the same value, but located indifferent areas of memory.

21. How are this() and super() used with constructors?

this() is used to invoke a constructor of the same class. super() is used to invoke a superclass constructor.

22. What class allows you to read objects directly from a stream?

The ObjectInputStream class supports the reading of objects from input streams.

23. What is the ResourceBundle class?

The ResourceBundle class is used to store locale-specific resources that can be loaded by a program to tailor the program's appearance to the particular locale in which it is being run.

24. What interface must an object implement before it can be written to a stream as an object?

An object must implement the Serializable or Externalizable interface before it can be written to a stream as an object.

25. What is Serialization and deserialization?

Serialization is the process of writing the state of an object to a byte stream. Deserialization is the process of restoring these objects.

26. What are the Object and Class classes used for?

The Object class is the highest-level class in the Java class hierarchy. The Class class is used to represent the classes and interfaces that are loaded by a Java program.

27. Can you write Java code for declaration of multiple inheritance in Java ?

Class C extends A implements B

{

}

28. What do you mean by multiple inheritance in C++ ?

Multiple inheritance is a feature in C++ by which one class can be of different types. Say class

teachingAssistant is inherited from two classes say teacher and Student.

29. Write the Java code to declare any constant (say gravitational constant) and to get its value.

Class ABC

{

static final float GRAVITATIONAL_CONSTANT = 9.8;

public void getConstant()

{

system.out.println("Gravitational_Constant: " + GRAVITATIONAL_CONSTANT);

}

30. What are the disadvantages of using threads?

DeadLock.

31. Given two tables Student(SID, Name, Course) and Level(SID, level) write the SQL statement to get the name and SID of the student who are taking course = 3

and at freshman level.

SELECT Student.name, Student.SID

FROM Student, Level

WHERE Student.SID = Level.SID

AND Level.Level = "freshman"

AND Student.Course = 3;

32. What do you mean by virtual methods?

virtual methods are used to use the polymorhism feature in C++. Say class A is inherited from class B. If we declare say fuction f() as virtual in class B and override the same function in class A then at runtime appropriate method of the class will be called depending upon the type of the object.

33. What do you mean by static methods?

By using the static method there is no need creating an object of that class to use that method. We can directly call that method on that class. For example, say class A has static function f(), then we can call f() function as A.f(). There is no need of creating an object of class A.

34. What do mean by polymorphism, inheritance, encapsulation?

Polymorhism: is a feature of OOPl that at run time depending upon the type of object the appropriate

method is called.

Inheritance: is a feature of OOPL that represents the "is a" relationship between different objects(classes).

Say in real life a manager is a employee. So in OOPL manger class is inherited from the employee class.

Encapsulation: is a feature of OOPL that is used to hide the information.

35. What are the advantages of OOPL?

Object oriented programming languages directly represent the real life objects. The features of OOPL as

inhreitance, polymorphism, encapsulation makes it powerful.

36. How many methods do u implement if implement the Serializable Interface?

The Serializable interface is just a "marker" interface, with no methods of its own to implement.

37. Are there any other 'marker' interfaces?

java.rmi.Remote

java.util.EventListener

38. What is the difference between instanceof and isInstance?

instanceof is used to check to see if an object can be cast into a specified type without throwing a cast class exception. isInstance() determines if the specified object is assignment-compatible with the object represented by this Class. This method is the dynamic equivalent of the Java language instanceof operator. The method returns true if the specified Object argument is nonnull and can be cast to the reference type represented by this Class object without raising a ClassCastException. It returns false otherwise.

39. why do you create interfaces, and when MUST you use one?

You would create interfaces when you have two or more functionalities talking to each other. Doing it this way help you in creating a protocol between the parties involved.

40. What's the difference between the == operator and the equals() method? What test does Object.equals() use, and why?

The == operator would be used, in an object sense, to see if the two objects were actually the same object. This operator looks at the actually memory address to see if it actually the same object. The equals() method is used to compare the values of the object respectively. This is used in a higher level to see if the object values are equal. Of course the the equals() method would be overloaded in a meaningful way for whatever object that you

were working with.

41. Discuss the differences between creating a new class, extending a class and implementing an interface; and when each would be appropriate.

* Creating a new class is simply creating a class with no extensions and no implementations. The signature is as follows

public class MyClass()

{

}

* Extending a class is when you want to use the functionality of another class or classes. The extended class inherits all of the functionality of the previous class. An example of this when you create your own applet class and extend from java.applet.Applet. This gives you all of the

functionality of the java.applet.Applet class. The signature would look like this

public class MyClass extends MyBaseClass

{

}

* Implementing an interface simply forces you to use the methods of the interface implemented. This gives you two advantages. This forces you to follow a standard(forces you to use certain methods) and in doing so gives you a channel for polymorphism. This isn’t the only way you can do polymorphism but this is one of the ways.

public class Fish implements Animal

{

}

42. Given a text file, input.txt, provide the statement required to open this file with the appropriate I/O stream to be able to read and process this file.

43. Name four methods every Java class will have.

public String toString();

public Object clone();

public boolean equals();

public int hashCode();

44. What does the "abstract" keyword mean in front of a method? A class?

Abstract keyword declares either a method or a class. If a method has a abstract keyword in front of it, it is called abstract method.Abstract method has no body. It has only arguments and return type. Abstract methods act as placeholder methods that are implemented in the subclasses. Abstract classes can't be instantiated.If a class is declared as abstract,no objects of that class can be created.If a class contains any abstract method it must be declared as abstract.

45. Does Java have destructors?

No garbage collector does the job working in the background

46. Are constructors inherited? Can a subclass call the parent's class constructor? When?

You cannot inherit a constructor. That is, you cannot create a instance of a subclass using a constructor of one of it's superclasses. One of the main reasons is because you probably don't want to overide the superclasses constructor, which would be possible if they were inherited. By giving the developer the ability to override a superclasses constructor you would erode the encapsulation abilities of the language.

47. What synchronization constructs does Java provide? How do they work?

48. Why "bytecode"? Can you reverse-engineer the code from bytecode?

49. Does Java have "goto"?

50. What does the "final" keyword mean in front of a variable? A method? A class?

FINAL for a variable : value is constant

FINAL for a method : cannot be overridden

FINAL for a class : cannot be derived

30 Java Interview Q & A

noreply@blogger.com (Nisheeth Ranjan) — Tue, 22 Jun 2010 12:20:00 +0530

30 Java Interview Questions

Q1. How could Java classes direct program messages to the system console, but error messages, say to a file?

A. The class System has a variable out that represents the standard output, and the variable err that represents the standard error device. By default, they both point at

the system console. This how the standard output could be re-directed:

Stream st = new Stream(new FileOutputStream("output.txt")); System.setErr(st); System.setOut(st);

Q2. What's the difference between an interface and an abstract class?

A. An abstract class may contain code in method bodies, which is not allowed in an interface. With abstract classes, you have to inherit your class from it and Java does

not allow multiple inheritance. On the other hand, you can implement multiple interfaces in your class.

Q3. Why would you use a synchronized block vs. synchronized method?

A. Synchronized blocks place locks for shorter periods than synchronized methods.

Q4. Explain the usage of the keyword transient?

A. This keyword indicates that the value of this member variable does not have to be serialized with the object. When the class will be de-serialized, this variable will be

initialized with a default value of its data type (i.e. zero for integers).

Q5. How can you force garbage collection?

A. You can't force GC, but could request it by calling System.gc(). JVM does not guarantee that GC will be started immediately.

Q6. How do you know if an explicit object casting is needed?

A. If you assign a superclass object to a variable of a subclass's data type, you need to do explicit casting. For example:

Object a; Customer b; b = (Customer) a;

When you assign a subclass to a variable having a supeclass type, the casting is performed automatically.

Q7. What's the difference between the methods sleep() and wait()

A. The code sleep(1000); puts thread aside for exactly one second. The code wait(1000), causes a wait of up to one second. A thread could stop waiting earlier if it

receives the notify() or notifyAll() call. The method wait() is defined in the class Object and the method sleep() is defined in the class Thread.

Q8. Can you write a Java class that could be used both as an applet as well as an application?

A. Yes. Add a main() method to the applet.

Q9. What's the difference between constructors and other methods?

A. Constructors must have the same name as the class and can not return a value. They are only called once while regular methods could be called many times.

Q10. Can you call one constructor from another if a class has multiple constructors

A. Yes. Use this() syntax.

Q11. Explain the usage of Java packages.

A. This is a way to organize files when a project consists of multiple modules. It also helps resolve naming conflicts when different packages have classes with the same

names. Packages access level also allows you to protect data from being used by the non-authorized classes.

Q12. If a class is located in a package, what do you need to change in the OS environment to be able to use it?

A. You need to add a directory or a jar file that contains the package directories to the CLASSPATH environment variable. Let's say a class Employee belongs to a package

com.xyz.hr; and is located in the file c:\dev\com\xyz\hr\Employee.java. In this case, you'd need to add c:\dev to the variable CLASSPATH. If this class contains the

method main(), you could test it from a command prompt window as follows:

c:\>java com.xyz.hr.Employee

Q13. What's the difference between J2SDK 1.5 and J2SDK 5.0?

A.There's no difference, Sun Microsystems just re-branded this version.

Q14. What would you use to compare two String variables - the operator == or the method equals()?

A. I'd use the method equals() to compare the values of the Strings and the == to check if two variables point at the same instance of a String object.

Q15. Does it matter in what order catch statements for FileNotFoundException and IOExceptipon are written?

A. Yes, it does. The FileNoFoundException is inherited from the IOException. Exception's subclasses have to be caught first.

Q16. Can an inner class declared inside of a method access local variables of this method?

A. It's possible if these variables are final.

Q17. What can go wrong if you replace && with & in the following code:

String a=null; if (a!=null && a.length()>10) {...}

A. A single ampersand here would lead to a NullPointerException.

Q18. What's the main difference between a Vector and an ArrayList

A. Java Vector class is internally synchronized and ArrayList is not.

Q19. When should the method invokeLater()be used?

A. This method is used to ensure that Swing components are updated through the event-dispatching thread.

Q20. How can a subclass call a method or a constructor defined in a superclass?

A. Use the following syntax: super.myMethod(); To call a constructor of the superclass, just write super(); in the first line of the subclass's constructor.

For senior-level developers:

Q21. What's the difference between a queue and a stack?

A. Stacks works by last-in-first-out rule (LIFO), while queues use the FIFO rule

Q22. You can create an abstract class that contains only abstract methods. On the other hand, you can create an interface that declares the same

methods. So can you use abstract classes instead of interfaces?

A. Sometimes. But your class may be a descendent of another class and in this case the interface is your only option.

Q23. What comes to mind when you hear about a young generation in Java?

A. Garbage collection.

Q24. What comes to mind when someone mentions a shallow copy in Java?

A. Object cloning.

Q25. If you're overriding the method equals() of an object, which other method you might also consider?

A. hashCode()

Q26. You are planning to do an indexed search in a list of objects. Which of the two Java collections should you use:

ArrayList or LinkedList?

A. ArrayList

Q27. How would you make a copy of an entire Java object with its state?

A. Have this class implement Cloneable interface and call its method clone().

Q28. How can you minimize the need of garbage collection and make the memory use more effective?

A. Use object pooling and weak object references.

Q29. There are two classes: A and B. The class B need to inform a class A when some important event has happened. What Java technique would

you use to implement it?

A. If these classes are threads I'd consider notify() or notifyAll(). For regular classes you can use the Observer interface.

Q30. What access level do you need to specify in the class declaration to ensure that only classes from the same directory can access it?

A. You do not need to specify any access level, and Java will use a default package access level.

Operating System

noreply@blogger.com (Nisheeth Ranjan) — Sat, 19 Jun 2010 21:05:00 +0530

Operating Systems

Following are a few basic questions that cover the essentials of OS:

1. Explain the concept of Reentrancy.

It is a useful, memory-saving technique for multiprogrammed timesharing systems. A Reentrant Procedure is one in which multiple users can share a single copy of a program during the same period. Reentrancy has 2 key aspects: The program code cannot modify itself, and the local data for each user process must be stored separately. Thus, the permanent part is the code, and the temporary part is the pointer back to the calling program and local variables used by that program. Each execution instance is called activation. It executes the code in the permanent part, but has its own copy of local variables/parameters. The temporary part associated with each activation is the activation record. Generally, the activation record is kept on the stack.

Note: A reentrant procedure can be interrupted and called by an interrupting program, and still execute correctly on returning to the procedure.

2. Explain Belady's Anomaly.

Also called FIFO anomaly. Usually, on increasing the number of frames allocated to a process' virtual memory, the process execution is faster, because fewer page faults occur. Sometimes, the reverse happens, i.e., the execution time increases even when more frames are allocated to the process. This is Belady's Anomaly. This is true for certain page reference patterns.

3. What is a binary semaphore? What is its use?

A binary semaphore is one, which takes only 0 and 1 as values. They are used to implement mutual exclusion and synchronize concurrent processes.

4. What is thrashing?

It is a phenomenon in virtual memory schemes when the processor spends most of its time swapping pages, rather than executing instructions. This is due to an inordinate number of page faults.

5. List the Coffman's conditions that lead to a deadlock.

Ø Mutual Exclusion: Only one process may use a critical resource at a time.

Ø Hold & Wait: A process may be allocated some resources while waiting for others.

Ø No Pre-emption: No resource can be forcible removed from a process holding it.

Ø Circular Wait: A closed chain of processes exist such that each process holds at least one resource needed by another process in the chain.

6. What are short-, long- and medium-term scheduling?

Long term scheduler determines which programs are admitted to the system for processing. It controls the degree of multiprogramming. Once admitted, a job becomes a process.

Medium term scheduling is part of the swapping function. This relates to processes that are in a blocked or suspended state. They are swapped out of real-memory until they are ready to execute. The swapping-in decision is based on memory-management criteria.

Short term scheduler, also know as a dispatcher executes most frequently, and makes the finest-grained decision of which process should execute next. This scheduler is invoked whenever an event occurs. It may lead to interruption of one process by preemption.

7. What are turnaround time and response time?

Turnaround time is the interval between the submission of a job and its completion. Response time is the interval between submission of a request, and the first response to that request.

8. What are the typical elements of a process image?

Ø User data: Modifiable part of user space. May include program data, user stack area, and programs that may be modified.

Ø User program: The instructions to be executed.

Ø System Stack: Each process has one or more LIFO stacks associated with it. Used to store parameters and calling addresses for procedure and system calls.

Ø Process control Block (PCB): Info needed by the OS to control processes.

9. What is the Translation Lookaside Buffer (TLB)?

In a cached system, the base addresses of the last few referenced pages is maintained in registers called the TLB that aids in faster lookup. TLB contains those page-table entries that have been most recently used. Normally, each virtual memory reference causes 2 physical memory accesses-- one to fetch appropriate page-table entry, and one to fetch the desired data. Using TLB in-between, this is reduced to just one physical memory access in cases of TLB-hit.

10. What is the resident set and working set of a process?

Resident set is that portion of the process image that is actually in real-memory at a particular instant. Working set is that subset of resident set that is actually needed for execution. (Relate this to the variable-window size method for swapping techniques.)

11. When is a system in safe state?

The set of dispatchable processes is in a safe state if there exists at least one temporal order in which all processes can be run to completion without resulting in a deadlock.

12. What is cycle stealing?

We encounter cycle stealing in the context of Direct Memory Access (DMA). Either the DMA controller can use the data bus when the CPU does not need it, or it may force the CPU to temporarily suspend operation. The latter technique is called cycle stealing. Note that cycle stealing can be done only at specific break points in an instruction cycle.

13. What is meant by arm-stickiness?

If one or a few processes have a high access rate to data on one track of a storage disk, then they may monopolize the device by repeated requests to that track. This generally happens with most common device scheduling algorithms (LIFO, SSTF, C-SCAN, etc). High-density multisurface disks are more likely to be affected by this than low density ones.

14. What are the stipulations of C2 level security?

C2 level security provides for:

Ø Discretionary Access Control

Ø Identification and Authentication

Ø Auditing

Ø Resource reuse

15. What is busy waiting?

The repeated execution of a loop of code while waiting for an event to occur is called busy-waiting. The CPU is not engaged in any real productive activity during this period, and the process does not progress toward completion.

16. Explain the popular multiprocessor thread-scheduling strategies.

Ø Load Sharing: Processes are not assigned to a particular processor. A global queue of threads is maintained. Each processor, when idle, selects a thread from this queue. Note that load balancing refers to a scheme where work is allocated to processors on a more permanent basis.

Ø Gang Scheduling: A set of related threads is scheduled to run on a set of processors at the same time, on a 1-to-1 basis. Closely related threads / processes may be scheduled this way to reduce synchronization blocking, and minimize process switching. Group scheduling predated this strategy.

Ø Dedicated processor assignment: Provides implicit scheduling defined by assignment of threads to processors. For the duration of program execution, each program is allocated a set of processors equal in number to the number of threads in the program. Processors are chosen from the available pool.

Ø Dynamic scheduling: The number of thread in a program can be altered during the course of execution.

17. When does the condition 'rendezvous' arise?

In message passing, it is the condition in which, both, the sender and receiver are blocked until the message is delivered.

18. What is a trap and trapdoor?

Trapdoor is a secret undocumented entry point into a program used to grant access without normal methods of access authentication. A trap is a software interrupt, usually the result of an error condition.

19. What are local and global page replacements?

Local replacement means that an incoming page is brought in only to the relevant process' address space. Global replacement policy allows any page frame from any process to be replaced. The latter is applicable to variable partitions model only.

20. Define latency, transfer and seek time with respect to disk I/O.

Seek time is the time required to move the disk arm to the required track. Rotational delay or latency is the time it takes for the beginning of the required sector to reach the head. Sum of seek time (if any) and latency is the access time. Time taken to actually transfer a span of data is transfer time.

21. Describe the Buddy system of memory allocation.

Free memory is maintained in linked lists, each of equal sized blocks. Any such block is of size 2^k. When some memory is required by a process, the block size of next higher order is chosen, and broken into two. Note that the two such pieces differ in address only in their kth bit. Such pieces are called buddies. When any used block is freed, the OS checks to see if its buddy is also free. If so, it is rejoined, and put into the original free-block linked-list.

22. What is time-stamping?

It is a technique proposed by Lamport, used to order events in a distributed system without the use of clocks. This scheme is intended to order events consisting of the transmission of messages. Each system 'i' in the network maintains a counter Ci. Every time a system transmits a message, it increments its counter by 1 and attaches the time-stamp Ti to the message. When a message is received, the receiving system 'j' sets its counter Cj to 1 more than the maximum of its current value and the incoming time-stamp Ti. At each site, the ordering of messages is determined by the following rules: For messages x from site i and y from site j, x precedes y if one of the following conditions holds....(a) if Ti (b) if Ti=Tj and i

23. How are the wait/signal operations for monitor different from those for semaphores?

If a process in a monitor signal and no task is waiting on the condition variable, the signal is lost. So this allows easier program design. Whereas in semaphores, every operation affects the value of the semaphore, so the wait and signal operations should be perfectly balanced in the program.

24. In the context of memory management, what are placement and replacement algorithms?

Placement algorithms determine where in available real-memory to load a program. Common methods are first-fit, next-fit, best-fit. Replacement algorithms are used when memory is full, and one process (or part of a process) needs to be swapped out to accommodate a new program. The replacement algorithm determines which are the partitions to be swapped out.

25. In loading programs into memory, what is the difference between load-time dynamic linking and run-time dynamic linking?

For load-time dynamic linking: Load module to be loaded is read into memory. Any reference to a target external module causes that module to be loaded and the references are updated to a relative address from the start base address of the application module.

With run-time dynamic loading: Some of the linking is postponed until actual reference during execution. Then the correct module is loaded and linked.

26. What are demand- and pre-paging?

With demand paging, a page is brought into memory only when a location on that page is actually referenced during execution. With pre-paging, pages other than the one demanded by a page fault are brought in. The selection of such pages is done based on common access patterns, especially for secondary memory devices.

27. Paging a memory management function, while multiprogramming a processor management function, are the two interdependent?

Yes.

28. What is page cannibalizing?

Page swapping or page replacements are called page cannibalizing.

29. What has triggered the need for multitasking in PCs?

Ø Increased speed and memory capacity of microprocessors together with the support fir virtual memory and

Ø Growth of client server computing

30. What are the four layers that Windows NT have in order to achieve independence?

Ø Hardware abstraction layer

Ø Kernel

Ø Subsystems

Ø System Services.

31. What is SMP?

To achieve maximum efficiency and reliability a mode of operation known as symmetric multiprocessing is used. In essence, with SMP any process or threads can be assigned to any processor.

32. What are the key object oriented concepts used by Windows NT?

Ø Encapsulation

Ø Object class and instance

33. Is Windows NT a full blown object oriented operating system? Give reasons.

No Windows NT is not so, because its not implemented in object oriented language and the data structures reside within one executive component and are not represented as objects and it does not support object oriented capabilities .

34. What is a drawback of MVT?

It does not have the features like

Ø ability to support multiple processors

Ø virtual storage

Ø source level debugging

35. What is process spawning?

When the OS at the explicit request of another process creates a process, this action is called process spawning.

36. How many jobs can be run concurrently on MVT?

15 jobs

37. List out some reasons for process termination.

Ø Normal completion

Ø Time limit exceeded

Ø Memory unavailable

Ø Bounds violation

Ø Protection error

Ø Arithmetic error

Ø Time overrun

Ø I/O failure

Ø Invalid instruction

Ø Privileged instruction

Ø Data misuse

Ø Operator or OS intervention

Ø Parent termination.

38. What are the reasons for process suspension?

Ø swapping

Ø interactive user request

Ø timing

Ø parent process request

39. What is process migration?

It is the transfer of sufficient amount of the state of process from one machine to the target machine

40. What is mutant?

In Windows NT a mutant provides kernel mode or user mode mutual exclusion with the notion of ownership.

41. What is an idle thread?

The special thread a dispatcher will execute when no ready thread is found.

42. What is FtDisk?

It is a fault tolerance disk driver for Windows NT.

43. What are the possible threads a thread can have?

Ø Ready

Ø Standby

Ø Running

Ø Waiting

Ø Transition

Ø Terminated.

44. What are rings in Windows NT?

Windows NT uses protection mechanism called rings provides by the process to implement separation between the user mode and kernel mode.

45. What is Executive in Windows NT?

In Windows NT, executive refers to the operating system code that runs in kernel mode.

46. What are the sub-components of I/O manager in Windows NT?

Ø Network redirector/ Server

Ø Cache manager.

Ø File systems

Ø Network driver

Ø Device driver

47. What are DDks? Name an operating system that includes this feature.

DDks are device driver kits, which are equivalent to SDKs for writing device drivers. Windows NT includes DDks.

48. What level of security does Windows NT meets?

C2 level security.

Computer Networks

noreply@blogger.com (Nisheeth Ranjan) — Sat, 19 Jun 2010 21:02:00 +0530

Computer Networks

1. What are the two types of transmission technology available?

(i) Broadcast and (ii) point-to-point

2. What is subnet?

A generic term for section of a large networks usually separated by a bridge or router.

3. Difference between the communication and transmission.

Transmission is a physical movement of information and concern issues like bit polarity, synchronisation, clock etc.

Communication means the meaning full exchange of information between two communication media.

4. What are the possible ways of data exchange?

(i) Simplex (ii) Half-duplex (iii) Full-duplex.

5. What is SAP?

Series of interface points that allow other computers to communicate with the other layers of network protocol stack.

6. What do you meant by "triple X" in Networks?

The function of PAD (Packet Assembler Disassembler) is described in a document known as X.3. The standard protocol has been defined between the terminal and the PAD, called X.28; another standard protocol exists between hte PAD and the network, called X.29. Together, these three recommendations are often called "triple X"

7. What is frame relay, in which layer it comes?

Frame relay is a packet switching technology. It will operate in the data link layer.

8. What is terminal emulation, in which layer it comes?

Telnet is also called as terminal emulation. It belongs to application layer.

9. What is Beaconing?

The process that allows a network to self-repair networks problems. The stations on the network notify the other stations on the ring when they are not receiving the transmissions. Beaconing is used in Token ring and FDDI networks.

10. What is redirector?

Redirector is software that intercepts file or prints I/O requests and translates them into network requests. This comes under presentation layer.

11. What is NETBIOS and NETBEUI?

NETBIOS is a programming interface that allows I/O requests to be sent to and received from a remote computer and it hides the networking hardware from applications.

NETBEUI is NetBIOS extended user interface. A transport protocol designed by microsoft and IBM for the use on small subnets.

12. What is RAID?

A method for providing fault tolerance by using multiple hard disk drives.

13. What is passive topology?

When the computers on the network simply listen and receive the signal, they are referred to as passive because they don’t amplify the signal in any way. Example for passive topology - linear bus.

14. What is Brouter?

Hybrid devices that combine the features of both bridges and routers.

15. What is cladding?

A layer of a glass surrounding the center fiber of glass inside a fiber-optic cable.

16. What is point-to-point protocol

A communications protocol used to connect computers to remote networking services including Internet service providers.

17. How Gateway is different from Routers?

A gateway operates at the upper levels of the OSI model and translates information between two completely different network architectures or data formats

18. What is attenuation?

The degeneration of a signal over distance on a network cable is called attenuation.

19. What is MAC address?

The address for a device as it is identified at the Media Access Control (MAC) layer in the network architecture. MAC address is usually stored in ROM on the network adapter card and is unique.

20. Difference between bit rate and baud rate.

Bit rate is the number of bits transmitted during one second whereas baud rate refers to the number of signal units per second that are required to represent those bits.

baud rate = bit rate / N

where N is no-of-bits represented by each signal shift.

21. What is Bandwidth?

Every line has an upper limit and a lower limit on the frequency of signals it can carry. This limited range is called the bandwidth.

22. What are the types of Transmission media?

Signals are usually transmitted over some transmission media that are broadly classified in to two categories.

a) Guided Media:

These are those that provide a conduit from one device to another that include twisted-pair, coaxial cable and fiber-optic cable. A signal traveling along any of these media is directed and is contained by the physical limits of the medium. Twisted-pair and coaxial cable use metallic that accept and transport signals in the form of electrical current. Optical fiber is a glass or plastic cable that accepts and transports signals in the form of light.

b) Unguided Media:

This is the wireless media that transport electromagnetic waves without using a physical conductor. Signals are broadcast either through air. This is done through radio communication, satellite communication and cellular telephony.

23. What is Project 802?

It is a project started by IEEE to set standards to enable intercommunication between equipment from a variety of manufacturers. It is a way for specifying functions of the physical layer, the data link layer and to some extent the network layer to allow for interconnectivity of major LAN

protocols.

It consists of the following:

Ø 802.1 is an internetworking standard for compatibility of different LANs and MANs across protocols.

Ø 802.2 Logical link control (LLC) is the upper sublayer of the data link layer which is non-architecture-specific, that is remains the same for all IEEE-defined LANs.

Ø Media access control (MAC) is the lower sublayer of the data link layer that contains some distinct modules each carrying proprietary information specific to the LAN product being used. The modules are Ethernet LAN (802.3), Token ring LAN (802.4), Token bus LAN (802.5).

Ø 802.6 is distributed queue dual bus (DQDB) designed to be used in MANs.

24. What is Protocol Data Unit?

The data unit in the LLC level is called the protocol data unit (PDU). The PDU contains of four fields a destination service access point (DSAP), a source service access point (SSAP), a control field and an information field. DSAP, SSAP are addresses used by the LLC to identify the protocol stacks on the receiving and sending machines that are generating and using the data. The control field specifies whether the PDU frame is a information frame (I - frame) or a supervisory frame (S - frame) or a unnumbered frame (U - frame).

25. What are the different type of networking / internetworking devices?

Repeater:

Also called a regenerator, it is an electronic device that operates only at physical layer. It receives the signal in the network before it becomes weak, regenerates the original bit pattern and puts the refreshed copy back in to the link.

Bridges:

These operate both in the physical and data link layers of LANs of same type. They divide a larger network in to smaller segments. They contain logic that allow them to keep the traffic for each segment separate and thus are repeaters that relay a frame only the side of the segment containing the intended recipent and control congestion.

Routers:

They relay packets among multiple interconnected networks (i.e. LANs of different type). They operate in the physical, data link and network layers. They contain software that enable them to determine which of the several possible paths is the best for a particular transmission.

Gateways:

They relay packets among networks that have different protocols (e.g. between a LAN and a WAN). They accept a packet formatted for one protocol and convert it to a packet formatted for another protocol before forwarding it. They operate in all seven layers of the OSI model.

26. What is ICMP?

ICMP is Internet Control Message Protocol, a network layer protocol of the TCP/IP suite used by hosts and gateways to send notification of datagram problems back to the sender. It uses the echo test / reply to test whether a destination is reachable and responding. It also handles both control and error messages.

27. What are the data units at different layers of the TCP / IP protocol suite?

The data unit created at the application layer is called a message, at the transport layer the data unit created is called either a segment or an user datagram, at the network layer the data unit created is called the datagram, at the data link layer the datagram is encapsulated in to a frame and finally transmitted as signals along the transmission media.

28. What is difference between ARP and RARP?

The address resolution protocol (ARP) is used to associate the 32 bit IP address with the 48 bit physical address, used by a host or a router to find the physical address of another host on its network by sending a ARP query packet that includes the IP address of the receiver.

The reverse address resolution protocol (RARP) allows a host to discover its Internet address when it knows only its physical address.

29. What is the minimum and maximum length of the header in the TCP segment and IP datagram?

The header should have a minimum length of 20 bytes and can have a maximum length of 60 bytes.

30. What is the range of addresses in the classes of internet addresses?

Class A 0.0.0.0 - 127.255.255.255

Class B 128.0.0.0 - 191.255.255.255

Class C 192.0.0.0 - 223.255.255.255

Class D 224.0.0.0 - 239.255.255.255

Class E 240.0.0.0 - 247.255.255.255

31. What is the difference between TFTP and FTP application layer protocols?

The Trivial File Transfer Protocol (TFTP) allows a local host to obtain files from a remote host but does not provide reliability or security. It uses the fundamental packet delivery services offered by UDP.

The File Transfer Protocol (FTP) is the standard mechanism provided by TCP / IP for copying a file from one host to another. It uses the services offer by TCP and so is reliable and secure. It establishes two connections (virtual circuits) between the hosts, one for data transfer and another for control information.

32. What are major types of networks and explain?

Ø Server-based network

Ø Peer-to-peer network

Peer-to-peer network, computers can act as both servers sharing resources and as clients using the resources.

Server-based networks provide centralized control of network resources and rely on server computers to provide security and network administration

33. What are the important topologies for networks?

Ø BUS topology:

In this each computer is directly connected to primary network cable in a single line.

Advantages:

Inexpensive, easy to install, simple to understand, easy to extend.

Ø STAR topology:

In this all computers are connected using a central hub.

Advantages:

Can be inexpensive, easy to install and reconfigure and easy to trouble shoot physical problems.

Ø RING topology:

In this all computers are connected in loop.

Advantages:

All computers have equal access to network media, installation can be simple, and signal does not degrade as much as in other topologies because each computer regenerates it.

34. What is mesh network?

A network in which there are multiple network links between computers to provide multiple paths for data to travel.

35. What is difference between baseband and broadband transmission?

In a baseband transmission, the entire bandwidth of the cable is consumed by a single signal. In broadband transmission, signals are sent on multiple frequencies, allowing multiple signals to be sent simultaneously.

36. Explain 5-4-3 rule?

In a Ethernet network, between any two points on the network ,there can be no more than five network segments or four repeaters, and of those five segments only three of segments can be populated.

37. What MAU?

In token Ring , hub is called Multistation Access Unit(MAU).

38. What is the difference between routable and non- routable protocols?

Routable protocols can work with a router and can be used to build large networks. Non-Routable protocols are designed to work on small, local networks and cannot be used with a router

39. Why should you care about the OSI Reference Model?

It provides a framework for discussing network operations and design.

40. What is logical link control?

One of two sublayers of the data link layer of OSI reference model, as defined by the IEEE 802 standard. This sublayer is responsible for maintaining the link between computers when they are sending data across the physical network connection.

41. What is virtual channel?

Virtual channel is normally a connection from one source to one destination, although multicast connections are also permitted. The other name for virtual channel is virtual circuit.

42. What is virtual path?

Along any transmission path from a given source to a given destination, a group of virtual circuits can be grouped together into what is called path.

43. What is packet filter?

Packet filter is a standard router equipped with some extra functionality. The extra functionality allows every incoming or outgoing packet to be inspected. Packets meeting some criterion are forwarded normally. Those that fail the test are dropped.

44. What is traffic shaping?

One of the main causes of congestion is that traffic is often busy. If hosts could be made to transmit at a uniform rate, congestion would be less common. Another open loop method to help manage congestion is forcing the packet to be transmitted at a more predictable rate. This is called traffic shaping.

45. What is multicast routing?

Sending a message to a group is called multicasting, and its routing algorithm is called multicast routing.

46. What is region?

When hierarchical routing is used, the routers are divided into what we will call regions, with each router knowing all the details about how to route packets to destinations within its own region, but knowing nothing about the internal structure of other regions.

47. What is silly window syndrome?

It is a problem that can ruin TCP performance. This problem occurs when data are passed to the sending TCP entity in large blocks, but an interactive application on the receiving side reads 1 byte at a time.

48. What are Digrams and Trigrams?

The most common two letter combinations are called as digrams. e.g. th, in, er, re and an. The most common three letter combinations are called as trigrams. e.g. the, ing, and, and ion.

49. Expand IDEA.

IDEA stands for International Data Encryption Algorithm.

50. What is wide-mouth frog?

Wide-mouth frog is the simplest known key distribution center (KDC) authentication protocol.

51. What is Mail Gateway?

It is a system that performs a protocol translation between different electronic mail delivery protocols.

52. What is IGP (Interior Gateway Protocol)?

It is any routing protocol used within an autonomous system.

53. What is EGP (Exterior Gateway Protocol)?

It is the protocol the routers in neighboring autonomous systems use to identify the set of networks that can be reached within or via each autonomous system.

54. What is autonomous system?

It is a collection of routers under the control of a single administrative authority and that uses a common Interior Gateway Protocol.

55. What is BGP (Border Gateway Protocol)?

It is a protocol used to advertise the set of networks that can be reached with in an autonomous system. BGP enables this information to be shared with the autonomous system. This is newer than EGP (Exterior Gateway Protocol).

56. What is Gateway-to-Gateway protocol?

It is a protocol formerly used to exchange routing information between Internet core routers.

57. What is NVT (Network Virtual Terminal)?

It is a set of rules defining a very simple virtual terminal interaction. The NVT is used in the start of a Telnet session.

58. What is a Multi-homed Host?

It is a host that has a multiple network interfaces and that requires multiple IP addresses is called as a Multi-homed Host.

59. What is Kerberos?

It is an authentication service developed at the Massachusetts Institute of Technology. Kerberos uses encryption to prevent intruders from discovering passwords and gaining unauthorized access to files.

60. What is OSPF?

It is an Internet routing protocol that scales well, can route traffic along multiple paths, and uses knowledge of an Internet's topology to make accurate routing decisions.

61. What is Proxy ARP?

It is using a router to answer ARP requests. This will be done when the originating host believes that a destination is local, when in fact is lies beyond router.

62. What is SLIP (Serial Line Interface Protocol)?

It is a very simple protocol used for transmission of IP datagrams across a serial line.

63. What is RIP (Routing Information Protocol)?

It is a simple protocol used to exchange information between the routers.

64. What is source route?

It is a sequence of IP addresses identifying the route a datagram must follow. A source route may optionally be included in an IP datagram header.

SQL

noreply@blogger.com (Nisheeth Ranjan) — Sat, 19 Jun 2010 20:59:00 +0530

SQL

1. Which is the subset of SQL commands used to manipulate Oracle Database structures, including tables?

Data Definition Language (DDL)

2. What operator performs pattern matching?

LIKE operator

3. What operator tests column for the absence of data?

IS NULL operator

4. Which command executes the contents of a specified file?

START or @

5. What is the parameter substitution symbol used with INSERT INTO command?

6. Which command displays the SQL command in the SQL buffer, and then executes it?

RUN

7. What are the wildcards used for pattern matching?

_ for single character substitution and % for multi-character substitution

8. State true or false. EXISTS, SOME, ANY are operators in SQL.

True

9. State true or false. !=, <>, ^= all denote the same operation.

True

10. What are the privileges that can be granted on a table by a user to others?

Insert, update, delete, select, references, index, execute, alter, all

11. What command is used to get back the privileges offered by the GRANT command?

REVOKE

12. Which system tables contain information on privileges granted and privileges obtained?

USER_TAB_PRIVS_MADE, USER_TAB_PRIVS_RECD

13. Which system table contains information on constraints on all the tables created?

USER_CONSTRAINTS

14. TRUNCATE TABLE EMP;

DELETE FROM EMP;

Will the outputs of the above two commands differ?

Both will result in deleting all the rows in the table EMP.

15. What is the difference between TRUNCATE and DELETE commands?

TRUNCATE is a DDL command whereas DELETE is a DML command. Hence DELETE operation can be rolled back, but TRUNCATE operation cannot be rolled back. WHERE clause can be used with DELETE and not with TRUNCATE.

16. What command is used to create a table by copying the structure of another table?

Answer :

CREATE TABLE .. AS SELECT command

Explanation :

To copy only the structure, the WHERE clause of the SELECT command should contain a FALSE statement as in the following.

CREATE TABLE NEWTABLE AS SELECT * FROM EXISTINGTABLE WHERE 1=2;

If the WHERE condition is true, then all the rows or rows satisfying the condition will be copied to the new table.

17. What will be the output of the following query?

SELECT REPLACE(TRANSLATE(LTRIM(RTRIM('!! ATHEN !!','!'), '!'), 'AN', '**'),'*','TROUBLE') FROM DUAL;

TROUBLETHETROUBLE

18. What will be the output of the following query?

SELECT DECODE(TRANSLATE('A','1234567890','1111111111'), '1','YES', 'NO' );

Answer :

Explanation :

The query checks whether a given string is a numerical digit.

19. What does the following query do?

SELECT SAL + NVL(COMM,0) FROM EMP;

This displays the total salary of all employees. The null values in the commission column will be replaced by 0 and added to salary.

20. Which date function is used to find the difference between two dates?

MONTHS_BETWEEN

21. Why does the following command give a compilation error?

DROP TABLE &TABLE_NAME;

Variable names should start with an alphabet. Here the table name starts with an '&' symbol.

22. What is the advantage of specifying WITH GRANT OPTION in the GRANT command?

The privilege receiver can further grant the privileges he/she has obtained from the owner to any other user.

23. What is the use of the DROP option in the ALTER TABLE command?

It is used to drop constraints specified on the table.

24. What is the value of ‘comm’ and ‘sal’ after executing the following query if the initial value of ‘sal’ is 10000?

UPDATE EMP SET SAL = SAL + 1000, COMM = SAL*0.1;

sal = 11000, comm = 1000

25. What is the use of DESC in SQL?

Answer :

DESC has two purposes. It is used to describe a schema as well as to retrieve rows from table in descending order.

Explanation :

The query SELECT * FROM EMP ORDER BY ENAME DESC will display the output sorted on ENAME in descending order.

26. What is the use of CASCADE CONSTRAINTS?

When this clause is used with the DROP command, a parent table can be dropped even when a child table exists.

27. Which function is used to find the largest integer less than or equal to a specific value?

FLOOR

28. What is the output of the following query?

SELECT TRUNC(1234.5678,-2) FROM DUAL;

1200

SQL – QUERIES

I. SCHEMAS

Table 1 : STUDIES

PNAME (VARCHAR), SPLACE (VARCHAR), COURSE (VARCHAR), CCOST (NUMBER)

Table 2 : SOFTWARE

PNAME (VARCHAR), TITLE (VARCHAR), DEVIN (VARCHAR), SCOST (NUMBER), DCOST (NUMBER), SOLD (NUMBER)

Table 3 : PROGRAMMER

PNAME (VARCHAR), DOB (DATE), DOJ (DATE), SEX (CHAR), PROF1 (VARCHAR), PROF2 (VARCHAR), SAL (NUMBER)

LEGEND :

PNAME – Programmer Name, SPLACE – Study Place, CCOST – Course Cost, DEVIN – Developed in, SCOST – Software Cost, DCOST – Development Cost, PROF1 – Proficiency 1

QUERIES :

Find out the selling cost average for packages developed in Oracle.
Display the names, ages and experience of all programmers.
Display the names of those who have done the PGDCA course.
What is the highest number of copies sold by a package?
Display the names and date of birth of all programmers born in April.
Display the lowest course fee.
How many programmers have done the DCA course.
How much revenue has been earned through the sale of packages developed in C.
Display the details of software developed by Rakesh.
How many programmers studied at Pentafour.
Display the details of packages whose sales crossed the 5000 mark.
Find out the number of copies which should be sold in order to recover the development cost of each package.
Display the details of packages for which the development cost has been recovered.
What is the price of costliest software developed in VB?
How many packages were developed in Oracle ?
How many programmers studied at PRAGATHI?
How many programmers paid 10000 to 15000 for the course?
What is the average course fee?
Display the details of programmers knowing C.
How many programmers know either C or Pascal?
How many programmers don’t know C and C++?
How old is the oldest male programmer?
What is the average age of female programmers?
Calculate the experience in years for each programmer and display along with their names in descending order.
Who are the programmers who celebrate their birthdays during the current month?
How many female programmers are there?
What are the languages known by the male programmers?
What is the average salary?
How many people draw 5000 to 7500?
Display the details of those who don’t know C, C++ or Pascal.
Display the costliest package developed by each programmer.
Produce the following output for all the male programmers

Programmer

Mr. Arvind – has 15 years of experience

KEYS:

SELECT AVG(SCOST) FROM SOFTWARE WHERE DEVIN = 'ORACLE';
SELECT PNAME,TRUNC(MONTHS_BETWEEN(SYSDATE,DOB)/12) "AGE", TRUNC(MONTHS_BETWEEN(SYSDATE,DOJ)/12) "EXPERIENCE" FROM PROGRAMMER;
SELECT PNAME FROM STUDIES WHERE COURSE = 'PGDCA';
SELECT MAX(SOLD) FROM SOFTWARE;
SELECT PNAME, DOB FROM PROGRAMMER WHERE DOB LIKE '%APR%';
SELECT MIN(CCOST) FROM STUDIES;
SELECT COUNT(*) FROM STUDIES WHERE COURSE = 'DCA';
SELECT SUM(SCOST*SOLD-DCOST) FROM SOFTWARE GROUP BY DEVIN HAVING DEVIN = 'C';
SELECT * FROM SOFTWARE WHERE PNAME = 'RAKESH';
SELECT * FROM STUDIES WHERE SPLACE = 'PENTAFOUR';
SELECT * FROM SOFTWARE WHERE SCOST*SOLD-DCOST > 5000;
SELECT CEIL(DCOST/SCOST) FROM SOFTWARE;
SELECT * FROM SOFTWARE WHERE SCOST*SOLD >= DCOST;
SELECT MAX(SCOST) FROM SOFTWARE GROUP BY DEVIN HAVING DEVIN = 'VB';
SELECT COUNT(*) FROM SOFTWARE WHERE DEVIN = 'ORACLE';
SELECT COUNT(*) FROM STUDIES WHERE SPLACE = 'PRAGATHI';
SELECT COUNT(*) FROM STUDIES WHERE CCOST BETWEEN 10000 AND 15000;
SELECT AVG(CCOST) FROM STUDIES;
SELECT * FROM PROGRAMMER WHERE PROF1 = 'C' OR PROF2 = 'C';
SELECT * FROM PROGRAMMER WHERE PROF1 IN ('C','PASCAL') OR PROF2 IN ('C','PASCAL');
SELECT * FROM PROGRAMMER WHERE PROF1 NOT IN ('C','C++') AND PROF2 NOT IN ('C','C++');
SELECT TRUNC(MAX(MONTHS_BETWEEN(SYSDATE,DOB)/12)) FROM PROGRAMMER WHERE SEX = 'M';
SELECT TRUNC(AVG(MONTHS_BETWEEN(SYSDATE,DOB)/12)) FROM PROGRAMMER WHERE SEX = 'F';
SELECT PNAME, TRUNC(MONTHS_BETWEEN(SYSDATE,DOJ)/12) FROM PROGRAMMER ORDER BY PNAME DESC;
SELECT PNAME FROM PROGRAMMER WHERE TO_CHAR(DOB,'MON') = TO_CHAR(SYSDATE,'MON');
SELECT COUNT(*) FROM PROGRAMMER WHERE SEX = 'F';
SELECT DISTINCT(PROF1) FROM PROGRAMMER WHERE SEX = 'M';
SELECT AVG(SAL) FROM PROGRAMMER;
SELECT COUNT(*) FROM PROGRAMMER WHERE SAL BETWEEN 5000 AND 7500;
SELECT * FROM PROGRAMMER WHERE PROF1 NOT IN ('C','C++','PASCAL') AND PROF2 NOT IN ('C','C++','PASCAL');
SELECT PNAME,TITLE,SCOST FROM SOFTWARE WHERE SCOST IN (SELECT MAX(SCOST) FROM SOFTWARE GROUP BY PNAME);

32.SELECT 'Mr.' || PNAME || ' - has ' || TRUNC(MONTHS_BETWEEN(SYSDATE,DOJ)/12) || ' years of experience' “Programmer” FROM PROGRAMMER WHERE SEX = 'M' UNION SELECT 'Ms.' || PNAME || ' - has ' || TRUNC (MONTHS_BETWEEN (SYSDATE,DOJ)/12) || ' years of experience' “Programmer” FROM PROGRAMMER WHERE SEX = 'F';

II . SCHEMA :

Table 1 : DEPT

DEPTNO (NOT NULL , NUMBER(2)), DNAME (VARCHAR2(14)),

LOC (VARCHAR2(13)

Table 2 : EMP

EMPNO (NOT NULL , NUMBER(4)), ENAME (VARCHAR2(10)),

JOB (VARCHAR2(9)), MGR (NUMBER(4)), HIREDATE (DATE),

SAL (NUMBER(7,2)), COMM (NUMBER(7,2)), DEPTNO (NUMBER(2))

MGR is the empno of the employee whom the employee reports to. DEPTNO is a foreign key.

QUERIES

1. List all the employees who have at least one person reporting to them.

2. List the employee details if and only if more than 10 employees are present in department no 10.

3. List the name of the employees with their immediate higher authority.

4. List all the employees who do not manage any one.

5. List the employee details whose salary is greater than the lowest salary of an employee belonging to deptno 20.

6. List the details of the employee earning more than the highest paid manager.

7. List the highest salary paid for each job.

8. Find the most recently hired employee in each department.

9. In which year did most people join the company? Display the year and the number of employees.

10. Which department has the highest annual remuneration bill?

11. Write a query to display a ‘*’ against the row of the most recently hired employee.

12. Write a correlated sub-query to list out the employees who earn more than the average salary of their department.

13. Find the nth maximum salary.

14. Select the duplicate records (Records, which are inserted, that already exist) in the EMP table.

15. Write a query to list the length of service of the employees (of the form n years and m months).

KEYS:

1. SELECT DISTINCT(A.ENAME) FROM EMP A, EMP B WHERE A.EMPNO = B.MGR; or SELECT ENAME FROM EMP WHERE EMPNO IN (SELECT MGR FROM EMP);

2. SELECT * FROM EMP WHERE DEPTNO IN (SELECT DEPTNO FROM EMP GROUP BY DEPTNO HAVING COUNT(EMPNO)>10 AND DEPTNO=10);

3. SELECT A.ENAME "EMPLOYEE", B.ENAME "REPORTS TO" FROM EMP A, EMP B WHERE A.MGR=B.EMPNO;

4. SELECT * FROM EMP WHERE EMPNO IN ( SELECT EMPNO FROM EMP MINUS SELECT MGR FROM EMP);

5. SELECT * FROM EMP WHERE SAL > ( SELECT MIN(SAL) FROM EMP GROUP BY DEPTNO HAVING DEPTNO=20);

6. SELECT * FROM EMP WHERE SAL > ( SELECT MAX(SAL) FROM EMP GROUP BY JOB HAVING JOB = 'MANAGER' );

7. SELECT JOB, MAX(SAL) FROM EMP GROUP BY JOB;

8. SELECT * FROM EMP WHERE (DEPTNO, HIREDATE) IN (SELECT DEPTNO, MAX(HIREDATE) FROM EMP GROUP BY DEPTNO);

9. SELECT TO_CHAR(HIREDATE,'YYYY') "YEAR", COUNT(EMPNO) "NO. OF EMPLOYEES" FROM EMP GROUP BY TO_CHAR(HIREDATE,'YYYY') HAVING COUNT(EMPNO) = (SELECT MAX(COUNT(EMPNO)) FROM EMP GROUP BY TO_CHAR(HIREDATE,'YYYY'));

10. SELECT DEPTNO, LPAD(SUM(12*(SAL+NVL(COMM,0))),15) "COMPENSATION" FROM EMP GROUP BY DEPTNO HAVING SUM( 12*(SAL+NVL(COMM,0))) = (SELECT MAX(SUM(12*(SAL+NVL(COMM,0)))) FROM EMP GROUP BY DEPTNO);

11. SELECT ENAME, HIREDATE, LPAD('*',8) "RECENTLY HIRED" FROM EMP WHERE HIREDATE = (SELECT MAX(HIREDATE) FROM EMP) UNION SELECT ENAME NAME, HIREDATE, LPAD(' ',15) "RECENTLY HIRED" FROM EMP WHERE HIREDATE != (SELECT MAX(HIREDATE) FROM EMP);

12. SELECT ENAME,SAL FROM EMP E WHERE SAL > (SELECT AVG(SAL) FROM EMP F WHERE E.DEPTNO = F.DEPTNO);

13. SELECT ENAME, SAL FROM EMP A WHERE &N = (SELECT COUNT (DISTINCT(SAL)) FROM EMP B WHERE A.SAL<=B.SAL);

14. SELECT * FROM EMP A WHERE A.EMPNO IN (SELECT EMPNO FROM EMP GROUP BY EMPNO HAVING COUNT(EMPNO)>1) AND A.ROWID!=MIN (ROWID));

15. SELECT ENAME "EMPLOYEE",TO_CHAR(TRUNC(MONTHS_BETWEEN(SYSDATE,HIREDATE)/12))||' YEARS '|| TO_CHAR(TRUNC(MOD(MONTHS_BETWEEN (SYSDATE, HIREDATE),12)))||' MONTHS ' "LENGTH OF SERVICE" FROM EMP;

1	10	35	28	3	12
18	27	2	11	34	29
9	36	17	30	13	4
26	19	8	33	22	31
7	16	21	24	5	14
20	25	6	15	32	23

1	14	63	40	3	24	57	42
62	39	2	13	60	41	4	23
15	64	61	36	25	56	43	58
38	33	12	55	52	59	22	5
11	16	37	26	35	54	51	44
32	29	34	53	50	47	6	21
17	10	27	30	19	8	45	48
28	31	18	9	46	49	20	7

	R	T	A	M	L	O	J	E
1	2	9	7	6	1	0	5	3
2	2	9	7	6	0	1	4	3
3	2	9	7	6	0	1	5	4
4	2	9	6	5	4	8	7	3
5	2	4	7	6	3	9	0	8
6	6	7	3	2	0	5	9	8
7	6	7	3	2	4	1	5	0

1	10	35	28	3	12
18	27	2	11	34	29
9	36	17	30	13	4
26	19	8	33	22	31
7	16	21	24	5	14
20	25	6	15	32	23

1	14	63	40	3	24	57	42
62	39	2	13	60	41	4	23
15	64	61	36	25	56	43	58
38	33	12	55	52	59	22	5
11	16	37	26	35	54	51	44
32	29	34	53	50	47	6	21
17	10	27	30	19	8	45	48
28	31	18	9	46	49	20	7

	R	T	A	M	L	O	J	E
1	2	9	7	6	1	0	5	3
2	2	9	7	6	0	1	4	3
3	2	9	7	6	0	1	5	4
4	2	9	6	5	4	8	7	3
5	2	4	7	6	3	9	0	8
6	6	7	3	2	0	5	9	8
7	6	7	3	2	4	1	5	0

1	10	35	28	3	12
18	27	2	11	34	29
9	36	17	30	13	4
26	19	8	33	22	31
7	16	21	24	5	14
20	25	6	15	32	23

1	14	63	40	3	24	57	42
62	39	2	13	60	41	4	23
15	64	61	36	25	56	43	58
38	33	12	55	52	59	22	5
11	16	37	26	35	54	51	44
32	29	34	53	50	47	6	21
17	10	27	30	19	8	45	48
28	31	18	9	46	49	20	7

	R	T	A	M	L	O	J	E
1	2	9	7	6	1	0	5	3
2	2	9	7	6	0	1	4	3
3	2	9	7	6	0	1	5	4
4	2	9	6	5	4	8	7	3
5	2	4	7	6	3	9	0	8
6	6	7	3	2	0	5	9	8
7	6	7	3	2	4	1	5	0