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CIVIL SERVICES' (I.A.S.) EXAMINATION

noreply@blogger.com (Mangesh Shinde) — Fri, 17 Jul 2015 09:21:00 +0000

The Union Public Service Commission (U.P.S.C.) conducts Civil Services' Examination once a year in two stages. The Preliminary Examination (Objective Type) for selection of candidates for the Main Examination is held in the month of May.

The Civil Services Main Examination is held in the months of October/November. Blank application forms and other particulars are published in the Employment News, generally in the month of December.

The last date for the submission of applications to the Secretary, Union Public Service Commission, Dholpur House, Shahjahan Road, NewDelhi-11001 1 is usually the last week of January of the year of examination.

The Combined Civil Services Examination is conducted for Recruitment to the following Services/Posts:

1. Indian Administrative Service.
2. Indian Foreign Service.
3. Indian Police Service.
4. Indian P & T Accounts & Finance Service, Group 'A'.
5. Indian Audit and Accounts Service, Group 'A'.
6. Indian Customs and Central Excise Service, Group 'A'.
7. Indian Defence Accounts Service, Group `A'.
8. Indian Revenue Service, Group 'A'.
9. Indian Ordnance Factories Service, Group `A' (Assistant Manager, Non-technical)
10. Indian Postal Service, Group `A'.
11. Indian Civil Accounts Service, Group `A'.
12. Indian Railway Traffic Service, Group 'A'.
13. Indian Railway Accounts Service, Group `A'.
14. Indian Railway Personnel Service, Group 'A'.
15. Posts of Assistant Security Officer, Group `A' in Railway Protection Force.
16. Indian Defence Estates Service, Group `A'.
17. Indian Information Service (Junior Grade), Group `A'.
18. Indian Trade Service, Group `A' (Grade III).
19. Posts of Assistant Commandant, Group' A' in the Central Industrial Security Force.
20. Posts of Deputy Superintendent of Police, Group `A' in the Central Bureau of      Investigation.
21. Central Secretariat Service, Group `B' (Section Officers Grade).
22. Railway Board Secretariat Service, Group `B' (Section Officers Grade).
23. Armed Forces Headquarters Civil Service, Group `B' [Assistant Civilian Staff Officers      Grade).
24. Customs Appraisers' Service, Group `B'.
25. Delhi and Andaman & Nicobar Islands, Lakshadweep, Daman & Diu and Dadar & Nagar      Haveli Civil Service, Group 'B'.
26. Pondicherry Civil Service, Group `B'.

Number of Chances: Every candidate, appearing in the examination, who is otherwise eligible, shall be permitted four chances at the examination.

Age:

A candidate must have attained the age of 21 years and must not have attained the age of 28 years on the 1st August of the year in which examination is being held.
The upper age limit may be relaxed in respect of the Scheduled Castes, Scheduled Tribes and Other Backward Classes and such other categories of persons as may from time to time be notified in this behalf by the Government of India to the extent and subject to the conditions notified in respect of each category.
Educational Qualifications:

A candidate must hold a degree of any of the Universities (any division) incorporated by an Act of the Central or State Legislature in India or other Educational Institutions established by an Act of Parliament or declared to be deemed as a University under Section 3 of the University Grants Commission Act, 1956, or possess an equivalent qualification.
A candidate who has appeared or intends to appear at an examination passing of which would render him eligible to appear at this examination, but has not been informed of the result, may also apply for admission to the examination.

Examination Fee: Rs. 40/- (Rupees Forty only). No fee for Scheduled Castes, Scheduled Tribes, etc. Fee should be paid through Central Recruitment Fee Stamps only.
Plan of the Examination: The Civil Services Examination comprises two successive stages:

1. Civil Services (Preliminary) Examination (Objective type) for the selection of candidates for Main Examination,

2. Civil Services (Main) Examination (Written and Interview) for the selection of candidates for the various services and posts.
The Preliminary Examination:

It will consist of two papers of Objective Type (multiple-choice questions) and carry a maximum of 450 marks. This examination is meant to serve as a screening test only; the marks obtained in the Preliminary Examination by the candidates who are declared qualified for admission to the Main Examination will not be counted for determining their final order of merit.

The number of candidates to be admitted to the Main Examination will be about twelve to thirteen times the total approximate number of vacancies to be filled in the year in the various Services and Posts.

Only those candidates, who are declared by the Commission to have qualified in the Preliminary Examination in a year will be eligible for admission to the Main Examination of that year, provided they are otherwise eligible for admission to the Main Examination.
The Preliminary Examination will consist of two papers:
Paper I- General Studies 150 marks
Paper II- One subject to be selected
from the list of Optional
subjects set out in Para 9
below 300 marks
------------
Total 450 marks
------------
List of Optional Subjects for Preliminary Examination:

1. Agriculture
2. Animal Husbandry and Veterinary Science
3. Botany
4. Chemistry
5. Civil Engineering
6. Commerce
7. Economics
8. Electrical Engineering
9. Geography
10. Geology
11. Indian History
12. Law
13. Mathematics
14. Mechanical Engineering
15. Medical Science
16. Philosophy
17. Physics
18. Political Science
19. Psychology
20. Public Administration
21. Sociology
22. Statistics
23. Zoology

Note: Both the question papers in the Prelim Test will be of the Objective Type (multiple-choice questions).
The question papers will be set both in Hindi and English.
The course content of the syllabi for the optional subjects will be of the degree level.
Each paper will be of two hours' duration.
Main Examination: The written examination will consist of the following papers:

Paper I One of the Indian Languages 300 marks
to be selected by the candidates from
the Languages included in the Eighth
Schedule to the Constitution
Paper II English 300 marks
Paper III Essay 200 marks
Papers IV&V General Studies 300 marks for each paper
Papers VI Any two subjects to be selected 300 marks
VII, VIII from the list of Optional for each paper
and IX subjects set out in Para 2 below.

Each subject will have two papers Interview Test will carry 300 marks.

Note: The papers on Indian Languages and English will be of Matriculation or equivalent standard and will be of qualifying nature; the marks obtained in these papers will not be counted for ranking.
The papers on Essay, General Studies and Optional Subjects. Of only such candidates will be evaluated as attain such minimum standard as may be fixed by the Commission in their discretion for the qualifying papers on Indian Language and English.
The Paper I of Indian Languages will not, however, be compulsory for candidates hailing from the North-Eastern States of Arunachal Pradesh, Meghalaya, Mizoram and Nagaland and also for candidates hailing from the State of Sikkim.
For the Language Papers, the script to be used by candidates will be as under:
Language Script
Assamese Assamese
Bengali Bengali
Gujarati Gujarati
Hindi Devanagari
Kannada Kannada
Kashmiri Persian
Konkani Devanagari
Malayalam Malayalam
Manipuri Bengali
Marathi Devanagari
Nepali Devanagari
Oriya Oriya
Punjabi Gurumukhi
Sanskrit Devanagari
Sindhi Devanagari or Arabic
Tamil Tamil
Telugu Telugu
Urdu Persian

AGRICULTURAL SCIENTIST RECRUITMENT BOARD

noreply@blogger.com (Mangesh Shinde) — Fri, 17 Jul 2015 09:18:00 +0000

Agricultural Research Service

National Eligibility Test/ Senior Research Fellowship Examination

The Agricultural Scientists Recruitment Board (ASRB) holds a Competitive Examination for recruiting Scientists of the ARS in the pay scale of Rs. 8,000-13,500 in the ICAR Institutes, combined with National Eligibility Test (NET) for recruitment of Lecturers and Assistant Professors by the State Agricultural Universities (SAUS) and for award of ICAR Senior Research Fellowships.

The selected candidates for Agricultural Research Service must serve in the institutes to which they are posted until they find appointment for higher positions through selection at other institutes.

(i) Candidates successful in ARS are appointed as Scientists in the Indian Council of Agricultural Research in the pay scale of Rs. 8,000-13,500.

(ii) Candidates clearing the National Eligibility Test are recommended to various State Agricultural Universities who will consider them for appointment as Lecturers or Assistant Professors after assessing their suitability through interview or such other tests as may be devised by the SAU concerned. Eligibility certificates are issued to these candidates by the ASRB to enable them to apply for vacancies to be notified or advertised by the SAUs.

(iii) For Senior Research Fellowship of ICAR, the Education Division prepares a merit list based on the following criteria:

(a) 60% marks for written examination
(b) 20% marks for Academic performance
(c) 20% marks for Viva-voce.

Note: The ASRB may increase or decrease the number of examination centers according to requirement. Candidates must select the examination center carefully. No request for change of center is entertained.

The number of candidates called for Viva-Voce is in the ratio of 1:5. The selection of candidates for Viva-Voce is made on the basis of marks obtained in the written examination and weightage is given to marks obtained by the candidate in Intermediate, Graduate and Post-graduate examinations.

Qualifications: Master's degree or equivalent with good academic record in the concerned subject.

Agricultural Research Service and National Eligibility Test:

(i) Ph.D, with at least a (II div) Master's degree (50%). OR
(ii) A candidate without a Ph.D degree should possess at least a high II div. Master's degree (55% and above) and I1 div. Bachelor's degree (50%).

Senior Research Fellowship Examination: The candidate must possess a high Second Class Master's degree (55% and above or equivalent OGPA) in the concerned discipline.

Age:

(a) Agricultural Research Service
21-30 yrs. on August 1. In-service candidates of ICAR/State Agricultural Universities less than 35 years in age are also eligible to appear subject to possession of prescribed qualifications.

(b) National Eligibility Test
21 years on August 1. No upper age limit.

(c) Senior Research Fellowship
21-30 yrs on August 1. In-service candidates of ICAR/State Agricultural Universities less than 45 years in age are also eligible to appear subject to possession of prescribed qualifications.

(d) The upper age limit is also relaxable for:

(i) SC or ST candidates upto a maximum of five years.
(ii) Other bonafide displaced persons/repatriates of Indian origin/Defense Services personnel/Border Security Force personnel etc. as per Government of India rules.

Subjects for Examination

Three papers in the written examination:

Paper	Subject Marks	Duration
1. General Knowledge 2. Professional Paper 1 3. Professional Paper 2	200 200 200	3 hrs. 3 hrs. 3 hrs.

Successful candidates are called for an Interview.

Apply to: Secretary, Agricultural Scientists Recruitment Board, Krishi Anusandhan Bhavan, Pusa, New Delhi-110012, on the prescribed application form.

Examination Centers: Anand, Bangalore, Bombay, Calcutta, Coimbatore, Cuttack, DehraDun, Delhi, Hissar, Hyderabad, Indore, Jabalpur, Jodhpur, Karnal, Lucknow, Ludhiana, Madras, Nagpur, Palampur, Port Blair, Ranchi, Shillong, Trivandrum, Varanasi.

Examination Fee: SC/ST candidates: Nil.

Others: Rs. 100/-.

What Questions to ask the HR

noreply@blogger.com (Mangesh Shinde) — Tue, 02 Dec 2014 19:56:00 +0000

Few questions you should ask the HR :

What kinds of assignments might I expect the first six months on the job?
How often are performance reviews given?
Please describe the duties of the job for me.
What products (or services) are in the development stage now?
Do you have plans for expansion?
What are your growth projections for next year?
Have you cut your staff in the last three years?
Are salary adjustments geared to the cost of living or job performance?
Does your company encourage further education?
How do you feel about creativity and individuality?
Do you offer flextime?
What is the usual promotional time frame?
Does your company offer either single or dual career-track programs?
What do you like best about your job/company?
Once the probation period is completed, how much authority will I have over decisions?
Has there been much turnover in this job area?
Do you fill positions from the outside or promote from within first?
Is your company environmentally conscious? In what ways?
In what ways is a career with your company better than one with your competitors?
Is this a new position or am I replacing someone?
What is the largest single problem facing your staff (department) now?
May I talk with the last person who held this position?
What qualities are you looking for in the candidate who fills this position?
What skills are especially important for someone in this position?
What characteristics do the achievers in this company seem to share?
Who was the last person that filled this position, what made them successful at it, where are they today, and how may I contact them?
Is there a lot of team/project work?
Will I have the opportunity to work on special projects?
Where does this position fit into the organizational structure?
How much travel, if any, is involved in this position?
What is the next course of action? When should I expect to hear from you or should I contact you?

Feel free to add more in comments section!! :)

Predict the output or error(s) for the following:

noreply@blogger.com (Mangesh Shinde) — Tue, 02 Dec 2014 19:33:00 +0000

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 tochange 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.

How do I "get" a null pointer in my programs?

noreply@blogger.com (Mangesh Shinde) — Sun, 30 Nov 2014 19:03:00 +0000

Answer: According to the language definition, a constant 0 in a pointer context is converted into a null pointer at compile time. That is, in an initialization, assignment, or comparison when one side is a variable or expression of pointer type, the compiler can tell that a constant 0 on the other side requests a null pointer, and generate the correctly-typed null pointer value. Therefore, the following fragments are perfectly legal:

char *p = 0;
if(p != 0)

However, an argument being passed to a function is not necessarily recognizable as a pointer context, and the compiler may not be able to tell that an unadorned 0 "means" a null pointer. For instance, the Unix system call "execl" takes a variable-length, null-pointer-terminated list of character pointer arguments. To generate a null pointer in a function call context, an explicit cast is typically required:

execl("/bin/sh", "sh", "-c", "ls", (char *)0);

If the (char *) cast were omitted, the compiler would not know to pass a null pointer, and would pass an integer 0 instead. (Note that many Unix manuals get this example wrong.)

When function prototypes are in scope, argument passing becomes an "assignment context," and most casts may safely be omitted, since the prototype tells the compiler that a pointer is required, and of which type, enabling it to correctly cast unadorned 0's. Function prototypes cannot provide the types for variable arguments in variable-length argument lists, however, so explicit casts are still required for those arguments. It is safest always to cast null pointer function arguments, to guard against varargs functions or those without prototypes, to allow interim use of non-ANSI compilers, and to demonstrate that you know what you are doing.

Summary:
Unadorned 0 okay: Explicit cast required:
initialization function call, no prototype in scope assignment variable argument in comparison arargs function call, prototype in scope, fixed argument

References: K&R I Sec. A7.7 p. 190, Sec. A7.14 p. 192; K&R II
Sec. A7.10 p. 207, Sec. A7.17 p. 209; H&S Sec. 4.6.3 p. 72; ANSI
Sec. 3.2.2.3 .

What is this infamous null pointer, anyway?

noreply@blogger.com (Mangesh Shinde) — Sun, 30 Nov 2014 17:20:00 +0000

Answer: The language definition states that for each pointer type, there is a special value -- the "null pointer" -- which is distinguishable from all other pointer values and which is not the address of any object. That is, the address-of operator & will never yield a null pointer, nor will a successful call to malloc. (malloc returns a null pointer when it fails, and this is a typical use of null pointers: as a "special" pointer value with some other meaning, usually "not allocated" or "not pointing anywhere yet.")

A null pointer is conceptually different from an uninitialized pointer. A null pointer is known not to point to any object; an uninitialized pointer might point anywhere. See also questions 49, 55, and 85.

As mentioned in the definition above, there is a null pointer for each pointer type, and the internal values of null pointers for different types may be different. Although programmers need not know the internal values, the compiler must always be informed which type of null pointer is required, so it can make the distinction if necessary (see below).

References: K&R I Sec. 5.4 pp. 97-8; K&R II Sec. 5.4 p. 102; H&S
Sec. 5.3 p. 91; ANSI Sec. 3.2.2.3 p. 38.

Aptitude Test Practice Questions - With Answers

noreply@blogger.com (Mangesh Shinde) — Sun, 23 Nov 2014 10:56:00 +0000

Question 1. Which of the following is least like the others?
A. cube
B. sphere
C. pyramid
D. circle

Anwser:
D (because the circle is the only two-dimensional figure)

Question 2. Consider a language which uses the following set of characters:
Small set: { a b c }
Large set: { A B C }
Punctuation set: { x y }
This language must follow the following rules:
1. A punctuation character must end all series.
2. A series can have up to but no more than 4 characters,including punctuation characters.

Does the following series follow all the rules of the language defined above?
axBy
A. Yes
B. No

Answer:
A (the series has only four characters and ends in a punctuation character)

Question 3. Consider the following flow chart for a customer:

The person in No.1 is:
A. Married, with children
B. Married, with at least one son
C. Unmarried, with at least one daughter
D. Unmarried, with at least one son
E. Unmarried, with no children

Answer:
D (married: no; children: yes; male child: yes)

Question 4. Susan can type 10 pages in 5 minutes. Mary can type 5 pages in 10 minutes. Working together, how many pages can they type in 30 minutes?
A. 15
B. 20
C. 25
D. 65
E. 75

Answer:
E (30/5=6; 6*10=60; Susan will type 60 pages in 30 min. 30/10=3; 5*3=15; Mary will type 15 pages in 30 min. 60+15=75)

Question 5. Consider the following series:
3, 4, 6, 9, 13, ____ What comes next?
A. 15
B. 16
C. 17
D. 18
E. 19

Answer:
D (3+1=4; 4+2=6; 6+3=9; 9+4=13; 13+5=18)

Army Institute of Technology, Pune : Addmition 2015 - Based on JEE

noreply@blogger.com (Mangesh Shinde) — Thu, 20 Nov 2014 10:52:00 +0000

Important Dates:
Last date for submission of completed AIT Application Form: April 30, 2015
Date of JEE (Main)-2014 (offline) : April 4, 2015
Date of JEE (Main)-2014 Online : April-10 - 11, 2015

Army Institute of Technology, Pune invites application to admission for the year 2015-2016 for the BE Courses commencing in July/August.

AIT admission is based on the merit of JEE. All applicants are therefore required to appear in JEE and also apply to AIT.

Eligibility:

The eligibility criteria for admission to AIT is given in detail in the Prospectus which is available in February every year. Brief outline is as below :-

Candidate must be a ward of one of the following :-

Serving Army Persons.
Serving medical officers of IN and IAF who have served in the Army for more than 10 yrs and are members of AOBF.
Retired Army pers who in receipt of regular pension, or former Army pers who retired after min service of 10 yrs.
Former Army pers who died during service/after retirement and family pension is granted.
TA pers who have completed 10 yrs of embodied service.
DSC pers who are serving or retired with pension.

War widow (in receipt of liberalized pension) and children in receipt of family pension (due to death of both parents) are also eligible.
Details of eligibility of adopted/step children are as in the prospectus.
Should have passed, or appeared for 10+2 exam, at the time of applying for admission to AIT. Minimum acceptable mark in PCM are 50%.
Age of the candidate must be between 16 to 21 years on 01 Jul 2014.
Must be an Indian citizen, or a PIO who obtains Indian citizenship before admission.

JEE Application Form:

Candidate can apply for JEE 2015 'Online'. For this candidates are advised to visit www.jeemain.nic.in. Details can be obtained from

Executive Director
JEE Main Secretariat
Central Board of Secondary Education
PS 1-2, Institutional Area, IP Extension
Patparganj, Delhi - 110 092
Tel: 91-11-22239177-80 Extn 110,151 &157
Fax: 011-22246095

AIT Application Form:

JEE Application Form should be filled before AIT Application Form. The JEE Roll No must be endorsed on the AIT Application Form before submitting it to AIT. Failure to do so would result in rendering the AIT Application Form null and void.

Prospectus and Application Form can be obtained by hand or by post from AIT as well as from HQ Commands and Army HQ (AWES Branch). The Prospectus can also be downloaded from the institute website.

Cost of Prospectus and Application form is as follows:-

By Post :- Rs 820/- through bank Demand Draft drawn in favour of Director, AIT payable at Pune.
By Hand :- Rs 750/- through bank Demand Draft drawn in favour of Director, AIT payable at Pune.
Down-loaded Forms:- Down-loaded Application Form should be accompanied with a bank Demand Draft for Rs 750/- drawn in favour of Director, AIT payable at Pune. The details of Bank Draft should be mentioned clearly in the Application Form.
Counseling at AIT, Pune, will be after one month from the date of declaration of JEE Results. Dates of counseling will be hosted on the AIT website

Note : Filling up of JEE form and AIT Form is compulsory. Failing to fill up any one of the form will lead to rejection of admission to AIT.

For more details refer official website

Contact Details
Address : Army Institute of Technology, Dighi Hills, Pune-411015
Phone : 020-27157534, 27157612
Fax : 020-27157534
Mobile :
E-mail : Contact I
Website : www.aitpune.com/

What is Encapsulation

noreply@blogger.com (Mangesh Shinde) — Fri, 04 Jul 2014 20:25:00 +0000

Encapsulation in the mechanism that binds together code and data and that leaps both safe from outside interference or misuse. It also allows the creation of an object. More simply, an object in a logical entity that encapsulate both data and the code that manipulators that data.

Within an object, some of the code and/ or data may be private to the objected and in accessible to anything outside the object. In this way and object provides a significant level of protection against some other unrelated part of the program accidentally modifying or incorrectly using the private parts of the object.

What is Data abstraction in oop

noreply@blogger.com (Mangesh Shinde) — Fri, 04 Jul 2014 20:23:00 +0000

The wrapping up of data and functions into a single unit (called class) is known as encapsulation. Data encapsulation is the most striking feature of a class. The data is not accessible to the outside world and only those functions which are wrapped in the class can access it. These functions provide the interface between the object’s data and the program. This insulation of the data from direct access by the program is called ‘data hiding‘.

Abstraction refers to the act of representing essential features without including the background details or explanations. Classes use the concept of abstraction and are defined as a list of abstract attributes such as size, weight and cost, and functions to operate on these attributes. They encapsulate al1 the essential properties of the objects that are to be created. Since the classes use the concept of data abstraction, they are known as Abstract Data Types (ADT).

Matrices: An introduction

noreply@blogger.com (Mangesh Shinde) — Fri, 04 Jul 2014 20:19:00 +0000

This tutorial discusses some of the ins and outs of matrices. Matrices can be fun, but more importantly, they can really save you time.

Introduction

A matrix is, by definition, a rectangular array of numeric or algebraic quantities which are subject to mathematical operations. Matrices can be defined in terms of their dimensions (number of rows and columns). Let us take a look at a matrix with 4 rows and 3 columns (we denote it as a 4×3 matrix and call it A):

Each individual item in a matrix is called a cell, and can be denoted by the particular row and column it resides in. For instance, in matrix A, element a32 can be found where the 3rd row and the 2nd column intersect.

What are they used for?

Matrices are used to represent complicated or time-consuming mathematical operations. A single matrix can hold an infinite number of calculations, which can then be applied to a number, vector, or another matrix. There are several operations that can be done on matrices, including addition, multiplication and inverse calculation; some of which will be discussed shortly. Operations done on one matrix can be transferred to another matrix simply by concatenating the two (by matrix multiplication). Matrices often find their use in 3 dimensional applications, were numerous identical operations are performed on thousands of vectors 30 or 40 times a second. Combining all these operations in one single matrix significantly improves the speed and functionality of a 3D rendering pipeline. Matrices are also used in financial processes (again, where a large number of data has to be processed in a similar fashion).

Operations on Matrices

1.Addition and Subtraction

Addition and subtraction operations can easily be performed on matrices, provided the matrices have the same dimensions. All that is required is to add or subtract the corresponding cells of each matrix involved in the operation. Let us take a look at the addition of two 2×3 matrices, A and B:

2.Scalar Multiplication

Multiplying a matrix by a scalar value involves multiplying every element of the matrix by that value. Here we multiply our 2×3 matrix A by a scalar value β:

3.Matrix Multiplication

The multiplication operation on matrices differs significantly from its real counterpart. One major difference is that multiplication can be performed on matrices with different dimensions. The first restriction is that the first matrix has to have the same amount of columns as the second has rows. The reason for this will become clear shortly. Another thing to note is that matrix multiplication is not commutative i.e., (CD) does not equal (DC).

The procedure for matrix multiplication is rather simple. First, we determine the dimensions of the resultant matrix. All we require is that there are as many columns in the first matrix as there are rows in the second. A simple way of determining is to look at the nearest and farthest dimensions of two matrix symbols written next to each other, for instance: C[2x3] D[3x2]. The nearest dimensions are both equal to 3, and so we know that the operation is possible. The farthest dimensions will give us the dimensions of the product matrix, so our result will be a 2×2 matrix. The general rule says that in order to perform the multiplication AB, where A is a (m x n) matrix and B a (k x l) matrix, we must have n=k. The result will be a (m x l) matrix.

Performing the operation product involves multiplying the cells of a particular rows in the first matrix by the cells of a particular column in the second matrix, adding the products, and storing the result in the cell of the resultant matrix whose coordinates correspond to the row of the first matrix and the column of the second matrix. For instance, in AB = C, if we want to find the value of c12, we must multiply the cells of row 1 in the first matrix by the cells of column 2 in the second matrix and sum the results.

4.Matrix/Vector Multiplication

We can also multiply matrices and vectors together, since a vector is nothing more than a 1-column matrix. Consider a 4×4 matrix M which comprises an arbitrary number of transformations (rotation, scaling, translation, etc.) which are to be applied to a 4×1 vector, S. The resultant 4×1 vector R obtained by multiplying M and S together is then a transformation of S.

5.Coding with matrices

How do we define a matrix in C++ you ask? Simple. If a matrix is a two dimensional array of numbers, by definition, then we should be able to define it in C++ that way too. Here’s how:

typedef float MATRIX[4][4]; // A 4×4 matrix

The simplest thing you would want to do is concatenate (multiply) two matrices. We perform the operation, as mentioned earlier, like so:

// matmult: Multiplies matrices A and B, storing the result in A
void matmult(MATRIX &A, MATRIX B)
{

MATRIX conc;
// Multiplies by rows and columns
for (int i = 0; i <4; i++)
for (int j = 0; j < 4; j++)
{

conc[i][j] =
(A[0][j] * B[i][0])+
(A[1][j] * B[i][1])+
(A[2][j] * B[i][2])+
(A[3][j] * B[i][3]);
}

// Copy result matrix into matrix A
for (int p = 0; p < 4; p++)
for (int q = 0; q < 4; q++)
a[p][q] = conc[p][q];
}

Note: The code references cell yx as cell xy. This may be changed in the near future for the sake of clarity.

You also need to initialize a matrix before you begin to work with it (unless you are simply copying from one matrix to another). Here’s the code to do so:

// matinit: Initializes a matrix, and sets to 'identity'
void matinit(MATRIX &a)
{

for (int i = 0; i < 4; i++)
for (int j = 0; j < 4; j++)
a[i][j] = 0.0f;
a[0][0] = 1.0f;

a[1][1] = 1.0f;
a[2][2] = 1.0f;
a[3][3] = 1.0f;
}

Note: The code references cell yx as cell xy. This may be changed in the near future for the sake of clarity.

Matrices are essentially useless unless you apply them to something. In 3D programming, you would want to apply them to vectors. By applying say, a rotation matrix, to a vector, we can rotate that vector around the origin in 3D space. 3D vectors are simply 3×1 matrices. Here is the code to do so:

// vecmatmult: Multiplies vector V and matrix B
void vecmatmult(VECTOR &v, MATRIX b)
{

float x, y, z, w = 1;
x = (v.x * b[0][0]) + (v.y * b[0][1]) + (v.z * b[0][2]) + (w * b[0][3]);
y = (v.x * b[1][0]) + (v.y * b[1][1]) + (v.z * b[1][2]) + (w * b[1][3]);
z = (v.x * b[2][0]) + (v.y * b[2][1]) + (v.z * b[2][2]) + (w * b[2][3]);

v.x = x;
v.y = y;
v.z = z;
}

Note: The code references cell yx as cell xy. This may be changed in the near future for the sake of clarity.

What is w you ask? The w is simply to accommodate the translation cells, since the vector is only 3×1 and the matrix is 4×4.

Useful matrices

Here are a couple of useful matrices that you can use. Most of them are for 3D geometry and are 4×4 matrices. I will add some more in the near future and also show how they are derived. Here they are:

The identity matrix is a matrix that does not affect the contents stored in another matrix when multiplied. It can be used as a basis for for constructing other matrices. An identity matrix can be any size, as long as it’s diagonal consists of 1’s and all other cells are filled with zeros, as shown:

The translation matrix is a matrix that can be used to translate vectors, such points, in n-dimensional space. Here, translations are for 3-dimensional space, where tx, ty and tz relate to displacements in the x, y and z dimensions respectively. Each component of the translation is added to it’s respective vector component. Notice how this matrix is based on the identity matrix:

The scaling matrix is a matrix that can be used to scale vectors, such points, in n-dimensional space. Here, scaling values are for 3-dimensional space, where sx, sy and sz relate to scaling factors in the x, y and z dimensions respectively. Notice, once again, how this matrix is based on the identity matrix:

The x-rotation matrix is a matrix that can be used to rotate vectors around the x-axis in 3-dimensional space. Theta (q) is the angle at which the vector is rotated. Lets look at how this matrix is derived:

The process of rotating a vector around the x-axis in 3D space is as follows:

x’ = x
y’ = (y*cosq) + (z*sinq)
z’ = (z*-sinq) - (y*cosq)

When applying this matrix to a vector (by means of a dot product), each of those transformations is only applied once, to the appropriate vector component. Again, these matrices are based on the identity matrix:

The y-rotation matrix is a matrix that can be used to rotate vectors around the y-axis in 3-dimensional space. Theta (q) is the angle at which the vector is rotated. Lets look at how this matrix is derived:

Here, the process of rotation is similar to that of the x-rotation matrix:

x’ = (x*cosq) - (z*sinq)
y’ = y
z’ = (x*sinq) + (z*cosq)

Here is the matrix:

The z-rotation matrix is a matrix that can be used to rotate vectors around the z-axis in 3-dimensional space. Again, theta (q) is the angle at which the vector is rotated. Lets look at a simple derivation:

Here, the process of rotation is similar to that of the x-rotation matrix:

x’ = (x*cosq) + (y*sinq)
y’ = -(x*sinq) + (y*cosq)
z’ = z

Here it is:

Two sticks and matchbox, measure exactly 45 minutes by burning these sticks

noreply@blogger.com (Mangesh Shinde) — Sat, 12 Apr 2014 20:20:00 +0000

Question: You have two sticks and matchbox. Each stick takes exactly an hour to burn from one end to the other. The sticks are not identical and do not burn at a constant rate. As a result, two equal lengths of the stick would not necessarily burn in the same amount of time. How would you measure exactly 45 minutes by burning these sticks?

Answer: This puzzle used to be asked in Wall Street interviews long time ago. It is very rare for this question to be asked now but it is a very good question to help you think a little outside the normal thought process.

The answer is really simple. Since the sticks do not burn at a constant rate, we can not use the length of the stick as any sort of measurement of time. If we light a stick, it takes 60 minutes to burn completely. What if we light the stick from both sides? It will take exactly half the original time, i.e. 30 minutes to burn completely.

0 minutes – Light stick 1 on both sides and stick 2 on one side.
30 minutes – Stick 1 will be burnt out. Light the other end of stick 2.
45 minutes – Stick 2 will be burnt out.

If you have any interesting questions, feel free to email them to support@goforexam.in

Aptitude Question - what is the proportion of boys to girls in the country?

noreply@blogger.com (Mangesh Shinde) — Sat, 12 Apr 2014 20:08:00 +0000

Question: In a country where everyone wants a boy, each family continues having babies till they have a boy. After some time, what is the proportion of boys to girls in the country? (Assuming probability of having a boy or a girl is the same)

Answer: This is a very simple probability question in a software interview. This question might be a little old to be ever asked again but it is a good warm up.

Assume there are C number of couples so there would be C boys. The number of girls can be calculated by the following method.

Number of girls = 0*(Probability of 0 girls) + 1*(Probability of 1 girl) + 2*(Probability of 2 girls) + …
Number of girls = 0*(C*1/2) + 1*(C*1/2*1/2) + 2*(C*1/2*1/2*1/2) + …
Number of girls = 0 + C/4 + 2*C/8 + 3*C/16 + …
Number of girls = C
(using mathematical formulas; it becomes apparent if you just sum up the first 4-5 terms)

Thus the proportion of boys to girls is 1 : 1.

Finding the sub-array with the largest sum.

noreply@blogger.com (Mangesh Shinde) — Sat, 12 Apr 2014 20:02:00 +0000

Question: You are given an array with integers (both positive and negative) in any random order. Find the sub-array with the largest sum.

Answer: This is an all-time favorite software interview question. The best way to solve this puzzle is to use Kadane’s algorithm which runs in O(n) time. The idea is to keep scanning through the array and calculating the maximum sub-array that ends at every position. The sub-array will either contain a range of numbers if the array has intermixed positive and negative values, or it will contain the least negative value if the array has only negative values. Here’s some code to illustrate.

void maxSumSubArray( int *array, int len, int *start, int *end, int *maxSum )
{
    int maxSumSoFar = -2147483648;
    int curSum = 0;
    int a = b = s = i = 0;
    for( i = 0; i < len; i++ ) {
        curSum += array[i];
        if ( curSum > maxSumSoFar ) {
            maxSumSoFar = curSum;
            a = s;
            b = i;
        }
        if( curSum < 0 ) {
            curSum = 0;
            s = i + 1;
        }
    }
    *start = a;
    *end = b;
    *maxSum = maxSumSoFar;
}

Think you have a better solution? We'd love to read your comments!

Reverse a Linked-list. Write code in C.

noreply@blogger.com (Mangesh Shinde) — Sat, 12 Apr 2014 19:57:00 +0000

Question : Reverse a Linked-list. Write code in C.

Answer: There are multiple ways to go about this. Let’s first look at a recursive solution.

Node * reverse( Node * ptr , Node * previous)
{
    Node * temp;
    if(ptr->next == NULL) {
        ptr->next = previous;
        return ptr;
    } else {
        temp = reverse(ptr->next, ptr);
        ptr->next = previous;
        return temp;
    }
}
reversedHead = reverse(head, NULL);

Now for a non-recursive solution.

Node * reverse( Node * ptr )
{
    Node * temp;
    Node * previous = NULL;
    while(ptr != NULL) {
        temp = ptr->next;
        ptr->next = previous;
        previous = ptr;
        ptr = temp;
    }
    return previous;
}

If anyone has any modifications or a better methods, please post it up. Comments are always welcome.

TCS Aptitude test, Placement Questions with Solution

noreply@blogger.com (Mangesh Shinde) — Sat, 12 Apr 2014 19:44:00 +0000

1) If log 0.317=0.3332 and log 0.318=0.3364 then find log 0.319 ?
Sol: log 0.317=0.3332 and log 0.318=0.3364, then
log 0.319=log0.318+(log0.318-log0.317) = 0.3396

2) A box of 150 packets consists of 1kg packets and 2kg packets. Total weight of box is 264kg. How many 2kg packets are there ?
Sol: x= 2 kg Packs
y= 1 kg packs
x + y = 150 .......... Eqn 1
2x + y = 264 .......... Eqn 2
Solve the Simultaneous equation; x = 114
so, y = 36
ANS : Number of 2 kg Packs = 114.

3) My flight takes of at 2am from a place at 18N 10E and landed 10 Hrs later at a place with coordinates 36N70W. What is the local time when my plane landed?
Options: a)6:00 am b) 6:40am c) 7:40 d) 7:00 e) 8:00
Sol: The destination place is 80 degree west to the starting place. Hence the time difference between these two places is 5 hour 20 min. (=24hr*80/360).
When the flight landed, the time at the starting place is 12 noon (2 AM + 10 hours).
Hence, the time at the destination place is 12 noon - 5:20 hours = 6: 40 AM

4) A plane moves from 9°N40°E to 9°N40°W. If the plane starts at 10 am and takes 8 hours to reach the destination, find the local arrival time ?
Sol: Since it is moving from east to west longitide we need to add both
ie,40+40=80
multiply the ans by 4
=>80*4=320min
convert this min to hours ie, 5hrs 33min
It takes 8hrs totally . So 8-5hr 30 min=2hr 30min
So the ans is 10am+2hr 30 min
=>ans is 12:30 it will reach

5) The size of the bucket is N kb. The bucket fills at the rate of 0.1 kb per millisecond. A programmer sends a program to receiver. There it waits for 10 milliseconds. And response will be back to programmer in 20 milliseconds. How much time the program takes to get a response back to the programmer, after it is sent? answer with explanation.
Sol: see it doesn't matter that wat the time is being taken to fill the bucket.after reaching program it waits there for 10ms and back to the programmer in 20 ms.then total time to get the response is 20ms +10 ms=30ms...it's so simple....

6) A fisherman's day is rated as good if he catches 9 fishes, fair if 7 fishes and bad if 5 fishes. He catches 53 fishes in a week n had all good, fair n bad days in the week. So how many good, fair n bad days did the fisher man had in the week
Ans: 4 good, 1 fair n 2 bad days
Sol: Go to river catch fish
4*9=36
7*1=7
2*5=10
36+7+10=53...
take what is given 53
good days means --- 9 fishes so 53/9=4(remainder=17) if u assume 5 then there is no chance for bad days.
fair days means ----- 7 fishes so remaining 17 --- 17/7=1(remainder=10) if u assume 2 then there is no chance for bad days.
bad days means -------5 fishes so remaining 10---10/5=2days.
Ans: 4 good, 1 fair, 2bad. ==== total 7 days.

x+y+z=7--------- eq1
9*x+7*y+5*z=53 -------eq2
multiply eq 1 by 9,
9*x+9*y+9*z=35 -------------eq3
from eq2 and eq3
2*y+4*z=10-----eq4
since all x,y and z are integer i sud put a integer value of y such that z sud be integer in eq 4 .....and ther will be two value y=1 or 3 then z = 2 or 1 from eq 4

for first y=1,z=2 then from eq1 x= 4
so 9*4+1*7+2*5=53.... satisfied
now for second y=3 z=1 then from eq1 x=3
so 9*3+3*7+1*5=53 ......satisfied
so finally there are two solution of this question
(x,y,z)=(4,1,2) and (3,3,1)...

7) Y catches 5 times more fishes than X. If total number of fishes caught by X and Y is 42, then number of fishes caught by X?
Sol: Let no. of fish x catches=p
no. caught by y =r
r=5p.
r+p=42
then p=7,r=35

8) Three companies are working independently and receiving the savings 20%, 30%, 40%. If the companies work combinely, what will be their net savings?
Sol: suppose total income is 100
so amount x is getting is 80
y is 70
z =60
total=210

but total money is 300
300-210=90
so they are getting 90 rs less
90 is 30% of 300 so they r getting 30% discount

9) The ratio of incomes of C and D is 3:4.the ratio of their expenditures is 4:5. Find the ratio of their savings if the savings of C is one fourths of his income?
Sol: incomes:3:4
expenditures:4:5
3x-4y=1/4(3x)
12x-16y=3x
9x=16y
y=9x/16
(3x-4(9x/16))/((4x-5(9x/16)))
ans:12/19

10) If G(0) = -1 G(1)= 1 and G(N)=G(N-1) - G(N-2) then what is the value of G(6)?
Ans: -1
bcoz g(2)=g(1)-g(0)=1+1=2
g(3)=1
g(4)=-1
g(5)=-2
g(6)=-1

Upcoming Exams Dates and Details- Government, Bank..

noreply@blogger.com (Mangesh Shinde) — Sat, 12 Apr 2014 10:08:00 +0000

Upcoming Notifications:

Banks :

Recruitment Board	Post name
SBI	Clerks
SBI Associate Bank	PO

UPSC :

Post Date	Exam Name	Last Date
17/05/2014	Civil Services (Preliminary) Exam 2014	16/06/2014
17/05/2014	Indian Forest Services (Preliminary) Exam 2014	16/06/2014
21/06/2014	NDA & NA Exam (II) 2014	21/07/2014
19/07/2014	CDS Exam(II) 2014	18/08/2014

SSC :

Post Date	Exam Name	Last Date
24/05/2014	Stenographer (Gr C&D) Exam 2014	20/06/2014
19/07/2014	Combined Higher Secondary (10+2) Exam 2014	15/08/2014
09/08/2014	Junior Translator (CSOLS)/Jr Hindi Translator Exam 2014	05/09/2014

Indian Army :

Notification Date	Name of The Course	Tentative Course of Commencement
.

TNPSC :

Post Date	Exam Name	Tentative Vacancies
1st Week of March 2014 (To be Notified Later)	Group VI Examination (Forest Apprentice)	26
3rd Week of March 2014 (To be Notified Later)	Assistant Works Manager	4
2nd Week of April 2014 (To be Notified Later)	Combined Civil Services Exam-II (Interview Posts) [Group-II Services]	.
4th Week of April 2014	Statistical Inspector	2
2nd Week of May 2014	Group VII-B Service Exam (Executive Officer Gr-III)	.
1st Week of June 2014	Group-IA (Asst Conservator of Forests)	.
2nd Week of June 2014	Group VIII Service Exam (Executive Officer Gr-IV)	.
3rd Week of June 2014	Research Assistant	4
1st Week of July 2014	Group IV Services Examination	.
1st Week of August 2014	Combined Civil Services-III Exam (Group III-A)	.
2nd Week of August 2014	Assistant Geologist	1
2nd Week of August 2014	Social Case Work Expert	3
1st Week of September 2014	Combined Civil Services Exam-IB (Group-IB Service)	.
2nd Week of September 2014	Chemist	3
1st Week of October 2014	Assistant Tester	2
1st Week of November 2014	General Foreman & Technical Assistant	5
1st Week of December 2014	Combined Civil Services Examination (Group-I)	.

Head First Java 2nd Edition - Popular Book

noreply@blogger.com (Mangesh Shinde) — Sat, 12 Apr 2014 09:55:00 +0000

Rs. 625 This is a book that is tailored for Java novices. Ideal for those who are interested in learning Java but have been put off by the complexities of learning the language, Head First Java explores a new way of teaching the same.

Head First Java is aimed at people who are complete novices when it comes to programming with the language, and the book makes the learning experience fun - one that?s filled with innovative and novel measures.

If you?re not a fan of wracking your brain with dull theoretical concepts that put you to sleep, Head First Java can be a welcome addition to your shelf.

The book starts from the fundamentals and progresses to extremely advanced levels, employing an easy-to-learn approach throughout. From distributed programming with RMI and network sockets, object oriented design, and object properties and methods, to graphical user interfaces, Java archives, network connectivity and Java 5.0, the book explores every facet of the programming language with mysteries, puzzles and visuals that will keep you engaged.

The unusual format is based on the theory that the brain needs stimulation to grasp complex issues. The adopted approach brings about a considerable reduction in the time required to grasp the language.

About The Author
Kathy Sierra is a well-known game developer and programming inspector. She has worked as a master trainer for Sun Microsystems. She?s also one of the founders of JavaRanch, the online community of Java programmers.

Other books of hers that she?s co-authored with Bert Bates include

Head First Design Patterns
Head First EJB
Head First Servlets and JSP

Bert Bates has been involved in the area of software development for more than 25 years. He has been the lead developer for many Java certification exams. He is also a consultant in the area of artificial intelligence. He has co-authored the Head First series of books with Kathy Sierra.

Table Of Contents
Introduction
I Breaking the Surface: a quick dip
2 A Trip to Objectville: yes there will be Object
3 Know Your Variables: primitives and references
4 How Objects Behave: object state affects method &Junior
5 Extra-Strength Methods: flow control operations, and more
6 Using the Java Library: so you don't have to write it all yourself
7 Better Living in Objectville: planning for the future
8 Serious Polymorphism: exploiting abstract Classes and interfaces
9 Life and Death of an Object: constructors and memory management
10 Numbers Matter: math, formatting, wrappers; and statics
11 Risky Behavior: exception handing'
12 A Very Graphic Story: intro to GUI, event handling, and inner classes
13, Work on Your Swing: layout managers to subcomponents
14 Saving Object: serialization and I/0
15 Make a Connection: networking sockets and multi-threading
16 Data Structures: collections and genetics
17 Release Your Code: packaging and deployment
18 Distributed Computing: RMI with a dash of serialize, EJE, and Jini
Appendix A: Fatal code kitchen
Appendix B: To 7th Mugs that &diet make it into the rest of the book
Index

Android device emulator shortcuts

noreply@blogger.com (Mangesh Shinde) — Tue, 08 Apr 2014 05:25:00 +0000

Escape - Back button
Home - Home button
F2, PageUp - Menu button
Shift-F2, PageDown - Start button
F3 - Call/Dial button
F4 - Hangup/EndCall button
F5 - Search button
F7 - Power button
Ctrl-F3, Ctrl-KEYPAD_5 - Camera button
Ctrl-F5, KEYPAD_PLUS - Volume up button
Ctrl-F6, KEYPAD_MINUS - Volume down button
KEYPAD_5 DPAD - center
KEYPAD_4, KEYPAD_6 DPAD left, DPAD - right
KEYPAD_8, KEYPAD_2 DPAD up, DPAD - down
F8 - Toggle cell network on/off
F9 - Toggle code profiling (when -trace set)
Alt-ENTER - Toggle fullscreen mode
Ctrl-T Toggle - trackball mode
Ctrl-F11, - KEYPAD_7
Ctrl-F12, - KEYPAD_9

Android Best Practices for User input, Background jobs, Performance, Security and privacy - Android

noreply@blogger.com (Mangesh Shinde) — Fri, 31 Jan 2014 11:08:00 +0000

There are some practices that you can follow while developing android application. These are suggested by the android itself and they keep on improving with respect to time. These best practices include interaction design features , performance , security and privacy , compatibility , testing , distributing and monetizing tips. They are narrowed down and are listed as below.

Best Practices - User input

Every text field is intented for a differnet job. For example, some textfields are for text and some are for numbers. If it is for numbers then it is better to dispaly the numeric keypad when that textfield is focused. Its syntax is.

<EditText
          android:id="@+id/phone"
          android:layout_width="fill_parent"
          android:layout_height="wrap_content"
          android:inputType="phone" />

Other then that if your field is for password , then it must show a password hint , so that the user can easily remember the password. It can be achieved as.

<EditText
         android:id="@+id/password"
         android:hint="@string/password_hint"
         android:inputType="textPassword" />

Best Practices - Background jobs

There are certain jobs in an application that are running in an application background. Their job might be to fetch some thing from the internet , playing music e.t.c. It is recommended that the long awaiting tasks should not be done in the UI thread and rather in the background by services or AsyncTask.

ASYNCTASK VS SERVICES.
Both are used for doing background tasks , but the service is not affected by most user inteface life cycle events, so it continues to run in circumstances that would shut down an AsyncTask.

Best Practices - Performance

Your application peformance should be upto the mark. But it should perform differently not on the front end , but on the back end when it the device is connected to a power source or charging. Charging could be of from USB and from wire cable.
When your device is charging itself , it is recommended to update your application settings if any, such as maximizing your refresh rate whenver the device is connected. It can be done as this.

IntentFilter ifilter = new IntentFilter(Intent.ACTION_BATTERY_CHANGED);
Intent batteryStatus = context.registerReceiver(null, ifilter);
// Are we charging / charged? Full or charging.
int status = batteryStatus.getIntExtra(BatteryManager.EXTRA_STATUS, -1);
// How are we charging? From AC or USB.
int chargePlug = batteryStatus.getIntExtra(BatteryManager.EXTRA_PLUGGED, -1);

Best Practices - Security and privacy

It is very important that your application should be secure and not only the application , but the user data and the application data should also be secured. The security can be increased by the following factors.
1. Use internal storage rather then external for storing applications files
2. Use content providers wherever possible
3. Use SSl when connecting to the web
4. Use appropriate permissions for accessing different functionalists of device

Java knowledge for android

noreply@blogger.com (Mangesh Shinde) — Thu, 30 Jan 2014 11:51:00 +0000

The Java programming language is one of the glorious tools that make programming Android a breeze compared with programming for other mobile platforms. Whereas other languages insist that you manage memory, deallocate and allocate bytes, and then shift bits around like a game of dominoes, Java has a little buddy called the Java Virtual Machine (JVM) that helps take care of that for you. The JVM allows you to focus on writing code to solve a business problem by using a clean, understandable programming language
(or to build that next really cool first-person shooter game you’ve been dreaming of) instead of focusing on the plumbing just to get the screens to show up.

You’re expected to understand the basics of the Java programming language before you write your first Android application. If you’re feeling a bit rusty and need a refresher course on Java, you can visit the Java tutorials site at http://java.sun.com/docs/books/tutorial. or you can find some tutorial here - JAVA

HSC Exam Time Table 2014 Arts, Commerce, Science

noreply@blogger.com (Mangesh Shinde) — Tue, 28 Jan 2014 10:38:00 +0000

Day & Date	Time	Subject
Thursday 20th February 2014	11.00 a.m. to 2.00 p.	Marathi (02) Gujarati (03) Kannada (06) Sindhi (07) Malayalam (08) Tamil (09) Telugu (10) Punjabi (11) Bengali (12)
Thursday 20th February 2014	3.00 p.m. to 6.00 p.m.	Urdu (05) French (13) Pali (35)
Friday, 21nd February 2014	11.00 a.m. to 2.00 p.m	English (01)
Saturday 22nd February 2014	11.00 a.m. to 2.00 p.m.	Hindi (04)
Saturday 22nd February 2014	3.00 p.m. to 6.00 p.m.	German (14) Ardhamagadhi (16) Persian (37) Avesta - Pahalavi (87)
Monday 24th February 2014	11.00 a.m. to 2.00 p.m.	Textiles (A/S) (44)
Monday 24th February 2014	3.00 p.m. to 6.00 p.m.	Marathi Lieterature (A/s/c)
Tuesday 25th February 2014	11.00 a.m. to 2.00 p.m	Secretarial Practice (C) Physics (S)
Tuesday 25th February 2014	3.00 p.m. to 6.00 p.m.	Political Science (A)
Wednesday 26th February 2014	11.00 a.m. to 2.00 p.m.	History & Development of Indian Music(A)
Wednesday 26th February 2014	3.00 p.m. to 6.00 p.m	Hindi Lieterature (A/s/c)
Friday28th February, 2014	11.00 a.m. to 2.00 p.m	Sociology (A/S)
Saturday 01st March, 2014	11.00 a.m. to 2.00 p.m.	Mathematics & Statistics (A/S) (40) Mathematics & Statistics Paper - I (C)
Tuesday 04th March, 2014	11.00 a.m. to 2. 00 p.m.	Economics (A/S/C)
Thurseday 06th March 2014	11.00 a.m. to 2. 00 p.m.	Organisation of Commerce & Management (C) (51) Chemistry (S)
Thurseday 06th March 2014	3.00 p.m. to 6.00 p.m	History (A)
Saturday, 08th March, 2014	11.00 a.m. to 2.00 p.m.	Geography (A/S/C)(39)
Saturday, 08th March, 2014	3.00 p.m. to 6.00 p.m.	Japanese (21)
Monday 10th March, 2014	11.00 a.m. to 2.00 p.m.	Book keeping & Accountancy (A/c) Biology (S) (56) ( As per Mar-2013 Syllabus)
Monday 10th March, 2014	3.00 p.m. to 6.00 p.m.	Philosophy (A)

Cafe Babe? Or, what's in a name?

noreply@blogger.com (Mangesh Shinde) — Tue, 28 Jan 2014 10:29:00 +0000

The name Java is not an acronym. In particular, it does not stand for Just Another Vague Acronym. The language was originally called Oak, but the language lawyers worried about the name Oak Technology (at the time a maker of video cards). So a meeting was held (varying accounts of that meeting were gathered up by JavaWorld), and the name Java made the short list. The language lawyers approved, and the rest is history.

Most kinds of executable file formats have some kind of "magic number" that identifies them. For example, old PDP-11 executables for the UNIX system used the "magic number" 0407 (the real magic there, and the putative origin of the term, is that 0407 was a machine instruction for a jump forward of 7 words, i.e., past the 8-word-long executable header, to the first executable statement in the program). Some versions of the Berkeley UNIX memory allocator use the hexadecimal string 0xDEADBEEF, which is considered unlikely to occur in normal operation(!), to uniquely identify a section of memory that has been allocated to the user. Not to be outdone, the Java Virtual Machine uses the string of hexadecimal numbers 0xCAFEBABE to identify valid JVM files, a kind of homage to the origin of the Java concept. Just who is, or was, this "cafe babe"? You'd have to ask James Gosling!

Finally, it's been pointed out that the Java magic number also forms the "smiley" "]~:>"when viewed in ASCII.

0000000   J   ~   :   > nul etx nul   - soh   4  bs soh   !  bs soh   -
         cafebabe        0003002d        01b40801        a10801ad
0000020

One final word: the "cafebabe" magic number had also been used years earlier at the Mach project (Mach was almost forgotten until Apple chose to incorporate it into Mac OS X) to indicate their executable file format:

$ grep cafebabe /etc/magic
0       belong          0xcafebabe      compiled Java class data,...
0       belong          0xcafebabe      Mach-O fat file

Top Ten Things Every Java Programmer Should Know

noreply@blogger.com (Mangesh Shinde) — Tue, 28 Jan 2014 10:16:00 +0000

These are in no particular order, but these are things that all Java programmers should probably know.

Who Invented Java, and when?
James Gosling, at Sun Labs, around 1992; the group was building a set-top box and started by "cleaning up" C++ and wound up with a new language and runtime.
What does Java stand for?
Java is not an acronym (not even Just Another Vague Acronym :-)). The language was first named Oak, after the tree outside James' window. The lawyers found another language called Oak so, legend has it, the gang went out to the local cafe to discuss names and wound up naming it after the beverage Java (which is in turn named after the island of Java). Possible confirmation of this theory.
What is the JLS?
JLS is The Java Language Specification. Every developer should buy or download (free) this specification and read it, a bit at a time.
How do changes get into Java?
JCP (Java Community Process).
Why is there no printf-like function in Java?
Actually there are! This was fixed in Java 5; see Java Cookbook (2nd Edition) Chapter 9. Java 5 (J2SE 1.5) includes printf (and scanf), String.format(), and lots more.
What is the GOF book?
The Gang Of Four book is entitled Design Patterns, by Erich Gamma, Richard Helm, Ralph Johnson and John Vlissides. This is a very good book. You should read it. Not when you're just learning Java, but when you've let it sink in for about six months.
What other Java book do I need?
- Most of the O'Reilly Java books.
- Effective Java, by Joshua Bloch
- Java Developer's Almanac
- My Java Cookbook (see below)
What is the Java Cookbook?
That's a book of Java recipes (for the programming language, not the coffee, but some bookstores still wind up listing it under Cooking).
What other Java sites do I need to know about?
- java.sun.com, Sun's main technology site
- java.net, a collaborative site (run by Sun)
- java.com, an advocacy/news site (run by Sun)
- developer.java.sun.com, Sun's main developer site
- www.javalobby.org, independent advocacy group
- www.javaworld.com, Java news
- www.theserverside.com, Java Review
- http://www.darwinsys.com/java/, my own Java site
What else do I need to know?
Everything! But nobody can know everything about Java - the subject is now too vast. Imagine somebody saying that they know everything about every single Microsoft product and technology. If someone like that calls me, I'm always out.

Amazon - Some interview quesions

noreply@blogger.com (Mangesh Shinde) — Sun, 19 Jan 2014 18:33:00 +0000

Interview Questions :

1. Given a Binary Search Tree, write a program to print the kth smallest element without using any static/global variable. You can?t pass the value k to any function also.

2. What are the 4 basics of OOP?

3. Define Data Abstraction. What is its importance?

4. Given an array of size n. It contains numbers in the range 1 to n. Each number is present at least once except for 2 numbers. Find the missing numbers.

5. Given an array of size n. It contains numbers in the range 1 to n. Find the numbers which aren?t present.

6. Given a string,find the first un-repeated character in it? Give some test cases

7. You are given a dictionary of all valid words. You have the following 3 operations permitted on a word: delete a character, insert a character, replace a character. Now given two words - word1 and word2 - find the minimum number of steps required to convert word1 to word2. (one operation counts as 1 step.)

8. Given a cube of size n*n*n (i.e made up of n^3 smaller cubes), find the number of smaller cubes on the surface. Extend this to k-dimension.

9. What is a C array and illustrate the how is it different from a list.

10. What is the time and space complexities of merge sort and when is it preferred over quick sort?

11. Write a function which takes as parameters one regular expression(only ? and * are the special characters) and a string and returns whether the string matched the regular expression.

12. Given n red balls and m blue balls and some containers, how would you distribute those balls among the containers such that the probability of picking a red ball is maximized, assuming that the user randomly chooses a container and then randomly picks a ball from that.

13. Find the second largest element in an array with minimum no of comparisons and give the minimum no of comparisons needed on an array of size N to do the same.

14. Given an array of size n, containing every element from 1 to n+1, except one. Find the missing element.

15. How do you convert a decimal number to its hex-decimal equivalent.Give a C code to do the same

16. Explain polymorphism. Provide an example.

17. Given an array all of whose elements are positive numbers, find the maximum sum of a sub sequence with the constraint that no 2 numbers in the sequence should be adjacent in the array. So 3 2 7 10 should return 13 (sum of 3 and 10) or 3 2 5 10 7 should return 15 (sum of 3, 5 and 7)

18. You are given some denominations of coins in an array (int denom[])and infinite supply of all of them. Given an amount (int amount), find the minimum number of coins required to get the exact amount. What is the method called?

19. Given an array of size n. It contains numbers in the range 1 to n. Each number is present at least once except for 1 number. Find the missing number