Cyczone

Discussion Text

2009-12-30T07:42:00.013+07:00

Definition of Discussion Text
Discussion text is a text which presents a problematic discourse. This problem will be discussed from different viewpoints. Discussion is commonly found in philosophical, historic and social text.

Generic structure of Discussion Text

Statement and issue; stating the issue which is to discussed

- Statement
- Preview

List of supporting points/Argument for; presenting the points in supporting the presented issue. List of contrastive points/Argument against; presenting the other points which disagree to supporting point

- Point; main idea
- Elaboration; supporting idea

Conclusion or Recommendation; stating the writer's recommendation of the discourse

Language Feature of Discussion Text

Focus on generic human and non-human participants
Using thinking verb; feel, hope, believe, etc
Using additive, contrastive and causal connection; similarly, the other hand, however, etc
Using modalities; must, should, could, etc
Using adverbial of manner; deliberately, hopefully, etc
Reasoning expressed as verbal and noun (abstraction)

Download examples of Discussion text here

Pedoman OSN Komputer

2009-12-29T12:08:00.004+07:00

Mungkin saat ini sudah banyak situs-situs di internet yang menyuguhkan tentang berbagai pedoman untuk menghadapi OSN (Olimpiade Sains Nasional). Banyak situs yang menyediakan berbagai macam soal-soal latihan untuk menghadapi OSN yang setiap tahunnya diadakan oleh pemerintah pusat guna menyaring siswa-siswi sekolah menengah yang mempunyai kemampuan di atas rata-rata.

Post ini hanya salah satu dari sekian banyak situs-situs penyedia soal-soal lain yang dimaksudkan untuk membantu Anda sekalian. Di sini saya baru bisa mem-post tentang 'Pedoman OSN' untuk bidang studi Komputer. Mungkin juga di lain kesempatan saya akan mem-post untuk bidang-bidang lain.

Pedoman OSN Komputer download here:

Conjunction

2009-12-29T11:55:00.004+07:00

Conjunction is a word who connect between a sentence and the other. This post just showing a little of a lot of kind of conjunctions which the writer know. I hope it will useful for you and for us. Amien

Download Conjunction.pdf here

Pascal - 3 (string)

2009-07-09T11:00:00.008+07:00

Question :

Note the following Pascal program

var s1, s2, s : string; i, ins : integer;
begin
s1 := 'TOKI';
s2 := '2005';
for i:=1 to length(s1) do if (ord(s1) < 10) then ins := 1;
insert(s2, s1, ins);
s := s1 + s2;
writeln(s);
end.

The output is:
a. runtime error
b. TOK2005I2005
c. no output
d. 2005TOKI2005
e. 200TOKI5TOKI

Discussion:

Oops ... just see the program has been a headache. But .. eh, do not faint first. "If there is a option Runtime Error, then check all the rules and syntax, may have errors. So that we do not need to do all the process in the program. " OK let's check it out!

Line 1 ... there is no problem, line 2 ... there is no problem, lines 3, 4 also does not have a problem. Line 5 ... if there is a problem? where there is a function Ord (). Function Ord () will return the integer that indicates ordinality from a number of diversified data Ordinal. View the Ord (s1) in the program. There seen that 's1' is a string data manifold.

Please note that the Ordinal type data is ascending type data. Type Ordinal data that include, among others: integer (and its derivative: shortint, longint, word, byte), boolean and char.

Eh ... why not real data types including Ordinal? because the real numbers can not be expressed in the form of a sequence. What is the real number after 1? if 2? whether the 1.5? 1.05? 1.000001? can not isn't it? because in between two real numbers there is another real number. Thus the real number can not be presented in Ascending. String is also not including Ordinal data type so that the program raises an error.

Source : Rosihan Ari’s Blog at blog.rosihanari.net (edited)

Pascal - 2 (string)

2009-07-08T10:50:00.012+07:00

Question:

Whether the output from this program?

var s : string[12];
begin
s := 'setia';
writeln('#', s, '#');
s := s + ' sampai akhir ';
writeln('#', s, '#');
end.

a. #setia# (and the runtime error occurred)
b. #setia         #
    #setia sampai#
c. #setia#
    #setia sampai akhir#
d. #setia#
    #setia sampai#
e. #setia#
    #sampai akhir#

Discussion:

If our analysis on the program, s bertipe data is variable length character string with a maximum of 12. S initial value is the string ’setia’ ((length of line 5). Means that there is still free space for the rest of the string s as much as 7 units. Remember ... an empty space in here not contain spaces, but the null character. So when printing writeln ('#', s,'#'); still appear ‘#setia#’ and not ‘#setia       #’.

Then on the next line profile with a long string ‘ sampai akhir ‘ (14 characters including spaces). Can the string plus the string? Runtime error? he ... 3x here the meaning of plus signs (+) does not mean the addition, but the string unification. Operator + can be given in the string. But do not try to use fraudulent operators -, * or / in the string operation. Computer can error.

So that the string s is the new, ’setia’ + ‘ sampai akhir ‘. Continue ... what is the result? whether the ’setia sampai akhir’? Eit .. whoa, how much capacity the number of characters that can be loaded on the s? Oops ... there are only 12 characters only. Thus the result is ’setia sampai’ (12 characters).

Next printing writeln ('#', s,'#'); and the result is ‘#setia sampai#‘. Thus the result is a D.

Source : Rosihan Ari’s Blog at blog.rosihanari.net (edited)

Pascal - 1

2009-07-08T10:46:00.005+07:00

Question:

If M (x, y) is the statement "x is greater than y", and there is a row in the following pseudo Pascal:

while M (x, y) do
begin
x: = x - 10;
y: = y + 2;
end;

with prices in the first and x = 70 y = 5, y was the price after the row out of the while loop?
A. 11
B. 15
C. 17
D. 21
E. 25

Discussion:

OK ... we will study the above questions.
Note that the while loop has a condition "x is greater than y".
So during these requirements met (Condition TRUE value) then the looping will continue to run.
And the looping process will stop when the condition is not met (value FALSE) Note the way the following:

x = 70, y = 5
Check (x> y) -> (70> 5) -> TRUE
x = x - 10 = 70 - 10 = 60
y = y + 2 = 5 + 2 = 7
Check (x> y) -> (60> 7) -> TRUE
x = x - 10 = 60 - 10 = 50
y = y + 2 = 7 + 2 = 9
Check (x> y) -> (50> 9) -> TRUE
x = x - 10 = 50 - 10 = 40
y = y + 2 = 9 + 2 = 11
Check (x> y) -> (40> 11) -> TRUE
x = x - 10 = 40 - 10 = 30
y = y + 2 = 11 + 2 = 13
Check (x> y) -> (30> 13) -> TRUE
x = x - 10 = 30 - 10 = 20
y = y + 2 = 13 + 2 = 15
Check (x> y) -> (20> 15) -> TRUE
x = x - 10 = 20 - 10 = 10
y = y + 2 = 15 + 2 = 17
Check (x> y) -> (10> 17) -> FALSE
STOP

After a while looping process is complete, the final value of y is 17 (Best in the right C).

Source : Rosihan Ari’s Blog at blog.rosihanari.net (edited)

Analitik - 5 (id)

2009-07-04T12:29:00.002+07:00

Soal :

Tentukan bilangan abcd jika 4.abcd = dcba !

(Dengan abcd masing-masing mewakili 1 bilangan bulat antara 1-9)

Pembahasan:

a yang paling mungkin adalah 1 atau 2. Karena jika a lebih dari 2, maka 4abcd = vwxyz (bilangan 5 digit). Misalkan saja a = 3, maka hasil dari d = 12. Padahal bilangan abcd hanya mewakili 1 digit angka dari 1-9. Sehingga nilai a yang paling memungkinkan adalah 1 atau 2.

Sekarang kita lihat bahwa dcba adalah bilangan genap. Karena bilangan apapun yang dikalikan dengan bilangan genap, pasti akan menghasilkan bilangan genap juga. Contoh : 5 x 4 = 20 (genap); 6 x 4 = 24 (genap),dsb. Nah karena dcba bilangan genap, maka nilai a yang memenuhi adalah 2.

Sekarang kita dapatkan 4.2bcd = dcb2.

Nah sekarang kita lihat bahwa hasil perkalian antara 4 dengan a akan menghasilkan d.

Dengan a = 2, maka 4 x 2 = 8. Namun d belum pasti = 8. Bisa jadi d bernilai 9 karena hasil perkalian sebelumnya yang masi belum diketahui. Jadi d bernilai 8 atau 9. Untuk memastikan nilai d, kita lihat hasil perkalian antara 4 dengan d akan berakhiran dengan 2. Maka dapat dipastikan nilai d = 8.

Sekarang kita dapatkan 4.2bc8 = 8cb2.

Kita ubah ke bentuk

8000 + 400b + 40c + 32 = 8000 + 100c + 10b + 2

390b + 30 = 60c

13b + 1 = 2c

Dari hasil di atas, b = 1.

Karena nilai 2 sudah ditempati oleh a dan jika b = 3, maka c akan > 9.

13.1 + 1 = 2c

c = 7

Sehingga bilangan abcd adalah 2178

Analitik - 4

2009-07-02T12:22:00.006+07:00

OSP Informathic 2008 :

Upik clever mathematics, but he only can write numbers 1 and 2. Therefore, when Upik want to write a number of more than 2, they write some of number 1 and number 2 a few. For example to write down the number 3, right Upik have 3 ways: 12, 21, or 111 (1 + 2 = 3, 2 + 1 = 3; 1 + 1 + 1 = 3). To write down the number 2, in fact Upik have 2 ways: 2 and 11 (2 = 2, 1 + 1 = 2), but there is only 1 way to write down the number 1. How many ways to write the number 8?

Discussion:

In fact, only the array fibonacci course we can already do. Is it only that? Let's see.
To write the number 1, requires a ways (1)
To write the number 2, requires 2 ways (11, 2)
To write a number 3, requires 3 ways (111, 12, 21)
To write the number 4, requires 5 ways (1111, 112, 211, 121, 22)
From way above, we have found the pattern of progression fibonacci.

1	2	3	4	5	6	7	8	9	10
1	2	3	5	8	13	21	34	55	89

From the table above we can see that:
To write the number 5, requires 8 ways.
To write the number 6, requires 13 ways.
To write the number 7, requires 21 ways. And,
To write the number 8, requires 34 ways

If you do not believe, do it yourself manually: P

So to write the number 8, there are 34 ways

Analitik - 3 (id)

2009-06-30T12:02:00.003+07:00

Soal :

	T	W	O
	T	W	O	+
F	O	U	R

T, W, O, F, U, dan R masing-masing mewakili sati digit bilangan bulat positif. Dan masing-masing mewakili bilangan yang berbeda. F dan T tidak sama dengan 0.

Jika diketahui R = 0, maka berapakah F + O + U + T ?

Pembahasan :

O pasti bernilai 5, karena 0 sudah menjadi milik R.

Dan T pasti bernilai ≥ 5, karena hasil penjumlahan T + T menyimpan 1 angka di depan,

yaitu F.

Karena T + T = F O, maka dipastikan T = 7 dengan menyimpan 1 angka dari penjumlahan di belakangnya.

	1	1
	7	W	5
	7	W	5	+
1	5	U	0

Sekarang tinggal mencari nilai dari W dan U.

W haruslah bernilai ≥ 5 karena hasil penjumlahan dari W akan menyimpan 1 di depan.

-	W ≠ 5 karena 5 sudah menjadi milik O.
-	W ≠ 7 karena 7 sudah menjadi milik T.
-	W ≠ 8 karena hasil penjumlahan 1 + W + W = 1 U, dan jika W diisi 8 maka hasilnya 1 + 8 + 8 = 17. Padahal angka 7 sudah menjadi milik T.
-	W ≠ 9 karena hasil penjumlahan 1 + W + W = 1 U, dan jika W diisi 9, maka hasilnya 1 + 9 + 9 = 19. Padahal syarat yang berlaku adalah 1 huruf mewakili 1 angka yang berbeda

Jadi nilai W yang paling mungkin adalah 6. ( 1 + 6 + 6 = 13)

Dan nilai U adalah 3.

Sekarang semua angka sudah kita dapatkan,

F + O + U + T = 1 + 5 + 3 + 7 = 16

Analitik - 2 (id)

2009-06-29T11:31:00.002+07:00

Soal :

Ada sebuah bilangan n, dimana n = 2²⁰⁰³. Berapakah digit terakhir dari n?

Pembahasan :

Pertanyaan bertipe seperti di atas, biasanya memiliki pola-pola tertentu.

Sekarang mari kita lihat seperti apakah pola yang dibentuk oleh soal di atas.

2¹ = 2	2⁵ = 32	2⁹ = 512
2² = 4	2⁶ = 64	2¹⁰ = 1024
2³ = 8	2⁷ = 128	2¹¹ = 2048
2⁴ = 16	2⁸= 256	2¹² = 4096

Nah sudah terlihat kan polanya….

Itu dia perulangan yang disebabkan oleh hasil perpangkatan dari angka 2.

Kita bisa mengambil kesimpulan bahwa terdapat 4 perulangan, yaitu :

Perulangan pertama : digit terakhir 2 ( 1 = 2 )

Perulangan kedua : digit terakhir 4 ( 2 = 4 )

Perulangan ketiga : digit terakhir 8 ( 3 = 8 )

Perulangan keempat : digit terakhir 6 ( 4 = 6 )

Sekarang kita lihat bahwa sisa hasil bagi antara pangkat dengan jumlah perulangan akan memberi hasil posisi perulangan keberapa. Contoh :

7 ÷ 4 = 1 sisa 3. Jadi digit terakhir dari 2⁷ = 8 (perulangan ketiga)

10 ÷ 4 = 2 sisa 2. Jadi digit terakhir dari 2¹⁰ = 4 (perulangan kedua)

Nah, sekarang bagaimana dengan 2²⁰⁰³ ?

Mari kita lihat sisa hasil bagi dari 2003 ÷ 4.

2003 ÷ 4 = 500 sisa 3. Jadi digit terakhir dari 2²⁰⁰³ = 8 (perulangan ketiga)

Analitik - 1

2009-06-28T11:05:00.006+07:00

Question:

3 digit number of the beginning of 2²⁰⁰² x 5²⁰⁰⁵ _is...

Discussion:

Still remember with aⁿx a^m= a^n+m and aⁿx bⁿ= abⁿ?

Should you need to remember back as we will apply them in this matter.

Ø 2²⁰⁰² x 5²⁰⁰⁵

Ø 2²⁰⁰²x 5²⁰⁰² x 5³

Ø 10²⁰⁰² x 5³

Ø 10²⁰⁰²x 125

Multiplication of 5³with10²⁰⁰²will give the effect of adding digits zero (0) behind 125.

So the number of 3 digits initial 2²⁰⁰²x 5²⁰⁰⁵is 1 + 2+ 5 = 8