이산수학 (discrete mathematics) 2.5 정수와 알고리즘 (integers and algorithms) 2006 년...

이산수학 이산수학 (Discrete Mathematics)(Discrete Mathematics)

2.5 2.5 정수와 알고리즘정수와 알고리즘(Integers and Algorithms)(Integers and Algorithms)

20062006 년 봄학기년 봄학기

문양세문양세강원대학교 컴퓨터과학과강원대학교 컴퓨터과학과

Discrete Mathematicsby Yang-Sae Moon

IntroductionIntroduction

Base-b representations of integers. (b 진법 표현 )

• Especially: binary, hexadecimal, octal.

• Also, two’s complement representation (2 의 보수 표현 )

Algorithms for computer arithmetic:

• Binary addition, multiplication, division.

Euclidean algorithm for finding GCD’s.

2.5 Integers and Algorithms


Base-Base-bb Representations of Integers Representations of Integers

If b is a positive integer greater than 1,then a given positive integer n can be uniquely represented as follows:

n = akbk + ak-1bk-1 + … + a1b1 + a0b0

where• k is a natural number.

• and a0, a1, …, and ak are a natural number less than b.

• ak 0.

Example:

• 165 = 1·102 + 6·101 + 5·100 = (165)10

• 165 = 2·82 + 4·81 + 5·80 = (245)8



Base-b Number SystemsBase-b Number Systems

Ordinarily we write base-10 representations of numbers (usi

ng digits 0-9).

However, 10 isn’t special; any base b>1 will work.

For any positive integers n, b, there is a unique sequence

akak-1… a1a0 of digits ai<b such that


k

i

iiban

0

The “base b expansion of n”

(n 의 밑수 b 전개 , n 의 b 진법 표현 )


Particular Bases of InterestParticular Bases of Interest

Base b=10 (decimal):

10 digits: 0,1,2,3,4,5,6,7,8,9.

Base b=2 (binary):

2 digits: 0,1. (“Bits”=“binary digits.”)

Base b=8 (octal):

8 digits: 0,1,2,3,4,5,6,7.

Base b=16 (hexadecimal):

16 digits: 0,1,2,3,4,5,6,7,8,9,A,B,C,D,E,F


Used internally inall modern computers

Octal digits correspond to

groups of 3 bits

Hex digits givegroups of 4

bits

Used only because we have 10 fingers


Converting to Base Converting to Base b b (1/2)(1/2)

Informal Algorithm

1. To convert any integer n to any base b>1:

2. To find the value of the rightmost (lowest-order) digit, simply compute n mod b. (n%b 로 가장 끝 자리 (digit) 를 찾는다 .)

3. Now replace n with the quotient n/b.( 다음 자리 (digit) 을 구하기 위하여 , 몫을 n 으로 삼는다 .)

4. Repeat above two steps to find subsequent digits, until n is gone (=0). ( 단계 2/3 을 n 이 0 이 될 때 까지 반복한다 .)


• (177130)10 = (?)16

− 177130 = 16·11070 + 10− 11070 = 16·691 + 14− 691 = 16·43 + 3− 43 = 16·2 + 11− 2 = 16·0 + 2

(177130)10 = (2B3EA)16

• (242)10 = (?)2

− 241 = 2·120 + 1, 120 = 2·60 + 0

− 60 = 2·30 + 0, 30 = 2·15 + 0

− 15 = 2·7 + 1, 7 = 2·3 + 1− 3 = 2·1 + 1, 1 = 2·0 + 1

(242)10 = (11110001)2


Converting to Base Converting to Base b b (2/2)(2/2)

Formal Algorithm


procedure base b expansion (n: positive integer)

q := n

k := 0

while q 0

begin

ak := q mod b {remainder}

q := q/b {quotient}

k := k + 1

end {the base b expansion of n is (akak-1… a1a0)b}


Addition of Binary NumbersAddition of Binary Numbers

Intuition (let a = (an-1… a1a0)2, b = (bn-1… b1b0)2)


a = an-1 an-2 . . . a2 a1 a0

b = bn-1 bn-2 . . . b2 b1 b0

a+b = sn sn-1 sn-2 . . . s2 s1 s0

cn-1 cn-2 . . . c2 c1 ci = (ai-1+bi-1)/2

si = (ai+bi)%2

Algorithmprocedure add(an 1− …a0, bn−1…b0: binary expressions of a,b)

c := 0 {c mean a carry}for i := 0 to n 1− {i means a bit index}begin

sum := ai + bi + c {2-bit sum}si := sum mod 2 {low bit of sum}c := sum/2 {high bit of sum}

endsn := c{the binary expression of the sum is (snsn-1… s1s0)2}

O(n)


2’s Complement (22’s Complement (2 의 보수의 보수 ) (1/2)) (1/2)

In binary, negative numbers can be conveniently represented using 2’s complement notation.( 실제로 , 컴퓨터에서는 음수를 2 의 보수로 표현한다 .)

In this scheme, a string of n bits can represent integers 2− n 1− ~ (2n 1− 1).−• 0 이상의 정수만 표현한다면… 0 ~ 2n 1− (unsigned int n)

• 음의 정수까지 표현한다면… − 2n ~ 2n 1− (int n)

The bit in the highest-order bit-position (n 1) represents a c−oefficient multiplying 2− n 1− ;( 왼쪽 첫 번째 bit 는 “ − 2n 1− ” 을 나타내며 , 흔히 부호 (+ or )− 를 의미한다 .)

• The other positions i < n 1 just represent 2− i, as before.



2’s Complement (22’s Complement (2 의 보수의 보수 ) (2/2)) (2/2)

The negation of any n-bit 2’s complement number

a(= an−1…a0) is given by an−1…a0 + 1.

Examples

• The negation of 1011 = −(0100 + 1) = −(0101) = −(5)10

• The negation of 0100 = (1011 + 1) = 10100 = (4)10


Bitwise logical complement of a


Subtraction of Binary NumbersSubtraction of Binary Numbers2.5 Integers and Algorithms

procedure subtract (an 1− …a0, bn−1…b0:

binary 2’s complement expressions of a,b)

return add(a, add(b, 1)) { a + (−b) }

Theorem: For an integer a represented in 2’s

complement notation, −a = a + 1. (a 가 2 보수로 표현된다면 ,

…)

Proof: Just try it by yourself!

Algorithm


Multiplication of Binary Numbers (1/2)Multiplication of Binary Numbers (1/2)

Intuition (let a = (an-1… a1a0)2, b = (bn-1… b1b0)2)


a = an-1 an-2 . . . a2 a1 a0

b = bn-1 bn-2 . . . b2 b1 b0

c0 = s(n-1,0)s(n-2,0). . . s(2,0) s(1,0) s(0,0)

c1 = s(n-1,1)s(n-2,1). . . s(2,1) s(1,1) s(0,1) 0

c2 = s(n-1,2)s(n-2,2). . . s(2,2) s(1,2) s(0,2) 0 0

. . . . . .a·b = cn-1 + cn-2 + ... + c2 + c1 + c0

s(i,j) = (if bj = 1 then ai else 0)

+)


Multiplication of Binary Numbers (Multiplication of Binary Numbers (22/2)/2)2.5 Integers and Algorithms

Algorithmprocedure multiply(an 1− …a0, bn−1…b0: binary expressions of a,b)

for j := 0 to n 1− {a bit index for b}begin

if bi = 1 then cj := a shifted j places {cj := a << j}then cj := a

end{c1, c2, …, cn-1 are the partial products.}p := 0for j := 0 to n−1

p := add(p, cj){p is the value of ab}

O(n2) O(n1.585) in $6.3


Division AlgorithmDivision Algorithm2.5 Integers and Algorithms

Example: 23/4?

procedure division(a, d: positive integer)q := 0r := |a|while r dbegin

r := r − dq := q + 1

end{q is the quotient(=a/d), r is the remainder(=a%d) }

Algorithm

r q23 − 4 = 19 119 − 4 = 15 215 − 4 = 11 311 − 4 = 7 47 − 4 = 3 5


Euclid’s Algorithm for GCDEuclid’s Algorithm for GCD

Finding GCDs by comparing prime factorizations can be difficult if the prime factors are unknown.( 소인수분해로 최대공약수를 구하는 것은 어렵다 . 특히 , 큰 수인 경우… )

Euclid discovered: For all integers a, b,gcd(a, b) = gcd((a mod b), b).

Sort a,b so that a>b, and then (given b>1) (a mod b) < a, so problem is simplified.

Examples

• gcd(372, 164) = gcd(372 mod 164, 164) [372%164 = 44]

• gcd(164, 44) = gcd(164 mod 44, 44) [164%44 = 32]

• gcd(44, 32) = gcd(44 mod 32, 32) = gcd(12, 32) = gcd(12, 8) = 4.



Proof of Proof of Euclid’s AlgorithmEuclid’s Algorithm

Prove gcd(a,b) = gcd(b,r) where r = a - bq

a 와 b 의 공약수는 b 와 r 의 공약수가 같음을 보이면 , 양쪽 공약수의 쌍이 같으므로 ( 최대공약수가 같아져 ) 증명이 이루어진다 .

d 를 a 와 b 의 공약수라 하자 .

• 그러면 , a = dqa, b = dqb 가 성립한다 .

• 정의에 의해 , r = dqa – dqbq = d(qa – qbq) 이 성립하므로 , d 는 r 의 약수이다 .

• 따라서 , d 는 b 와 r 의 공약수이다 .

d 를 b 와 r 의 공약수라 하자 .

• 그러면 , b = dqb, r = dqr 이 성립한다 .

• 정의에 의해 , a = dqr + dqbq = d(qr + qbq) 이 성립하므로 , d 는 a 의 약수이다 .

• 따라서 , d 는 a 와 b 의 공약수이다 .



Euclid’s AlgorithmEuclid’s Algorithmss2.5 Integers and Algorithms

procedure gcd(a, b: positive integer)while b 0begin

r := a mod b {r = a % b}a := bb := r

return a

int gcd(int a, int b) /* assume a > b */{

if(b==0) return a;else return gcd(b, a%b);

}

Algorithm in Pseudocode

Algorithm in C (using recursive calls)

이산수학 (discrete mathematics) 2.5 정수와 알고리즘 (integers and algorithms) 2006 년...

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