이산수학 (Discrete Mathematics) 2.5 정수와 알고리즘 (Integers and Algorithms)

# 이산수학 (Discrete Mathematics) 2.5 정수와 알고리즘 (Integers and Algorithms)

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## 이산수학 (Discrete Mathematics) 2.5 정수와 알고리즘 (Integers and Algorithms)

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1. 이산수학 (Discrete Mathematics) 2.5 정수와 알고리즘 (Integers and Algorithms) 2006년 봄학기 문양세 강원대학교 컴퓨터과학과

2. Introduction 2.5 Integers and Algorithms • Base-b representations of integers. (b진법 표현) • Especially: binary, hexadecimal, octal. • Also, two’s complement representation(2의 보수 표현) • Algorithms for computer arithmetic: • Binary addition, multiplication, division. • Euclidean algorithm for finding GCD’s.

3. Base-b Representations of Integers 2.5 Integers and Algorithms • 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

4. Base-b Number Systems 2.5 Integers and Algorithms Ordinarily we write base-10 representations of numbers (using 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… a1a0of digitsai<b such that The “base b expansion of n” (n의 밑수 b전개, n의 b진법 표현)

5. Particular Bases of Interest 2.5 Integers and Algorithms Used only because we have 10 fingers 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 in all modern computers Octal digits correspond to groups of 3 bits Hex digits give groups of 4 bits

6. Converting to Base b (1/2) 2.5 Integers and Algorithms • 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%b로 가장 끝 자리(digit)를 찾는다.) 3. Now replace n with the quotient n/b.(다음 자리(digit)을 구하기 위하여, 몫을 n으로 삼는다.) 4. Repeat above two steps to find subsequent digits, until n is gone (=0). (단계 2/3을 n이 0이 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

7. Converting to Base b (2/2) 2.5 Integers and Algorithms Formal Algorithm procedurebase b expansion (n: positive integer) q := n k := 0 whileq 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}

8. Addition of Binary Numbers 2.5 Integers and Algorithms Intuition (let a = (an-1… a1a0)2, b = (bn-1… b1b0)2) ci = (ai-1+bi-1)/2 cn-1 cn-2 . . . c2 c1 a = an-1 an-2 . . . a2 a1 a0 b = bn-1 bn-2 . . . b2 b1 b0 si = (ai+bi)%2 a+b = sn sn-1 sn-2 . . . s2 s1 s0 Algorithm procedureadd(an−1…a0, bn−1…b0: binary expressions of a,b) c := 0 {c mean a carry} fori := 0 to n−1 {i means a bit index} begin sum := ai+ bi + c {2-bit sum} si := summod 2 {low bit of sum} c := sum/2 {high bit of sum} end sn := c {the binary expression of the sum is (snsn-1… s1s0)2} O(n)

9. 2’s Complement (2의 보수) (1/2) 2.5 Integers and Algorithms • In binary, negative numbers can be conveniently represented using 2’s complement notation.(실제로, 컴퓨터에서는 음수를 2의 보수로 표현한다.) • In this scheme, a string of n bits can represent integers −2n−1 ~ (2n−1−1). • 0이상의 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 coefficient multiplying −2n−1;(왼쪽 첫 번째 bit는 “ −2n−1”을 나타내며, 흔히 부호(+ or −)를 의미한다.) • The other positions i < n−1 just represent 2i, as before.

10. 2’s Complement (2의 보수) (2/2) 2.5 Integers and Algorithms • The negation of any n-bit 2’s complement numbera(= 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

11. Subtraction of Binary Numbers 2.5 Integers and Algorithms Theorem: For an integer a represented in 2’s complement notation, −a = a + 1. (a가 2보수로 표현된다면, …) Proof: Just try it by yourself! Algorithm proceduresubtract (an−1…a0, bn−1…b0: binary 2’s complement expressions of a,b) returnadd(a, add(b, 1)){ a + (−b) }

12. Multiplication of Binary Numbers (1/2) 2.5 Integers and Algorithms 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) 00 . . . . . . a·b = cn-1 + cn-2 + ... + c2 + c1 + c0 +) s(i,j) = (ifbj = 1 then ai else 0)

13. Multiplication of Binary Numbers (2/2) 2.5 Integers and Algorithms Algorithm proceduremultiply(an−1…a0, bn−1…b0: binary expressions of a,b) forj := 0 to n−1{a bit index for b} begin ifbi= 1 thencj:= a shifted j places {cj:= a << j} thencj:= a end {c1, c2, …,cn-1 are the partial products.} p := 0 forj := 0 ton−1 p := add(p, cj) {p is the value of ab} O(n2)  O(n1.585) in \$6.3

14. Division Algorithm 2.5 Integers and Algorithms Example: 23/4? r q 23 − 4 = 19 1 19 − 4 = 15 2 15 − 4 = 11 3 11 − 4 = 7 4 7 − 4 = 35 Algorithm proceduredivision(a, d: positive integer) q := 0 r := |a| whilerd begin r := r−d q := q + 1 end {q is the quotient(=a/d), r is the remainder(=a%d)}

15. Euclid’s Algorithm for GCD 2.5 Integers and Algorithms • 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.

16. Proof of Euclid’s Algorithm 2.5 Integers and Algorithms Prove gcd(a,b) = gcd(b,r) where r = a - bq • a와 b의 공약수는 b와 r의 공약수가 같음을 보이면, 양쪽 공약수의 쌍이 같으므로 (최대공약수가 같아져) 증명이 이루어진다. • d를 a와 b의 공약수라 하자. • 그러면, a = dqa, b = dqb가 성립한다. • 정의에 의해, r = dqa – dqbq = d(qa – qbq)이 성립하므로, d는 r의 약수이다. • 따라서, d는 b와 r의 공약수이다. • d를 b와 r의 공약수라 하자. • 그러면, b = dqb, r = dqr이 성립한다. • 정의에 의해, a = dqr + dqbq = d(qr + qbq)이 성립하므로, d는 a의 약수이다. • 따라서, d는 a와 b의 공약수이다.

17. Euclid’s Algorithms 2.5 Integers and Algorithms Algorithm in Pseudocode proceduregcd(a, b: positive integer) whileb 0 begin r := amodb {r = a % b} a := b b := r return a Algorithm in C (using recursive calls) int gcd(int a, int b) /* assume a > b */ { if(b==0) return a; else return gcd(b, a%b); }