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2014 년 봄학기 강원대학교 컴퓨터과학전공 문양세

이산수학 (Discrete Mathematics) 분할과 정복 (Divide and Conquer). 2014 년 봄학기 강원대학교 컴퓨터과학전공 문양세. Time to break problem up into sub-problems. Time for each sub-problem. Divide & Conquer Recurrence Relations. Divide and Conquer. Main points so far:

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2014 년 봄학기 강원대학교 컴퓨터과학전공 문양세

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  1. 이산수학(Discrete Mathematics) 분할과 정복 (Divide and Conquer) 2014년 봄학기 강원대학교 컴퓨터과학전공 문양세

  2. Time to break problem up into sub-problems Time for each sub-problem Divide & Conquer Recurrence Relations Divide and Conquer Main points so far: Many types of problems are solvable by reducing a problem of size n into some number a of independent sub-problems, each of size n/b, where a1 and b>1.(많은 경우에 있어서, 크기 n의 문제를 a개의 크기 n/b의 작은 문제로 바꾸어 처리할 수 있다.) The time complexity to solve such problems is given by a recurrence relation: T(n) = a·T(n/b) + g(n)(이런 경우의 시간 복잡도는 T(n)의 점화 관계로 나타낼 수 있다.)

  3. Divide & Conquer Examples Divide and Conquer • Binary search:Break list into 1 sub-problem (smaller list) (so a=1) of size n/2 (so b=2). • So T(n) = T(n/2)+c (g(n)=c constant) • Merge sort:Break list of length n into 2 sub-lists (a=2), each of size n/2 (so b=2), then merge them,in g(n) = (n) time. • So T(n) = 2T(n/2) + cn(roughly, for some c)

  4. Fast Multiplication Example (1/3) Divide and Conquer • The ordinary grade-school algorithm takes (n2) steps to multiply two n-digit numbers.(학교에서 배운 방법에 따르면, n자리인 두 수의 곱은 (n2) 스텝이 필요하다.) • This seems like too much work! • So, let’s find a faster multiplication algorithm! • To find the product cd of two 2n-digit base-b numbers, c=(c2n-1c2n-2…c0)b and d=(d2n-1d2n-2…d0)b,first, we break c and d in half:c=bnC1+C0, d=bnD1+D0, (e.g., 4321 = 10243+21)and then... (see next slide)

  5. Zero Three multiplications, each with n-digit numbers Fast Multiplication Example (2/3) Divide and Conquer (Factor last polynomial)

  6. Fast Multiplication Example (3/3) Divide and Conquer • Notice that the time complexity T(n) of the fast multiplication algorithm obeys the recurrence: • T(2n)=3T(n)+(n)i.e., • T(n)=3T(n/2)+(n) So a=3, b=2. ( The order will be reduced …) Time to do the needed adds & subtracts of n-digit and 2n-digit numbers

  7. The Master Theorem Divide and Conquer Consider a function f(n) that, for all n=bk for all kZ+,satisfies the recurrence relation: (n=bk일 때, 다음 점화 관계가 성립하면) f(n) = af(n/b) + cnd with a≥1, integer b>1, real c>0, d≥0. Then: Proof of the theorem is …. omitted.

  8. Master Theorem Examples (1/3) Divide and Conquer • Recall that complexity of fast multiplication was: T(n)=3T(n/2)+(n) • Thus, a=3, b=2, d=1. So a > bd, so case 3 of the master theorem applies, so: which is O(n1.58…), so the new algorithm is strictly faster than ordinary (n2) multiply!

  9. Master Theorem Examples (2/3) Divide and Conquer • 예제(Binary Search):이진 탐색의 복잡도는 얼마인가(비교 수를 추정하라)? • 이진 탐색의 점화 관계: T(n) = T(n/2)+c (n이 짝수라 가정) • 매스터 정리로 보면, a = 1, b = 2, d = 0으로서,a = 1 = bd인 두 번째 경우에 해당한다. • 결국, 다음과 같은 과정에 의해 O(logn)이 된다.

  10. Master Theorem Examples (3/3) Divide and Conquer • 예제(Merge Sort):합병 정렬의 복잡도는 얼마인가(비교 수를 추정하라)? • 이진 탐색의 점화 관계: T(n) = 2T(n/2)+cn (n이 짝수라 가정) • 매스터 정리로 보면, a = 2, b = 2, d = 1로서,a = 2 = bd인 두 번째 경우에 해당한다. • 결국, 다음과 같은 과정에 의해 O(nlogn)이 된다.

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