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Introduction to Algorithms

Introduction to Algorithms. Chapter 4: Recurrences. Solving Recurrences. A recurrence is an equation that describes a function in terms of itself by using smaller inputs The expression: Describes the running time for a function contains recursion. Solving Recurrences. Examples:

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Introduction to Algorithms

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  1. Introduction to Algorithms Chapter 4: Recurrences

  2. Solving Recurrences • A recurrence is an equation that describes a function in terms of itself by using smaller inputs • The expression: • Describes the running time for a function contains recursion.

  3. Solving Recurrences • Examples: • T(n) = 2T(n/2) + (n)  T(n) = (n lg n) • T(n) = 2T(n/2) + n  T(n) = (n lg n) • T(n) = 2T(n/2) + 17 + n  T(n) = (n lg n) • Three methods for solving recurrences • Substitution method • Iteration method • Master method

  4. Recurrence Examples

  5. Substitution Method • The substitution method • “making a good guess method” • Guess the form of the answer, then • use induction to find the constants and show that solution works • Our goal: show that T(n) = 2T(n/2) + n = O(n lg n)

  6. Substitution MethodT(n) = 2T(n/2) + n = O(n lg n) • Thus, we need to show that T(n)  c n lg n with an appropriate choice of c • Inductive hypothesis: assume T(n/2)  c (n/2) lg (n/2) • Substitute back into recurrence to show thatT(n)  c n lg n follows, when c  1 • T(n) = 2 T(n/2) + n  2 (c (n/2) lg (n/2)) + n = cn lg(n/2) + n = cn lg n – cn lg 2 + n = cn lg n – cn + n  cn lg n for c  1 = O(n lg n) for c  1

  7. Iteration Method • Iteration method: • Expand the recurrence k times • Work some algebra to express as a summation • Evaluate the summation

  8. T(n) = c + T(n-1) = c + c + T(n-2) = 2c + T(n-2) = 2c + c + T(n-3) = 3c + T(n-3) … kc + T(n-k) = ck + T(n-k) • So far for nk we have • T(n) = ck + T(n-k) • To stop the recursion, we should have • n - k = 0  k = n • T(n) = cn + T(0) = cn • Thus in general T(n) = O(n)

  9. T(n) = n + T(n-1) = n + n-1 + T(n-2) = n + n-1 + n-2 + T(n-3) = n + n-1 + n-2 + n-3 + T(n-4) = … = n + n-1 + n-2 + n-3 + … + (n-k+1) + T(n-k) = for nk • To stop the recursion, we should have n - k = 0  k = n

  10. T(n) = 2 T(n/2) + c 1 = 2(2 T(n/2/2) + c) + c 2 = 22 T(n/22) + 2c + c = 22(2 T(n/22/2) + c) + (22-1)c 3 = 23 T(n/23) + 4c + 3c = 23 T(n/23) + (23-1)c = 23(2 T(n/23/2) + c) + 7c 4 = 24 T(n/24) + (24-1)c = … = 2kT(n/2k) + (2k - 1)ck

  11. So far for nk we have • T(n) = 2k T(n/2k) + (2k - 1)c • To stop the recursion, we should have • n/2k = 1  n = 2k k = lg n • T(n) = n T(n/n) + (n - 1)c = nT(1) + (n-1)c = nc + (n-1)c = nc + nc – c = 2cn – c T(n) = 2cn – c = O(n)

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