1 / 18

Recurrences

Recurrences. It continues…. Recurrences. When an algorithm calls itself recursively, its running time is described by a recurrence. A recurrence describes itself in terms of its value on smaller inputs

lisle
Download Presentation

Recurrences

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Recurrences It continues… Jeff Chastine

  2. Recurrences • When an algorithm calls itself recursively, its running time is described by a recurrence. • A recurrence describes itself in terms of its value on smaller inputs • There are four methods of solving these: substitution, iteration, recursion tree, or master method Jeff Chastine

  3. What it looks like • This is the recurrence of MERGE-SORT • What this says is that the time involved is 1 if n = 1, • Else, the time involved is 2 times calling on half the size of the array, plus n time to merge sorted sub-arrays { Θ(1) if n = 1 T(n) = 2T(n/2) + Θ(n) if n > 1 Jeff Chastine

  4. Substitution • Similar to induction • Guess solution, and prove it holds true for next call • Powerful method, but can sometimes be difficult to guess the solution! Jeff Chastine

  5. Substitution • Example: • T (n) = 2T (n/2) + n • Guess that T (n) = O(n lg n) • We must prove thatT (n)  cn lg n | c > 0 • Assume it holds for n/2 as well T (n/2) = c(n/2) lg (n/2) + n // Rewrite using n/2 T (n)  2 (c(n/2) lg (n/2)) + n // Substitution = cn lg (n/2)) + n // Math stuff = cn (lg n – lg 2) + n = cn lg n – cn + n  cn lg n  c  1 Note:  means ‘for all’ Jeff Chastine

  6. Subtleties • Let T (n) = T (n/2) + T(n/2) + 1 • Assume that T (n) = O(n) • ThenT (n/2) = c(n/2) T (n)  c(n/2) + c(n/2) + 1 = cn + 1 Which does not implyT (n)  cn • Here, we’re correct, but off by a constant! Jeff Chastine

  7. Subtleties • We strengthen our guess: T (n)  cn – b T (n)  (c(n/2) – b) + (c(n/2) – b) + 1 = cn – 2b + 1  cn – b  b > 1 Jeff Chastine

  8. One Last Example Jeff Chastine

  9. Iteration Method • Expand out the recurrence • T(n) = 3T(n/4) + n • T(n) = 3(3T(n/16) + (n/4)) + n • T(n) = 3((3T(n/64) + 3(n/16))+ (n/4))) + n • T(n) < n + 3n/4 + 9n/16 + 27n/64 + … ∞ 1 1 n Σ i ( ) 3 = n 1 = n = 4n 1 - 3/4 4 4 i=0 Jeff Chastine

  10. Recursion Tree Method • A recursion tree is built • We sum up each level, • Total cost = number of levels * cost at each level • Usually used to generate a good guess for the substitution method • Could still be used as direct proof • Example: T (n) = 3T (n/4) + (n2) Jeff Chastine

  11. T(n) Jeff Chastine

  12. cn2 T(n/4) T(n/4) T(n/4) Jeff Chastine

  13. cn2 c(n/4)2 c(n/4)2 c(n/4)2 T(n/16) T(n/16) T(n/16) T(n/16) T(n/16) T(n/16) T(n/16) T(n/16) T(n/16) Jeff Chastine

  14. cn2 c(n/4)2 c(n/4)2 c(n/4)2 c(n/16)2 c(n/16)2 c(n/16)2 c(n/16)2 c(n/16)2 c(n/16)2 c(n/16)2 c(n/16)2 c(n/16)2 T(1) T(1) T(1) T(1) T(1) T(1) T(1) T(1) T(1) T(1) T(1) T(1) T(1) T(1) T(1) T(1) T(1) Jeff Chastine

  15. cn2 cn2 c(n/4)2 c(n/4)2 c(n/4)2 3/16cn2 c(n/16)2 c(n/16)2 c(n/16)2 c(n/16)2 c(n/16)2 c(n/16)2 c(n/16)2 (3/16)2cn2 c(n/16)2 c(n/16)2 (nlog43) T(1) T(1) T(1) T(1) T(1) T(1) T(1) T(1) T(1) T(1) T(1) T(1) T(1) T(1) T(1) T(1) T(1) Jeff Chastine

  16. Questions • How many levels does this tree have? • The subproblem at depth i is n/4i • When does the subproblem hit size 1? • n/4i = 1 • n = 4i • lg4n = i • Therefore, the tree has lg4n levels • The cost at each level is 3ic(n/4i)2 • The last level has 3log4n nodes = nlog43 Jeff Chastine

  17. The Master's Method When it has this form: T(n) = aT(n/b) + f(n) • If f (n) = Ο(nlogba-ε) for some constant ε>0, then T (n) = Θ (nlogba) • If f (n) = Θ(nlogba), then T (n) = Θ (nlogbalgn) • If f (n) = Ω (nlogba+ε) for some constant ε>0, then T (n) = Θ (f (n)) Jeff Chastine

  18. T(n) = 9T(n/3) + n • a = 9, b = 3 thus nlogba =nlog3 9=n2 • T(n) = T(2n/3) + 1 • a = 1, b = 3/2 thus nlogba =nlog3/2 1 =n0 =1 • Problem: • T(n) = 2T(n/2) + nlgn Jeff Chastine

More Related