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CS200: Algorithm Analysis

Delve into the speed of comparison sorting algorithms, unveiling the mysteries behind decision trees and lower bounds. Discover the intricate workings of Insertion Sort and Merge Sort, and explore the optimal sorting algorithms. Dive deep into theorem 8.1 and corollary 8.2 for comprehensive insights.

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CS200: Algorithm Analysis

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  1. CS200: Algorithm Analysis

  2. MORE ON SORTING • How fast can we sort n items if we use a comparison sort?____________ • Comparison sorts require comparison of elements to induce order: Mergesort, Quicksort, Heapsort, Insertion sort...etc. • A comparison sort must do at least n comparisons in order to compare n elements: but what about the large gap between n and n log n? Can we do better than n log n?

  3. Why comparison sorts must be at least n log n? • Argument uses a Decision Tree abstraction that represents the comparisons performed by a sorting algorithm on an input of size n. • Show the decision tree for insertion sort on A = <a1,a2,a3> (next slide). Note that there are n! leaves in the tree corresponding to all possible permutations of n input elements. • Each interior node is annotated with the relationship (≤, >) between ai and aj; elements in the set to be sorted. • Each leaf is annotated by a permutation of n.

  4. Tracing Insertion sort requires finding a path from the root to a leaf in the decision tree. • Trace Insertion sort for A= <9,2,6>. • The longest path for a decision tree that models Insertion sort is Q(n2) and for MergeSort is Q(n log n). This represents the worst-case number of comparisons. • A lower bound on the height of a decision tree thus gives a lower bound on the running time of a comparison sort.

  5. Theorem: Any decision tree that sorts n elements has a height of h = W(n log n). Proof: a decision tree has n! leaves. A decision tree must be a binary tree because a split at a sub-root depends on the two relationships; ≤ and >. A binary tree of height h has no more than 2h leaves (by definition, see Appendix B, done in lecture). n! ≤ 2h, lemma proved next, see Appendix B. Take log of both sides, log(n!) ≤ log(2h) h ≥ log(n!) for n! use, n! > (n/e)n By Stirlings approx. h ≥ log(n/e)n log(n/e)n = n log n - n log e n log e is a lower order term. h = W(n log n)

  6. Lemma - Now to prove the lemma: Proof: By induction on h. Basis: h = 0. Tree is just one node, which is a leaf. 2h= 1. Inductive step: Assume true for height = h − 1. Extend tree of height h − 1 by making as many new leaves as possible. Each leaf becomes parent to two new leaves. # of leaves for height h = 2·(# of leaves for height h−1) = 2 · 2h-1 (ind. hypothesis) • • • • = 2h.

  7. By Theorem the following holds: mergesort, heapsort and median partitioned quicksort are optimal sorting algorithms.

  8. Summary • Lower bound on sorts • Theorem 8.1 and corollary 8.2

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