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September 22, Algorithms and Data Structures Lecture III Simonas Šaltenis Aalborg University

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September 22, This Lecture Divide-and-conquer technique for algorithm design. Example problems: Tiling Searching (binary search) Sorting (merge sort).

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September 22, Tiling A tromino tile: And a 2 n x2 n board with a hole: A tiling of the board with trominos:

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September 22, Tiling: Trivial Case (n = 1) Trivial case (n = 1): tiling a 2x2 board with a hole: Idea – try somehow to reduce the size of the original problem, so that we eventually get to the 2x2 boards which we know how to solve…

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September 22, Tiling: Dividing the Problem To get smaller square boards lets divide the original board into for boards Great! We have one problem of the size 2 n-1 x2 n-1 ! But: The other three problems are not similar to the original problems – they do not have holes!

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September 22, Tiling: Dividing the Problem Idea: insert one tromino at the center to get three holes in each of the three smaller boards Now we have four boards with holes of the size 2 n-1 x2 n-1. Keep doing this division, until we get the 2x2 boards with holes – we know how to tile those

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September 22, Tiling: Algorithm INPUT: n – the board size (2 n x2 n board), L – location of the hole. OUTPUT: tiling of the board Tile(n, L) if n = 1 then Trivial case Tile with one tromino return Divide the board into four equal-sized boards Place one tromino at the center to cut out 3 additional holes Let L1, L2, L3, L4 denote the positions of the 4 holes Tile(n-1, L1) Tile(n-1, L2) Tile(n-1, L3) Tile(n-1, L4) INPUT: n – the board size (2 n x2 n board), L – location of the hole. OUTPUT: tiling of the board Tile(n, L) if n = 1 then Trivial case Tile with one tromino return Divide the board into four equal-sized boards Place one tromino at the center to cut out 3 additional holes Let L1, L2, L3, L4 denote the positions of the 4 holes Tile(n-1, L1) Tile(n-1, L2) Tile(n-1, L3) Tile(n-1, L4)

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September 22, Divide and Conquer Divide-and-conquer method for algorithm design: If the problem size is small enough to solve it in a straightforward manner, solve it. Else: Divide: Divide the problem into two or more disjoint subproblems Conquer: Use divide-and-conquer recursively to solve the subproblems Combine: Take the solutions to the subproblems and combine these solutions into a solution for the original problem

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September 22, Tiling: Divide-and-Conquer Tiling is a divide-and-conquer algorithm: Just do it trivially if the board is 2x2, else: Divide the board into four smaller boards (introduce holes at the corners of the three smaller boards to make them look like original problems) Conquer using the same algorithm recursively Combine by placing a single tromino in the center to cover the three introduced holes

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September 22, Find a number in a sorted array: Just do it trivially if the array is of one element Else divide into two equal halves and solve each half Combine the results Binary Search INPUT: A[1..n] – a sorted (non-decreasing) array of integers, s – an integer. OUTPUT: an index j such that A[j] = s. NIL, if j (1 j n): A[j] s Binary-search(A, p, r, s): if p = r then if A[p] = s then return p else return NIL q (p+r)/2 ret Binary-search(A, p, q, s) if ret = NIL then return Binary-search(A, q+1, r, s) else return ret INPUT: A[1..n] – a sorted (non-decreasing) array of integers, s – an integer. OUTPUT: an index j such that A[j] = s. NIL, if j (1 j n): A[j] s Binary-search(A, p, r, s): if p = r then if A[p] = s then return p else return NIL q (p+r)/2 ret Binary-search(A, p, q, s) if ret = NIL then return Binary-search(A, q+1, r, s) else return ret

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September 22, Recurrences Running times of algorithms with Recursive calls can be described using recurrences A recurrence is an equation or inequality that describes a function in terms of its value on smaller inputs Example: Binary Search

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September 22, Binary Search (improved) INPUT: A[1..n] – a sorted (non-decreasing) array of integers, s – an integer. OUTPUT: an index j such that A[j] = s. NIL, if j (1 j n): A[j] s Binary-search(A, p, r, s): if p = r then if A[p] = s then return p else return NIL q (p+r)/2 if A[q] s then return Binary-search(A, p, q, s) else return Binary-search(A, q+1, r, s) INPUT: A[1..n] – a sorted (non-decreasing) array of integers, s – an integer. OUTPUT: an index j such that A[j] = s. NIL, if j (1 j n): A[j] s Binary-search(A, p, r, s): if p = r then if A[p] = s then return p else return NIL q (p+r)/2 if A[q] s then return Binary-search(A, p, q, s) else return Binary-search(A, q+1, r, s) T(n) = (n) – not better than brute force! Clever way to conquer: Solve only one half!

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September 22, Running Time of BS T(n) = (lg n) !

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September 22, Merge Sort Algorithm Divide: If S has at least two elements (nothing needs to be done if S has zero or one elements), remove all the elements from S and put them into two sequences, S 1 and S 2, each containing about half of the elements of S. (i.e. S 1 contains the first n/2 elements and S 2 contains the remaining n/2 elements). Conquer: Sort sequences S 1 and S 2 using Merge Sort. Combine: Put back the elements into S by merging the sorted sequences S 1 and S 2 into one sorted sequence

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September 22, Merge Sort: Algorithm Merge-Sort(A, p, r) if p < r then q (p+r)/2 Merge-Sort(A, p, q) Merge-Sort(A, q+1, r) Merge(A, p, q, r) Merge-Sort(A, p, r) if p < r then q (p+r)/2 Merge-Sort(A, p, q) Merge-Sort(A, q+1, r) Merge(A, p, q, r) Take the smallest of the two topmost elements of sequences A[p..q] and A[q+1..r] and put into the resulting sequence. Repeat this, until both sequences are empty. Copy the resulting sequence into A[p..r]. Merge(A, p, q, r) Take the smallest of the two topmost elements of sequences A[p..q] and A[q+1..r] and put into the resulting sequence. Repeat this, until both sequences are empty. Copy the resulting sequence into A[p..r].

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September 22, Merge Sort Summarized To sort n numbers if n=1 done! recursively sort 2 lists of numbers n/2 and n/2 elements merge 2 sorted lists in (n) time Strategy break problem into similar (smaller) subproblems recursively solve subproblems combine solutions to answer

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September 22, Running time of MergeSort Again the running time can be expressed as a recurrence:

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September 22, Repeated Substitution Method Lets find the running time of merge sort (lets assume that n=2 b, for some b).

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September 22, Example: Finding Min and Max Given an unsorted array, find a minimum and a maximum element in the array INPUT: A[l..r] – an unsorted array of integers, l r. OUTPUT: (min, max) such that j (l j r): A[j] min and A[j] max MinMax(A, l, r): if l = r then return (A[l], A[r]) Trivial case q (l+r)/2 Divide (minl, maxl) MinMax(A, l, q) (minr, maxr) MinMax(A, q+1, r) if minl < minr then min = minl else min = minr if maxl > maxr then max = maxl else max = maxr return (min, max) INPUT: A[l..r] – an unsorted array of integers, l r. OUTPUT: (min, max) such that j (l j r): A[j] min and A[j] max MinMax(A, l, r): if l = r then return (A[l], A[r]) Trivial case q (l+r)/2 Divide (minl, maxl) MinMax(A, l, q) (minr, maxr) MinMax(A, q+1, r) if minl < minr then min = minl else min = minr if maxl > maxr then max = maxl else max = maxr return (min, max) Conquer Combine

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September 22, Next Week Analyzing the running time of recursive algorithms (such as divide-and-conquer)

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