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COSO 1030 Section 4 Recursion

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What is about Towers of Hanoi Divide and Conquer Strategy Recursion and Induction Thinking Recursively Recursion Pitfalls Analyze Efficiency of Recursion Tail Recursion Elimination

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Towers of Hanoi Move n disks from one peg to another with a help peg, can’t put a big disk on a small Mark pegs as source, destination and spare Or simply 1, 2 and 3 Print each move by saying “Move disk k from peg i to j”.

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Divide and Conquer If there is only one disk, just move it. If we can move n-1 disks from source to spare, then we can move the biggest one from source to destination. And the problem size reduced by 1, we only need to move n-1 disks from spare to destination using source as helper

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Pseudo Code (level 1) /** * Towers of Hanoi * @param n int – n disks * @param I int – source peg * @param j int – destination peg * @param k int – spare peg * print out each move */ Void hanoi(int n, int I, int j, int k)

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{ /* divide */ if( n == 1) { print(“Move ” + n + “th disk from ”+ i + “ to ” + j); } else { // 1. move n-1 disks from i to k using j for help // 2. move nth disk from i to j // 3. move n-1 disks from k to j using i. } The sub problem 2. is easy to solve print(“Move ” + n + “th disk from ”+ i + “ to ” + j); To the 1 and 3, problem size is n-1

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Conquer Solve Sub Problems Recursively To the 1: – Move n-1 disks from i to k using j – hanoi(n-1, i, k, j); To the 3: – Move n-1 disks from k to j using I – hanoi(n-1, k, j, I);

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void hanoi(int n, int i, int j, int k) { /* merge */ if( n == 1) { print(“Move ” + n + “th disk from ”+ i + “ to ” + j); } else { // 1. move n-1 disks from i to k using j for help hanoi(n-1, i, k, j); // 2. move nth disk from i to j print(“Move ” + n + “th disk from ”+ i + “ to ” + j); // 3. move n-1 disks from k to j using i. hanoi(n-1, k, j, i); }

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Recursion A function call it self directly or indirectly Recursive Function Termination – If it is defined on Natural numbers (N) A sequence with a minimal value – Base - the minimal value – N with 0 as base – Can’t recursively calculate the function on base – Any other well defined sequence can be mapped to a sub set of N

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Recursion and Induction Inductive definition of String – String ::= EMPTY – String ::= String + CHAR Recursively counts space characters in a string int numOfSpaceCh(String string) { if(string == null) { return 0; // base } else { int n = numOfSpaceCh(string.substring(1)); // recursion return ((string.charAt(0) == ‘ ’) ? 1: 0) + n; // merge }

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Think Recursively Divide into small problem(s) – Half and half – Head and tail – Random divide – Cut one at end Solve base directly Recursively solve sub problem(s) Merge the result

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Recursion Pitfalls Infinite recursion – Not well ordered …, -n, -(n-1), …, -2, -1, 0 – hanoi(0, 1, 2, 3) Exponent complexity class – Hanoi(n) is O(2^n) – Fibonacci(n) could be O(2^n)

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Analyze Recursion Efficiency Recursion Efficiency depends on – Number of recursive calls – Size reduced Hanoi(n) – Has 2 recursive calls – Size reduced by 1 – O(2*O(Hanoi(n-1))) – O(2^n)

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Inductive Proof There exists K and n0 such that for any n >= n0, K 2^n <= Hanoi(n) There exists K’ and n0’ such that for any n >= n0’, Hanoi(n) <= K’ 2^n Let K = 1 and n0 = 2 Base n = n0: if; Hanoi(1); print; Hanoi(1); 2^2 = 4 = Steps of Hanoi(2) Use SofH(n) for Steps of Hanoi(n)

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Induction Assume for any m = n - 1, m >= n0, 2^m <= SofH(m) SofH(m + 1) are no less than SofH(m) + 1 + SofH(m). By assumption 2^m <= SofH(m), we get SofH(m+1)<= 2 * 2^m = 2 ^(m+1) Conclusion: for any n > n0 Steps of Hanoi(n) >= 2 ^ n.

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O(numOfSpaceCh()) One recursive call Length reduced by 1 It is O(n) where n is the length of a string 1 * Steps of numOfSpaceCh(n-1)

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Tail Recursion A recursive function is tail recursion, if – It has only one recursive call – And the call is the last statement of the function numOfSpaceCh is a tail recursive function Iteration version int n = 0; for (int I = 0; I < string.length; I ++) { if (string.charAt(I) == ‘ ’) n++; }; return n;

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Tail Recursion Elimination Recursive Version F(n) { if (n == 0) base; else { m = non-recursive(n); F(m); // m < n } Iteration Version F(n) { base; while (n != 0) { n = non-recursive(n); }

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Why Eliminate Recursion Efficiency – Tail Recursion O(n) Time, O(n) Stack Space – Iteration O(n) Time, O(1) Stack Space

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Summary Recursion – A Powerful Tool Apply Divide and Conquer Strategy Inductive Definition, Proof and Recursive Computation How to Efficiently Divide Problems Analyze Computation Complexity Avoidable Pitfalls Eliminate Tail Recursion

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