# Fall 2007CS 2251 Proof by Induction. Fall 2007CS 2252 Proof by Induction There are two forms of induction Standard Induction –Prove the theorem is true.

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Fall 2007CS 2251 Proof by Induction

Fall 2007CS 2252 Proof by Induction There are two forms of induction Standard Induction –Prove the theorem is true for some base case (n=0 or n=1) –Show that if the theorem is assumed true for n, then it must be true for n+1 Complete Induction –Show that if the theorem is assumed true for all k<n, then it must be true for n+1

Fall 2007CS 2253 Example 1 Prove that for all n>=1, S(n) is true S(n): 1/(1x2) + 1/(2x3) +... + 1/(n(n+1)) = n/(n+1)

Fall 2007CS 2254 Proof of Basis Show S(1) is true. – The left-hand side is just 1/1x2) which is 1/2. – The right-hand side is 1/(1+1) which is 1/2 so the statement is true when n=1.

Fall 2007CS 2255 Induction Step Assume S(k) is true for some k >= 1: 1/(1x2) +... + 1/(k(k+1)) = k/(k+1) Show S(k+1) is true: 1/(1x2) +... + 1/((k+1)(k+2)) = (k+1)/(k+2) –The left-hand side of S(k+1) is 1/(1x2) +... + 1/(k(k+1)) + 1/((k+1)(k+2)) = [1/(1x2) +... + 1/(k(k+1))] + 1/((k+1)(k+2)) –By our induction assumption, the above equals [k/(k+1)] + 1/((k+1)(k+2)) = (1/(k+1)) [k + 1/(k+2)] = (1/(k+1)) [ (k(k+2) + 1))/(k+2) ] = (1/(k+1)) [ (k^2 + 2k + 1)/(k+2) ] = (1/(k+1)) [ (k + 1)^2 / (k+2) ] = (k+1)/(k+2)

Fall 2007CS 2256 Conclusion Under our induction hypothesis, we have shown that the left- and right-hand sides of S(k+1) are true. Therefore, the equation is valid for all n >= 1.

Fall 2007CS 2257 Towers of Hanoi A method for the Towers of Hanoi problem is: public static void Towers(int n, int from, int to, int other) { /*1 */ if (n > 1) /* 2 */ Towers(n-1, from, other, to); /* 3 */ System.out.println("Move disk " + n + " to pole " + to); /* 4 */ if (n > 1) /* 5*/ Towers(n-1, other, to, from); } Show that if n has the value k, k>= 1, then the total number of calls to Towers is 2 k - 1.

Fall 2007CS 2258 Basis The value of n = 1. If Towers is called and the value of n is 1, then the tests at (1) and (4) fail so there is no recursion and hence the number of calls is 1. But 1 = 2 1 - 1 so the assertion is true for the basis case.

Fall 2007CS 2259 Induction Step Assume the assertion is true if n has the value k for some k>=1: the number of calls is 2 k - 1. Assume Towers is called and the value of the formal paramter n is k+1. Show the total number of calls is 2 k+1 - 1. Since k>=1, (k+1)>1 so the calls at (2) and (5) occurwith actual parameter values of k in each case. By our induction assumption, these calls each require 2 k - 1 calls for a total of 2 k - 1 + 2 k - 1 + 1 = 2x2 k - 1 = 2 k+1 - 1 The number of calls is thus 2 k - 1, if the value of n is k for any k>=1.

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