MS 101: Algorithms Instructor Neelima Gupta

Table Of Contents Proving Correctness of Algorithms

Proving the correctness of Algorithm Sequential Search 1. index = 1; 2. While index ≤ n and L[index] ≠ x do index = index + 1; 3. if index > n then index = 0;

Defining the I/O of the algorithm Input : Given an array L containing n items (n ≥ 0) and given x, Output: –the sequential search algorithm terminates –with index = first occurrence of x in L, if found –and, index = 0 otherwise.

Loop Invariant Hypothesis: For 1 ≤ k ≤ n + 1, L(k): when the control reaches the test in line 2 for kth time –index = k and, –for 1 ≤ i ≤ k-1, L[i] ≠ x. Prove the above hypothesis by induction

Proving the Hypothesis by induction Base Case : H[1] is true vacuously. Let H(k) be true We will prove that H(k+1) is also true.

control reaches the test in line 2 for (k+1)th time only if L(k) ≠ x ………….….. (i) Also since H(k) is true 1 ≤ i ≤ k-1, L[i] ≠ x ………….. (ii) from (i) and (ii) we get, 1 ≤ i ≤ k, L[i] ≠ x ·˙· it holds for index = k+1 Thus by induction our hypothesis is true.

Correctness contd.. Suppose the test condition is executed exactly k times. i.e. body of the loop is executed k-1 times. –Case 1: k = n+1, by loop invariant hypothesis, index = n+ 1 and for 1 ≤ i ≤ n, L[i] ≠ x. Since index = n+ 1, line 3 sets index to 0 and by second condition above x is not in the array. So correct. –Case 2: k ≤ n, => index =k and loop terminated because L[k] = x. Thus index is the position of the first occurrence of x in the array. Hence the algorithm is correct in either case.

Binary Search Input : Given an array L containing n items (n ≥ 0) ordered such that L(1) <= L(2) <= L(3) <=… <=L(n) and given x, Output: The binary search algorithm terminates –with index = an occurrence of x in L, if found –and, index = 0 otherwise. Binary search can only be applied if the array to be searched is already sorted.

Search for the no

Search for the no

1.index_first = 1; 2.index_last=n; 3.While index_first<=index_last do 4.index_mid= floor((index_first+index_last)/2); 5.if L[index_mid]=x 6.Exit loop 7.if L[index_mid]>x 8.index_last=index_mid-1; 9.if L[index_mid]<x 10.index_first=index_mid+1; If index_first>index_last then 13.index_mid=0;

Proof of correctness The Loop Invariant Let r be the maximum number of times the loop starting from line 3 will run. Hypothesis: For 1 ≤ k ≤ r + 1, H(k): when the control reaches the test in line 3 for k th time –L[i] ≠ x for every i<first & for every i>last Prove the above hypothesis by induction

Correctness of the algorithm assuming the hypothesis (the loop invariant) Suppose the test condition is executed t times. –Case 1: first<=last. We exit from the loop because L[m t ]=x. Thus, mid is a position of occurrence of x as mid=m t. –Case 2: If first>last, we exit from the loop starting at line 3. Line 13 sets the value of mid to 0 since first>last. By loop invariant hypothesis, for i last, L[i] ≠ x i.e. x is not found in the array. The algorithm correctly returns 0 in mid. Hence assuming that the statement H(k) is correct the algorithm works correctly.

Proof of Induction Hypothesis Let f i and l i be the values of first and last when the test condition at line 3 is executed for the i th time with f 1 and l 1 being 1 and n respectively. Assume that the statement is true for H(k) We will prove that it is true for H(k+1) In the k th iteration either first was set to (f k + l k )/2 +1 i.e. f k+1 = (f k + l k )/2 +1 or last was set to (f k + l k )/2 -1 i.e l k+1 = (f k + l k )/2 -1

Proof of Induction Hypothesis contd.. Case 1: f k+1 > l k+1 ( only if f k = l k and L[(f k + l k )/2] ≠ x) By induction hypothesis L[i] ≠ x  i< f k L[i] ≠ x  i> l k (= f k ) And, also L[f k ] = L[(f k + l k )/2] ≠ x So x is not present, hence trivially L[i] ≠ x  i< f k+1 L[i] ≠ x  i> l k+1 Thus H(k+1) is true.

Proof of Induction Hypothesis contd.. Case 2: f k+1 <= l k+1 a:) when L[(f k + l k )/2] <x then f k+1 = (f k + l k )/2 +1  i< f k +1, L[i] ≤ L[(f k + l k )/2]<x (Why?)

Proof of Induction Hypothesis contd.. Case 2: f k+1 <= l k+1 a:) when L[(f k + l k )/2] <x then f k+1 = (f k + l k )/2 +1  i< f k +1, L[i] ≤ L[(f k + l k )/2]<x (input is sorted) i.e. L[i] ≠ x  i<= f k+1 – 1 or  I < f k+1 Also, by induction hypothesis L[i] ≠ x  i> l k (= l k+1 ) Thus H(k+1) is true.

Proof of Induction Hypothesis contd.. b:) When L[(f k + l k )/2] >x then l k+1 =(f k + l k )/2 -1  i> l k+1, L[i] ≥ L[(f k + l k )/2]>x (Why?) i.e. L[i] ≠ x  i> l k+1

Proof of Induction Hypothesis contd.. b:) When L[(f k + l k )/2] >x then l k+1 =(f k + l k )/2 -1  i> l k+1, L[i] ≥ L[(f k + l k )/2]>x (Why?) i.e. L[i] ≠ x  i> l k+1 Also, by induction hypothesis L[i] ≠ x  i< f k+1 (= f k ) Hence H(k+1) is true. This proves that our hypothesis is correct

Assignment 5 Show that Insertion sort works correctly by using Induction.

Up Next Introduction to some tools to designing algorithms through Sorting

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