By Aho A.V., Ullman J.D.

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J0 . Then we must do the following two steps: ✦ 1. Prove each of the basis cases, the statements S(i0 ), S(i0 + 1), . . , S(j0 ). 2. As an inductive hypothesis, assume all of S(i0 ), S(i0 + 1), . . , S(n) hold, for some n ≥ j0 , and prove S(n + 1). 7. Our first example of a complete induction is a simple one that uses multiple basis cases. As we shall see, it is only “complete” in a limited sense. To prove S(n + 1) we do not use S(n) but we use S(n − 1) only. In more general complete inductions to follow, we use S(n), S(n − 1), and many other instances of the statement S.

Triangular number a) n i=1 b) n 2 i=1 i = n(n + 1)(2n + 1)/6. c) n 3 i=1 i = n2 (n + 1)2 /4. d) n i=1 i = n(n + 1)/2. 1/i(i + 1) = n/(n + 1). 1(a) is tn marbles. For example, bowling pins are arranged in a triangle 4 on a side and there are t4 = 4 × 5/2 = 10 pins. Show by induction on n that nj=1 tj = n(n + 1)(n + 2)/6. 4: Suppose we use three digits — say 0, 1, and 2 — to code symbols. A set of strings C formed from 0’s, 1’s, and 2’s is error detecting if no two strings in C differ in only one position.

2 ITERATION 33 Line (3) calls SelectionSort to sort the array. Lines (4) and (5) print the integers in sorted order. = EOF; n++) ; SelectionSort(A,n); /* sort A */ for (i = 0; i < n; i++) printf("%d\n", A[i]); /* print A */ } void SelectionSort(int A[], int n) { int i, j, small, temp; for (i = 0; i < n-1; i++) { small = i; for (j = i+1; j < n; j++) if (A[j] < A[small]) small = j; temp = A[small]; A[small] = A[i]; A[i] = temp; } } Fig. 3. A sorting program using selection sort. 1: Simulate the function SelectionSort on an array containing the elements a) b) c) 6, 8, 14, 17, 23 17, 23, 14, 6, 8 23, 17, 14, 8, 6 How many comparisons and swaps of elements are made in each case?

### Foundations of computer science, C version by Aho A.V., Ullman J.D.

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