Data Structures and Algorithms (NTU, Class 01, Spring 2020) instructor: Hsuan-Tien Lin
Correctness of getMinIndex
1 Question
The following getMinIndex algorithm has been introduced in the class. Prove that the algorithm is correct. That is, the index to the minimum element will be returned.
getMinIndex(integer array arr, integer len) minpos ← 0
for i ← 1 to len − 1 do if arr[i] < arr[minpos] then
minpos ← i end if end for
return minpos
2 Answer
Claim 0: Upon existing the loop at i = k for any k = 1, 2, . . . , len − 1, arr[minpos] ≤ arr[j] for j = 0, 1, 2, . . . , k.
Proof : Let mk denote the value of minpos when existing the loop at i = k.
1. The claim is true when i = 1, because either
• arr[1] < arr[0], which means that the if is true (i is 1 and minpos was assigned to 0 in the first line) and m1 is then assigned to 1, making arr[m1] ≤ arr[j] for j = 0, 1, or
• arr[1] ≥ arr[0], which keeps m1= 0 and thus arr[m1] ≤ arr[j] for j = 0, 1.
2. Assume that when i = t − 1, the claim is true. That is,
arr[mt−1] ≤ arr[j] for j = 0, 1, 2, . . . , t − 1. (1) Then, when i = t, there are two cases
• the if is true, which means arr[t] < arr[mt−1]. Combining the inequality with (1),
arr[t] < arr[j] for j = 0, 1, 2, . . . , t − 1. (2) In this case, mtgets updated to t. Combining the trivial arr[t] ≤ arr[t] with (2), we get
arr[mt] ≤ arr[j] for j = 0, 1, 2, . . . , t. (3)
• the if is false, which means arr[t] ≥ arr[mt−1]. Note that mtkeeps the value of mt−1 here.
We can then combine the inequality with (1), and get
arr[mt] ≤ arr[j] for j = 0, 1, 2, . . . , t. (4) So in both cases, the claim is true when i = t. By mathematical induction, the claim is true for any i = 1, 2, . . . , k.
Claim 1: Upon existing the algorithm, arr[minpos] ≤ arr[j] for j = 0, 1, 2, . . . , len − 1 for any positive len. That is, the algorithm is correct.
Proof : Claim 1 is trivially true for len = 1, where minpos stays at 0 and is surely ≤ arr[j] for j = 0.
For other len, apply Claim 0 with k = len − 1 and we see that Claim 1 is also true.
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