The following algorithms were given for computing the nth power of a positive integer a.

Homework Help: Questions and Answers: The following algorithms were given for computing the n^th power of a positive integer a. Use induction to prove that each of these algorithms is correct. (b) Hint: you need cases due to last two if-statements.

Solution: Step by Step Answering

To prove the correctness of the given algorithms using induction, we need to follow these steps:

Base Case: Show that the algorithm works for the smallest input value, typically n = 0.
Inductive Step: Assume the algorithm works for some arbitrary value k, and then show it works for k + 1.

Part (a)

Algorithm:

def pow(a, n):
    if n == 0:
        return 1
    return pow(a, n-1) * a

Base Case

For n = 0:
pow (a, 0) = 1
This is correct because a⁰ = 1 for any positive integer a.

Inductive Step

Assume the algorithm works for n = k, i.e., assume:
pow (a, k) = a^k

We need to show that the algorithm works for n = k + 1:
pow (a, k + 1) = pow (a, (k+1) – 1) . a = pow (a, k). a

By the induction hypothesis:
pow (a, k) = a^k

Therefore:
pow (a, k + 1) = a^k . a = a^{k + 1}

Since both the base case and the inductive step are proven, by mathematical induction, the algorithm correctly computes aⁿor all non-negative integers n.

Part (b)

Algorithm:

def pow(a, n):
    if n == 0:
        return 1
    x = pow(a, floor(n/2))
    if n % 2 == 0:
        return x * x
    if n % 2 != 0:
        return a * x * x

Base Case

For n = 0:
pow (a, 0) = 1
This is correct because a⁰=1 for any positive integer a.

Inductive Step

Assume the algorithm works for n = k, i.e., assume:
pow (a, k) = a^k

We need to show that the algorithm works for n = k + 1:

If k + 1 is even:
Let k + 1 = 2m for some integer m
pow (a, 2m) = pow (a, m) . pow (a, m)
By the induction hypothesis:
pow (a, m) = a^m
Therefore:
pow (a, 2m) = a^m . a^m = a^2m
If k + 1 is odd:
Let k + 1 = 2m + 1 for some integer m.
pow (a, 2m + 1) = a. pow (a, m). pow (a, m)
By the induction hypothesis:
pow (a, m) = a^m
Therefore:
pow (a, 2m + 1) = a . a^m . a^m = a . a^2m = a^{2m + 1}

Since both the base case and the inductive step are proven, by mathematical induction, the algorithm correctly computes aⁿ for all non-negative integers n.

This completes the proof for both algorithms using mathematical induction.

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