**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 a^{0} = 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 a^{n }or 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 a^{0}=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 a^{n} for all non-negative integers n.

This completes the proof for both algorithms using mathematical induction.

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