the algorithm is intended to compute m^n for any positive integer m, n. How do i show the correctness of this algorithm through induction on n.
long exp(long m, int n) {
if(n == 0) return 1;
if(n == 1) return m;
if(n % 2 == 0) return exp(m*m, n/2);
else return exp(m*m, n/2) * m;
}
Let P(i) be the statement:
exp(m, i)computes mi for any integer m
For the base case, it is clear that P(0) is true as we have exp(m, 0) = 1 = m^0.
For the inductive step, we assume that P(0), P(1), ..., P(k-1) are all true, and we claim that P(k) is also true. We have to consider the following three cases:
1.) If k = 1, then P(1) is clearly true as we have exp(m, 1) = m = m^1;
2.) If k > 1 and k % 2 == 0, then by the definition of exp we have:
exp(m, k) = exp(m * m, k / 2)
By the induction hypothesis, we have exp(m * m, k / 2) = (m * m)^(k / 2) = m^k, hence P(k) is true in this case.
3.) If k > 1 and k % 2 == 1, then by the definition of exp we have:
exp(m, k) = exp(m * m, k / 2) * m
By the induction hypothesis, we have exp(m * m, k / 2) = (m * m)^(k / 2). Since k % 2 == 1, we have k / 2 = (k - 1) / 2. Hence we have:
exp(m * m, k / 2) = (m * m)^(k / 2) = (m * m)^((k-1) / 2) = m^(k-1)
Thus:
exp(m, k) = exp(m * m, k / 2) * m = m^k
So P(k) is also true in this case.
