Question

use a loop invariant to prove that the following program segment for computing the nth power, where is a positive integer, of a real number x is correct.

$\begin{array}{r}\mathbf{\text{power}}\mathbf{:=}\mathbf{1}\\ \mathbf{:=}\mathbf{1}\\ \mathbf{\text{while}}\mathbf{i}\mathbf{\le }\mathbf{n}\\ \mathbf{\text{power}}\mathbf{:=}{\mathbf{\text{power}}}^{\mathbf{\ast }}\mathbf{x}\\ \mathbf{:=}\mathbf{i}\mathbf{+}\mathbf{1}\end{array}$

Short answer

yes, it is correct.

Step-by-step solution

Step 1:loop invariant.

A loop invariant is a condition [between program variables] that is necessarily true immediately before and immediately after each iteration of the loop. (Note that this doesn't say anything about its trueness or falsehood over the course of an iteration.)

A loop invariant is some condition that applies to each iteration of the loop.

Explanation

In the given question a real number x is correct

$\begin{array}{r}\text{power}:=1\\ :=1\\ \text{while}\mathrm{i}\le \mathrm{n}\\ \text{power}:={\text{power}}^{\ast }\mathrm{x}\\ :=\mathrm{i}+1\end{array}$

Step 3:solution

Let p be the loop invariant.

Where, $\text{power}=x^{i-1}\text{and}\mathrm{i}\le \mathrm{n}+1$

Here, must show that p is true before the loop, after an iteration of the loop, and after the termination of the loop.

Before the loop$\mathrm{i}=1\text{and power}={\mathrm{x}}^{\mathrm{i}-1}={\mathrm{x}}^0=1$

Now here is the enter loop, in the given question i < n.

Since i is incremented by 1.

Here i < n + 1 after the iteration of the loop.

Here is power,

that ${\text{power}}_{\text{initial}}=x^{i-1}$.

Then the new value is,

$\begin{array}{r}{\text{power}}_{\text{new}}={\text{power}}_{\text{initial}}\times x\\ =x^{i-1}\times x\\ x^i=x^{(i+1)-1}\end{array}$

however because i = i + 1 from before

${\text{power}}_{\text{new}}=x^{i-1}$

After the termination of the loop,

The given question i < n is false

So, i = i + 1

$\begin{array}{r}\text{power}=x^{i-1}\\ =x^{(i+1)-1}\\ =x^n\end{array}$

Yes, it is correct.