Question

This is a question from a past exam paper.

Why does the loop invariant say i<=n when the loop test says i<n.

Is an appropriate answer: It says i<=n as i will equal n on the failing condition of the while loop. Therefore the 6th iteration of i will equal the n value 6 on the failing condition. However, the while loop itself states i<n as i starts at 0 and will finish looping once i is equal to 5.

private int n =6;

public int fact(){
    int i = 0;
    int f = 1;

    /**loop invariant
     * 0<=i<=n
     * f=i!
     */
    while(i<n){//loop test
        i=i+1;
        f=f*i;
    }

    return f;
}

Answer 1

Because Post-Condition is i==n when the loop is left. Pre-Condition when entering the loop is i==0. Inside the loop i is counting up towards n. So the invariant is 0 <= i <= n.

I omitted the invariant parts for f in my exlanation. This is not really sufficient since the invariant must capture the correctness and the meaning of the loop.

private int n = 6;

public int fact(){
    int i = 0;
    int f = 1;

    /* loop invariant: 0 <= i <= n && f == i! */
    /* PRE: i == 0 && f == i! */
    while (i < n) {
        i = i + 1;
        f = f * i;
    }
    /* POST: i == n && f == i! */

    return f;
}