Question

l = -1; u = n;
while l+1 != u
    m = l + (u-l)/2;
    if x[m] < t
        l = m;
    else
        u = m;
p = u;
if p >= n || x[p] != t
     p = -1;

We assume x[-1] < t and x[n] >= t and n >= 0 in the above code.
The above code is a modified binary search which can return the first occurrence of the integer t in the integer array x[0..n-1] instead of returning a random one.

My question is like this:

Why do the above code always halt? Can anyone explain it or prove it?

Thanks,

Answer 1

Because on every iteration, the gap between l and u halves, within the constraints of integer arithmetic. All sequences of (positive) integer halving must eventually reach 1, which is the termination condition.