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I have the following simple algorithm to find the smallest element of an array $A$ of numbers:

m = A[0];
For i = 0 to A.length() do
    if m < A[i] then
        m = A[i];
    end
end
return m;

The complexity of this algorithm is $O(n)$ where $n$ is the length of $A$.

I would like to prove the correctness of this algorithm using an invariant. Yet I don't know which invariant is nice here. Maybe something like: $m$ is the minimum of the subarray $A[1..k]$.

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  • 2
    $\begingroup$ I think that invariant works perfectly for this algorithm. What are you stuck on exactly? $\endgroup$ – Nathan Feb 21 at 14:03
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The invariant is good (once you fix lots of imprecise things, like the 1..k which is obvious nonsense, and wouldn’t proof correctness. Precision is important. ) The code is broken, so your proof should fail - because it would be an invariant of a correct algorithm, but not your code posted here.

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