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    <p>You need to form a theory for why you think the algorithm does terminate. Ideally, prove the theory as a mathematical theorem.</p> <p>You can look for a function of the problem state that does reduce on each recursive call. For example, see the following discussion of Ackermann's function, from <a href="http://en.wikipedia.org/wiki/Ackermann_function#Definition_and_properties" rel="nofollow">Wikipedia</a></p> <blockquote> <p>It may not be immediately obvious that the evaluation of A(m, n) always terminates. However, the recursion is bounded because in each recursive application either m decreases, or m remains the same and n decreases. Each time that n reaches zero, m decreases, so m eventually reaches zero as well. (Expressed more technically, in each case the pair (m, n) decreases in the lexicographic order on pairs, which is a well-ordering, just like the ordering of single non-negative integers; this means one cannot go down in the ordering infinitely many times in succession.) However, when m decreases there is no upper bound on how much n can increase — and it will often increase greatly.</p> </blockquote> <p>That is the type of reasoning you should be thinking of applying to your algorithm.</p> <p>If you cannot find any way to prove your algorithm terminates, consider looking for a variation whose termination you can prove. It is not always possible to decide whether an arbitrary program terminates or not. The trick is to write algorithms you can prove terminate.</p>
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    1. CO+1, Nice answer.
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