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Anonymous Monk
<p>Sounds like you've got a very straightforward algorithm laid out there. It is difficult to be both "greater" and "equal" at the same time, but I presume you meant to say "or".</p><p>You may have also noticed how repetitive it is, and that it can be reworded as a for loop, using <c>N</c>, and <c>(N+1)%$numTidePoints</c></p>
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