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Erm. Isn't that covered by the bit that says: #When determining "betweeness", treat any cases where $x or $y equals a value in @z as if $x or $y was actually less than that value.? By my (probably flawed) reading of that, your first case would be a fail because the $x would be judged less than the second element of @z And the second case would be a pass as both $x and $y would be judges as being lower than the first valeue of @z Examine what is said, not who speaks. In reply to Re: Re: Golf Challenge - does $x and $y fit into the same slot on @z?
by BrowserUk
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