Because my original version of the algorithm was wrong, that's why! :) I've fixed it. I don't know if it is any clearer. The basic idea is that the tests for points and quota are not independent; they follow the form: 

$points > 2000*X && $quota < 15+X

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for some integer 0 < X < 10. One can eliminate the redundancy between these two tests. Algebraically, the above is equivalent to 
$points/2000 > X && X > $quota-15

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...which can be further reduced to 
$points/2000 + 15 > $quota

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...except that there are edge cases (basically, X must be an integer strictly between 0 and 10) which make the algorithm a bit more complex.

Specifically, the last pair of inequalities can be written as 

ceil( $points/2000 ) > X && X > $quota - 15

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but since X is an integer, we can write 
ceil( $points/2000 ) - 1 >= X && X > $quota - 15

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Now we can eliminate the X: 
( ceil( $points/2000 ) - 1 ) + 15 > $quota

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This manipulation takes care of enforcing the constraint that X must be an integer.

Lastly, to take care of the limits on X, we use min and max. The final test is then: 

max( min( 24, ( ceil( $points/2000 ) - 1 ) + 15 ), 15 ) > $quota

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the lowliest monk

In reply to Re^3: alternatives to if and series of elsif by tlm
in thread alternatives to if and series of elsif by kiat

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