if you allow both increments at the same time, then any value after 3 ( = 5 * 2 - 5 - 2 ) is representable as a sum of 2's and 5's, so you might as well go the brute force way, with (0..100) without 1 and 3, cut up to current sum ( sorry, unclear here - basically, use, jdporter's solution, with generate_solutions(1, 100 ...I'd ideally like to be able to chose to only allow increments of 2 or +5
if you only want to allow 1 type of increment, say 5, then you might as well generate all partitions of 20 and multiply them by 5. The above solutions either give you only one increment, or, as in Ovid's allow only one of the two to be true, ie, 7 won't be a valid element..return if ($i == 1 || $i == 3)
Rethinking further the part about both increments allowed. What about starting with an array (100,0,0,0,0), and then, from the right hand side, try to move either a 2 or 5 to the right, with backtracking recursion, printing any time you succeed without making a negative number. Hmm, but that would make it possible to shift a 2 from a 5, with a remaining 3, which is bad...
Well, ok, I'll stop now.. the above idea, with just shifting 1's, should easily generate all unrestricted partitions of a number in a lexicographic order, so if you only needed 1 increment, it should suffice.. You can probably write the actual code for it faster than I.. in the meantime, I'll start thinking on generating weird restricted partitions...
In reply to Re: How to determine & record all possible variations for 5 items?
by ivancho
in thread How to determine & record all possible variations for 5 items?
by Stenyj
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