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Re^2: Not A Magic Square But Similar (Finite Geometry)

by kennethk (Abbot)
on Sep 06, 2013 at 20:59 UTC ( #1052767=note: print w/replies, xml ) Need Help??

in reply to Re: Not A Magic Square But Similar (Finite Geometry)
in thread Not A Magic Square But Similar

For the 3 square, you are correct; we have 6 constraints, as opposed to the magic square's 8 (or really 5 as opposed to 7, given arbitrary normalization). However, for 4, we have 24 constraints as opposed to a magic square's 10; in general, constraint count grows factorially as opposed to the magic square's linear. Of course, I'm out of my depth w.r.t. geometry here, so maybe there's another mapping I'm missing.

#11929 First ask yourself `How would I do this without a computer?' Then have the computer do it the same way.

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Re^3: Not A Magic Square But Similar (Finite Geometry)
by LanX (Bishop) on Sep 06, 2013 at 21:37 UTC
    Nope I was too lazy to update the cases of non-parallel "sections", only mentioned it in reply to hdb (whose solution makes mine obsolete).

    But I think that hdb's solution resp. your generalization already describe the complete solution room and that all possible magic square can be found there by mapping rows to diagonals.

    So since the other way round works you will always find a back-projection from magic to "limbic" square.

    Too tired to dig into proving it, but looking into the literature for magic squares should show it's trivial.

    (unproven opinion)

    Cheers Rolf

    ( addicted to the Perl Programming Language)


    ) and I'm still not sure if L~R really wants them.

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[Eily]: if you want to store in a structure with the coordinates as key, arrays might do, since the keys are going to be 0..n
[LanX]: (Pascale path)
[Eily]: paths like that
[Discipulus]: yes Eily++ (very keen) I want to integrate my project with a 17th experiments. I want to colorize in sequence all paths
[oiskuu]: Yeah, modifry the recursive func combinations() to return not the number, but the paths themselves.
[Eily]: Discipulus I'd do that by starting from the bottom node I think. That way it can inherit the paths from the two nodes above (and so on, recursively)
[LanX]: oh I meant fixed amount
[LanX]: every path must have l left and r right edges and l and r are fixed and l+r is the height
[LanX]: simple recursive function which goes left and right till l or r are exhausted
LanX #solved

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