in reply to Re: help with nesting
in thread help with nesting

Thanks, and I apologize if I was not specific about the problem. Lets say I have a piece of paper 8.5x11 and a set of rectangles {(x,y,n),(x1,y1,n1),(x2,y2,n2)} x=base, y=height, n=number of these. I am trying to fit as many of these rectangles onto the paper as I can.

I guess that I also need to add in a "profit" aspect, because I would like to have the least amount of cuts possible. Thanks again and any help is greatly appreciated.

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Re: Re: Re: help with nesting
by dragonchild (Archbishop) on Oct 01, 2003 at 17:29 UTC
    That problem statement is a variation of the Knapsack problem. Your initial constraint is the 8.5x11 size and you have an additional constraint of aligning the sides as much as possible. The code I posted does accidentally align the sides of the rectangles, but provides no constraints on the size.

    I might play around with it a bit to provide the paper-size constraint, but you'd be better off following the googlinks that others have provided.

    Update: You can try the following algorithm to see if it helps:

    1. Get your paper size
    2. Get your list of rectangles. Collapse it to (x,y,n) if you don't already have it that way.
    3. Eliminate all rectangles that cannot fit on the paper. E.g., 1x12 won't fit on 8.5x11, but will fit on 8.5x14
    4. Reduce the number of rectangles of a given size to the most that can fit on your paper. For example, two 6x11 rectangles cannot fit on the same 8.5x11 piece of paper
    5. Expand the list of rectangles so that (x,y,n) => (x,y) x n
    6. Order the remaining rectangles by area, smallest first, then by X, smallest first
    7. Start tiling in the X direction, filling the Y direction and see what comes out
    8. Re-order the rectangles by area, then Y (smallest first in both values)
    9. Repeat, but in the Y direction filling the X direction.
    10. Repeat the two tilings, swapping each particular rectangle's orientation. (($x, $y) = ($y, $x);) This will result in N! tilings, where N is the total number of rectangles (of all sizes) you're working with
  • This problem is O(n!), which is pretty crappy. It's also (at least) NP-hard, so you can't really prove this algorithm will provide the best solution. It might provide a good base for a human to take and improve upon, but that's the best it could possibly do.

    ------
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    The idea is a little like C++ templates, except not quite so brain-meltingly complicated. -- TheDamian, Exegesis 6

    Please remember that I'm crufty and crochety. All opinions are purely mine and all code is untested, unless otherwise specified.

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