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Anonymous
Mondrian art puzzle 0 is impossible proof 2016-11-17 12:24:36 Post No. 8479692
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Mondrian art puzzle 0 is impossible proof 2016-11-17 12:24:36 Post No. 8479692
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I believe I have a proof shows its impossible to score 0 in the Mondrian art puzzle.
For those who don't know what that is the rules are as follows
>start with a square of side N
>split the square into rectangles
>no two squares can have the same side lengths (if you have 3x2 you can't have another 3x2 or a 2x3)
>your score is the largest rectangle's area minus the smallest rectangle's area (lowest score wins)
Therefore a score of 0 requires all rectangles to be the same area. And thus no two rectangles can share a side length.
From this point on I will need to use terms which will be greentext to describe things.
>bisect
To bisect a rectangle is to split it with one line spanning one edge to the other.
Bisections are not adequect to achieve a score of 0 as any bisect ion will leave two rectangles sharing a side. Therefore we need a new way to cut a rectangle that doesn't require a bisection, and that brings us to the
>Aperture cut
An aperture cut is made by placing a rectangle in the rectangle to be cut and then drawing a line from each corner to a different edge of the outer rectangle (this shape reminds me of an aperture). An example of an aperture cut could be 4 2x1 rectangles arranged around a 1x1 rectangle to make a 3x3 rectangle. The centre of an aperture cut is the
>core
The core is something that occurs in all possible divisions of a rectangle that avoid a bisection. By placing other rectangles around an aperture cut and then stretching one or more of the outer rectangles of the aperture cut you can make a
>mutation
Mutations can have 2 or 3 more rectangles than their original cut, which may itself be a mutation (thus all numbers from 7 onwards can be achieved by mutation).
Now that the terms are out of the way onto proving that aperture cuts and mutations can't be used to score 0.
Any given rectangle can be skewed along one of its axis to change its shape while keeping its components areas ratios constant (cont)
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(Cont)
Thus by skewing an aperture cut you can always make the core a square. And solving for other rectangles of the aperture cut you'll find the only possible side lengths that result in a shared area are phi (1.618) and 1/phi. Further solving for the side lengths of a mutation requires at least one to have a length of 0 or for one to have a different area which is not permissible. It is impossible to split a rectangle without a bisection or a core and thus the only methods for splitting a rectangle result in
1 a shared side length
2 an irrational side length
3 at least one rectangle with a different area
So that should cover everything. Did I miss anything.
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I suppose for clarity I should mention multi cored rectangles and nested cores
>multi cored
Refers to a cut with 2 or more cores. and this retains the same irrational side length problem
>nested core
Should be self explanatory
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Not enough math symbols.
Post it on math stackexchange, overflow, or /r/maths.
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Is saying:
They cant have the same area by definition, so the difference in area is always non zero.
Considered a proof? I'm not a mathematician, just curious.
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