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Old 2003-07-26, 16:06   #1
Kevin
 
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Aug 2002
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Default Checkerboard Problem

Another good puzzle out of my logic book.

http://www.mersenneforum.org/jpg/che...ardproblem.jpg
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Old 2003-07-26, 18:45   #2
ET_
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"Luigi"
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For the first puzzle, note that each domino has a black square and a white square, while the deleted corner squares are both white.

Each domino will cover exactly one black and one white square, leaving out (uncovered) a couple of black squares.

Luigi
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Old 2003-07-26, 20:03   #3
AllPrimesAreOdd
 
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Default solution

ET nailed the first one.

The second problem is solved with very similar logic.

Note that the 4 squares that are removed consist of 2 white and 2 black. This leaves 60 squares: 30 white, 30 black.
Each T covers 4 squares. THus, 60/4 = 15 T's must be used. Since half a T cannot be used, we cannot use an equal number of each T.
T1 covers 3W + 1B, T2 covers 1W + 3B. If we use A T1's and C T2's, then A + C = 15 => C +15 - A.
So, A(3W + B) + (15 - A)(W + 3B) = 3AW + AB + 15W +45 B - AW - 3B = (2A + 15)W + (A + 42)B = 60. (Note: W and B are not values. they are units.)
(2A + 15)W = 30W because there are 30 white squares.
thus, 2A = 15 => A=7.5 impossible because we have to use whole T's.
Thus, the board cannot be covered.

This proof could be a lot more compact, but I wanted to be fairly thorough.
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Old 2003-07-26, 21:39   #4
Kevin
 
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AllPrimesAreOdd- You saw the solution in class, that's cheating. Let the other people have a go at it, first.
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Old 2003-07-27, 01:11   #5
AllPrimesAreOdd
 
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Default I DID NOT

I did not just see the problem in class. I solved it.
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