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BMAD

Connect the squares

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Within the outlined rectangle and Using vertical and horizontal line segments as a chain connect the squares of the same color but do not let the chains that connect common squares cross. All white squares must be occupied.

post-53485-0-20335300-1376702833_thumb.j

Edited by BMAD

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We have a 7x6 chessboard, so let's paint the squares black and white starting with black top-left square.


Observe, that a path always consists of squares with alternating colors.
Therefore each path connecting squares of different color has equal number of black and white squares,
and each path connecting two squares of the same color contains one more square in color of it's endpoints.
Observe, that blue and green endpoints are all black, while third path has it's endpoints of different colors (one white and one black).
Covering the whole chessboard with three such paths would mean that the chessboard has two black squares more than white.
But clearly black and white squares of the chessboard are equal in number, therefore such three paths cannot exist.

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i'm not seeing any possible solution.

post-51880-0-09697600-1376879583_thumb.p

the top most square is whats the minimum absolute necessary.

the problem comes in with the two lettered squares. if i cover A with blue, i have no way of getting A and C in the left most pic. if i cover B, i either cant connect to my already existing chain, or can do so but skip a square or two.

if i cover neither A nor B; the green doesn't work. i cant cover both A, B and connect to my errant chain.

i can't wait to see the solution if there is one.

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