Two Colors in Two Rows II

Elsewhere we discussed the following problem:

Consider two rows of dots with an even number of columns. Color the dots, each one red or blue, subject to the condition that in each row half the dots are red and half are blue. Now, if two points are adjacent in a column and like-colored, join them by an edge of their color. Show that the number of the blue segments is equal to the number of red segments.

Presently, we'll turn to a natural generalization. As it happens, some stipulations of the problem are not important. For example, it is not necessary that the number of dots in a row be even, or that the two rows have the same number of, say, red dots. On the other hand, it is clear, or becomes so after brief experimentation, that we can't do away with all the conditions of the problem. But if those two are removed, what then remains? Implicit in the formulation is the requirement that the number of the red dots in the first row (originally M) equals the number of the blue dots in the second row (also M) and the number of the blue dots in the first row (M) equals the number of the red dots in the second row (also M).

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Two Colors in Two Rows II

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Instead of one quantity M, we now have two: M₁, the number of red dots in the first row which is also the number of the blue dots in the second row, and M₂, the number of blue dots in the first row which is also the number of the red dots in the second row. The number of dots in a row N = M₁ + M₂ need not be even. Assume, as before, that

x - the number of red segments
u - the number of blue segments

The number of unmatched red dots in the first row is M₁ - x. These are not matched by the red dots in the second row, but are located in the columns occupied by the unmatched blue dots in the second row. The latter are M₁ - u in number. Therefore,

M₁ - x = M₁ - u,

or x = u, as required.

Obviously, we may start with identifying the number of unmatched blue dots in the first row, M₂ - u. This is equal to the number of unmatched red dots in the second row, M₂ - x. Since the unmatched blue dots in the first row are located directly above the unmatched red dots in the second,

M₂ - u = M₂ - x,

with the same result.

Note

The generalized problem reminds me of an old conundrum. Given two glasses of not necessarily equal capacities, one filled with water, the other with wine. Pour a spoonful of water into the wine glass, stir diligently, and then pour a spoonful of the mix back into the water glass. At this point, is there more water in wine than wine in water? What if you repeat the exercise several times?

Think in molecular terms and color water molecules red, wine molecules blue. Transfer a certain amount of red molecules into the blue container. Mix and then transfer the same amount of molecules from the blue container into the red one. Since the number of molecules in each container did not change the blue molecules in the red container came at the expense of red molecules in the blue container, so that the two amounts must be the same.

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