Riddles!

[TheGreatErenan]

I think this is one among several solutions:


Label the balls B1, R1, W1, and B2, R2, and W2.

Weigh B1R1 vs B2W1.

If the two sides are equal:

We know that B1 != B2, so R1 != W1. Furthermore, a light and a heavy must be on each side, so B1 = W1 and B2 = R1.

So if we weigh B1 vs B2, we instantly learn the weight relationship between B1 and B2. Since B1 = W1, then B2 = W2. Since B2 = R1, then B1 = R2. Thus, we know all balls' weight relationships. In other words, if B1 < B2, then W1 < W2 and R1 > R2. Or if B1 > B2, then W1 > W2 and R1 < R2.



If B1R1 < B2W1:

This could not happen if B1 > B2. So we know B1 < B2.

Observation 1: Either R1 = W1 or R1 < W1, since R1 > W1 and B1 < B2 implies B1R1 = B2W1.

Now, weigh R2 vs W1.

If R2 < W1, then R1 > R2 and W1 > W2. Therefore B1 R2 W2 are light, and B2 R1 W1 are heavy.

If R2 > W1, then R1 < R2 and W1 < W2. Therefore B1 R1 W1 are light, and B2 R2 W2 are heavy.

If R2 = W1, then R1 = W1 is false, and from observation 1, we know that R1 < W1. Therefore, B1 R1 W2 are light, and B2 R2 W1 are heavy.



If B1R1 > B2W1:

This could not happen if B1 < B2. So we know B1 > B2.

Observation 2: Either R1 = W1 or R1 > W1, since R1 < W1 and B1 > B2 implies B1R1 = B2W1.

Now, weigh R2 vs W1.

If R2 < W1, then R1 > R2 and W1 > W2. Therefore B2 R2 W2 are light, and B1 R1 W1 are heavy.

If R2 > W1, then R1 < R2 and W1 < W2. Therefore B2 R1 W1 are light, and B1 R2 W2 are heavy.

If R2 = W1, then R1 = W1 is false, and from observation 2, we know that R1 > W1. Therefore, B2 R2 W1 are light, and B1 R1 W2 are heavy.