Weighing 6 Balls
Here's a nice variation on the classic weighing puzzle genre. You have a balance and six balls: two red, two yellow and two blue. In each colour there is a lighter ball and a heavier ball, with all the lighter balls weighing the same and all the heavier balls weighing the same.
By loading up the balance and using it twice, how can you determine the identity of each ball?
To guarantee a solution which will always work within two weighings we need to get a big case based. Let's name the balls r1, r2, y1, y2, b1 and b2.
Weighing 1: r1 + y1 against r2 + b1
If they are equal then y1 can't have the same weight as b1. We can go further with this. y1, r2 and b2 must all have the same weight and y2, r1 and b1 must also be the same as each other. The second weighing of any ball from the first list against any ball from the second list will determine which set is heavier.
Otherwise, if r1 + y1 is heavier then r1 is definitely the heavier of the red balls, but we don't necessarily know that y1 is the heavier yellow ball. It could be that both y1 and b1 where the lighter balls of their colours.
Weighing 2: Using the same four balls but switching the sides a bit let's measure r1 + r2 against y1 + b1.
If y1 + b1 is heavier then they must both be the heavier balls of their colours.
If both sides are equal then exactly one of y1 and b1 is light, but we know from the first weighing that it must be b1.
If y1 + b1 is lighter then they are both the lighter balls of their colours.
Finally, the last case is that in the first weighing r1 + y1 is lighter which follows the mirror image of the above working.