Post your best riddles.
To start: have 12 identical-looking balls. One of these balls has a different weight from all the others. You also have a two-pan balance for comparing weights. Using the balance in the smallest number of times possible, determine which ball has the unique weight, and also determine whether it is heavier or lighter than the others.
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I think you can do it in 5?
6v6, select heaviest group
3v3, if both are the same weight do 3v3 the lighter group
Now do 1v1 with each of the 3 in the anomalous group against a known correct weight ball from the other 6.
Worst case: ball is lighter, ball is last or second last tested: 6v6, 3v3, 3v3, 1v1, 1v1
That's the best I could come up with.
That's ideal; there are 2*12=24 possible scenarios, which is more than the 2^4=16 outcomes which could be determined in 4 weighings.
Just take two balls from the last three and weigh them against each other.
If the scale is uneven replace one ball with the third.
If the scale is even replace one ball with the third.
Either way its 2 measurements instead of 3 and you will know which ball and if it's heavier or lighter
You're given a lock with three rotating dials, each labeled 0 through 9, and wish to crack it in the fewest possible tries. Ordinarily, this would take up to 10*10*10=1000 attempts, but you know that one of the dials on this lock is broken. However, you aren't sure which, and there's no way to tell other than by analysis of which settings open the lock and which don't. How many different times do you have to try to crack the lock before you can be sure of opening it?
Except each weighing can have 3 outcomes:
>Left side is heavier
>Right side is heavier
>Both have equal weight
This is why 3 weighings are enough.
Split into three sets of 4: Weigh twice to isolate which set of 4 has the unique weight.
Split the set of 4 into two sets of 2: Weigh once with 2 balls from one of the other sets to isolate which set of 2 has the unique weight.
Split the set of 2 and weigh one of the balls individually with a ball from the other set. Since the weight needs to be determined as greater or less, both balls need to be weighed even though only one weigh is required to identify it.
Total weighs: 5
Shit, my bad. 3-weighing solution:
Weigh two groups of 4.
If they're equal:
Look at the other 4. Weigh 2 of the 4 against one of the 4 and one of the normal.
If they're equal:
Weigh the remaining one against a regular coin.
If (WLOG) the 2 are heavier:
Weigh one of the potentially-heavy and one of the potentially-light coins against 2 normal-weight ones.
If coins A,B,C,D are heavier than a,b,c,d:
Weigh A,B,c and a,C,D.
If equal:
Weigh d against b.
If A,B,c heavier:
Weigh A,c against 2 normal coins.
If a,C,D heavier:
Weigh c,C against 2 normal coins.
Given 32 stones of distinct weights and a pan balance, what is the least number of weighings needed to identify the heaviest and second-heaviest stones?