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Today's riddle is harder than all the others so far... I figure that Thanksgiving break is a good time for it.
You 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.
Remember to put <font color="white"> and </font> around any spoilers. I thought I had the answer, but then I found a flaw in my solution, so I need to alter some things. Oh yeah, it can be done in three measurements. Enjoy!
Update: I solved it.
Small Hint/Observation: Each measurement can have THREE outcomes: balanced, left is heavier, or right is heavier. Since there are then 33 possible sets of events (three measurements with three possibilities each), that means 27 possible events. The number of possible ball and weight selections is 12 * 2 = 24 (since you need to choose one ball and whether it's lighter or heavier). This means that almost every measurement must have a different outcome based upon all three possibiliites. Not the strongest hint, but it's important to keep in mind.
Suggested Strategy: You can number the balls and use H and L (heavy and light) to list the possible outcomes each step of the way (1H, 1L, 2H, 2L, 3H, etc). Set up a tree where all 24 possibilities are at the top and after each comparison outcome, list which possibilities remain.
Slightly Bigger Hint: Your last measurement will almost always choose between THREE balls, though it'll only measure two (not quite always, though).
November 27 2005, 16:51:32 UTC 6 years ago
step 1. Weigh four unknown balls against four other unknowns with four held in reserve.- This means you have four possible heavy balls (4H), four possible light balls (4L), and the four balls you left out are known to be normal weight (4N).
- step 2. Weigh 3H + 2L in the left pan against 4N+1H in the right. Keep 2L in reserve.
- case 2a -- the left pan is heavier.
- This means that one of the 3H in the left pan is heavy and everything else is normal.
- step 3. Put 1H in each pan and hold one in reserve. If the left pan is heavier, that's the heavy ball. If the right is heavier, that's the one. If the pans are even, the ball in reserve is heavy.
- case 2b -- the right pan is heavier.
- The culprit is one of 2L in the left pan or 1H in the right; everything else is normal.
- step 3. Put one of the 2L in each pan, hold the 1H in reserve. If one pan is lighter then it has the light ball. If they're even then the reserve ball is heavy.
- case 2c -- the pans are even.
- This means that one of the 2L in reserve is light.
- step 3. Put one of the 2L in each pan. Whichever one is lighter is the light ball.
- That means you have 8 normal weight balls (8N) in the balance and 4 unknowns (4U) in reserve.
- step 2. Weigh 3U in the left pan against 3N in the right.
- case 2a -- the left pan is heavier.
- You've got 3 possible heavies (3H) and the one in reserve is normal.
- step 3. Weigh 1H versus 1H with 1H in reserve. If one pan is heavier, that's the heavy ball. If they're even, then the reserve ball is heavy.
- case 2b -- the left pan is lighter.
- You've got 3 possible lights and the one in reserve is normal.
- step 3. Weigh 1L versus 1L with 1L in reserve. If one pan is lighter, that's the light ball. If they're even, then the reserve ball is light.
- case 2c -- the two pans are even.
- That means the reserve ball is still unknown, and all other balls are normal weight.
- step 3. Weigh the unknown against any other ball.
case 1a -- the balance is uneven.
case 1b -- The (original 4:4) balance is even.
November 29 2005, 05:52:20 UTC 6 years ago
Your solution had the same number of measurements, but approached the problem differently, which was cool to read. My solution was similar enough that I'm not going to bother to type it out, unless someone really wants to see it.