The gentleman in the next office has been writhing over this one, and he's a mathematician. Can anyone help?

You have 12 blocks which look identical; 11 of them weigh the same amount, the twelfth is either heavier or lighter. You also have a balance.

The question is: what is the fewest number of weighings required to determine which is the odd block and whether it's too heavy or too light?

You can balance any group of blocks against any other group (1 vs 1, 2 vs 2, . . ., 6 vs 6) and mix them any way you like.

He's been told that you can do it in three weighings, but cannot figure out how.

 

this is a classic binary search problem.

weigh them 6 and 6. the heavier side has the heavier block, discard the rest.

weigh them 3 and 3. the heavier side has the heavier block, discard the rest.

of the remaining 3, pick two and weigh them against each other. if they are equal, the third block is heavier. if one side is heavier, then that is the heavy block.

 

Can I have a bowl of water? Then it would be easy. I have been puzzling this particular puzzle for the past 10 minutes.


The closest we can get here at work is to put 4 on each side of the scale. If the scales are even, then you know it is not any of those eight. Put the remaining 4 on and whichever way the scales tip will tell you whether or not the block is heavier or lighter, but then you still don't know which block it is because there are 4 on the scale. Ideas anyone? Am I on the right track?

 

[QUOTE]Originally posted by josh3io
[B]this is a classic binary search problem.

weigh them 6 and 6.

 

Do we get weights to measure the weight of the blocks?

 

I think it WILL work. The issue is NOT the odd block being heavier or lighter, it's only which is the ODD block. The other criteria is which is the LEAST number of weighings, you must assume that you're running some luck here and the best case scenario is that you always pick the side that in fact has the "odd" block. Now if the question was to say what is the least number of weighings to "guarantee" you find the "odd" block AND to determine if this block is heavier or lighter, then we have a different problem. Josh3io's rationale is sound. His wording is too specific.

 

Nope! You have everything you need: 12 blocks and a balance. (OK, maybe a six-pack of Pepsi (or whatever) would be useful, too.)

 

[QUOTE]Originally posted by xviper
[B]
I think it WILL work.

 

[QUOTE]Originally posted by xviper
[B]
I think it WILL work.

 

[QUOTE]Originally posted by skitz
[B]

Ok you put them on the scales.