Logic Puzzle

[aimee]

Help! You have twelve blocks all of uniform appearance. However, one block is lighter/heaver than the others, which are of equal mass. You also have a set of balancing scales. Using only 3 turns on the scale, work out which block has the different mass and say if it is heavier or lighter. Apparently the quickest time this has been solved in is 4 hours.

[incitatus]

or four minutes. assuming the anomalous block is lighter... divide the blocks into two (6 and 6). balancethem onn the scales. (turn one) the block must be on the lighter side, so discard the six from the heavier side. take three of the blocks from the lighter side and put them in the empty pan (turn two) one side is lighter. Discard the heavy side. Of the three blocks left, put one in each pan and one in your hand. if the pans balance (turn three) then the one in your hand is the light one. If they do not then the one on the light side is.

[wildwood]

But Incitatus, how do you decide with the first weigh whether the odd block is lighter or heavier?
The best I can do is 4 moves, but will work on it a bit more.

[jenstar]

I solved this in twenty minutes but answerbank rejects the answer for being too long. You'll have to fil in the fine detail for yourself, but basically you divide them into three groups of four. Weigh group A against group B - if they balance, weigh three from group C against three from group A (which you now know to be normal). That will tell you whether the dodgy block is heavier or lighter and you then weigh two of the three against each other to establish which one (if the second weighing balanced, then you know the fourth block from group C is dodgy and need only weigh it against any other block to establish whether it's heavier or lighter.)

[jenstar]

CONTINUED... if groups A and B didn't balance, then call the blocks on the heavier side H and those on the lighter side L. On the second weighing, make up two groups of 4 as follows: 3H + 1L against 1H and three "normal" blocks from group C. If the first side is heavier, then one of the 3 "H" blocks is heavy, so weigh two against each other. If the other side is heavier, then one of the 3 "L" blocks you set aside this time round must be too light - again, weigh two against each other to establish which one of the three it is. IF THEY BALANCE on the second weighing, then either the 1H must be too heavy or the 1L must be too light. Keeping track of which is which, weigh them both - together - against any two other blocks, if the side with 1H+1L is too heavy then 1H is the culprit, if too light then 1L is the culprit. BINGO!

[jenstar]

I hope that made sense. I rather suspect that the four hours is not so much how long it takes to solve, but how long it takes to explain the answer...

[wildwood]

Jenstar, you need 3 stars just for explaining the procedure :-)

[jenstar]

Oh hang on... I made a mistake. In the second post I should have said that *if the 1H+3 normal blocks are heavier* then either the 1H or the 1L is the culprit, and *if the two sides balance* then one of the 3 "L" blocks set aside must be too light. Just remember it's that way round and not the way I originally said, and it does work, I promise you! Blimey.

[treaclefight]

jenstar... if you hadnt worked that out i would have had to, i could not have let it lie, and i really dont need to do that sort of thing right now....thankyou so much :-D