OT - Interesting Mind Teaser

[Gramps]

Ok, my friend sprung me on this tonight when (a) I was already semi-drunk (on the way to being fully drunk), and (b) without a pen and pad in front of me to help figure it out. He works with some Stat PhDs who decided to use this quesiton when interviewing other Stat PhDs who they were thinking of hiring.

I love these mind teaser things, kind of like an LSAT Logic Game type of thing (on steroids), so I couldn't get it off my mind (I was sitting at the SF Giants vs. Atlanta Braves game using my fingers to try and figure it out). I finally got if after a few innings and about 7 incorrect answers. If you know it already, keep quiet - no posting the answer to make you look smart if you've had prior help. If not, enjoy (if you like these sort of things)...

...You have 12 stones. 1 of the 12 weighs more or less than the other 12. You have to figure out which one it is, and whether it weighs more or less. How to figure this out? You have a balancing scale, and can only use it 3 times. How do you get the above task done?

[pineapple888]

[ QUOTE ]
Ok, my friend sprung me on this tonight when (a) I was already semi-drunk (on the way to being fully drunk), and (b) without a pen and pad in front of me to help figure it out. He works with some Stat PhDs who decided to use this quesiton when interviewing other Stat PhDs who they were thinking of hiring.

I love these mind teaser things, kind of like an LSAT Logic Game type of thing (on steroids), so I couldn't get it off my mind (I was sitting at the SF Giants vs. Atlanta Braves game using my fingers to try and figure it out). I finally got if after a few innings and about 7 incorrect answers. If you know it already, keep quiet - no posting the answer to make you look smart if you've had prior help. If not, enjoy (if you like these sort of things)...

...You have 12 stones. 1 of the 12 weighs more or less than the other 12. You have to figure out which one it is, and whether it weighs more or less. How to figure this out? You have a balancing scale, and can only use it 3 times. How do you get the above task done?

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I saw this once, and it's a huge pain in the ass, with one of the cases about three pages long. The obvious solutions don't work.

[Gramps]

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I saw this once, and it's a huge pain in the ass, with one of the cases about three pages long. The obvious solutions don't work.

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The jist is simpler than that, it might take a long time to explain every scenario, but that main key is a bit of a twist that doesn't jump out right away (though what it accomplishes logically can be explained succinctly).

[NoahSD]

This is gonna kill me.

[FlyWf]

So tired, but I'll try. Break the stones into 3 groups of 4. Weigh 2 of the groups. If they weigh the same, you know the stone is in the 3rd group.

If they weigh the same(call them 1, 2, 3, so if 1=2) put 2 stones from 3 on one side(call them 3a and 3b) and one stone from 3(3c) and one stone from 1 on the other side. If they are equal again, the 4th stone from group 3 is the different one. Weigh it against a stone from group 1.

If the side with the stone from 1 is heavier, either 3c is heavy or one of 3a or 3b is light. Weigh 3a and 3b. If they are equal, 3c is the heavy stone. If they aren't equal, the lighter of the two is the light stone.

If the side with 2 stones from 3 is heavier, either 3c is light or one of 3a or 3b is heavy. Weight 3a and 3b. If they are equal, 3c is the light stone. If they aren't equal, the heavier one is the heavy stone.

Uh, if 1|=2, I dunno. Takes more thought.

[FlyWf]

After a whopping minute or two, I have no [censored] idea what do if the different one is in 1 or 2. You only have 2 weighings left. You know 7 of them are normal, and you know where 4 normal stones are. I suspect it involves swapping in some known normal stones, but how many and to what side escapes my feeble intellect.

[Gramps]

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I suspect it involves swapping in some known normal stones

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You are wiser than you think young grasshopper...

...It's bad enough to tackle the problem drunk with time on your hands, tired and about to go to bed is worse I think...

[bones]

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I suspect it involves swapping in some known normal stones

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You are wiser than you think young grasshopper...

...It's bad enough to tackle the problem drunk with time on your hands, tired and about to go to bed is worse I think...

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At least you're not trying to tackle it with Yugo yelling in your ear after he's just beaten you HU.

[FlyWf]

Alright, I tried to go to bed and this [censored] kept me up but I think I got it.

Same notation as above, if 1<2, either one of group 1 is light or one of group 2 is heavy. Take 1d and 2d and place them off the scale. The lighter initial side will have 1a, 1b, 2a, and 2b on it. The heavier side will have 2c, 1c, 3a, and 3b on it. 1d and 2d you remove. Reweigh. If they are equal, weigh 1d against a 3. If it is equal, 2d is heavy, if not it is light.

If the heavier initial side is now lighter, one of the balls that changed position(either 2a, 2b, or 1c) is the different one. Weigh 2a vs 2b. If they are equal 1c is light. If they aren't, the heavier one is the heavy. If the heavier initial side is still heavy, either 1a, 1b, or 2c is the different one. Weigh 1a and 1b against each other as above but reversed. You can also reverse the whole thing if 1>2 initially. And yeah, my notation system was a bad idea but it would be too confusing to change now.

[Quetzo]

second: 2a 2b 1a - 2c 2d 1b

if same: 1c or 1d is light.
third 1c - 1d, light one goes up.

if left side heavy: 2a or 2b is heavy or 1b is light.
third 2a - 2b.
if equal 1b is light, otherwice the heavy of 2a and 2b goes down.