3
$\begingroup$

This question already has an answer here:

A brain teaser posed by the character Captain Ray Holt of 'Brooklyn Nine-Nine' played by Andre Braugher during episode 18 of season 2:

"There are twelve(12) men on an island, eleven(11) weigh exactly the same amount, but one of them is slightly lighter or heavier, [the object is to] figure out [whether he is lighter or heavier]. A standard see-saw must be used and only three times."

(Rephrased for clarity, below is a verbatim transcript of 'Holts' dialogue)

"There are twelve men on an island, eleven weigh exactly the same amount, but one of them is slightly lighter or heavier, you must figure out which. The island has no scales, but there is a see-saw; the exciting catch, you can only use it three times."

$\endgroup$

marked as duplicate by Ross Millikan, Mark N, Tryth, Len, leoll2 May 24 '15 at 7:58

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ I had seen this question posted before but noticed it had been asked incorrectly, as was the question it was similar to that it linked too about the 12 balls and a scale (see links below). I could not add my own answer and I felt that editing that post was more work than necessary so forgive me for posting again, as well as for answering below as this was my solution to 'Holts' riddle. Thankyou for reading and understanding. ((puzzling.stackexchange.com/questions/9979/… )) ((puzzling.stackexchange.com/questions/183/… )) $\endgroup$ – Rocco Ruscitti May 24 '15 at 4:41
  • $\begingroup$ Please explain your claim that is was not asked properly. I believe it was in puzzling.stackexchange.com/questions/183/… $\endgroup$ – Ross Millikan May 24 '15 at 5:02
  • $\begingroup$ @RoccoRuscitti - Here is a video of Holt's solution. This should help to clarify the intent of his question as well as to explain his answer. $\endgroup$ – Len May 24 '15 at 6:06
6
$\begingroup$

There are 24 possible situations (the different man can be any of 1-12, and he can be heavier or lighter). Thus we need to log224 bits of information to solve the puzzle. You can weigh three combinations of men on the see-saw. Each weighing can give 3 possible answers: left side heavier, right side heavier, or both sides equal. Thus in principle we can get log227 bits from the three comparisons. So in principle, we should be able to solve the problem. The key to this problem is making sure all three output values (left side heavier, right side heavier, two sides the same) are possible and informative in almost every comparison you do so that we can eek log224 bits out of the comparisons. Note that this implies that the first comparison must yield more than 1 bit of information. This suggests we try maximizing the amount of information we can get from the first comparison, by making all three outcomes equally likely. Comparing (1,2,3,4) to (5,6,7,8) does exactly this. Similar logic will help us design all further comparisons.

Here is one solution:

Number the men 1,2,3...12. First weigh 1,2,3,4 against 5,6,7,8. One of two things will happen:

1) They are equal. Now we know that the different man is among {9,10,11,12}. Weigh 9,10,11 against 1,2,3. If these are equal, the different man is 12. Weigh 12 against 1 to find out whether 12 is heaver or lighter. If the 9,10,11 differs from 1,2,3, then weigh 9 against 10. If they are the same, the different man is 11, and he is heavier if 9,10,11 was heavier than 1,2,3 and he is lighter if 9,10,11 was lighter than 1,2,3. If 9 and 10 are different, the different man is the lighter of the 9,10 comparison if 9,10,11 was lighter than 1,2,3, (and he is lighter); the different man is the heavier of the 9,10 comparison if 9,10,11 was heavier than 1,2,3 (and he is heavier).

2) They are different. Without loss of generality suppose that 1,2,3,4 is heavier than 5,6,7,8. (We could always relabel the men so that this is true). We know {9,10,11,12} all weigh the same.

Weigh 1,2,5,6,7 against 8,9,10,11,12:

a) If 1,2,5,6,7 is heavier, then either 1 or 2 heavier, or 8 is lighter. Weigh 1 against 2. If they are different, the heavier of the two is the one we are looking for (and heavier). If they are the same, 8 is the one we are looking for (and lighter).

b) If 1,2,5,6,7 is lighter, then one of 5,6,7 is different and lighter. Weigh 5 against 6. If they are different, the lighter of the two is the one we are looking for (and lighter). If they are the same, 7 is different (and lighter).

c) If they are the same, then one of 3,4 is different. Weigh them against each other. The one who is heavier is the different man (and heavier).

$\endgroup$
  • $\begingroup$ I concede that my previous hypothesis about the validity of the question was false. @Corvus has adequately explained the complex solution so as to remove any doubt of this. $\endgroup$ – Rocco Ruscitti May 24 '15 at 7:25
0
$\begingroup$

The solution:

Split the men into two(2) groups 'abcdef' and '123456'.

Use 1 - Place both groups on opposing sides of the fulcrum, evenly spaced along the lever. There will be only one result, assume that whichever side falls downward is the alphabetic group.

Use 2 - Remove six(6) men from the see-saw, three(3) from both groups. Let's say 'abc' and '456'. There are two possible results. A_ the equilibrium of the see-saw remains unchanged, therefore the man of a different weight is now amongst group 'def123' or B_ the see-saw becomes level with the ground, therefore the man of a different weight is standing with group 'abc456'. Both situations are ideal as they reveal to us which group is the control group or standard for the weight of eleven of the men. Which brings us to...

Use 3 - Place both new groups 'def123' and 'abc456' on the teeter-totter again as we did in the beginning. Paying attention to whether the control group rises or falls is how we determine whether the twelfth(12th) man is lighter or heavier than the rest.

$\endgroup$
  • 1
    $\begingroup$ One problem - you have to figure out which person it is too. $\endgroup$ – Deusovi May 24 '15 at 4:51
  • $\begingroup$ Thankyou for your input but I believe you are wrong because it is my understanding of Holts dialogue that leads me to conclude it is a simple riddle with a simple solution. $\endgroup$ – Rocco Ruscitti May 24 '15 at 5:01
  • $\begingroup$ Agree with Rocco here but only because it is this interpretation of the riddle that is described in the OP. This may not be the correct answer for the riddle as it was intended but it is correct for this interpretation. $\endgroup$ – Lord Jebus VII Jul 5 '18 at 15:53

Not the answer you're looking for? Browse other questions tagged or ask your own question.