Riddle

[peter_at]

micro's procedure may fail if the scales are in balance in the first step, while it will work if they are out of balance. Why? In the latter case we have some information about the system that we are missing in the first case, namely whether the left or the right scale is heavier. Thus we should not divide the twelve sacs into two equal groups, but in groups of different size. Well, in fact the only reasonable choice is BogPoet's choice. Four sacs on one scale, four on the other one, and four are put aside. If the scales are in balance we know that the different sac is one of the four that have been put aside, and the different one can be found by two additional measurements (check it!).

However, if the scales are out of balance, the situation is now more complicated than in micro's procedure, since we only know that the different sac is one out of eight ones (instead of six, namely those on the scales). So we must try to exploit all we know about the sacs and we must exploit all the possibilities of a pair of scales! Let us summarize what all procedures so far have had in common: You divide the sacs into two groups, one group is put onto the scales and one is put aside. We look whether the scales are in equilibrium or not and infer in which of the two groups the different sac must be. But this is not efficient enough even if we can find out whether the different sac is heavier or lighter, as micro did. So reconsider my previous statement:

we may leave some sacs on their scale

What will happen with the scales, if the different sac is in this group?

we may transfer some sacs from one scale to the other one

(and vice versa): What will happen with the pair of scales if the different sac is in this group?

we may put some other sacs aside

What will happen with the pair of scales if the different sac is in this group?

We may use some sacs that we have put aside in the first step

You might need them to assure that there is the same number of sacs on both scales.

A small remark on micro's assertion (do not consider this as a hint!! it could easily mislead you!!)

(3) ... If the scale is out of balance, and the group was Y, and the scale was out of balance in (1), the fake is on the scale and you know on which side it is.

This only works if the three Fs you put on the scale in step 2 have been on the same scale in step 1, otherwise you would not know. (of course, you must remember which sacs have been on which scale in (1) and which scale was down and which was up - this assertion may be used as a hint)

Do not think binary. There are more possibilities for the scales!

[€€F]

I noticed a good hint!

[micro]

Yesterday I also tried to start with 3 groups of 4 but didn't succeed.

[Guest]

I think this works:

Take three groups of 4 (ABCD) (EFGH) and (IJKL).

Weigh one group of 4 (ABCD, left side of scale) against a second group of 4 (EFGH right side of scale).

If the scales balance, then the odd sack is in the third group of 4 (IJKL) and ABCDEFGH all weigh the same. So, take AB against IJ. If they balance, the odd sack is K or L. So weight K against A - they do not balance, K is the odd sack; if A and K balance, L is the odd sack.

But, if AB and IJ do not balance, the odd sack is I or J. Again, weigh one of these (I) against A. If they do not balance, I is the odd sack; if A and I balance, J is the odd sack.

4 down, 8 to go!

If ABCD and EFGH do not balance, then the odd sack is one of ABCDEFGH and IJKL are all normal. So mix and weigh the sacks as follows:

ABGH (left side) against CEJK (right side). If they balance, it is either sack D or sack F that is the odd sack. Weigh sack D against sack A - if it does not balance D is the odd sack, if it does balance F is the odd sack.

6 down, 6 to go

However, if ABGH and CEJK do not balance, and the unbalance stays on the same side, the odd sack is one that has not moved (i.e. A, B or E). So weigh A against B - if they balance, E is the odd sack. If they do not balance, either A or B is the odd sack (and, as we know the odd sack was on the left side originally, we can determine if it is A or B by comparing the way A v B is unbalanced with the original unbalance... )

9 down, 3 to go

Finally, if ABGH and CEJK do not balance, and the unbalance moves to the other side, the odd sack is one that has moved (C, G or H). Weigh G against H, if they balance, the odd sack is C, if they do not balance the odd sack is G or H (and we can work out which, using the same argument above).



Having just re-read that, I realise that it makes almost no sense whatsoever!

[Guest]

I quite like the following riddle:

Three people are given a challenge. They are told that there are 5 pieces of card - three black, two white. Each person then has one piece of card fixed the their back, and they are asked to enter a room together and the winner will be the first to correctly identify the colour of the card on their back, with a rational explanation of how they know. All three people have the same chance to win. They may not look at their own backs, but can look at those of others; they may not ask what colour card is on their own back.

Within a minute, one person comes out and correctly identifies the colour. What colour was their card, and what was the explanation?

[€€F]

If they do not balance, either A or B is the odd sack (and, as we know the odd sack was on the left side originally, we can determine if it is A or B by comparing the way A v B is unbalanced with the original unbalance... )

So ; you have to weigh a fourth time... or not??

Weigh G against H, if they balance, the odd sack is C, if they do not balance the odd sack is G or H (and we can work out which, using the same argument above).

also..

but you are on the right way to solve the problem..

[€€F]

I quite like the following riddle:

Three people are given a challenge. They are told that there are 5 pieces of card - three black, two white. Each person then has one piece of card fixed the their back, and they are asked to enter a room together and the winner will be the first to correctly identify the colour of the card on their back, with a rational explanation of how they know. All three people have the same chance to win. They may not look at their own backs, but can look at those of others; they may not ask what colour card is on their own back.

Within a minute, one person comes out and correctly identifies the colour. What colour was their card, and what was the explanation?

seems simple ; one guy has a black card on his back , two have a white card. the guy with the black card sees the two white cards , so he knows his card must be black.. thr guys with white cards only see one white and one black card , so are unaware of the card on their back..

[bocky]

If two of them have a black card and one has a white card, the black card guys can win too: for exemple, if guy A sees guy B with a black card and C with a white card, he knows that if B doesn't say anything, that's because A isn't white. If it was white, B would immediatly know he had a black card (as FVS explained), since there are only two white cards, and would exit the room. So A knows his card is black.

[peter_at]

If they do not balance, either A or B is the odd sack (and, as we know the odd sack was on the left side originally, we can determine if it is A or B by comparing the way A v B is unbalanced with the original unbalance... )

So ; you have to weigh a fourth time... or not??

No. As "Guest" points out sac A and B have been on the left scale in step two, so you know whether the odd sac is heavier or lighter, and thus you know whether A or B is the odd sac. But you do not need to care whether the odd sac is heavier or lighter. Let us put A on the left scale and B on the right one (we can also interchange the roles of A and B, it does not matter). If the same scale goes down in step 3 and in step 2, the odd sac is the one on the left scale (hence it is A), if the other scale goes down the odd sac is on the right scale (hence it is B). So you do not need to weigh a forth time.

That' s the solution I had found, too. In fact, you can find two similar solutions. In step 2, "Guest" has left 3 sacs on their scales, 3 have changed their scale, and 2 have been put aside. Alternatively, you can also leave 3 on their scales, interchange only 2 and put 3 aside. Or you leave only 2 on their scales, move 3 and put 3 aside. It is just important that in any of these three groups at least one sac was on the left scale and one was on the right scale in step 1. And in step 3 you must compare those two sacs that have been on the same scale in step 1.

BTW, if we have found out which one is the odd sac, we know whether it is heavier or lighter if it was on the scales in at least one of the three steps. Only if the scales have been balanced in all three steps, we cannot find out whether the odd sac is heavier or lighter. Thus we can always find the odd sac, and in 11 out of 12 cases we also find out whether it is heavier or lighter.

[peter_at]

@ Card riddle: I know this riddle, so I want to make it a bit more precise:
None of the three persons is dull (bocky has assumed that implicitly), in fact they are all equally intelligent (and they know it). Nobody finds the solution immediately, it takes them all about one minute (of course there are no mirrors in the room ).

Please put all further riddles into this thread.

[lazza]

Sorry folks - "Guest" was me in both cases.

None of the three persons is dull (bocky has assumed that implicitly), in fact they are all equally intelligent

Thanks for adding this comment - it is an important piece of info (but I always assume everyone is intelligent....

[micro]

There are three possibilities:
- black, white, white
- black, black, white
- black, black, black


Consequently, there are three possibilities for me as I enter the room:

(1) If I see two white cards I know instantly that my card is black.

(2) If I see a white and a black card and no one leaves instantly (as in (1)), I know after a moment that my card is black. (But this is also true for the other guy with the black card so we would probably collide in the doorway.)

(3) If I see two black cards, my card could be white. But then the two other guys would leave after a moment as in (2). If they don't, I know after another moment that my card is black. (But this is also true for the other two guys so all three would probably collide in the doorway.)

It all depends on what I and the other guys think how long a 'moment' is. In cases (2) and (3) the fastest guy wins, and they all know that. So all of them would try to rush and actually everything would end up in chaos and chance.

Or did I miss something?

[lazza]

Micro - a good explanation...

Another answer also explained to me, which relies more on human nature than logic, is that all three must have a black card to make the competition fair....

[androl]

To the original problem: we still can't find out if the sac is too heavy or too light.
But it is possible:

You have 12 sacs ABCDEFGHIJKL

(1) Weigh ABCD against ...oops, EFGH of course

If ABCD = EFGH, then the odd sack is in the third group of 4 (IJKL) and ABCDEFGH all weigh the same. So, take IJ against KA.
If IJ = KA, the odd sack is L. So weight L against A to find out if L is heavier or lighter.
If IJ > KA, either I or J is heavier, or K is lighter, so weigh I against J to find out.
If IJ < KA, either I or J is lighter, or K is heavier, so weigh I against J to find out.

If ABCD > EFGH, then either one of ABCD is heavier, or one of EFGH is lighter.
So weigh ABE against CDF
If ABE = CDF, weigh G against H to find out the lighter one
If ABE < CDF, either C or D is heavier, or E is lighter, so weigh C against D
If ABE > CDF, either A or B is heavier, or F is lighter, so weigh A against B

If ABCD < EFGH, then either one of ABCD is lighter, or one of EFGH is heavier.
So weigh ABE against CDF
If ABE = CDF, weigh G against H to find out the heavier one
If ABE > CDF, either C or D is lighter, or E is heavier, so weigh C against D
If ABE < CDF, either A or B is lighter, or F is heavier, so weigh A against B

I think it's also possible to do it with 13 sacs, but then you need a 14th sac of which you know that it has the correct weight.

[€€F]

If ABCD > EFGH, then either one of ABCD is heavier, or one of EFGH is lighter.
So weigh ABE against CDF
If ABE = CDF, weigh G against H to find out the lighter one

Sorry ; but you can also find out ( in an other way) if the sac is heavier or lighter..