[Guest]

You are presented with a set of 6 suspect coins. Of these, it is known that the number of counterfeit coins is either 0, 1 or 2. Each of the counterfeit coins, if any, is known to have come from one of two batches of coins, one of which is too light and the other of which is too heavy. It is also known that a light coin and a heavy coin, taken together, weigh exactly the same as two normal coins. You must identify the counterfeit coins, if any, after 4 weighings or less.

[Guest]

Weigh 123 vs 456

if equal then there are none or both in same group

Weigh 1 vs 2, if equal then none in 123

Weigh 4 vs 5, if equal then there are no fakes

If not equal then all fakes are in 456. call 4 the heavy one of 45,

weigh 45 vs 16

If equal then 4 heavy, 5 light

if 45 heavy, 4 is heavy and 6 light

if 45 light, 5 light and 6 heavy

Going back to 2nd weigh, if not equal, call 1 the heavy and solve as 45 above, changing 123 and 456 around

Going back to first weigh, if not equal, call 123 the heavy side

We know there are 1 or 2 fakes, and 2 can not be in same group.

Weigh 12 vs 45

if equal, then fakes are 3 or 6 or both 3 and 6

weigh 36 vs 45. if equal then fakes are 3 and 6

if 36 heavy, 3 is heavy, if 36 light then 6 light.

If 12 vs 45 was not equal, 12 has to be heavy side and there are

1 or 2 fakes not in same group. either 1,2,4,5,14,15,16,24,25,26,34,35

and we have used 2 weighs and have 2 left

Weigh 146 vs 235

case equal, possibles are 14,16,25

weigh 6 vs 5. if equal, fakes are 14. If heavy, fakes are 25. If light, fakes are 16.

case 146 is heavy, possibles are 1 or 5 or both 1 and 5

weigh 15 vs 24. if equal, fakes are 1 and 5. if heavy, 1. if light, 5

case 146 is light, possibles are 2 or 4 or both 2 and 4

weigh 24 vs 15. equal then 2 and 4 fake, if heavy then 2, if light, 4

[k-man]

Weigh 123 vs 456

if equal then there are none or both in same group

Weigh 1 vs 2, if equal then none in 123

Weigh 4 vs 5, if equal then there are no fakes

If not equal then all fakes are in 456. call 4 the heavy one of 45,

weigh 45 vs 16

If equal then 4 heavy, 5 light

if 45 heavy, 4 is heavy and 6 light

if 45 light, 5 light and 6 heavy

Going back to 2nd weigh, if not equal, call 1 the heavy and solve as 45 above, changing 123 and 456 around

Going back to first weigh, if not equal, call 123 the heavy side

We know there are 1 or 2 fakes, and 2 can not be in same group.

Weigh 12 vs 45

if equal, then fakes are 3 or 6 or both 3 and 6

weigh 36 vs 45. if equal then fakes are 3 and 6

if 36 heavy, 3 is heavy, if 36 light then 6 light.

If 12 vs 45 was not equal, 12 has to be heavy side and there are

1 or 2 fakes not in same group. either 1,2,4,5,14,15,16,24,25,26,34,35

and we have used 2 weighs and have 2 left

Weigh 146 vs 235

case equal, possibles are 14,16,25

weigh 6 vs 5. if equal, fakes are 14. If heavy, fakes are 25. If light, fakes are 16.

case 146 is heavy, possibles are 1 or 5 or both 1 and 5

weigh 15 vs 24. if equal, fakes are 1 and 5. if heavy, 1. if light, 5

case 146 is light, possibles are 2 or 4 or both 2 and 4

weigh 24 vs 15. equal then 2 and 4 fake, if heavy then 2, if light, 4

Sorry, but you're not quite there yet.

You didn't detect 3 cases of 2 fakes: [2,6],[3,4] and [3,5]. Generally speaking, when you have 2 weighings left and 12 possibilities, you should rethink your approach. 2 weighings will produce 9 distinct results, so you cannot distinguish between 12 different possible scenarios.

As a side note, I think you simplified the problem by assuming that if there are 2 fakes then one is heavy and one is light. I don't interpret the problem this way. In my interpretation both fakes can be heavy or both can be light. In this interpretation we are faced with 73 possible scenarios. 4 weighings are sufficient to not only identify the fakes, but also to state for each fake whether it's heavy or light.

In your interpretation you only have 43 scenarios, which still requires 4 weighings to determine the fakes and their relative weight, althogh it's MUCH easier. However, if we only need to identify the fakes without knowing whether they are light or heavy we MIGHT be able to do it in only 3 weighings.

[Guest]

Oh right, I accidentally ruled those 3 out.

[Guest]

Oh right, I accidentally ruled those 3 out.

I think the previous solution is close, but not quite correct but I believe I know why that is.

A Heavy and Light coin on the same side of the balance equals 2 normal coins so there's really more possible outcomes than it seems. (this is probably what threw the last solution off)

Said that it could be 0 1 or 2 counterfeit coins as well.

So just to try and make it a bit more clear what I'm saying assume that any coin that I mark as X is normal weight, H is heavy, and L is light.

XXX vs XHL would balance, even though a light and heavy are on same side.

XXH vs XXH would balance, since one H is on each side

XXL vs XXL would balance, since one L is on each side

XXX vs XXX represents the possibility of 0 fakes

XXL vs XXX represents the possibility of 1 fake light coin

XXH vs XXX represents the possibility of 1 fake heavy coin

So pretty much, gotta compare em in a way that you know which coins are fake, whether they are both light, both heavy, one of each Light and Heavy, only a single fake being Light or Heavy, or all the coins being normal weight.

[Guest]

Weigh coins 123

Weigh coins 234

Weigh coins 345

Weigh coins 456

By knowing if the scale goes up or down between the weighings, by the end you can tell if there was one or two fake coins. If all 4 scale readings are the same, no coins are fake. I would look for the discrepancy between weighings to find the fakes. For example, if 123 > 234, then either 1 is heavy and 4 is light, just 1 is heavy, or 2 is heavy and 3 is light (or vise versa). By the end of the weighings, you would be able to figure this out because you would know how much to expect the reading to go up or down for one fake coin. I'd like to see someone take this further!

[Guest]

Weigh coins 123

Weigh coins 234

Weigh coins 345

Weigh coins 456

By knowing if the scale goes up or down between the weighings, by the end you can tell if there was one or two fake coins. If all 4 scale readings are the same, no coins are fake. I would look for the discrepancy between weighings to find the fakes. For example, if 123 > 234, then either 1 is heavy and 4 is light, just 1 is heavy, or 2 is heavy and 3 is light (or vise versa). By the end of the weighings, you would be able to figure this out because you would know how much to expect the reading to go up or down for one fake coin. I'd like to see someone take this further!

Can't weigh 123 against 234, can't have the same coin on both sides of the scales at the same time. Other than that you couldn't narrow this down with certainty exactly which coin(s) if any are fake.

[Guest]

Can't weigh 123 against 234, can't have the same coin on both sides of the scales at the same time. Other than that you couldn't narrow this down with certainty exactly which coin(s) if any are fake.

I was assuming that the scale was digital. As in you would weigh 123 at time A and you would weigh 234 at time B. Did you intend for this riddle to only be solved using an old school scale in which you are always weighing one mass against a different mass?

[plainglazed]

Some clarification, please. I read this line: "Each of the counterfeit coins, if any, is known to have come from one of two batches of coins, one of which is too light and the other of which is too heavy." to mean that if two coins were counterfeit, one would be heavy and one would be light. Is that correct or can you have two heavy coins or two light coins?

[Guest]

Some clarification, please. I read this line: "Each of the counterfeit coins, if any, is known to have come from one of two batches of coins, one of which is too light and the other of which is too heavy." to mean that if two coins were counterfeit, one would be heavy and one would be light. Is that correct or can you have two heavy coins or two light coins?

I'm working on this same riddle as njblob and from my trials and errors I've seen solutions where its listed as being 2light fakes, 2heavy fakes, 1light and 1heavy fakes. So yes there can be two heavy, yes there can be two light. A light and heavy could be on the same side of the scale and equal the weight of 2 normal coins so as appearing that they're all normal as well.

[Guest]

I think the previous solution is close, but not quite correct but I believe I know why that is.

A Heavy and Light coin on the same side of the balance equals 2 normal coins so there's really more possible outcomes than it seems. (this is probably what threw the last solution off)

Said that it could be 0 1 or 2 counterfeit coins as well.

So just to try and make it a bit more clear what I'm saying assume that any coin that I mark as X is normal weight, H is heavy, and L is light.

XXX vs XHL would balance, even though a light and heavy are on same side.

XXH vs XXH would balance, since one H is on each side

XXL vs XXL would balance, since one L is on each side

XXX vs XXX represents the possibility of 0 fakes

XXL vs XXX represents the possibility of 1 fake light coin

XXH vs XXX represents the possibility of 1 fake heavy coin

So pretty much, gotta compare em in a way that you know which coins are fake, whether they are both light, both heavy, one of each Light and Heavy, only a single fake being Light or Heavy, or all the coins being normal weight.

missing

XXH vs XXL

[Guest]

Just didn't add XXH vs XXL cause if its the first weighing and you have 3 coins vs 3 coins it always balances out, so only way that'd happen and still have a H and L is if it was XXXvsXHL since a H and L equal two normal coins.

[Guest]

Has anyone been able to solve it completely yet? I know its tough but its for sure solvable with 4 weighings.

[bonanova]

Some clarification, please. I read this line: "Each of the counterfeit coins, if any, is known to have come from one of two batches of coins, one of which is too light and the other of which is too heavy." to mean that if two coins were counterfeit, one would be heavy and one would be light. Is that correct or can you have two heavy coins or two light coins?

I think the OP allows for the set of counterfeits to be {null} {H} {L} {HH} {HL} or {LL}.

It doesn't say only one coin can come from each of the batches, nor, if there are two, that both must come from one of the two batches.

[Guest]

Lets call the 6 coins [ 1 2 3 4 5 6 ]

We will seperate out coins 5 and 6

Weighing only (1 2) against (3 4) [ w1]

Irregardless of the result, we will proceed with the 2nd weighing (1 3) against (2 4) [ w2]

There will be a total of 3 possible results, W1 balanced, W2 balanced

[Guest]

Lets call the 6 coins [ 1 2 3 4 5 6 ]

We will seperate out coins 5 and 6

Weighing only (1 2) against (3 4) [ w1]

Irregardless of the result, we will proceed with the 2nd weighing (1 3) against (2 4) [ w2]

There will be a total of 3 possible results, W1 balanced, W2 balanced

W1 balanced, W2 not balanced

W1 not balanced, W2 balanced

W1 not balanced, W2 not balanced

In the event that W1 and W2 are both balanced, we may hence conclude that the counterfeits are likely to be 5 and 6. In which we shall exhaust the remaining 2 trys on balancing coin5 and 6 with a standard coin.

W1 not balanced, W2 no balanced : For simplistic sake, we will assume the following sceniro

W1 : (1 2) against (3 4) with ( coin 1 2) Heavier

W2 : (1 3) against (2 4) with ( coin 1 3) Heavier

* Just call the coins whatever names, john jack jill or james. play with the names and u will find that this is the only possible sceniro, any other sceniro are the result of naming convention.

We can now conclude that coin 3 and coin 2 are standard coins.

W3 : ( 1 4 6) against ( 2 3 5)

If balanced: We will know that coin 5 and 6 are also standard coins. We can then proceed in finding out which coin is heavy and which coint is light by weighing 1 against 4.

If unbalanced: ( 1 4 6 ) Heavier

We know that the max number of counterfeit are 2, hence coin 4 can be taken to be standard, Coin 1 to be counterfeit. Coin 6 to be either standard or heavy fake, coin 5 to be either standard or light.

We can then proceed to weight coin 5 6 against 2 standard coins.

If balanced, coin 5 6 are standard,

If unbalanced, depending on which side is heavier, we can infer which coin is the fake.

[Guest]

The issue with W1 B , W2 NB is the same as W1 NB , W2 B.

With B = balanced and NB not balanced

For simplistic sake, we assume this is the following sceniro

W1 : coin 1 2 against 3 4 Balanced

W2 : coin 1 3 against 2 4 Not balanced

In this situation, we are able to affirm that coin 5 and 6 are standard coins.

W3 : ( 1 2 3) against ( 4 5 6)

If balanced: We will know that coin 3 and 4 are also standard coins. We can then proceed in finding out which coin is heavy and which coint is light by weighing 1 against 2.

If unbalanced: ( 1 2 3 ) Heavier

We know that the max number of counterfeit are 2, hence coin 1 and 2 can be taken to be standard, and coin 3 4 to be non standard, Depending on which side is heavier, that coin is the heavy fake. same logic applys for light fake.

for some reason i am unable to edit my post, hence the 3 post

[k-man]

Sorry, swongy, but your solution is not complete. I didn't go through the whole thing since I found a bad assumption right in the beginning:

In the event that W1 and W2 are both balanced, we may hence conclude that the counterfeits are likely to be 5 and 6. In which we shall exhaust the remaining 2 trys on balancing coin5 and 6 with a standard coin.

If both 2 and 3 are heavy (or light), or both 1 and 4 are heavy (or light) you will not discover them using your method.

Since no one has solved this yet I will work on the solution and post it soon.

[plainglazed]

if the the set of counterfeit coins is either one heavy, one light, one heavy and one light, two heavy, or two light; am thinking there is no way to determine the counterfeit(s) in four weighings. thinking there are only 73 total possibilities but with the added criterion that one heavy and one light = two normal coins; one can not reduce the number of remaining possibilities to 27 or less after one weighing where that weighing shows that the two sides of the scale are equal. there are three possibilities for the first weighing: compare one coin on each pan, compare two coins on each pan, or compare three coins on each pan. with three coins on each pan showing equal there are 49 remaining possibilities by my count. with one coin on each pan there are 37 remaining possibilities. and with two coins on each pan there are 33 remaining possibilities. with three remaining weighings you can only differentiate 27 possibilities.

[k-man]

if the the set of counterfeit coins is either one heavy, one light, one heavy and one light, two heavy, or two light; am thinking there is no way to determine the counterfeit(s) in four weighings.  thinking there are only 73 total possibilities but with the added criterion that one heavy and one light = two normal coins; one can not reduce the number of remaining possibilities to 27 or less after one weighing where that weighing shows that the two sides of the scale are equal.  there are three possibilities for the first weighing: compare one coin on each pan, compare two coins on each pan, or compare three coins on each pan.  with three coins on each pan showing equal there are 49 remaining possibilities by my count.  with one coin on each pan there are 37 remaining possibilities.  and with two coins on each pan there are 33 remaining possibilities.  with three remaining weighings you can only differentiate 27 possibilities.

I think you miscalculated. I'm almost done with the solution (it's actually taking me longer to type than to figure it out) Starting out with 2 coins on each side of the scale breaks the possibilities down nicely - 21:26:26. I have to stop for today, so I'm posting what I've got so far and I will try to finish it tomorrow.

Number the coins 1 through 6.

There are 73 possible arrangements of coins given the conditions of the puzzle. I split them into groups by category and for further reference will use below codes. The number in parenthesis means the number of arrangements in the category:

NO. No fakes (1)

1H. One heavy fake (6)

1L. One light fake (6)

LH. One light fake preceding one heavy fake (e.g. #2 is light and #4 is heavy) (15)

HL. One heavy fake preceding one light fake (e.g. #2 is heavy and #4 is light) (15)

HH. Two heavy fakes (15)

LL. Two light fakes (15)

For the first weighing let's compare coins 1,2 against 3,4.

If the scale is balanced then we are down to 21 possibilities:

NO. (1)

1H. #5 or #6 (2)

1L. #5 or #6 (2)

LH. [1,2],[3,4] or [5,6] (3)

HL. [1,2],[3,4] or [5,6] (3)

HH. [1,3],[1,4],[2,3],[2,4],[5,6] (5)

LL. [1,3],[1,4],[2,3],[2,4],[5,6] (5)

For the second weighing we weigh odd numbers against even - 1,3,5 vs. 2,4,6

If the scale is balanced we have 7 possibilities:

NO. (1)

1H. (0)

1L. (0)

LH. (0)

HL. (0)

HH. [1,4],[2,3],[5,6] (3)

LL. [1,4],[2,3],[5,6] (3)

So, for the third and fourth weighings we should compare #1 against #2 and #1 against #5.

If both weighings are balanced then there are no fakes. ----END----

It's easy to see how other outcomes deteremine which of the 3 possible pairs is fake and whether it's heavy or light, so I won't spell it out. ----END----

Now, if the second weighing was heavy on the left side (odd coins are heavier) then we have these 7 possibilities:

NO. (0)

1H. #5 (1)

1L. #6 (1)

LH. (0)

HL. [1,2],[3,4] or [5,6] (3)

HH. [1,3] (1)

LL. [2,4] (1)

So, for the third weighing we will compare 1,6 vs. 2,5. If the scale is balanced then we solved it after 3 weighings - #3 is heavy and #4 is light. ----END----

If 1,6 is heavier than 2,5 then 3 possilibities remain:

HL. [1,2]

HH. [1,3]

LL. [2,4]

We know that 5 and 6 are good coins, so compare 1,2 vs. 5,6. If balanced, then #1 is heavy and #2 is light. If left is heavy then [1,3] is heavy pair. Otherwise, [2,4] is a light pair. ----END----

If in the third weighing 1,6 is lighter than 2,5, then 3 possibilities remain. Either #5 is heavy, #6 is light or both.

1H. #5

1L. #6

HL. [5,6]

4th weighing is the same as before - compare 1,2 vs. 5,6. If balanced the #5 is heavy and #6 is light. If left is heavy then #6 is light. If right is heavy then #5 is heavy. ----END----

Going back to the second weighing. If the second weighing was heavy on the right side (even coins are heavier) then the possibilities are:

NO. (0)

1H. #6 (1)

1L. #5 (1)

LH. [1,2],[3,4] or [5,6] (3)

HL. (0)

HH. [2,4] (1)

LL. [1,3] (1)

As you can see, it's symmetrical to the previous case, so the same exact weighings allows us to figure it out, but the conclusions are the opposite from the above. ----END----

If the scale is not balanced then let's say 1,2 is heavier than 3,4. We are down to 26 possibilities:

NO. (0) - there must be fakes here

1H. #1 or #2 (2)

1L. #3 or #4 (2)

LH. Either #3 or #4 is light and either #5 or #6 is heavy (4)

HL. Either #1 or #2 is heavy and one of the remaining coins #3 through #6 is light (8)

HH. [1,2],[1,5],[1,6],[2,5],[2,6] (5)

LL. [3,4],[3,5],[3,6],[4,5],[4,6] (5)

Compare 3,5 vs. 4,6.

If balanced, then there are 8 possibilities:

NO. (0)

1H. #1 or #2 (2)

1L. (0)

LH. [3,5] or [4,6] (2)

HL. (0)

HH. [1,2] (1)

LL. [3,4],[3,6] or [4,5] (3)

In this case, the third weighing is 1,3,6 vs. 2,4,5. If balanced, then either [1,2] pair is heavy or [3,4] pair is light. Weigh #1 vs. #5 to resolve. ----END----

If 1,3,6 is heavier than 2,4,5, then there are 3 possibilities:

1H. #1

LH. [4,6]

LL. [4,5]

Compare 1,4 vs. 5,6. If balanced, then [4,5] is a light pair. If left is heavy, then #1 is heavy. Otherwise, #4 is light and #6 is heavy. ----END----

If 1,3,6 is lighter than 2,4,5, then 3 possibilities remain:

1H. #2

LH. [3,5]

LL. [3,6]

Similarly to above, compare 2,3 vs. 5,6. ----END----

If the second weighing was not in balance and the left side (3,5) was heavier than the right side (4,6) we have 9 possibilities:

NO. (0)

1H. (0)

1L. #4 (1)

LH. [4,5] (1)

HL. [1,4],[1,6],[2,4],[2,6] (4)

HH. [1,5] or [2,5] (2)

LL. [4,6] (1)

... to be continued

Edited July 28, 2011 by k-man

[plainglazed]

oops, right on k-man. * ... 2 + 6 carry the one ... *

[Guest]

Ok so I only have this half figured out how to work these puzzles. But I worked with the first 2 weighings provided earlier by I believe K-Man and from there I was able to find a solution. The way I went about this was to work each possible combination, which was kind of time consuming and rough but it led to success. So first I didn't know if the next weighing needed to be 3coinsv3coins, so I went through those combinations.

So weighings would look like this,

w1 1 2 v 3 4

left

w2 3 5 v 4 6

left

coins by weight HHHNLL

coins by number 125346

3x3 combos

HHHvLLN 125v463fail ***************

HHNvLLH 123v465fail 153v462fail 523v461fail***********

HLHvHLN 142v563fail, 145v263fail, 542v163failed

HLHvHLN 162v543fail 165v243fail 562v143fail*************

So basically once I'd look at a specific weighing, I would then compare it to the possible solutions provided earlier in the thread. I'd run each scenario of the scale tipping left, staying balanced in middle, or going right. If a combination left you with more than 3 possible solutions then it wasn't able to be solved, no matter if they were revealed balanced, right or left scenario.

I moved to 2x2 combinations, and repeated same process and it finally revealed a correct solution.

HHvHN 12v53F, 15v23F, 25v13F

HHvHL 12v54F 15v24F, 25v14F

2HHvHL 12v56F 15v26F 25v16F

HHvLL 12v46F, 15v46F 25v46F

HHvLN 12v43F, 15v43F, 25v43F

2HHvLN 12v63F 15v63F 25v63F

LLvHN 46v13F 46v23F 46v53F

LHvLN ***SUCCESS41v63SUCCESS*** 42v63 45v63

2LHvLN 61v43 62v43 65v43

w3 41v63

balances middle

HL. [1,4],

LL. [4,6]

HH 2,5

possH 1 2 5, PossL 4 6

1v5

r

HL. [1,4],

LL. [4,6]

HH 2,5

2,5 heavy is answer