The odd ball puzzle

[Alan McDougall]

This puzzle is only for those who have not come across it before. You can get the solution from the internet but this would defeat the purpose of this post which having a little fun!!

 

The Problem 12 Odd Pool Balls Puzzle

The Problem:

You have 12 balls identical in size and appearance but 1 is an odd weight (could be either light or heavy).

 

You have a set scales (balance) which will give 3 possible readings: Left = Right, Left > Right or Left < Right (i.e. Left and Right have equal weight, Left is Heavier, or Left is Lighter).

 

You have only 3 chances to weigh the balls in any combination using the scales. Determine which ball is the odd one and if it's heavier or lighter than the rest. How do you do it?

[uncool]

This puzzle is only for those who have not come across it before. You can get the solution from the internet but this would defeat the purpose of this post which having a little fun!!

 

The Problem 12 Odd Pool Balls Puzzle

The Problem:

You have 12 balls identical in size and appearance but 1 is an odd weight (could be either light or heavy).

 

You have a set scales (balance) which will give 3 possible readings: Left = Right, Left > Right or Left < Right (i.e. Left and Right have equal weight, Left is Heavier, or Left is Lighter).

 

You have only 3 chances to weigh the balls in any combination using the scales. Determine which ball is the odd one and if it's heavier or lighter than the rest. How do you do it?

Reasoning backwards from the fact that there's an answer:

 

 

First, we note that each weighing has 3 possible outcomes, and therefore that there are a total of 3^3 possible outcomes of 3 weighings. There are 24 possible original setups, so we can only afford to "waste" 3 of the weighing outcomes.

 

The first weighing is of two sets of the same number of balls; call it n. If the weighing is Left = Right, then the odd ball is in the ones that are left, giving 2*(12 - 2n) possibilities. This must be bounded above by 9, so n is at least 4. The other possibilities are evenly split between Left < Right and Left > Right, meaning that there are 2n possibilities for each, so n is bounded above by 4, so the first step must be to weigh 4 balls against each other.

 

We first consider the "odd man out": the case where Left = Right. In this case, we have 8 balls that are certainly normal, and 4 balls, one of which is odd in one of the two directions. The second weighing can't involve all 4, as in that case the weighing will certainly not be Left = Right, so at least one must be left out. If we leave two out, and we get a weighing of Left = Right on what's left, then there are 4 situations, which cannot be decided by a single weighing, so there is at most one left out. The easiest way to do this is to weigh 3 of the 4 against 3 of the 8; say the 3 of the 4 are on the left. If we get Left = Right, then we know that the odd ball is the last one, so we can weigh the last ball against any of the normal balls. If we get Left < Right, then we know that the odd ball is lighter. We can then weigh 2 of the 3 against each other; if we get Left < Right, then the left ball is the odd one out and is lighter. If we get Left > Right, then the right ball is the odd one out and is lighter. If we get Left = Right, then the third ball is the odd one out and is lighter.

 

Note: We could also have weight 2 of the 4 and 1 of the 8 against 1 of the 4 and 2 of the 8.

 

We now turn to the case where we get Left < Right. The possibilities are that one of the 4 on the left is lighter, or that one of the 4 on the right is heavier. If in the second weighing, we get Left = Right, the odd ball out will be one of the balls not used in the second weighing, so we must use at least 5 balls in the second weighing and at most 6; this clearly includes at least one ball from each side. We also only know of 4 normal balls, so we know that there is at least one of the 8 on each side of the second weighing. This implies that no matter what, at least one pair of balls on opposite sides in the first weighing will be on opposite sides in the second weighing. We can further see by inspection that there will be at least one triplet of balls such that two were on the same side in the first weighing and are on the same side again in the second weighing, and the third was on the opposite side of the other two in the first weighing and is again on the opposite side. We can then see that no more of the same kind can be added to either side. This gives us the obvious solution:

 

If the balls on the left in the first weighing are labeled 1-4 and the balls on the right 5-8, weigh 1, 2, 5, and 6 against 3, 7, 9, and 10. If Left < Right for the first weighing and Left < Right for the second weighing, weigh 1 against 2 to determine which of 1, 2, and 7 it is. If Left < Right and Left > Right, weigh 5 against 6 to determine which of 3, 5, and 6 it is. If Left < Right and Left = Right, weigh 4 against 9 to determine which of 4 and 8 it is. A similar answer works for Left > Right, so we are done.

 

 

=Uncool-

Edited July 11, 2012 by uncool

[phillip1882]

 

 

weigh 4 vs 4.

if equal, all normal, one of the 4 remaining are odd. weigh 1 vs 1.

if equal, again both normal. weigh 1 of the remaining against a normal.

if 1v1 are unequal, weigh one of them against a normal.

now for the hard case. if 4v4 are unequal, you have 4 potential heavies and 4 potential lights.

weigh HHL against HHL. if equal, one of the 2 lights are odd, 1 more weighing determines which.

if unequal, two of the heavies and the opposite light might be odd. weigh the two heavies. done.

 

 

[Alan McDougall]

Reasoning backwards from the fact that there's an answer:

 

 

First, we note that each weighing has 3 possible outcomes, and therefore that there are a total of 3^3 possible outcomes of 3 weighings. There are 24 possible original setups, so we can only afford to "waste" 3 of the weighing outcomes.

 

The first weighing is of two sets of the same number of balls; call it n. If the weighing is Left = Right, then the odd ball is in the ones that are left, giving 2*(12 - 2n) possibilities. This must be bounded above by 9, so n is at least 4. The other possibilities are evenly split between Left < Right and Left > Right, meaning that there are 2n possibilities for each, so n is bounded above by 4, so the first step must be to weigh 4 balls against each other.

 

We first consider the "odd man out": the case where Left = Right. In this case, we have 8 balls that are certainly normal, and 4 balls, one of which is odd in one of the two directions. The second weighing can't involve all 4, as in that case the weighing will certainly not be Left = Right, so at least one must be left out. If we leave two out, and we get a weighing of Left = Right on what's left, then there are 4 situations, which cannot be decided by a single weighing, so there is at most one left out. The easiest way to do this is to weigh 3 of the 4 against 3 of the 8; say the 3 of the 4 are on the left. If we get Left = Right, then we know that the odd ball is the last one, so we can weigh the last ball against any of the normal balls. If we get Left < Right, then we know that the odd ball is lighter. We can then weigh 2 of the 3 against each other; if we get Left < Right, then the left ball is the odd one out and is lighter. If we get Left > Right, then the right ball is the odd one out and is lighter. If we get Left = Right, then the third ball is the odd one out and is lighter.

 

Note: We could also have weight 2 of the 4 and 1 of the 8 against 1 of the 4 and 2 of the 8.

 

We now turn to the case where we get Left < Right. The possibilities are that one of the 4 on the left is lighter, or that one of the 4 on the right is heavier. If in the second weighing, we get Left = Right, the odd ball out will be one of the balls not used in the second weighing, so we must use at least 5 balls in the second weighing and at most 6; this clearly includes at least one ball from each side. We also only know of 4 normal balls, so we know that there is at least one of the 8 on each side of the second weighing. This implies that no matter what, at least one pair of balls on opposite sides in the first weighing will be on opposite sides in the second weighing. We can further see by inspection that there will be at least one triplet of balls such that two were on the same side in the first weighing and are on the same side again in the second weighing, and the third was on the opposite side of the other two in the first weighing and is again on the opposite side. We can then see that no more of the same kind can be added to either side. This gives us the obvious solution:

 

If the balls on the left in the first weighing are labeled 1-4 and the balls on the right 5-8, weigh 1, 2, 5, and 6 against 3, 7, 9, and 10. If Left < Right for the first weighing and Left < Right for the second weighing, weigh 1 against 2 to determine which of 1, 2, and 7 it is. If Left < Right and Left > Right, weigh 5 against 6 to determine which of 3, 5, and 6 it is. If Left < Right and Left = Right, weigh 4 against 9 to determine which of 4 and 8 it is. A similar answer works for Left > Right, so we are done.

 

 

=Uncool-

 

Reasoning backwards from the fact that there's an answer:

 

 

First, we note that each weighing has 3 possible outcomes, and therefore that there are a total of 3^3 possible outcomes of 3 weighings. There are 24 possible original setups, so we can only afford to "waste" 3 of the weighing outcomes.

 

The first weighing is of two sets of the same number of balls; call it n. If the weighing is Left = Right, then the odd ball is in the ones that are left, giving 2*(12 - 2n) possibilities. This must be bounded above by 9, so n is at least 4. The other possibilities are evenly split between Left < Right and Left > Right, meaning that there are 2n possibilities for each, so n is bounded above by 4, so the first step must be to weigh 4 balls against each other.

 

We first consider the "odd man out": the case where Left = Right. In this case, we have 8 balls that are certainly normal, and 4 balls, one of which is odd in one of the two directions. The second weighing can't involve all 4, as in that case the weighing will certainly not be Left = Right, so at least one must be left out. If we leave two out, and we get a weighing of Left = Right on what's left, then there are 4 situations, which cannot be decided by a single weighing, so there is at most one left out. The easiest way to do this is to weigh 3 of the 4 against 3 of the 8; say the 3 of the 4 are on the left. If we get Left = Right, then we know that the odd ball is the last one, so we can weigh the last ball against any of the normal balls. If we get Left < Right, then we know that the odd ball is lighter. We can then weigh 2 of the 3 against each other; if we get Left < Right, then the left ball is the odd one out and is lighter. If we get Left > Right, then the right ball is the odd one out and is lighter. If we get Left = Right, then the third ball is the odd one out and is lighter.

 

Note: We could also have weight 2 of the 4 and 1 of the 8 against 1 of the 4 and 2 of the 8.

 

We now turn to the case where we get Left < Right. The possibilities are that one of the 4 on the left is lighter, or that one of the 4 on the right is heavier. If in the second weighing, we get Left = Right, the odd ball out will be one of the balls not used in the second weighing, so we must use at least 5 balls in the second weighing and at most 6; this clearly includes at least one ball from each side. We also only know of 4 normal balls, so we know that there is at least one of the 8 on each side of the second weighing. This implies that no matter what, at least one pair of balls on opposite sides in the first weighing will be on opposite sides in the second weighing. We can further see by inspection that there will be at least one triplet of balls such that two were on the same side in the first weighing and are on the same side again in the second weighing, and the third was on the opposite side of the other two in the first weighing and is again on the opposite side. We can then see that no more of the same kind can be added to either side. This gives us the obvious solution:

 

If the balls on the left in the first weighing are labeled 1-4 and the balls on the right 5-8, weigh 1, 2, 5, and 6 against 3, 7, 9, and 10. If Left < Right for the first weighing and Left < Right for the second weighing, weigh 1 against 2 to determine which of 1, 2, and 7 it is. If Left < Right and Left > Right, weigh 5 against 6 to determine which of 3, 5, and 6 it is. If Left < Right and Left = Right, weigh 4 against 9 to determine which of 4 and 8 it is. A similar answer works for Left > Right, so we are done.

 

 

=Uncool-

 

You must use a balance scale as indicated in a following post, not the solution yet!

[uncool]

You must use a balance scale as indicated in a following post, not the solution yet!

My solution is using a balance scale in all cases.

 

 

1. 4v4

a. Same: 1 from last 4 v 1 from last 4

i. Same: 1 of first 10 v 1 of last 2

ii. Different: 1 of first 8 v 1 of the 2 just weighed

b. Different: LLHH vs LHUU where L means light-side ball, H heavy-side ball, U unweighed

i. Left < Right: 1 of the 2 L's on the left vs other L on the left

ii. Left > Right: 1 of the 2 R's on the left vs other R on the left

iii. Left = Right: Last L vs an unweighed

 

=Uncool-

[J.C.MacSwell]

This puzzle is only for those who have not come across it before. You can get the solution from the internet but this would defeat the purpose of this post which having a little fun!!

 

The Problem 12 Odd Pool Balls Puzzle

The Problem:

You have 12 balls identical in size and appearance but 1 is an odd weight (could be either light or heavy).

 

You have a set scales (balance) which will give 3 possible readings: Left = Right, Left > Right or Left < Right (i.e. Left and Right have equal weight, Left is Heavier, or Left is Lighter).

 

You have only 3 chances to weigh the balls in any combination using the scales. Determine which ball is the odd one and if it's heavier or lighter than the rest. How do you do it?

 

Sorry, don't know how to add spoiler (could someone help?)

 

1.Balance 4 and 4, if they balance the remaining 4 have the odd ball, and you have 8 that are known to be standard

 

2a. Measure 3 of the remaining 4 against 3 standard, if they balance the final ball is the odd ball.

 

3a. Balance the odd ball against a standard ball to determine if it is heavy or light

 

2b. If measure #2a did not balance you know that the 3 are heavy or light...

 

3b. Balance 2 of the 3 to find the odd ball (heavy or light) or if they balance the third of the 3 is the odd ball

 

If measure #1 does not balance you have 4 balls known standard, 4 which may contain an odd heavy ball, and 4 which may contain an odd light ball

 

2c. Measure the four standard balls against 3 possibly heavy and 1 possibly light, if they balance you have 1 possibly heavy and 3 possibly light

 

3c. Measure the 1 possibly heavy and 1 possibly light against 1 possibly light and one standard, if they balance the remaining possibly light is the odd ball (if not you can pick out the odd ball and know whether it is heavy or light)

 

If measure #2c does not balance and the unknown side is light, you have your 1 light odd ball in that group and do not require a third weighing

 

If measure #2c does not balance and the unknown side is heavy, you have 3 possibly heavy balls

 

3d. measure 1 possibly heavy against 1 possibly heavy, if they balance the third possibly heavy is the heavy odd ball

 

If they do not balance the heavy side is the heavy odd ball.

Edited July 13, 2012 by J.C.MacSwell

[Spyman]

Sorry, don't know how to add spoiler (could someone help?)

 

Hidden text.

 

[J.C.MacSwell]

 

Hidden text.

 

 

Thanks. Can no longer edit but will know for next time.

[Alan McDougall]

Sorry, don't know how to add spoiler (could someone help?)

 

 

 

Sorry, don't know how to add spoiler (could someone help?)

 

1.Balance 4 and 4, if they balance the remaining 4 have the odd ball, and youhave 8 that are known to be standard

 

2a. Measure 3 of the remaining 4 against 3 standard, if they balance the finalball is the odd ball. Correct!

 

3a. Balance the odd ball against a standard ball to determine if it is heavy orlight Correct!

2b. If measure #2a did not balance you know that the 3 are heavy or light... Correct!

 

3b. Balance 2 of the 3 to find the odd ball (heavy or light) or if they balancethe third of the 3 is the odd ball Correct!

 

If measure #1 does not balance you have 4 balls known standard, 4 which maycontain an odd heavy ball, and 4 which may contain an odd light ball Correct!

 

2c. Measure the four standard balls against 3 possibly heavy and 1 possiblylight, if they balance you have 1 possibly heavy and 3 possibly light or heavy? Explain how you would know which ballswere possibly heavy or possibly light?

3c. Measure the 1 possibly heavy and 1 possibly light against 1 possibly lightand one standard, if they balance the remaining possibly light is the odd ball(if not you can pick out the odd ball and know whether it is heavy or light) Or it could be a heavy ball?

 

If measure #2c does not balance and the unknown side is light, you have your 1light odd ball in that group and do not require a third weighing. Incorrect how do you know the unknown ballis not heavy?

If measure #2c does not balance and the unknown side is heavy, you have 3possibly heavy balls Incorrect theUnknown ball might be a light one on the other pan of the scale

 

3d. measure 1 possibly heavy against 1 possibly heavy, if they balance thethird possibly heavy is the heavy odd ball Incorrect!

 

If they do not balance the heavy side is the heavy odd ball. "Wrong the ball might be a lighter one on the other pan of the scale"

 

 

 

You made a great effort to solvethe problem but did not solve in completely!

 

Regards

 

 

Alan

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[IceFall]

 

 

First measure 4 vs 4 balls with 4 ball not being measured. (Times balance used: 1)

Possible outcomes:

1. Balance doesn't incline

2. Balance inclines to a side.

 

Solving 1:

All eight balls are standard meaning the odd ball is within the 4 previously not measured.

Measure 3 of the 8 known standard balls vs 3 out of the 4 balls not measured. (Times balance used: 2)

Possible outcomes:

1a. Balance doesn't incline

1b. Balance inclines.

 

Solving 1a:

If there is no difference that means that the ball left out of the group of 4 balls is the odd one.

To determine whether it is heavier or lighter just use another standard ball, and depending the outcome you can tell if the odd ball is heavier or lighter. (Times: 3)

 

Solving 1b:

Depending the outcome you can tell whether the odd ball is lighter of heavier. Measure 1v1 of the bunch that was odd (Times: 3). If there is no difference the ball not measured is the odd ball.

If there is a difference, you can deduce which one is the odd ball since you already know if the odd ball is either light or heavy.

 

Solving 2:

The balance with move to one side. Meaning that the side that weights more has a heavy ball OR the lighter bunch has a light ball. Which would will give us 4 standards balls (SB) (the ones not measured), 4 balls of the bunch that could have a heavy ball (HB) and 4 with a possible light ball (LB). Note that HBs can only contain a odd heavy ball and LBs can only contain a light odd ball and not viceversa.

Now measure 3 HB + 1 LB vs 4 SB, leaving 3 LB outside. (Times: 2)

Possible outcomes:

2a. There is no difference.

2b. It tilts toward the 4 SB

2c. It tilts toward the 3 HB + 1 LB.

 

Solving 2a:

No difference, meaning that the odd ball is within the 3 LBs not measured. Measure 1v1 of those 3 leaving one out (Times: 3). If there is no difference then that one out is the light ball.

If there is a difference, whichever ball is recorded to be lighter than the other is the odd ball.

 

Solving 2b:

If the balance tilts toward the 4 SB, it means that the LB from the 3 HB + 1 LB is the odd one. Since only the LB could be a light odd ball. Thus making the 4 SB heavier.

 

Solving 2c:

If it balances toward the 3 HB + 1 LB. Then the LB is a actually a SB since only the 3 HB could be an odd heavy ball. Use the same method as in (2a) but instead look for the heavy ball. (Times: 3)

 

 

[GlassPilot]

I didn't read any of the posts after the OP so sorry if this has been guessed already.

 

 

A. put any 6 balls on the scale, three on each side.

- if they balance throw them out...the heavy ball (HB) isn't there. Proceed to B.

- if they don't balance throw out the others (the six you didn't weigh) as the HB isn't there. Also throw out the three balls on the light side. Proceed to C.

 

B. put the other 6 balls one the scale and find the set of three that are heavy. Throw out the light three.

 

C take any two balls and weigh them.

- if they balance then the one not weighed is the HB. If they don't balance then the lower one is the HB.

 

Edited October 21, 2012 by GlassPilot