ODD BALL LOGIC PROBLEM

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ODD BALL LOGIC PROBLEM

Postby Alan McDougall on March 26th, 2016, 10:46 am 

HI,

I am new, the solution to this problem can be found on the internet, but please don't spoil it be going there and posting the solution here!

Work it out, yourselves, please. (It is a really fun puzzle to try!)

The Problem

You have a balance type Scale you often see depicted in law courts, with pans on either side to hold the balls

You can only use this device to work out your solution (Not, for instance, a bathroom type scale)

You can weigh any combination of balls.

BUT YOU MUST SOLVE THE PROBLEM IN JUST THREE COMPARISON EFFORTS

For example, you could use 6 and 6 balls and one side will go up and the other down?

You have 12 identical balls in volume, shape and color.

However , one ball, namely the oddball differs in mass/weight in such a tiny amount that you need a way to find by logic means exactly which one it is

The oddball might be heavier or lighter than the 11 other balls of average weight ?

You must locate the oddball and establish whether it is lighter or heavier than the others ?

Thus, there is a two-paths to a full solution.

Alan
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Re: ODD BALL LOGIC PROBLEM

Postby Braininvat on March 26th, 2016, 2:15 pm 

Nice one. I think I see a way, but will wait a while. What makes it tricky is not knowing which way the differential goes WRT to the average balls.
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Re: ODDBALL

Postby Faradave on March 26th, 2016, 8:31 pm 

I like the title.

It was fun, though I usually don't do games.

I don't think matters, solving it alone or looking it up. A posted solution spoils it for others.

Mum's the word.

Limited access (to the scale) is analogous to a limited number of resources. Efficiency is key to maximizing utility.
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Re: ODD BALL LOGIC PROBLEM

Postby Alan McDougall on March 27th, 2016, 5:10 am 

Process of elimination is the way to go.
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Re: ODD BALL LOGIC PROBLEM

Postby Watson on March 31st, 2016, 1:02 pm 

I seem to need the 4th comparison, or use some guess work.
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Re: ODD BALL LOGIC PROBLEM

Postby Braininvat on March 31st, 2016, 1:17 pm 

Think about the logical implications of each comparison, insofar as the ball groups not measured are concerned. SPOILER AHEAD


Your thought on piles of three was right on. Remember after the first comparison, you have a "control" group that consists entirely of normal balls. That's the key later.
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Re: ODD BALL LOGIC PROBLEM

Postby Watson on April 1st, 2016, 12:38 pm 

Yes that seems to get me to the third weigh in, which may disclose the odd ball of the 3 balls. If not, I have 2 balls to check with the 4th try. Then I either find it, or the last ball must be it. Previous weighings would have disclosed if the odd ball was more or less in weight.
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Re: ODD BALL LOGIC PROBLEM

Postby Braininvat on April 1st, 2016, 1:12 pm 

If you reached the 3rd weighing with 3 balls left, then you don't need a 4th weighing. Here's why:

SPOILERS AHEAD


You take two balls from the final group of 3, and weigh them. Either they balance, in which case the other ball is the oddball, or they don't, and if they don't the one that goes in the direction of the oddball (already determined) is the oddball.

If steps 1 and 2 were done properly, I think that should work.
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Re: Solution process

Postby Faradave on April 2nd, 2016, 1:09 am 

SEMI-SPOILER:

I also started with three groups. Did anyone else end by a swap of two balls between scale pans?
Alan referred to two solutions.
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Re: ODD BALL LOGIC PROBLEM

Postby Watson on April 2nd, 2016, 12:52 pm 

I think so, but isn't that an extra turn at the scales?
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Re: One solution

Postby Faradave on April 2nd, 2016, 11:38 pm 

Yes. Swapping counts as one turn.
.
.
SPOILER (of the simpler case).
.
.
This is the simpler case but it can work in either if you extend the method.

1. Put four balls on each of two scale pans. (1st measure)

2. If they are equal (simple case), all those balls are eliminated as "odd".

3. With two remaining balls on each scale pan, say the left side is lighter. (2nd measure)

4. Swap a ball from each pan, keeping track of them. (3rd measure)

If the left side stays light, you know an odd light ball is on the left, the ball not swapped.
If the right side is now lighter, a light odd ball the one that moved to the right side.

The works similarly, if the odd ball is heavier.

See if you can use this method when the original pans of 4 balls are unequal.
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Re: ODD BALL LOGIC PROBLEM (Solution)

Postby Alan McDougall on May 14th, 2017, 4:03 am 

Solution


You are given 12 seemingly identical metal balls, but one of them weighs slightly more or less than the others. With a balance scale and three weighings only, how can it be determined which is the oddball and whether it is heavier or lighter?

This is one those puzzles that'll likely take a fair amount of time to work out, but the solution (at least the only one I know of) is ingenious and very satisfying to discover. No special math or logic skills required, just perseverance and insight.

Solution to Twelve Balls puzzle

This one requires a little work and inspiration to solve. I know of only this one solution here. If anyone knows of another, I'd be happy to see it.

Divide the balls into three groups of four each; label these groups AAAA, BBBB and CCCC.

Weigh AAAA_BBBB. The possible results are:
1. If they balance: One of the C's is heavy or light. Therefore:
weigh CCC_AAA (remember, all A's are now known to be standard):
a. If they balance: The 4th C is the oddball. Therefore, weigh the 4th C against any other ball.
i. If the 4th C falls: The 4th C is heavy.
ii. If the 4th C rises: The 4th C is light.
b. If the CCC side falls: One of the C's is heavy. (Remember, the A's are known to be standard.) Therefore, weigh C_C:
i. If they balance: The other C is heavy.
ii. If one side falls: The fallen C is heavy.
c. If the CCC side rises: One of the C's is light. Therefore, weigh C_C.
i. If they balance: The other C is light.
ii. If one side rises: The risen C is light.
2. If the AAAA side falls: The oddball is either a heavy A or a light B and the C's are all standard. Therefore, arrange the balls into three new groups like so: AAAC BBBA CCCB. (This re-arrangement step, and the one like it below in step 3, are the key to solving this puzzle.) Weigh BBBA_CCCB:
a. If they balance: The oddball is in AAAC. Therefore, weigh A_A.
i. If they balance: The other A in AAAC is heavy.
ii. If one side falls: The fallen side has the heavy A.
b. If the left side (BBBA) falls: The A in BBBA is heavy or the B in CCCB is light. Therefore, weigh A_C (C is known to be standard).
i. If they balance: The B in CCCB is light.
ii. If the A side falls: A is heavy.
iii. If the C falls: Not possible.
c. If the right side (CCCB) falls: The a B in BBBA is light. Therefore, from the BBBA group weigh B_B.
i. If they balance: The other B in BBBA is light.
ii. If the left side falls: The B on the right is light.
iii. If the right side falls: The B on the left is light.
3. If the BBBB side falls: The oddball is either a heavy B or a light A and the C's are all standard. Therefore, arrange the balls into three new groups like so: AAAB BBBC CCCA. Weigh AAAB_CCCA:
a. If they balance: The Oddball is in BBBC. Therefore, weigh B_B.
i. If they balance: The other B in BBBC is heavy.
ii. If one side falls: The fallen side has the heavy B.
b. If the left side (AAAB) falls: The B in AAAB is heavy or the A in CCCA is light. Therefore, weigh B_C (C is known to be standard).
i. If they balance: The A in CCCA is light.
ii. If the B side falls: B is heavy.
iii. If the C side falls: Not possible.
c. If the right side (CCCA) falls: An A in AAAB is light. Therefore, from AAAB weigh A_A.
i. If they balance: The other A in AAAB is light.
ii. If the left side falls: The A on the right is light.
iii. If the right side falls: The A on the left is light.
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