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Weighing IV.


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35 replies to this topic

#1 rookie1ja

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Posted 30 March 2007 - 05:53 PM

Weighing IV. - Back to the Water and Weighing Puzzles
One of twelve tennis balls is a bit lighter or heavier (you do not know which) than the others. How would you identify this odd ball if you could use an old two-pan balance scale only 3 times?
You can only balance one set of balls against another, so no reference weights and no weight measurements.

This old topic is locked since it was answered many times. You can check solution in the Spoiler below.
Pls visit New Puzzles section to see always fresh brain teasers.


Spoiler for Solution


Spoiler for old wording

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#2 tambay

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Posted 20 June 2007 - 01:57 AM

I have a shorter solution.
Put six balls on each side of the scale. Whichever is lighter, the lighter ball is in that group of six.
From these balls, put three balls on each side of the scale.
Whichever is lighter, the lighter ball is in that group of three.
From the three balls, put a ball on each side of the scale.
Case1: If the scales balance, the third ball is the lighter ball.
Case2: If one is lighter, then that is the ball...
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#3 rookie1ja

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Posted 20 June 2007 - 09:18 AM

I have a shorter solution.
Put six balls on each side of the scale. Whichever is lighter, the lighter ball is in that group of six.
From these balls, put three balls on each side of the scale.
Whichever is lighter, the lighter ball is in that group of three.
From the three balls, put a ball on each side of the scale.
Case1: If the scales balance, the third ball is the lighter ball.
Case2: If one is lighter, then that is the ball...


Just a little detail you missed:
One of twelve pool balls is a bit lighter or heavier (you do not know)
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#4 mjkrism

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Posted 20 June 2007 - 07:06 PM

I have another solution. Use the scale just twice. Place 6 balls in each. One of the scale is down at this point. Keep removing one ball from each scale. At one point the scale will balance. Now we know that one of the two balls is lighter or heavier than the rest. Clear the scale and place one of the 2 balls in one scale and another ball from the remaining group on the other. if they balance, then the other ball is the lighter or heavier one, if not that is the one !
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#5 Camshaft

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Posted 02 July 2007 - 07:55 PM

The very least amount of times you could weigh would be 2. You pick two balls number 1 and number 2, you weigh them and they happen to be different weights. You know that one of these balls is the one you are looking for. You weigh balls number 2 and number 3, if they are equal in weight than it is ball number 1, if they are different in weight than it is ball number 2.
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#6 stephcorbin

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Posted 08 July 2007 - 05:54 AM

I think the best way to resolve this problem is by building a deterministic scale array. Every combinations of balls should not be evaluated more then once against another group, then 3 balls will appear once, 3 will appear 3 times and the other 6 balls will be tested 2 times.

A sample array will be as follow:

1 2 3 7 | 8 9 10 4
3 4 5 8 | 9 7 11 6
5 6 1 9 | 7 8 12 2

Then to find any combination, you simply evaluate the 3 balance results:

Ball 1 heavier:
1 2 3 7 > 8 9 10 4
3 4 5 8 = 9 7 11 6
5 6 1 9 > 7 8 12 2

Ball 2 heavier:
1 2 3 7 > 8 9 10 4
3 4 5 8 = 9 7 11 6
5 6 1 9 < 7 8 12 2

Ball 3 heavier:
1 2 3 7 > 8 9 10 4
3 4 5 8 > 9 7 11 6
5 6 1 9 = 7 8 12 2

Ball 7 lighter:
1 2 3 7 < 8 9 10 4
3 4 5 8 > 9 7 11 6
5 6 1 9 > 7 8 12 2

Ball 10 heavier:
1 2 3 7 < 8 9 10 4
3 4 5 8 = 9 7 11 6
5 6 1 9 = 7 8 12 2

Another virtue to this method is that it is easily programmable.
.
/S
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#7 bonanova

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Posted 19 July 2007 - 06:24 PM

Each weighing has three outcomes:
Left side is [lighter than] [equal to] [heavier than] Right side.
Three weighings can thus discern among [3]x[3]x[3]= 27 cases.

We have only 24 cases:
one of 12 balls is heavier or lighter than the rest.
So we can solve the problem, so long as ...
[1] The first weighing reduces the cases to no more than 9.
[2] The second weighing reduces the cases to no more than 3.
[3] The third weighing then distinguishes among 3 or fewer cases.

First weighing:
Set aside four balls. Why?
Because, if the first weighing balances, we have 8 [fewer than 9] cases:
One of the 4 excluded balls is heavier or lighter.

[1] Weigh 1 2 3 4 <=> 5 6 7 8

Outcome 1[a] 1 2 3 4 balances 5 6 7 8.
Since only one ball is odd, 1 2 3 4 5 6 7 8 must all be normal.
We have 8 cases: 9 10 11 or 12 is H or L
[shorthand: H=heavier; L=lighter]

[2] Weigh 1 2 3 <=> 9 10 11 [we know 1 2 3 are normal]

If this balances, 9 10 11 are also normal, and 12 is H or L.

[3] Weigh 12 <=> [any other ball]
If 12 rises, then 12L; if 12 falls, then 12H.

If [2] 9 10 11 rises, we have 3 cases: 9L, 10L or 11L

[3] Weigh 9 <=> 10.
If [3] balances, then 11L;
If 10 rises, then 10L; If 9 rises, then 9L

If [2] 9 10 11 falls, we have 3 cases: 9H, 10H or 11H

[3] Weigh 9 <=> 10.
If [3] balances, then 11H;
If 10 falls, then 10H; If 9 falls, then 9H

end of Outcome 1[a]: balance.

Outcome 1: 1 2 3 4 falls, and 5 6 7 8 rises.
This gives 8 cases: 1H, 2H, 3H, 4H, 5L, 6L, 7L, or 8L

The second weighing in this case becomes tricky.
Remember each of its three outcomes can lead to no
more than 3 cases for the third weighing to resolve.

Again, we exclude a number of balls and involve the others.
We exclude any three of the balls. Why? Because if the
other [included] balls balance, we have exactly 3 cases.
Without loss of generality we exclude balls 1 2 3.

Since that leaves an odd number of balls, 4 5 6 7 8,
we need to use one of the normal balls.
Finally we choose which three to weigh against the others.

And here's the only hard part of this problem.
We must mix some of the possibly light balls with some
of the possibly heavy balls. Otherwise, one of the
outcomes of the second weighing will leave us with more
than 3 cases, and the third weighing will not resolve this.

[2] weigh 4[H] 5[L] 6[L] <=> 1[normal] 7[L] 8[L]
in parentheses I've indicated the POSSIBLE cases
that we have determined:
4[H] means 4 is heavier if it's not normal.

Outcome 2[a]: 4 5 6 balances 1 7 8
These balls are all normal.
We have 3 cases: 1H 2H or 3H.

[3] weigh 1 <=> 2
If [3] balances, then 1 and 2 are normal, and 3H
if 2 falls, then 2H
if 1 falls, then 1H

Outcome 2b: 4 5 6 falls, and 1 7 8 rises.
We have 3 cases: 4H 7L or 8L

[3] weigh 7 <=> 8
if [3] balances, then 7 and 8 are normal, and 4H
if 7 rises, then 7L
if 8 rises, then 8L

Outcome 2[c]: 4 5 6 rises, and 1 7 8 falls.
We have 2 cases: 5L or 6L.

[3] weigh 1[normal] <=> 5[L]
If 5 rises, then 5L
If balance, then 5 is normal, and 6L

Now we can go back and take the remaining case
Outcome 1[c]: 1 2 3 4 rises, and 5 6 7 8 falls.
to distinguish among the remaining 8 cases:
1L 2L 3L 4L 5H 6H 7H and [b]8H

exactly as we analyzed Outcome 1b.

Simply substitute H for L and v.v.

Problem solved.
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#8 erictheread

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Posted 13 August 2007 - 01:39 PM

1 weigh 4 balls on scale - groups a,b & c- if a & b balance the oddball is in c, go to step 3
2 if unbalanced hold a and balance b- 2 on each side- if equal discard
3 balance a- repeat 2 & 3 until you have 1 ball on each side. the oddball and another
4 pick any other ball to balance off the finalists-if equal discard
if your lucky you can confirm the oddball in as few as 4 tries
cai
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#9 celinerocks

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Posted 10 September 2007 - 12:20 PM

i think that this would be easier,

1st weighing - weigh 6 balls on each pan and the oddball is in the uneven one.
2nd weighing - weigh 3 balls in each pan and the oddball is in the uneven one.
3rd weighing - weigh any two balls the oddball is in the un even one. if they are even, the oddball is the remaining.

right?

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#10 Martini

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Posted 12 September 2007 - 06:45 PM

I have another solution. Use the scale just twice. Place 6 balls in each. One of the scale is down at this point. Keep removing one ball from each scale.


Each time you remove a ball and look at the scales, it's considered a separate weighing.


The very least amount of times you could weigh would be 2.


Actually, the least amount of times you could weigh to identify which ball is different is once. This, however, is not what the riddle is asking for. It's asking (bolding mine): "At least how many times do you have to use an old pair of scales to identify this ball (and be always 100% sure)?"


1st weighing - weigh 6 balls on each pan and the oddball is in the uneven one.


If you're weighing two groups of balls, there is no oddball group; they are both different from one another. You won't know which group to separate for the second weighing.
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