## 8 identical weights

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Jynnantonix
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Joined: Tue May 18, 2010 9:49 pm UTC

### 8 identical weights

This is a fairly simple problem but I didn't find it anywhere on this forum so I thought I would post it.

You have a scale and 8 weights that look identical. However, one of these weights is lighter than the rest. What is the fewest number of times you would need to use the scale to figure out which is the lighter one and why?

If you want the answer but not the reasoning behind it:
Spoiler:
You would only need to use the scale twice.

jaap
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### Re: 8 identical weights

As with most such problems, you have a set of weighing scales, i.e. a balance.
Spoiler:
You can do 9 weights with only 2 weighings, provided you are sure that exactly one weight is too light.
Weigh two sets of 3 against each other. From this you can deduce which triplet the light weight is in.
Weigh two weights from that triplet against each other. From this you can deduce which weight of that triplet is the light weight one.

One somewhat unsatisfying possibility is that both weighings are balanced, meaning that the lighter weight must be the one that you never placed on the scales. If you start with only 8 weights you can avoid that case, and even allow for the possibility that none of the weights are light.

levantis
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### Re: 8 identical weights

Spoiler:
one. you take 2 balls, scale them, if the scales read unequal, you`ve found it )

back to being serious, how much do you need to find TWO lighter balls?

redrogue
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### Re: 8 identical weights

levantis wrote:back to being serious, how much do you need to find TWO lighter balls?

Spoiler:
The best I can get is four. I visualized it by organizing the weights in a 2x2x2 cube, split across the x, y, and z planes. Weigh 4 vs 4 each time.

Three weighings are needed if the two are "adjacent" two each other in the cube, but a fourth is necessary if they are diagonal in the same half or if they are in originally completely opposite corners.

Tirian
Posts: 1891
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### Re: 8 identical weights

redrogue wrote:
levantis wrote:back to being serious, how much do you need to find TWO lighter balls?

Spoiler:
The best I can get is four. I visualized it by organizing the weights in a 2x2x2 cube, split across the x, y, and z planes. Weigh 4 vs 4 each time.

Three weighings are needed if the two are "adjacent" two each other in the cube, but a fourth is necessary if they are diagonal in the same half or if they are in originally completely opposite corners.

Spoiler:
And four would be the minimum possible, as there are C(8,2) = 28 possible ways to choose two light balls and three weighings could only distinguish between 3^3 = 27 cases.

Nitrodon
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### Re: 8 identical weights

redrogue wrote:
levantis wrote:back to being serious, how much do you need to find TWO lighter balls?

Spoiler:
The best I can get is four. I visualized it by organizing the weights in a 2x2x2 cube, split across the x, y, and z planes. Weigh 4 vs 4 each time.

Three weighings are needed if the two are "adjacent" two each other in the cube, but a fourth is necessary if they are diagonal in the same half or if they are in originally completely opposite corners.

Spoiler:
I'm not sure I follow. If they are in opposite corners of the cube, then the first three weighings will all be balanced if I'm inferring your strategy correctly. You wouldn't be able to distinguish between the four possible pairs of opposite corners in the final weighing.

redrogue
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### Re: 8 identical weights

Nitrodon wrote:
Spoiler:
I'm not sure I follow. If they are in opposite corners of the cube, then the first three weighings will all be balanced if I'm inferring your strategy correctly. You wouldn't be able to distinguish between the four possible pairs of opposite corners in the final weighing.

Now that I'm returning to it, you are correct.

Here's a an accurate four weighing solution:

Spoiler:

Code: Select all

`Number them 1-8.Weigh 123 vs 456.If both sides are even (either 7&8 are light, or there's one light ball on each side) [  Weigh 345 (left) vs 678 (right)   If even (light balls must be 3 & 6) [       3 and 6 are light.   ]   If left side is lighter (narrowed to 14, 15, 24, 25, 34, or 35) [      Weigh 1 vs 2 (to determine if 1, 2, or 3 is lighter)      Weigh 4 vs 5 (to determine if 4 or 5 is lighter)   ]   If right side is lighter (either 16, 26, or 78 [       Weigh 1 vs 2       If 1 is lighter, 1 and 6 are light.       If 2 is lighter, 2 and 6 are light.       If even, 7 & 8 are light.  ]]If one side is lighter (either one or both light balls are on the light side) [   Weigh 7 vs 8.   If even (both light balls were in the light group) [     Weigh two balls from light side of original weighing to determine two light balls from that group.   ]   else (you've just determined which of 7 or 8 is light, one ball is light in the original light group) [      Weigh two balls from light side of original weighing to determine sole light ball from that group.   ]]`

What about the case where the two counterfeit balls are both either heavy or light (i.e.: two light balls or two heavy balls)?

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### Re: 8 identical weights

Spoiler:
2.
Put 3 on each side, and if they match, then you know the lighter weight is one of the remaining two. So weigh them, and the lighter one is the lighter one.

attempt
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### Re: 8 identical weights

Spoiler:
2.
Put 3 on each side, and if they match, then you know the lighter weight is one of the remaining two. So weigh them, and the lighter one is the lighter one.

Spoiler:
you seem to be forgetting the possiblity that the 6 on the scale may not be match. you are using probability, not an exact method.I believe he is looking for a failsafe method to find the light weight.

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### Re: 8 identical weights

attempt wrote:
Spoiler:
2.
Put 3 on each side, and if they match, then you know the lighter weight is one of the remaining two. So weigh them, and the lighter one is the lighter one.

Spoiler:
you seem to be forgetting the possiblity that the 6 on the scale may not be match. you are using probability, not an exact method.I believe he is looking for a failsafe method to find the light weight.

Spoiler:
Least I can come up with then is three.
Put two on each side, if they match, take them off, and put the other set of four on. Take one off of each end, and if they match then, then the one you took off of the lighter side is the lighter ball, if they don't match, then the lighter side has the lighter ball.
If they don't match originally, take one off of each side, and if they still don't match the lighter side has the lighter ball, and if they do match, then the ball you took off the formerly lighter side is lighter.

Chen
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### Re: 8 identical weights

attempt wrote:
Spoiler:
2.
Put 3 on each side, and if they match, then you know the lighter weight is one of the remaining two. So weigh them, and the lighter one is the lighter one.

Spoiler:
you seem to be forgetting the possiblity that the 6 on the scale may not be match. you are using probability, not an exact method.I believe he is looking for a failsafe method to find the light weight.

Spoiler:
Are we still talking about the original problem or the variants posted? Because the original can be done in 2. If the 3 on each side does do not balance you simply take the lighter side and weigh 2 of the 3. If they balance, the light one is the one you didn't weigh. If they don't balance you've determined which one is lighter.

redrogue
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### Re: 8 identical weights

Chen wrote:Are we still talking about the original problem or the variants posted?

Hopefully just the latest variant I posted. I'm confident in my (second) solution to 8 weights where 2 are light, though it is welcome to scrutiny.

There's a "twelve coin" variation on the original problem where there are 12 coins and a counterfeit that is either heavier or lighter (but you don't know which). See here for details (and a solution): http://en.wikipedia.org/wiki/Balance_puzzle.

Which inspired this:

What about the case where the two counterfeit balls are both either heavy or light (i.e.: six standard balls plus either two light balls or two heavy balls)?

Qaanol
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### Re: 8 identical weights

With a total of 8 balls, of which 6 weigh the same as each other and the remaining 2 weigh the same as each other but different from the 6, there are 56 possibilities, so I believe this can be solved in 4 weighings. I will use +AB to denote that balls A and B are heavier than the others, and -AB to denote that they are lighter.

Spoiler:
Here is the beginning of an attempt. On the first weighing, put 123 v 456. On the second weighing, put 147 v 258. Now the possibilities remaining in each case, where I write “Even” if the scales are balanced and otherwise write the name of the heavier side, are:

Even Even: +15, -14, +24, -24, +36, -36, +78, -78
Even Left: +14, +16, +34, -25, -26, -35
Even Right: -14, -16, -34, +25, +26, +35

Left Even: +12, +18, +27, -45, -48, -57
Left Left: +13, +17, +37, -46, -58, -68
Left Right: +23, +28, +38, -46, -47, -67

Right Even: -12, -18, -27, +45, +48, +57
Right Left: -13, -17, -37, +46, +58, +68
Right Right: -23, -28, -38, +46, +47, +67

So after two weighings there are only 6 or 8 possibilities remaining. I leave it to others to show that each case can or cannot be solved in 2 weighings.
wee free kings

Jammerjoint
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### Re: 8 identical weights

Spoiler:
Weigh 3 against 3. If they're equal, then weigh the last two against each other. If not, weigh two from the lighter group against each other. If those two are equal, then the last one is lightest.

crazyiscool899
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Joined: Sat Jul 31, 2010 8:36 pm UTC

### Re: 8 identical weights

I can get 3 Times.
Spoiler:
You start with four weights measured against the other four. Whichever side is lighter take 2 of the weights then measure them against the other two weights. Whichever side is lighter measure those 2 weights against each other.

Sorry if someone already posted this and/or it didn't make sense I didn't want to read the whole thread and forget how i solved it.

drkfalco
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### Re: 8 identical weights

Th jammerjoint answer is the correct guys...
Spoiler:
3 & 3 if them weight the same, then just measure the last 2 and you see whcih one...
if theres one of those 3 groups lighter, took itand meassure only 2 bags o the 3...

you would have 2 heavy and one lighter, so, you just weight 2 of them and you got the lighter, if both weight the same, the last one is the lighter...
good1

WarDaft
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### Re: 8 identical weights

Slightly more interestingly, what is the minimum number of weighings to assure all n weights are the same weight? There is the n-1 trivial solution, but is there a better one?
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jaap
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### Re: 8 identical weights

WarDaft wrote:Slightly more interestingly, what is the minimum number of weighings to assure all n weights are the same weight? There is the n-1 trivial solution, but is there a better one?

Spoiler:
I don't think so. There are n unknowns, and each weighing gives one linear equation. You need at least n equations to solve the variables. One equation arises from the fact that we can choose our unit of measurement, so we can arbitrarily choose any weight and say it has a weight of 1 unit. To determine that the other n-1 variables are also 1 we need n-1 more equations which must then come from n-1 weighings.