9 weights problem

I'm sure many of you have heard of the 12-weight problem. There is a version of it on the wikipedia page:

https://en.wikipedia.org/wiki/Balance_puzzle

My problem is similar but only involves 9 weights. You have 9 identical looking weights all of which are the same weight except 2, one of them is slightly lighter than the rest and the other one is slightly heavier than the rest by the same amount. In other words 7 of them weigh x grams, one of them weighs x-e grams and the other one x+e grams.

How can you determine which weight is the heavier one and which is the lighter one in just 4 weighing on a set of balance scales?

It seems impossible because any group of 2 or more could have both the heavy and the light weight which cancel each other out and are therefore undetectable.

While it is true that, on the first weighing, you cannot avoid having the heavy and the light weight on the same side of the scales that does not mean you cannot tell them apart on subsequent weighings.

Originally posted by Zuggy
While it is true that, on the first weighing, you cannot avoid having the heavy and the light weight on the same side of the scales that does not mean you cannot tell them apart on subsequent weighings.

The point is that you don't glean muhc infromation from each weighing.

Lets assume you divide the 9 weights into three groups: a1a2a3 b1b2b3 c1c2c3.

Now you weigh group a against b.

If they are the same it could mean:

all a's and b's have the same weight, that means the orther two are in the c-group.

OR
both different weights are in the a group

OR both are ikn the c-group.

Summary: All you learned with that result is that both sought weights are in the same group, but you still have to do a lot mor weighing. Since now you must check through all of the groups.

1 weighing a1 vs. a2 if they are the same

weighing b1 vs. b2 if they are the same

weighing c1 against c2 [you already are at the fourth weighing here]
IF c1> c2 you still don't know which one is the normal (if any)
so weigh c1 against c3 and you still have to weigh c2 against c3 to establish the order [7 weighings]

back to the first weighing:
If a>b the wither one of the a's is the heavy weight OR one of the b's is the light weight. OR one of the a#s is the heavy AND one of the b's is the light.

so you weigh a vs. c a>c you now know that one of the a's is the heavy one. *
If c>a you know that one of the c's is the heavy one and all a's are normal ones.

On the second case you have to weigh c1 vs. c2 and b1 vs. b2 to find the sought weigths. Making it four weighings at the least.

On case * you still have to weigh c aginst b to find out which batch is the normal one. and you have to weigh then the groups with the sought weights to establish which one is they are.


In Sum you have to make 4 weighings if you are lucky and 7 if you are unlucky.

There might be a simpler solution, but I doubt that.

There definitely is a simpler solution. It can be guaranteed to be done in four weighings. I'm going to give a clue that should help push people in the right direction.

You initially have 72 possibilities (9 choices for the heavier weight and 8 for the lighter one). There are 81 possible different outcomes that the 4 weighings could give you (3 on each weighing 3*3*3*3=81). This suggests that it should be possible (I appreciate this is not a proof). Now it is a good idea on each weighing to try to divide the remaining possibilities into 3 roughly equal groups. For example on the first weighing it would be nice if you were left with 24 remaining possibilities if the left side is heavier, 24 remaining possibilities if the right side is heavier and 24 remaining possibilities if they balance. Such an even split might not be possible but you must keep all 3 groups less than or equal to 27 (you cannot distinguish between 28 possibilities in just three weighings). This should suggest at least what the first weighing should be. Then try to continue in this in this vein on subsequent weighings.

I think this is a much more practical method for attempting to solve it than to try and come up with a long term strategy from the beginning.

Ok, me again.

I keep the premise to distribute in the groups a, b, c and denoting further with 1,2,3

First experiment a vs. b I consider a<>b. One further weighing a vs c or b vs. c is sufficient to establish the order e.g. a>b>c...

Now you only have to wiegh two of the heavy fraction against each other which gives the heavy (assmue a is the heavy fraction if a1=a2 then a3 is the heavy, otherwise the heavier is), and two of the light fraction.

For the case a=b I have found the solution also:

weigh 1 ggainst 2 and 2 against 3 to establish the group in which the heavy and the light are. Then weigh from the heavy or the light two of one number group.

Assume 1>2>3 Then weigh a1 against b1 if they are the same then c1 is the heavy and c3 the light.

Originally posted by Ponderable
First experiment a vs. b I consider a<>b. One further weighing a vs c or b vs. c is sufficient to establish the order e.g. a>b>c...

This is not true. Say a<b and a<c.

Originally posted by Ponderable
weigh 1 ggainst 2 and 2 against 3 to establish the group in which the heavy and the light are.

Again this doesn't work. What if 1<2 and 3<2?

Impossible.

For example, with only three weights it takes three attempts to solve and you expect to solve with nine weights in four attempts?

Originally posted by iChopWoodForFree
Impossible.

For example, with only three weights it takes three attempts to solve and you expect to solve with nine weights in four attempts?

I'm not sure how to answer such a post other than to present the answer. I assure you it is possible. It is not easy and no one's answers have even come close yet but I did give a hint as to how to attack the problem a few posts back. If no one gets close soon then I'll start to give bigger clues.

Originally posted by Zuggy
I'm not sure how to answer such a post other than to present the answer. I assure you it is possible. It is not easy and no one's answers have even come close yet but I did give a hint as to how to attack the problem a few posts back. If no one gets close soon then I'll start to give bigger clues.

Is the balance such that I can tell the difference between:

1 regular vs. 1 light
1 heavy vs. 1 light

?

If so, I might have an answer.

Hmm, from re-reading the thread, it looks like the answer to my question is "no". 😞

Each step I will choose the outcome that gives the highest number of possibilities.

*1.abcd < efgh

Possibilities:

ae af ag ah ai be bf bg bh bi ce cf cg ch ci de df dg dh di ei fi gi hi

We know that if any of the odd weights is abc or d then that is the light weight and if it is def or g then that is the heavy weight. This is why there are only 24 possibilities instead of 48.

2.abfg>icde

cf cg ch df dg dh ei fi gi hi

3.cef>dig(if = then weigh c against d or weigh f against g to answer. If < then c must be the light weight, finding the heavy weight between fgh is trivial)

ei fd fi

4.ed<ab then d is light f is heavy
4.ed=ab then i is light and f is heavy
4.ed>ab then i is light and e is heavy.

*abcd=defg also gives 24 possibilities. However, there are only 12 different pairs so the method is much simpler. This is why I decided not to include it in my solution.


I guess i owe you an apology.

Since you were so certain it could be done I found myself compelled to look deeper.