Again the puzzle of balls and scales

pixel01 · 2007-06-11 11:02:20

There are 4 balls which look identical. But their weights are different : 1kg, 2kg, 3kg and 4kg each. Can you use just 3 times of balancing and identify the balls?

Last edited by pixel01 (2007-06-11 11:14:27)

John E. Franklin · 2007-06-12 11:08:34

A scale with left and right sides that says even or greater or less.
Label balls A B C D.
Put AB vs. CD.
Put AD vs. BC.
Put AC vs. BD.
Based on results, I guess you can figure it out. But do I have to go through it??
1+2<3+4 2+1<3+4 2+1<4+3 1+2 < 4+3
1+4 = 2+3 4+1 = 3+2 4+1 = 2+3 1+4=3+2
1+3 < 2 + 4 3+1<4+2 3+1<2+4 1+3<4+2
2+1<4+3
2+4>1+3
1+3<4+2
1+2<3+4
Yes!!!! I got it!!!
You know which one is the one pounder because the one pounder
comes up less than the other two when with 1+2 and 1+3.
So the one is identified!
Then the 4pounder is known because it was with the one pounder in the
equal balance.
But still you don't know which one is the 2 or the 3.
Sorry, I still don't know how to do it. :Sad

Last edited by John E. Franklin (2007-06-12 11:19:12)

mathsyperson · 2007-06-13 07:24:01

This puzzle is annoying me.

There are 4! = 24 possible configurations of balls and 3³ = 27 possible combinations of weighing results, which means that there should be a way of doing it, but it will need to be very efficient. I really can't figure out what it could be though.

Edit: I've managed to convince myself that it's impossible. My proof's kind of clumsy, but I'm fairly sure it's valid.

Maelwys · 2007-06-14 07:57:31

As far as I can tell, it's impossible. There are 24 (4!) possible solutions for which ball is which. Each weighing can result in 3 possible outcomes for that particular comparison (>, =, <). So even using the most efficient possible combinations to weigh, there's always a chance that there are two balls you don't know. To illustrate:
Start: 24 combinations. Define one test to perform, the solution is either > = or <. If you chose the most economic test, then this will narrow your possibilities down to 8 combinations. Again, choose the most economic test, and you'll either get > = or <. But breaking down 8 possibilities into 3 categories means that the number of possibilities per result is 2, 2, or 4. If you get a result with only 2 possible answers you can use the 3rd weighing to determine which of those answers is correct. If you get an answer that leaves 4 possibilities than your 3rd weighing will give you possible answers of < = or >, with possibilities per result of 1, 1, and 2. So there's a 50% chance here that you'll know the answer, but still a chance that you narrow it down to only two possible answers.

To further illustrate, here's a logic table for the first (or 3) branches of this test. The other two branches also have 3 possibilities, so they'll look very similar to this.
If AB < CD
If D < AB
If A < B
A = 1, B = 3, C = 4, D = 2
If A > B
A = 3, B = 1, C = 4, D = 2
If D = AB
If AC > BD
A = 2, B = 1, C = 4, D = 3
If AC = BD
A = 1, B = 2, C = 4, D = 3
**OR** A = 3, B = 1, C = 2, D = 4
If AC < BD
A = 1, B = 3, C = 2, D = 4

If D > AB
If A < B
A = 1, B = 2, C = 3, D = 4
If A > B
A = 2, B = 1, C = 3, D = 4

As you can see, there's one test in the middle that still leaves us with two possibilities. So it's not possible to answer 100% of the time, because in 1 test out of 8 (12.5% of the time) you'll need a 4th weighing to find the final answer.

pixel01 · 2007-06-14 18:26:44

Thank you all for reading and trying solving it. When my friend gave it to me last week, I had been trying but failed. Now it is clear that the puzzle is not doable. Thanks to Math and Mael.

John E. Franklin · 2007-06-19 09:18:50

Here's another go at it.

Start with 2 balls on left and 1 on right.
If 2 balls equal 1 ball then it is 2 + 1 = 3 or 1 + 3 = 4. So weigh remaining
unweighed ball against heavy weighed ball, and you know 4 and 3 now. Then
weigh two light balls against each other for 1 and 2.

Now imagine a totally new scenario and we start again with 2 on left and 1 on
the right. This time the 2 balls weigh more than the one ball of right. The
cases for this are 4 + 3 > 1 or 4 + 3 > 2 or 3 + 2 > 1 or 3 + 1 > 2 or 4 + 2 > 3
or 4 + 2 > 1 or 4 + 1 > 3 or 4 + 1 > 2 or 3 + 2 > 4. So we label the two balls
that read heavier and label them each HP for heavy pair. And we label the one
that was lighter on the right LO for light one. Next we take one of the HP
balls and put it on the left against the unweighed ball with the LO ball on both
on the right. There are 3 possible weighing results, HP<LO+lastball, which I'll
call resultA, or HP=LO+lastball, which I'll call resultB, or HP>LO+lastball,
which I'll call resultC.
Now for resultA, it could be 3 < 1 + 4 or 2 < 1 + 4 or 3 < 2 + 4 or 1 < 2 + 4 or
2 < 3 + 1 or 4 < 3 + 2 or 1 < 3 + 2.
I give up... again...

pi man · 2007-06-19 14:54:01

John E. Franklin wrote:

Here's another go at it.
Start with 2 balls on left and 1 on right.
If 2 balls equal 1 ball then it is 2 + 1 = 3 or 1 + 3 = 4. So weigh remaining
unweighed ball against heavy weighed ball, and you know 4 and 3 now. Then
weigh two light balls against each other for 1 and 2.

In the case where 2 balls (A and B) weigh the same as 1 ball (C), the remaining, un-weighed ball (D) has to be either 2 or 4 kg. Weighing D against C will then give one of the following results:
1) D is heavier so it must be 4 kg and C is 3 kg. You can then weigh A against B to figure out which is 1 kg and which is 2 kg.

2) D is lighter so it must be 2 kg and C is 4. You can then weith A against B to figure out which is 1 kg and which is 3 kg.

I haven't even looked at the inequalities yet.

I personally think you should be able to figure out the wieghts of each ball with ZERO weighings. We're talking KILOGRAMS here!

mathsyperson · 2007-06-19 23:11:37

Heh, true. Then again, I suppose technically that would still count as weighing, you would just be using your hands instead of the scales.

Math Is Fun Forum

#1 2007-06-11 11:02:20

Again the puzzle of balls and scales

#2 2007-06-12 11:08:34

Re: Again the puzzle of balls and scales

#3 2007-06-13 07:24:01

Re: Again the puzzle of balls and scales

#4 2007-06-14 07:57:31

Re: Again the puzzle of balls and scales

#5 2007-06-14 18:26:44

Re: Again the puzzle of balls and scales

#6 2007-06-19 09:18:50

Re: Again the puzzle of balls and scales

#7 2007-06-19 14:54:01

Re: Again the puzzle of balls and scales

#8 2007-06-19 23:11:37

Re: Again the puzzle of balls and scales

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