Another Doozy for the Dawg.

[sussexscooterhead]

Sorry Dawg, didn't mean to spoil the Einstein riddle for you.

Here's another one for you....and anyone else that likes a challenge.



There are 12 cannon balls. 11 of the balls have the same weight. This means, of course, that 1 does not weigh the same as all the rest.

You have a ballance scale which you are allowed to use ONLY 3 TIMES.

The challenge is to discover which ball is the ODD ball and decide whether it is heavier or lighter than the others.

This is not a trick, it can be done.

Remember, you only have 3 chances with the scale.





Good Luck!

[riker140]

OK put six balls on each scale the lighter on will go high
discard the heavy six take the six on the light scale Put three the on each scale discarded the heavy on.
Now put one one each scale is they are equal then the on you didn't put on is the light one. If the are not in balance then the light one is on the scale.

Dave

[medman1952]

I thought of that also, but what if the odd ball was a heavier one?

[sussexscooterhead]

Cool to see someone's actually trying this!

Maybe I'm weird but I like these kinds of puzzles.

[Admin]

Do you know how agonizing this is if you have a mild hangover?

[sussexscooterhead]

Just wondering if anyone's got this yet?

[Admin]

I think we all admit defeat.

[Admin]

So, are you going to tell us the answer or just wait until somebody gets it?

[sussexscooterhead]

LOL

I was waiting for ask, otherwise it'd just be a lot of typing for nothing.

I'll post it soon, it'll take me a while to word it well enough to make sense.

There are at least 2 ways of solving it and I was hoping someone (like an accountant or something) would post the other method, which, if I remember correctly, is all mathematical. The solution I came up with is just a series of logical steps intended to maximize the number of variables for each use of the scale.

I don't know if this will help anyone solve it between now and when I post the answer but the balls need to be arranged in groups of 4, you'll need to keep track of which side each ball was on, either the heavy or light side and at some point take one ball per side and trade sides.

But yeah, I'll post the answer.

[sussexscooterhead]

Okay, Ill try to make this as un-confusing as possible.

In order to keep track of the balls we’ll label them with letters A, B, C, D, E and so on until we have 12 letters, the last one being L.

Divide the balls into groups of four. ABCD - EFGH - IJKL

Take 2 groups and put 1 group on either side of the scale. You can imagine that now we have ABCD on one side and EFGH on the other. IJK and L are sitting off to the side.

If the scale balances you know the odd ball is one of the other 4 balls (I, J, K or L). Go to step XXX.

If the scale DOES NOT balance you know the odd ball is one of these 8. Continue following along.

Lets say the side with ABCD is heavy and the side with EFGH is light. It’s important to remember this. The fact that ABCD is heavy isn’t yet significant, it could just as easily have been the other way around, the important thing is to maintain the association between the letter of the ball and whichever characteristic it is associated with – heavy or light.

This is where things can get confusing.

Remove D from the scale and set it to the side – remembering that it came from the heavy side.

Remove G and H – remembering they came from the light side.

Trade paces with balls C and E – keeping in mind that C was heavy and E was light.

Now you have 3 balls on one side of the scale, ABE and only 2 balls on the other, CF – to this side add on of the balls that has not yet been on the scale, let’s say I. This enters a known “good” ball into the process.

This is the second use of the scale. The first was ABCD against EFGH and now we are comparing ABE against CFI.

If the scale balances we know that the odd ball is among the 3 we removed from the scale in the previous step. Go to step 444.

If the scale DOES NOT balance there are 2 possible outcomes:


a. The scale is NOT balanced but has NOT changed, meaning that balls AB are still on the Heavy side. See solution a.


OR……


b. The scale is NOT balanced but IS changed and now balls AB are on the light side. See solution b.



Solution a.

So, we did all that swapping around and removing of balls and the scale didn’t move, this means one of the 3 balls that was left in it’s original place is the odd ball – either A or B from the heavy side or F from the light side.

To get to the bottom of this once and for all, with our 3rd use, compare ball A to B – one on each side of the scale.

If they balance we know that ball F is lighter than all the rest.

If they DO NOT balance, the odd ball is the one that is on the heavy side – both balls came from the heavy side of the scale in previous weighings and now we know which one is responsible for that.

You can see know why it was important to remember what state relates to each ball – heavy or light.



Solution b.

Ok, the only reason for the scale to now be in the opposite orientation is the 2 balls that swapped places – C and E. Don’t forget C is associated with a heavy condition and E with a light condition.

Compare C with any ball other than E.

If it balances we know E is lighter than every other ball.

If does not balance, C will show itself to be heavier than the other balls.



Step 444.

Alright. We had removed a total of 3 balls from the scale (D, G and H), swapped a couple balls and added one from the group that hadn’t been weighed and the result was that the scale balanced. This means that 1 of the 3 we removed is the odd ball.

G and H came from the same side of the scale and if you recall it was the light side.

For our 3rd and final weigh-in, let’s put 1 on each side of the scale – G against H.

If they balance we know D is heavier than the rest.

If they do not balance, the 1 that is lighter is the odd ball.


Step XXX.

So, in our first use of the scale, we compared 2 groups of 4 balls and found that the scale balanced. This means the odd ball is among the remaining 4 – either I, J, K or L.

For our second use of the scale we’re going to put I and J on one side of the scale.and on the other, we’ll put K and A.

Now we know from our first use of the scale that A is not the odd ball. I could have chosen any 1 of them to serve as the “good” ball here.

If the scale is now balanced we know that L is the odd ball and by comparing it to any of the others, with our 3rd and last use of the scale, will determine whether it is heavier or lighter.

If the scale is NOT balanced we, again, have to take note of which sides the balls are on.

For the sake of argument lets say I and J are on the heavy side.

Compare I against J. This would be the third use of the scale.

If they balance we know K is lighter than all the other balls.

If they DO NOT balance the heavier one is the odd ball.



So that’s it.

The important thing in this riddle is to keep track of as much information as you can and introduce as many variables as you can for each use of the scale. That’s why I stressed that you somehow record which state (heavy or light) is associated with each ball and also why known “good” balls were introduced into the plot.

It’s tough to explain the reasons for each move but if you have a question I’d be happy to answer it.

Oh, and if you feel I missed a step or left something out just let me know that too.