ratsy wrote:

M57 wrote:

ratsy wrote:

This one is for the mathematicians among you:  

You have 12 balls identical in size and appearance but 1 is an odd weight (could be either light or heavy). 

You have a set scales (balance) which will give 3 possible readings: Left = Right, Left > Right or Left < Right (ie Left and Right have equal weight, Left is Heavier, or Left is Lighter). 

You have only 3 chances to weigh the balls in any combination using the scales. Determine which ball is the odd one and if it's heavier or lighter than the rest. How do you do it?

This is more of a logic problem. I have been unable to solve it.  I can do it with 9 balls but even there, I'm having trouble distinguishing if the odd ball is heavier or lighter.  There are some cases where I'm sure, but..

It took me the better part of a year to finally figure this out.  Don't give up too quick!

I have solved the problem, in part by reducing it to 4 balls with 2 weightings.

Ozyman wrote:

AttilaTheHun wrote: I have long ears and hop when I travel, but require no sustenance. What am I?

Some kind of radio signal or  network?  They can use 'rabbit ear' antennas.  Each bounce or re-transmission is referred to as a hop.

 

Either that or Zombie Bunnies.

Lol, better than my guesses, but the initial broadcast would need energy AKA sustenance.

It took me a second, but I can picture the shoelaces now. =)

My belt holds my pants up, but the belt loops hold my belt up, and my pants hold the belt loops.  Which one is the hero?

If the VP is such a VIP, shouldn't we keep the PC on the QT, because if it leaks to the VC he might end up an MIA and we'd all be put on KP?

itsnotatumor wrote:

My belt holds my pants up, but the belt loops hold my belt up, and my pants hold the belt loops.  Which one is the hero?

The unsung heroes are the buttons and zippers that make all of that even possible.

M57 wrote:

ratsy wrote:

This one is for the mathematicians among you:  

You have 12 balls identical in size and appearance but 1 is an odd weight (could be either light or heavy). 

...

I have solved the problem, in part by reducing it to 4 balls with 2 weightings.

Interesting.    I don't see how you can go from four balls to one with one weighing, even if you know the heavy/light attribute, so I'd like to see your solution.

I think I also have a solution, but in mine after the second weighing the outlier is known to be one of a group of 1, 2, or 3 balls depending on the outcomes.

Are we ready for spoilers?  If not, maybe you could PM your solution to me M, and I'll return the favor if you're interested.

NewlyIdle wrote:

M57 wrote:

ratsy wrote:

This one is for the mathematicians among you:  

You have 12 balls identical in size and appearance but 1 is an odd weight (could be either light or heavy). 

...

I have solved the problem, in part by reducing it to 4 balls with 2 weightings.

...after the second weighing the outlier is known to be one of a group of 1, 2, or 3 balls depending on the outcomes.

This is what more what My solution looks like...

itsnotatumor wrote:

If the VP is such a VIP, shouldn't we keep the PC on the QT, because if it leaks to the VC he might end up an MIA and we'd all be put on KP?

Adrian Cronour.  +12

AttilaTheHun wrote: I have long ears and hop when I travel, but require no sustenance. What am I?

I'm thinking change it a bit and it's brilliant, "I have long ears, a little bow mouth, and hop and flop when I travel, but require no sustenance."

NewlyIdle wrote:

M57 wrote:  I have solved the problem, in part by reducing it to 4 balls with 2 weightings.

Interesting.    I don't see how you can go from four balls to one with one weighing, even if you know the heavy/light attribute, so I'd like to see your solution.

I think I also have a solution, but in mine after the second weighing the outlier is known to be one of a group of 1, 2, or 3 balls depending on the outcomes.

Are we ready for spoilers?  If not, maybe you could PM your solution to me M, and I'll return the favor if you're interested.

What I meant is I made up a reduced problem .. solve if there were only 4 balls and you had only  2 weightings, which sounds the same as yours.  Of course, that is only one scenario.

Would you mathematicians get a room already?! :-D

M57 wrote:

M57 wrote:  I have solved the problem, in part by reducing it to 4 balls with 2 weightings.

What I meant is I made up a reduced problem .. solve if there were only 4 balls and you had only  2 weightings, which sounds the same as yours.  Of course, that is only one scenario.

Ah, I didn't understand what you meant.  But since I can't easily solve 4 balls with two weighings, nor reduce 12 balls to 4 with one weighing, I'd still like to see your solution.  I don't think it can be the same as mine.

the problem (i think) is that it could be heavier OR lighter. easy to do if you know the ball is one or the other, but with both possibilities it seems to add an extra step no matter what you do.

weathertop wrote:

the problem (i think) is that it could be heavier OR lighter. easy to do if you know the ball is one or the other, but with both possibilities it seems to add an extra step no matter what you do.

Here's the Four Ball Solution, which is really a subset of the 12 Ball solution.  Balls in question are 1, 2, 3, and 4. Relative weight of the defective ball is not known. You only get two weightings.

First Weighing  123 vs 567 (remember, you have an entire set at your disposal).

If they are equal, then 4 is your ball. Weigh it against any other ball to determine if it's heavy or light.

Otherwise, you've narrowed your suspects down to three AND you now know the relative disposition of the defective ball.  In this case, let's say we now know one of them is heavier.

Second Weighing 1 v 2 to determine which is a heavy ball.

If they're equal, 3 is a heavy ball.

that doesn't make any sense to me. first, it ignores 8-12 completely as if they're already known to not contain the odd ball. second, if the weighing was not equal, you'd have no clue as to which side had the odd ball (OP said could be heavier OR lighter).

on the other hand you COULD do something like this:
First weighing: 1-4 v 9-12
if unequal then 5-8 v 1-4 (to determine which is the odd set)

if equal then you're down to your 4ball weighing
5,6 v 7,8
use the 'heavier' of the two to compare with two of the known balls from the first weighing
you'll then narrow down to which two, and if the odd one is heavier or lighter, from there one more weighing solves it.

WT.. I'm not solving the 12 Ball problem - where 1 of 12 balls is defective, and you have three weightings.  

I'm solving a separate 4 Ball problem where 1 of 4 balls is defective, and the nature of its defect is not known (heavy or light), and you only get two weightings.  This is actually one of the sub-problems you need to solve in order to solve the larger 12-ball problem.

In order to solve the 4-ball problem, you need balls 567 (which would be known to be non-defective after the first weighing if you were solving the 12-ball problem).

'Top your close... but always think about your weightings as information gathering. You get info from the two things you compare AND the control group (what's not on the scale).

M57: you can't have a 4 ball problem with 3 extra 'known' balls...that'd be 7 ball problem

Ratsy, thought i did. if you knew the ball was EITHER heavy OR light, you could do 12 balls in 3 weighs. without  knowing which, you have to add 1 more weigh (or 2 s depending on result of first weigh) 

it takes you 1 if you're lucky - 2 if you're not - to get down to 4 balls. from there i can't see a way to get it in 2 weighs, you need 3 (unless you know the heavy/light or you get lucky on the 2nd weigh). which means you need to get lucky twice - and i can't even do that in a year...

you could do 2v2, then 1v1 & 1v1 or

1v1, if equal same1vanother1, if unlucky to be equal again, you'd need a third with the last ball

M57: you can't have a 4 ball problem with 3 extra 'known' balls...that'd be 7 ball problem

I don't think that M is proposing a solution to an original 4ball problem, he's working the problem backwards.
He's starting from the assumption that he found the unique ball, and now he's expanding the "original set size" by adding more and more weighings, and trying to find his way up to the original 12 ball set.
It's not really a completely unique problem, he's solving step 2 of the original problem:
Step 1:  Narrow the 12 ball set to 4 -> which he hasn't solved yet
Step 2: Find the 1 in the 4, but to do so he's reincorporate 3 of the identical balls from the 8 he'd already removed.

it takes you 1 if you're lucky - 2 if you're not 

If we're going to use a solution that luck at least once, and maybe twice, why not just use a 1:12 luck and randomly pick a ball.  :)

Both M57 and WT are very close.
I HATED this problem when I was first presented with it.

weathertop wrote:

M57: you can't have a 4 ball problem with 3 extra 'known' balls...that'd be 7 ball problem

OK, Let me re-phrase.  You are playing a game of pool.  There are 15 balls on the pool table.  You are told that either the 1, 2, 3, or 4 ball is defective/illegal. You get two chances to use the scale and you need to identify the illegal ball AND identify it as being either under or overweight.

Now some might call that a 15 ball problem, but I'm calling it a 4 ball problem.  Now I suppose technically, you still might insist on calling it a 7 ball problem, because in order to solve it you need to use three balls other than the 1 2 3 and 4, but to call it a 7 ball problem would be giving a considerable hint as to how to solve it; nowhere in the problem is the number seven mentioned.

Solving this problem is critical to solving the 12 ball problem.