12 balls, logical problem (bit like a game), can you solve it?

See if you can solve this logic problem. You'll probably need a pen and paper for this one...

You have 12 balls. 11 balls are the same weight. 1 is either heavier or lighter than the rest (you don't know which). You have a set of scales, you can use 3 times to compare one set of balls with another (they tell you if set A is heavier or lighter than set B, or if set A and set B are the same weight).

So, using the scales only 3 times, work out which of the 12 balls is the odd one, and also, if it is heavier or lighter than the other 11.

Go! (no cheating with Google)...

Jamie wrote:See if you can solve this logic problem. You'll probably need a pen and paper for this one...

You have 12 balls. 11 balls are the same weight. 1 is either heavier or lighter than the rest (you don't know which). You have a set of scales, you can use 3 times to compare one set of balls with another (they tell you if set A is heavier or lighter than set B, or if set A and set B are the same weight).

So, using the scales only 3 times, work out which of the 12 balls is the odd one, and also, if it is heavier or lighter than the other 11.

Go! (no cheating with Google)...

Easy ! Pick up all the Balls and the Scales and throw them at Jamie as hard as you can, you wont know which is heavier/lighter but will feel less frustrated and not being able to work out the stupid puzzle

Jamie wrote:CAN YOU SOLVE IT?

NOPE

It's definitely solvable, without any trickery! You'd probably need a pen and paper for it though (you would have to super genius to do it all in your head). Everyone give up?...

Jamie wrote:It's definitely solvable, without any trickery! You'd probably need a pen and paper for it though (you would have to super genius to do it all in your head). Everyone give up?...

No, will solve later - beware spoilers!

Mattel wrote:

Jamie wrote:It's definitely solvable, without any trickery! You'd probably need a pen and paper for it though (you would have to super genius to do it all in your head). Everyone give up?...

No, will solve later - beware spoilers!

OK, I think I have a solution and lots of crossing out on a piece of paper.

4 groups of 3

Pick any 2 groups of 3... And weigh

3vs3 same (move to B) or pick heavy group (move to A)

A - pick 2 to weigh - if balanced 1 left out is heaviest... Otherwise scales tip to heaviest. (2 turns total)

B - use other 2 groups find heaviest group and move to C (2 turns so far)

C - pick 2 of that group to weigh - if balanced 1 left out is heaviest... Otherwise scales tip to heaviest. (3 turns total)

Ah just seen what I missed - you don't know it's heavier... Back to thinking...

Yep, you need to also work out if the oddball is heavier or lighter; it's a bit fiendish!!

Bloody hell Jamie! I've got better things to do than solving your problems...

I've figured it using up to 4 uses of the scale, just need to iron out the wrinkle...

Paul wrote:Bloody hell Jamie! I've got better things to do than solving your problems...

I've figured it using up to 4 uses of the scale, just need to iron out the wrinkle...

hehe I love this kind of puzzle Paul.

As a bit of clue; you should try and do weighings, such that you can get 3 results; 1. Scales tip one way. 2. Scales tip the other way. 3. Scales balance.

I think I've got it.

Should I post my solution now, or wait a little longer...

Not that my solution is at all elegantly written. Will probably just confuse everyone more...

Paul wrote:Not that my solution is at all elegantly written. Will probably just confuse everyone more...

I had the same problem mate. Working it out is one thing. Explaining your solution, in a way that makes sense to other people, is another thing altogether!

Feel free to share solution though.

Please do, Paul, not knowing is killing me.

Here's what I came up with...

Paul wrote: -Take 2 balls from the lighter set and one ball from the heavier set and weigh them against one other ball from each set (2)

I can't process this.

I did warn you...

2 balls from the 4 that were on the lighter side of the scale and one of the four from the heavier side of the scale should be weighed against one ball from each side of the scale plus one unweighed ball.

Better?

Nice one Paul! Haven't read your full solution yet, but your solution is different from mine at this point...

-Take 2 balls from the lighter set and one ball from the heavier set and weigh them against one other ball from each set (2)

Are you weighing 3 balls against 2? (presumably you add one of the known to be good balls?)

At that point, I was weighing 2 balls from the lighter set, plus 1 ball from the heavier set; against the other 2 balls from the lighter set, plus one known to be good ball (from set C in the original weighing).

It's possible there are multiple solutions though!

No, not 3 balls against 2, when I said one from each set, I meant one from A, one from B and one from C. I.e. One from the heavy side, one from the light side and one that wasn't weighed.

Paul wrote:I did warn you...

2 balls from the 4 that were on the lighter side of the scale and one of the four from the heavier side of the scale should be weighed against one ball from each side of the scale plus one unweighed ball.

Better?

Does that leave you with two potential heavies and one potential light off of the scales? How do you determine one of the heavy balls off the scales isn't the rogue?

My solution...

12 balls problem and solution

Problem: 12 balls. 1 is an oddball, find it out, using 3 weighings, also determine if oddball is heavy or light (compared to other balls).

Each weighing needs to reveal as much information as possible. So, we should compare 2 sets of balls, A and B, such that there are 3 possible results (A heavy, B heavy, A and B balanced).

Notation / short-hand I used:
0 = control ball (known not to be the oddball).
1-12 = ball number 1 to 12.
[w?] = weighing number 1 to 3.

There are 24 possible results, 2 (heavy or light) for each of the 12 balls. I’ve checked and all 24 eventualities are accounted for in the logical result paths below.



1.2.3.4 v 5.6.7.8 [w1]

result of w1: balanced ...

9.10.11 v 0.0.0 [w2]

result: balanced (12 is the oddball) ...
12 v 0 [w3]
12 heavy = 12 is heavy
12 light = 12 is light

result: 9.10.11 heavy ...
9 v 10 [w3]
balanced = 11 is heavy
9 heavy = 9 is heavy
10 heavy = 10 is heavy

result: 9.10.11 light ...
9 v 10 [w3]
balanced = 11 is light
9 light = 9 is light
10 light = 10 is light

result of w1: 1.2.3.4 heavy ...

1.2.5 v 3.4.0 [w2]

result: balanced (6.7.8 is light) ...
6 v 7 [w3]
balanced = 8 is light
6 light = 6 is light
7 light = 7 is light

result: 1.2.5 heavy (1.2 is heavy) ...
1 v 0 [w3]
balanced = 2 is heavy
1 heavy = 1 is heavy

result: 3.4.0 heavy (3.4 is heavy OR 5 is light) ...
3.5 v 0.0 [w3]
balanced = 4 is heavy
3.5 heavy = 3 is heavy
3.5 light = 5 is light

result of w1: 1.2.3.4 light...

1.2.5 v 3.4.0 [w2]

result: balanced (6.7.8 is heavy) ...
6 v 7 [w3]
balanced = 8 is heavy
6 heavy = 6 is heavy
7 heavy = 7 is heavy

result: 1.2.5 light (1.2 is light) ...
1 v 0 [w3]
balanced = 2 is light
1 light = 1 is light

result: 3.4.0 light (3.4 is light OR 5 is heavy) ...
3.5 v 0.0 [w3]
balanced = 4 is light
3.5 light = 3 is light
3.5 heavy = 5 is heavy

Sam wrote:

Paul wrote:I did warn you...

2 balls from the 4 that were on the lighter side of the scale and one of the four from the heavier side of the scale should be weighed against one ball from each side of the scale plus one unweighed ball.

Better?

Does that leave you with two potential heavies and one potential light off of the scales? How do you determine one of the light balls off the scales isn't the rogue?

No, leaves you with just one potential heavy and one potential light off the scale.

Paul wrote:No, not 3 balls against 2, when I said one from each set, I meant one from A, one from B and one from C. I.e. One from the heavy side, one from the light side and one that wasn't weighed.

That's different to what I did.

I also had 3 vs 3 on my second weighing (assuming first weighing isn't balanced). So, first weighing, I have 3 sets of 4 balls, and weigh set A vs set B. If they don't balance, for weighing 2, on one side, I have 2 from set A and 1 from set B. On the other side I have the other 2 from set A and 1 from set C.

Im clearly as good at explaining logic problems as I am at teaching games.

Paul wrote:
No, leaves you with just one potential heavy and one potential light off the scale.

You couldn't have two balls from the first weighing off the scales if you replaced a weighed ball with an unweighed ball in the second scale. You must have three. You've only put two heavies back on the scales of the original set of four. If the scales balance, which they might do as the spy is in deep cover on the table, you can't determine from one weighing which of the three balls is the rogue if you don't also already know if the rogue is a Sandor or a Tyrion.

You got weigh closer than I did (!).