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

SamVS Wed 27 May 2015, 23:55

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 (!).

Jamie Wed 27 May 2015, 23:59

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

You did better than me! Full credit to you Paul for solving it! Smile

By the way, I think your solution may be incomplete. Assuming the 1st weighing is unbalanced; what happens if the second weighing is balanced? Pretty straight forward and trivial really. It means one of the balls that made weighing 1, and put aside for weighing 2, is the culprit. Which is just 3 balls, and we know 1 is a potential light rouge, and the other 2 are potential heavy rouges; which you'd just weigh against each other, and Bob's your uncle!

PaulC Thu 28 May 2015, 00:01

Yes, lost track of my balls there.

With 2 pot. Heavy and one pot. Light, left off the scales, if they balance you with the two pot. Heavies. If THEY balance the pot. Light is the rogue otherwise it's the heavier of the pot. Heavies.

Jamie Thu 28 May 2015, 00:02

Also, interesting that there are multiple solutions to this! Smile

PaulC Thu 28 May 2015, 00:04

Jamie wrote:
Paul wrote:Im clearly as good at explaining logic problems as I am at teaching games.

You did better than me! Full credit to you Paul for solving it!

By the way, I think your solution may be incomplete. Assuming the 1st weighing is unbalanced; what happens if the second weighing is balanced? Pretty straight forward and trivial really. It means one of the balls that made weighing 1, and put aside for weighing 2, is the culprit. Which is just 3 balls, and we know 1 is a potential light rouge, and the other 2 are potential heavy rouges; which you'd just weigh against each other, and Bob's your uncle!

Yup, that was Sam's worry too. Luckily our solutions were the same. Smile

Do I get to work for Google now?

SamVS Thu 28 May 2015, 00:09

Oh, duh, I get it now.

SamVS Thu 28 May 2015, 00:12

Give us another one!

Jamie Thu 28 May 2015, 00:14

Paul wrote:Yup, that was Sam's worry too. Luckily our solutions were the same.

They seemed a little different to me; on the second weighing, I use all 4 balls from set A (2 on each side of the scale); you're only using 3 balls from set A (2 on one side, 1 on the other). Very close though! Smile

Paul wrote:Do I get to work for Google now?

Nope; far too clever for them!

I've got more logic puzzles; may have to wait for next Tuesday night to unleash them though; don't want to be responsible for Sam blowing a fuse! Wink

Jamie Thu 28 May 2015, 00:15

Sam wrote:Give us another one!

Okay, you asked for it! I'll post on another thread. Here you go Sam: https://www.sheffieldboardgamers.com/t612-eye-colour-problem

PaulC Thu 28 May 2015, 09:08

Jamie wrote:
Paul wrote:Yup, that was Sam's worry too. Luckily our solutions were the same.

They seemed a little different to me; on the second weighing, I use all 4 balls from set A (2 on each side of the scale); you're only using 3 balls from set A (2 on one side, 1 on the other). Very close though!

Paul wrote:Do I get to work for Google now?

Nope; far too clever for them!

I've got more logic puzzles; may have to wait for next Tuesday night to unleash them though; don't want to be responsible for Sam blowing a fuse!

I meant my solution to the sub-problem highlighted by Sam was the same as yours. Overall, yes, our solutions were different. Smile

PaulC Thu 28 May 2015, 09:14

Jamie wrote:
Paul wrote:Yup, that was Sam's worry too. Luckily our solutions were the same.

They seemed a little different to me; on the second weighing, I use all 4 balls from set A (2 on each side of the scale); you're only using 3 balls from set A (2 on one side, 1 on the other). Very close though!

Paul wrote:Do I get to work for Google now?

Nope; far too clever for them!

I've got more logic puzzles; may have to wait for next Tuesday night to unleash them though; don't want to be responsible for Sam blowing a fuse!

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