Puzzle time!

[fluffymark]

Insomnia makes me do odd things. Last night I decided to twist a well known puzzle to make it more evil, and then solve it, which I did. And this stopped me thinking of monsters under the bed and other scaries (I read the Nightmares and Fairy tales graphic novel yesterday, and it may have gotten to me...ooops), so was actually a good thing, I'm trying to tell myself.

Anyway, lets see which of you clever people solves this first: (all comments initially screened)

You have a balance scale and a big bag of marbles. The marbles are all identical in weight. You take 39 of the identical marbles and mix in another special marble (so 40 marbles now), which may be heavier or lighter than the others, or may not, but looks identical, so can be identified by weight alone. Allowing only 4 weighings, how can you identify whether the special marble is heavier or lighter than the others, and if so, which marble it is?

Update: Congratulations to the clever ixwin who solved it by adding more marbles from the bag. So, lets make the problem HARDER. Assume the bag only has one more marble left in it. You have the original 40 marbles (39 identical and one special) and a bag with one more marble in it. Now solve it.

Comments

[ixwin.livejournal.com]

Okay, having done some work...

Weighing 1

Weigh 14 of the unknown against 13 of the unknown and the 1 known.

If they balance, you know it's one of the 13 not on the scales, and you can solve as the previous case (as you now know that all of the marbles on the scales are of standard weight).

If they don't balance (and assuming the side with 14 of known weight was heavier - obviously just reverse light and heavy if the reverse is true),call the ones on the heavy side, possibly-heavy marbles, and the ones on the light side possibly-light marbles. You have 14 possibly-heavy, 13 possibly-light and 14 of known standard weight.

Weighing 2

Put 5 possibly-heavy marbles and 9 possibly light marbles on the left side of the scales, and 4 possibly-light and 10 known-weight marbles on the right side. This leaves 9 possibly-heavy marbles not in this weighing.

If the scales balance, you know the odd marble is heavier and one of the 9 not currently on the scales. It reduces to the previously solved case, where you are choosing between 9 and the direction is known and you have 2 weighings left.

If the left side of the scales rises, you similarly know that the odd marble is lighter, and one of the 9 possibly lighter ones on the left side. Again, you've reduced it to the previous case.

If the right side of the scales rises, you know it's either one of the 5 possibly-heavy ones on the left hand side, or one of the 4 possibly-light ones on the right hand side. In which case, proceed to weighing 3.

Weighing 3
Put 3 of the possibly-heavy and 3 of the possibly-light on one side of the scale, and 6 of known weight on the other.

If the left side rises you know it's one of the 3 possibly-light ones on the left hand side, and likewise if it falls you know it's one of the three possibly-heavy ones on the left hand side. (either way, reducing it again to the previously solved case for the final weighing)

If it balances, you know it's one of the two possibly-heavy ones, or the one possibly-light one that were left off the scales for weighing 3. In which case, weighing 4 comprises one possibly-heavy, and one possibly-light against 2 of known weight, which gives you your final answer by the same process of deduction as used for weighing 3.

[fluffymark]

We have a winner. :) Easy when you know how. Very clever Liz! :)

And for my next trick, take 120 identical marbles and the special one, and 1 more in the bag, and solve it with 5 weighings...