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        riddles >> hard >> 4 coins, one different
        
(Message started by: Grimbal on Nov 18^th, 2005, 7:54am)

Title: 4 coins, one different
Post by Grimbal on Nov 18^th, 2005, 7:54am

The usual story. You have 4 coins. All of them have the same standard weight, except for a fake coin that is too heavy or too light.

How can you find which is the fake coin in 2 weighings?

Your weighing aparatus is a ruler. It rests on a pencil placed exactly in the middle. You can put coins anywhere along its length. You have graduations to place them exactly at any distance from the middle. The further a coin is from the middle, the more it affects the balance. But you only can tell whether the coins balance exactly and, if not, on which side the ruler tilts.

Can you find the fake coin?

Same question with 13 coins and 3 weighings?

Title: Re: 4 coins, one different
Post by beatsamurai on Nov 18^th, 2005, 9:09am

Ok...think i have it for 4 coins.
[hide]
Number coins 1,2,3,4.
In the first weighing, the two coins on one side are
stacked at half the distance from center as the other coin.

First weighing -> 1 vs 2,3
If balanced, then 4 is fake and you can use 2nd weighing (1 vs. 4) to find if heavy or light.

Case1 (tilts to left)
2nd weighing 2 vs. 3
If tilts to left -> 3 = L
If tilts to right -> 2 = L
If balances -> 1 = H

Case2 (tilts to right)
2nd weighing 2 vs. 3
If tilts to left -> 2 = H
If tilts to right -> 3 = H
If balances -> 1 = L[/hide]

Hope I didn't overlook something :)

Title: Re: 4 coins, one different
Post by Grimbal on Nov 19^th, 2005, 8:44am

No, it is right. It is easier than I thought.
Followup: In the solution I had in mind, the program of weighings is decided at the start. Can you do that?

Title: Re: 4 coins, one different
Post by ChunkTug on Nov 19^th, 2005, 3:27pm

If you wanted you could extend the question to include "at most one is fake." (for 4 coins at least)

[hideb]Weighed with the single coin twice as far from the pencil than the pair of coins.
1 vs 23
2 vs 34[/hideb]

[hideb]

[hide]Left,Left[/hide]	[hide]3L[/hide]
[hide]Left,Even[/hide]	[hide]1H[/hide]
[hide]Left,Right[/hide]	[hide]2L[/hide]
[hide]Right,Left[/hide]	[hide]2H[/hide]
[hide]Right,Even[/hide]	[hide]1L[/hide]
[hide]Right,Right[/hide]	[hide]3H[/hide]
[hide]Even,Left[/hide]	[hide]4L[/hide]
[hide]Even,Right[/hide]	[hide]4H[/hide]
[hide]Even,Even[/hide]	[hide]None[/hide]

[/hideb]

Title: Re: 4 coins, one different
Post by ChunkTug on Nov 19^th, 2005, 4:12pm

Ok, I have a solution to 13. All weighings placed with coins stacked at a distance s.t. they would be balanced if all weighed the same.

[hideb]First group the coins as such...
#1 = {1,5,9}
#2 = {2,6,10}
#3 = {3,7,11}
#4 = {4,8,12}

And perform the wieghings...
#1 vs #2#3
#2 vs #3#4

From the solution for 4 we will know either:
(Case 1): Which group (#1,#2,#3, or #4) has the fake and whether it is heavy or light.
-or-
(Case 2): 13 is the fake (or possibly fake if "at most one is fake" and we've just eliminated 1-12 as possibilities)

The solution extends naturally to 13 if we add
1,2,3,4 vs 5,6,7,8,13

(Case 1): We know we have a fake and whether it is heavy/light. So this weighing will tell us which range 1-4,5-8,9-12 it is in. Groups #1,#2,#3,#4 contain exactly one coin in each of these ranges.

(Case 2): We've determined the only possible fake is 13, this will tell us whether it is heavy, light, or not fake (if balanced).[/hideb]