How is your logic?

Question

Among twelve similar coins there is one counterfeit. It is not known whether it is lighter or heavier than a genuine one (all genuine coins weigh the same). Using three weighings on a pan balance, how can the counterfeit be identified and in the process determined to be lighter or heavier than a genuine coin?

Answer 1

You most certainly can determine which is counterfeit and whether it is heavier or lighter. In fact, for a number of weighings W, you can determine this for up to (3^W-3)/2 coins. So 2 weighings would work for up to 3 coins, 4 weighings for up to a whopping 39 coins, etc. Here is how to do it for 12 coins:
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Divide the coins into 3 stacks of 4 each. Attempting to list the remaining steps descriptively will simply obfuscate the solution, so let's letter all the coins (ABCDEFGHIJKL). So now we have 3 stacks of coins: (ABCD), (EFGH), and (IJKL).

Weight (ABCD) against (EFGH).

IF THE FIRST WEIGHING IS BALANCED:

The counterfeight is obviously within (IJKL). Take any known good coin (say, A) and pair it with an unknown (I), and weigh these two against two more unknowns (J,K).

...if (AI) balances with (JK), then (L) is obviously the counterfeit, so simply weigh it against any other coin to determine if it is counterfeit-light or counterfeit-heavy.

...if (AI) is heavier than (JK), then either (I) is counterfeit-heavy or one of (JK) is counterfeit-light. Remove (AI) and weight (J) against (K). If they are balanced, then (I) is counterfeit-heavy. If they are not balanced, then whichever is lighter is counterfeit-light.

...if (AI) is lighter than (JK), then either (I) is counterfeit-light or one of (JK) is counterfeit-heavy. Remove (AI) and weigh (J) against (K). If they are balanced, then (I) was counterfeit-light. If unbalanced, then whichever is heavier is counterfeit-heavy.

IF THE FIRST WEIGHING IS IMBALANCED, and (ABCD) IS HEAVIER THAN (EFGH):

Either (ABCD) contains a counterfeit-heavy coin, or (EFGH) contains a counterfeit-light coin.

Take any known good coin (say, I) from the stack of known good coins and pair it with one coin from each of the two unknown stacks (AE). Weigh this against two coins from one unknown stack (BC) and one coin from the other unknown stack (F).

...if (AEI) balances with (BCF), then the counterfeit must be within (DGH). We already know that if it is (D) then it must be counterfeit-heavy, but either of (GH) can be counterfeit-light, so we weigh (G) against (H). If they are balanced, then (D) is counterfeit-heavy. If they are unbalanced, then whichever is lighter is counterfeit-light.

...if (AEI) is heavier than (BCF), then either (A) is counterfeit-heavy, or (F) is counterfeit-light (remember from the first weighing, either one of [ABCD] is heavy, or one of [EFGH] is light). Simply pick one of the two (say, A) and weigh it against a known good coin (I). If they are balanced, then (F) is counterfeit-light. If they are not balanced, then (A) will be shown as counterfeit-heavy.

...if (AEI) is lighter than (BCF), then either (E) is counterfeit-light, or one of (BC) is counterfeit-heavy (remember from the first weighing, either one of [ABCD] is heavy, or one of [EFGH] is light). We already know that if it is (E) then it must be counterfeit-light, but either of (BC) can be counterfeit-heavy, so we weigh (B) against (C). If they are balanced, then (E) is counterfeit-light. If they are unbalanced, then whichever is heavier is counterfeit-heavy.

IF THE FIRST WEIGHING IS IMBALANCED, and (ABCD) IS LIGHTER THAN (EFGH):

Either (ABCD) contains a counterfeit-light coin, or (EFGH) contains a counterfeit-heavy coin.

Take any known good coin (say, I) from the stack of known good coins and pair it with one coin from each of the two unknown stacks (AE). Weigh this against two coins from one unknown stack (BC) and one coin from the other unknown stack (F).

...if (AEI) balances with (BCF), then the counterfeit must be within (DGH). We already know that if it is (D) then it must be counterfeit-light, but either of (GH) can be counterfeit-heavy, so we weigh (G) against (H). If they are balanced, then (D) is counterfeit-light. If they are unbalanced, then whichever is heavier is counterfeit-heavy.

...if (AEI) is lighter than (BCF), then either (A) is counterfeit-light, or (F) is counterfeit-heavy (remember from the first weighing, either one of [ABCD] is light, or one of [EFGH] is heavy). Simply pick one of the two (say, A) and weigh it against a known good coin (I). If they are balanced, then (F) is counterfeit-heavy. If they are not balanced, then (A) will be shown as counterfeit-light.

...if (AEI) is heavier than (BCF), then either (E) is counterfeit-heavy, or one of (BC) is counterfeit-light (remember from the first weighing, either one of [ABCD] is light, or one of [EFGH] is heavy). We already know that if it is (E) then it must be counterfeit-heavy, but either of (BC) can be counterfeit-light, so we weigh (B) against (C). If they are balanced, then (E) is counterfeit-heavy. If they are unbalanced, then whichever is lighter is counterfeit-light.

Answer 2

You can't. Weighing is a binary valued decision (equal, not-equal) and not-equal decomposes into greater-than and less-than so, in three weighings, you can get ot one of 2*2^3 = 2^4 = 16 possible outcomes. (Why not 3^3? Because the greater-than or less-than is a joint distribution which *only* occurs if the case is not-equal)

But the bad coin may be one of 12 and there 2 ways for it to be bad so there are 2*12 = 24 possible outcomes to be tested.

You can't get there from here. In some cases you *will* be able to do it in three but, in some cases, you will require 4 weighings.


Doug

For Iberius
And if the balance is equal, yes, you can tell where the 'off' coin is in the remaining 2 groups of 2 coins with 2 weighings.

But if they *don't* ballance, your screwed because you don't know if it's the heavier side or the lighter side that has the phoney coin.

Answer 3

First of all put six coin in one pan and six on the other. You will see one is either lighter and the other is heavier. Choose six coins which are heavier. Out of these six put three in one pan and three on the other. If it weigh equal than the required coin is lighter, otherwise one pan out of the two will be heavier. Similarly take the 6 coins which were lighter, out of this put three coins in one pain and three on the other. If it weigh equal then the required coin is heavy otherwise lighter

Answer 4

1.weigh 6,6 at a time.
2.whichever pan goes down weigh 3,3.
3.if they balance we know the counterfeit coin is lighter.
4.weigh 1,1.if it balances the left out coin is the culprit.if it doesn't we know the heavier one is the culprit
5.however if after step 2 it doesn't balance we know the counterfeit coin is lighter
6.weight 1,1
7.if they balance the left out coin is the culprit,if they don't the one which is lighter is the culprit

Answer 5

well, doug has a limited intelect, and no imagination. i won't give it all awFirst, let's deal with 12 marbles, call them ABCDEFGHIJKL:

First weighing: Compare ABCD and EFGH.

If ABCD=EFGH, you know the special marble is among IJKL. Use your 2nd weighing to compare AI and JK (you know A is ordinary):
If AI=JK, you know L is the odd marble. You may compare L and A in the 3rd weighing to determine if L is heavy or light.
If AI and JK are not equal, you know L is ordinary. Use your 3rd weighing to compare J and K.
If J=K, you know I is the special marble (the result of the second weighing tells you if it's heavy or light).
Otherwise, the special one is either J or K and you can tell which: It's the heavier one if we had AIJK.
If ABCD is not equal to EFGH, you know IJKL are ordinary (we may use L as a known ordinary marble in the rest of the procedure). Also, you will always know from this first weighing whether the special marble is heavy or light, once you've determined which it is.
Use the 2nd weighing to compare ABE and CFL :
If ABE=CFL, you know the special marble is among DGH. You may use the 3rd weighing to compare G and H :
If G=H, then D is the special marble.
If G and H are not equal, the special marble is the heavier one if we had ABCD If ABE and CFL are not equal, the special marble is among ABCEF and we may distinguish four cases :
ABCD>EFGH and ABE>CFL :
Either F is light or one of AB is heavy.
Compare A and B in the 3rd weighing to find out.
ABCD>EFGH and ABE Either E is light or C is heavy.
Compare C and L in the 3rd weighing to find out.
ABCD Either F is heavy or one of AB is light.
Compare A and B in the 3rd weighing to find out.
ABCDCFL :
Either E is heavy or C is light.
Compare C and L in the 3rd weighing to find out.
ay, but the first weighing has only four coins on each side

Answer 6

by weighing them all you would be able to tell by the one that is either lighter than the others or heavier!

Answer 7

Uhh, the one that off-balances the scale is the conterfeit. WOW.

Answer 8

You can't

Answer 9

i dont know - how it can be done in only 3 weighings - tell us please:)