Talk:Puzzles/Decision puzzles/12 Coins

For the 12 coins I would split them into 3 groups of four \and call them group A, B, and C Then I would perform 3 weighings. A-B A-C B-C. The two groups that do not contain the fake will balance so the fake is in the other one. Also since I know that the other two groups are genuine i can easilly tell if the fake is lighter or heavier by seeing which way the balancve went when the fake group was weighed against one of the other two.

I now have four coins, and know if the fake is heavier or lighter. So I split in half and weigh then split in half and weigh again.

This gives 5 weighings in total. Theresa knott 09:19, 6 Oct 2003 (UTC)


 * Good, this is a valid solution! If you want you can try and improve it - the optimum is just 3 weighings. Thomas
 * OK I'm thinking out loud here so let's see what I xcome up with:

Split into three groups ABC as before. WEIGH A-B they will either balance or not.

scenario 1 A and B balance

Theresa knott
 * The fake is therefore in C.
 * label the coins s,t,u,v
 * weigh s,t, against u,v. and note which side goes down
 * Now weight s,u, against t,v.
 * from this you can work out the fake.
 * Total 3 weighings

scenario 2 A and B do not balance
 * Note which group A or B was heavier.
 * Split A into two subgroups w,x each with two coins in Split B into y, z.
 * rearrange As wy, xz and reweigh. note which side was heavier.
 * From this you can work out wich pair of coins contains the fake and if the fake is heavier or lighter.
 * Let's say x comes out as heavier than the other three groups, weigh the two coins of x against one another to find the fake
 * Total 3 weighings

(Man I'm proud of myself. That one was tricky)Theresa knott 08:41, 7 Oct 2003 (UTC)


 * You're on the right way... However, there are some little issues still there. For scenario 1 you could for example observe that s,t is heavier than u,v and that s,u is heavier than t,v. But there are still two possibilities left: s could be fake and heavier OR v could be fake and lighter. Similarly, in the second weighing of scenario 2 you will for instance not know whether a coin of w is fake and heavier or a coin of z is fake and lighte. Keep working on it, you're really close! Thomas

Lord Emsworth: Now restaing "A and B do not Balance"


 * Note whether A is heavier or B is heavier.
 * The Heavy coins will be referred to as Hi while the light coins will be referred to as Lt. The coins themselves will become H1, H2, H3, or H4, and also L1, L2, L3, and L4.
 * Obviously, group C has only normal coins: the abnormality is in Hi or Lt. (C is also counted as C1, C2, C3, and C4)
 * Take H1 and put it in Lt, and take L1 and put it in C, creating [1 group: three Lt + one Hi] and [1 group: three C + 1 Lt].
 * NOTE: H1 cannot be lighter than C & L1 cannot be heavier than C
 * IF [three Lt + one Hi] = [three C + 1 Lt] THEN [H2 or H3 or H4] is counterfit and heavier. C is assumed normal. If the two sides are equal, then both sides are normal. This only leaves H2, H3, and H4.
 * IF [three Lt + one Hi] < [three C + 1 Lt] THEN [L2 or L3 or L4] is counterfit and lighter. ''The side with H1 is lighter, so H1 cannot be heavy. Therefore, H1 is normal. The side with L1 is heavier, so L1 cannot be light. Therefore L1 is normal. C is assumed normal. This only leaves L2, L3, or L4 as light and counterfit.
 * IF [three Lt + one Hi] > [three C + 1 Lt] THEN L1 is counterfit and lighter or H1 is counterfit and heavier. Since the Cs are defined as normal but are on the lighter side, two possibilities exist: one coin with the C's is light or one coin with the Lt's is heavy. Thus, either H1 is heavy and counterfit or L1 is light and counterfit.


 * Next Step:
 * [H2 or H3 or H4] is heavier: Compare H3 and H4. If they are equal, then H2 is counterfit and heavy. If H3 is heavier than H4, H3 is counterfit and heavy. If H4 is heavier than H3, H4 is counterfit and heavy.
 * [L2 or L3 or L4] is lighter: Compare L3 and L4. If they are equal, then L2 is counterfit and light. If L3 is lighter than L4, L3 is counterfit and light. If L4 is lighter than L3, L4 is counterfit and light.
 * H1 is counterfit and heavy or L1 is counterfit and light: Compare L1 and C. If L1 is equal to C, then H1 is counterfit and heavy. If L1 is lighter than C, then L1 is counterfit and light.

QED(Is this the right acronym?) Lord Emsworth 00:26, 11 Oct 2003 (UTC)


 * very good explanation Lord Elmsworth! This solves the case where A and B are not in balance. For A and B in balance we will need a similar trick to find the fake coin. Thomas

using scenario 1 A and B balance (Note: if the scale balances we know which is the fake but we do not know if it was lighter or heavier, if the scale tips in either direction the weight difference will be known) *Total 3 weighings
 * The fake is therefore in C.
 * label the coins in pile C {S,T,U,V} and label all the other coins N (for normal weight)
 * weigh S vs T
 * If they weigh the same then S and T can be marked/changed to N
 * If they don't weigh the same then U and V can be marked/changed to N
 * This should leave you with two unknowns (either {S,T} or (U,V))
 * Then weigh either of the two unknowns against one of the N coin
 * If the scale balances the other is the fake if not the one being weighed is the fake