[Guest]

Forgive me if this is already posted, I found a few counterfeit coin weighing post, but I believe this is a new twist on it.

Given the conditions

You can make a maximum of 3 comparisons

There is exactly 1 coin out of 'X' coins that is a different weight

That coin can be either heavier or lighter

What is the maximum number of coins you can absolutely predict the odd one out from?

Can you propose a formula that may support this? I.E. to determine with 'n' comparisons a maximum of 'x' coins can be used?

12 coins

[Guest]

1 weighing: 1 coin

2 weighings: 4 coins

n>2 weighings: (3^n-3)/2 coins

[Guest]

i'd be very interested in seeing the logic for 12 coins in 3 weighings, my best is 10.

remember the coin can be lighter or heavier, we aren't sure which.

my formula would simply be 3^(n-1)+1

[Guest]

i got 12.

weigh 4 against 4. if equal, then easy.

if not, then weigh 2 light and 1 heavy against 2 light 1 heavy.

the rest is cake.

[Guest]

okay, i think i got 13.

weigh 4 against 4 as before. if equal you have 5 left, with 8 normal coins.

weigh 3 normal against 3 unknown. if heavy or light, you know the special coin is that., and with three coins, one more weigh is all you need. if equal, you have 2 coins left. weigh 1 coin against a normal coin.

is 40 possible with 4 weighings? I suspect so, I'll work on it.

[Guest]

weigh 13 against 13.

if uneven, weigh 4 light 5 heavy against 4 light 5 heavy.

---if uneven, weigh 2 light 1 heavy against 2 light 1 heavy.

------if uneven, weigh two light against one another

------if even, weigh 2 heavy.

---if even, basically same as above.

if even, you have 14 unknown coins. weigh 9 normal against 9 unknown. easy from there.

yeah i know, that was fast :-).

[Guest]

okay, i think i got 13.

weigh 4 against 4 as before. if equal you have 5 left, with 8 normal coins.

weigh 3 normal against 3 unknown. if heavy or light, you know the special coin is that., and with three coins, one more weigh is all you need. if equal, you have 2 coins left. weigh 1 coin against a normal coin.

is 40 possible with 4 weighings? I suspect so, I'll work on it.

The problem is that last "if"

if you are left with 5 coins, 2 weighings can determine which is the odd, but not if it is lighter or heavier. The last "if you have 2 left over" is the kicker. Weighing them against themselves wont determine heavy or light.

[Guest]

I would appreciate any input towards determining some sort of formula, here's what I've deduced.

1 weigh - 0 coins

2 weighs - 3 coins

3 weighs - 12 coins

4 weighs - 36 coins

I am pretty sure 36 is max for 4 weighings, here's my solution for that appending - or + for potential to be normal or light/heavy

Step1: weigh 12 vs 12

If uneven

___Step 2: weigh 1-2-3-4-1+2+3+4+ vs 5-6-7-8-5+6+7+8+

_________if uneven (well say first group was low on scale)

_________Step3: 1+2+1- vs 3+4+2-

_______________if even last step between 3-4-

_______________if uneven weigh whichever 2 potential +

_________if even

_________Step3: 9+10+9- vs 11+12+10-

_________Step4: weigh +'s from low side if uneven else 11-12-

if even

Step2:

You have 3 steps, follow solution for 12 with 3 steps

Once I find some more pennies I'll start working on 5 steps

[Guest]

hey sleeping. the goal isn't necessarily to determine if the odd coin is heavy or light, only point out the odd one.

therefore my solution for 13 works.

[CaptainEd]

Link to relevant thread

If you know the bad coin is light (or heavy), it's a "restricted" problem, and the max coins is 3^n in n weighings.

If you don't know whether it's light or heavy, it's an "unrestricted" problem, and the max coins is (3^n+1)/2 in n weighings, if we can use an additional known good coin for the first weighing, otherwise (3^n-1)/2