Warrior General Games

Originally posted by wallysmith
3 weighings minimum, and 4 max:

1st weighing: Weigh 6 of 12, 3 on each side
If equal: These are all real; set aside and grab other 6 coins
If imbalanced: The counterfeit is in these 6; use in next weighing

2nd weighing: Weigh 4 of 6, 2 on each side
If equal: These 4 are real; one of the other two is counterfeit (see option 1)
If unequal: One of these 4 is counterfeit; the other two are real (lets call these two pairs A and B; see option 2)


Option 1:
3rd weighing: Take one guaranteed real coin (coin Z) in with the pair that has the counterfeit. Weigh coin Z with coin A of the counterfeit pair; leave coin B out
If equal: Coin B is counterfeit
If unequal: Coin A is counterfeit


Option 2:
3rd weighing: Weigh pair A, 1 on each side
If equal: These are both real; the counterfeit is in pair B
If imbalanced: The counterfeit is in pair A

4th weighing: Take one guaranteed real coin (coin Z) in with the pair that has the counterfeit. Weigh coin Z with coin A of the counterfeit pair; leave coin B out
If equal: Coin B is counterfeit
If unequal: Coin A is counterfeit


This was more difficult than anticipated. I've heard of something like this before (which definitely helped), but it was 10 balls, one of them is heavier, and you try to figure out which one.

So the answer (which I hope is right, and I'm staying late at work to figure it out dammit), is a minimum of 3 weighings, and a max of 4.

great try!

but, it is not the lowest max you can find. keep looking and i will post a hint later if y'all like.

djf;lasdjfa;lsjfa;ljasj

ok bck to the drawing board it looks like

You can do it, wally. I have faith.

I got to go POOP

Dont weigh them at all. Take them to a vending machine and put them in. The one that weighs different wont be accepted. Hit coin return and get your money back.

the minimum number possible is 2. You choose 2 coins and get lucky and one of them is the couterfeit. Then you test one of those coins against another coin and get lucky again and you find out the weight.

Originally posted by kimimar0
the minimum number possible is 2. You choose 2 coins and get lucky and one of them is the couterfeit. Then you test one of those coins against another coin and get lucky again and you find out the weight.

lol, the problem is to find the minimum maximum for the solution set

That means you are looking for the smallest number of weighs that guarantee you will have the answer

Originally posted by AWolf02
Dont weigh them at all. Take them to a vending machine and put them in. The one that weighs different wont be accepted. Hit coin return and get your money back.

lol, not that kind of problem, plus i said all you had were the balance, the coins, and yourself.

Vending machines are not available :-/

what a sad, sad world i created

My money is on Wally, this is too much thinking for me..

I still think the vending machine was a great answer

This damn riddle will NOT leave my brain! I see where my initial flaws were, but now I can't close the deal. And it's killing me a;kldjfasl;fjasl;fjas;

By the way, please no hints... I have faith in myself (but it's slowly shrinking)

The wording of the problem bothers me because of the word 'guarantee'.

You have 12 coins and the fake has two possible states, heavier or lighter. That means that there are 24 possible outcomes.

A single weighing of an equal number of coins can produce 3 results, left side tilt, right side tilt or balanced. A second weighing produces a possible 9 results and third weighing produces 27 results.

So in theory, you could arrive at the correct answer in 3 weighings. However, it doesn't guarantee that you will. In order for a 3 weighing result, you would need to have had a degree of luck in choosing your coins.

So I am stumped. I see no way to guarantee a result of 3 or less.

2 is the lowest...

3 Groups of 4 (A, B, C)

If A=B, then C contains the counterfeit.

Take 3 good coins from A and weigh against 3 from C.

If they are equal the other C coin is the counterfeit.


If A doesnt = B then A or B = C

Weigh A against C, If A=C then the counterfeit is in B but if A doesnt =C the counterfeit is in A.

Take 3 good coins and weigh against 3 from the counterfeit group. If equal then the other coin is counterfeit but if not rotate the counterfeit group until equal.

MAX of 6

You see, this is what happens when the options on the defensive AI aren't working yet, my DC putting riddles to you lot to keep you active.

My brain melted after 10 minutes of trying to work out where Wally went wrong and so I gave up and found a bucket of iced water to put my head in.