Warrior General Games

Originally posted by pabloex
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.

Nope, there is a process to guarantee it, where luck is not a factor. The hard part is finding that process

Unless you already know the process ... but then I have stayed out of this one for a reason to date. The answer is 2 to 3 weighings. Not going to show my work yet though for the others.

Originally posted by AWolf02
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




Ok, Wally, don't read this beyond the next paragraph

Don't forget to find out whether the coin is heavier or lighter, so in your first section it would take a third weighing to determine that

First hint is that the smallest max is definitely less than 6. The second is that your first step was correct.

Originally posted by fnordish
Unless you already know the process ... but then I have stayed out of this one for a reason to date. The answer is 2 to 3 weighings. Not going to show my work yet though for the others.

haha, thats cool. how long did it take you to get it the first time though?

I can't remember if I figured it out on my own or not. It was in a book I read. Hero did the solve in 3-4 method, and got cocky. Someone I forget who put them in thier place with the 2-3 method. Was around 20 years ago I read it. Story also included the 3 missionaries and 3 canabals and the boat that holds 2, and the towers of hanoi puzzles.

I think I figured it out....answer coming when I get it into a way to explain it.

Ok so you have the 12 coins

Divide into 3 stacks of 4 (A,B,C)

Weigh #1

Compare A to B, if equal you know it is in C...go to option 1

If unequal to go option 2.

Option 1


Weigh #2
Option 1
Weigh three of the c’s against three of the a’s. If they are unequal you know one of the c’s is the fake, and you know if it is heavier or lighter…if it is equal you know the other c is fake

Weigh #3
If they were equal then you take the remaining c and compare it to an a, if it is heavier it is the fake and it is heavier; if lighter its lighter and the fake

If they were unequal then you know if the fake is lighter or heavier…so compare 2 of the three c’s, if they are unequal you have your fake, if they are equal then the fake is the remaining c


Option 2 (lets say a was heavier…if b was heavier then change all the b’s to a’s and a’s to b’s that follow)

Weigh #2

Compare the combination of 2a’s and 2b’s with the combination of 1a1b and 2c’s.

If the 2a2b side is heavier then one of the a’s on that side is fake heavy or the opposite side 1b is fake light
If the 2a2b side is lighter then one of the b’s on that side is fake light or the opposite a is fake heavy

If equal then you know either the remaining a is heavy or the remaining b is light so simply compare the a with a c and you have your answer (that would be 3 weighs)


If they were unequal and the 2a2b side was heavier then compare the two a’s. Whichever is heavier is fake, and if equal then the b from the other side is fake light.

If they were unequal and the 2a2b side was light then compare the two b’s. Whichever is lighter is fake, and if equal then the a from the other side fake heavy.

That was tough...if any of you guys saw it before I edited it I came up with what I thought was right once and was wrong...i'm going to hope this is right and i'm not doing bad math again.


I wouldn't have gotten the answer (would have stopped at assuming 4 was the answer) if you guys hadn't said in this thread that 3 was the right answer...so unfortunately I don't pass the test haha

Originally posted by NYUfresh
Ok so you have the 12 coins

Divide into 3 stacks of 4 (A,B,C)

Weigh #1

Compare A to B, if equal you know it is in C...go to option 1

If unequal to go option 2.

Option 1


Weigh #2
Option 1
Weigh three of the c’s against three of the a’s. If they are unequal you know one of the c’s is the fake, and you know if it is heavier or lighter…if it is equal you know the other c is fake

Weigh #3
If they were equal then you take the remaining c and compare it to an a, if it is heavier it is the fake and it is heavier; if lighter its lighter and the fake

If they were unequal then you know if the fake is lighter or heavier…so compare 2 of the three c’s, if they are unequal you have your fake, if they are equal then the fake is the remaining c


Option 2 (lets say a was heavier…if b was heavier then change all the b’s to a’s and a’s to b’s that follow)

Weigh #2

Compare the combination of 2a’s and 2b’s with the combination of 1a1b and 2c’s.

If the 2a2b side is heavier then one of the a’s on that side is fake heavy or the opposite side 1b is fake light
If the 2a2b side is lighter then one of the b’s on that side is fake light or the opposite a is fake heavy

If equal then you know either the remaining a is heavy or the remaining b is light so simply compare the a with a c and you have your answer (that would be 3 weighs)


If they were unequal and the 2a2b side was heavier then compare the two a’s. Whichever is heavier is fake, and if equal then the b from the other side is fake light.

If they were unequal and the 2a2b side was light then compare the two b’s. Whichever is lighter is fake, and if equal then the a from the other side fake heavy.




very good, and don't worry. When i first did it i assumed that 4 was the was the lowest also. So i didn't get it right the first time either.

good job!

i'll post another one in a few days

If you would have taken the weight of 3 and 7 and divided it by 16 to the 3rd power and put it in the wash machine you would have got the same answer.

well done!

i think i just mentally gave up on this one, haha

huh?

this is boring..

nah - gimme a SWOT - anallysis or something business related

man some of youg guys sure dont have a lot of spare time activities..

peace

Originally posted by Dix-11
huh?

this is boring..

nah - gimme a SWOT - anallysis or something business related

man some of youg guys sure dont have a lot of spare time activities..

peace

haha, what kind of analysis are you referring to?

definitely not interested in business. and the kind of analysis i am used to i find boring compared to logic puzzles and my favored subjects in math.

Strength, Weakness, Opportunities, Threats.

Most of the SWOT analysis I have seen people do is largely a waste of time. It is mostly just "duh".

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