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Thread: Puzzle

102903, 01:29 #1Registered User
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Unanswered: Puzzle
Hi,
There is a puzzle..
Please try to solve it ....
A person has 12 coins among which one is of different weight (HEAVY/LIGHT). The proble is that that person want to find out the faulty coin by weighing only thrice....
so how he can find the faulty coin in 3 weighs.....
Regards
SudeeshSudeesh
www.GaloisTech.com

102903, 05:15 #2Registered User
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 Aug 2003
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how many weighing scale do u have?

102903, 05:36 #3Registered User
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traditional scale only/balance
Originally posted by ynoel
how many weighing scale do u have?Sudeesh
www.GaloisTech.com

102903, 05:45 #4Registered User
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 Sep 2003
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Re: Puzzle
Originally posted by sudeesh
Hi,
There is a puzzle..
Please try to solve it ....
A person has 12 coins among which one is of different weight (HEAVY/LIGHT). The proble is that that person want to find out the faulty coin by weighing only thrice....
so how he can find the faulty coin in 3 weighs.....
Regards
Sudeesh
2) split the group with the faulty coin into 2 groups of 2 each, repeat the procedure, remove the group with the faulty coin
3) compare the 2 coins on the scale, and select the faulty coin...rgs,
Ghostman

102903, 06:19 #5Registered User
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 india, delhi
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Re: Puzzle
Originally posted by GhostMan
1  split the coins into 3 groups of four and weigh 2 groups. If they weigh the same, then you know the 3rd group has the faulty coin. If they dont weigh the same, then you can identify the group with the faulty coin.
2) split the group with the faulty coin into 2 groups of 2 each, repeat the procedure, remove the group with the faulty coin
3) compare the 2 coins on the scale, and select the faulty coin...
Please keep in mind that we don't know whether the faulty coin is heavy or light...Sudeesh
www.GaloisTech.com

103003, 02:07 #6Registered User
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 Oct 2003
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 india, delhi
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Solution to Puzzle
Puzzle:
A person has 12 coins among which one is of different weight (HEAVY/LIGHT). The proble is that that person want to find out the faulty coin by weighing only thrice....
so how he can find the faulty coin in 3 weighs.....
Solution:
I find a solution though its not complete but I could do upto this level only. It has branches . Step are below:
Distribute the 12 coins into
3 groups say I, II, III each
containing 4 coins.
Weigh the any two groupssay I n II,
two cases arise
A) Two groups are same:
then the group III contain
the disimilar coin. say coins are
a, b, c, d.
Weigh any 2 coins say a & b gain
two cases:
i) a = b then either c or d
is disimilar.
Now compare a and c
if a = c then d is faulty.
else c is faulty coin.
ii) a != b then either a or b
is disimilar.
Now compare a and c
if a = c then b is disimilar
else a is disimilar coin.
B) Two bunches are not same:
Then we have 8 coins containing
faulty one.
Again make three bunches say IV,
V, VI containing 3, 3, 2 coins
Compare IV and V, two cases:
i) Both are not same:
V bunch contains disimilar one.
say coins are a, b
Now, compare a with any other coin
if both are same then b is disimilar
else a is disimilar one.
ii) Both are same
Now we have 6 coins containing
faulty one
AND I'M UNABLE TO FIND FAULT COIN
AMONG 6 IN A SINGLE WEIGH. I NEED
2 WEIGHs.
If you can do this please tell me.
Regards
SudeeshSudeesh
www.GaloisTech.com

103003, 09:43 #7Moderator.
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 Sep 2002
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Provided Answers: 1Re: Solution to Puzzle
I have done this before a few years ago, but it's still difficult.
The key to solving it is that you just have to find out which coin is different, you don't have to know whether it was heavier or lighter (you may or may not find that out).
I think I have managed to do it (again)  the difficult part is how to write down the solution! But here goes anyway...
First, let's label the coins A,B,C,D,E,F,G,H,I,J,K,L
1) 1st weighing: A,B,C,D against E,F,G,H
1.1) If both sides same then we know the odd coin is one of I,J,K,L
2nd weighing: I,J,K against A,B,C
If same, L is the odd coin and we are done.
If different, say I,J,K lighter, then we know the odd coin is one of I,J,K AND we know it is lighter.
3rd weighing: I against J
If same, K is the odd coin  done.
If I < J then I is odd coin, else J is odd coin  done.
1.2) If result of first weighing is A,B,C,D < E,F,G,H then we know the odd coin is one of these 8, but don't know whether heavier or lighter.
Now we mix them up a bit and include some of the "reference" coins I,J,K,L (which we know are not odd):
2nd weighing: A,B,C,E against D,I,J,K
1.2.1) If same, odd coin is one of F,G,H and it must be heavy.
3rd weighing: F against G
If same, H is odd coin  done.
If different, whichever is heaviest is odd coin  done.
1.2.2) If A,B,C,E < D,I,J,K then odd coin has not swapped sides, so is one of A,B,C and is light.
3rd weighing: A against B
If same, C is odd coin  done
If diferent, lightest is odd coin  done.
1.2.3) If A,B,C,E > D,I,J,K then odd coin HAS swapped sides, so is either D or E
3rd weighing: D against A
If same, E is odd coin  done.
Otherwise, D is odd coin  done.
1.3) If result of 2st weighing is A,B,C,D > E,F,G,H then relabel:
A <> E
B <> F
C <> G
D <> H
Now A,B,C,D < E,F,G,H, so go to step 1.2Tony Andrews
http://tinyurl.com/tonyandrews