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Thread: Puzzle

  1. #1
    Join Date
    Oct 2003
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    india, delhi
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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
    Sudeesh
    Sudeesh
    www.GaloisTech.com

  2. #2
    Join Date
    Aug 2003
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    25
    how many weighing scale do u have?

  3. #3
    Join Date
    Oct 2003
    Location
    india, delhi
    Posts
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    Thumbs up traditional scale only/balance

    Originally posted by ynoel
    how many weighing scale do u have?
    The have only one traditional scale / balance, which has two plates tied with a stick...
    Sudeesh
    www.GaloisTech.com

  4. #4
    Join Date
    Sep 2003
    Posts
    156

    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
    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...
    rgs,

    Ghostman

  5. #5
    Join Date
    Oct 2003
    Location
    india, delhi
    Posts
    18

    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

  6. #6
    Join Date
    Oct 2003
    Location
    india, delhi
    Posts
    18

    Thumbs up 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
    Sudeesh
    Sudeesh
    www.GaloisTech.com

  7. #7
    Join Date
    Sep 2002
    Location
    UK
    Posts
    5,171
    Provided Answers: 1

    Re: 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 1.2

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