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    Unknown008's Avatar
    Unknown008 Posts: 8,076, Reputation: 723
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    #1

    Oct 28, 2011, 09:14 AM
    [Solved] Small Exercise
    Okay, I've done those 'weighing puzzles', but this one stumps me as there seems to be no solution to it... Google gave me wrong solutions, as there were eventualities that seemed like nobody considered... Anyway, here it is ;)

    There are 12 oranges, all but one have the same mass (or weight). When it comes to the last orange, we don't know if it is heavier or lighter than the rest. How can we get the 'odd' orange out, using only thrice, a balance.

    And by balance, I mean this sort of balance without any scale:


    I tried dividing into 3 groups of 4.
    1 2 3 4 | 5 6 7 8 | 9 10 11 12

    First try: taking first two groups:
    - If they balance, the odd orange is in the last group of 4.
    - If not, it's among the eight used.

    Taking the worst scenario, they are in the 8.
    Divide into 4 more groups: 1 2 | 3 4 | 5 6 | 7 8
    Second try: take 1st and 2nd:
    - If they balance, it's between the 4 left.
    - If not, they are between the first 4.

    And with one remaining try, I don't see how I can get the odd orange from 4 oranges :(

    =====================================

    I tried taking 4 groups of 3.
    1 2 3 | 4 5 6 | 7 8 9 | 10 11 12

    First try: taking first two groups:
    - If they balance, the odd orange is in the first 6.
    - If not, it's among the last 6.

    Well, I suppose now dividing into groups of 2.
    1 2 | 3 4 | 5 6
    Second try: taking first two:
    - If they balance, the odd orange is in the last group of 2.
    - If not, it's among the 4 used.

    And worse case scenario, I can't get the orange from 4...

    So, the aim is to get down to 2 oranges with one last weighing. So that on the last try, we take one orange at random from the 'good' oranges and compare with any of the two:
    - If they balance, the one not picked is the odd orange.
    - If they don't, the one picked is the odd orange.

    Any idea? :)
    ebaines's Avatar
    ebaines Posts: 12,131, Reputation: 1307
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    #2

    Oct 28, 2011, 12:25 PM
    Quote Originally Posted by Unknown008 View Post
    I tried dividing into 3 groups of 4.
    1 2 3 4 | 5 6 7 8 | 9 10 11 12

    First try: taking first two groups:
    - If they balance, the odd orange is in the last group of 4.
    - If not, it's among the eight used.

    Taking the worst scenario, they are in the 8.
    Divide into 4 more groups: 1 2 | 3 4 | 5 6 | 7 8
    A classic puzzle. Your first thought about dividing into the three groups is good: so put 1 2 3 & 4 against 5 6 7 & 8. Now for the trick - if they don't balance the 2nd measurement needs to be a mix of oranges taken from the left and right side: for example you could put 1, 2, & 5 against 3, 4, and 6. Suppose in your first test the balance tilted to the left. If in the 2nd test it tilts to the left you know either 1 is heavy, 2 is heavy, or 6 is light, and the final measurement would be 1 versus 2; if it tilts to the right either 3 is heavy, 4 is heavy, or 5 is light and your final measurement is 3 versus 4; and if it balances either 7 or 8 is light so your final measurement is 7 versus 8.

    Similarly if the result of the first test is tilt to the right you then put 1 2 & 5 against 3 4 & 6; and follow essentially the same logic as above.

    If the result of the first test is balance then you can put 1 2 & 3 (which are all known to be good) against 9 10 and 11. If it tilts left then either 9, 10, or 11 is heavy; if it tilts right then either 9, 10, or 11 is light; and if it balances then 12 is either light or heavy. Again, one more measurement and you get your answer.

    Below is the complete solution. I labelled the oranges with letters rather than numbers. The gray boxes show which oranges to test. Start at the top and follow the arrows either down to the left, center, or right depending on whether the scale tilted to the left, stayed in balance, or tilted to the right. The final result of the tree is shown in the orange boxes with the letter of the orange and either a 'H' or 'L' subscript to indicate whether it is heavy or light.
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    jcaron2's Avatar
    jcaron2 Posts: 986, Reputation: 204
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    #3

    Oct 28, 2011, 01:37 PM
    Can't give you a greenie, EB, until I spread it around some more. ;) Anyway, nice question and nice answer. It gave me something to think about during a long boring work meeting I just sat through. I managed to work it out in my head, then came back to my desk to give the answer, only to find (as usual) that EB had beat me to it. LOL
    ebaines's Avatar
    ebaines Posts: 12,131, Reputation: 1307
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    #4

    Oct 28, 2011, 03:19 PM
    Thanks JC - it took me my entire lunch break to work it out, and I was afraid that someone was going to beat me to it!
    Unknown008's Avatar
    Unknown008 Posts: 8,076, Reputation: 723
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    #5

    Oct 28, 2011, 10:47 PM
    Oh! Really nice! I was almost thinking it was impossible lol.

    I never thought about mixing them together and as soon as they got mixed, I needed to read it more. It made more sense to me when I labelled them a second time through 'h' and 'l' like that:

    1h 2h 3h 4h 5l 6l 7l 8l

    H being potentially heavy, l being potentially light.

    And taking it to the second try gave this:

    1h 2h 5l v/s 3h 4h 6l

    And from there, observing which group was heavier was easy to understand.

    Thanks a lot ebaines! :)

    Sorry to have taken your lunch break :o

    And Josh, I'm sure you'd have done a similar good job :)

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