Try out the ODDBALL logic test?

Discussion in 'Intelligence & Machines' started by Alan McDougall, Jul 15, 2010.

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  1. Stryder Keeper of "good" ideas. Valued Senior Member

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    Solution working

    The HTML page uses Javascript, it's not particular wonderful, however I've created it to be visual. It goes on the basis that: (you only weigh 4 on each side of the scales each time for all three times). The visual outcome is done so you can attempt to work out which ball is the one with the weight difference and whether it's heavier or lighter than the others.

    There is also a box for you to textselect with the actual randomised letter shown and it's weight in relationship to the others, so you can see if you worked it out correctly.

    Just reload the page for re-rolling the random allocation of the weigh change.

    Incidentally:
    If it's [sub]ABCDE[/sub] \ [sup]FGHIJ[/sup], then the Left side is heavier and the right side lighter.
    If it's [sup]ABCDE[/sup] / [sub]FGHIJ[/sub], the the Left side is lighter and the right side is heavier.
    If it's ABCDE FGHIJ it's balanced and therefore struck those letters from being the one you are after. (process of elimination. You'll find sometimes the left overs are struck if the scales aren't balanced.)

    Just mention if you spot a mistake (I noticed some of my letters were the wrong way around originally so there were occasions where their would be a couple of balls that would be paired heavy and light, but you couldn't work out which one with the solution)
     
    Last edited: Jul 16, 2010
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  3. Captain Kremmen All aboard, me Hearties! Valued Senior Member

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    Test 1:A B C D/E F G H

    Test 2:A F G H /K D L I

    Test 3:K G D J / I B L E

    Say it was Ball L which was heavier.
    Using your three weighings it could be either L or I which was the ODDBALL.

    In either case on weighings two and three the right side would go down.
     
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  5. Alan McDougall Alan McDougall Registered Senior Member

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    Your approach is correct only by weighing 4 balls against 4 balls is it possible to solve this problem

    But what if your 4 balls balance, then you know the oddball must be among the others left out on the side, but think again your attempt at a solution is wrong as soon as you assumed a certain balls were heavier , they might be lighter!!




    All you have achieved is that you now know in which THAT THE ODDBALL IS AMONG THE ABCDE ^ FGIJ weight You cant get a solution by having the oddball among 8 balls. If you are lucky and your balls balance then the oddball is among KLMNO it is easier to get a solution if this happens but you are basing your solution on luck. But starting to balance 4 ^ 4 is the correct approach

    Note this is quite a difficult test of logic and I knew about this test of logic many years ago and outside the INTERNET no one has come up with a complete and full answer (Luck)must not play a part)

    The solution should look like this

    It is ball * and it is heavier or it is ball# and it is lighter
     
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  7. Captain Kremmen All aboard, me Hearties! Valued Senior Member

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    Bit of confusion here.
    That wasn't my solution, it was my comment on Stryder's solution, saying that it didn't cover every possibility.

    My own solution was in post 4.
    I've now found a flaw in it.

    First Weighing
    Weigh 4 of the balls against another 4.


    If they balance in the first weighing the odd ball is in the other 4. Call these C Balls.
    Call the balls that balance "Standard balls".
    Weigh any 3 of the C balls against 3 standard balls.
    This will tell you whether they are heavier or lighter.

    If there is a balance, then the remaining C ball is the Oddball, and can be weighed against a Standard Ball to see whether it is heavy or light.
    Otherwise, because you are down to three balls and know whether the oddball is heavy or light, finding the oddball in the third weighing is easy.
    Just weigh any two of them against each other.

    If there is an imbalance in the first weighing, call the lighter set of 4 balls "Set A" and the heavier set " Set B". Call the remaining balls "Standard Balls"

    Weigh 3 Balls from Set A and one ball from set B against 4 standard balls. If they balance, the ball is heavy and is one of the other 3 balls from Set B.
    (Wrong. It could be the remaining light ball from set A.

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    )

    If they are lighter, the ball is lighter, and one of the 3 Balls from Set A.
    Weigh any two of these to find which is the odd ball. If they balance, then it is the other one.

    If they are heavier, the Ball is the Ball which was chosen from set B and is heavier.
     
    Last edited: Jul 17, 2010
  8. Captain Kremmen All aboard, me Hearties! Valued Senior Member

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    Stryder.
    If you get your weighings right, you will find the right solution
     
    Last edited: Jul 17, 2010
  9. Cifo Day destroys the night, Registered Senior Member

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    685

    Ok then, I’ve returned with my [blatantly wrong] original post as best as I can remember it.)


    Unfortunately, 8 of the balls (A through H) require 4 weighings. It's all very logical, but it's too formal and not sophisticated.


    Wow, was there another member who's real name was also Gordon Freeman?

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  10. Ellie Banned Banned

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    The answer is:

    It is impossible. There is no answer and you will never know.
     
  11. tablariddim forexU2 Valued Senior Member

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    4,793
    Now that we got a clue, I believe this is the solution: Note The words 'right' and 'heavier' are just for example, it could very well be 'left' or 'lighter', in which case you'd adjust your weighings and still find the answer.

    ABCDxEFGH= heavier on right
    EFxGH = heavier on right
    EGxFH = heavier on right
    H is heavier

    OR

    ABCDxEFGH = Equal
    IJxKL = heavier on right
    IKxJL = heavier on right

    L is heavier

    Of course, you could have this situation

    ABCDxEFGH = heavier on right
    EFxGH =EQUAL
    ABxCD = heavier on right

    A or B is lighter, but you'd need 4 weighings to find out...hmmm.
     
    Last edited: Jul 17, 2010
  12. tablariddim forexU2 Valued Senior Member

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    4,793
    it does seem impossible, but i think i got it

    maybe not
     
    Last edited: Jul 17, 2010
  13. Ellie Banned Banned

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    Heavier...but the one in the left group ABCD can also be lighter...As an example.
     
  14. Ellie Banned Banned

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    424
    What i think is that the guy in the OP THOUGHT he had the answer but i bet he does not.
     
  15. Cifo Day destroys the night, Registered Senior Member

    Messages:
    685
    Let’s look at: a) the data obtained from the weighings, and b) the number of state to solve for.

    Each weighing can have one of three results: 1) tip left (L), 2) balanced (O), and 3) tip right (R), and the combination of three weighings with three results each = 3³ = 27 outcomes. One of 12 balls having two states (lighter, heavier) = 12 × 2 = 24 states to solve for. So far, 27 outcomes is greater than 24 states, thus it may be possible to solve.

    The 27 outcomes are:
    OOO, OOL, OOR, OLO, OLL, OLR, ORO, ORL, ORR,
    LOO, LOL, LOR, LLO, LLL, LLR, LRO, LRL, LRR,
    ROO, ROL, ROR, RLO, RLL, RLR, RRO, RRL, and RRR

    Of these outcomes, the first (OOO) is invalid because it means the ball was never weighed, and so, even if only was one ball not weighed, we could identify the ball but wouldn’t know if it was lighter or heavier. Thus, our useable outcomes are reduced by one down to 26.

    It also seems plausible (although not absolutely clear to me) that, if the ball is weighed only once, we could not discern if it were lighter or heavier. So, OOL, OOR, OLO, ORO, LOO, and ROO might also be outcomes without enough information. Thus, our useable outcomes are reduced by six more down to 20.

    With 20 outcomes to determine 24 states, this seems impossible to solve.

    ###################################################

    Or, from another perspective, let’s look at weighing each ball the same number of times as any other ball (ie, equal treatment), with each weighing involving the same number of balls.

    If we weigh 1 × 1, then we weigh only 6 balls — not all the balls get weighed.

    If we weigh 2 × 2, then we weigh 12 balls once — not enough information.

    If we weigh 3 × 3, then 18 balls are weighed — not an even multiple of 12, and not enough information.

    If we weigh 4 × 4, then we can weigh 12 balls twice — yes, enough information.

    If we weigh 5 × 5, then we weigh 30 balls — not an even multiple of 12.

    If we weigh 6 × 6, then every ball is always on the scale, resulting in only 8 outcomes.

    Additionally, looking at the 27 outcomes above, we see that a third of the individual outcomes are balanced (O), meaning that, for one in three weighings, the ball in question (ie, any given ball) is not on the scales. This also points toward the method of weighing 4 × 4 balls each time.

    ###################################################

    Or, let’s look at the 20 possible scale placements with at least two weighings:
    (L = left side, O = not weighed, R = right side)

    OLL, OLR, LOL, LOR, LLO, LLL, LLR, LRO, LRL, LRR
    ORR, ORL, ROR, ROL, RRO, RRR, RRL, RLO, RLR, RLL

    I have shown them above as “mirror images” … that is, a heavy ball placed in the OLL placement scheme will produce the same results as a light ball in the ORR placement scheme. This means that these 20 placements reduce to 10 non-ambiguous placement schemes. Ten placements is less than 12 balls, so the puzzle seems not solvable.

    ###################################################

    I think this puzzle is not solvable. :scratchin:
     
  16. Captain Kremmen All aboard, me Hearties! Valued Senior Member

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    Too tough for me.
    I've done a google check on it, and it does have an answer.
    It's up to the poster to decide when he wants to reveal it.
    I can't see how you would get to the complete answer logically though.
     
  17. Alan McDougall Alan McDougall Registered Senior Member

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    410
    Hey Guys,

    The solution is not impossible and some of you might have got is right or near right, but to establish that I have to painstakingly follow your individual logic.

    I will not come back with the solution yet, why not give you palls a try at this fascinating puzzle

    Alan
     
  18. Captain Kremmen All aboard, me Hearties! Valued Senior Member

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    The answer I've seen by googling doesn't explain how it is arrived at.
    That's not logic, it's just correct.

    I think you are on the right route to a logical solution Cifo.
    I wonder if you can get to the right answer logically by working backwards, ie by assigning the outcomes to the states.
    So you could start with OOL and assign that to A-heavy, then put that into the weighings.

    So A-heavy would fit in like this:

    First Weighing ^
    Second Weighing ^
    Third Weighing ^A

    A-light would have to do the opposite
    A^ on the third weighing.

    You would then have to fit in the rest of the outcomes one by one, avoiding conflicts as you went.


    @Alan.
    Have you seen an answer that gives the route taken to get to the answer.?
    I wonder if there are other solutions that use a different number of balls in the weighings.
     
    Last edited: Jul 18, 2010
  19. Emil Valued Senior Member

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    2,789

    I think I need to spoil your little pleasure.
    No solution.
    With two measurements,maximum even number of balls is 4 and only if you have another 3 standards. (Six can not resolve with two measurements).
    Results,the first measurement is four plus four balls.
    What can not be solved unless you have another six standards that you do not have, have only four.
    If you succeed, with 2 measurements to detemin, which of the four balls is heavier or lighter,then I managed to determine with three measurements from 12 balls.

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    Last edited: Jul 18, 2010
  20. Cifo Day destroys the night, Registered Senior Member

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    In the back of my mind, I'm thinking this might require a 27×27 matrix (placements × outcomes) for solving it … along with a lot of sifting through combinations. :bugeye:
     
  21. Emil Valued Senior Member

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    Balls are: A,B,C,D,E,F,G,H,I,J,K,L
    1)I put [ABCD/EFGH]
    ---If ABCD=EFGH result that is in IJKL , and A=B=C=D=E=F=G=H
    ---------1.2)I put [ ABC/IJK ]
    ------------------ If ABC=IJK results it is L
    ------------------------------1.2.1)I put [A/L] and result the solution.
    ------------------- If not equal, I know that is one of IJK, and that is harder or easier as ABC are equal.
    ------------------------------1.2.2)I put [I/J]
    -------------------------------------- If egal rsults is K ,
    ---------------------------------------if not egal result is I or J,dependent as it is easier or harder as determined from measurement 1.2) result solution.

    --If ABCD not egal EFGH (and retain the group which is harder and which is more easier)
    ----2)I put [ABCEFG/IJKL+2etalon] (But I have not another two equal balls.)
    -----------if egal result D or H
    -----------------------2.1)put [D/I]
    ----- -----------------------if egal,results it is H,harder or easier from the 1)measurement
    ----------------------------if not egal results it is D
    -----------if not egal,results one of the groups ABC and EFG ,depends on 1)measurement
    -----------------------2.1 put [A/B] or [E/F]
    --------------------------------------if egal results C or G
    --------------------------------------if not egal results what scales show



     
    Last edited: Jul 19, 2010
  22. Randwolf Ignorance killed the cat Valued Senior Member

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    4,121
    I found my own solution some 25 years ago - symbology (without formal training), was a bitch then and turned out to be a bitch now. Hopefully, someone can follow this and let me know if it is correct....

    I solved it again, from scratch, just to see if I could. Once again, describing the solution in English / Math determined to be the hardest part. If you have any questions, PM me or just ask here - I have tried to be clear, but translating to symbology is very difficult for me, apparently I did not have the proper training (or don't remember it). I don't even remember my logic symbols...

    Definitions - sorry, don't know any formal way of designating this stuff so I had to make up my own (cumbersome at best):

    • "ball" - any of twelve steel balls, regardless of weight
    • "normal" - any ball that is part of the set of equivalent weighted balls
    • "undetermined" - any set of balls not known to contain the odd ball and/or or with unknown parity
    • "odd ball" - refers to ball of anomalous weight
    • "parity" - refers to weight of odd ball relative to normal balls, either lighter or heavier
    • "weighing" - process of comparing weights of groups of balls on a balance scale
    • "result" - observation of results of weighing and deductions obtained regarding odd ball and parity

    -----------------------------------------------------------------------------------------------------------------------

    Assign each ball an imaginary reference number, i.e. 01,02,03,04,05,06,07,08,09,10,11,12 ---> all balls undetermined
    1st weighing round is always the same - 3 vs 3 with 3 left out...

    -----------------------------------------------------------------------------------------------------------------------

    1st round: weigh 01-04 (undetermined) vs 05-08 (undetermined), set aside 09-12 (undetermined)
    1st result: scales balance: 01-08 (normal), 09-12 (undetermined)

    2nd round: weigh 01-03 (normal) vs 09-11 (undetermined), set aside 04-08 (normal), set aside 12 (undetermined)
    2nd result: scales balanced: 01-11 (normal), 12 (odd, parity unknown)

    3rd round: weigh any one of 01-11 (normal) vs 12 (odd, parity unknown)
    3rd result: scales can not logically balance - if 12 is heavier, 12=odd,heavy else 12=odd,light ---> solved

    -----------------------------------------------------------------------------------------------------------------------

    1st round: weigh 01-04 (undetermined) vs 05-08 (undetermined), set aside 09-12 (undetermined)
    1st result: scales balanced: 01-08 (normal), 09-12 (undetermined)

    2nd round: weigh 01-03 (normal) vs 09-11 (undetermined), set aside 04-08 (normal) and 12 (undetermined)
    2nd result: scales unbalanced: 01-08 (normal), 09-11 (undetermined), 12 (normal)
    • *** Note - This process identifies another normal ball - 12 - which was excluded from Round #2 of this permutation.
    • It is a little tricky to explain, but we know that the odd ball is contained in 09-12 from results of Round #1.
    • After excluding 12 from testing in this round and determining that the scales are unbalanced it necessarily means that 12 is normal, since the odd ball is contained in 09-11. This leaves us with a collection of 01-08+12 normal and 09-11 undetermined, but parity "group" known, either heavier or lighter within 09-11.

    3rd round: Weigh any two of the three remaining undetermined balls 09-11, one against the other. This would be 09 vs 10, 09 vs 11 or 10 vs 11.
    3rd result: We already know that the odd ball is isolated within this group of three balls from the 2nd round results. We also know that the odd ball's parity is determined from the results of the 2nd round. So...

    If the scales balance, no matter which two balls are weighed, the remaining ball must be the odd one, and must have known parity as explained above. E.g., if the results from the 2nd round show the undetermined set, 09-11, as heavier than the control group 01-03, than the odd ball determined in the 3rd round must be heavier as well. ---> solved

    Of course, the converse is also true, if group 09-11 is lighter than group 01-03, than the odd ball is lighter. ---> solved

    When the scales do not balance in the 3rd round, proceed as follows:
    Determine which parity (heavy or light) has been established for group 09-11 in the 2nd round, than the ball with the corresponding parity in Round 3 is the odd ball. ---> solved

    -------------------------------------------------------------------------------------------------

    1st round: Weigh 01-04 (undetermined) vs 05-08 (undetermined), set aside 09-12 (undetermined)
    1st result: Scales unbalanced -
    01-08 (undetermined), 09-12 (normal) *** Note - Eight balls, 01-08, remain unknown, although parity is known for groups 01-04 and 05-08 - (we know which set is heavier and which set is lighter). Parity of the odd ball will be known once it is determined which set of four contains said "odd" ball.
    It will be very important to track parity throughout completion of this permutation.


    2nd round: Weigh 01-03 (undetermined, parity known) vs 05-07 (undetermined, parity known), leave aside 04 (undetermined, parity known), 08 (undetermined, parity known), 09-12 (normal) [/i]As will be seen, it is important to continue to track parity of each ball / set of balls throughout.[/i]
    2nd result: Scales balance -
    01-03 (normal), 05-07 (normal), 04 (determined), parity known), 08 (undetermined, parity known), 09-12 (normal)


    3rd round: Weigh 04 (undetermined, parity known) vs 09 (known)
    3rd result: Scales do not balance -
    *** 04 is odd and parity is lighter or heavier (known) depending on determination from 1st round. ---> solved
    Scales balance -
    *** 08 is odd and parity is lighter or heavier (known) depending on determination from 1st round. ---> solved
     
  23. Emil Valued Senior Member

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    Correct,like me.






    2nd round:Weigh 01-03 (undetermined, parity known) vs 05-07 (undetermined, parity known), leave aside 04 (undetermined, parity known), 08 (undetermined, parity known), 09-12 (normal) [/i]As will be seen, it is important to continue to track parity of each ball / set of balls throughout.[/i]
    2nd result: Scales unbalanced ? and Weigh 01-03 (undetermined, parity known) and 05-07 (undetermined, parity known)


    My hypothetical solution was:
    2nd round:Weigh 01-03 (undetermined, parity known) + 05-07 (undetermined, parity known), vs 09-12 (normal)+2 (normal) (But I have not)
     
    Last edited: Jul 19, 2010
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