Try out the ODDBALL logic test?

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

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  1. Alan McDougall Alan McDougall Registered Senior Member

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    You can do what you like with the balls except weigh them on a bathroom type scale to get to the solution
     
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  3. Maika Registered Member

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    that sounds naughty.

    why not a bathroom scale? i guess they'd roll off.
     
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  5. Maika Registered Member

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    what does that mean?
     
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  7. Dywyddyr Penguinaciously duckalicious. Valued Senior Member

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

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    Anything? You sure about that?

    What if you made them orbit a planet?
    One of them would have a larger or small orbit, or go quicker or slower.

    Or.
    You could put them in a salt solution, and slowly add salt until one or all except one of them floated.

    No.
    The only thing that you can do is weigh them three times using the pan scales.
    A very light marking in felt tip is permissible to mark them if you wish.
     
    Last edited: Aug 10, 2010
  9. nirakar ( i ^ i ) Registered Senior Member

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    It took me less than two hours but probably more than one hour to solve the puzzle. I needed about six tries to find a method that worked.

    That was a good puzzle.


    After solving the puzzle I started reading posts 2 through- .
    Trying to read and understand what people were trying to say was more difficult than the puzzle.

    I have looked at Stryder's hidden stuff. Thats not the way I solved the puzzle. I don't understand what he is saying.

    None of the people's proposed solutions look like mine. I wonder if they are wrong. Mine seems to pass the tests I gave it.

    Solution in below in blank box: You'll have to mouse over to see it
    Code:
    [color=white] My way weighs four balls against four balls in the first weighing, Three balls against three balls in the second weighing, and one ball against one ball in the third weighing.
    
    I am calling the balls 1 through 12. Following the image of a balance scale with two pans I refer to one pan as left and one as right just for visualization sake.  It does not affect the puzzle. 
    
    You can test this. Decide which ball is heavier or lighter and then just follow the lines according to what you chose.  
    
    Line 1, First weighing: place balls {1,2,3,4} in left pan and balls {5,6,7,8} in the right pan and note which pan was heavier.
    Line 2, If the balls {1,2,3,4} in the left pan were heavier than balls {3,4,5,6} then proceed to Line 5.
    Line 3, If the balls {1,2,3,4} in the left pan were lighter than the balls {5,6,7,8} in the right pan then proceed to Line 15
    Line 4, If the balls {1,2,3,4} in the left pan were equal to the balls {5,6,7,8} in the right pan then proceed to Line 25
    
    Line 5, Second weighing: place balls {1,2,5} in the left pan and balls {3,6,9} in the right pan and note which pan was heavier.
    Line 6, If the balls {1,2,5} in the left pan were heavier than balls {3,6,9} in the right pan then proceed to Line 9.
    Line 7, If the balls {1,2,5} in the left pan were lighter than balls {3,6,9} in the right pan then proceed to Line 11.
    Line 8, If the balls {1,2,5} in the left pan were equal to balls {3,6,9} in the right pan then proceed to Line 13
    
    Line 9, Third weighing: place ball {1} in the left pan and ball {2} in the right pan and note which pan was heavier.
    Line 10, Answer, the heavier of ball 1 and ball 2 is heavier than the other 11 balls but if they weigh the same then ball 6 is lighter than the other 11 balls.
    
    Line 11, Third weighing: place ball {5} in the left pan and ball {9} in the right pan and note which pan was heavier.
    Line 12, Answer, if ball 5 is lighter than ball 9 then ball 5 is lighter than the other 11 but if they weigh the same then ball 3 is heavier than the other 11 balls Ball 9 can not be lighter than ball 5.
    
    Line 13 Third weighing: place ball {7} in the left pan and ball {8} in the right pan and note which pan was heavier.
    Line 14, Answer, the lighter of ball 7 and ball 8 is lighter than the other 11 balls but if they weigh the same then ball 4 is heavier than the other 11 balls.
    
    
    
    Line 15, Second weighing: place balls {1,2,5} in the left pan and balls {3,6,9} in the right pan and note which pan was heavier.
    Line 16, If the balls {1,2,5} in the left pan were lighter than balls {3,6,9} in the right pan then proceed to Line 19.
    Line 17, If the balls {1,2,5} in the left pan were heavier than balls {3,6,9} in the right pan then proceed to Line 21.
    Line 18, If the balls {1,2,5} in the left pan were equal to balls {3,6,9} in the right pan then proceed to Line 23
    
    Line 19, Third weighing: place ball {1} in the left pan and ball {2} in the right pan and note which pan was lighter.
    Line 20, Answer, the lighter of ball 1 and ball 2 is lighter than the other 11 balls but if they weigh the same then ball 6 is heavier than the other 11 balls.
    
    Line 21, Third weighing: place ball {5} in the left pan and ball {9} in the right pan and note which pan was heavier.
    Line 22, Answer, if ball 5 is heavier than ball 9 then ball 5 is heavier than the other 11 but if they weigh the same then ball 3 is lighter than the other 11 balls Ball 9 can not be heavier than ball 5.
    
    Line 23, Third weighing: place ball {7} in the left pan and ball {8} in the right pan and note which pan was heavier.
    Line 24, Answer, the heavier of ball 7 and ball 8 is heavier than the other 11 balls but if they weigh the same then ball 4 is lighter than the other 11 balls.
    
    
    
    Line 25, Second weighing: place balls {9,10} in the left pan and balls {11,2} in the right pan and note which pan was heavier.
    Line 26, If the balls {9,10} in the left pan were heavier than balls {11,2} in the right pan then proceed to Line 29.
    Line 27, If the balls {9,10} in the left pan were lighter than balls {11,2} in the right pan then proceed to Line 31 .
    Line 28, If the balls {9,10} in the left pan were equal to balls {11,2} in the right pan then proceed to Line 33
    
    Line 29, Third weighing: place ball {9} in the left pan and ball {10} in the right pan and note which pan was heavier.
    Line 30, Answer, the heavier of ball 9 and ball 10 is heavier than the other 11 balls but if they weigh the same then ball 11 is lighter than the other 11 balls.
    
    Line 31, Third weighing: place ball {9} in the left pan and ball {10} in the right pan and note which pan was lighter.
    Line 32, Answer, the lighter of ball 9 and ball 10 is lighter than the other 11 balls but if they weigh the same then ball 11 is heavier than the other 11 balls.
    
    Line 33, Third weighing: It is knowing that ball 12 is the odd ball prior to this weighing but compare ball 12 to any of the other balls to determine whether ball 12 is heavier or lighter than the other 11 balls.
    
    My method of solving the puzzle was primarily about eliminating methods that could not work.
    [/color]
    
     
    Last edited: Aug 24, 2010
  10. John99 Banned Banned

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    I just think it is two entirely different puzzles, if you can mark them.
     
  11. dsdsds Valued Senior Member

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    I haven’t solved it yet but it would be cool to create some code to solve the puzzle – not only solve but to automatically create the method (groups, etc.. ) to solve it. Actually the problem would be restated to “find the minimum required weigh-ins to find the oddball”. – and if that is too easy, then make it so the user enters the number of balls AND number of oddballs.
     
  12. nirakar ( i ^ i ) Registered Senior Member

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    I sort of stylized my telling of my solution given three posts above as if it was programming code. I had one course an BASIC decades ago.
     
  13. John99 Banned Banned

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    Actually a few are very much like yours.

    Not necessarily wrong but wishful thinking.
     
  14. nirakar ( i ^ i ) Registered Senior Member

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    That approach is incorrect because if the scales are unbalanced on your first and second tests you will not be able to construct a third test that can identify the oddball for all six remaining scenarios as to which ball is the oddball.
     
  15. nirakar ( i ^ i ) Registered Senior Member

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    Alan McDougall post 61 used the same technique I did. I did not read it before.
     
  16. John99 Banned Banned

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    Will it work ALL the time = NO
    Will it work most of the time = YES
    Will it work some of the time = YES
     
  17. nirakar ( i ^ i ) Registered Senior Member

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    Mine will work all the time. Since Alan McDougall was working the same way I assume his will work all the time but I did not check his third weighings or how he handled the easier balanced scenario because the second weighing was the tricky part of that method. Unless he made some silly mistake his will work because he got past the hard part.

    Post 62 that Captain Kremmen got from the internet also works all the time. It is a completely different technique from what I and Alan Mcdougall came up with.

    Now that I see Captain Kremmen post 62 the stuff Stryder was putting in white text is no longer so nonsensical to me.

    I wonder if there is a radically different third approach that will work.
     
  18. John99 Banned Banned

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    No, it wont work all the time. Your method is similar to some of the others and none of them will work 100% of the time and max. 98% of the time is only possibility. 98% is according to my calculation and has margin of error of 1-2% in the negative.
     
  19. nirakar ( i ^ i ) Registered Senior Member

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    You can test mine with all 24 scenarios. It will work every time.

    There is a chance that I could have gotten something wrong in the transferring from my flow chart to my lines of instructions but the approach works 100% not 98%. Abandoning symmetry by adding just one known ball on the second test was the key difference between what I and McDougall did and other 4-4 3-3 approaches. There was something addictive about symmetry.
     
    Last edited: Aug 25, 2010
  20. John99 Banned Banned

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    One thing that surprises me is peoples reaoning for using less than 6+6 for first weighing if you can number the balls.
     
  21. John99 Banned Banned

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    The problem is lighter or heavier equals 2, you will be left with 2 balls and there are two places to put the last two balls.

    Please Register or Log in to view the hidden image!

     
  22. nirakar ( i ^ i ) Registered Senior Member

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    You learn less from 6+ 6 than you learn from 4 + 4. With 4 plus four you eliminate 4 or 8 balls with your first weighing.

    Doing 6+6 all twelve balls could still be the odd ball after the first weighing.

    Because you can only weigh the ball three times you need to rule out as many balls as you can with every weighing.
     
  23. John99 Banned Banned

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    is it lighter or heavier?
     
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