Try out the ODDBALL logic test?

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Maika said:
you guys are making this too complex. i want those balls so i can try it out.

and why can't you just put the balls in different places, you don't need to mark them.

chill and have a drink. :p
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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
 
Alan McDougall said:
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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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.
 
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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.
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Originally Posted by Maika
you guys are making this too complex. i want those balls so i can try it out.

and why can't you just put the balls in different places, you don't need to mark them.

chill and have a drink.
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I just think it is two entirely different puzzles, if you can mark them.
 
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.
 
dsdsds said:
I haven’t solved it yet but it would be cool to create some code to solve the puzzle
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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.
 
Randwolf said:
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

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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...

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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

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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

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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
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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.
 
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.
 
nirakar said:
Mine will work all the time.
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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.
 
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.
 
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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.
 
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.;)
 
John99 said:
One thing that surprises me is peoples reaoning for using less than 6+6 for first weighing if you can number the balls.
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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.
 
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