OK Folks, my competitiveness really came out here trying to be the 1st to get it. Must say the volatility today has helped, I'm much less inclided to take on positions and play with the puzzle instead...

this was DIFFICLULt... anyway, here is my second, rather less c0cky attempt at solving it. I havent covered all the possible permutations, but the methodology is there (i.e. if 1a 1b 1c 1d < 2a 2b 2c 2d, do the same thing except the other way around. Or just re-label the balls).

Throughout, > means "is heavier than" and < means "is lighter than"... e.g. Car > Dog, duck < feather

OK here goes.....

split the balls into 3 groups of 4;

Group 1 = [1a, 1b, 1c, 1d]

Group 2 = [2a, 2b, 2c, 2d]

Group 3 = [3a, 3b, 3c, 3d]

then, weigh Group 1 vs. Group 2:

If they are EQUAL, then group 3 must contain the odd ball (If we keep an equal number of balls on each side, for every "go", if they balance the ODD ball must NOT be present, and if they DONT balance then the ODD ball MUST be present).

Then collect group 1 (which we know are all EVEN) and compare them with group 3 (which we know one of which is ODD)...

remove one ball from each group, and weigh:

1a 1b 1c vs. 3a 3b 3c

__Scenario 1:__

If they Balance, we know ball 3d must be the ODD one, so compare it with 1d (which we know is EVEN) to determine whether is it heavy or light.

__Scenario 2:__

1a 1b 1c > 3a 3b 3c

therefore we know 3d must be EVEN, and one of 3a 3b 3c is LIGHT...

weigh 3a vs 3b:

if 3a > 3b, then 3b is LIGHT

if 3a < 3b, then 3a is LIGHT

if 3a = 3b, then 3c is LIGHT

reverse the process if 1a 1b 1c < 3a 3b 3c, except the odd ball must be HEAVY

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Now it gets harder...

back to weighing go 1, where 1a 1b 1c 1d vs. 2a 2b 2c 2d

say 1a 1b 1c 1d > 2a 2b 2c 2d

COMMENT:

We can do 3 changes to the arrangement of the balls. we can...

perform [X] = remove an equal number of balls from either side

perform [Y] = replace an unknown ball with a known ball

perform [Z] = switch any number of balls from one side to the other

** note: re-name all group 3 balls to E (e.g. 3a = 3b = 3c = 3d = E) as we know they are all EVEN, because the odd ball MUST be present for the scales NOT to balance **

now, after our 1st weighing of balls, we have

1a 1b 1c 1d > 2a 2b 2c 2d

then, perform [X] and remove ONE ball from either side, 1d and 2d, leaves us with

1a 1b 1c vs. 2a 2b 2c

then perform [Y], replace ONE UNKNOWN ball with a KNOWN ball E from the 3rd group, say 2c, leaves us with

1a 1b 1c vs. 2a 2b E

and lastly perform [Z]. swap one ball on each side, say swap 1a with 2a, leaves us with

2a 1b 1c vs. 1a 2b E

now weight them, our 2nd "go"......

__Scenario 1 part 1__

2a 1b 1c > 1a 2b E

then either 1b or 1c is HEAVY, as they have remained on the heavy side

OR

2b is LIGHT, as it has remained on the LIGHT side

in which case.....

__Scenario 1 part 2__

remove 2b, which is either LIGHT or EVEN, and weigh 1b vs 1c....

if 1b > 1c then 1b is HEAVY

if 1b < 1c then 1c is HEAVY

if 1b = 1c then 2b is LIGHT

__Scenario 2 part 1__

2a 1b 1c = 1a 2b E

then we have removed the ODD ball, so either

1d is HEAVY (ball removed from heavier side)

OR

2c or 2d is LIGHT (the balls removed from the LIGHT side)

then....

__Scenario 2 part 2__

remove 1d, which must be HEAVY or EVEN...

weigh 2c vs 2d

if 2c > 2d then 2d is LIGHT

if 2c < 2d then 2c is LIGHT

if 2c = 2d then 1d is HEAVY

__Scenario 3__

2a 1b 1c < 1a 2b E (i.e. the heavier and lighter side have swapped over)

then we have removed 2 balls much must be EVEN, as the ODD ball MUST be present.

then either 1a is HEAVY (swapped over, keeping on the heavy side)

OR

2a is LIGHT (swapped over, keeping on the light side)

then

__Scenario 3 part 2__

weigh 1a vs E

if 1a > E then 1a is HEAVY

if 1a = E then 2a is LIGHT

note 1a < E is impossible

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thats it... if the scales go the other way at any stage, just flip the notation around.

I would scan the workings out, but I did them all on A3 paper and only got an A4 scanner