12-Ball Riddle

beats the crap outta me

Welcome Sabron,

If you have asked this question correctly I can't see how it can be done in THREE weighings?????

First we must correct your version of the riddle:
You have 12 small balls. All appear to be exactly the same, but one of the balls is an exception: it is either lighter or heavier, but looks the same.
You are given a double-sided weighing machine (ie. a ballance) to find out which ball is the exception, and whether it is lighter or heavier than the others. You have 3 uses of the weighing machine.

step 1. place four balls on each side of the scale; if they ballance (A), mark the other four '?'.
If they do not balance (B), mark four 'L', and four 'H' (according to the scale).

step 2(A). weigh three '?' against three '=' balls;if they are equal the other one is the odd ball (weigh it against a normal ball to see if it is heavy or light. If they do not balance mark the three balls 'H', or 'L' as the scale indicates.

step 3(A). weigh one of the three remaining balls against another, if they balance the other one is the odd ball; if they do not balance the designation of 'H', or 'L' from step 2 will indicate which ball is 'odd'.

step 2(B). weigh three 'L' balls and three 'H' balls against six '=' balls; if they ballance the odd ball is one of the two others (C); if they do do not ballance the odd ball is one of the three 'H' balls, or one of the three 'L' balls as the scale indicates heavier than = or lighter (D).

step 3(C). weigh the 'H' ball against an '=' ball; if they are equal the odd ball is the 'L' ball; if they are not equal the odd ball is the "H" ball.

step 3(D). weigh one of the three 'suspect' (either the three 'H' balls, OR the three 'L' balls from step 2) balls against another; if they are equal, it is the other one; if they are not equal the scale will indicate which one it is according to its label.

well now that took a while!

can someone please decipher

BoGoWo: I must correct some small glitches in your process.



Step 1 is OK, and steps 2A and 3A are correct also.


Now the step 2B is a bit problematic:


The thing is, you don't have SIX other balls; just 4H,4L and 4=.
step 2: So instead of this , measure L1H1L2 against H2L3= and leave H3H4L4 off.
3A) If equal, measure H3 against H4 to get the ball. Heavier of those is heavy.
3B) if L1H1L2 is heavy, measure H1 against =. If heavy, it is H1, else L3 is light.
3C) if H2L3= is heavy, then either H2 is heavy or L1 / L2 is light. So measure L1 against L2 : if equal it is H2 heavy, else the lighter is light.

Relative

The issue with the L1H1L2... et cetera, is that the inconsistant ball was never stated to be a multiple of the consistant balls. If the "heavy ball" was guaranteed to be 2x and a regular ball 1x (or if the ball is lighter, .5x) then stacking 3 against 2 and getting an equal result would tell you volumes. However, (assuming heavy) if the heavy ball is 1.5x, and it was on the side of the scale with 2 balls, then you have 2.5x on the 2 ball side and 3x on the 3 ball side. This tells you nothing except that 3 balls are indeed heavier than 2.

I agree. Steps 1, 2(A) and 3(A) are correct. If the 4 and 4 do not balance, you indicate them 'H' and 'L' according to the scale, and you also indicate the other 4 '=' (equal). You grab 3 'H' balls and 2 'L' balls and throw them on one side. You then grab the other 4 which you know are all equal with the remainder 'H' ball and throw them on the other side, leaving the 2 remainder 'L' balls outside. Hence on the scale you now have H1H2H3L1L2 weighing against 4 '=' and H4. And you also have L3 and L4 outside.

A: If H1H2H3L1L2 weighing against 4 '=' and H4 are equal
Step 1: L3 or L4 you left outside is the 1 you are looking for. You grab L3 and weigh it with a '='.
Step 2: If those 2 are equal then L4 is the ball you are looking for. If they are not equal, then L3 is the 1 you need.

B: If H1H2H3L1L2 is heavier than 4 '=' and H4.
Step 1: Obviously H4 is not the one. So you grab the H1H2H3 balls and weigh H1 and H2 against each other.
Step 2: If they are equal then your ball is H3. If H1 and H2 are not equal then the heavier 1 is your ball.

C: If H1H2H3L1L2 is lighter than 4 '=' and H4.
Step 1: The ball is L1, L2 or H4 (hopefully this is obvious to everyone). You grab L1 and L2 and you weigh them against each other.
Step 2: If they are equal then your ball is H4. If L1 and L2 are not equal then the lighter 1 is your ball.

I tried to make the solution as clear as possible so if there is anything "wrong", then it is either your comprehension skills or my lack of skill to explain better. The solution has been tested so many times. This is it.
Thank you! Thank you very much