balance scale problem for 13 (not 12) items
The way to solve problems of this sort is to think about how much information you have, which is equivalent to how few possibilities remain.
Each weighing can have one of three possible outcomes: the left pan is heavier, ther right pan is heavier, or they weigh the same. You want each outcome to correspond to (roughly) the same number of remaining possibilities.
Initially, you have 26 or 27 possibilities: one item is heavier (and there are 13 choices for which that item that is), one item is lighter (out of 13), or maybe they're all the same. Since 27 = 3 * 3 * 3, you might hope that three weighings will suffice even if it's possible that they're all the same.
But for that to happen, the first weighing has to split the possibilities into three sets of exactly 9 each.
If on the first weighing you weigh 4 items against 4 others, the split is:
Left pan is heavier means one item in the left pan is heavier or one item in the right pan is lighter. That's 8 possibilities.
Right pan is heavier means one item in the left pan is lighter or one item in the right pan is heavier. Again 8 possibilities.
The pans balance. This must cover the 10 or 11 remaining possibilities. We cannot resolve these in only two more weighings. (Two weighings can have only 3 * 3 = 9 outcomes.)
So, four against four won't work. What about 5 against 5? This splits the 26 or 27 possibilities into sets of size 10, 10, and 6 or 7. Again, we cannot answer the question in only 2 more weighings.
More than 4 in each pan on the first weighing leaves too many possibilities for when the scales do not balance. Fewer than 5 leaves too many possibilities when they do balance.
The problem cannot be solved in only three weighings, even if you know there is an odd item. (Unless you know something else, like the color of the odd item, or that the odd item has a different density and there is some water we can immerse the apparatus in, or we can tie them together bolo-fashion, spin them, and observe how the center of mass moves. Or the scale has more than two pans.)
UPDATE: I just realized you only need to identify the odd item, not whether it's heavy or light. That means you start with only 13 possibilities. I'll be back.
I'M BACK... Ignore the 13th item. In three weighings, you can tell if one of 12 items is odd, and if so whether it's light or heavy. I won't re-iterate this well-known solution. Re-interpret the "none of the 12 is odd" as "the so-far-ignored 13th item is odd".
You lose the ability to tell that there is an odd item, and if the odd item is the 13th you lose the ability to tell if it's light or heavy, but if you can assume there is an odd item you can always tell which it is, and sometimes (read usually) tell whether it's heavy or light.
You can do it in three pre-determined weighings, by numbering the coins from 1 to 13 in the style 1, -2, 3, -4, 5, &c.
1 001 6 M10 11 11M
2 0M1 7 1M1 12 MM0
3 010 8 M01 13 111
4 0MM 9 100
5 1MM 10 M0M KG M1M
You then weigh according to these rules. For each column, put coins numbered 1 in the left pan, and M in the right pan. If the '1' pan goes down, write '1', if the M pan goes down, write 'M'.
After three weighings, you should have a three-place sequence, like 01M. This abc gives 9a+3b+c, where M=-1. If it is an even number, reverse the sign, and that's the coin that is not fair, and whether it's overweight or underweight. So our 01M gives 9.0 + 3.1 + 1.-1 = 2, being even gives -2, so the coin numbered '2' is underweight.
This is different to the usual answer, since we always have seven coins on each pan, (you can't do it with simply six, because at any stage, you are testing for a trinary digit, and it needs here 9 coins, which is odd.
When you do the same for four or five weighings, the process is the same, but the known good is always opposite the half-point, ie 1M1M or M1M1M. You can get to 40 or 121 coins on 4 and 5 weighings, but if you are not allowed to add a known good, then you need to drop the half-coin (7 or 20 or 61) from the list.