S=SET, 1=A~B+C, 2=A~B+D, 3=A~C+D, 4=B~C+D, 5=A+B~C+D, 6=A+C~B+D (1 to 6 are weighing configurations).
~ (twiddles) means "is balanced against"; < means "lighter than"; > means "heavier than"; = means "balances".
S A B C D 1 2 3 4 5 6
1 3 5 8 11 < < < < < <
1 5 8 11 3 < < < < < >
1 8 11 3 5 < < = > > <
1 11 3 5 8 > = < < > >
2 3 5 11 8 < < < < < >
2 5 11 8 3 < < < = > <
2 11 8 3 5 = < > = > >
2 8 3 5 11 = < < < < <
3 3 8 5 11 < < < < < <
3 8 5 11 3 < = < < < >
3 5 11 3 8 < < < = > <
3 11 3 8 5 = > < < > >
4 3 8 11 5 < < < < < >
4 8 11 5 3 < < = > > <
4 11 5 3 8 > < = < > >
4 5 3 8 11 < < < < < <
5 3 11 5 8 < < < < > <
5 11 5 8 3 < > = < > >
5 5 8 3 11 < < < < < <
5 8 3 11 5 < = < < < >
6 3 11 8 5 < < < < > <
6 11 8 5 3 < = > = > >
6 8 5 3 11 = < < < < <
6 5 3 11 8 < < < < < >
The table shows all possible arrangements of the four weighted objects, together with various weighing results. Before carrying out any weighings, we first label the four objects so that we can't get them mixed up. The labels are weightless. We carry out weighings 5 and 6 first. We take the four objects and put two in one scale and two in the other and note the results of the two different weighings. There are four possible results. Note how the table is arranged. Each set is ordered and the order is rotated ABCD, BCDA, CDAB and DABC to make up the set. Note that each set is circular: DABC is followed by ABCD, and ABCD is preceded by DABC. This is important for later.
After weighing the pairs, we will have identified one arrangement in each set that would produce the weighing results we've just obtained. We have two weighings left. Using these we should arrive at the unique set of weights.
Let's use an example. Suppose the first two weighings gave the result < and < (the pair in the left scale pan was lighter than the pair in the right scale pan in each case). That gives us 6 possible values for A, B, C and D, one group from each of the six sets: 3 5 8 11, 8 3 5 11, 3 8 5 11, 5 3 8 11, 5 8 3 11 and 8 5 3 11. Note that D has already been found to weigh 11 ounces, because it's common to all. If we look at the table at the configuration immediately preceding each of the six groups we've identified in each set and look at the weighing results 2 and 3, we can see that the results are unique. For example, the arrangement preceding 3 5 8 11 is 11 3 5 8 and the weighing results are = and <; the arrangement preceding 5 8 3 11 is 11 5 8 3 and the results are > and =. All we have to do is carry out the weighings 2 and 3 on this preceding arrangement and note the results. Let's continue with the example: let's suppose that the weighing results are actually > and <. The only ABCD line satisfying this is 3 8 5 11. So these are the weights of the labelled objects in our example. (The arrangement for weighings 2 and 3 is 11 3 8 5.)
Another example: weighings 5 and 6 are < and >. This implies 5 8 11 3, 3 5 11 8, 8 5 11 3, 3 8 11 5, 8 3 11 5, 5 3 11 8. This time C=11. However, we can't use the weighings of the previous arrangements, because the results aren't unique. But if we move down or up two arrangements, we have uniqueness. Suppose we carry out weighings 2 and 3 on this other arrangement and get the results > and =, then ABCD is 8 3 11 5. (The arrangement for weighings 2 and 3 is 11 5 8 3.)
Sometimes you won't need to look for other arrangements to establish uniqueness, because the arrangements you find from weighings 5 and 6 already contain the uniqueness you need. And it doesn't always have to be weighings 2 and 3, it could be any two of the weighings between 1 and 4 that are used for the weighing test, so long as it's always the same two weighings that are used for the six possibilities. Hint: = and > are much less common than <, so = and > in the weighings between 1 and 4, so look out for them when establishing uniqueness.