This has taken quite some time to get my head around and is a bit tricky to write-up, but I believe I now finally have the answer. This is an analysis of a puzzle along the following lines:
I am thinking of two numbers whose sum is less than 16. I tell Alice their product and Bob their sum, and challenge them to tell me the two numbers.
Alice: "I don't know the two numbers."
Bob: "I knew that."
Alice: "In that case, I do know the numbers!"
Bob: "Oh, then now I do too!"
Solution in practice:
I told Alice that the product of my two numbers is 4, and I told Bob that their sum is 5.
Alice (who knows the product is 4) knows that the numbers must be 1 and 4 (in which case Bob was told sum 5) or 2 and 2 (in which case Bob was told sum 4). She cannot tell which of these two is correct, and so says "I do not know".
Bob (who knows the sum is 5) knows that the numbers must be 1 and 4 (in which case Alice was told product 4) or 2 and 3 (in which case Alice was told product 6). Bob reasons that if Alice was told product 4 she had two options: 1-4 and 2-2, and wouldn't be able to narrow them down; he also reasons that if Alice was told product 6 she had two options: 1-6 and 2-3, again unable to narrow them down. Whatever Alice was told, Bob knows she could not know the answer for sure!
This, however, gives Alice new information... she now knows Bob must have been told sum 5, since if he knew she couldn't possibly know, then for every possible number pair that summed to his known sum there must have existed another number pair with the same product (thereby preventing Alice from knowing for sure). If Bob was told sum 4 the two possible number pairs were 1-3 or 2-2 - but no pair other than 1-3 has a product of 3. In contrast, with Bob being told sum 5 the two possible number pairs are 1-4 and 2-3, which each have companion number pairs with the same product: 2-2 and 1-6. Bob must therefore have been told sum 5, and thus (as per her original logic) the answer must be 1-4.
Bob - knowing that Alice now knows the answer - also now knows the answer must be 1-4. After all, if it was 2-3 (his other option), Alice would have been told product 6 and would still have no way of differentiating between 1-6 and 2-3, unable to make the deduction she did in the previous step.
Explaining the diagrams:
With this in mind, we can now appreciate which steps of this process are reflected in the diagrams.
First thing is to note that the diagrams reflect our logical deduction process to find the end solution. We do not have the benefit of knowing the product or the sum - we just know that Alice knows the product while Bob knows the sum... Thus we must consider all options that meet the initial criteria of summing to less than 16.
A reflects the initial state of play - we need to deduce which of the number pairs marked with a '?' is the correct answer. We have no information before anybody has made any statements.
B1 displays, for each possible number pair, how many number pairs share the same product (e.g. 1-14 and 2-7 both show '2', since these are the 2 solutions with product 14). Alice saying she doesn't know the answer means we can remove all pairs who have a unique product (those marked with a '1'), which is what is depicted in B2.
B2. * -- . .
6 -- 3 . . .
5 -- . 2 . 2 . .
4 -- 2 2 3 . . 3 2 .
3 -- 2 3 . 2 . 3 . 2 . 3
2 -- 2 2 2 2 3 2 2 2 2 . 3 .
1 -- . . . 2 . 2 . 2 2 2 . 3 . 2
/ / / / / / / / / / / / / /
* 2 3 4 5 6 7 8 9 10 * 12 * 14
C1 and C2 show the results of Bob's assertion that Bob already knew Alice couldn't possibly know - in particular, C2 depicts us being able to reduce the possibilities further to only the three sum-values (5, 7 and 9) for which at least one other number pair existed with the same product, thereby rendering Alice's hoped of solving the puzzle initially impossible.
C2. 5 7 9
`. `. `.
`.* --`. `. .
*`.- `. `. . .
4 --`. `. . `? . `. .
3 -- `. . `? `? . `. .
2 -- . `? `. `? . `? . `. . .
1 -- `? `? `? `. `. . .
/ / / / / / / / / / /
* 3 4 5 6 7 8 * * * *
D1 now shows the products of the remaining number-pair possibilities. Since Alice was able to solve it at this point we know that the product cannot have been 6. D2 depicts the two number-pairs with this product being chalked off.
D1. 4 -- 20 D2. 4 -- 20
3 -- 12 `18 3 -- 12 `18
2 -- 6 `10 `14 2 -- . `10 `14
1 -- `4 `6 `8 1 -- `4 `. `8
/ / / / / / / / / / / /
3 4 5 6 7 8 * 4 5 6 7 8
E1 now shows the remaining possibilities for Bob. Since he is able to solve it at this point we need the number-pair which has a unique sum among those remaining - that which sums to 5 (i.e. 1-4).
E1. 4 -- ? E2. & F! * -- .
3 -- ? `? * -- . `.
2 -- `? `? * -- `. `.
1 -- ? `? 1 ----> ! `.
/ / / / / / / / / /
4 5 6 7 8 4 * * * *
Finally, this is corroborated in F! - we have our answer at last! Phew!