Task[edit]
Solve the "Impossible Puzzle":


X and Y are two different whole numbers greater than 1. Their sum is no greater than 100, and Y is greater than X. S and P are two mathematicians (and consequently perfect logicians); S knows the sum X+Y and P knows the product X*Y. Both S and P know all the information in this paragraph.
The following conversation occurs:

 S says "P does not know X and Y."
 P says "Now I know X and Y."
 S says "Now I also know X and Y!"
What are X and Y?


Guidance
It can be hard to wrap one's head around what the three lines of dialog between S (the "sum guy") and P (the "product guy") convey about the values of X and Y.

So for your convenience, here's a break-down:





 Quote

 Implied fact


 1)

 S says "P does not know X and Y."

 For every possible sum decomposition of the number X+Y, the product has in turn more than one product decomposition.


 2)

 P says "Now I know X and Y."

 The number X*Y has only one product decomposition for which fact 1 is true.


 3)

 S says "Now I also know X and Y."

 The number X+Y has only one sum decomposition for which fact 2 is true.

Terminology:

 "sum decomposition" of a number = Any pair of positive integers (A, B) so that A+B equals the number. Here, with the additional constraint 2 ≤ A < B.
 "product decomposition" of a number = Any pair of positive integers (A, B) so that A*B equals the number. Here, with the additional constraint 2 ≤ A < B.


Your program can solve the puzzle by considering all possible pairs (X, Y) in the range 2 ≤ X < Y ≤ 98, and then successively eliminating candidates based on the three facts. It turns out only one solution remains!

See the Python example for an implementation that uses this approach with a few optimizations.

See also
   Wikipedia:   Sum and Product Puzzle

F counter(arr)
   DefaultDict[Int, Int] d
   L(a) arr
      d[a]++
   R d
 
F decompose_sum(s)
   R (2 .< Int(s / 2 + 1)).map(a -> (a, @s - a))
 
Set[(Int, Int)] all_pairs_set
L(a) 2..99
   L(b) a + 1 .< 100
      I a + b < 100
         all_pairs_set.add((a, b))
V all_pairs = Array(all_pairs_set)
 
V product_counts = counter(all_pairs.map((c, d) -> c * d))
V unique_products = Set(all_pairs.filter((a, b) -> :product_counts[a * b] == 1))
V s_pairs = all_pairs.filter((a, b) -> all(decompose_sum(a + b).map((x, y) -> (x, y) !C :unique_products)))
 
product_counts = counter(s_pairs.map((c, d) -> c * d))
V p_pairs = s_pairs.filter((a, b) -> :product_counts[a * b] == 1)
 
V sum_counts = counter(p_pairs.map((c, d) -> c + d))
V final_pairs = p_pairs.filter((a, b) -> :sum_counts[a + b] == 1)
 
print(final_pairs)