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

import Data.List (intersect)
 
s1, s2, s3, s4 :: [(Int, Int)]
s1 = [(x, y) | x <- [1 .. 100], y <- [1 .. 100], 1 < x && x < y && x + y < 100]
 
add, mul :: (Int, Int) -> Int
add (x, y) = x + y
mul (x, y) = x * y
 
sumEq, mulEq :: (Int, Int) -> [(Int, Int)]
sumEq p = filter (\q -> add q == add p) s1
mulEq p = filter (\q -> mul q == mul p) s1
 
s2 = filter (\p -> all (\q -> (length $ mulEq q) /= 1) (sumEq p)) s1
s3 = filter (\p -> length (mulEq p `intersect` s2) == 1) s2
s4 = filter (\p -> length (sumEq p `intersect` s3) == 1) s3
 
main = print s4