Sum and Product Puzzle

From Rosetta Code
Task
Sum and Product Puzzle
You are encouraged to solve this task according to the task description, using any language you may know.
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

AWK[edit]

 
# syntax: GAWK -f SUM_AND_PRODUCT_PUZZLE.AWK
BEGIN {
for (s=2; s<=100; s++) {
if ((a=satisfies_statement3(s)) != 0) {
printf("%d (%d+%d)\n",s,a,s-a)
}
}
exit(0)
}
function satisfies_statement1(s, a) { # S says: P does not know the two numbers.
# Given s, for all pairs (a,b), a+b=s, 2 <= a,b <= 99, true if at least one of a or b is composite
for (a=2; a<=int(s/2); a++) {
if (is_prime(a) && is_prime(s-a)) {
return(0)
}
}
return(1)
}
function satisfies_statement2(p, i,j,winner) { # P says: Now I know the two numbers.
# Given p, for all pairs (a,b), a*b=p, 2 <= a,b <= 99, true if exactly one pair satisfies statement 1
for (i=2; i<=int(sqrt(p)); i++) {
if (p % i == 0) {
j = int(p/i)
if (!(2 <= j && j <= 99)) { # in range
continue
}
if (satisfies_statement1(i+j)) {
if (winner) {
return(0)
}
winner = 1
}
}
}
return(winner)
}
function satisfies_statement3(s, a,b,winner) { # S says: Now I know the two numbers.
# Given s, for all pairs (a,b), a+b=s, 2 <= a,b <= 99, true if exactly one pair satisfies statements 1 and 2
if (!satisfies_statement1(s)) {
return(0)
}
for (a=2; a<=int(s/2); a++) {
b = s - a
if (satisfies_statement2(a*b)) {
if (winner) {
return(0)
}
winner = a
}
}
return(winner)
}
function is_prime(x, i) {
if (x <= 1) {
return(0)
}
for (i=2; i<=int(sqrt(x)); i++) {
if (x % i == 0) {
return(0)
}
}
return(1)
}
 

Output:

17 (4+13)

C#[edit]

using System;
using System.Linq;
using System.Collections.Generic;
 
public class Program
{
public static void Main()
{
const int maxSum = 100;
var pairs = (
from X in 2.To(maxSum / 2 - 1)
from Y in (X + 1).To(maxSum - 2).TakeWhile(y => X + y <= maxSum)
select new { X, Y, S = X + Y, P = X * Y }
).ToHashSet();
 
Console.WriteLine(pairs.Count);
 
var uniqueP = pairs.GroupBy(pair => pair.P).Where(g => g.Count() == 1).Select(g => g.Key).ToHashSet();
 
pairs.ExceptWith(pairs.GroupBy(pair => pair.S).Where(g => g.Any(pair => uniqueP.Contains(pair.P))).SelectMany(g => g));
Console.WriteLine(pairs.Count);
 
pairs.ExceptWith(pairs.GroupBy(pair => pair.P).Where(g => g.Count() > 1).SelectMany(g => g));
Console.WriteLine(pairs.Count);
 
pairs.ExceptWith(pairs.GroupBy(pair => pair.S).Where(g => g.Count() > 1).SelectMany(g => g));
Console.WriteLine(pairs.Count);
 
foreach (var pair in pairs) Console.WriteLine(pair);
}
}
 
public static class Extensions
{
public static IEnumerable<int> To(this int start, int end) {
for (int i = start; i <= end; i++) yield return i;
}
 
public static HashSet<T> ToHashSet<T>(this IEnumerable<T> source) => new HashSet<T>(source);
}
Output:
2352
145
86
1
{ X = 4, Y = 13, S = 17, P = 52 }

D[edit]

Translation of: Scala
void main() {
import std.stdio, std.algorithm, std.range, std.typecons;
 
const s1 = cartesianProduct(iota(1, 101), iota(1, 101))
.filter!(p => 1 < p[0] && p[0] < p[1] && p[0] + p[1] < 100)
.array;
 
alias P = const Tuple!(int, int);
enum add = (P p) => p[0] + p[1];
enum mul = (P p) => p[0] * p[1];
enum sumEq = (P p) => s1.filter!(q => add(q) == add(p));
enum mulEq = (P p) => s1.filter!(q => mul(q) == mul(p));
 
const s2 = s1.filter!(p => sumEq(p).all!(q => mulEq(q).walkLength != 1)).array;
const s3 = s2.filter!(p => mulEq(p).setIntersection(s2).walkLength == 1).array;
s3.filter!(p => sumEq(p).setIntersection(s3).walkLength == 1).writeln;
}
Output:
[const(Tuple!(int, int))(4, 13)]

With an older version of the LDC2 compiler replace the cartesianProduct line with:

 
const s1 = iota(1, 101).map!(x => iota(1, 101).map!(y => tuple(x, y))).joiner
 

The .array turn the lazy ranges into arrays. This is a necessary optimization because D lazy Ranges aren't memoized as Haskell lazy lists.

Run-time: about 0.43 seconds with dmd, 0.08 seconds with ldc2.

Elixir[edit]

Translation of: Ruby
defmodule Puzzle do
def sum_and_product do
s1 = for x <- 2..49, y <- x+1..99, x+y<100, do: {x,y}
s2 = Enum.filter(s1, fn p ->
Enum.all?(sumEq(s1,p), fn q -> length(mulEq(s1,q)) != 1 end)
end)
s3 = Enum.filter(s2, fn p -> only1?(mulEq(s1,p), s2) end)
Enum.filter(s3, fn p -> only1?(sumEq(s1,p), s3) end) |> IO.inspect
end
 
defp add({x,y}), do: x + y
 
defp mul({x,y}), do: x * y
 
defp sumEq(s, p), do: Enum.filter(s, fn q -> add(p) == add(q) end)
 
defp mulEq(s, p), do: Enum.filter(s, fn q -> mul(p) == mul(q) end)
 
defp only1?(a, b) do
MapSet.size(MapSet.intersection(MapSet.new(a), MapSet.new(b))) == 1
end
end
 
Puzzle.sum_and_product
Output:
[{4, 13}]

Go[edit]

package main
 
import "fmt"
 
type pair struct{ x, y int }
 
func main() {
//const max = 100
// Use 1685 (the highest with a unique answer) instead
// of 100 just to make it work a little harder :).
const max = 1685
var all []pair
for a := 2; a < max; a++ {
for b := a + 1; b < max-a; b++ {
all = append(all, pair{a, b})
}
}
fmt.Println("There are", len(all), "pairs where a+b <", max, "(and a<b)")
products := countProducts(all)
 
// Those for which no sum decomposition has unique product to are
// S mathimatician's possible pairs.
var sPairs []pair
pairs:
for _, p := range all {
s := p.x + p.y
// foreach a+b=s (a<b)
for a := 2; a < s/2+s&1; a++ {
b := s - a
if products[a*b] == 1 {
// Excluded because P would have a unique product
continue pairs
}
}
sPairs = append(sPairs, p)
}
fmt.Println("S starts with", len(sPairs), "possible pairs.")
//fmt.Println("S pairs:", sPairs)
sProducts := countProducts(sPairs)
 
// Look in sPairs for those with a unique product to get
// P mathimatician's possible pairs.
var pPairs []pair
for _, p := range sPairs {
if sProducts[p.x*p.y] == 1 {
pPairs = append(pPairs, p)
}
}
fmt.Println("P then has", len(pPairs), "possible pairs.")
//fmt.Println("P pairs:", pPairs)
pSums := countSums(pPairs)
 
// Finally, look in pPairs for those with a unique sum
var final []pair
for _, p := range pPairs {
if pSums[p.x+p.y] == 1 {
final = append(final, p)
}
}
 
// Nicely show any answers.
switch len(final) {
case 1:
fmt.Println("Answer:", final[0].x, "and", final[0].y)
case 0:
fmt.Println("No possible answer.")
default:
fmt.Println(len(final), "possible answers:", final)
}
}
 
func countProducts(list []pair) map[int]int {
m := make(map[int]int)
for _, p := range list {
m[p.x*p.y]++
}
return m
}
 
func countSums(list []pair) map[int]int {
m := make(map[int]int)
for _, p := range list {
m[p.x+p.y]++
}
return m
}
 
// not used, manually inlined above
func decomposeSum(s int) []pair {
pairs := make([]pair, 0, s/2)
for a := 2; a < s/2+s&1; a++ {
pairs = append(pairs, pair{a, s - a})
}
return pairs
}
Output:

For x + y < 100 (max = 100):

There are 2304 pairs where a+b < 100 (and a<b)
S starts with 145 possible pairs.
P then has 86 possible pairs.
Answer: 4 and 13

For x + y < 1685 (max = 1685):

There are 706440 pairs where a+b < 1685 (and a<b)
S starts with 50485 possible pairs.
P then has 17485 possible pairs.
Answer: 4 and 13

Run-time ~1 msec and ~600 msec respectively. Could be slightly faster if the slices and maps were given an estimated capacity to start (e.g. (max/2)² for all pairs) to avoid re-allocations (and resulting copies).

Haskell[edit]

Translation of: D
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
Output:
[(4,13)]

Run-time: about 1.97 seconds.

Or, to illustrate some of the available variants, it turns out that we can double performance by slightly rearranging the filters in sumEq and mulEq. It also proves fractionally faster to shed some of the of outer list comprehension sugaring, using >>= or concatMap directly.

For a further doubling of performance, we can redefine add and mul as uncurried versions of (+) and (*).

The y > x condition can usefully be moved upstream – dropping it from the test, and redefining the range of y as [x + 1 .. 100] from the start. (The 1 < x test can also be moved out of the test and into the initial generator).

Finally, as we expect and need only one solution, Haskell's lazy evaluation strategy will avoid wasted tests if we request only the first item from the possible solution stream.

import Data.List (intersect)
 
s1, s2, s3, s4 :: [(Int, Int)]
s1 =
[2 .. 100] >>=
\x ->
[x + 1 .. 100] >>=
\y ->
[ (x, y)
| x + y < 100 ]
 
add, mul :: (Int, Int) -> Int
add = uncurry (+)
 
mul = uncurry (*)
 
sumEq, mulEq :: (Int, Int) -> [(Int, Int)]
sumEq p = filter ((add p ==) . add) s1
 
mulEq p = filter ((mul p ==) . mul) s1
 
s2 = filter (all ((1 /=) . length . mulEq) . sumEq) s1
 
s3 = filter ((1 ==) . length . (`intersect` s2) . mulEq) s2
 
s4 = filter ((1 ==) . length . (`intersect` s3) . sumEq) s3
 
-- TEST -----------------------------------------------------------------------
main :: IO ()
main = print $ take 1 s4
Output:
[(4,13)]

Java[edit]

package org.rosettacode;
 
import java.util.ArrayList;
import java.util.List;
 
 
/**
* This program applies the logic in the Sum and Product Puzzle for the value
* provided by systematically applying each requirement to all number pairs in
* range. Note that the requirements: (x, y different), (x < y), and
* (x, y > MIN_VALUE) are baked into the loops in run(), sumAddends(), and
* productFactors(), so do not need a separate test. Also note that to test a
* solution to this logic puzzle, it is suggested to test the condition with
* maxSum = 1685 to ensure that both the original solution (4, 13) and the
* additional solution (4, 61), and only these solutions, are found. Note
* also that at 1684 only the original solution should be found!
*/

public class SumAndProductPuzzle {
private final long beginning;
private final int maxSum;
private static final int MIN_VALUE = 2;
private List<int[]> firstConditionExcludes = new ArrayList<>();
private List<int[]> secondConditionExcludes = new ArrayList<>();
 
public static void main(String... args){
 
if (args.length == 0){
new SumAndProductPuzzle(100).run();
new SumAndProductPuzzle(1684).run();
new SumAndProductPuzzle(1685).run();
} else {
for (String arg : args){
try{
new SumAndProductPuzzle(Integer.valueOf(arg)).run();
} catch (NumberFormatException e){
System.out.println("Please provide only integer arguments. " +
"Provided argument " + arg + " was not an integer. " +
"Alternatively, calling the program with no arguments " +
"will run the puzzle where maximum sum equals 100, 1684, and 1865.");
}
}
}
}
 
public SumAndProductPuzzle(int maxSum){
this.beginning = System.currentTimeMillis();
this.maxSum = maxSum;
System.out.println("Run with maximum sum of " + String.valueOf(maxSum) +
" started at " + String.valueOf(beginning) + ".");
}
 
public void run(){
for (int x = MIN_VALUE; x < maxSum - MIN_VALUE; x++){
for (int y = x + 1; y < maxSum - MIN_VALUE; y++){
 
if (isSumNoGreaterThanMax(x,y) &&
isSKnowsPCannotKnow(x,y) &&
isPKnowsNow(x,y) &&
isSKnowsNow(x,y)
){
System.out.println("Found solution x is " + String.valueOf(x) + " y is " + String.valueOf(y) +
" in " + String.valueOf(System.currentTimeMillis() - beginning) + "ms.");
}
}
}
System.out.println("Run with maximum sum of " + String.valueOf(maxSum) +
" ended in " + String.valueOf(System.currentTimeMillis() - beginning) + "ms.");
}
 
public boolean isSumNoGreaterThanMax(int x, int y){
return x + y <= maxSum;
}
 
public boolean isSKnowsPCannotKnow(int x, int y){
 
if (firstConditionExcludes.contains(new int[] {x, y})){
return false;
}
 
for (int[] addends : sumAddends(x, y)){
if ( !(productFactors(addends[0], addends[1]).size() > 1) ) {
firstConditionExcludes.add(new int[] {x, y});
return false;
}
}
return true;
}
 
public boolean isPKnowsNow(int x, int y){
 
if (secondConditionExcludes.contains(new int[] {x, y})){
return false;
}
 
int countSolutions = 0;
for (int[] factors : productFactors(x, y)){
if (isSKnowsPCannotKnow(factors[0], factors[1])){
countSolutions++;
}
}
 
if (countSolutions == 1){
return true;
} else {
secondConditionExcludes.add(new int[] {x, y});
return false;
}
}
 
public boolean isSKnowsNow(int x, int y){
 
int countSolutions = 0;
for (int[] addends : sumAddends(x, y)){
if (isPKnowsNow(addends[0], addends[1])){
countSolutions++;
}
}
return countSolutions == 1;
}
 
public List<int[]> sumAddends(int x, int y){
 
List<int[]> list = new ArrayList<>();
int sum = x + y;
 
for (int addend = MIN_VALUE; addend < sum - addend; addend++){
if (isSumNoGreaterThanMax(addend, sum - addend)){
list.add(new int[]{addend, sum - addend});
}
}
return list;
}
 
public List<int[]> productFactors(int x, int y){
 
List<int[]> list = new ArrayList<>();
int product = x * y;
 
for (int factor = MIN_VALUE; factor < product / factor; factor++){
if (product % factor == 0){
if (isSumNoGreaterThanMax(factor, product / factor)){
list.add(new int[]{factor, product / factor});
}
}
}
return list;
}
}
Output:
Run with maximum sum of 100 started at 1492436207694.
Found solution x is 4 y is 13 in 7ms.
Run with maximum sum of 100 ended in 54ms.
Run with maximum sum of 1684 started at 1492436207748.
Found solution x is 4 y is 13 in 9084ms.
Run with maximum sum of 1684 ended in 8234622ms.
Run with maximum sum of 1685 started at 1492444442371.
Found solution x is 4 y is 13 in 8922ms.
Found solution x is 4 y is 61 in 8939ms.
Run with maximum sum of 1685 ended in 8013991ms.

JavaScript[edit]

ES5[edit]

Translation of: Haskell
(function () {
'use strict';
 
// GENERIC FUNCTIONS
 
// concatMap :: (a -> [b]) -> [a] -> [b]
var concatMap = function concatMap(f, xs) {
return [].concat.apply([], xs.map(f));
},
 
// curry :: ((a, b) -> c) -> a -> b -> c
curry = function curry(f) {
return function (a) {
return function (b) {
return f(a, b);
};
};
},
 
// intersectBy :: (a - > a - > Bool) - > [a] - > [a] - > [a]
intersectBy = function intersectBy(eq, xs, ys) {
return xs.length && ys.length ? xs.filter(function (x) {
return ys.some(curry(eq)(x));
}) : [];
},
 
// range :: Int -> Int -> Maybe Int -> [Int]
range = function range(m, n, step) {
var d = (step || 1) * (n >= m ? 1 : -1);
return Array.from({
length: Math.floor((n - m) / d) + 1
}, function (_, i) {
return m + i * d;
});
};
 
// PROBLEM FUNCTIONS
 
// add, mul :: (Int, Int) -> Int
var add = function add(xy) {
return xy[0] + xy[1];
},
mul = function mul(xy) {
return xy[0] * xy[1];
};
 
// sumEq, mulEq :: (Int, Int) -> [(Int, Int)]
var sumEq = function sumEq(p) {
var addP = add(p);
return s1.filter(function (q) {
return add(q) === addP;
});
},
mulEq = function mulEq(p) {
var mulP = mul(p);
return s1.filter(function (q) {
return mul(q) === mulP;
});
};
 
// pairEQ :: ((a, a) -> (a, a)) -> Bool
var pairEQ = function pairEQ(a, b) {
return a[0] === b[0] && a[1] === b[1];
};
 
// MAIN
 
// xs :: [Int]
var xs = range(1, 100);
 
// s1 s2, s3, s4 :: [(Int, Int)]
var s1 = concatMap(function (x) {
return concatMap(function (y) {
return 1 < x && x < y && x + y < 100 ? [
[x, y]
] : [];
}, xs);
}, xs),
 
s2 = s1.filter(function (p) {
return sumEq(p).every(function (q) {
return mulEq(q).length > 1;
});
}),
 
s3 = s2.filter(function (p) {
return intersectBy(pairEQ, mulEq(p), s2).length === 1;
}),
 
s4 = s3.filter(function (p) {
return intersectBy(pairEQ, sumEq(p), s3).length === 1;
});
 
return s4;
})();
 
Output:
[[4, 13]]

(Finished in 0.69s)


ES6[edit]

Translation of: Haskell
(() => {
'use strict';
 
// GENERIC FUNCTIONS
 
// concatMap :: (a -> [b]) -> [a] -> [b]
let concatMap = (f, xs) => [].concat.apply([], xs.map(f)),
 
// curry :: ((a, b) -> c) -> a -> b -> c
curry = f => a => b => f(a, b),
 
// intersectBy :: (a - > a - > Bool) - > [a] - > [a] - > [a]
intersectBy = (eq, xs, ys) => (xs.length && ys.length) ?
xs.filter(x => ys.some(curry(eq)(x))) : [],
 
// range :: Int -> Int -> Maybe Int -> [Int]
range = (m, n, step) => {
let d = (step || 1) * (n >= m ? 1 : -1);
return Array.from({
length: Math.floor((n - m) / d) + 1
}, (_, i) => m + (i * d));
};
 
// PROBLEM FUNCTIONS
 
// add, mul :: (Int, Int) -> Int
let add = xy => xy[0] + xy[1],
mul = xy => xy[0] * xy[1];
 
// sumEq, mulEq :: (Int, Int) -> [(Int, Int)]
let sumEq = p => {
let addP = add(p);
return s1.filter(q => add(q) === addP);
},
mulEq = p => {
let mulP = mul(p)
return s1.filter(q => mul(q) === mulP);
};
 
// pairEQ :: ((a, a) -> (a, a)) -> Bool
let pairEQ = (a, b) => (a[0] === b[0]) && (a[1] === b[1]);
 
 
// MAIN
 
// xs :: [Int]
let xs = range(1, 100);
 
// s1 s2, s3, s4 :: [(Int, Int)]
let s1 = concatMap(x =>
concatMap(y =>
((1 < x) && (x < y) && (x + y) < 100) ? [
[x, y]
] : [],
xs), xs),
s2 = s1.filter(
p => sumEq(p)
.every(
q => mulEq(q)
.length > 1
)
),
s3 = s2.filter(
p => intersectBy(
pairEQ, mulEq(p), s2
)
.length === 1
);
 
return s3.filter(
p => intersectBy(
pairEQ, sumEq(p), s3
)
.length === 1
);
 
})();
Output:
[[4, 13]]

(Finished in 0.77s)


Julia[edit]

From the awk/sidef version. It is also possible to use filters as in the Scala solution, but although less verbose, using filters would be much slower in Julia, which often favors fast for loops over lists for speed.

 
function satisfies1(x)
primeslt100list = [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43,
47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97]
for i in 2:Int(floor(x/2))
if(in(i, primeslt100list) && in(x - i, primeslt100list))
return false
end
end
true
end
 
function satisfies2(x)
oncesatisfied = false
for i in 2:Int(floor(sqrt(x)))
if(x % i == 0)
j = Int(floor(x/i))
if(j > 2 && j <= 99 && satisfies1(i+j))
if(oncesatisfied)
return false
end
oncesatisfied = true
end
end
end
oncesatisfied
end
 
function satisfiesboth(x)
if(!satisfies1(x))
return 0
end
found = 0
for i in 2:Int(floor(x/2))
if(satisfies2(i *(x - i)))
if(found > 0)
return 0
end
found = i
end
end
found
end
 
for i in 2:99
if((j = satisfiesboth(i)) > 0)
println("Solution: ($j, $(i - j))")
end
end
 
Output:

Solution: (4, 13)

Kotlin[edit]

// version 1.1.4-3
 
data class P(val x: Int, val y: Int, val sum: Int, val prod: Int)
 
fun main(args: Array<String>) {
val candidates = mutableListOf<P>()
for (x in 2..49) {
for (y in x + 1..100 - x) {
candidates.add(P(x, y, x + y, x * y))
}
}
 
val sums = candidates.groupBy { it.sum }
val prods = candidates.groupBy { it.prod }
 
val fact1 = candidates.filter { sums[it.sum]!!.all { prods[it.prod]!!.size > 1 } }
val fact2 = fact1.filter { prods[it.prod]!!.intersect(fact1).size == 1 }
val fact3 = fact2.filter { sums[it.sum]!!.intersect(fact2).size == 1 }
print("The only solution is : ")
for ((x, y, _, _) in fact3) println("x = $x, y = $y")
}
Output:
The only solution is : x = 4, y = 13

ooRexx[edit]

version 1[edit]

Translation of: REXX

for comments see REXX version 4.

all      =.set~new
Call time 'R'
cnt.=0
do a=2 to 100
do b=a+1 to 100-2
p=a b
if a+b>100 then leave b
all~put(p)
prd=a*b
cnt.prd+=1
End
End
Say "There are" all~items "pairs where X+Y <=" max "(and X<Y)"
 
spairs=.set~new
Do Until all~items=0
do p over all
d=decompositions(p)
If take Then
spairs=spairs~union(d)
dif=all~difference(d)
Leave
End
all=dif
end
Say "S starts with" spairs~items "possible pairs."
 
sProducts.=0
Do p over sPairs
Parse Var p x y
prod=x*y
sProducts.prod+=1
End
 
pPairs=.set~new
Do p over sPairs
Parse Var p xb yb
prod=xb*yb
If sProducts.prod=1 Then
pPairs~put(p)
End
Say "P then has" pPairs~items "possible pairs."
 
Sums.=0
Do p over pPairs
Parse Var p xc yc
sum=xc+yc
Sums.sum+=1
End
 
final=.set~new
Do p over pPairs
Parse Var p x y
sum=x+y
If Sums.sum=1 Then
final~put(p)
End
 
si=0
Do p Over final
si+=1
sol.si=p
End
Select
When final~items=1 Then Say "Answer:" sol.1
When final~items=0 Then Say "No possible answer."
Otherwise Do; Say final~items "possible answers:"
Do p over final
Say p
End
End
End
Say "Elapsed time:" time('E') "seconds"
Exit
 
decompositions: Procedure Expose cnt. take spairs
epairs=.set~new
Use Arg p
Parse Var p aa bb
s=aa+bb
take=1
Do xa=2 To s/2
ya=s-xa
pp=xa ya
epairs~put(pp)
prod=xa*ya
If cnt.prod=1 Then
take=0
End
return epairs
Output:
There are 2352 pairs where X+Y <= MAX (and X<Y)
S starts with 145 possible pairs.
P then has 86 possible pairs.
Answer: 4 13
Elapsed time: 0.016000 seconds

version 2[edit]

Uses objects for storing the number pairs. Note the computed hash value and the == mathod (required to make the set difference work)

all      =.set~new
Call time 'R'
cnt.=0
do a=2 to 100
do b=a+1 to 100-2
p=.pairs~new(a,b)
if p~sum>100 then leave b
all~put(p)
prd=p~prod
cnt.prd+=1
End
End
Say "There are" all~items "pairs where X+Y <=" max "(and X<Y)"
 
spairs=.set~new
Do Until all~items=0
do p over all
d=decompositions(p)
If take Then
spairs=spairs~union(d)
dif=all~difference(d)
Leave
End
all=dif
end
Say "S starts with" spairs~items "possible pairs."
 
sProducts.=0
Do p over sPairs
prod=p~prod
sProducts.prod+=1
End
 
pPairs=.set~new
Do p over sPairs
prod=p~prod
If sProducts.prod=1 Then
pPairs~put(p)
End
Say "P then has" pPairs~items "possible pairs."
 
Sums.=0
Do p over pPairs
sum=p~sum
Sums.sum+=1
End
 
final=.set~new
Do p over pPairs
sum=p~sum
If Sums.sum=1 Then
final~put(p)
End
 
si=0
Do p Over final
si+=1
sol.si=p
End
Select
When final~items=1 Then Say "Answer:" sol.1~string
When final~items=0 Then Say "No possible answer."
Otherwise Do; Say final~items "possible answers:"
Do p over final
Say p~string
End
End
End
Say "Elapsed time:" time('E') "seconds"
Exit
 
decompositions: Procedure Expose cnt. take spairs
epairs=.set~new
Use Arg p
s=p~sum
take=1
Do xa=2 To s/2
ya=s-xa
pp=.pairs~new(xa,ya)
epairs~put(pp)
prod=pp~prod
If cnt.prod=1 Then
take=0
End
return epairs
 
::class pairs
::attribute a -- allow access to attribute
::attribute b -- allow access to attribute
::attribute sum -- allow access to attribute
::attribute prod -- allow access to attribute
 
-- only the strict equality form is needed for the collection classes,
::method "=="
expose a b
use strict arg other
return a == other~a & b == other~b
 
-- not needed to make the set difference work, but added for completeness
::method "\=="
expose a b
use strict arg other
return a \== other~a | b \== other~b
 
::method hashCode
expose hash
return hash
 
::method init -- create pair, calculate sum, product
-- and index (blank delimited values)
expose hash a b sum prod oid
use arg a, b
hash = a~hashCode~bitxor(b~hashCode) -- create hash value
sum =a+b -- sum
prod=a*b -- product
 
::method string -- this creates the string to be shown
expose a b
return "[x="||a",y="||b"]"
Output:
There are 2352 pairs where X+Y <= MAX (and X<Y)
S starts with 145 possible pairs.
P then has 86 possible pairs.
Answer: [x=4,y=13]
Elapsed time: 0.079000 seconds

Perl 6[edit]

Translation of: Python
Works with: Rakudo version 2016.07
sub grep-unique (&by, @list) { @list.classify(&by).values.grep(* == 1).map(*[0]) }
sub sums ($n) { ($_, $n - $_ for 2 .. $n div 2) }
sub sum ([$x, $y]) { $x + $y }
sub product ([$x, $y]) { $x * $y }
 
my @all-pairs = (|($_ X $_+1 .. 98) for 2..97);
 
# Fact 1:
my %p-unique := Set.new: map ~*, grep-unique &product, @all-pairs;
my @s-pairs = @all-pairs.grep: { none (%p-unique{~$_} for sums sum $_) };
 
# Fact 2:
my @p-pairs = grep-unique &product, @s-pairs;
 
# Fact 3:
my @final-pairs = grep-unique &sum, @p-pairs;
 
printf "X = %d, Y = %d\n", |$_ for @final-pairs;
Output:
X = 4, Y = 13

Phix[edit]

Translation of: AWK

Runs in 0.03s

function is_prime(integer x)
if x>3 then
for i=2 to floor(sqrt(x)) do
if mod(x,i)=0 then
return 0
end if
end for
end if
return 1
end function
 
function satisfies_statement1(integer s)
-- S says: P does not know the two numbers.
-- Given s, for /all/ pairs (a,b), a+b=s, 2<=a,b<=99, at least one of a or b is composite
for a=2 to floor(s/2) do
if is_prime(a) and is_prime(s-a) then
return 0
end if
end for
return 1
end function
 
function satisfies_statement2(integer p)
-- P says: Now I know the two numbers.
-- Given p, for /all/ pairs (a,b), a*b=p, 2<=a,b<=99, exactly one pair satisfies statement 1
integer winner = 0
for i=2 to floor(sqrt(p)) do
if mod(p,i)=0 then
integer j = floor(p/i)
if 2<=j and j<=99 then
if satisfies_statement1(i+j) then
if winner then return 0 end if
winner = 1
end if
end if
end if
end for
return winner
end function
 
function satisfies_statement3(integer s)
-- S says: Now I know the two numbers.
-- Given s, for /all/ pairs (a,b), a+b=s, 2<=a,b<=99, exactly one pair satisfies statements 1 and 2
integer winner = 0
if satisfies_statement1(s) then
for a=2 to floor(s/2) do
if satisfies_statement2(a*(s-a)) then
if winner then return 0 end if
winner = a
end if
end for
end if
return winner
end function
 
for s=2 to 100 do
integer a = satisfies_statement3(s)
if a!=0 then
printf(1,"%d (%d+%d)\n",{s,a,s-a})
end if
end for
Output:
17 (4+13)

Python[edit]

Based on the Python solution from Wikipedia:

#!/usr/bin/env python
 
from collections import Counter
 
def decompose_sum(s):
return [(a,s-a) for a in range(2,int(s/2+1))]
 
# Generate all possible pairs
all_pairs = set((a,b) for a in range(2,100) for b in range(a+1,100) if a+b<100)
 
# Fact 1 --> Select pairs for which all sum decompositions have non-unique product
product_counts = Counter(c*d for c,d in all_pairs)
unique_products = set((a,b) for a,b in all_pairs if product_counts[a*b]==1)
s_pairs = [(a,b) for a,b in all_pairs if
all((x,y) not in unique_products for (x,y) in decompose_sum(a+b))]
 
# Fact 2 --> Select pairs for which the product is unique
product_counts = Counter(c*d for c,d in s_pairs)
p_pairs = [(a,b) for a,b in s_pairs if product_counts[a*b]==1]
 
# Fact 3 --> Select pairs for which the sum is unique
sum_counts = Counter(c+d for c,d in p_pairs)
final_pairs = [(a,b) for a,b in p_pairs if sum_counts[a+b]==1]
 
print(final_pairs)
Output:
[(4, 13)]

Racket[edit]

Translation of: D
To calculate the results faster this program use memorization. So it has a modified version of sum= and mul= to increase the chances of reusing the results.
#lang racket
(define-syntax-rule (define/mem (name args ...) body ...)
(begin
(define cache (make-hash))
(define (name args ...)
(hash-ref! cache (list args ...) (lambda () body ...)))))
 
(define (sum p) (+ (first p) (second p)))
(define (mul p) (* (first p) (second p)))
 
(define (sum= p s) (filter (lambda (q) (= p (sum q))) s))
(define (mul= p s) (filter (lambda (q) (= p (mul q))) s))
 
(define (puzzle tot)
(printf "Max Sum: ~a\n" tot)
(define s1 (for*/list ([x (in-range 2 (add1 tot))]
[y (in-range (add1 x) (- (add1 tot) x))])
(list x y)))
(printf "Possible pairs: ~a\n" (length s1))
 
(define/mem (sumEq/all p) (sum= p s1))
(define/mem (mulEq/all p) (mul= p s1))
 
(define s2 (filter (lambda (p) (andmap (lambda (q)
(not (= (length (mulEq/all (mul q))) 1)))
(sumEq/all (sum p))))
s1))
(printf "Initial pairs for S: ~a\n" (length s2))
 
(define s3 (filter (lambda (p) (= (length (mul= (mul p) s2)) 1))
s2))
(displayln (length s3))
(printf "Pairs for P: ~a\n" (length s3))
 
(define s4 (filter (lambda (p) (= (length (sum= (sum p) s3)) 1))
s3))
(printf "Final pairs for S: ~a\n" (length s4))
 
(displayln s4))
 
(puzzle 100)
Output:
Max Sum: 100
Possible pairs: 2352
Initial pairs for S: 145
Pairs for P: 86
Final pairs for S: 1
((4 13))

REXX[edit]

version 1[edit]

I tried hard to understand/translate the algorithms shown so far (16 Oct 2016) Unfortunately to no avail (not knowing the semantics of the used languages). Finally I was successful by implementing the rules referred to in Wikipedia http://www.win.tue.nl/~gwoegi/papers/freudenthal1.pdf which had a very clear description.

debug=0
If debug Then Do
oid='sppn.txt'; 'erase' oid
End
Call time 'R'
all_pairs=''
cnt.=0
i=0
/* first take all possible pairs 2<=x<y with x+y<=100 */
/* and compute the respective sums and products */
/* count the number of times a sum or product occurs */
Do x=2 To 98
Do y=x+1 To 100-x
x=right(x,2,0)
y=right(y,2,0)
all_pairs=all_pairs x'/'y
i=i+1
x.i=x
y.i=y
sum=x+y
prd=x*y
cnt.0s.sum=cnt.0s.sum+1
cnt.0p.prd=cnt.0p.prd+1
End
End
n=i
/* now compute the possible pairs for each sum sum_d.sum */
/* and product prd_d.prd */
/* also the list of possible sums and products suml, prdl*/
sum_d.=''
prd_d.=''
suml=''
prdl=''
Do i=1 To n
x=x.i
y=y.i
x=right(x,2,0)
y=right(y,2,0)
sum=x+y
prd=x*y
cnt.0s.x.y=cnt.0s.sum
cnt.0p.x.y=cnt.0p.prd
sum_d.sum=sum_d.sum x'/'y
prd_d.prd=prd_d.prd x'/'y
If wordpos(sum,suml)=0 Then suml=suml sum
If wordpos(prd,prdl)=0 Then prdl=prdl prd
End
Say n 'possible pairs'
Call o 'SUM'
suml=wordsort(suml)
prdl=wordsort(prdl)
sumlc=suml
si=0
pi=0
Do While sumlc>''
Parse Var sumlc sum sumlc
si=si+1
sum.si=sum
si.sum=si
If sum=17 Then sx=si
temp=prdl
Do While temp>''
Parse Var temp prd temp
If si=1 Then Do
pi=pi+1
prd.pi=prd
pi.prd=pi
If prd=52 Then px=pi
End
A.prd.sum='+'
End
End
sin=si
pin=pi
Call o 'SUM'
Do si=1 To sin
Call o f5(si) f3(sum.si)
End
Call o 'PRD'
Do pi=1 To pin
Call o f5(pi) f6(prd.pi)
End
a.='-'
Do pi=1 To pin
prd=prd.pi
Do si=1 To sin
sum=sum.si
Do sj=1 To words(sum_d.sum)
If wordpos(word(sum_d.sum,sj),prd_d.prd)>0 Then
Parse Value word(sum_d.sum,sj) with x '/' y
prde=x*y
sume=x+y
pa=pi.prde
sa=si.sume
a.pa.sa='+'
End
End
End
Call show '1'
 
Do pi=1 To pin
prow=''
cnt=0
Do si=1 To sin
If a.pi.si='+' Then Do
cnt=cnt+1
pj=pi
sj=si
End
End
If cnt=1 Then
a.pj.sj='1'
End
Call show '2'
 
Do si=1 To sin
Do pi=1 To pin
If a.pi.si='1' Then Leave
End
If pi<=pin Then Do
Do pi=1 To pin
If a.pi.si='+' Then
a.pi.si='2'
End
End
End
Call show '3'
 
Do pi=1 To pin
prow=''
Do si=1 To sin
prow=prow||a.pi.si
End
If count('+',prow)>1 Then Do
Do si=1 To sin
If a.pi.si='+' Then
a.pi.si='3'
End
End
End
Call show '4'
 
Do si=1 To sin
scol=''
Do pi=1 To pin
scol=scol||a.pi.si
End
If count('+',scol)>1 Then Do
Do pi=1 To pin
If a.pi.si='+' Then
a.pi.si='4'
End
End
End
Call show '5'
 
sol=0
Do pi=1 To pin
Do si=1 To sin
If a.pi.si='+' Then Do
Say sum.si prd.pi
sum=sum.si
prd=prd.pi
sol=sol+1
End
End
End
Say sol 'solution(s)'
Say ' possible pairs'
Say 'Product='prd prd_d.52
Say ' Sum='sum sum_d.17
Say 'The only pair in both lists is 04/13.'
Say 'Elapsed time:' time('E') 'seconds'
Exit
show:
If debug Then Do
Call o 'show' arg(1)
Do pi=1 To 60
ol=''
Do si=1 To 60
ol=ol||a.pi.si
End
Call o ol
End
Say 'a.'px'.'sx'='a.px.sx
End
Return
 
Exit
o: Return lineout(oid,arg(1))
f3: Return format(arg(1),3)
f4: Return format(arg(1),4)
f5: Return format(arg(1),5)
f6: Return format(arg(1),6)
 
count: Procedure
Parse Arg c,s
s=translate(s,c,c||xrange('00'x,'ff'x))
s=space(s,0)
Return length(s)
Output:
2352 possible pairs
17 52
1 solution(s)
            possible pairs
Product=52  02/26 04/13
    Sum=17  02/15 03/14 04/13 05/12 06/11 07/10 08/09
The only pair in both lists is 04/13.
Elapsed time: 4.891000 seconds

version 2[edit]

Translation of: AWK
Call time 'R'
Do s=2 To 100
a=satisfies_statement3(s)
If a>0 Then Do
p=a*(s-a)
Say a'/'||(s-a) 's='s 'p='p
End
End
Say 'Elapsed time:' time('E') 'seconds'
Exit
 
satisfies_statement1: Procedure
Parse Arg s
Do a=2 To s/2
If is_prime(a) & is_prime(s-a) Then
Return 0
End
Return 1
 
satisfies_statement2: Procedure
Parse Arg p
winner=0
Do i=2 By 1 While i**2<p
If p//i=0 Then Do
j=p%i
If 2<=j & j<=99 Then Do
if satisfies_statement1(i+j) Then Do
if winner Then
Return 0
winner=1
End
End
End
End
Return winner
 
satisfies_statement3: Procedure
Parse Arg s
winner=0
If satisfies_statement1(s)=0 Then
Return 0
Do a=2 To s/2
b=s-a
If satisfies_statement2(a*b) Then Do
If winner>0 Then
Return 0
winner=a
End
End
Return winner
 
is_prime: Procedure
call Trace 'O'
Parse Arg x
If x<=3 Then Return 1
i=2
Do i=2 By 1 While i**2<=x
If datatype(x/i,'W') Then Return 0
End
Return 1
Output:
4/13 s=17 p=52
Elapsed time: 0.078000 seconds

version 3[edit]

Translation of: GO
/*---------------------------------------------------------------------
* 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?
*--------------------------------------------------------------------*/

Call time 'R'
max=100
Products.=0
all=''
Do x=2 To max
Do y=x+1 To max-2
If x+y<=100 Then Do
all=all x'/'y
prod=x*y; Products.prod=Products.prod+1
End
End
End
Say "There are" words(all) "pairs where X+Y <=" max "(and X<Y)"
/*---------------------------------------------------------------------
* First eliminate all pairs where the product is unique:
* For each pair we look at the decompositions of the sum (x+y).
* If for any of these decompositions (xa/ya) the product is unique
* then x/y cannot be the solution of the puzzle and we eliminate it
* from the list of possible pairs
*--------------------------------------------------------------------*/

sPairs=''
Do i=1 To words(all)
xy=word(all,i)
Parse Var xy x '/' Y
Parse Var xy xx '/' Yy
s=x+y
take=1
Do xa=2 To s/2
ya=s-xa
prod=xa*ya
If products.prod=1 Then Do
take=0
Iterate i
End
End
If take Then
sPairs=sPairs xy
End
Say "S starts with" words(sPairs) "possible pairs."
 
/*---------------------------------------------------------------------
* From the REMAINING pairs take only these where the product is unique:
* For each pair we look at the decompositions of the known product.
* If for any of these decompositions (xb/yb) the product is unique
* then xb/yb can be the solution of the puzzle and we add it
* to the list of possible pairs.
*--------------------------------------------------------------------*/

sProducts.=0
Do i=1 To words(sPairs)
xy=word(sPairs,i)
Parse Var xy x '/' y
prod=x*y
sProducts.prod=sProducts.prod+1
End
pPairs=''
Do i=1 To words(sPairs)
xy=word(sPairs,i)
Parse Var xy x '/' y
prod=x*y
If sProducts.prod=1 Then
pPairs=pPairs xy
End
Say "P then has" words(pPairs) "possible pairs."
 
/*---------------------------------------------------------------------
* From the now REMAINING pairs take only these where the sum is unique
* Now we look at all possible pairs and find the one (xc/yc)
* with a unique sum which must be the sum we knew from the beginning.
* The pair xc/yc is then the solution
*--------------------------------------------------------------------*/

Sums.=0
Do i=1 To words(pPairs)
xy=word(pPairs,i)
Parse Var xy x '/' y
sum=x+y
Sums.sum=Sums.sum+1
End
final=''
Do i=1 To words(pPairs)
xy=word(pPairs,i)
Parse Var xy x '/' y
sum=x+y
If Sums.sum=1 Then
final = final xy
End
Select
When words(final)=1 Then Say "Answer:" strip(final)
When words(final)=0 Then Say "No possible answer."
Otherwise Do; Say words(final) "possible answers:"
Say strip(final)
End
End
Say "Elapsed time:" time('E') "seconds"
Exit
Output:
There are 2352 pairs where X+Y <= 100 (and X<Y)
S starts with 145 possible pairs.
P then has 86 possible pairs.
Answer: 4/13
Elapsed time: 0.045000 seconds

version 4[edit]

Now that I have understood the logic (I am neither S nor P) I have created an alternative to version 3.

/*---------------------------------------------------------------------
* 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?
*--------------------------------------------------------------------*/

Call time 'R'
max=100
Products.=0
all=''
Do x=2 To max
Do y=x+1 To max-2
If x+y<=100 Then Do
all=all x'/'y
prod=x*y; Products.prod=Products.prod+1
End
End
End
Say "There are" words(all) "pairs where X+Y <=" max "(and X<Y)"
/*---------------------------------------------------------------------
* First eliminate all pairs where the product is unique:
* For each pair we look at the decompositions of the sum (x+y).
* If for any of these decompositions (xa/ya) the product is unique
* then the given sum cannot be the sum of the pair we are looking for
* Otherwise all pairs in the sum's decompositions are eligible.
*--------------------------------------------------------------------*/

sPairs=''
done.=0
Do i=1 To words(all)
xy=word(all,i)
If done.xy Then Iterate
Parse Var xy x '/' y
s=x+y
take=1
el=''
Do xa=2 To s/2
ya=s-xa
m=xa'/'ya
done.m=1
el=el m
prod=xa*ya
If products.prod=1 Then
take=0
End
If take Then
sPairs=sPairs el
End
Say "S starts with" words(sPairs) "possible pairs."
 
/*---------------------------------------------------------------------
* From the REMAINING pairs take only these where the product is unique:
* For each pair we look at the decompositions of the known product.
* If for any of these decompositions (xb/yb) the product is unique
* then xb/yb can be the solution of the puzzle and we add it
* to the list of possible pairs.
*--------------------------------------------------------------------*/

sProducts.=0
Do i=1 To words(sPairs)
xy=word(sPairs,i)
Parse Var xy x '/' y
prod=x*y
sProducts.prod=sProducts.prod+1
End
pPairs=''
Do i=1 To words(sPairs)
xy=word(sPairs,i)
Parse Var xy xb '/' yb
prod=xb*yb
If sProducts.prod=1 Then
pPairs=pPairs xy
End
Say "P then has" words(pPairs) "possible pairs."
 
/*---------------------------------------------------------------------
* From the now REMAINING pairs take only these where the sum is unique
* Now we look at all possible pairs and find the one (xc/yc)
* with a unique sum which must be the sum we knew from the beginning.
* The pair xc/yc is then the solution
*--------------------------------------------------------------------*/

Sums.=0
Do i=1 To words(pPairs)
xy=word(pPairs,i)
Parse Var xy xc '/' yc
sum=xc+yc
Sums.sum=Sums.sum+1
End
final=''
Do i=1 To words(pPairs)
xy=word(pPairs,i)
Parse Var xy x '/' y
sum=x+y
If Sums.sum=1 Then
final = final xy
End
Select
When words(final)=1 Then Say "Answer:" strip(final)
When words(final)=0 Then Say "No possible answer."
Otherwise Do; Say words(final) "possible answers:"
Say strip(final)
End
End
Say "Elapsed time:" time('E') "seconds"
Exit
Output:
There are 2352 pairs where X+Y <= 100 (and X<Y)
S starts with 145 possible pairs.
P then has 86 possible pairs.
Answer: 4/13
Elapsed time: 0.032000 seconds

version 5[edit]

/*REXX program solves the  Sum and Product Puzzle (also known as the Impossible Puzzle).*/
parse arg H; if H=='' | H=="," then H=100 /*obtain optional HIGH number from CL*/
@.=0; call genPrimes H /*generate all primes ≤ H. */
do s=2 for H-1; if C1(s)==0 then iterate /*find and display the puzzle solution.*/
$=0; do m=2 to s%2 /* [↓] check for uniqueness of product*/
if C2(m * (s-m)) then do; if $>0 then iterate s; $=m; end
end /*m*/
if $>0 then say "The numbers are: " $ " and " s-$
end /*s*/
exit /*stick a fork in it, we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
C1: procedure expose @.; parse arg s /*validate the first puzzle condition. */
do a=2 to s%2; if @.a then do; _=s-a; if @._ then return 0; end; end; return 1
/*──────────────────────────────────────────────────────────────────────────────────────*/
C2: procedure expose @. H; parse arg p; $=0 /*validate the second puzzle condition.*/
do j=2 while j*j<p /*perform up to the squart root of P. */
if p//j==0 then do; q=p%j /* // is REXX's remainder operator.*/
if q>=2 then if q<=H then if C1(j+q) then do; if $ then return 0
$=1
end
end
end /*j*/; return $
/*──────────────────────────────────────────────────────────────────────────────────────*/
genPrimes: do j=3 by 2 to arg(1) /*find all odd primes ≤ 1st argument.*/
do k=3 until k*k>j; if @.k==0 then iterate; if j//k==0 then iterate j
end /*k*/; @.j=1 /*found prime number: J */
end /*j*/; @.2=1; return /*assign the even prime.*/
output   when using the default input:
The numbers are:   4  and  13

Ruby[edit]

Translation of: D
def add(x,y) x + y end
def mul(x,y) x * y end
 
def sumEq(s,p) s.select{|q| add(*p) == add(*q)} end
def mulEq(s,p) s.select{|q| mul(*p) == mul(*q)} end
 
s1 = (a = *2...100).product(a).select{|x,y| x<y && x+y<100}
s2 = s1.select{|p| sumEq(s1,p).all?{|q| mulEq(s1,q).size != 1} }
s3 = s2.select{|p| (mulEq(s1,p) & s2).size == 1}
p s3.select{|p| (sumEq(s1,p) & s3).size == 1}
Output:
[[4, 13]]

Scala[edit]

object ImpossiblePuzzle extends App {
type XY = (Int, Int)
val step0 = for {
x <- 1 to 100
y <- 1 to 100
if 1 < x && x < y && x + y < 100
} yield (x, y)
 
def sum(xy: XY) = xy._1 + xy._2
def prod(xy: XY) = xy._1 * xy._2
def sumEq(xy: XY) = step0 filter { sum(_) == sum(xy) }
def prodEq(xy: XY) = step0 filter { prod(_) == prod(xy) }
 
val step2 = step0 filter { sumEq(_) forall { prodEq(_).size != 1 }}
val step3 = step2 filter { prodEq(_).intersect(step2).size == 1 }
val step4 = step3 filter { sumEq(_).intersect(step3).size == 1 }
println(step4)
}
Output:
Vector((4,13))

Run-time: about 3.82 seconds.

Scheme[edit]

 
(import (scheme base)
(scheme cxr)
(scheme write)
(srfi 1))
 
;; utility method to find unique sum/product in given list
(define (unique-items lst key)
(let ((all-items (map key lst)))
(filter (lambda (i) (= 1 (count (lambda (p) (= p (key i)))
all-items)))
lst)))
 
;; list of all (x y x+y x*y) combinations with y > x
(define *xy-pairs*
(apply append
(map (lambda (i)
(map (lambda (j)
(list i j (+ i j) (* i j)))
(iota (- 98 i) (+ 1 i))))
(iota 96 2))))
 
;; S says "P does not know X and Y"
(define *products* ; get products which have multiple decompositions
(let ((all-products (map fourth *xy-pairs*)))
(filter (lambda (p) (> (count (lambda (i) (= i p)) all-products) 1))
all-products)))
 
(define *fact-1* ; every x+y has x*y in *products*
(filter (lambda (i)
(every (lambda (p) (memq (fourth p) *products*))
(filter (lambda (p) (= (third i) (third p))) *xy-pairs*)))
*xy-pairs*))
 
;; P says "Now I know X and Y"
(define *fact-2* ; find the unique X*Y
(unique-items *fact-1* fourth))
 
;; S says "Now I also know X and Y"
(define *fact-3* ; find the unique X+Y
(unique-items *fact-2* third))
 
(display (string-append "Initial pairs: " (number->string (length *xy-pairs*)) "\n"))
(display (string-append "After S: " (number->string (length *fact-1*)) "\n"))
(display (string-append "After P: " (number->string (length *fact-2*)) "\n"))
(display (string-append "After S: " (number->string (length *fact-3*)) "\n"))
(display (string-append "X: "
(number->string (caar *fact-3*))
" Y: "
(number->string (cadar *fact-3*))
"\n"))
 
Output:
Initial pairs: 4656
After S: 145
After P: 86
After S: 1
X: 4 Y: 13

Sidef[edit]

Translation of: Perl 6
func grep_uniq(a, by) { a.group_by{ .(by) }.values.grep{.len == 1}.map{_[0]} }
func sums (n) { 2 .. n//2 -> map {|i| [i, n-i] } }
 
var pairs = (2..97 -> map {|i| ([i] ~X (i+1 .. 98))... })
 
var p_uniq = Hash()
p_uniq{grep_uniq(pairs, :prod).map { .to_s }...} = ()
 
var s_pairs = pairs.grep {|p| sums(p.sum).all { !p_uniq.contains(.to_s) } }
var p_pairs = grep_uniq(s_pairs, :prod)
var f_pairs = grep_uniq(p_pairs, :sum)
 
f_pairs.each { |p| printf("X = %d, Y = %d\n", p...) }
Output:
X = 4, Y = 13

zkl[edit]

Damn it Jim, I'm a programmer, not a logician. So I translated the python code found in https://qmaurmann.wordpress.com/2013/08/10/sam-and-polly-and-python/ but I don't understand it. It does seem quite a bit more efficient than the Scala code, on par with the Python code.

mul:=Utils.Helpers.summer.fp1('*,1); //-->list.reduce('*,1), multiply list items
var allPairs=[[(a,b); [2..100]; { [a+1..100] },{ a+b<100 }; ROList]]; // 2,304 pairs
 
sxys,pxys:=Dictionary(),Dictionary(); // hashes of allPairs sums and products: 95,1155
foreach xy in (allPairs){ sxys.appendV(xy.sum(),xy); pxys.appendV(xy:mul(_),xy) }
 
sOK:= 'wrap(s){ (not sxys[s].filter1('wrap(xy){ pxys[xy:mul(_)].len()<2 })) };
pOK:= 'wrap(p){ 1==pxys[p].filter('wrap([(x,y)]){ sOK(x+y) }).len() };
sOK2:='wrap(s){ 1==sxys[s].filter('wrap(xy){ pOK(xy:mul(_)) }).len() };
allPairs.filter('wrap([(x,y)]){ sOK(x+y) and pOK(x*y) and sOK2(x+y) })
.println();

[[ ]] denotes list comprehension, filter1 returns (and stops at) the first thing that is "true", 'wrap creates a closure so the "wrapped" code/function can see local variables (read only). In a [function] prototype, the "[(x,y)]xy]" notation says xy is a list like thing, assign the parts to x & y (xy is optional), used here to just to do it both ways. The ":" says take the LHS and stuff it into the "_".

Output:
L(L(4,13))