Sum and product puzzle: Difference between revisions
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17 (4+13) |
17 (4+13) |
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</pre> |
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=={{header|Picat}}== |
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<lang Picat> |
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import cp. |
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main => |
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N = 98, |
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PD = new_array(N*N), % PD[I] = no. of product decompositions of I |
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foreach(I in 1..N*N) PD[I] = 0 end, |
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foreach(X in 2..N-1, Y in X+1..N) PD[X * Y] := PD[X * Y] + 1 end, |
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% Fact 1: S says "P does not know X and Y.", i.e. |
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% For every possible sum decomposition of the number X+Y, the product has in turn more than one product decomposition: |
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Solutions1 = [[X,Y] : X in 2..N-1, Y in X+1..100-X, foreach(XX in 2..X+Y-3) PD[XX * (X+Y-XX)] #> 1 end], |
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% Fact 2: P says "Now I know X and Y.", i.e. |
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% The number X*Y has only one product decomposition for which fact 1 is true: |
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Solutions2 = [[X,Y] : [X,Y] in Solutions1, foreach([XX,YY] in Solutions1, XX * YY = X * Y) XX = X, YY = Y end], |
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% Fact 3: S says "Now I also know X and Y.", i.e. |
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% The number X+Y has only one sum decomposition for which fact 2 is true. |
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Solutions3 = [[X,Y] : [X,Y] in Solutions2, foreach([XX,YY] in Solutions2, XX + YY = X + Y) XX = X, YY = Y end], |
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println(Solutions3). |
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</lang> |
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Output: |
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<pre>[[4,13]]</pre> |
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=={{header|Python}}== |
=={{header|Python}}== |
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Revision as of 08:18, 1 June 2022
You are encouraged to solve this task according to the task description, using any language you may know.
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?
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.
- Wikipedia: Sum and Product Puzzle
11l
<lang 11l>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)</lang>
- Output:
[(4, 13)]
AWK
<lang AWK>
- 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)
} </lang>
Output:
17 (4+13)
C
<lang c>#include <stdbool.h>
- include <stdio.h>
- include <stdlib.h>
typedef struct node_t {
int x, y; struct node_t *prev, *next;
} node;
node *new_node(int x, int y) {
node *n = malloc(sizeof(node)); n->x = x; n->y = y; n->next = NULL; n->prev = NULL; return n;
}
void free_node(node **n) {
if (n == NULL) {
return;
}
(*n)->prev = NULL; (*n)->next = NULL;
free(*n);
*n = NULL;
}
typedef struct list_t {
node *head; node *tail;
} list;
list make_list() {
list lst = { NULL, NULL };
return lst;
}
void append_node(list *const lst, int x, int y) {
if (lst == NULL) {
return;
}
node *n = new_node(x, y);
if (lst->head == NULL) {
lst->head = n;
lst->tail = n;
} else {
n->prev = lst->tail;
lst->tail->next = n;
lst->tail = n;
}
}
void remove_node(list *const lst, const node *const n) {
if (lst == NULL || n == NULL) {
return;
}
if (n->prev != NULL) {
n->prev->next = n->next;
if (n->next != NULL) {
n->next->prev = n->prev;
} else {
lst->tail = n->prev;
}
} else {
if (n->next != NULL) {
n->next->prev = NULL;
lst->head = n->next;
}
}
free_node(&n);
}
void free_list(list *const lst) {
node *ptr;
if (lst == NULL) {
return;
}
ptr = lst->head;
while (ptr != NULL) {
node *nxt = ptr->next;
free_node(&ptr);
ptr = nxt;
}
lst->head = NULL; lst->tail = NULL;
}
void print_list(const list *lst) {
node *it;
if (lst == NULL) {
return;
}
for (it = lst->head; it != NULL; it = it->next) {
int sum = it->x + it->y;
int prod = it->x * it->y;
printf("[%d, %d] S=%d P=%d\n", it->x, it->y, sum, prod);
}
}
void print_count(const list *const lst) {
node *it; int c = 0;
if (lst == NULL) {
return;
}
for (it = lst->head; it != NULL; it = it->next) {
c++;
}
if (c == 0) {
printf("no candidates\n");
} else if (c == 1) {
printf("one candidate\n");
} else {
printf("%d candidates\n", c);
}
}
void setup(list *const lst) {
int x, y;
if (lst == NULL) {
return;
}
// numbers must be greater than 1
for (x = 2; x <= 98; x++) {
// numbers must be unique, and sum no more than 100
for (y = x + 1; y <= 98; y++) {
if (x + y <= 100) {
append_node(lst, x, y);
}
}
}
}
void remove_by_sum(list *const lst, const int sum) {
node *it;
if (lst == NULL) {
return;
}
it = lst->head;
while (it != NULL) {
int s = it->x + it->y;
if (s == sum) {
remove_node(lst, it);
it = lst->head;
} else {
it = it->next;
}
}
}
void remove_by_prod(list *const lst, const int prod) {
node *it;
if (lst == NULL) {
return;
}
it = lst->head;
while (it != NULL) {
int p = it->x * it->y;
if (p == prod) {
remove_node(lst, it);
it = lst->head;
} else {
it = it->next;
}
}
}
void statement1(list *const lst) {
short *unique = calloc(100000, sizeof(short)); node *it, *nxt;
for (it = lst->head; it != NULL; it = it->next) {
int prod = it->x * it->y;
unique[prod]++;
}
it = lst->head;
while (it != NULL) {
int prod = it->x * it->y;
nxt = it->next;
if (unique[prod] == 1) {
remove_by_sum(lst, it->x + it->y);
it = lst->head;
} else {
it = nxt;
}
}
free(unique);
}
void statement2(list *const candidates) {
short *unique = calloc(100000, sizeof(short)); node *it, *nxt;
for (it = candidates->head; it != NULL; it = it->next) {
int prod = it->x * it->y;
unique[prod]++;
}
it = candidates->head;
while (it != NULL) {
int prod = it->x * it->y;
nxt = it->next;
if (unique[prod] > 1) {
remove_by_prod(candidates, prod);
it = candidates->head;
} else {
it = nxt;
}
}
free(unique);
}
void statement3(list *const candidates) {
short *unique = calloc(100, sizeof(short)); node *it, *nxt;
for (it = candidates->head; it != NULL; it = it->next) {
int sum = it->x + it->y;
unique[sum]++;
}
it = candidates->head;
while (it != NULL) {
int sum = it->x + it->y;
nxt = it->next;
if (unique[sum] > 1) {
remove_by_sum(candidates, sum);
it = candidates->head;
} else {
it = nxt;
}
}
free(unique);
}
int main() {
list candidates = make_list();
setup(&candidates); print_count(&candidates);
statement1(&candidates); print_count(&candidates);
statement2(&candidates); print_count(&candidates);
statement3(&candidates); print_count(&candidates);
print_list(&candidates);
free_list(&candidates); return 0;
}</lang>
- Output:
2352 candidates 145 candidates 86 candidates one candidate [4, 13] S=17 P=52
C++
<lang cpp>#include <algorithm>
- include <iostream>
- include <map>
- include <vector>
std::ostream &operator<<(std::ostream &os, std::vector<std::pair<int, int>> &v) {
for (auto &p : v) {
auto sum = p.first + p.second;
auto prod = p.first * p.second;
os << '[' << p.first << ", " << p.second << "] S=" << sum << " P=" << prod;
}
return os << '\n';
}
void print_count(const std::vector<std::pair<int, int>> &candidates) {
auto c = candidates.size();
if (c == 0) {
std::cout << "no candidates\n";
} else if (c == 1) {
std::cout << "one candidate\n";
} else {
std::cout << c << " candidates\n";
}
}
auto setup() {
std::vector<std::pair<int, int>> candidates;
// numbers must be greater than 1
for (int x = 2; x <= 98; x++) {
// numbers must be unique, and sum no more than 100
for (int y = x + 1; y <= 98; y++) {
if (x + y <= 100) {
candidates.push_back(std::make_pair(x, y));
}
}
}
return candidates;
}
void remove_by_sum(std::vector<std::pair<int, int>> &candidates, const int sum) {
candidates.erase(std::remove_if(
candidates.begin(), candidates.end(),
[sum](const std::pair<int, int> &pair) {
auto s = pair.first + pair.second;
return s == sum;
}
), candidates.end());
}
void remove_by_prod(std::vector<std::pair<int, int>> &candidates, const int prod) {
candidates.erase(std::remove_if(
candidates.begin(), candidates.end(),
[prod](const std::pair<int, int> &pair) {
auto p = pair.first * pair.second;
return p == prod;
}
), candidates.end());
}
void statement1(std::vector<std::pair<int, int>> &candidates) {
std::map<int, int> uniqueMap;
std::for_each(
candidates.cbegin(), candidates.cend(),
[&uniqueMap](const std::pair<int, int> &pair) {
auto prod = pair.first * pair.second;
uniqueMap[prod]++;
}
);
bool loop;
do {
loop = false;
for (auto &pair : candidates) {
auto prod = pair.first * pair.second;
if (uniqueMap[prod] == 1) {
auto sum = pair.first + pair.second;
remove_by_sum(candidates, sum);
loop = true;
break;
}
}
} while (loop);
}
void statement2(std::vector<std::pair<int, int>> &candidates) {
std::map<int, int> uniqueMap;
std::for_each(
candidates.cbegin(), candidates.cend(),
[&uniqueMap](const std::pair<int, int> &pair) {
auto prod = pair.first * pair.second;
uniqueMap[prod]++;
}
);
bool loop;
do {
loop = false;
for (auto &pair : candidates) {
auto prod = pair.first * pair.second;
if (uniqueMap[prod] > 1) {
remove_by_prod(candidates, prod);
loop = true;
break;
}
}
} while (loop);
}
void statement3(std::vector<std::pair<int, int>> &candidates) {
std::map<int, int> uniqueMap;
std::for_each(
candidates.cbegin(), candidates.cend(),
[&uniqueMap](const std::pair<int, int> &pair) {
auto sum = pair.first + pair.second;
uniqueMap[sum]++;
}
);
bool loop;
do {
loop = false;
for (auto &pair : candidates) {
auto sum = pair.first + pair.second;
if (uniqueMap[sum] > 1) {
remove_by_sum(candidates, sum);
loop = true;
break;
}
}
} while (loop);
}
int main() {
auto candidates = setup(); print_count(candidates);
statement1(candidates); print_count(candidates);
statement2(candidates); print_count(candidates);
statement3(candidates); print_count(candidates);
std::cout << candidates;
return 0;
}</lang>
- Output:
2352 candidates 145 candidates 86 candidates one candidate [4, 13] S=17 P=52
C#
<lang csharp>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);
}</lang>
- Output:
2352
145
86
1
{ X = 4, Y = 13, S = 17, P = 52 }
Common Lisp
Version 1
<lang lisp>
- Calculate all x's and their possible y's.
(defparameter *x-possibleys*
(loop for x from 2 to 49
collect (cons x (loop for y from (- 100 x) downto (1+ x)
collect y)))
"For every x there are certain y's, with respect to the rules of the puzzle")
(defun xys-operation (op x-possibleys)
"returns an alist of ((x possible-y) . (op x possible-y))" (let ((x (car x-possibleys))
(ys (cdr x-possibleys)))
(mapcar #'(lambda (y) (cons (list x y) (funcall op x y))) ys)))
(defun sp-numbers (op x-possibleys)
"generates all possible sums or products of the puzzle"
(loop for xys in x-possibleys
append (xys-operation op xys)))
(defun group-sp (sp-numbers)
"sp: Sum or Product"
(loop for sp-number in (remove-duplicates sp-numbers :key #'cdr)
collect (cons (cdr sp-number)
(mapcar #'car (remove-if-not #'(lambda (sp) (= sp (cdr sp-number))) sp-numbers :key #'cdr)))))
(defun statement-1a (sum-groups)
"remove all sums with a single possible xy" (remove-if #'(lambda (xys) (= (list-length xys) 1)) sum-groups :key #'cdr))
(defun statement-1b (x-possibleys)
"S says: P does not know X and Y." (let ((multi-xy-sums (statement-1a (group-sp (sp-numbers #'+ x-possibleys))))
(products (group-sp (sp-numbers #'* x-possibleys))))
(flet ((sum-has-xy-which-leads-to-unique-prod (sum-xys)
;; is there any product with a single possible xy? (some #'(lambda (prod-xys) (= (list-length (cdr prod-xys)) 1)) ;; all possible xys of the sum's (* x ys) (mapcar #'(lambda (xy) (assoc (apply #'* xy) products)) (cdr sum-xys)))))
;; remove sums with even one xy which leads to a unique product
(remove-if #'sum-has-xy-which-leads-to-unique-prod multi-xy-sums))))
(defun remaining-products (remaining-sums-xys)
"P's number is one of these"
(loop for sum-xys in remaining-sums-xys
append (loop for xy in (cdr sum-xys)
collect (apply #'* xy))))
(defun statement-2 (remaining-sums-xys)
"P says: Now I know X and Y." (let ((remaining-products (remaining-products remaining-sums-xys))) (mapcar #'(lambda (a-sum-unit)
(cons (car a-sum-unit) (mapcar #'(lambda (xy) (list (count (apply #'* xy) remaining-products) xy)) (cdr a-sum-unit)))) remaining-sums-xys)))
(defun statement-3 (remaining-sums-with-their-products-occurrences-info)
"S says: Now I also know X and Y."
(remove-if
#'(lambda (sum-xys)
;; remove those sums which have more than 1 product, that
;; appear only once amongst all remaining products
(> (count 1 sum-xys :key #'car) 1))
remaining-sums-with-their-products-occurrences-info
:key #'cdr))
(defun solution (survivor-sum-and-its-xys)
"Now we know X and Y too :-D" (let* ((sum (caar survivor-sum-and-its-xys))
(xys (cdar survivor-sum-and-its-xys)) (xy (second (find 1 xys :key #'car))))
(pairlis '(x y sum product)
(list (first xy) (second xy) sum (apply #'* xy)))))
(solution
(statement-3 (statement-2 (statement-1b *x-possibleys*)))) ;; => ((PRODUCT . 52) (SUM . 17) (Y . 13) (X . 4))
</lang>
Version 2
<lang lisp>
- Algorithm of Rosetta code
- All possible pairs
(defparameter *all-possible-pairs*
(loop for i from 2 upto 100
append (loop for j from (1+ i) upto 100
if (<= (+ i j) 100) collect (list i j))))
(defun oncep (item list)
(eql 1 (count item list)))
- Terminology
(defun sum-decomp (n)
(loop for x from 2 below (/ n 2)
for y = (- n x)
collect (list x y)))
(defun prod-decomp (n)
(loop for x from 2 below (sqrt n)
for y = (/ n x)
if (and (>= 100 (+ x y)) (zerop (rem n x)))
collect (list x y)))
- For every possible sum decomposition of the number X+Y, the product has in turn more than one product decomposition
(defun fact-1 (n)
"n = x + y" (flet ((premise (pair)
(> (list-length (prod-decomp (apply #'* pair))) 1))) (every #'premise (sum-decomp n))))
- The number X*Y has only one product decomposition for which fact 1 is true
(defun fact-2 (n)
"n = x * y" (oncep t (mapcar (lambda (pair) (fact-1 (apply #'+ pair))) (prod-decomp n))))
- The number X+Y has only one sum decomposition for which fact 2 is true
(defun fact-3 (n)
"n = x + y" (oncep t (mapcar (lambda (pair) (fact-2 (apply #'* pair))) (sum-decomp n))))
(defun find-xy (all-possible-pairs)
(remove-if-not
#'(lambda (p) (fact-3 (apply #'+ p)))
(remove-if-not
#'(lambda (p) (fact-2 (apply #'* p)))
(remove-if-not
#'(lambda (p) (fact-1 (apply #'+ p)))
all-possible-pairs))))
(find-xy *all-possible-pairs*) ;; => ((4 13)) </lang>
D
<lang d>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;
}</lang>
- Output:
[const(Tuple!(int, int))(4, 13)]
With an older version of the LDC2 compiler replace the cartesianProduct line with:
<lang d>
const s1 = iota(1, 101).map!(x => iota(1, 101).map!(y => tuple(x, y))).joiner
</lang>
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
<lang elixir>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</lang>
- Output:
[{4, 13}]
Factor
A loose translation of D. <lang factor>USING: combinators.short-circuit fry kernel literals math math.ranges memoize prettyprint sequences sets tools.time ; IN: rosetta-code.sum-and-product
CONSTANT: s1 $[
2 100 [a,b] dup cartesian-product concat
[ first2 { [ < ] [ + 100 < ] } 2&& ] filter
]
- quot-eq ( pair quot -- seq )
[ s1 ] 2dip tuck '[ @ _ @ = ] filter ; inline
MEMO: sum-eq ( pair -- seq ) [ first2 + ] quot-eq ; MEMO: mul-eq ( pair -- seq ) [ first2 * ] quot-eq ;
- s2 ( -- seq )
s1 [ sum-eq [ mul-eq length 1 = not ] all? ] filter ;
- only-1 ( seq quot -- newseq )
over '[ @ _ intersect length 1 = ] filter ; inline
- sum-and-product ( -- )
[ s2 [ mul-eq ] [ sum-eq ] [ only-1 ] bi@ . ] time ;
MAIN: sum-and-product</lang>
- Output:
{ { 4 13 } }
Running time: 0.241637693 seconds
Go
<lang go>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 }</lang>
- 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
<lang haskell>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</lang>
- 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. <lang Haskell>import Data.List (intersect)
SUM AND PRODUCT PUZZLE ----------------
s1, s2, s3, s4 :: [(Int, Int)] s1 =
[2 .. 100]
>>= \x ->
[succ x .. 100]
>>= \y ->
[ (x, y)
| x + y < 100
]
s2 = filter (all ((1 /=) . length . mulEq) . sumEq) s1
s3 = filter ((1 ==) . length . (`intersect` s2) . mulEq) s2
s4 = filter ((1 ==) . length . (`intersect` s3) . sumEq) s3
sumEq, mulEq :: (Int, Int) -> [(Int, Int)] sumEq p = filter ((add p ==) . add) s1 mulEq p = filter ((mul p ==) . mul) s1
add, mul :: (Int, Int) -> Int add = uncurry (+) mul = uncurry (*)
TEST -------------------------
main :: IO () main = print $ take 1 s4</lang>
- Output:
[(4,13)]
Java
<lang Java>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;
}
}</lang>
- 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
ES5
<lang JavaScript>(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;
})(); </lang>
- Output:
<lang JavaScript>4, 13</lang> (Finished in 0.69s)
ES6
<lang JavaScript>(() => {
"use strict";
// ------------- SUM AND PRODUCT PUZZLE --------------
// main :: IO ()
const main = () => {
const
// xs :: [Int]
xs = enumFromTo(1)(100),
// s1 s2, s3, s4 :: [(Int, Int)]
s1 = xs.flatMap(
x => xs.flatMap(y =>
(1 < x) && (x < y) && 100 > (x + y) ? [
[x, y]
] : []
)
),
s2 = s1.filter(
p => sumEq(p, s1).every(
q => 1 < mulEq(q, s1).length
)
),
s3 = s2.filter(
p => 1 === intersectBy(pairEQ)(
mulEq(p, s1)
)(s2).length
);
return s3.filter(
p => 1 === intersectBy(pairEQ)(
sumEq(p, s1)
)(s3).length
);
};
// ---------------- PROBLEM FUNCTIONS ----------------
// add, mul :: (Int, Int) -> Int
const
add = xy => xy[0] + xy[1],
mul = xy => xy[0] * xy[1],
// sumEq, mulEq :: (Int, Int) ->
// [(Int, Int)] -> [(Int, Int)]
sumEq = (p, s) => {
const addP = add(p);
return s.filter(q => add(q) === addP);
},
mulEq = (p, s) => {
const mulP = mul(p);
return s.filter(q => mul(q) === mulP);
},
// pairEQ :: ((a, a) -> (a, a)) -> Bool
pairEQ = a => b => (
a[0] === b[0]
) && (a[1] === b[1]);
// ---------------- GENERIC FUNCTIONS ----------------
// enumFromTo :: Int -> Int -> [Int]
const enumFromTo = m =>
n => Array.from({
length: 1 + n - m
}, (_, i) => m + i);
// intersectBy :: (a -> a -> Bool) -> [a] -> [a] -> [a]
const intersectBy = eqFn =>
// The intersection of the lists xs and ys
// in terms of the equality defined by eq.
xs => ys => xs.filter(
x => ys.some(eqFn(x))
);
// MAIN --- return main();
})();</lang>
- Output:
<lang JavaScript>4, 13</lang>
jq
Works with: jq
Works with: gojq (the Go implementation of jq)
A transcription from the problem statement, with these helper functions: <lang jq>
- For readability:
def collect(c): map(select(c));
- stream-oriented checks:
def hasMoreThanOne(s): [limit(2;s)] | length > 1;
def hasOne(s): [limit(2;s)] | length == 1;
def prod: .[0] * .[1];
- A stream of admissible [x,y] values
def xy:
[range(2;50) as $x # 1 < X < Y < 100 | range($x+1; 101-$x) as $y | [$x, $y] ] ;
- The stream of [x,y] pairs matching "S knows the sum is $sum"
def sumEq($sum): select( $sum == add );
- The stream of [x,y] pairs matching "P knows the product is $prod"
def prodEq($p): select( $p == prod );
- The solver:
def solve:
xy as $s0
- S says P does not know:
| $s0
| collect(add as $sum
| all( $s0[]|sumEq($sum);
prod as $p
| hasMoreThanOne($s0[] | prodEq($p)))) as $s1
- P says: Now I know:
| $s1 | collect(prod as $prod | hasOne( $s1[]|prodEq($prod)) ) as $s2
- S says: Now I also know
| $s2[] | select(add as $sum | hasOne( $s2[] | sumEq($sum)) ) ;
solve </lang>
- Output:
[4,13]
Julia
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.
<lang julia> using Primes
function satisfy1(x::Integer)
prmslt100 = primes(100)
for i in 2:(x ÷ 2)
if i ∈ prmslt100 && x - i ∈ prmslt100
return false
end
end
return true
end
function satisfy2(x::Integer)
once = false
for i in 2:isqrt(x)
if x % i == 0
j = x ÷ i
if 2 < j < 100 && satisfy1(i + j)
if once return false end
once = true
end
end
end
return once
end
function satisfyboth(x::Integer)
if !satisfy1(x) return 0 end
found = 0
for i in 2:(x ÷ 2)
if satisfy2(i * (x - i))
if found > 0 return 0 end
found = i
end
end
return found
end
for i in 2:99
if (j = satisfyboth(i)) > 0
println("Solution: ($j, $(i - j))")
end
end</lang>
- Output:
Solution: (4, 13)
Kotlin
<lang scala>// 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
() 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") }</lang>
- Output:
The only solution is : x = 4, y = 13
Lua
<lang lua>function print_count(t)
local cnt = 0
for k,v in pairs(t) do
cnt = cnt + 1
end
print(cnt .. ' candidates')
end
function make_pair(a,b)
local t = {}
table.insert(t, a) -- 1
table.insert(t, b) -- 2
return t
end
function setup()
local candidates = {}
for x = 2, 98 do
for y = x + 1, 98 do
if x + y <= 100 then
local p = make_pair(x, y)
table.insert(candidates, p)
end
end
end
return candidates
end
function remove_by_sum(candidates, sum)
for k,v in pairs(candidates) do
local s = v[1] + v[2]
if s == sum then
table.remove(candidates, k)
end
end
end
function remove_by_prod(candidates, prod)
for k,v in pairs(candidates) do
local p = v[1] * v[2]
if p == prod then
table.remove(candidates, k)
end
end
end
function statement1(candidates)
local unique = {}
for k,v in pairs(candidates) do
local prod = v[1] * v[2]
if unique[prod] ~= nil then
unique[prod] = unique[prod] + 1
else
unique[prod] = 1
end
end
local done
repeat
done = true
for k,v in pairs(candidates) do
local prod = v[1] * v[2]
if unique[prod] == 1 then
local sum = v[1] + v[2]
remove_by_sum(candidates, sum)
done = false
break
end
end
until done
end
function statement2(candidates)
local unique = {}
for k,v in pairs(candidates) do
local prod = v[1] * v[2]
if unique[prod] ~= nil then
unique[prod] = unique[prod] + 1
else
unique[prod] = 1
end
end
local done
repeat
done = true
for k,v in pairs(candidates) do
local prod = v[1] * v[2]
if unique[prod] > 1 then
remove_by_prod(candidates, prod)
done = false
break
end
end
until done
end
function statement3(candidates)
local unique = {}
for k,v in pairs(candidates) do
local sum = v[1] + v[2]
if unique[sum] ~= nil then
unique[sum] = unique[sum] + 1
else
unique[sum] = 1
end
end
local done
repeat
done = true
for k,v in pairs(candidates) do
local sum = v[1] + v[2]
if unique[sum] > 1 then
remove_by_sum(candidates, sum)
done = false
break
end
end
until done
end
function main()
local candidates = setup() print_count(candidates)
statement1(candidates) print_count(candidates)
statement2(candidates) print_count(candidates)
statement3(candidates) print_count(candidates)
for k,v in pairs(candidates) do
local sum = v[1] + v[2]
local prod = v[1] * v[2]
print("a=" .. v[1] .. ", b=" .. v[2] .. "; S=" .. sum .. ", P=" .. prod)
end
end
main()</lang>
- Output:
2352 candidates 145 candidates 86 candidates 1 candidates a=4, b=13; S=17, P=52
Nim
<lang Nim>import sequtils, sets, sugar, tables
var
xycandidates = toSeq(2..98)
sums = collect(initHashSet, for s in 5..100: {s}) # Set of possible sums.
factors: Table[int, seq[(int, int)]] # Mapping product -> list of factors.
- Build the factor mapping.
for i in 0..<xycandidates.high:
let x = xycandidates[i] for j in (i + 1)..xycandidates.high: let y = xycandidates[j] factors.mgetOrPut(x * y, @[]).add (x, y)
iterator terms(n: int): (int, int) =
## Yield the possible terms (x, y) of a given sum. for x in 2..(n - 1) div 2: yield (x, n - x)
- S says "P does not know X and Y."
- => For every decomposition of S, there is no product with a single decomposition.
for s in toSeq(sums):
for (x, y) in s.terms():
let p = x * y
if factors[p].len == 1:
sums.excl s
break
- P says "Now I know X and Y."
- => P has only one decomposition with sum in "sums".
for p in toSeq(factors.keys):
var sums = collect(initHashSet):
for (x, y) in factors[p]:
if x + y in sums: {x + y}
if card(sums) > 1: factors.del p
- S says "Now I also know X and Y."
- => S has only one decomposition with product in "factors".
for s in toSeq(sums):
var prods = collect(initHashSet):
for (x, y) in s.terms():
if x * y in factors: {x * y}
if card(prods) > 1: sums.excl s
- Now, combine the sums and the products.
for s in sums:
for (x, y) in s.terms: if x * y in factors: echo (x, y)</lang>
- Output:
(4, 13)
ooRexx
version 1
for comments see REXX version 4. <lang oorexx>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</lang>
- 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
Uses objects for storing the number pairs. Note the computed hash value and the == method (required to make the set difference work) <lang oorexx>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"]"</lang>
- 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
<lang perl>use List::Util qw(none);
sub grep_unique {
my($by, @list) = @_;
my @seen;
for (@list) {
my $x = &$by(@$_);
$seen[$x]= defined $seen[$x] ? 0 : join ' ', @$_;
}
grep { $_ } @seen;
}
sub sums {
my($n) = @_; my @sums; push @sums, [$_, $n - $_] for 2 .. int $n/2; @sums;
}
sub sum { $_[0] + $_[1] } sub product { $_[0] * $_[1] }
for $i (2..97) {
push @all_pairs, map { [$i, $_] } $i + 1..98
}
- Fact 1:
%p_unique = map { $_ => 1 } grep_unique(\&product, @all_pairs); for my $p (@all_pairs) {
push @s_pairs, [@$p] if none { $p_unique{join ' ', @$_} } sums sum @$p;
}
- Fact 2:
@p_pairs = map { [split ' ', $_] } grep_unique(\&product, @s_pairs);
- Fact 3:
@final_pair = grep_unique(\&sum, @p_pairs);
printf "X = %d, Y = %d\n", split ' ', $final_pair[0];</lang>
- Output:
X = 4, Y = 13
Phix
Runs in 0.03s
with javascript_semantics 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 false end if end for return true 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 bool winner = false 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 false end if winner = true 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)
Picat
<lang Picat> import cp. main =>
N = 98, PD = new_array(N*N), % PD[I] = no. of product decompositions of I foreach(I in 1..N*N) PD[I] = 0 end, foreach(X in 2..N-1, Y in X+1..N) PD[X * Y] := PD[X * Y] + 1 end, % Fact 1: S says "P does not know X and Y.", i.e. % For every possible sum decomposition of the number X+Y, the product has in turn more than one product decomposition: Solutions1 = [[X,Y] : X in 2..N-1, Y in X+1..100-X, foreach(XX in 2..X+Y-3) PD[XX * (X+Y-XX)] #> 1 end], % Fact 2: P says "Now I know X and Y.", i.e. % The number X*Y has only one product decomposition for which fact 1 is true: Solutions2 = [[X,Y] : [X,Y] in Solutions1, foreach([XX,YY] in Solutions1, XX * YY = X * Y) XX = X, YY = Y end], % Fact 3: S says "Now I also know X and Y.", i.e. % The number X+Y has only one sum decomposition for which fact 2 is true. Solutions3 = [[X,Y] : [X,Y] in Solutions2, foreach([XX,YY] in Solutions2, XX + YY = X + Y) XX = X, YY = Y end], println(Solutions3).
</lang> Output:
[[4,13]]
Python
Based on the Python solution from Wikipedia: <lang python>#!/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)</lang>
- Output:
[(4, 13)]
Racket
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>#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)</lang>
- Output:
Max Sum: 100 Possible pairs: 2352 Initial pairs for S: 145 Pairs for P: 86 Final pairs for S: 1 ((4 13))
Raku
(formerly Perl 6)
<lang perl6>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;</lang>
- Output:
X = 4, Y = 13
REXX
version 1
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. <lang rexx>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)
wordsort: Procedure /**********************************************************************
- Sort the list of words supplied as argument. Return the sorted list
- /
Parse Arg wl
wa.=
wa.0=0
Do While wl<>
Parse Var wl w wl
Do i=1 To wa.0
If wa.i>w Then Leave
End
If i<=wa.0 Then Do
Do j=wa.0 To i By -1
ii=j+1
wa.ii=wa.j
End
End
wa.i=w
wa.0=wa.0+1
End
swl=
Do i=1 To wa.0
swl=swl wa.i
End
/* Say swl */
Return strip(swl)</lang>
- 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
<lang rexx>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</lang>
- Output:
4/13 s=17 p=52 Elapsed time: 0.078000 seconds
version 3
<lang rexx>/*---------------------------------------------------------------------
- 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</lang>
- 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
Now that I have understood the logic (I am neither S nor P) I have created an alternative to version 3. <lang rexx>/*---------------------------------------------------------------------
- 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</lang>
- 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, fast
This REXX version is over ten times faster than the previous REXX version. <lang rexx>/*REXX program solves the Sum and Product Puzzle (also known as the Impossible Puzzle).*/ @.= 0; h= 100; @.3= 1 /*assign array default; assign high P.*/
do j=5 by 2 to h /*find all odd primes ≤ 1st argument.*/
do k=3 while k*k<=j; if j//k==0 then iterate j /*J ÷ by K ? */
end /*k*/; @.j= 1 /*found a net prime number: J */
end /*j*/
@.2=1 /*assign the even prime, ex post facto.*/
do s=2 for h-1; if C1(s)==0 then iterate /*find and display the puzzle solution.*/
$= 0; do m=2 for s%2 -1 /* [↓] 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*/
if $==0 then say 'No solution found.' exit 0 /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ C1: procedure expose @.; parse arg s /*validate the first puzzle condition. */
do a=2 for s%2-1; if @.a then do; _= s - a; if @._ then return 0; end
end; /*a*/; return 1
/*──────────────────────────────────────────────────────────────────────────────────────*/ C2: procedure expose @. h; parse arg p; $= 0 /*validate the second puzzle condition.*/
do j=2 while j*j
=2 then if q<=h then if C1(j+q) then if $ then return 0 else $= 1 end end /*j*/; return $</lang>
- output when using the default internal input:
The numbers are: 4 and 13
Ruby
<lang ruby>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}</lang>
- Output:
[[4, 13]]
Scala
<lang scala>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)
}</lang>
- Output:
Vector((4,13))
Run-time: about 3.82 seconds.
Scheme
<lang scheme> (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"))
</lang>
- Output:
Initial pairs: 4656 After S: 145 After P: 86 After S: 1 X: 4 Y: 13
Sidef
<lang ruby>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...) }</lang>
- Output:
X = 4, Y = 13
Wren
<lang ecmascript>import "/dynamic" for Tuple import "/seq" for Lst
var P = Tuple.create("P", ["x", "y", "sum", "prod"])
var intersect = Fn.new { |l1, l2|
var l3 = (l1.count < l2.count) ? l1 : l2 var l4 = (l3 == l1) ? l2 : l1 var l5 = [] for (e in l3) if (l4.contains(e)) l5.add(e) return l5
}
var candidates = [] for (x in 2..49) {
for (y in x + 1..100 - x) {
candidates.add(P.new(x, y, x + y, x * y))
}
}
var sumGroups = Lst.groups(candidates) { |c| c.sum } var prodGroups = Lst.groups(candidates) { |c| c.prod } var sumMap = {} for (sumGroup in sumGroups) {
sumMap[sumGroup[0]] = sumGroup[1].map { |l| l[0] }.toList
} var prodMap = {} for (prodGroup in prodGroups) {
prodMap[prodGroup[0]] = prodGroup[1].map { |l| l[0] }.toList
} var fact1 = candidates.where { |c| sumMap[c.sum].all { |c| prodMap[c.prod].count > 1 } }.toList var fact2 = fact1.where { |c| intersect.call(prodMap[c.prod], fact1).count == 1 }.toList var fact3 = fact2.where { |c| intersect.call(sumMap[c.sum], fact2).count == 1 }.toList System.write("The only solution is : ") for (p in fact3) System.print("x = %(p.x), y = %(p.y)")</lang>
- Output:
The only solution is : x = 4, y = 13
zkl
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. <lang zkl>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();</lang> [[ ]] 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))