Cartesian product of two or more lists

Show one or more idiomatic ways of generating the Cartesian product of two arbitrary lists in your language.

Task

Demonstrate that your function/method correctly returns:

{1, 2} × {3, 4} = {(1, 3), (1, 4), (2, 3), (2, 4)}

and, in contrast:

{3, 4} × {1, 2} = {(3, 1), (3, 2), (4, 1), (4, 2)}

Also demonstrate, using your function/method, that the product of an empty list with any other list is empty.

{1, 2} × {} = {}

{} × {1, 2} = {}

For extra credit, show or write a function returning the n-ary product of an arbitrary number of lists, each of arbitrary length. Your function might, for example, accept a single argument which is itself a list of lists, and return the n-ary product of those lists.

Use your n-ary Cartesian product function to show the following products:

{1776, 1789} × {7, 12} × {4, 14, 23} × {0, 1}

{1, 2, 3} × {30} × {500, 100}

{1, 2, 3} × {} × {500, 100}

11l

Translation of: Go

<lang 11l>F cart_prod(a, b)

  V p = [(0, 0)] * (a.len * b.len)
  V i = 0
  L(aa) a
     L(bb) b
        p[i++] = (aa, bb)
  R p

print(cart_prod([1, 2], [3, 4])) print(cart_prod([3, 4], [1, 2])) [Int] empty_array print(cart_prod([1, 2], empty_array)) print(cart_prod(empty_array, [1, 2]))</lang>

Alternative version

<lang 11l>F cart_prod(a, b)

  R multiloop(a, b, (aa, bb) -> (aa, bb))</lang>

Output:

[(1, 3), (1, 4), (2, 3), (2, 4)]
[(3, 1), (3, 2), (4, 1), (4, 2)]
[]
[]

APL

APL has a built-in outer product operator: X ∘.F Y will get you an ⍴X-by-⍴Y matrix containing every corresponding value of x F y for all x∊X, y∊Y.

The Cartesian product can therefore be expressed as ∘.,, but as that would return a matrix, and the task is asking for a list, you also need to ravel the result.

Output:

      1 2 cart 3 4
 1 3  1 4  2 3  2 4 
      3 4 cart 1 2
 3 1  3 2  4 1  4 2 
      1 2 cart ⍬   ⍝ empty output

      ⍬ cart 1 2   ⍝ empty output again

This can be reduced over a list of lists to generate the Cartesian product of an arbitrary list of lists.

Output:

The items are listed on separate lines (using ↑) for clarity.

      ↑nary_cart (1776 1789)(7 12)(4 14 23)(0 1)
1776  7  4 0
1776  7  4 1
1776  7 14 0
1776  7 14 1
1776  7 23 0
1776  7 23 1
1776 12  4 0
1776 12  4 1
1776 12 14 0
1776 12 14 1
1776 12 23 0
1776 12 23 1
1789  7  4 0
1789  7  4 1
1789  7 14 0
1789  7 14 1
1789  7 23 0
1789  7 23 1
1789 12  4 0
1789 12  4 1
1789 12 14 0
1789 12 14 1
1789 12 23 0
1789 12 23 1
      ↑nary_cart(1 2 3)(,30)(50 100)
1 30  50
1 30 100
2 30  50
2 30 100
3 30  50
3 30 100
      ↑nary_cart(1 2 3)(⍬)(50 100)  ⍝ empty output

AppleScript

<lang AppleScript>-- CARTESIAN PRODUCTS ---------------------------------------------------------

-- Two lists:

-- cartProd :: [a] -> [b] -> [(a, b)] on cartProd(xs, ys)

   script
       on |λ|(x)
           script
               on |λ|(y)
                   x, y
               end |λ|
           end script
           concatMap(result, ys)
       end |λ|
   end script
   concatMap(result, xs)

end cartProd

-- N-ary – a function over a list of lists:

-- cartProdNary :: a -> a on cartProdNary(xss)

   script
       on |λ|(accs, xs)
           script
               on |λ|(x)
                   script
                       on |λ|(a)
                           {x & a}
                       end |λ|
                   end script
                   concatMap(result, accs)
               end |λ|
           end script
           concatMap(result, xs)
       end |λ|
   end script
   foldr(result, {{}}, xss)

end cartProdNary

-- TESTS ---------------------------------------------------------------------- on run

   set baseExamples to unlines(map(show, ¬
       [cartProd({1, 2}, {3, 4}), ¬
           cartProd({3, 4}, {1, 2}), ¬
           cartProd({1, 2}, {}), ¬
           cartProd({}, {1, 2})]))
   
   set naryA to unlines(map(show, ¬
       cartProdNary([{1776, 1789}, {7, 12}, {4, 14, 23}, {0, 1}])))
   
   set naryB to show(cartProdNary([{1, 2, 3}, {30}, {500, 100}]))
   
   set naryC to show(cartProdNary([{1, 2, 3}, {}, {500, 100}]))
   
   intercalate(linefeed & linefeed, {baseExamples, naryA, naryB, naryC})

end run

-- GENERIC FUNCTIONS ----------------------------------------------------------

-- concatMap :: (a -> [b]) -> [a] -> [b] on concatMap(f, xs)

   set lst to {}
   set lng to length of xs
   tell mReturn(f)
       repeat with i from 1 to lng
           set lst to (lst & |λ|(item i of xs, i, xs))
       end repeat
   end tell
   return lst

end concatMap

-- foldr :: (a -> b -> a) -> a -> [b] -> a on foldr(f, startValue, xs)

   tell mReturn(f)
       set v to startValue
       set lng to length of xs
       repeat with i from lng to 1 by -1
           set v to |λ|(v, item i of xs, i, xs)
       end repeat
       return v
   end tell

end foldr

-- intercalate :: Text -> [Text] -> Text on intercalate(strText, lstText)

   set {dlm, my text item delimiters} to {my text item delimiters, strText}
   set strJoined to lstText as text
   set my text item delimiters to dlm
   return strJoined

end intercalate

-- map :: (a -> b) -> [a] -> [b] on map(f, xs)

   tell mReturn(f)
       set lng to length of xs
       set lst to {}
       repeat with i from 1 to lng
           set end of lst to |λ|(item i of xs, i, xs)
       end repeat
       return lst
   end tell

end map

-- Lift 2nd class handler function into 1st class script wrapper -- mReturn :: Handler -> Script on mReturn(f)

   if class of f is script then
       f
   else
       script
           property |λ| : f
       end script
   end if

end mReturn

-- show :: a -> String on show(e)

   set c to class of e
   if c = list then
       script serialized
           on |λ|(v)
               show(v)
           end |λ|
       end script
       
       "[" & intercalate(", ", map(serialized, e)) & "]"
   else if c = record then
       script showField
           on |λ|(kv)
               set {k, ev} to kv
               "\"" & k & "\":" & show(ev)
           end |λ|
       end script
       
       "{" & intercalate(", ", ¬
           map(showField, zip(allKeys(e), allValues(e)))) & "}"
   else if c = date then
       "\"" & iso8601Z(e) & "\""
   else if c = text then
       "\"" & e & "\""
   else if (c = integer or c = real) then
       e as text
   else if c = class then
       "null"
   else
       try
           e as text
       on error
           ("«" & c as text) & "»"
       end try
   end if

end show

-- unlines :: [String] -> String on unlines(xs)

   intercalate(linefeed, xs)

end unlines</lang>

Output:

[[1, 3], [1, 4], [2, 3], [2, 4]]
[[3, 1], [3, 2], [4, 1], [4, 2]]
[]
[]

[1776, 7, 4, 0]
[1776, 7, 4, 1]
[1776, 7, 14, 0]
[1776, 7, 14, 1]
[1776, 7, 23, 0]
[1776, 7, 23, 1]
[1776, 12, 4, 0]
[1776, 12, 4, 1]
[1776, 12, 14, 0]
[1776, 12, 14, 1]
[1776, 12, 23, 0]
[1776, 12, 23, 1]
[1789, 7, 4, 0]
[1789, 7, 4, 1]
[1789, 7, 14, 0]
[1789, 7, 14, 1]
[1789, 7, 23, 0]
[1789, 7, 23, 1]
[1789, 12, 4, 0]
[1789, 12, 4, 1]
[1789, 12, 14, 0]
[1789, 12, 14, 1]
[1789, 12, 23, 0]
[1789, 12, 23, 1]

[[1, 30, 500], [1, 30, 100], [2, 30, 500], [2, 30, 100], [3, 30, 500], [3, 30, 100]]

[]

Bracmat

<lang Bracmat>( ( mul

 =   R a b A B
   .   :?R
     & !arg:(.?A) (.?B)
     & (   !A
         :   ?
             ( %@?a
             &   !B
               :   ?
                   ( (%@?b|(.?b))
                   & !R (.!a !b):?R
                   & ~
                   )
                   ?
             )
             ?
       | (.!R)
       )
 )

& ( cartprod

 =   a
   .   !arg:%?a %?arg&mul$(!a cartprod$!arg)
     | !arg
 )

& out

 $ ( cartprod
   $ ( (.1776 1789)
       (.7 12)
       (.4 14 23)
       (.0 1)
     )
   )

& out$(cartprod$((.1 2 3) (.30) (.500 100))) & out$(cartprod$((.1 2 3) (.) (.500 100))) )</lang>

.   (.1776 7 4 0)
    (.1776 7 4 1)
    (.1776 7 14 0)
    (.1776 7 14 1)
    (.1776 7 23 0)
    (.1776 7 23 1)
    (.1776 12 4 0)
    (.1776 12 4 1)
    (.1776 12 14 0)
    (.1776 12 14 1)
    (.1776 12 23 0)
    (.1776 12 23 1)
    (.1789 7 4 0)
    (.1789 7 4 1)
    (.1789 7 14 0)
    (.1789 7 14 1)
    (.1789 7 23 0)
    (.1789 7 23 1)
    (.1789 12 4 0)
    (.1789 12 4 1)
    (.1789 12 14 0)
    (.1789 12 14 1)
    (.1789 12 23 0)
    (.1789 12 23 1)

.   (.1 30 500)
    (.1 30 100)
    (.2 30 500)
    (.2 30 100)
    (.3 30 500)
    (.3 30 100)
.

C

Recursive implementation for computing the Cartesian product of lists. In the pursuit of making it as interactive as possible, the parsing function ended up taking the most space. The product set expression must be supplied enclosed by double quotes. Prints out usage on incorrect invocation. <lang C>

include<string.h>
include<stdlib.h>
include<stdio.h>

void cartesianProduct(int** sets, int* setLengths, int* currentSet, int numSets, int times){ int i,j;

if(times==numSets){ printf("("); for(i=0;i<times;i++){ printf("%d,",currentSet[i]); } printf("\b),"); } else{ for(j=0;j<setLengths[times];j++){ currentSet[times] = sets[times][j]; cartesianProduct(sets,setLengths,currentSet,numSets,times+1); } } }

void printSets(int** sets, int* setLengths, int numSets){ int i,j;

printf("\nNumber of sets : %d",numSets);

for(i=0;i<numSets+1;i++){ printf("\nSet %d : ",i+1); for(j=0;j<setLengths[i];j++){ printf(" %d ",sets[i][j]); } } }

void processInputString(char* str){ int **sets, *currentSet, *setLengths, setLength, numSets = 0, i,j,k,l,start,counter=0; char *token,*holder,*holderToken;

for(i=0;str[i]!=00;i++) if(str[i]=='x') numSets++;

if(numSets==0){ printf("\n%s",str); return; }

currentSet = (int*)calloc(sizeof(int),numSets + 1);

setLengths = (int*)calloc(sizeof(int),numSets + 1);

sets = (int**)malloc((numSets + 1)*sizeof(int*));

token = strtok(str,"x");

while(token!=NULL){ holder = (char*)malloc(strlen(token)*sizeof(char));

j = 0;

for(i=0;token[i]!=00;i++){ if(token[i]>='0' && token[i]<='9') holder[j++] = token[i]; else if(token[i]==',') holder[j++] = ' '; } holder[j] = 00;

setLength = 0;

for(i=0;holder[i]!=00;i++) if(holder[i]==' ') setLength++;

if(setLength==0 && strlen(holder)==0){ printf("\n{}"); return; }

setLengths[counter] = setLength+1;

sets[counter] = (int*)malloc((1+setLength)*sizeof(int));

k = 0;

start = 0;

for(l=0;holder[l]!=00;l++){ if(holder[l+1]==' '||holder[l+1]==00){ holderToken = (char*)malloc((l+1-start)*sizeof(char)); strncpy(holderToken,holder + start,l+1-start); sets[counter][k++] = atoi(holderToken); start = l+2; } }

counter++; token = strtok(NULL,"x"); }

printf("\n{"); cartesianProduct(sets,setLengths,currentSet,numSets + 1,0); printf("\b}");

}

int main(int argC,char* argV[]) { if(argC!=2) printf("Usage : %s <Set product expression enclosed in double quotes>",argV[0]); else processInputString(argV[1]);

return 0; } </lang> Invocation and output :

C:\My Projects\threeJS>cartesianProduct.exe "{1,2} x {3,4}"

{(1,3),(1,4),(2,3),(2,4)}
C:\My Projects\threeJS>cartesianProduct.exe "{3,4} x {1,2}"

{(3,1),(3,2),(4,1),(4,2)}
C:\My Projects\threeJS>cartesianProduct.exe "{1,2} x {}"

{}
C:\My Projects\threeJS>cartesianProduct.exe "{} x {1,2}"

{}
C:\My Projects\threeJS>cartesianProduct.exe "{1776, 1789} x {7, 12} x {4, 14, 23} x {0, 1}"

{(1776,7,4,0),(1776,7,4,1),(1776,7,14,0),(1776,7,14,1),(1776,7,23,0),(1776,7,23,1),(1776,12,4,0),(1776,12,4,1),(1776,12,14,0),(1776,12,14,1),(1776,12,23,0),(1776,12,23,1),(1789,7,4,0),(1789,9,12,14,1),(1789,12,23,0),(1789,12,23,1)}
C:\My Projects\threeJS>cartesianProduct.exe "{1, 2, 3} x {30} x {500, 100}"

{(1,30,500),(1,30,100),(2,30,500),(2,30,100),(3,30,500),(3,30,100)}
C:\My Projects\threeJS>cartesianProduct.exe "{1, 2, 3} x {} x {500, 100}"

{}

C#

<lang csharp>using System; public class Program {

   public static void Main()
   {
       int[] empty = new int[0];
       int[] list1 = { 1, 2 };
       int[] list2 = { 3, 4 };
       int[] list3 = { 1776, 1789 };
       int[] list4 = { 7, 12 };
       int[] list5 = { 4, 14, 23 };
       int[] list6 = { 0, 1 };
       int[] list7 = { 1, 2, 3 };
       int[] list8 = { 30 };
       int[] list9 = { 500, 100 };
       
       foreach (var sequenceList in new [] {
           new [] { list1, list2 },
           new [] { list2, list1 },
           new [] { list1, empty },
           new [] { empty, list1 },
           new [] { list3, list4, list5, list6 },
           new [] { list7, list8, list9 },
           new [] { list7, empty, list9 }
       }) {
           var cart = sequenceList.CartesianProduct()
               .Select(tuple => $"({string.Join(", ", tuple)})");
           Console.WriteLine($"{{{string.Join(", ", cart)}}}");
       }
   }

}

public static class Extensions {

   public static IEnumerable<IEnumerable<T>> CartesianProduct<T>(this IEnumerable<IEnumerable<T>> sequences) {
       IEnumerable<IEnumerable<T>> emptyProduct = new[] { Enumerable.Empty<T>() };
       return sequences.Aggregate(
           emptyProduct,
           (accumulator, sequence) =>
           from acc in accumulator
           from item in sequence
           select acc.Concat(new [] { item }));
   }

}</lang>

Output:

{(1, 3), (1, 4), (2, 3), (2, 4)}
{(3, 1), (3, 2), (4, 1), (4, 2)}
{}
{}
{(1776, 7, 4, 0), (1776, 7, 4, 1), (1776, 7, 14, 0), (1776, 7, 14, 1), (1776, 7, 23, 0), (1776, 7, 23, 1), (1776, 12, 4, 0), (1776, 12, 4, 1), (1776, 12, 14, 0), (1776, 12, 14, 1), (1776, 12, 23, 0), (1776, 12, 23, 1), (1789, 7, 4, 0), (1789, 7, 4, 1), (1789, 7, 14, 0), (1789, 7, 14, 1), (1789, 7, 23, 0), (1789, 7, 23, 1), (1789, 12, 4, 0), (1789, 12, 4, 1), (1789, 12, 14, 0), (1789, 12, 14, 1), (1789, 12, 23, 0), (1789, 12, 23, 1)}
{(1, 30, 500), (1, 30, 100), (2, 30, 500), (2, 30, 100), (3, 30, 500), (3, 30, 100)}
{}

If the number of lists is known, LINQ provides an easier solution: <lang csharp>public static void Main() {

   ///...
   var cart1 =
       from a in list1
       from b in list2
       select (a, b); // C# 7.0 tuple
   Console.WriteLine($"{{{string.Join(", ", cart1)}}}");
       
   var cart2 =
       from a in list7
       from b in list8
       from c in list9
       select (a, b, c);
   Console.WriteLine($"{{{string.Join(", ", cart2)}}}");

}</lang>

Output:

{(1, 3), (1, 4), (2, 3), (2, 4)}
{(1, 30, 500), (1, 30, 100), (2, 30, 500), (2, 30, 100), (3, 30, 500), (3, 30, 100)}

C++

include <iostream>
include <vector>
include <algorithm>

void print(const std::vector<std::vector<int>>& v) {

 std::cout << "{ ";
 for (const auto& p : v) {
   std::cout << "(";
   for (const auto& e : p) {
     std::cout << e << " ";
   }
   std::cout << ") ";
 }
 std::cout << "}" << std::endl;

}

auto product(const std::vector<std::vector<int>>& lists) {

 std::vector<std::vector<int>> result;
 if (std::find_if(std::begin(lists), std::end(lists), 
   [](auto e) -> bool { return e.size() == 0; }) != std::end(lists)) {
   return result;
 }
 for (auto& e : lists[0]) {
   result.push_back({ e });
 }
 for (size_t i = 1; i < lists.size(); ++i) {
   std::vector<std::vector<int>> temp;
   for (auto& e : result) {
     for (auto f : lists[i]) {
       auto e_tmp = e;
       e_tmp.push_back(f);
       temp.push_back(e_tmp);
     }
   }
   result = temp;
 }
 return result;

}

int main() {

 std::vector<std::vector<int>> prods[] = {
   { { 1, 2 }, { 3, 4 } },
   { { 3, 4 }, { 1, 2} },
   { { 1, 2 }, { } },
   { { }, { 1, 2 } },
   { { 1776, 1789 }, { 7, 12 }, { 4, 14, 23 }, { 0, 1 } },
   { { 1, 2, 3 }, { 30 }, { 500, 100 } },
   { { 1, 2, 3 }, { }, { 500, 100 } }
 };
 for (const auto& p : prods) {
   print(product(p));
 }
 std::cin.ignore();
 std::cin.get();
 return 0;

}</lang>

Output:

{ (1 3) (1 4) (2 3) (2 4) }
{ (3 1) (3 2) (4 1) (4 2) }
{ }
{ }
{ (1776 7 4 0) (1776 7 4 1) (1776 7 14 0) (1776 7 14 1) (1776 7 23 0) (1776 7 23 1) (1776 12 4 0) (1776 12 4 1) (1776 12 14 0) (1776 12 14 1) (1776 12 23 0) (1776 12 23 1) (1789 7 4 0) (1789 7 4 1) (1789 7 14 0) (1789 7 14 1) (1789 7 23 0) (1789 7 23 1) (1789 12 4 0) (1789 12 4 1) (1789 12 14 0) (1789 12 14 1) (1789 12 23 0) (1789 12 23 1) }
{ (1 30 500) (1 30 100) (2 30 500) (2 30 100) (3 30 500) (3 30 100) }
{ }

Clojure

(ns clojure.examples.product

(:gen-class) (:require [clojure.pprint :as pp]))

(defn cart [colls]

 "Compute the cartesian product of list of lists"
 (if (empty? colls)
   '(())
   (for [more (cart (rest colls))
         x (first colls)]
     (cons x more))))

</lang> Output <lang clojure> (doseq [lst [ [[1,2],[3,4]],

               [[3,4],[1,2]], [[], [1, 2]], 
               [[1, 2], []],
               [[1776, 1789],  [7, 12], [4, 14, 23], [0, 1]],
               [[1, 2, 3], [30,], [500, 100]],
               [[1, 2, 3], [], [500, 100]]
           ]
       ]
   (println lst "=>")
   (pp/pprint (cart lst)))

</lang>

[[1 2] [3 4]] =>
((1 3) (2 3) (1 4) (2 4))
[[3 4] [1 2]] =>
((3 1) (4 1) (3 2) (4 2))
[[] [1 2]] =>
()
[[1 2] []] =>
()
[[1776 1789] [7 12] [4 14 23] [0 1]] =>
((1776 7 4 0)
 (1789 7 4 0)
 (1776 12 4 0)
 (1789 12 4 0)
 (1776 7 14 0)
 (1789 7 14 0)
 (1776 12 14 0)
 (1789 12 14 0)
 (1776 7 23 0)
 (1789 7 23 0)
 (1776 12 23 0)
 (1789 12 23 0)
 (1776 7 4 1)
 (1789 7 4 1)
 (1776 12 4 1)
 (1789 12 4 1)
 (1776 7 14 1)
 (1789 7 14 1)
 (1776 12 14 1)
 (1789 12 14 1)
 (1776 7 23 1)
 (1789 7 23 1)
 (1776 12 23 1)
 (1789 12 23 1))
[[1 2 3] [30] [500 100]] =>
((1 30 500) (2 30 500) (3 30 500) (1 30 100) (2 30 100) (3 30 100))
[[1 2 3] [] [500 100]] =>
()

Common Lisp

<lang lisp>(defun cartesian-product (s1 s2)

 "Compute the cartesian product of two sets represented as lists"
 (loop for x in s1

nconc (loop for y in s2 collect (list x y)))) </lang>

Output

<lang lisp> CL-USER> (cartesian-product '(1 2) '(3 4)) ((1 3) (1 4) (2 3) (2 4)) CL-USER> (cartesian-product '(3 4) '(1 2)) ((3 1) (3 2) (4 1) (4 2)) CL-USER> (cartesian-product '(1 2) '()) NIL CL-USER> (cartesian-product '() '(1 2)) NIL </lang>

Extra credit:

<lang lisp>(defun n-cartesian-product (l)

 "Compute the n-cartesian product of a list of sets (each of them represented as list).
  Algorithm:
    If there are no sets, then produce an empty set of tuples;
    otherwise, for all the elements x of the first set, concatenate the sets obtained by
    inserting x at the beginning of each tuple of the n-cartesian product of the remaining sets."
 (if (null l)
     (list nil)
     (loop for x in (car l)
           nconc (loop for y in (n-cartesian-product (cdr l))  
                       collect (cons x y)))))</lang>

Output:

<lang lisp>CL-USER> (n-cartesian-product '((1776 1789) (7 12) (4 14 23) (0 1))) ((1776 7 4 0) (1776 7 4 1) (1776 7 14 0) (1776 7 14 1) (1776 7 23 0) (1776 7 23 1) (1776 12 4 0) (1776 12 4 1) (1776 12 14 0) (1776 12 14 1) (1776 12 23 0) (1776 12 23 1) (1789 7 4 0) (1789 7 4 1) (1789 7 14 0) (1789 7 14 1) (1789 7 23 0) (1789 7 23 1) (1789 12 4 0) (1789 12 4 1) (1789 12 14 0) (1789 12 14 1) (1789 12 23 0) (1789 12 23 1)) CL-USER> (n-cartesian-product '((1 2 3) (30) (500 100))) ((1 30 500) (1 30 100) (2 30 500) (2 30 100) (3 30 500) (3 30 100)) CL-USER> (n-cartesian-product '((1 2 3) () (500 100))) NIL </lang>

D

<lang D>import std.stdio;

void main() {

   auto a = listProduct([1,2], [3,4]);
   writeln(a);

   auto b = listProduct([3,4], [1,2]);
   writeln(b);

   auto c = listProduct([1,2], []);
   writeln(c);

   auto d = listProduct([], [1,2]);
   writeln(d);

}

auto listProduct(T)(T[] ta, T[] tb) {

   struct Result {
       int i, j;

       bool empty() {
           return i>=ta.length
               || j>=tb.length;
       }

       T[] front() {
           return [ta[i], tb[j]];
       }

       void popFront() {
           if (++j>=tb.length) {
               j=0;
               i++;
           }
       }
   }

   return Result();

}</lang>

Output:

[[1, 3], [1, 4], [2, 3], [2, 4]]
[[3, 1], [3, 2], [4, 1], [4, 2]]
[]
[]

F#

The Task

<lang fsharp> //Nigel Galloway February 12th., 2018 let cP2 n g = List.map (fun (n,g)->[n;g]) (List.allPairs n g) </lang>

Output:

cP2 [1;2] [3;4] -> [[1; 3]; [1; 4]; [2; 3]; [2; 4]]
cP2 [3;4] [1;2] -> [[3; 1]; [3; 2]; [4; 1]; [4; 2]]
cP2 [1;2] []    -> []
cP2 [] [1;2]    -> []

Extra Credit

<lang fsharp> //Nigel Galloway August 14th., 2018 let cP ng=Seq.foldBack(fun n g->[for n' in n do for g' in g do yield n'::g']) ng [[]] </lang>

Output:

cP [[1;2];[3;4]] -> [[1; 3]; [1; 4]; [2; 3]; [2; 4]]
cP [[3;4];[1;2]] -> [[3; 1]; [3; 2]; [4; 1]; [4; 2]]
cP [[3;4];[]] ->[]
cP [[];[1;2]] ->[]
cP [[1776;1789];[7;12];[4;14;23];[0;1]] -> [[1776; 7; 4; 0]; [1776; 7; 4; 1]; [1776; 7; 14; 0]; [1776; 7; 14; 1];
                                            [1776; 7; 23; 0]; [1776; 7; 23; 1]; [1776; 12; 4; 0]; [1776; 12; 4; 1];
                                            [1776; 12; 14; 0]; [1776; 12; 14; 1]; [1776; 12; 23; 0]; [1776; 12; 23; 1];
                                            [1789; 7; 4; 0]; [1789; 7; 4; 1]; [1789; 7; 14; 0]; [1789; 7; 14; 1];
                                            [1789; 7; 23; 0]; [1789; 7; 23; 1]; [1789; 12; 4; 0]; [1789; 12; 4; 1];                                                                                                        
                                            [1789; 12; 14; 0]; [1789; 12; 14; 1]; [1789; 12; 23; 0]; [1789; 12; 23; 1]]
cP [[1;2;3];[30];[500;100]] -> [[1; 30; 500]; [1; 30; 100]; [2; 30; 500]; [2; 30; 100]; [3; 30; 500]; [3; 30; 100]]
cP [[1;2;3];[];[500;100]] -> []

Factor

<lang Factor>IN: scratchpad { 1 2 } { 3 4 } cartesian-product . { { { 1 3 } { 1 4 } } { { 2 3 } { 2 4 } } } IN: scratchpad { 3 4 } { 1 2 } cartesian-product . { { { 3 1 } { 3 2 } } { { 4 1 } { 4 2 } } } IN: scratchpad { 1 2 } { } cartesian-product . { { } { } } IN: scratchpad { } { 1 2 } cartesian-product . { }</lang>

FreeBASIC

I'll leave the extra credit part for someone else. It's just going to amount to repeatedly finding Cartesian products and flattening the result, so considerably less interesting than Cartesian products where the list items themselves can be lists.

<lang freebasic>#define MAXLEN 64

type listitem ' An item of a list may be a number

   is_num as boolean       ' or another list, so I have to account
   union                      ' for both, implemented as a union.
       list as any ptr        ' FreeBASIC is twitchy about circularly
       num as uinteger        ' defined types, so one workaround is to
   end union                  ' use a generic pointer that I will cast

end type ' later.

type list

   length as uinteger              'simple, fixed storage length lists
   item(1 to MAXLEN) as listitem   'are good enough for this example

end type

sub print_list( list as list )

   print "{";
   if list.length = 0 then print "}"; : return
   for i as uinteger = 1 to list.length
       if list.item(i).is_num then
           print str(list.item(i).num);
       else     'recursively print sublist
           print_list( *cast(list ptr, list.item(i).list) )
       end if
   if i<list.length then print ", "; else print "}";   'handle comma
   next i                                              'gracefully
   return

end sub

function cartprod( A as list, B as list ) as list

   dim as uinteger i, j
   dim as list C
   dim as list ptr inner  'for brevity
   C.length = 0
   for i = 1 to A.length
       for j = 1 to B.length
           C.length += 1
           C.item(C.length).is_num = false   'each item of the new list is a list itself
           inner = allocate( sizeof(list) )     'make space for it
           C.item(C.length).list = inner
           inner->length = 2                    'each inner list contains two items
           inner->item(1) = A.item(i)           'one from the first list
           inner->item(2) = B.item(j)           'and one from the second
       next j
   next i
   return C

end function

dim as list EMPTY, A, B, R EMPTY.length = 0 A.length = 2 A.item(1).is_num = true : A.item(1).num = 1 A.item(2).is_num = true : A.item(2).num = 2 B.length = 2 B.item(1).is_num = true : B.item(1).num = 3 B.item(2).is_num = true : B.item(2).num = 4

R = cartprod(A, B) print_list(R) : print 'print_list does not supply a final newline R = cartprod(B, A) : print_list(R) : print R = cartprod(A, EMPTY) : print_list(R) : print R = cartprod(EMPTY, A) : print_list(R) : print</lang>

Output:

{{1, 3}, {1, 4}, {2, 3}, {2, 4}}
{{3, 1}, {3, 2}, {4, 1}, {4, 2}}
{}
{}

Fōrmulæ

In this page you can see the solution of this task.

Fōrmulæ programs are not textual, visualization/edition of programs is done showing/manipulating structures but not text (more info). Moreover, there can be multiple visual representations of the same program. Even though it is possible to have textual representation —i.e. XML, JSON— they are intended for transportation effects more than visualization and edition.

The option to show Fōrmulæ programs and their results is showing images. Unfortunately images cannot be uploaded in Rosetta Code.

Go

Basic Task <lang go>package main

import "fmt"

type pair [2]int

func cart2(a, b []int) []pair {

   p := make([]pair, len(a)*len(b))
   i := 0
   for _, a := range a {
       for _, b := range b {
           p[i] = pair{a, b}
           i++
       }
   }
   return p

}

func main() {

   fmt.Println(cart2([]int{1, 2}, []int{3, 4}))
   fmt.Println(cart2([]int{3, 4}, []int{1, 2}))
   fmt.Println(cart2([]int{1, 2}, nil))
   fmt.Println(cart2(nil, []int{1, 2}))

}</lang>

Output:

[[1 3] [1 4] [2 3] [2 4]]
[[3 1] [3 2] [4 1] [4 2]]
[]
[]

Extra credit 1

This solution minimizes allocations and computes and fills the result sequentially. <lang go>package main

import "fmt"

func cartN(a ...[]int) [][]int {

   c := 1
   for _, a := range a {
       c *= len(a)
   }
   if c == 0 {
       return nil
   }
   p := make([][]int, c)
   b := make([]int, c*len(a))
   n := make([]int, len(a))
   s := 0
   for i := range p {
       e := s + len(a)
       pi := b[s:e]
       p[i] = pi
       s = e
       for j, n := range n {
           pi[j] = a[j][n]
       }
       for j := len(n) - 1; j >= 0; j-- {
           n[j]++
           if n[j] < len(a[j]) {
               break
           }
           n[j] = 0
       }
   }
   return p

}

func main() {

   fmt.Println(cartN([]int{1, 2}, []int{3, 4}))
   fmt.Println(cartN([]int{3, 4}, []int{1, 2}))
   fmt.Println(cartN([]int{1, 2}, nil))
   fmt.Println(cartN(nil, []int{1, 2}))

   fmt.Println()
   fmt.Println("[")
   for _, p := range cartN(
       []int{1776, 1789},
       []int{7, 12},
       []int{4, 14, 23},
       []int{0, 1},
   ) {
       fmt.Println(" ", p)
   }
   fmt.Println("]")
   fmt.Println(cartN([]int{1, 2, 3}, []int{30}, []int{500, 100}))
   fmt.Println(cartN([]int{1, 2, 3}, []int{}, []int{500, 100}))

   fmt.Println()
   fmt.Println(cartN(nil))
   fmt.Println(cartN())

}</lang>

Output:

[[1 3] [1 4] [2 3] [2 4]]
[[3 1] [3 2] [4 1] [4 2]]
[]
[]

[
  [1776 7 4 0]
  [1776 7 4 1]
  [1776 7 14 0]
  [1776 7 14 1]
  [1776 7 23 0]
  [1776 7 23 1]
  [1776 12 4 0]
  [1776 12 4 1]
  [1776 12 14 0]
  [1776 12 14 1]
  [1776 12 23 0]
  [1776 12 23 1]
  [1789 7 4 0]
  [1789 7 4 1]
  [1789 7 14 0]
  [1789 7 14 1]
  [1789 7 23 0]
  [1789 7 23 1]
  [1789 12 4 0]
  [1789 12 4 1]
  [1789 12 14 0]
  [1789 12 14 1]
  [1789 12 23 0]
  [1789 12 23 1]
]
[[1 30 500] [1 30 100] [2 30 500] [2 30 100] [3 30 500] [3 30 100]]
[]

[]
[[]]

Extra credit 2

Code here is more compact, but with the cost of more garbage produced. It produces the same result as cartN above. <lang go>func cartN(a ...[]int) (c [][]int) {

   if len(a) == 0 {
       return [][]int{nil}
   }
   r := cartN(a[1:]...)
   for _, e := range a[0] {
       for _, p := range r {
           c = append(c, append([]int{e}, p...))
       }
   }
   return

}</lang> Extra credit 3

This is a compact recursive version like Extra credit 2 but the result list is ordered differently. This is still a correct result if you consider a cartesian product to be a set, which is an unordered collection. Note that the set elements are still ordered lists. A cartesian product is an unordered collection of ordered collections. It draws attention though to the gloss of using list representations as sets. Any of the functions here will accept duplicate elements in the input lists, and then produce duplicate elements in the result. <lang go>func cartN(a ...[]int) (c [][]int) {

   if len(a) == 0 {
       return [][]int{nil}
   }
   last := len(a) - 1
   l := cartN(a[:last]...)
   for _, e := range a[last] {
       for _, p := range l {
           c = append(c, append(p, e))
       }
   }
   return

}</lang>

Groovy

Solution:
The following CartesianCategory class allows for modification of regular Iterable interface behavior, overloading Iterable's multiply (*) operator to perform a Cartesian Product when the second operand is also an Iterable. <lang groovy>class CartesianCategory {

   static Iterable multiply(Iterable a, Iterable b) {
       assert [a,b].every { it != null }
       def (m,n) = [a.size(),b.size()]
       (0..<(m*n)).inject([]) { prod, i -> prod << [a[i.intdiv(n)], b[i%n]].flatten() }
   }

}</lang> Test:
The mixin method call is necessary to make the multiply (*) operator work. <lang groovy>Iterable.metaClass.mixin CartesianCategory

println "\nCore Solution:" println "[1, 2] × [3, 4] = ${[1, 2] * [3, 4]}" println "[3, 4] × [1, 2] = ${[3, 4] * [1, 2]}" println "[1, 2] × [] = ${[1, 2] * []}" println "[] × [1, 2] = ${[] * [1, 2]}"

println "\nExtra Credit:" println "[1776, 1789] × [7, 12] × [4, 14, 23] × [0, 1] = ${[1776, 1789] * [7, 12] * [4, 14, 23] * [0, 1]}" println "[1, 2, 3] × [30] × [500, 100] = ${[1, 2, 3] * [30] * [500, 100]}" println "[1, 2, 3] × [] × [500, 100] = ${[1, 2, 3] * [] * [500, 100]}"

println "\nNon-Numeric Example:" println "[John,Paul,George,Ringo] × [Emerson,Lake,Palmer] × [Simon,Garfunkle] = [" ( ["John","Paul","George","Ringo"] * ["Emerson","Lake","Palmer"] * ["Simon","Garfunkle"] ).each { println "\t${it}," } println "]"</lang> Output:

Core Solution:
[1, 2] × [3, 4] = [[1, 3], [1, 4], [2, 3], [2, 4]]
[3, 4] × [1, 2] = [[3, 1], [3, 2], [4, 1], [4, 2]]
[1, 2] × [] = []
[] × [1, 2] = []

Extra Credit:
[1776, 1789] × [7, 12] × [4, 14, 23] × [0, 1] = [[1776, 7, 4, 0], [1776, 7, 4, 1], [1776, 7, 14, 0], [1776, 7, 14, 1], [1776, 7, 23, 0], [1776, 7, 23, 1], [1776, 12, 4, 0], [1776, 12, 4, 1], [1776, 12, 14, 0], [1776, 12, 14, 1], [1776, 12, 23, 0], [1776, 12, 23, 1], [1789, 7, 4, 0], [1789, 7, 4, 1], [1789, 7, 14, 0], [1789, 7, 14, 1], [1789, 7, 23, 0], [1789, 7, 23, 1], [1789, 12, 4, 0], [1789, 12, 4, 1], [1789, 12, 14, 0], [1789, 12, 14, 1], [1789, 12, 23, 0], [1789, 12, 23, 1]]
[1, 2, 3] × [30] × [500, 100] = [[1, 30, 500], [1, 30, 100], [2, 30, 500], [2, 30, 100], [3, 30, 500], [3, 30, 100]]
[1, 2, 3] × [] × [500, 100] = []

Non-Numeric Example:
[John,Paul,George,Ringo] × [Emerson,Lake,Palmer] × [Simon,Garfunkle] = [
	[John, Emerson, Simon],
	[John, Emerson, Garfunkle],
	[John, Lake, Simon],
	[John, Lake, Garfunkle],
	[John, Palmer, Simon],
	[John, Palmer, Garfunkle],
	[Paul, Emerson, Simon],
	[Paul, Emerson, Garfunkle],
	[Paul, Lake, Simon],
	[Paul, Lake, Garfunkle],
	[Paul, Palmer, Simon],
	[Paul, Palmer, Garfunkle],
	[George, Emerson, Simon],
	[George, Emerson, Garfunkle],
	[George, Lake, Simon],
	[George, Lake, Garfunkle],
	[George, Palmer, Simon],
	[George, Palmer, Garfunkle],
	[Ringo, Emerson, Simon],
	[Ringo, Emerson, Garfunkle],
	[Ringo, Lake, Simon],
	[Ringo, Lake, Garfunkle],
	[Ringo, Palmer, Simon],
	[Ringo, Palmer, Garfunkle],
]

Haskell

Various routes can be taken to Cartesian products in Haskell. For the product of two lists we could write: <lang Haskell>cartProd :: [a] -> [b] -> [(a, b)] cartProd xs ys =

 [ (x, y)
 | x <- xs 
 , y <- ys ]</lang>

more directly: <lang Haskell>cartProd :: [a] -> [b] -> [(a, b)] cartProd xs ys = xs >>= \x -> ys >>= \y -> [(x, y)]</lang>

applicatively: <lang Haskell>cartProd :: [a] -> [b] -> [(a, b)] cartProd xs ys = (,) <$> xs <*> ys</lang>

parsimoniously: <lang Haskell>cartProd :: [a] -> [b] -> [(a, b)] cartProd = (<*>) . fmap (,)</lang>

We might test any of these with: <lang haskell>main :: IO () main =

 mapM_ print $
 uncurry cartProd <$>
 [([1, 2], [3, 4]), ([3, 4], [1, 2]), ([1, 2], []), ([], [1, 2])]</lang>

Output:

[(1,3),(1,4),(2,3),(2,4)]
[(3,1),(3,2),(4,1),(4,2)]
[]
[]

For the n-ary Cartesian product of an arbitrary number of lists, we could apply the Prelude's standard sequence function to a list of lists, <lang haskell>cartProdN :: a -> a cartProdN = sequence

main :: IO () main = print $ cartProdN [[1, 2], [3, 4], [5, 6]]</lang>

Output:

[[1,3,5],[1,3,6],[1,4,5],[1,4,6],[2,3,5],[2,3,6],[2,4,5],[2,4,6]]

or we could define ourselves an equivalent function over a list of lists in terms of a fold, for example as: <lang haskell>cartProdN :: a -> a cartProdN = foldr (\xs as -> xs >>= (<$> as) . (:)) [[]]</lang> or, equivalently, as: <lang haskell>cartProdN :: a -> a cartProdN = foldr

   (\xs as ->
       [ x : a
       | x <- xs
       , a <- as ])
   [[]]</lang>

testing any of these with something like: <lang haskell>main :: IO () main = do

 mapM_ print $ 
   cartProdN [[1776, 1789], [7,12], [4, 14, 23], [0,1]]
 putStrLn ""
 print $ cartProdN [[1,2,3], [30], [500, 100]]
 putStrLn ""
 print $ cartProdN [[1,2,3], [], [500, 100]]</lang>

Output:

[1776,7,4,0]
[1776,7,4,1]
[1776,7,14,0]
[1776,7,14,1]
[1776,7,23,0]
[1776,7,23,1]
[1776,12,4,0]
[1776,12,4,1]
[1776,12,14,0]
[1776,12,14,1]
[1776,12,23,0]
[1776,12,23,1]
[1789,7,4,0]
[1789,7,4,1]
[1789,7,14,0]
[1789,7,14,1]
[1789,7,23,0]
[1789,7,23,1]
[1789,12,4,0]
[1789,12,4,1]
[1789,12,14,0]
[1789,12,14,1]
[1789,12,23,0]
[1789,12,23,1]

[[1,30,500],[1,30,100],[2,30,500],[2,30,100],[3,30,500],[3,30,100]]

[]

J

The J primitive catalogue { forms the Cartesian Product of two or more boxed lists. The result is a multi-dimensional array (which can be reshaped to a simple list of lists if desired). <lang j> { 1776 1789 ; 7 12 ; 4 14 23 ; 0 1 NB. result is 4 dimensional array with shape 2 2 3 2 ┌────────────┬────────────┐ │1776 7 4 0 │1776 7 4 1 │ ├────────────┼────────────┤ │1776 7 14 0 │1776 7 14 1 │ ├────────────┼────────────┤ │1776 7 23 0 │1776 7 23 1 │ └────────────┴────────────┘

┌────────────┬────────────┐ │1776 12 4 0 │1776 12 4 1 │ ├────────────┼────────────┤ │1776 12 14 0│1776 12 14 1│ ├────────────┼────────────┤ │1776 12 23 0│1776 12 23 1│ └────────────┴────────────┘

┌────────────┬────────────┐ │1789 7 4 0 │1789 7 4 1 │ ├────────────┼────────────┤ │1789 7 14 0 │1789 7 14 1 │ ├────────────┼────────────┤ │1789 7 23 0 │1789 7 23 1 │ └────────────┴────────────┘

┌────────────┬────────────┐ │1789 12 4 0 │1789 12 4 1 │ ├────────────┼────────────┤ │1789 12 14 0│1789 12 14 1│ ├────────────┼────────────┤ │1789 12 23 0│1789 12 23 1│ └────────────┴────────────┘

  { 1 2 3 ; 30 ; 50 100    NB. result is a 2-dimensional array with shape 2 3

┌───────┬────────┐ │1 30 50│1 30 100│ ├───────┼────────┤ │2 30 50│2 30 100│ ├───────┼────────┤ │3 30 50│3 30 100│ └───────┴────────┘

  { 1 2 3 ;  ; 50 100    NB. result is an empty 3-dimensional array with shape 3 0 2

</lang>

Java

Works with: Java Virtual Machine version 1.8

<lang Java> import static java.util.Arrays.asList; import static java.util.Collections.emptyList; import static java.util.Optional.of; import static java.util.stream.Collectors.toList;

import java.util.List;

public class CartesianProduct {

   public List<?> product(List<?>... a) {
       if (a.length >= 2) {
           List<?> product = a[0];
           for (int i = 1; i < a.length; i++) {
               product = product(product, a[i]);
           }
           return product;
       }

       return emptyList();
   }

   private <A, B> List<?> product(List<A> a, List b) {
       return of(a.stream()
               .map(e1 -> of(b.stream().map(e2 -> asList(e1, e2)).collect(toList())).orElse(emptyList()))
               .flatMap(List::stream)
               .collect(toList())).orElse(emptyList());
   }

} </lang>

Using a generic class with a recursive function <lang Java> import java.util.ArrayList; import java.util.Arrays; import java.util.List;

public class CartesianProduct<V> {

public List<List<V>> product(List<List<V>> lists) { List<List<V>> product = new ArrayList<>();

// We first create a list for each value of the first list product(product, new ArrayList<>(), lists);

return product; }

private void product(List<List<V>> result, List<V> existingTupleToComplete, List<List<V>> valuesToUse) { for (V value : valuesToUse.get(0)) { List<V> newExisting = new ArrayList<>(existingTupleToComplete); newExisting.add(value);

// If only one column is left if (valuesToUse.size() == 1) { // We create a new list with the exiting tuple for each value with the value // added result.add(newExisting); } else { // If there are still several columns, we go into recursion for each value List<List<V>> newValues = new ArrayList<>(); // We build the next level of values for (int i = 1; i < valuesToUse.size(); i++) { newValues.add(valuesToUse.get(i)); }

product(result, newExisting, newValues); } } }

public static void main(String[] args) { List<Integer> list1 = new ArrayList<>(Arrays.asList(new Integer[] { 1776, 1789 })); List<Integer> list2 = new ArrayList<>(Arrays.asList(new Integer[] { 7, 12 })); List<Integer> list3 = new ArrayList<>(Arrays.asList(new Integer[] { 4, 14, 23 })); List<Integer> list4 = new ArrayList<>(Arrays.asList(new Integer[] { 0, 1 }));

List<List<Integer>> input = new ArrayList<>(); input.add(list1); input.add(list2); input.add(list3); input.add(list4);

CartesianProduct<Integer> cartesianProduct = new CartesianProduct<>(); List<List<Integer>> product = cartesianProduct.product(input); System.out.println(product); } } </lang>

JavaScript
ES6
Functional

Cartesian products fall quite naturally out of concatMap (Array.flatMap), and its argument-flipped twin bind.

For the Cartesian product of just two lists: <lang JavaScript>(() => {

// CARTESIAN PRODUCT OF TWO LISTS ---------------------
// cartProd :: [a] -> [b] -> a, b const cartProd = xs => ys => xs.flatMap(x => ys.map(y => [x, y]))

// TEST ----------------------------------------------- return [ cartProd([1, 2])([3, 4]), cartProd([3, 4])([1, 2]), cartProd([1, 2])([]), cartProd([])([1, 2]), ].map(JSON.stringify).join('\n');

})();</lang>

Output:
[[1,3],[1,4],[2,3],[2,4]] [[3,1],[3,2],[4,1],[4,2]] [] []

Abstracting a little more, we can define the cartesian product quite economically in terms of a general applicative operator: <lang Javascript>(() => {

// CARTESIAN PRODUCT OF TWO LISTS ---------------------
// cartesianProduct :: [a] -> [b] -> [(a, b)] const cartesianProduct = xs => ap(xs.map(Tuple));

// GENERIC FUNCTIONS ----------------------------------
// e.g. [(*2),(/2), sqrt] <*> [1,2,3] // --> ap([dbl, hlf, root], [1, 2, 3]) // --> [2,4,6,0.5,1,1.5,1,1.4142135623730951,1.7320508075688772]
// Each member of a list of functions applied to each // of a list of arguments, deriving a list of new values.
// ap (<*>) :: [(a -> b)] -> [a] -> [b] const ap = fs => xs => // The sequential application of each of a list // of functions to each of a list of values. fs.flatMap( f => xs.map(f) );
// Tuple (,) :: a -> b -> (a, b) const Tuple = a => b => [a, b];
// TEST ----------------------------------------------- return [ cartesianProduct([1, 2])([3, 4]), cartesianProduct([3, 4])([1, 2]), cartesianProduct([1, 2])([]), cartesianProduct([])([1, 2]), ] .map(JSON.stringify) .join('\n');

})();</lang>

Output:
[[1,3],[1,4],[2,3],[2,4]] [[3,1],[3,2],[4,1],[4,2]] [] []

For the n-ary Cartesian product over a list of lists: <lang JavaScript>(() => {

const main = () => { // n-ary Cartesian product of a list of lists.
// cartProdN :: a -> a const cartProdN = foldr( xs => as => bind(as)( x => bind(xs)( a => [ [a].concat(x) ] ) ) )([ [] ]);
// TEST ------------------------------------------- return intercalate('\n\n')([ map(show)( cartProdN([ [1776, 1789], [7, 12], [4, 14, 23], [0, 1] ]) ).join('\n'), show(cartProdN([ [1, 2, 3], [30], [50, 100] ])), show(cartProdN([ [1, 2, 3], [], [50, 100] ])) ]) };
// GENERIC FUNCTIONS ----------------------------------
// bind :: [a] -> (a -> [b]) -> [b] const bind = xs => f => xs.flatMap(f);
// foldr :: (a -> b -> b) -> b -> [a] -> b const foldr = f => a => xs => xs.reduceRight((a, x) => f(x)(a), a);
// intercalate :: String -> [a] -> String const intercalate = s => xs => xs.join(s);
// map :: (a -> b) -> [a] -> [b] const map = f => xs => xs.map(f);
// show :: a -> String const show = x => JSON.stringify(x);
return main();

})();</lang>

Output:
[1776,7,4,0] [1776,7,4,1] [1776,7,14,0] [1776,7,14,1] [1776,7,23,0] [1776,7,23,1] [1776,12,4,0] [1776,12,4,1] [1776,12,14,0] [1776,12,14,1] [1776,12,23,0] [1776,12,23,1] [1789,7,4,0] [1789,7,4,1] [1789,7,14,0] [1789,7,14,1] [1789,7,23,0] [1789,7,23,1] [1789,12,4,0] [1789,12,4,1] [1789,12,14,0] [1789,12,14,1] [1789,12,23,0] [1789,12,23,1] [[1,30,50],[1,30,100],[2,30,50],[2,30,100],[3,30,50],[3,30,100]] []
Imperative

Imperative implementations of Cartesian products are inevitably less compact and direct, but we can certainly write an iterative translation of a fold over nested applications of bind or concatMap:

<lang JavaScript>(() => {

// n-ary Cartesian product of a list of lists // ( Imperative implementation )
// cartProd :: [a] -> [b] -> a, b const cartProd = lists => { let ps = [], acc = [ [] ], i = lists.length; while (i--) { let subList = lists[i], j = subList.length; while (j--) { let x = subList[j], k = acc.length; while (k--) ps.push([x].concat(acc[k])) }; acc = ps; ps = []; }; return acc.reverse(); };
// GENERIC FUNCTIONS ------------------------------------------------------
// intercalate :: String -> [a] -> String const intercalate = (s, xs) => xs.join(s);
// map :: (a -> b) -> [a] -> [b] const map = (f, xs) => xs.map(f);
// show :: a -> String const show = x => JSON.stringify(x);
// unlines :: [String] -> String const unlines = xs => xs.join('\n');
// TEST ------------------------------------------------------------------- return intercalate('\n\n', [show(cartProd([ [1, 2], [3, 4] ])), show(cartProd([ [3, 4], [1, 2] ])), show(cartProd([ [1, 2], [] ])), show(cartProd([ [], [1, 2] ])), unlines(map(show, cartProd([ [1776, 1789], [7, 12], [4, 14, 23], [0, 1] ]))), show(cartProd([ [1, 2, 3], [30], [50, 100] ])), show(cartProd([ [1, 2, 3], [], [50, 100] ])) ]);

})();</lang>

Output:
[[1,4],[1,3],[2,4],[2,3]] [[3,2],[3,1],[4,2],[4,1]] [] [] [1776,12,4,1] [1776,12,4,0] [1776,12,14,1] [1776,12,14,0] [1776,12,23,1] [1776,12,23,0] [1776,7,4,1] [1776,7,4,0] [1776,7,14,1] [1776,7,14,0] [1776,7,23,1] [1776,7,23,0] [1789,12,4,1] [1789,12,4,0] [1789,12,14,1] [1789,12,14,0] [1789,12,23,1] [1789,12,23,0] [1789,7,4,1] [1789,7,4,0] [1789,7,14,1] [1789,7,14,0] [1789,7,23,1] [1789,7,23,0] [[1,30,50],[1,30,100],[2,30,50],[2,30,100],[3,30,50],[3,30,100]] []
jq

jq is stream-oriented and so we begin by defining a function that will emit a stream of the elements of the Cartesian product of two arrays: <lang jq> def products: .[0][] as $x | .[1][] as $y | [$x,$y]; </lang>

To generate an array of these arrays, one would in practice most likely simply write `[products]`, but to comply with the requirements of this article, we can define `product` as: <lang jq> def product: [products]; </lang>

For the sake of brevity, two illustrations should suffice:

[ [1,2], [3,4] ] | products

produces the stream:

[1,3] [1,4] [2,3] [2,4]

And <lang jq> [[1,2], []] | product </lang> produces:

[]
n-way Cartesian Product

Given an array of two or more arrays as input, `cartesians` as defined here produces a stream of the components of their Cartesian product:

<lang jq> def cartesians:

if length <= 2 then products else .[0][] as $x | (.[1:] | cartesians) as $y | [$x] + $y end;

</lang>

Again for brevity, in the following, we will just show the number of items in the Cartesian products:

Run in REPL. <lang julia>

Product {1, 2} × {3, 4}

collect(Iterators.product([1, 2], [3, 4]))

Product {3, 4} × {1, 2}

collect(Iterators.product([3, 4], [1, 2]))

Product {1, 2} × {}

collect(Iterators.product([1, 2], []))

Product {} × {1, 2}

collect(Iterators.product([], [1, 2]))

Product {1776, 1789} × {7, 12} × {4, 14, 23} × {0, 1}

collect(Iterators.product([1776, 1789], [7, 12], [4, 14, 23], [0, 1]))

Product {1, 2, 3} × {30} × {500, 100}

collect(Iterators.product([1, 2, 3], [30], [500, 100]))

Product {1, 2, 3} × {} × {500, 100}

collect(Iterators.product([1, 2, 3], [], [500, 100])) </lang>

Kotlin

<lang scala>// version 1.1.2

fun flattenList(nestList: List<Any>): List<Any> {

val flatList = mutableListOf<Any>()
fun flatten(list: List<Any>) { for (e in list) { if (e !is List<*>) flatList.add(e) else @Suppress("UNCHECKED_CAST") flatten(e as List<Any>) } }
flatten(nestList) return flatList

}

operator fun List<Any>.times(other: List<Any>): List<List<Any>> {

val prod = mutableListOf<List<Any>>() for (e in this) { for (f in other) { prod.add(listOf(e, f)) } } return prod

}

fun nAryCartesianProduct(lists: List<List<Any>>): List<List<Any>> {

require(lists.size >= 2) return lists.drop(2).fold(lists[0] * lists[1]) { cp, ls -> cp * ls }.map { flattenList(it) }

}

fun printNAryProduct(lists: List<List<Any>>) {

println("${lists.joinToString(" x ")} = ") println("[") println(nAryCartesianProduct(lists).joinToString("\n ", " ")) println("]\n")

}

fun main(args: Array<String>) {

println("[1, 2] x [3, 4] = ${listOf(1, 2) * listOf(3, 4)}") println("[3, 4] x [1, 2] = ${listOf(3, 4) * listOf(1, 2)}") println("[1, 2] x [] = ${listOf(1, 2) * listOf()}") println("[] x [1, 2] = ${listOf<Any>() * listOf(1, 2)}") println("[1, a] x [2, b] = ${listOf(1, 'a') * listOf(2, 'b')}") println() printNAryProduct(listOf(listOf(1776, 1789), listOf(7, 12), listOf(4, 14, 23), listOf(0, 1))) printNAryProduct(listOf(listOf(1, 2, 3), listOf(30), listOf(500, 100))) printNAryProduct(listOf(listOf(1, 2, 3), listOf<Int>(), listOf(500, 100))) printNAryProduct(listOf(listOf(1, 2, 3), listOf(30), listOf('a', 'b')))

}</lang>

Output:
[1, 2] x [3, 4] = [[1, 3], [1, 4], [2, 3], [2, 4]] [3, 4] x [1, 2] = [[3, 1], [3, 2], [4, 1], [4, 2]] [1, 2] x [] = [] [] x [1, 2] = [] [1, a] x [2, b] = [[1, 2], [1, b], [a, 2], [a, b]] [1776, 1789] x [7, 12] x [4, 14, 23] x [0, 1] = [ [1776, 7, 4, 0] [1776, 7, 4, 1] [1776, 7, 14, 0] [1776, 7, 14, 1] [1776, 7, 23, 0] [1776, 7, 23, 1] [1776, 12, 4, 0] [1776, 12, 4, 1] [1776, 12, 14, 0] [1776, 12, 14, 1] [1776, 12, 23, 0] [1776, 12, 23, 1] [1789, 7, 4, 0] [1789, 7, 4, 1] [1789, 7, 14, 0] [1789, 7, 14, 1] [1789, 7, 23, 0] [1789, 7, 23, 1] [1789, 12, 4, 0] [1789, 12, 4, 1] [1789, 12, 14, 0] [1789, 12, 14, 1] [1789, 12, 23, 0] [1789, 12, 23, 1] ] [1, 2, 3] x [30] x [500, 100] = [ [1, 30, 500] [1, 30, 100] [2, 30, 500] [2, 30, 100] [3, 30, 500] [3, 30, 100] ] [1, 2, 3] x [] x [500, 100] = [ ] [1, 2, 3] x [30] x [a, b] = [ [1, 30, a] [1, 30, b] [2, 30, a] [2, 30, b] [3, 30, a] [3, 30, b] ]
langur

We could use mapX() to map each set of values to a function, but this assignment only requires an array of arrays, so we use the X() function.

Works with: langur version 0.8.3

<lang langur>writeln X([1, 2], [3, 4]) == [[1, 3], [1, 4], [2, 3], [2, 4]] writeln X([3, 4], [1, 2]) == [[3, 1], [3, 2], [4, 1], [4, 2]] writeln X([1, 2], []) == [] writeln X([], [1, 2]) == [] writeln()

writeln X [1776, 1789], [7, 12], [4, 14, 23], [0, 1] writeln()

writeln X [1, 2, 3], [30], [500, 100] writeln()

writeln X [1, 2, 3], [], [500, 100] writeln()</lang>

Output:
true true true true [[1776, 7, 4, 0], [1776, 7, 4, 1], [1776, 7, 14, 0], [1776, 7, 14, 1], [1776, 7, 23, 0], [1776, 7, 23, 1], [1776, 12, 4, 0], [1776, 12, 4, 1], [1776, 12, 14, 0], [1776, 12, 14, 1], [1776, 12, 23, 0], [1776, 12, 23, 1], [1789, 7, 4, 0], [1789, 7, 4, 1], [1789, 7, 14, 0], [1789, 7, 14, 1], [1789, 7, 23, 0], [1789, 7, 23, 1], [1789, 12, 4, 0], [1789, 12, 4, 1], [1789, 12, 14, 0], [1789, 12, 14, 1], [1789, 12, 23, 0], [1789, 12, 23, 1]] [[1, 30, 500], [1, 30, 100], [2, 30, 500], [2, 30, 100], [3, 30, 500], [3, 30, 100]] []
Lua
Functional

An iterator is created to output the product items. <lang lua> local pk,upk = table.pack, table.unpack

local getn = function(t)return #t end local const = function(k)return function(e) return k end end local function attachIdx(f)-- one-time-off function modifier local idx = 0 return function(e)idx=idx+1 ; return f(e,idx)end end local function reduce(t,acc,f) for i=1,t.n or #t do acc=f(acc,t[i])end return acc end local function imap(t,f) local r = {n=t.n or #t, r=reduce, u=upk, m=imap} for i=1,r.n do r[i]=f(t[i])end return r end
local function prod(...) local ts = pk(...) local limit = imap(ts,getn) local idx, cnt = imap(limit,const(1)), 0 local max = reduce(limit,1,function(a,b)return a*b end) local function ret(t,i)return t[idx[i]] end return function() if cnt>=max then return end -- no more output if cnt==0 then -- skip for 1st cnt = cnt + 1 else cnt, idx[#idx] = cnt + 1, idx[#idx] + 1 for i=#idx,2,-1 do -- update index list if idx[i]<=limit[i] then break -- no further update need else -- propagate limit overflow idx[i],idx[i-1] = 1, idx[i-1]+1 end end end return cnt,imap(ts,attachIdx(ret)):u() end end

--- test

for i,a,b in prod({1,2},{3,4}) do print(i,a,b) end print() for i,a,b in prod({3,4},{1,2}) do print(i,a,b) end

</lang>

Output:
1 1 3 2 1 4 3 2 3 4 2 4 1 3 1 2 3 2 3 4 1 4 4 2
Using coroutines

I have not benchmarked this, but I believe that this should run faster than the functional implementation and also likely the imperative implementation, it has significantly fewer function calls per iteration, and only the stack changes during iteration (no garbage collection during iteration). On the other hand due to avoiding garbage collection, result is reused between returns, so mutating the returned result is unsafe.

It is possible that specialising descend by depth may yield a further improvement in performance, but it would only be able to eliminate the lookup of sets[depth] and the if test, because the reference to result[depth] is required; I doubt the increase in complexity would be worth the (potential) improvement in performance. <lang lua>local function cartesian_product(sets)

local result = {} local set_count = #sets

--[[ I believe that this should make the below go very slightly faster, because it doesn't need to lookup yield in coroutine each time it

yields, though perhaps the compiler optimises the lookup away? ]] local yield = coroutine.yield local function descend(depth) if depth == set_count then for k,v in pairs(sets[depth]) do result[depth] = v yield(result) end else for k,v in pairs(sets[depth]) do result[depth] = v descend(depth + 1) end end end return coroutine.wrap(function() descend(1) end)

end

--- tests local test_cases = {

{{1, 2}, {3, 4}}, {{3, 4}, {1, 2}}, {{1776, 1789}, {7, 12}, {4, 14, 23}, {0,1}}, {{1,2,3}, {30}, {500, 100}}, {{1,2,3}, {}, {500, 100}}

}

local function format_nested_list(list)

if list[1] and type(list[1]) == "table" then local formatted_items = {} for i, item in ipairs(list) do formatted_items[i] = format_nested_list(item) end return format_nested_list(formatted_items) else return "{" .. table.concat(list, ",") .. "}" end

end

for _,test_case in ipairs(test_cases) do

print(format_nested_list(test_case)) for product in cartesian_product(test_case) do print(" " .. format_nested_list(product)) end

end</lang>

Imperative iterator

The functional implementation restated as an imperative iterator, also adjusted to not allocate a new result table on each iteration; this saves time, but makes mutating the returned table unsafe. <lang lua>local function cartesian_product(sets)

local item_counts = {} local indices = {} local results = {} local set_count = #sets local combination_count = 1 for set_index=set_count, 1, -1 do local set = sets[set_index] local item_count = #set item_counts[set_index] = item_count indices[set_index] = 1 results[set_index] = set[1] combination_count = combination_count * item_count end local combination_index = 0 return function() if combination_index >= combination_count then return end -- no more output
if combination_index == 0 then goto skip_update end -- skip first index update indices[set_count] = indices[set_count] + 1 for set_index=set_count, 1, -1 do -- update index list local set = sets[set_index] local index = indices[set_index] if index <= item_counts[set_index] then results[set_index] = set[index] break -- no further update needed else -- propagate item_counts overflow results[set_index] = set[1] indices[set_index] = 1 if set_index > 1 then indices[set_index - 1] = indices[set_index - 1] + 1 end end end ::skip_update:: combination_index = combination_index + 1 return combination_index, results end

end --- tests local test_cases = {

{{1, 2}, {3, 4}}, {{3, 4}, {1, 2}}, {{1776, 1789}, {7, 12}, {4, 14, 23}, {0,1}}, {{1,2,3}, {30}, {500, 100}}, {{1,2,3}, {}, {500, 100}}

}

local function format_nested_list(list)

if list[1] and type(list[1]) == "table" then local formatted_items = {} for i, item in ipairs(list) do formatted_items[i] = format_nested_list(item) end return format_nested_list(formatted_items) else return "{" .. table.concat(list, ",") .. "}" end

end

for _,test_case in ipairs(test_cases) do

print(format_nested_list(test_case)) for i, product in cartesian_product(test_case) do print(i, format_nested_list(product)) end

end</lang>

Maple

<lang Maple> cartmulti := proc ()

local m, v; if [] in {args} then return []; else

m := Iterator:-CartesianProduct(args);

for v in m do printf("%{}a\n", v); end do; end if; end proc;

</lang>

Mathematica

<lang Mathematica>cartesianProduct[args__] := Flatten[Outer[List, args], Length[{args}] - 1]</lang>

Modula-2

<lang modula2>MODULE CartesianProduct; FROM FormatString IMPORT FormatString; FROM Terminal IMPORT WriteString,WriteLn,ReadChar;

PROCEDURE WriteInt(a : INTEGER); VAR buf : ARRAY[0..9] OF CHAR; BEGIN

FormatString("%i", buf, a); WriteString(buf)

END WriteInt;

PROCEDURE Cartesian(a,b : ARRAY OF INTEGER); VAR i,j : CARDINAL; BEGIN

WriteString("["); FOR i:=0 TO HIGH(a) DO FOR j:=0 TO HIGH(b) DO IF (i>0) OR (j>0) THEN WriteString(","); END; WriteString("["); WriteInt(a[i]); WriteString(","); WriteInt(b[j]); WriteString("]") END END; WriteString("]"); WriteLn

END Cartesian;

TYPE

AP = ARRAY[0..1] OF INTEGER; E = ARRAY[0..0] OF INTEGER;

VAR

a,b : AP;

BEGIN

a := AP{1,2}; b := AP{3,4}; Cartesian(a,b);
a := AP{3,4}; b := AP{1,2}; Cartesian(a,b);
(* If there is a way to create an empty array, I do not know of it *)
ReadChar

END CartesianProduct.</lang>

Nim
Task: product of two lists

To compute the product of two lists (Nim arrays or sequences), we use an iterator. Obtaining a sequence from an iterator is easily done using "toSeq" from the module “sequtils” of the standard library.

The procedure allows to mix sequences of different types, for instance integers and characters.

In order to display the result using mathematical formalism, we have created a special procedure “$$” for the sequences and have overloaded the procedure “$” for tuples.

<lang Nim>iterator product[T1, T2](a: openArray[T1]; b: openArray[T2]): tuple[a: T1, b: T2] =

# Yield the element of the cartesian product of "a" and "b". # Yield tuples rather than arrays as it allows T1 and T2 to be different. for x in a: for y in b: yield (x, y)
———————————————————————————————————————————————————————————————————————————————————————————————————

when isMainModule:

from seqUtils import toSeq import strformat from strutils import addSep
#-------------------------------------------------------------------------------------------------
proc `$`[T1, T2](t: tuple[a: T1, b: T2]): string = ## Overloading of `$` to display a tuple without the field names. &"({t.a}, {t.b})"
proc `$$`[T](s: seq[T]): string = ## New operator to display a sequence using mathematical set notation. result = "{" for item in s: result.addSep(", ", 1) result.add($item) result.add('}')
-------------------------------------------------------------------------------------------------
const Empty = newSeq[int]() # Empty list of "int".
for (a, b) in [(@[1, 2], @[3, 4]), (@[3, 4], @[1, 2]), (@[1, 2], Empty ), ( Empty, @[1, 2])]:
echo &"{$$a} x {$$b} = {$$toSeq(product(a, b))}"</lang>

Output:
1, 2} x {3, 4} = {(1, 3), (1, 4), (2, 3), (2, 4)} {3, 4} x {1, 2} = {(3, 1), (3, 2), (4, 1), (4, 2)} {1, 2} x {} = {} {} x {1, 2} = {}
Extra credit: product of n list
Recursive procedure

As iterators cannot be recursive, we have to use a procedure which returns the whole sequence. And as we don’t know the number of sequences, we use a “varargs”. So all the sequences must contain the same type of values, values which are returned as sequences and not tuples.

Note that there exists in the standard module “algorithm” a procedure which computes the product of sequences of a same type. It is not recursive and, so, likely more efficient that the following version.

<lang Nim>proc product[T](a: varargs[seq[T]]): seq[seq[T]] =

## Return the product of several sets (sequences).
if a.len == 1: for x in a[0]: result.add(@[x]) else: for x in a[0]: for s in product(a[1..^1]): result.add(x & s)
———————————————————————————————————————————————————————————————————————————————————————————————————

when isMainModule:

import strformat let a = @[1, 2] b = @[3, 4] c = @[5, 6] echo &"{a} x {b} x {c} = {product(a, b, c)}"</lang>

Output:
@[1, 2] x @[3, 4] x @[5, 6] = @[@[1, 3, 5], @[1, 3, 6], @[1, 4, 5], @[1, 4, 6], @[2, 3, 5], @[2, 3, 6], @[2, 4, 5], @[2, 4, 6]]
Using a macro

Another way to compute the product consists to use a macro. It would be possible to create an iterator but it’s somewhat easier to produce the code to build the whole sequence. No recursion here: we generate nested loops, so the algorithm is the simplest possible.

With a macro, we are able to mix several value types: the “varrags” is no longer a problem as being used at compile time it may contain sequences of different types. And we are able to return tuples of n values instead of sequences of n values.

<lang Nim>import macros

macro product(args: varargs[typed]): untyped =

## Macro to generate the code to build the product of several sequences.
let t = args[0].getType() if t.kind != nnkBracketExpr or t[0].kind != nnkSym or $t[0] != "seq": error("Arguments must be sequences", args)
# Build the result type i.e. a tuple with "args.len" elements. # Fields are named "f0", "f1", etc. let tupleTyNode = newNimNode(nnkTupleTy) for idx, arg in args: let identDefsNode = newIdentDefs(ident('f' & $idx), arg.getType()[1]) tupleTyNode.add(identDefsNode)
# Build the nested for loops with counter "i0", "i1", etc. var stmtListNode = newStmtList() let loopsNode = nnkForStmt.newTree(ident("i0"), ident($args[0]), stmtListNode) var idx = 0 for arg in args[1..^1]: inc idx let loopNode = nnkForStmt.newTree(ident('i' & $idx), ident($arg)) stmtListNode.add(loopNode) stmtListNode = newStmtList() loopNode.add(stmtListNode)
# Build the instruction "result.add(i1, i2,...)". let parNode = newPar() let addNode = newCall(newDotExpr(ident("result"), ident("add")), parNode) for i, arg in args: parNode.add(ident('i' & $i)) stmtListNode.add(addNode)
# Build the tree. result = nnkStmtListExpr.newTree( nnkVarSection.newTree( newIdentDefs( ident("result"), nnkBracketExpr.newTree(ident("seq"), tupleTyNode))), loopsNode, ident("result"))
———————————————————————————————————————————————————————————————————————————————————————————————————

when isMainModule:

import strformat import strutils
#-------------------------------------------------------------------------------------------------
proc `$`[T: tuple](t: T): string = ## Overloading of `$` to display a tuple without the field names. result = "(" for f in t.fields: result.addSep(", ", 1) result.add($f) result.add(']')
proc `$$`[T](s: seq[T]): string = ## New operator to display a sequence using mathematical set notation. result = "{" for item in s: result.addSep(", ", 1) result.add($item) result.add('}')
#-------------------------------------------------------------------------------------------------
var a = @[1, 2] var b = @['a', 'b'] var c = @[false, true] echo &"{$$a} x {$$b} x {$$c} = {$$product(a, b, c)}"</lang>

Output:
{1, 2} x {a, b} x {false, true} = {(1, a, false], (1, a, true], (1, b, false], (1, b, true], (2, a, false], (2, a, true], (2, b, false], (2, b, true]}
OCaml

The double semicolons are necessary only for the toplevel

Naive but more readable version<lang ocaml>let rec product l1 l2 =

match l1, l2 with | [], _ | _, [] -> [] | h1::t1, h2::t2 -> (h1,h2)::(product [h1] t2)@(product t1 l2)

product [1;2] [3;4];; (*- : (int * int) list = [(1, 3); (1, 4); (2, 3); (2, 4)]*) product [3;4] [1;2];; (*- : (int * int) list = [(3, 1); (3, 2); (4, 1); (4, 2)]*) product [1;2] [];; (*- : (int * 'a) list = []*) product [] [1;2];; (*- : ('a * int) list = []*)</lang>

Implementation with a bit more tail-call optimization, introducing a helper function. The order of the result is changed but it should not be an issue for most uses. <lang ocaml>let product' l1 l2 =

let rec aux ~acc l1' l2' = match l1', l2' with | [], _ | _, [] -> acc | h1::t1, h2::t2 -> let acc = (h1,h2)::acc in let acc = aux ~acc t1 l2' in aux ~acc [h1] t2 in aux [] l1 l2

product' [1;2] [3;4];; (*- : (int * int) list = [(1, 4); (2, 4); (2, 3); (1, 3)]*) product' [3;4] [1;2];; (*- : (int * int) list = [(3, 2); (4, 2); (4, 1); (3, 1)]*) product' [1;2] [];; (*- : (int * 'a) list = []*) product' [] [1;2];; (*- : ('a * int) list = []*)</lang> Implemented using nested folds: <lang ocaml>let cart_prod l1 l2 =

List.fold_left (fun acc1 ele1 -> List.fold_left (fun acc2 ele2 -> (ele1,ele2)::acc2) acc1 l2) [] l1 ;;

cart_prod [1; 2; 3] ['a'; 'b'; 'c'] ;; (*- : (int * char) list = [(3, 'c'); (3, 'b'); (3, 'a'); (2, 'c'); (2, 'b'); (2, 'a'); (1, 'c'); (1, 'b'); (1, 'a')]*) cart_prod [1; 2; 3] [] ;; (*- : ('a * int) list = [] *)</lang>

Extra credit function. Since in OCaml a function can return only one type, and because tuples of different arities are different types, this returns a list of lists rather than a list of tuples. Since lists are homogeneous this version is restricted to products over a single type, eg integers. <lang ocaml>let rec product l =

(* We need to do the cross product of our current list and all the others * so we define a helper function for that *) let rec aux ~acc l1 l2 = match l1, l2 with | [], _ | _, [] -> acc | h1::t1, h2::t2 -> let acc = (h1::h2)::acc in let acc = (aux ~acc t1 l2) in aux ~acc [h1] t2 (* now we can do the actual computation *) in match l with | [] -> [] | [l1] -> List.map (fun x -> [x]) l1 | l1::tl -> let tail_product = product tl in aux ~acc:[] l1 tail_product

product [[1;2];[3;4]];; (*- : int list list = [[1; 4]; [2; 4]; [2; 3]; [1; 3]]*) product [[3;4];[1;2]];; (*- : int list list = [[3; 2]; [4; 2]; [4; 1]; [3; 1]]*) product [[1;2];[]];; (*- : int list list = []*) product [[];[1;2]];; (*- : int list list = []*) product [[1776; 1789];[7; 12];[4; 14; 23];[0; 1]];; (* - : int list list =

[[1776; 7; 4; 1]; [1776; 12; 4; 1]; [1776; 12; 14; 1]; [1776; 12; 23; 1];

[1776; 12; 23; 0]; [1776; 12; 14; 0]; [1776; 12; 4; 0]; [1776; 7; 14; 1]; [1776; 7; 23; 1]; [1776; 7; 23; 0]; [1776; 7; 14; 0]; [1789; 7; 4; 1]; [1789; 12; 4; 1]; [1789; 12; 14; 1]; [1789; 12; 23; 1]; [1789; 12; 23; 0]; [1789; 12; 14; 0]; [1789; 12; 4; 0]; [1789; 7; 14; 1]; [1789; 7; 23; 1]; [1789; 7; 23; 0]; [1789; 7; 14; 0]; [1789; 7; 4; 0]; [1776; 7; 4; 0]]
)

product [[1; 2; 3];[30];[500; 100]];; (* - : int list list =

[[1; 30; 500]; [2; 30; 500]; [3; 30; 500]; [3; 30; 100]; [2; 30; 100];

[1; 30; 100]]
)

product [[1; 2; 3];[];[500; 100]];; (*- : int list list = []*)</lang>

Better type

In the latter example, our function has this signature: <lang ocaml>val product : 'a list list -> 'a list list = <fun></lang> This lacks clarity as those two lists are not equivalent since one replaces a tuple. We can get a better signature by creating a tuple type: <lang ocaml>type 'a tuple = 'a list

let rec product (l:'a list tuple) =

(* We need to do the cross product of our current list and all the others * so we define a helper function for that *) let rec aux ~acc l1 l2 = match l1, l2 with | [], _ | _, [] -> acc | h1::t1, h2::t2 -> let acc = (h1::h2)::acc in let acc = (aux ~acc t1 l2) in aux ~acc [h1] t2 (* now we can do the actual computation *) in match l with | [] -> [] | [l1] -> List.map ~f:(fun x -> ([x]:'a tuple)) l1 | l1::tl -> let tail_product = product tl in aux ~acc:[] l1 tail_product

type 'a tuple = 'a list val product : 'a list tuple -> 'a tuple list = <fun></lang>

Perl
Iterative

Nested loops, with a short-circuit to quit early if any term is an empty set. <lang perl>sub cartesian {

my $sets = shift @_; for (@$sets) { return [] unless @$_ }
my $products = [[]]; for my $set (reverse @$sets) { my $partial = $products; $products = []; for my $item (@$set) { for my $product (@$partial) { push @$products, [$item, @$product]; } } } $products;

}

sub product {

my($s,$fmt) = @_; my $tuples; for $a ( @{ cartesian( \@$s ) } ) { $tuples .= sprintf "($fmt) ", @$a; } $tuples . "\n";

}

print product([[1, 2], [3, 4] ], '%1d %1d' ). product([[3, 4], [1, 2] ], '%1d %1d' ). product([[1, 2], [] ], '%1d %1d' ). product([[], [1, 2] ], '%1d %1d' ). product([[1,2,3], [30], [500,100] ], '%1d %1d %3d' ). product([[1,2,3], [], [500,100] ], '%1d %1d %3d' ). product([[1776,1789], [7,12], [4,14,23], [0,1]], '%4d %2d %2d %1d')</lang>

Output:
(1 3) (1 4) (2 3) (2 4) (3 1) (3 2) (4 1) (4 2) (1 30 500) (1 30 100) (2 30 500) (2 30 100) (3 30 500) (3 30 100) (1776 7 4 0) (1776 7 4 1) (1776 7 14 0) (1776 7 14 1) (1776 7 23 0) (1776 7 23 1) (1776 12 4 0) (1776 12 4 1) (1776 12 14 0) (1776 12 14 1) (1776 12 23 0) (1776 12 23 1) (1789 7 4 0) (1789 7 4 1) (1789 7 14 0) (1789 7 14 1) (1789 7 23 0) (1789 7 23 1) (1789 12 4 0) (1789 12 4 1) (1789 12 14 0) (1789 12 14 1) (1789 12 23 0) (1789 12 23 1)
Glob

This being Perl, there's more than one way to do it. A quick demonstration of how glob, more typically used for filename wildcard expansion, can solve the task. <lang perl>$tuples = [ map { [split /:/] } glob '{1,2,3}:{30}:{500,100}' ];

for $a (@$tuples) { printf "(%1d %2d %3d) ", @$a; }</lang>

Output:
(1 30 500) (1 30 100) (2 30 500) (2 30 100) (3 30 500) (3 30 100)
Modules

A variety of modules can do this correctly for an arbitrary number of lists (each of independent length). Arguably using modules is very idiomatic Perl. <lang perl>use ntheory qw/forsetproduct/; forsetproduct { say "@_" } [1,2,3],[qw/a b c/],[qw/@ $ !/];

use Set::Product qw/product/; product { say "@_" } [1,2,3],[qw/a b c/],[qw/@ $ !/];

use Math::Cartesian::Product; cartesian { say "@_" } [1,2,3],[qw/a b c/],[qw/@ $ !/];

use Algorithm::Loops qw/NestedLoops/; NestedLoops([[1,2,3],[qw/a b c/],[qw/@ $ !/]], sub { say "@_"; });</lang>

Phix

<lang Phix>function cart(sequence s) sequence res = {}

for n=2 to length(s) do for i=1 to length(s[1]) do for j=1 to length(s[2]) do res = append(res,s[1][i]&s[2][j]) end for end for if length(s)=2 then exit end if s[1..2] = {res} res = {} end for return res

end function

?cart({{1,2},{3,4}}) ?cart({{3,4},{1,2}}) ?cart({{1,2},{}}) ?cart({{},{1,2}}) ?cart({{1776, 1789},{7, 12},{4, 14, 23},{0, 1}}) ?cart({{1, 2, 3},{30},{500, 100}}) ?cart({{1, 2, 3},{},{500, 100}})</lang>

Output:
{{1,3},{1,4},{2,3},{2,4}} {{3,1},{3,2},{4,1},{4,2}} {} {} {{1776,7,4,0},{1776,7,4,1},{1776,7,14,0},{1776,7,14,1},{1776,7,23,0},{1776,7,23,1}, {1776,12,4,0},{1776,12,4,1},{1776,12,14,0},{1776,12,14,1},{1776,12,23,0},{1776,12,23,1}, {1789,7,4,0},{1789,7,4,1},{1789,7,14,0},{1789,7,14,1},{1789,7,23,0},{1789,7,23,1}, {1789,12,4,0},{1789,12,4,1},{1789,12,14,0},{1789,12,14,1},{1789,12,23,0},{1789,12,23,1}} {{1,30,500},{1,30,100},{2,30,500},{2,30,100},{3,30,500},{3,30,100}} {}
Phixmonti

<lang Phixmonti>include ..\Utilitys.pmt

def cart

( ) var res -1 get var ta -1 del -1 get var he -1 del ta "" != he "" != and if he len nip for he swap get var h drop ta len nip for ta swap get var t drop ( h t ) flatten res swap 0 put var res endfor endfor len if res 0 put cart endif endif

enddef

/# ---------- MAIN ---------- #/

( ( 1 2 ) ( 3 4 ) ) cart drop res print nl nl

( ( 1776 1789 ) ( 7 12 ) ( 4 14 23 ) ( 0 1 ) ) cart drop res print nl nl

( ( 1 2 3 ) ( 30 ) ( 500 100 ) ) cart drop res print nl nl

( ( 1 2 ) ( ) ) cart drop res print nl nl</lang>

PicoLisp

<lang PicoLisp>(de 2lists (L1 L2)

(mapcan '((I) (mapcar '((A) ((if (atom A) list cons) I A)) L2 ) ) L1 ) )

(de reduce (L . @)

(ifn (rest) L (2lists L (apply reduce (rest)))) )

(de cartesian (L . @)

(and L (rest) (pass reduce L)) )

(println

(cartesian (1 2)) )

(println

(cartesian NIL (1 2)) )

(println

(cartesian (1 2) (3 4)) )

(println

(cartesian (3 4) (1 2)) )

(println

(cartesian (1776 1789) (7 12) (4 14 23) (0 1)) )

(println

(cartesian (1 2 3) (30) (500 100)) )

(println

(cartesian (1 2 3) NIL (500 100)) )</lang>

Output:
NIL NIL ((1 3) (1 4) (2 3) (2 4)) ((3 1) (3 2) (4 1) (4 2)) ((1776 7 4 0) (1776 7 4 1) (1776 7 14 0) (1776 7 14 1) (1776 7 23 0) (1776 7 23 1) (1776 12 4 0) (1776 12 4 1) (1776 12 14 0) (1776 12 14 1) (1776 12 23 0) (1776 12 23 1) (1789 7 4 0) (1789 7 4 1) (1789 7 14 0) (1789 7 14 1) (1789 7 23 0) (1789 7 23 1) (1789 12 4 0) (1789 12 4 1) (1789 12 14 0) (1789 12 14 1) (1789 12 23 0) (1789 12 23 1)) ((1 30 500) (1 30 100) (2 30 500) (2 30 100) (3 30 500) (3 30 100)) NIL
Prolog

<lang Prolog> product([A|_], Bs, [A, B]) :- member(B, Bs). product([_|As], Bs, X) :- product(As, Bs, X). </lang>

Output:
?- findall(X, product([1,2],[3,4],X), S). S = [[1, 3], [1, 4], [2, 3], [2, 4]]. ?- findall(X, product([3,4],[1,2],X), S). S = [[3, 1], [3, 2], [4, 1], [4, 2]]. ?- findall(X, product([1,2,3],[],X), S). S = []. ?- findall(X, product([],[1,2,3],X), S). S = [].
Python
Using itertools

<lang python>import itertools

def cp(lsts):

return list(itertools.product(*lsts))

if __name__ == '__main__':

from pprint import pprint as pp for lists in [[[1,2],[3,4]], [[3,4],[1,2]], [[], [1, 2]], [[1, 2], []], ((1776, 1789), (7, 12), (4, 14, 23), (0, 1)), ((1, 2, 3), (30,), (500, 100)), ((1, 2, 3), (), (500, 100))]: print(lists, '=>') pp(cp(lists), indent=2)

</lang>

Output:
[[1, 2], [3, 4]] => [(1, 3), (1, 4), (2, 3), (2, 4)] [[3, 4], [1, 2]] => [(3, 1), (3, 2), (4, 1), (4, 2)] [[], [1, 2]] => [] [[1, 2], []] => [] ((1776, 1789), (7, 12), (4, 14, 23), (0, 1)) => [ (1776, 7, 4, 0), (1776, 7, 4, 1), (1776, 7, 14, 0), (1776, 7, 14, 1), (1776, 7, 23, 0), (1776, 7, 23, 1), (1776, 12, 4, 0), (1776, 12, 4, 1), (1776, 12, 14, 0), (1776, 12, 14, 1), (1776, 12, 23, 0), (1776, 12, 23, 1), (1789, 7, 4, 0), (1789, 7, 4, 1), (1789, 7, 14, 0), (1789, 7, 14, 1), (1789, 7, 23, 0), (1789, 7, 23, 1), (1789, 12, 4, 0), (1789, 12, 4, 1), (1789, 12, 14, 0), (1789, 12, 14, 1), (1789, 12, 23, 0), (1789, 12, 23, 1)] ((1, 2, 3), (30,), (500, 100)) => [ (1, 30, 500), (1, 30, 100), (2, 30, 500), (2, 30, 100), (3, 30, 500), (3, 30, 100)] ((1, 2, 3), (), (500, 100)) => []
Using the 'Applicative' abstraction

This task calls for alternative approaches to defining cartesian products, and one particularly compact alternative route to a native cartesian product (in a more mathematically reasoned idiom of programming) is through the Applicative abstraction (see Applicative Functor), which is slightly more general than the possibly better known monad structure. Applicative functions are provided off-the-shelf by languages like Agda, Idris, Haskell and Scala, and can usefully be implemented in any language, including Python, which supports higher-order functions.

If we write ourselves a re-usable Python ap function for the case of lists (applicative functions for other 'data containers' can also be written – this one applies a list of functions to a list of values):

<lang python># ap (<*>) :: [(a -> b)] -> [a] -> [b] def ap(fs):

return lambda xs: foldl( lambda a: lambda f: a + foldl( lambda a: lambda x: a + [f(x)])([])(xs) )([])(fs)</lang>

then one simple use of it will be to define the cartesian product of two lists (of possibly different type) as:

<lang python>ap(map(Tuple, xs))</lang>

where Tuple is a constructor, and xs is bound to the first of two lists. The returned value is a function which can be applied to a second list.

For an nAry product, we can then use a fold (catamorphism) to lift the basic function over two lists cartesianProduct :: [a] -> [b] -> [(a, b)] to a function over a list of lists:

<lang python># nAryCartProd :: [[a], [b], [c] ...] -> [(a, b, c ...)] def nAryCartProd(xxs):

return foldl1(cartesianProduct)( xxs )</lang>

For example:

<lang python># Two lists -> list of tuples

cartesianProduct :: [a] -> [b] -> [(a, b)]

def cartesianProduct(xs):

return ap(map(Tuple, xs))

List of lists -> list of tuples
nAryCartProd :: [[a], [b], [c] ...] -> [(a, b, c ...)]

def nAryCartProd(xxs):

return foldl1(cartesianProduct)( xxs )

main :: IO ()

def main():

# Product of lists of different types print ( 'Product of two lists of different types:' ) print( cartesianProduct(['a', 'b', 'c'])( [1, 2] ) )
# TESTS OF PRODUCTS OF TWO LISTS
print( '\nSpecified tests of products of two lists:' ) print( cartesianProduct([1, 2])([3, 4]), ' <--> ', cartesianProduct([3, 4])([1, 2]) ) print ( cartesianProduct([1, 2])([]), ' <--> ', cartesianProduct([])([1, 2]) )
# TESTS OF N-ARY CARTESIAN PRODUCTS
print('\nSpecified tests of nAry products:') for xs in nAryCartProd([[1776, 1789], [7, 12], [4, 14, 23], [0, 1]]): print(xs)
for xs in ( map_(nAryCartProd)( [ [[1, 2, 3], [30], [500, 100]], [[1, 2, 3], [], [500, 100]] ] ) ): print( xs )
GENERIC -------------------------------------------------

Applicative function for lists
ap (<*>) :: [(a -> b)] -> [a] -> [b]

def ap(fs):

return lambda xs: foldl( lambda a: lambda f: a + foldl( lambda a: lambda x: a + [f(x)])([])(xs) )([])(fs)

foldl :: (a -> b -> a) -> a -> [b] -> a

def foldl(f):

def go(v, xs): a = v for x in xs: a = f(a)(x) return a return lambda acc: lambda xs: go(acc, xs)

foldl1 :: (a -> a -> a) -> [a] -> a

def foldl1(f):

return lambda xs: foldl(f)(xs[0])( xs[1:] ) if xs else None

map :: (a -> b) -> [a] -> [b]

def map_(f):

return lambda xs: list(map(f, xs))

Tuple :: a -> b -> (a, b)

def Tuple(x):

return lambda y: ( x + (y,) ) if tuple is type(x) else (x, y)

TEST ----------------------------------------------------

if __name__ == '__main__':

main()</lang>

Output:
Product of two lists of different types: [('a', 1), ('a', 2), ('b', 1), ('b', 2), ('c', 1), ('c', 2)] Specified tests of products of two lists: [(1, 3), (1, 4), (2, 3), (2, 4)] <--> [(3, 1), (3, 2), (4, 1), (4, 2)] [] <--> [] Specified tests of nAry products: (1776, 7, 4, 0) (1776, 7, 4, 1) (1776, 7, 14, 0) (1776, 7, 14, 1) (1776, 7, 23, 0) (1776, 7, 23, 1) (1776, 12, 4, 0) (1776, 12, 4, 1) (1776, 12, 14, 0) (1776, 12, 14, 1) (1776, 12, 23, 0) (1776, 12, 23, 1) (1789, 7, 4, 0) (1789, 7, 4, 1) (1789, 7, 14, 0) (1789, 7, 14, 1) (1789, 7, 23, 0) (1789, 7, 23, 1) (1789, 12, 4, 0) (1789, 12, 4, 1) (1789, 12, 14, 0) (1789, 12, 14, 1) (1789, 12, 23, 0) (1789, 12, 23, 1) [(1, 30, 500), (1, 30, 100), (2, 30, 500), (2, 30, 100), (3, 30, 500), (3, 30, 100)] []
R

<lang R> one_w_many <- function(one, many) lapply(many, function(x) c(one,x))

Let's define an infix operator to perform a cartesian product.

"%p%" <- function( a, b ) {

p = c( sapply(a, function (x) one_w_many(x, b) ) ) if (is.null(unlist(p))) list() else p}

display_prod <-

function (xs) { for (x in xs) cat( paste(x, collapse=", "), "\n" ) }

fmt_vec <- function(v) sprintf("(%s)", paste(v, collapse=', '))

go <- function (...) {

cat("\n", paste( sapply(list(...),fmt_vec), collapse=" * "), "\n") prod = Reduce( '%p%', list(...) ) display_prod( prod ) }

</lang>

Simple tests:

<lang R> > display_prod( c(1, 2) %p% c(3, 4) ) 1, 3 1, 4 2, 3 2, 4 > display_prod( c(3, 4) %p% c(1, 2) ) 3, 1 3, 2 4, 1 4, 2 > display_prod( c(3, 4) %p% c() ) > </lang>

Tougher tests:

Output:
(1776, 1789) * (7, 12) * (4, 14, 23) * (0, 1) 1776, 7, 4, 0 1776, 7, 4, 1 1776, 7, 14, 0 1776, 7, 14, 1 1776, 7, 23, 0 1776, 7, 23, 1 1776, 12, 4, 0 1776, 12, 4, 1 1776, 12, 14, 0 1776, 12, 14, 1 1776, 12, 23, 0 1776, 12, 23, 1 1789, 7, 4, 0 1789, 7, 4, 1 1789, 7, 14, 0 1789, 7, 14, 1 1789, 7, 23, 0 1789, 7, 23, 1 1789, 12, 4, 0 1789, 12, 4, 1 1789, 12, 14, 0 1789, 12, 14, 1 1789, 12, 23, 0 1789, 12, 23, 1 (1, 2, 3) * (30) * (500, 100) 1, 30, 500 1, 30, 100 2, 30, 500 2, 30, 100 3, 30, 500 3, 30, 100 (1, 2, 3) * () * (500, 100)
Racket

Racket has a built-in "cartesian-product" function:

<lang>#lang racket/base (require rackunit

;; usually, included in "racket", but we're using racket/base so we ;; show where this comes from (only-in racket/list cartesian-product))

these tests will pass silently

(check-equal? (cartesian-product '(1 2) '(3 4))

'((1 3) (1 4) (2 3) (2 4)))

(check-equal? (cartesian-product '(3 4) '(1 2))

'((3 1) (3 2) (4 1) (4 2)))

(check-equal? (cartesian-product '(1 2) '()) '()) (check-equal? (cartesian-product '() '(1 2)) '())

these will print

(cartesian-product '(1776 1789) '(7 12) '(4 14 23) '(0 1)) (cartesian-product '(1 2 3) '(30) '(500 100)) (cartesian-product '(1 2 3) '() '(500 100))</lang>

Output:
'((1776 7 4 0) (1776 7 4 1) (1776 7 14 0) (1776 7 14 1) (1776 7 23 0) (1776 7 23 1) (1776 12 4 0) (1776 12 4 1) (1776 12 14 0) (1776 12 14 1) (1776 12 23 0) (1776 12 23 1) (1789 7 4 0) (1789 7 4 1) (1789 7 14 0) (1789 7 14 1) (1789 7 23 0) (1789 7 23 1) (1789 12 4 0) (1789 12 4 1) (1789 12 14 0) (1789 12 14 1) (1789 12 23 0) (1789 12 23 1)) '((1 30 500) (1 30 100) (2 30 500) (2 30 100) (3 30 500) (3 30 100)) '()
Raku

(formerly Perl 6)

Works with: Rakudo version 2017.06

The cross meta operator X will return the cartesian product of two lists. To apply the cross meta-operator to a variable number of lists, use the reduce cross meta operator [X].

<lang perl6># cartesian product of two lists using the X cross meta-operator say (1, 2) X (3, 4); say (3, 4) X (1, 2); say (1, 2) X ( ); say ( ) X ( 1, 2 );

cartesian product of variable number of lists using

the [X] reduce cross meta-operator

say [X] (1776, 1789), (7, 12), (4, 14, 23), (0, 1); say [X] (1, 2, 3), (30), (500, 100); say [X] (1, 2, 3), (), (500, 100);</lang>

Output:
((1 3) (1 4) (2 3) (2 4)) ((3 1) (3 2) (4 1) (4 2)) () () ((1776 7 4 0) (1776 7 4 1) (1776 7 14 0) (1776 7 14 1) (1776 7 23 0) (1776 7 23 1) (1776 12 4 0) (1776 12 4 1) (1776 12 14 0) (1776 12 14 1) (1776 12 23 0) (1776 12 23 1) (1789 7 4 0) (1789 7 4 1) (1789 7 14 0) (1789 7 14 1) (1789 7 23 0) (1789 7 23 1) (1789 12 4 0) (1789 12 4 1) (1789 12 14 0) (1789 12 14 1) (1789 12 23 0) (1789 12 23 1)) ((1 30 500) (1 30 100) (2 30 500) (2 30 100) (3 30 500) (3 30 100)) ()
REXX
version 1

This REXX version isn't limited by the number of lists or the number of sets within a list. <lang rexx>/*REXX program calculates the Cartesian product of two arbitrary-sized lists. */ @.= /*assign the default value to @. array*/ parse arg @.1 /*obtain the optional value of @.1 */ if @.1= then do; @.1= "{1,2} {3,4}" /*Not specified? Then use the defaults*/

@.2= "{3,4} {1,2}" /* " " " " " " */ @.3= "{1,2} {}" /* " " " " " " */ @.4= "{} {3,4}" /* " " " " " " */ @.5= "{1,2} {3,4,5}" /* " " " " " " */ end /* [↓] process each of the @.n values*/ do n=1 while @.n \= /*keep processing while there's a value*/ z= translate( space( @.n, 0), , ',') /*translate the commas to blanks. */ do #=1 until z== /*process each elements in first list. */ parse var z '{' x.# '}' z /*parse the list (contains elements). */ end /*#*/ $= do i=1 for #-1 /*process the subsequent lists. */ do a=1 for words(x.i) /*obtain the elements of the first list*/ do j=i+1 for #-1 /* " " subsequent lists. */ do b=1 for words(x.j) /* " " elements of subsequent list*/ $=$',('word(x.i, a)","word(x.j, b)')' /*append partial Cartesian product ──►$*/ end /*b*/ end /*j*/ end /*a*/ end /*i*/ say 'Cartesian product of ' space(@.n) " is ───► {"substr($, 2)'}' end /*n*/ /*stick a fork in it, we're all done. */</lang>

output when using the default lists:
Cartesian product of {1,2} {3,4} is ───► {(1,3),(1,4),(2,3),(2,4)} Cartesian product of {3,4} {1,2} is ───► {(3,1),(3,2),(4,1),(4,2)} Cartesian product of {1,2} {} is ───► {} Cartesian product of {} {3,4} is ───► {} Cartesian product of {1,2} {3,4,5} is ───► {(1,3),(1,4),(1,5),(2,3),(2,4),(2,5)}
version 2

<lang rexx>/* REXX computes the Cartesian Product of up to 4 sets */ Call cart '{1, 2} x {3, 4}' Call cart '{3, 4} x {1, 2}' Call cart '{1, 2} x {}' Call cart '{} x {1, 2}' Call cart '{1776, 1789} x {7, 12} x {4, 14, 23} x {0, 1}' Call cart '{1, 2, 3} x {30} x {500, 100}' Call cart '{1, 2, 3} x {} x {500, 100}' Exit

cart:

Parse Arg sl Say sl Do i=1 By 1 while pos('{',sl)>0 Parse Var sl '{' list '}' sl Do j=1 By 1 While list<> Parse Var list e.i.j . ',' list End n.i=j-1 If n.i=0 Then Do /* an empty set */ Say '{}' Say Return End End n=i-1 ct2.=0 Do i=1 To n.1 Do j=1 To n.2 z=ct2.0+1 ct2.z=e.1.i e.2.j ct2.0=z End End If n<3 Then Return output(2) ct3.=0 Do i=1 To ct2.0 Do k=1 To n.3 z=ct3.0+1 ct3.z=ct2.i e.3.k ct3.0=z End End If n<4 Then Return output(3) ct4.=0 Do i=1 To ct3.0 Do l=1 To n.4 z=ct4.0+1 ct4.z=ct3.i e.4.l ct4.0=z End End Return output(4)

output:

Parse Arg u Do v=1 To value('ct'u'.0') res='{'translate(value('ct'u'.'v),',',' ')'}' Say res End Say ' ' Return 0</lang>

Output:
{1, 2} x {3, 4} {1,3} {1,4} {2,3} {2,4} {3, 4} x {1, 2} {3,1} {3,2} {4,1} {4,2} {1, 2} x {} {} {} x {1, 2} {} {1776, 1789} x {7, 12} x {4, 14, 23} x {0, 1} {1776,7,4,0} {1776,7,4,1} {1776,7,14,0} {1776,7,14,1} {1776,7,23,0} {1776,7,23,1} {1776,12,4,0} {1776,12,4,1} {1776,12,14,0} {1776,12,14,1} {1776,12,23,0} {1776,12,23,1} {1789,7,4,0} {1789,7,4,1} {1789,7,14,0} {1789,7,14,1} {1789,7,23,0} {1789,7,23,1} {1789,12,4,0} {1789,12,4,1} {1789,12,14,0} {1789,12,14,1} {1789,12,23,0} {1789,12,23,1} {1, 2, 3} x {30} x {500, 100} {1,30,500} {1,30,100} {2,30,500} {2,30,100} {3,30,500} {3,30,100} {1, 2, 3} x {} x {500, 100} {}
Ring

Project : Cartesian product of two or more lists

list1 = [[1,2],[3,4]] list2 = [[3,4],[1,2]] cartesian(list1) cartesian(list2)

func cartesian(list1)

for n = 1 to len(list1[1]) for m = 1 to len(list1[2]) see "(" + list1[1][n] + ", " + list1[2][m] + ")" + nl next next see nl

</lang> Output:

(1, 3) (1, 4) (2, 3) (2, 4) (3, 1) (3, 2) (4, 1) (4, 2)
Ruby

"product" is a method of arrays. It takes one or more arrays as argument and results in the Cartesian product: <lang ruby>p [1, 2].product([3, 4]) p [3, 4].product([1, 2]) p [1, 2].product([]) p [].product([1, 2]) p [1776, 1789].product([7, 12], [4, 14, 23], [0, 1]) p [1, 2, 3].product([30], [500, 100]) p [1, 2, 3].product([], [500, 100]) </lang>

Output:
[[1, 3], [1, 4], [2, 3], [2, 4]]
[[3, 1], [3, 2], [4, 1], [4, 2]] [] [] [[1776, 7, 4, 0], [1776, 7, 4, 1], [1776, 7, 14, 0], [1776, 7, 14, 1], [1776, 7, 23, 0], [1776, 7, 23, 1], [1776, 12, 4, 0], [1776, 12, 4, 1], [1776, 12, 14, 0], [1776, 12, 14, 1], [1776, 12, 23, 0], [1776, 12, 23, 1], [1789, 7, 4, 0], [1789, 7, 4, 1], [1789, 7, 14, 0], [1789, 7, 14, 1], [1789, 7, 23, 0], [1789, 7, 23, 1], [1789, 12, 4, 0], [1789, 12, 4, 1], [1789, 12, 14, 0], [1789, 12, 14, 1], [1789, 12, 23, 0], [1789, 12, 23, 1]] [[1, 30, 500], [1, 30, 100], [2, 30, 500], [2, 30, 100], [3, 30, 500], [3, 30, 100]] []

Rust

<lang rust>fn cartesian_product(lists: &Vec<Vec<u32>>) -> Vec<Vec<u32>> {

let mut res = vec![];
let mut list_iter = lists.iter(); if let Some(first_list) = list_iter.next() { for &i in first_list { res.push(vec![i]); } } for l in list_iter { let mut tmp = vec![]; for r in res { for &el in l { let mut tmp_el = r.clone(); tmp_el.push(el); tmp.push(tmp_el); } } res = tmp; } res

}

fn main() {

let cases = vec![ vec![vec![1, 2], vec![3, 4]], vec![vec![3, 4], vec![1, 2]], vec![vec![1, 2], vec![]], vec![vec![], vec![1, 2]], vec![vec![1776, 1789], vec![7, 12], vec![4, 14, 23], vec![0, 1]], vec![vec![1, 2, 3], vec![30], vec![500, 100]], vec![vec![1, 2, 3], vec![], vec![500, 100]], ]; for case in cases { println!( "{}\n{:?}\n", case.iter().map(|c| format!("{:?}", c)).collect::<Vec<_>>().join(" × "), cartesian_product(&case) ) }

} </lang>

Output:
[1, 2] × [3, 4]
[[1, 3], [1, 4], [2, 3], [2, 4]]
[3, 4] × [1, 2] [[3, 1], [3, 2], [4, 1], [4, 2]]
[1, 2] × [] []
[] × [1, 2] []
[1776, 1789] × [7, 12] × [4, 14, 23] × [0, 1] [[1776, 7, 4, 0], [1776, 7, 4, 1], [1776, 7, 14, 0], [1776, 7, 14, 1], [1776, 7, 23, 0], [1776, 7, 23, 1], [1776, 12, 4, 0], [1776, 12, 4, 1], [1776, 12, 14, 0], [1776, 12, 14, 1], [1776, 12, 23, 0], [1776, 12, 23, 1], [1789, 7, 4, 0], [1789, 7, 4, 1], [1789, 7, 14, 0], [1789, 7, 14, 1], [1789, 7, 23, 0], [1789, 7, 23, 1], [1789, 12, 4, 0], [1789, 12, 4, 1], [1789, 12, 14, 0], [1789, 12, 14, 1], [1789, 12, 23, 0], [1789, 12, 23, 1]]
[1, 2, 3] × [30] × [500, 100] [[1, 30, 500], [1, 30, 100], [2, 30, 500], [2, 30, 100], [3, 30, 500], [3, 30, 100]]
[1, 2, 3] × [] × [500, 100] []

Scala

Function returning the n-ary product of an arbitrary number of lists, each of arbitrary length:

<lang scala>def cartesianProduct[T](lst: List[T]*): List[List[T]] = {

/** * Prepend single element to all lists of list * @param e single elemetn * @param ll list of list * @param a accumulator for tail recursive implementation * @return list of lists with prepended element e */ def pel(e: T, ll: List[List[T]], a: List[List[T]] = Nil): List[List[T]] = ll match { case Nil => a.reverse case x :: xs => pel(e, xs, (e :: x) :: a ) }
lst.toList match { case Nil => Nil case x :: Nil => List(x) case x :: _ => x match { case Nil => Nil case _ => lst.par.foldRight(List(x))( (l, a) => l.flatMap(pel(_, a)) ).map(_.dropRight(x.size)) } }

}</lang> and usage: <lang scala>cartesianProduct(List(1, 2), List(3, 4))

.map(_.mkString("(", ", ", ")")).mkString("{",", ","}")</lang>

Output:
{(1, 3), (1, 4), (2, 3), (2, 4)}

<lang scala>cartesianProduct(List(3, 4), List(1, 2))

.map(_.mkString("(", ", ", ")")).mkString("{",", ","}")</lang>

Output:
{(3, 1), (3, 2), (4, 1), (4, 2)}

<lang scala>cartesianProduct(List(1, 2), List.empty)

.map(_.mkString("(", ", ", ")")).mkString("{",", ","}")</lang>

Output:
{}

<lang scala>cartesianProduct(List.empty, List(1, 2))

.map(_.mkString("(", ", ", ")")).mkString("{",", ","}")</lang>

Output:
{}

<lang scala>cartesianProduct(List(1776, 1789), List(7, 12), List(4, 14, 23), List(0, 1))

.map(_.mkString("(", ", ", ")")).mkString("{",", ","}")</lang>

Output:
{(1776, 7, 4, 0), (1776, 7, 4, 1), (1776, 7, 14, 0), (1776, 7, 14, 1), (1776, 7, 23, 0), (1776, 7, 23, 1), (1776, 12, 4, 0), (1776, 12, 4, 1), (1776, 12, 14, 0), (1776, 12, 14, 1), (1776, 12, 23, 0), (1776, 12, 23, 1), (1789, 7, 4, 0), (1789, 7, 4, 1), (1789, 7, 14, 0), (1789, 7, 14, 1), (1789, 7, 23, 0), (1789, 7, 23, 1), (1789, 12, 4, 0), (1789, 12, 4, 1), (1789, 12, 14, 0), (1789, 12, 14, 1), (1789, 12, 23, 0), (1789, 12, 23, 1)}

<lang scala>cartesianProduct(List(1, 2, 3), List(30), List(500, 100))

.map(_.mkString("(", ", ", ")")).mkString("{",", ","}")</lang>

Output:
{(1, 30, 500), (1, 30, 100), (2, 30, 500), (2, 30, 100), (3, 30, 500), (3, 30, 100)}

<lang scala>cartesianProduct(List(1, 2, 3), List.empty, List(500, 100))

.map(_.mkString("[", ", ", "]")).mkString("\n")</lang>

Output:
{}
Sidef

In Sidef, the Cartesian product of an arbitrary number of arrays is built-in as Array.cartesian(): <lang ruby>cartesian([[1,2], [3,4], [5,6]]).say cartesian([[1,2], [3,4], [5,6]], {|*arr| say arr })</lang>

Alternatively, a simple recursive implementation: <lang ruby>func cartesian_product(*arr) {

var c = [] var r = []
func { if (c.len < arr.len) { for item in (arr[c.len]) { c.push(item) __FUNC__() c.pop } } else { r.push([c...]) } }()
return r

}</lang>

Completing the task: <lang ruby>say cartesian_product([1,2], [3,4]) say cartesian_product([3,4], [1,2])</lang>

Output:
[[1, 3], [1, 4], [2, 3], [2, 4]] [[3, 1], [3, 2], [4, 1], [4, 2]]

The product of an empty list with any other list is empty: <lang ruby>say cartesian_product([1,2], []) say cartesian_product([], [1,2])</lang>

Output:
[] []

Extra credit: <lang ruby>cartesian_product([1776, 1789], [7, 12], [4, 14, 23], [0, 1]).each{ .say }</lang>

Output:
[1776, 7, 4, 0] [1776, 7, 4, 1] [1776, 7, 14, 0] [1776, 7, 14, 1] [1776, 7, 23, 0] [1776, 7, 23, 1] [1776, 12, 4, 0] [1776, 12, 4, 1] [1776, 12, 14, 0] [1776, 12, 14, 1] [1776, 12, 23, 0] [1776, 12, 23, 1] [1789, 7, 4, 0] [1789, 7, 4, 1] [1789, 7, 14, 0] [1789, 7, 14, 1] [1789, 7, 23, 0] [1789, 7, 23, 1] [1789, 12, 4, 0] [1789, 12, 4, 1] [1789, 12, 14, 0] [1789, 12, 14, 1] [1789, 12, 23, 0] [1789, 12, 23, 1]

<lang ruby>say cartesian_product([1, 2, 3], [30], [500, 100]) say cartesian_product([1, 2, 3], [], [500, 100])</lang>

Output:
[[1, 30, 500], [1, 30, 100], [2, 30, 500], [2, 30, 100], [3, 30, 500], [3, 30, 100]] []
SQL

If we create lists as tables with one column, cartesian product is easy. <lang sql>-- set up list 1 create table L1 (value integer); insert into L1 values (1); insert into L1 values (2); -- set up list 2 create table L2 (value integer); insert into L2 values (3); insert into L2 values (4); -- get the product select * from L1, L2;</lang>

Output:
VALUE VALUE ---------- ---------- 1 3 1 4 2 3 2 4

You should be able to be more explicit should get the same result:<lang sql>select * from L1 cross join L2;</lang>

Product with an empty list works as expected (using the tables created above): <lang sql>delete from L2; select * from L1, L2;</lang>

Output:
no rows selected

I don't think "extra credit" is meaningful here because cartesian product is so hard-baked into SQL, so here's just one of the extra credit examples (again using the tables created above):<lang sql>insert into L1 values (3); insert into L2 values (30); create table L3 (value integer); insert into L3 values (500); insert into L3 values (100); -- product works the same for as many "lists" as you'd like select * from L1, L2, L3;</lang>

Output:
VALUE VALUE VALUE ---------- ---------- ---------- 1 30 500 2 30 500 3 30 500 1 30 100 2 30 100 3 30 100
Standard ML

<lang sml>fun prodList (nil, _) = nil

| prodList ((x::xs), ys) = map (fn y => (x,y)) ys @ prodList (xs, ys)

fun naryProdList zs = foldl (fn (xs, ys) => map op:: (prodList (xs, ys))) [[]] (rev zs)</lang>

Output:
- prodList ([1, 2], [3, 4]); val it = [(1,3),(1,4),(2,3),(2,4)] : (int * int) list - prodList ([3, 4], [1, 2]); val it = [(3,1),(3,2),(4,1),(4,2)] : (int * int) list - prodList ([1, 2], []); stdIn:8.1-8.22 Warning: type vars not generalized because of value restriction are instantiated to dummy types (X1,X2,...) val it = [] : (int * ?.X1) list - naryProdList [[1776, 1789], [7, 12], [4, 14, 23], [0, 1]]; val it = [[1776,7,4,0],[1776,7,4,1],[1776,7,14,0],[1776,7,14,1],[1776,7,23,0], [1776,7,23,1],[1776,12,4,0],[1776,12,4,1],[1776,12,14,0],[1776,12,14,1], [1776,12,23,0],[1776,12,23,1],[1789,7,4,0],[1789,7,4,1],[1789,7,14,0], [1789,7,14,1],[1789,7,23,0],[1789,7,23,1],[1789,12,4,0],[1789,12,4,1], [1789,12,14,0],[1789,12,14,1],[1789,12,23,0],[1789,12,23,1]] : int list list - naryProdList [[1, 2, 3], [30], [500, 100]]; val it = [[1,30,500],[1,30,100],[2,30,500],[2,30,100],[3,30,500],[3,30,100]] : int list list - naryProdList [[1, 2, 3], [], [500, 100]]; val it = [] : int list list
Stata

In Stata, the command fillin may be used to expand a dataset with all combinations of a number of variables. Thus it's easy to compute a cartesian product.

<lang stata>. list

+-------+ | a b | |-------| 1. | 1 3 | 2. | 2 4 | +-------+

. fillin a b . list

+-----------------+ | a b _fillin | |-----------------| 1. | 1 3 0 | 2. | 1 4 1 | 3. | 2 3 1 | 4. | 2 4 0 | +-----------------+</lang>

The other way around:

<lang stata>. list

+-------+ | a b | |-------| 1. | 3 1 | 2. | 4 2 | +-------+

. fillin a b . list

+-----------------+ | a b _fillin | |-----------------| 1. | 3 1 0 | 2. | 3 2 1 | 3. | 4 1 1 | 4. | 4 2 0 | +-----------------+</lang>

Note, however, that this is not equivalent to a cartesian product when one of the variables is "empty" (that is, only contains missing values).

<lang stata>. list

+-------+ | a b | |-------| 1. | 1 . | 2. | 2 . | +-------+

. fillin a b . list

+-----------------+ | a b _fillin | |-----------------| 1. | 1 . 0 | 2. | 2 . 0 | +-----------------+</lang>

This command works also if the varaibles have different numbers of nonmissing elements. However, this requires additional code to remove the observations with missing values.

<lang stata>. list

+-----------+ | a b c | |-----------| 1. | 1 4 6 | 2. | 2 5 . | 3. | 3 . . | +-----------+

. fillin a b c . list

+---------------------+ | a b c _fillin | |---------------------| 1. | 1 4 6 0 | 2. | 1 4 . 1 | 3. | 1 5 6 1 | 4. | 1 5 . 1 | 5. | 1 . 6 1 | |---------------------| 6. | 1 . . 1 | 7. | 2 4 6 1 | 8. | 2 4 . 1 | 9. | 2 5 6 1 | 10. | 2 5 . 0 | |---------------------| 11. | 2 . 6 1 | 12. | 2 . . 1 | 13. | 3 4 6 1 | 14. | 3 4 . 1 | 15. | 3 5 6 1 | |---------------------| 16. | 3 5 . 1 | 17. | 3 . 6 1 | 18. | 3 . . 0 | +---------------------+

. foreach var of varlist _all {

quietly drop if missing(`var') }

. list

+---------------------+ | a b c _fillin | |---------------------| 1. | 1 4 6 0 | 2. | 1 5 6 1 | 3. | 2 4 6 1 | 4. | 2 5 6 1 | 5. | 3 4 6 1 | |---------------------| 6. | 3 5 6 1 | +---------------------+</lang>
Swift
Translation of: Scala

<lang swift>func + <T>(el: T, arr: [T]) -> [T] {

var ret = arr
ret.insert(el, at: 0)
return ret

}

func cartesianProduct<T>(_ arrays: [T]...) -> T {

guard let head = arrays.first else { return [] }
let first = Array(head)
func pel( _ el: T, _ ll: T, _ a: T = [] ) -> T { switch ll.count { case 0: return a.reversed() case _: let tail = Array(ll.dropFirst()) let head = ll.first!
return pel(el, tail, el + head + a) } }
return arrays.reversed() .reduce([first], {res, el in el.flatMap({ pel($0, res) }) }) .map({ $0.dropLast(first.count) })

}

print(cartesianProduct([1, 2], [3, 4])) print(cartesianProduct([3, 4], [1, 2])) print(cartesianProduct([1, 2], [])) print(cartesianProduct([1776, 1789], [7, 12], [4, 14, 23], [0, 1])) print(cartesianProduct([1, 2, 3], [30], [500, 100])) print(cartesianProduct([1, 2, 3], [], [500, 100])</lang>

Output:
[[1, 3], [1, 4], [2, 3], [2, 4]] [[3, 1], [3, 2], [4, 1], [4, 2]] [] [[1776, 7, 4, 0], [1776, 7, 4, 1], [1776, 7, 14, 0], [1776, 7, 14, 1], [1776, 7, 23, 0], [1776, 7, 23, 1], [1776, 12, 4, 0], [1776, 12, 4, 1], [1776, 12, 14, 0], [1776, 12, 14, 1], [1776, 12, 23, 0], [1776, 12, 23, 1], [1789, 7, 4, 0], [1789, 7, 4, 1], [1789, 7, 14, 0], [1789, 7, 14, 1], [1789, 7, 23, 0], [1789, 7, 23, 1], [1789, 12, 4, 0], [1789, 12, 4, 1], [1789, 12, 14, 0], [1789, 12, 14, 1], [1789, 12, 23, 0], [1789, 12, 23, 1]] [[1, 30, 500], [1, 30, 100], [2, 30, 500], [2, 30, 100], [3, 30, 500], [3, 30, 100]] []
Tailspin

<lang tailspin> templates cartesianProduct

{ product: [$(1)... -> [$]], rest: $(2..last) } -> # when <{ rest: <[](0)> }> do $.product ! otherwise def m: $.rest(1); { product: [$.product... -> $def n: $; $m... -> [$n..., $] !$], rest: $.rest(2..last) } -> #

end cartesianProduct

'{1,2}x{3,4} = $:[[1,2],[3,4]] -> cartesianProduct; ' -> !OUT::write

'{3,4}x{1,2} = $:[[3,4],[1,2]] -> cartesianProduct; ' -> !OUT::write

'{1,2}x{} = $:[[1,2],[]] -> cartesianProduct; ' -> !OUT::write

'{}x{1,2} = $:[[],[1,2]] -> cartesianProduct; ' -> !OUT::write

'{1776, 1789} × {7, 12} × {4, 14, 23} × {0, 1} = $:[[1776, 1789], [7, 12], [4, 14, 23], [0, 1]] -> cartesianProduct; ' -> !OUT::write

'{1, 2, 3} × {30} × {500, 100} = $:[[1, 2, 3], [30], [500, 100]] -> cartesianProduct; ' -> !OUT::write

'{1, 2, 3} × {} × {500, 100} = $:[[1, 2, 3], [], [500, 100]] -> cartesianProduct; ' -> !OUT::write </lang>

Output:
{1,2}x{3,4} = [[1, 3], [1, 4], [2, 3], [2, 4]] {3,4}x{1,2} = [[3, 1], [3, 2], [4, 1], [4, 2]] {1,2}x{} = [] {}x{1,2} = [] {1776, 1789} × {7, 12} × {4, 14, 23} × {0, 1} = [[1776, 7, 4, 0], [1776, 7, 4, 1], [1776, 7, 14, 0], [1776, 7, 14, 1], [1776, 7, 23, 0], [1776, 7, 23, 1], [1776, 12, 4, 0], [1776, 12, 4, 1], [1776, 12, 14, 0], [1776, 12, 14, 1], [1776, 12, 23, 0], [1776, 12, 23, 1], [1789, 7, 4, 0], [1789, 7, 4, 1], [1789, 7, 14, 0], [1789, 7, 14, 1], [1789, 7, 23, 0], [1789, 7, 23, 1], [1789, 12, 4, 0], [1789, 12, 4, 1], [1789, 12, 14, 0], [1789, 12, 14, 1], [1789, 12, 23, 0], [1789, 12, 23, 1]] {1, 2, 3} × {30} × {500, 100} = [[1, 30, 500], [1, 30, 100], [2, 30, 500], [2, 30, 100], [3, 30, 500], [3, 30, 100]] {1, 2, 3} × {} × {500, 100} = []
Tcl

<lang tcl> proc cartesianProduct {l1 l2} {

set result {} foreach el1 $l1 { foreach el2 $l2 { lappend result [list $el1 $el2] } } return $result

}

puts "simple" puts "result: [cartesianProduct {1 2} {3 4}]" puts "result: [cartesianProduct {3 4} {1 2}]" puts "result: [cartesianProduct {1 2} {}]" puts "result: [cartesianProduct {} {3 4}]"

proc cartesianNaryProduct {lists} {

set result {{}} foreach l $lists { set res {} foreach comb $result { foreach el $l { lappend res [linsert $comb end $el] } } set result $res } return $result

}

puts "n-ary" puts "result: [cartesianNaryProduct {{1776 1789} {7 12} {4 14 23} {0 1}}]" puts "result: [cartesianNaryProduct {{1 2 3} {30} {500 100}}]" puts "result: [cartesianNaryProduct {{1 2 3} {} {500 100}}]"

</lang>

Output:
simple result: {1 3} {1 4} {2 3} {2 4} result: {3 1} {3 2} {4 1} {4 2} result: result: n-ary result: {1776 7 4 0} {1776 7 4 1} {1776 7 14 0} {1776 7 14 1} {1776 7 23 0} {1776 7 23 1} {1776 12 4 0} {1776 12 4 1} {1776 12 14 0} {1776 12 14 1} {1776 12 23 0} {1776 12 23 1} {1789 7 4 0} {1789 7 4 1} {1789 7 14 0} {1789 7 14 1} {1789 7 23 0} {1789 7 23 1} {1789 12 4 0} {1789 12 4 1} {1789 12 14 0} {1789 12 14 1} {1789 12 23 0} {1789 12 23 1} result: {1 30 500} {1 30 100} {2 30 500} {2 30 100} {3 30 500} {3 30 100} result:
Visual Basic .NET
Translation of: C#

<lang vbnet>Imports System.Runtime.CompilerServices

Module Module1

<Extension()> Function CartesianProduct(Of T)(sequences As IEnumerable(Of IEnumerable(Of T))) As IEnumerable(Of IEnumerable(Of T)) Dim emptyProduct As IEnumerable(Of IEnumerable(Of T)) = {Enumerable.Empty(Of T)} Return sequences.Aggregate(emptyProduct, Function(accumulator, sequence) From acc In accumulator From item In sequence Select acc.Concat({item})) End Function
Sub Main() Dim empty(-1) As Integer Dim list1 = {1, 2} Dim list2 = {3, 4} Dim list3 = {1776, 1789} Dim list4 = {7, 12} Dim list5 = {4, 14, 23} Dim list6 = {0, 1} Dim list7 = {1, 2, 3} Dim list8 = {30} Dim list9 = {500, 100}
For Each sequnceList As Integer()() In { ({list1, list2}), ({list2, list1}), ({list1, empty}), ({empty, list1}), ({list3, list4, list5, list6}), ({list7, list8, list9}), ({list7, empty, list9}) } Dim cart = sequnceList.CartesianProduct().Select(Function(tuple) $"({String.Join(", ", tuple)})") Console.WriteLine($"{{{String.Join(", ", cart)}}}") Next End Sub

End Module</lang>

Output:
{(1, 3), (1, 4), (2, 3), (2, 4)} {(3, 1), (3, 2), (4, 1), (4, 2)} {} {} {(1776, 7, 4, 0), (1776, 7, 4, 1), (1776, 7, 14, 0), (1776, 7, 14, 1), (1776, 7, 23, 0), (1776, 7, 23, 1), (1776, 12, 4, 0), (1776, 12, 4, 1), (1776, 12, 14, 0), (1776, 12, 14, 1), (1776, 12, 23, 0), (1776, 12, 23, 1), (1789, 7, 4, 0), (1789, 7, 4, 1), (1789, 7, 14, 0), (1789, 7, 14, 1), (1789, 7, 23, 0), (1789, 7, 23, 1), (1789, 12, 4, 0), (1789, 12, 4, 1), (1789, 12, 14, 0), (1789, 12, 14, 1), (1789, 12, 23, 0), (1789, 12, 23, 1)} {(1, 30, 500), (1, 30, 100), (2, 30, 500), (2, 30, 100), (3, 30, 500), (3, 30, 100)} {}
zkl

Cartesian product is build into iterators or can be done with nested loops. <lang zkl>zkl: Walker.cproduct(List(1,2),List(3,4)).walk().println(); L(L(1,3),L(1,4),L(2,3),L(2,4)) zkl: foreach a,b in (List(1,2),List(3,4)){ print("(%d,%d) ".fmt(a,b)) } (1,3) (1,4) (2,3) (2,4)

zkl: Walker.cproduct(List(3,4),List(1,2)).walk().println(); L(L(3,1),L(3,2),L(4,1),L(4,2))</lang>

The walk method will throw an error if used on an empty iterator but the pump method doesn't. <lang zkl>zkl: Walker.cproduct(List(3,4),List).walk().println(); Exception thrown: TheEnd(Ain't no more)

zkl: Walker.cproduct(List(3,4),List).pump(List).println(); L() zkl: Walker.cproduct(List,List(3,4)).pump(List).println(); L()</lang> <lang zkl>zkl: Walker.cproduct(L(1776,1789),L(7,12),L(4,14,23),L(0,1)).walk().println(); L(L(1776,7,4,0),L(1776,7,4,1),L(1776,7,14,0),L(1776,7,14,1),L(1776,7,23,0),L(1776,7,23,1),L(1776,12,4,0),L(1776,12,4,1),L(1776,12,14,0),L(1776,12,14,1),L(1776,12,23,0),L(1776,12,23,1),L(1789,7,4,0),L(1789,7,4,1),L(1789,7,14,0),L(1789,7,14,1),L(1789,7,23,0),L(1789,7,23,1),L(1789,12,4,0),L(1789,12,4,1),...)

zkl: Walker.cproduct(L(1,2,3),L(30),L(500,100)).walk().println(); L(L(1,30,500),L(1,30,100),L(2,30,500),L(2,30,100),L(3,30,500),L(3,30,100))

zkl: Walker.cproduct(L(1,2,3),List,L(500,100)).pump(List).println(); L()</lang>