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A set is a collection (container) of certain values, without any particular order, and no repeated values. It corresponds with a finite set in mathematics. A set can be implemented as an associative array (partial mapping) in which the value of each key-value pair is ignored.

Given a set S, the power set (or powerset) of S, written P(S), or 2^S, is the set of all subsets of S.

Task : By using a library or built-in set type, or by defining a set type with necessary operations, write a function with a set S as input that yields the power set 2^S of S.

For example, the power set of {1,2,3,4} is {{}, {1}, {2}, {1,2}, {3}, {1,3}, {2,3}, {1,2,3}, {4}, {1,4}, {2,4}, {1,2,4}, {3,4}, {1,3,4}, {2,3,4}, {1,2,3,4}}.

For a set which contains n elements, the corresponding power set has 2ⁿ elements, including the edge cases of empty set.

The power set of the empty set is the set which contains itself (2⁰ = 1):

${\mathcal {P}}$ ( $\varnothing$ ) = { $\varnothing$ }

And the power set of the set which contains only the empty set, has two subsets, the empty set and the set which contains the empty set (2¹ = 2):

${\mathcal {P}}$ ({ $\varnothing$ }) = { $\varnothing$ , { $\varnothing$ } }

Ada

This example is incomplete. It shows no proof of the cases with empty sets. Please ensure that it meets all task requirements and remove this message.

This solution prints the power set of words read from the command line.<lang ada>with Ada.Text_IO, Ada.Command_Line;

procedure Power_Set is

  type List is array  (Positive range <>) of Positive;
  Empty: List(1 .. 0);
  
  procedure Print_All_Subsets(Set: List; Printable: List:= Empty) is

     procedure Print_Set(Items: List) is

First: Boolean := True;

     begin

Ada.Text_IO.Put("{ "); for Item of Items loop if First then First := False; -- no comma needed else Ada.Text_IO.Put(", "); -- comma, to separate the items end if; Ada.Text_IO.Put(Ada.Command_Line.Argument(Item)); end loop; Ada.Text_IO.Put_Line(" }");

     end Print_Set;
     
     Tail: List := Set(Set'First+1 .. Set'Last);
     
  begin
     if Set = Empty then

Print_Set(Printable);

     else

Print_All_Subsets(Tail, Printable & Set(Set'First)); Print_All_Subsets(Tail, Printable);

     end if;
  end Print_All_Subsets;
  
  Set: List(1 .. Ada.Command_Line.Argument_Count);

begin

  for I in Set'Range loop -- initialize set
     Set(I) := I;
  end loop;
  Print_All_Subsets(Set); -- do the work

end Power_Set;</lang>

Output:

>./power_set cat dog mouse
{ cat, dog, mouse }
{ cat, dog }
{ cat, mouse }
{ cat }
{ dog, mouse }
{ dog }
{ mouse }
{  }
>./power_set 1 2
{ 1, 2 }
{ 1 }
{ 2 }

{  }

ALGOL 68

This example is incomplete. It shows no proof of the cases with empty sets. Please ensure that it meets all task requirements and remove this message.

Works with: ALGOL 68 version Revision 1 - no extensions to language used

Works with: ALGOL 68G version Any - tested with release 1.18.0-9h.tiny

Requires: ALGOL 68g mk14.1+<lang algol68>MODE MEMBER = INT;

PROC power set = ([]MEMBER s)[][]MEMBER:(

 [2**UPB s]FLEX[1:0]MEMBER r;
 INT upb r := 0;
 r[upb r +:= 1] := []MEMBER(());
 FOR i TO UPB s DO
   MEMBER e = s[i];
   FOR j TO upb r DO
     [UPB r[j] + 1]MEMBER x;
     x[:UPB x-1] := r[j];
     x[UPB x] := e; # append to the end of x #
     r[upb r +:= 1] := x # append to end of r #
   OD
 OD;
 r[upb r] := s;
 r

);

Example: #

test:(

 [][]MEMBER set = power set((1, 2, 4));
 FOR member TO UPB set DO
   INT upb = UPB set[member];
   FORMAT repr set = $"("f( upb=0 | $$ | $n(upb-1)(d", ")d$ )");"$;
   printf(($"set["d"] = "$,member, repr set, set[member],$l$))
 OD

)</lang>

Output:

set[1] = ();
set[2] = (1);
set[3] = (2);
set[4] = (1, 2);
set[5] = (4);
set[6] = (1, 4);
set[7] = (2, 4);

set[8] = (1, 2, 4);

AutoHotkey

This example is incomplete. It shows no proof of the cases with empty sets. Please ensure that it meets all task requirements and remove this message.

ahk discussion<lang autohotkey>a = 1,a,-- ; elements separated by commas

StringSplit a, a, `, ; a0 = #elements, a1,a2,... = elements of the set

t = { Loop % (1<<a0) { ; generate all 0-1 sequences

  x := A_Index-1
  Loop % a0
     t .= (x>>A_Index-1) & 1 ? a%A_Index% "," : ""
  t .= "}`n{"         ; new subsets in new lines

} MsgBox % RegExReplace(SubStr(t,1,StrLen(t)-1),",}","}")</lang>

BBC BASIC

This example is incomplete. It shows no proof of the cases with empty sets. Please ensure that it meets all task requirements and remove this message.

The elements of a set are represented as the bits in an integer (hence the maximum size of set is 32).<lang bbcbasic> DIM list$(3) : list$() = "1", "2", "3", "4"

     PRINT FNpowerset(list$())
     END
     
     DEF FNpowerset(list$())
     IF DIM(list$(),1) > 31 ERROR 100, "Set too large to represent as integer"
     LOCAL i%, j%, s$
     s$ = "{"
     FOR i% = 0 TO (2 << DIM(list$(),1)) - 1
       s$ += "{"
       FOR j% = 0 TO DIM(list$(),1)
         IF i% AND (1 << j%) s$ += list$(j%) + ","
       NEXT
       IF RIGHT$(s$) = "," s$ = LEFT$(s$)
       s$ += "},"
     NEXT i%

= LEFT$(s$) + "}"</lang>

Output:

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

Bracmat

This example is incomplete. It shows no proof of the cases with empty sets. Please ensure that it meets all task requirements and remove this message.

<lang bracmat>( ( powerset

 =   done todo first
   .   !arg:(?done.?todo)
     & (   !todo:%?first ?todo
         & (powerset$(!done !first.!todo),powerset$(!done.!todo))
       | !done
       )
 )

& out$(powerset$(.1 2 3 4))

);</lang>

Output:

1 2 3 4

, 1 2 3 , 1 2 4 , 1 2 , 1 3 4 , 1 3 , 1 4 , 1 , 2 3 4 , 2 3 , 2 4 , 2 , 3 4 , 3 , 4

,

Burlesque

This example is incomplete. It shows no proof of the cases with empty sets. Please ensure that it meets all task requirements and remove this message.

<lang burlesque>blsq ) {1 2 3 4}R@

{{} {1} {2} {1 2} {3} {1 3} {2 3} {1 2 3} {4} {1 4} {2 4} {1 2 4} {3 4} {1 3 4} {2 3 4} {1 2 3 4}}</lang>

C

This example is incomplete. It shows no proof of the cases with empty sets. Please ensure that it meets all task requirements and remove this message.

<lang c>#include <stdio.h>

struct node { char *s; struct node* prev; };

void powerset(char **v, int n, struct node *up) { struct node me;

if (!n) { putchar('['); while (up) { printf(" %s", up->s); up = up->prev; } puts(" ]"); } else { me.s = *v; me.prev = up; powerset(v + 1, n - 1, up); powerset(v + 1, n - 1, &me); } }

int main(int argc, char **argv) { powerset(argv + 1, argc - 1, 0); return 0;

}</lang>

Output:

% ./a.out 1 2 3
[ ]
[ 3 ]
[ 2 ]
[ 3 2 ]
[ 1 ]
[ 3 1 ]
[ 2 1 ]

[ 3 2 1 ]

C++

This example is incomplete. It shows no proof of the cases with empty sets. Please ensure that it meets all task requirements and remove this message.

Non-recursive version

<lang cpp>#include <iostream>

include <set>
include <vector>
include <iterator>
include <algorithm>

typedef std::set<int> set_type; typedef std::set<set_type> powerset_type;

powerset_type powerset(set_type const& set) {

 typedef set_type::const_iterator set_iter;
 typedef std::vector<set_iter> vec;
 typedef vec::iterator vec_iter;

 struct local
 {
   static int dereference(set_iter v) { return *v; }
 };

 powerset_type result;

 vec elements;
 do
 {
   set_type tmp;
   std::transform(elements.begin(), elements.end(),
                  std::inserter(tmp, tmp.end()),
                  local::dereference);
   result.insert(tmp);
   if (!elements.empty() && ++elements.back() == set.end())
   {
     elements.pop_back();
   }
   else
   {
     set_iter iter;
     if (elements.empty())
     {
       iter = set.begin();
     }
     else
     {
       iter = elements.back();
       ++iter;
     }
     for (; iter != set.end(); ++iter)
     {
       elements.push_back(iter);
     }
   }
 } while (!elements.empty());

 return result;

}

int main() {

 int values[4] = { 2, 3, 5, 7 };
 set_type test_set(values, values+4);

 powerset_type test_powerset = powerset(test_set);

 for (powerset_type::iterator iter = test_powerset.begin();
      iter != test_powerset.end();
      ++iter)
 {
   std::cout << "{ ";
   char const* prefix = "";
   for (set_type::iterator iter2 = iter->begin();
        iter2 != iter->end();
        ++iter2)
   {
     std::cout << prefix << *iter2;
     prefix = ", ";
   }
   std::cout << " }\n";
 }

}</lang>

Output:


{  }
{ 2 }
{ 2, 3 }
{ 2, 3, 5 }
{ 2, 3, 5, 7 }
{ 2, 3, 7 }
{ 2, 5 }
{ 2, 5, 7 }
{ 2, 7 }
{ 3 }
{ 3, 5 }
{ 3, 5, 7 }
{ 3, 7 }
{ 5 }
{ 5, 7 }
{ 7 }

Recursive version

<lang cpp>#include <iostream>

include <set>

template<typename Set> std::set<Set> powerset(const Set& s, size_t n) {

   typedef typename Set::const_iterator SetCIt;
   typedef typename std::set<Set>::const_iterator PowerSetCIt;
   std::set<Set> res;
   if(n > 0) {
       std::set<Set> ps = powerset(s, n-1);
       for(PowerSetCIt ss = ps.begin(); ss != ps.end(); ss++)
           for(SetCIt el = s.begin(); el != s.end(); el++) {
               Set subset(*ss);
               subset.insert(*el);
               res.insert(subset);
           }
       res.insert(ps.begin(), ps.end());
   } else
       res.insert(Set());
   return res;

} template<typename Set> std::set<Set> powerset(const Set& s) {

   return powerset(s, s.size());

}</lang>

C#

This example is incomplete. It shows no proof of the cases with empty sets. Please ensure that it meets all task requirements and remove this message.

<lang csharp>public IEnumerable<IEnumerable<T>> GetPowerSet<T>(List<T> list)

{

   return from m in Enumerable.Range(0, 1 << list.Count)
                 select
                     from i in Enumerable.Range(0, list.Count)
                     where (m & (1 << i)) != 0
                     select list[i];

}

public void PowerSetofColors() {

   var colors = new List<KnownColor> { KnownColor.Red, KnownColor.Green, 
       KnownColor.Blue, KnownColor.Yellow };
   
   var result = GetPowerSet(colors);
   
   Console.Write( string.Join( Environment.NewLine, 
       result.Select(subset => 
           string.Join(",", subset.Select(clr => clr.ToString()).ToArray())).ToArray()));

}</lang>

 Red
 Green
 Red,Green
 Blue
 Red,Blue
 Green,Blue
 Red,Green,Blue
 Yellow
 Red,Yellow
 Green,Yellow
 Red,Green,Yellow
 Blue,Yellow
 Red,Blue,Yellow
 Green,Blue,Yellow
 Red,Green,Blue,Yellow

An alternative implementation for an arbitrary number of elements: <lang csharp> public IEnumerable<IEnumerable<T>> GetPowerSet<T>(IEnumerable<T> input) {

   var seed = new List<IEnumerable<T>>() { Enumerable.Empty<T>() }
     as IEnumerable<IEnumerable<T>>;

   return input.Aggregate(seed, (a, b) =>
     a.Concat(a.Select(x => x.Concat(new List<T>() { b }))));
 }</lang>

Clojure

<lang Clojure>(use '[clojure.math.combinatorics :only [subsets] ])

(def S #{1 2 3 4})

user> (subsets S) (() (1) (2) (3) (4) (1 2) (1 3) (1 4) (2 3) (2 4) (3 4) (1 2 3) (1 2 4) (1 3 4) (2 3 4) (1 2 3 4))</lang>

Alternate solution, with no dependency on third-party library: <lang Clojure>(defn powerset [coll]

 (reduce (fn [a x]
           (->> a
                (map #(set (concat #{x} %)))
                (concat a)
                set))
         #{#{}} coll))

(powerset #{1 2 3})</lang>

#{#{} #{1} #{2} #{1 2} #{3} #{1 3} #{2 3} #{1 2 3}}

CoffeeScript

This example is incomplete. It shows no proof of the cases with empty sets. Please ensure that it meets all task requirements and remove this message.

<lang coffeescript>print_power_set = (arr) ->

 console.log "POWER SET of #{arr}"
 for subset in power_set(arr)
   console.log subset

power_set = (arr) ->

 result = []
 binary = (false for elem in arr)
 n = arr.length
 while binary.length <= n
   result.push bin_to_arr binary, arr
   i = 0
   while true
     if binary[i]
       binary[i] = false
       i += 1
     else
       binary[i] = true
       break
   binary[i] = true
 result

bin_to_arr = (binary, arr) ->

 (arr[i] for i of binary when binary[arr.length - i  - 1])

print_power_set [] print_power_set [4, 2, 1]

print_power_set ['dog', 'c', 'b', 'a']</lang>

Output:

> coffee power_set.coffee
POWER SET of 
[]
POWER SET of 4,2,1
[]
[ 1 ]
[ 2 ]
[ 2, 1 ]
[ 4 ]
[ 4, 1 ]
[ 4, 2 ]
[ 4, 2, 1 ]
POWER SET of dog,c,b,a
[]
[ 'a' ]
[ 'b' ]
[ 'b', 'a' ]
[ 'c' ]
[ 'c', 'a' ]
[ 'c', 'b' ]
[ 'c', 'b', 'a' ]
[ 'dog' ]
[ 'dog', 'a' ]
[ 'dog', 'b' ]
[ 'dog', 'b', 'a' ]
[ 'dog', 'c' ]
[ 'dog', 'c', 'a' ]
[ 'dog', 'c', 'b' ]

[ 'dog', 'c', 'b', 'a' ]

ColdFusion

This example is incomplete. It shows no proof of the cases with empty sets. Please ensure that it meets all task requirements and remove this message.

Port from the JavaScript version, compatible with ColdFusion 8+ or Railo 3+<lang javascript>public array function powerset(required array data)

{

 var ps = [""];
 var d = arguments.data;
 var lenData = arrayLen(d);
 var lenPS = 0;
 for (var i=1; i LTE lenData; i++)
 {
   lenPS = arrayLen(ps);
   for (var j = 1; j LTE lenPS; j++)
   {
     arrayAppend(ps, listAppend(ps[j], d[i]));
   }
 }
 return ps;

}

var res = powerset([1,2,3,4]);</lang>

Output:

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

Common Lisp

This example is incomplete. It shows no proof of the cases with empty sets. Please ensure that it meets all task requirements and remove this message.

<lang lisp>(defun power-set (s)

 (reduce #'(lambda (item ps)
             (append (mapcar #'(lambda (e) (cons item e))
                             ps)
                     ps))
         s
         :from-end t

:initial-value '(())))</lang>

Output:

>(power-set '(1 2 3))
((1 2 3) (1 2) (1 3) (1) (2 3) (2) (3) NIL)

Alternate, more recursive (same output): <lang lisp>(defun powerset (l)

 (if (null l)
     (list nil)
     (let ((prev (powerset (cdr l))))

(append (mapcar #'(lambda (elt) (cons (car l) elt)) prev) prev))))</lang> Imperative-style using LOOP: <lang lisp>(defun powerset (xs)

 (loop for i below (expt 2 (length xs)) collect
      (loop for j below i for x in xs if (logbitp j i) collect x)))</lang>

Output:

>(powerset '(1 2 3)
(NIL (1) (2) (1 2) (3) (1 3) (2 3) (1 2 3))

Yet another imperative solution, this time with dolist. <lang lisp>(defun power-set (list)

   (let ((pow-set (list nil)))
     (dolist (element (reverse list) pow-set)
       (dolist (set pow-set)
         (push (cons element set) pow-set)))))</lang>

Output:

>(power-set '(1 2 3))
((1) (1 3) (1 2 3) (1 2) (2) (2 3) (3) NIL)

D

This example is incomplete. It shows no proof of the cases with empty sets. Please ensure that it meets all task requirements and remove this message.

Version using just arrays (it assumes the input to contain distinct items):<lang d>T[][] powerSet(T)(in T[] s) pure nothrow @safe {

   auto r = new typeof(return)(1, 0);
   foreach (e; s) {
       typeof(return) rs;
       foreach (x; r)
           rs ~= x ~ [e];
       r ~= rs;
   }
   return r;

}

void main() {

   import std.stdio;

   [1, 2, 3].powerSet.writeln;

}</lang>

Output:

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

Lazy Version

Compile with -version=power_set2_main to run the main. <lang d>auto powerSet(T)(T[] xs) pure nothrow @safe {

   static struct Result {
       T[] xsLocal, output;
       size_t len;
       size_t bits;

       this(T[] xs_) pure nothrow @safe {
           this.xsLocal = xs_;
           this.output.length = xs_.length;
           this.len = 1U << xs_.length;
       }

       @property empty() const pure nothrow @safe {
           return bits == len;
       }

       void popFront() pure nothrow @safe { bits++; }
       @property save() pure nothrow @safe { return this; }

       T[] front() pure nothrow @safe {
           size_t pos = 0;
           foreach (immutable size_t i; 0 .. xsLocal.length)
               if (bits & (1 << i))
                   output[pos++] = xsLocal[i];
           return output[0 .. pos];
       }
   }

   return Result(xs);

}

version (power_set2_main) {

   void main() {
       import std.stdio;
       [1, 2, 3].powerSet.writeln;
   }

}</lang>Same output.

A set implementation and its power set function.

Déjà Vu

This example is incomplete. It shows no proof of the cases with empty sets. Please ensure that it meets all task requirements and remove this message.

In Déjà Vu, sets are dictionaries with all values true and the default set to false.<lang dejavu>powerset s:

local :out [ set{ } ] for value in keys s: for subset in copy out: local :subset+1 copy subset set-to subset+1 value true push-to out subset+1 out

!. powerset set{ 1 2 3 4 }</lang>

Output:

[ set{ } set{ 4 } set{ 3 4 } set{ 3 } set{ 2 3 } set{ 2 3 4 } set{ 2 4 } set{ 2 } set{ 1 2 } set{ 1 2 4 } set{ 1 2 3 4 } set{ 1 2 3 } set{ 1 3 } set{ 1 3 4 } set{ 1 4 } set{ 1 } ]

E

This example is incomplete. It shows no proof of the cases with empty sets. Please ensure that it meets all task requirements and remove this message.

<lang e>pragma.enable("accumulator")

def powerset(s) {

 return accum [].asSet() for k in 0..!2**s.size() {
   _.with(accum [].asSet() for i ? ((2**i & k) > 0) => elem in s {
     _.with(elem)
   })
 }

}</lang>It would also be possible to define an object which is the powerset of a provided set without actually instantiating all of its members immediately.

Erlang

This example is incomplete. It shows no proof of the cases with empty sets. Please ensure that it meets all task requirements and remove this message.

Generates all subsets of a list with the help of binary:

For [1 2 3]:
    [     ] | 0 0 0 | 0
    [    3] | 0 0 1 | 1
    [  2  ] | 0 1 0 | 2
    [  2 3] | 0 1 1 | 3
    [1    ] | 1 0 0 | 4
    [1   3] | 1 0 1 | 5
    [1 2  ] | 1 1 0 | 6
    [1 2 3] | 1 1 1 | 7
    ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

<lang erlang>powerset(Lst) ->

   N = length(Lst),
   Max = trunc(math:pow(2,N)),
   [[lists:nth(Pos+1,Lst) || Pos <- lists:seq(0,N-1), I band (1 bsl Pos) =/= 0]

|| I <- lists:seq(0,Max-1)].</lang>

Which outputs:

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

Alternate shorter and more efficient version

<lang erlang>powerset([]) -> [[]];

powerset([H|T]) -> PT = powerset(T),

 [ [H|X] || X <- PT ] ++ PT.</lang>or even more efficient version:<lang erlang>powerset([]) -> [[]];

powerset([H|T]) -> PT = powerset(T),

 powerset(H, PT, PT).

powerset(_, [], Acc) -> Acc; powerset(X, [H|T], Acc) -> powerset(X, T, [[X|H]|Acc]).</lang>

F#

This example is incomplete. It shows no proof of the cases with empty sets. Please ensure that it meets all task requirements and remove this message.

almost exact copy of OCaml version

<lang fsharp> let subsets xs = List.foldBack (fun x rest -> rest @ List.map (fun ys -> x::ys) rest) xs [[]]</lang>alternatively with list comprehension<lang fsharp>let rec pow =

   function
   | [] -> [[]]
   | x::xs -> [for i in pow xs do yield! [i;x::i]]</lang>

Factor

This example is incomplete. It shows no proof of the cases with empty sets. Please ensure that it meets all task requirements and remove this message.

We use hash sets, denoted by HS{ } brackets, for our sets. members converts from a set to a sequence, and <hash-set> converts back.<lang factor>USING: kernel prettyprint sequences arrays sets hash-sets ;

IN: powerset

add ( set elt -- newset ) 1array <hash-set> union ;; powerset ( set -- newset ) members { HS{ } } [ dupd [ add ] curry map append ] reduce <hash-set> ;</lang>
Usage: ( scratchpad ) HS{ 1 2 3 4 } powerset .

HS{ HS{ 1 2 3 4 } HS{ 1 2 } HS{ 1 3 } HS{ 2 3 } HS{ 1 2 3 } HS{ 1 4 } HS{ 2 4 } HS{ } HS{ 1 } HS{ 2 } HS{ 3 } HS{ 4 } HS{ 1 2 4 } HS{ 3 4 } HS{ 1 3 4 } HS{ 2 3 4 } }

Forth

This example is incomplete. It shows no proof of the cases with empty sets. Please ensure that it meets all task requirements and remove this message.

Works with: 4tH version 3.61.0

.

Translation of: C

<lang forth>: ?print dup 1 and if over args type space then ;

.set begin dup while ?print >r 1+ r> 1 rshift repeat drop drop ;

.powerset 0 do ." ( " 1 i .set ." )" cr loop ;

check-none dup 2 < abort" Usage: powerset [val] .. [val]" ;

check-size dup /cell 8 [*] >= abort" Set too large" ;

powerset 1 argn check-none check-size 1- lshift .powerset ;

powerset</lang>

Output:

$ 4th cxq powerset.4th 1 2 3 4
( )
( 1 )
( 2 )
( 1 2 )
( 3 )
( 1 3 )
( 2 3 )
( 1 2 3 )
( 4 )
( 1 4 )
( 2 4 )
( 1 2 4 )
( 3 4 )
( 1 3 4 )
( 2 3 4 )

( 1 2 3 4 )

Frink

This example is incomplete. It shows no proof of the cases with empty sets. Please ensure that it meets all task requirements and remove this message.

Frink's set and array classes have built-in subsets[] methods that return all subsets. If called with an array, the results are arrays. If called with a set, the results are sets.<lang frink>

a = new set[1,2,3,4] a.subsets[]</lang>

FunL

FunL uses Scala type scala.collection.immutable.Set as it's set type, which has a built-in method subsets returning an (Scala) iterator over subsets.<lang funl>def powerset( s ) = s.subsets().toSet()</lang> The powerset function could be implemented in FunL directly as:<lang funl>def

 powerset( {} ) = {{}}
 powerset( s ) =
   acc = powerset( s.tail() )
   acc + map( x -> {s.head()} + x, acc )</lang>

or, alternatively as:

<lang funl>import lists.foldr

def powerset( s ) = foldr( \x, acc -> acc + map( a -> {x} + a, acc), {{}}, s )

println( powerset({1, 2, 3, 4}) )</lang>

Output:

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

GAP

This example is incomplete. It shows no proof of the cases with empty sets. Please ensure that it meets all task requirements and remove this message.

<lang gap># Built-in

Combinations([1, 2, 3]);

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

Note that it handles duplicates

Combinations([1, 2, 3, 1]);

[ [ ], [ 1 ], [ 1, 1 ], [ 1, 1, 2 ], [ 1, 1, 2, 3 ], [ 1, 1, 3 ], [ 1, 2 ], [ 1, 2, 3 ], [ 1, 3 ],
[ 2 ], [ 2, 3 ], [ 3 ] ]</lang>

Go

This example is incomplete. It shows no proof of the cases with empty sets. Please ensure that it meets all task requirements and remove this message.

No native set type in Go. While the associative array trick mentioned in the task description works well in Go in most situations, it does not work here because we need sets of sets, and converting a general set to a hashable value for a map key is non-trivial.

Instead, this solution uses a simple (non-associative) slice as a set representation. To ensure uniqueness, the element interface requires an equality method, which is used by the set add method. Adding elements with the add method ensures the uniqueness property.

The power set method implemented here does not need the add method though. The algorithm ensures that the result will be a valid set as long as the input is a valid set. This allows the more efficient append function to be used. <lang go>package main

import (

   "bytes"
   "fmt"
   "strconv"

)

// types needed to implement general purpose sets are element and set

// element is an interface, allowing different kinds of elements to be // implemented and stored in sets. type elem interface {

   // an element must be distinguishable from other elements to satisfy
   // the mathematical definition of a set.  a.eq(b) must give the same
   // result as b.eq(a).
   Eq(element) bool
   // String result is used only for printable output.  Given a, b where
   // a.eq(b), it is not required that a.String() == b.String().
   fmt.Stringer

}

// integer type satisfying element interface type Int int

func (i Int) Eq(e elem) bool {

   j, ok := e.(Int)    
   return ok && i == j

}

func (i Int) String() string {

   return strconv.Itoa(int(i))

}

// a set is a slice of elem's. methods are added to implement // the element interface, to allow nesting. type set []elem

// uniqueness of elements can be ensured by using add method func (s *set) add(e elem) {

   if !s.has(e) {
       *s = append(*s, e)
   }

}

func (s *set) has(e elem) bool {

   for _, ex := range *s {
       if e.Eq(ex) {
           return true
       }
   }
   return false

}

// elem.Eq func (s set) Eq(e elem) bool {

   t, ok := e.(set)
   if !ok {
       return false
   }
   if len(s) != len(t) {
       return false
   }
   for _, se := range s {
       if !t.has(se) {
           return false
       }
   }
   return true

}

// elem.String func (s set) String() string {

   var buf bytes.Buffer
   buf.WriteRune('{')
   for _, e := range s {
       if len(r) > 1 {
           buf.WriteRune(' ')
       }
       buf.WriteString(e.String())
   }
   buf.WriteRune('}')
   return buf.String()

}

// method required for task func (s set) powerSet() set {

   r := set{set{}}
   for _, es := range s {
       var u set
       for _, er := range r {
           u = append(u, append(er.(set), es))
       }
       r = append(r, u...)
   }
   return r

}

func main() {

   var s set
   for _, i := range []Int{1, 2, 2, 3, 4, 4, 4} {
       s.add(i)
   }
   fmt.Println(s)
   fmt.Println("length =", len(s))
   ps := s.powerSet()
   fmt.Println(ps)
   fmt.Println("length =", len(ps))

}</lang>

Output:

{1 2 3 4}
length = 4
{{} {1} {2} {1 2} {3} {1 3} {2 3} {1 2 3} {4} {1 4} {2 4} {1 2 4} {3 4} {1 3 4} {2 3 4} {1 2 3 4}}

length = 16

Groovy

This example is incomplete. It shows no proof of the cases with empty sets. Please ensure that it meets all task requirements and remove this message.

Builds on the Combinations solution. Sets are not a "natural" collection type in Groovy. Lists are much more richly supported. Thus, this solution is liberally sprinkled with coercion from Set to List and from List to Set.<lang groovy>def comb

comb = { m, List list ->

   def n = list.size()
   m == 0 ?
       [[]] :
       (0..(n-m)).inject([]) { newlist, k ->
           def sublist = (k+1 == n) ? [] : list[(k+1)..<n] 
           newlist += comb(m-1, sublist).collect { [list[k]] + it }
       }

}

def powerSet = { set ->

   (0..(set.size())).inject([]){ list, i ->  list + comb(i,set as List)}.collect { it as LinkedHashSet } as LinkedHashSet

}</lang>Test program:<lang groovy>def vocalists = [ "C", "S", "N", "Y" ] as LinkedHashSet println "${vocalists}" println powerSet(vocalists)</lang>

Output:

[C, S, N, Y]
[[], [C], [S], [N], [Y], [C, S], [C, N], [C, Y], [S, N], [S, Y], [N, Y], [C, S, N], [C, S, Y], [C, N, Y], [S, N, Y], [C, S, N, Y]]

Note: In this example, LinkedHashSet was used throughout for Set coercion. This is because LinkedHashSet preserves the order of input, like a List. However, if order does not matter you could replace all references to LinkedHashSet with Set.

Haskell

This example is incomplete. It shows no proof of the cases with empty sets. Please ensure that it meets all task requirements and remove this message.

<lang Haskell>import Data.Set

import Control.Monad

powerset :: Ord a => Set a -> Set (Set a) powerset = fromList . fmap fromList . listPowerset . toList

listPowerset :: [a] -> a listPowerset = filterM (const [True, False])</lang>listPowerset describes the result as all possible (using the list monad) filterings (using filterM) of the input list, regardless (using const) of each item's value. powerset simply converts the input and output from lists to sets.

Alternate Solution <lang Haskell>powerset [] = [[]] powerset (head:tail) = acc ++ map (head:) acc where acc = powerset tail</lang> or <lang Haskell>powerset = foldr (\x acc -> acc ++ map (x:) acc) [[]]</lang> Examples:

*Main> listPowerset [1,2,3]
[[1,2,3],[1,2],[1,3],[1],[2,3],[2],[3],[]]
*Main> powerset (Data.Set.fromList [1,2,3])
{{},{1},{1,2},{1,2,3},{1,3},{2},{2,3},{3}}

Works with: GHC version 6.10

Prelude> import Data.List
Prelude Data.List> subsequences [1,2,3]
[[],[1],[2],[1,2],[3],[1,3],[2,3],[1,2,3]]

Alternate solution

A method using only set operations and set mapping is also possible. Ideally, Set would be defined as a Monad, but that's impossible given the constraint that the type of inputs to Set.map (and a few other functions) be ordered.<lang Haskell>import qualified Data.Set as Set type Set=Set.Set unionAll :: (Ord a) => Set (Set a) -> Set a unionAll = Set.fold Set.union Set.empty

--slift is the analogue of liftA2 for sets. slift :: (Ord a, Ord b, Ord c) => (a->b->c) -> Set a -> Set b -> Set c slift f s0 s1 = unionAll (Set.map (\e->Set.map (f e) s1) s0)

--a -> {{},{a}} makeSet :: (Ord a) => a -> Set (Set a) makeSet = (Set.insert Set.empty) . Set.singleton.Set.singleton

powerSet :: (Ord a) => Set a -> Set (Set a) powerSet = (Set.fold (slift Set.union) (Set.singleton Set.empty)) . Set.map makeSet</lang>

Usage:

Prelude Data.Set> powerSet fromList [1,2,3]
fromList [fromList [], fromList [1], fromList [1,2], fromList [1,2,3], fromList [1,3], fromList [2], fromList [2,3], fromList [3]]

Icon and Unicon

This example is incomplete. It shows no proof of the cases with empty sets. Please ensure that it meets all task requirements and remove this message.

The two examples below show the similarities and differences between constructing an explicit representation of the solution, i.e. a set containing the powerset, and one using generators. The basic recursive algorithm is the same in each case, but wherever the first stores part of the result away, the second uses 'suspend' to immediately pass the result back to the caller. The caller may then decide to store the results in a set, a list, or dispose of each one as it appears.

Set building

The following version returns a set containing the powerset:<lang Icon>procedure power_set (s)

 result := set ()
 if *s = 0 
   then insert (result, set ()) # empty set
   else {
     head := set(?s) # take a random element
     # and find powerset of remaining part of set
     tail_pset := power_set (x -- head)
     result ++:= tail_pset # add powerset of remainder to results
     every ps := !tail_pset do # and add head to each powerset from the remainder
       insert (result, ps ++ head)
   }
 return result

end</lang>To test the above procedure:<lang Icon>procedure main ()

 every s := !power_set (set(1,2,3,4)) do { # requires '!' to generate items in the result set
   writes ("[ ")
   every writes (!s || " ")
   write ("]")
 }

end</lang>

Output:

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

Generator

An alternative version, which generates each item in the power set in turn:<lang Icon>procedure power_set (s)

 if *s = 0 
   then suspend set ()
   else {
     head := set(?s)
     every ps := power_set (s -- head) do {
       suspend ps
       suspend ps ++ head
     }
   }

end

procedure main ()

 every s := power_set (set(1,2,3,4)) do { # power_set's values are generated by 'every'
   writes ("[ ")
   every writes (!s || " ")
   write ("]")
 }

end</lang>

J

This example is incomplete. It shows no proof of the cases with empty sets. Please ensure that it meets all task requirements and remove this message.

There are a number of ways to generate a power set in J. Here's one:

<lang j>ps =: #~ 2 #:@i.@^ #</lang> For example: <lang j> ps 'ACE'

E C CE A AE AC ACE</lang>In the typical use, this operation makes sense on collections of unique elements.<lang J> ~.1 2 3 2 1 1 2 3

  #ps 1 2 3 2 1

32

  #ps ~.1 2 3 2 1

8</lang>In other words, the power set of a 5 element set has 32 sets where the power set of a 3 element set has 8 sets. Thus if elements of the original "set" were not unique then sets of the power "set" will also not be unique sets.

Java

This example is incomplete. It shows no proof of the cases with empty sets. Please ensure that it meets all task requirements and remove this message.

Works with: Java version 1.5+

Recursion

This implementation sorts each subset, but not the whole list of subsets (which would require a custom comparator). It also destroys the original set.<lang java5>public static ArrayList<String> getpowerset(int a[],int n,ArrayList<String> ps)

   {
       if(n<0)
       {
           return null;
       }
       if(n==0)
       {
           if(ps==null)
               ps=new ArrayList<String>();
           ps.add(" ");
           return ps;
       }
       ps=getpowerset(a, n-1, ps);
       ArrayList<String> tmp=new ArrayList<String>();
       for(String s:ps)
       {
           if(s.equals(" "))
               tmp.add(""+a[n-1]);
           else
               tmp.add(s+a[n-1]);
       }
       ps.addAll(tmp);
       return ps;
   }</lang>

Iterative

The iterative implementation of the above idea. Each subset is in the order that the element appears in the input list. This implementation preserves the input.<lang java5>public static <T> List<List<T>> powerset(Collection<T> list) {

 List<List<T>> ps = new ArrayList<List<T>>();
 ps.add(new ArrayList<T>());   // add the empty set

 // for every item in the original list
 for (T item : list) {
   List<List<T>> newPs = new ArrayList<List<T>>();

   for (List<T> subset : ps) {
     // copy all of the current powerset's subsets
     newPs.add(subset);

     // plus the subsets appended with the current item
     List<T> newSubset = new ArrayList<T>(subset);
     newSubset.add(item);
     newPs.add(newSubset);
   }

   // powerset is now powerset of list.subList(0, list.indexOf(item)+1)
   ps = newPs;
 }
 return ps;

}</lang>

Binary String

This implementation works on idea that each element in the original set can either be in the power set or not in it. With n elements in the original set, each combination can be represented by a binary string of length n. To get all possible combinations, all you need is a counter from 0 to 2ⁿ - 1. If the k^th bit in the binary string is 1, the k^th element of the original set is in this combination.<lang java5>public static <T extends Comparable<? super T>> LinkedList<LinkedList<T>> BinPowSet( LinkedList<T> A){ LinkedList<LinkedList<T>> ans= new LinkedList<LinkedList<T>>(); int ansSize = (int)Math.pow(2, A.size()); for(int i= 0;i< ansSize;++i){ String bin= Integer.toBinaryString(i); //convert to binary while(bin.length() < A.size()) bin = "0" + bin; //pad with 0's LinkedList<T> thisComb = new LinkedList<T>(); //place to put one combination for(int j= 0;j< A.size();++j){ if(bin.charAt(j) == '1')thisComb.add(A.get(j)); } Collections.sort(thisComb); //sort it for easy checking ans.add(thisComb); //put this set in the answer list } return ans; }</lang>

JavaScript

This example is incomplete. It shows no proof of the cases with empty sets. Please ensure that it meets all task requirements and remove this message.

Uses a JSON stringifier from http://www.json.org/js.html

Works with: SpiderMonkey

<lang javascript>function powerset(ary) {

   var ps = [[]];
   for (var i=0; i < ary.length; i++) {
       for (var j = 0, len = ps.length; j < len; j++) {
           ps.push(ps[j].concat(ary[i]));
       }
   }
   return ps;

}

var res = powerset([1,2,3,4]);

load('json2.js');

print(JSON.stringify(res));</lang>

Output:

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

Julia

This example is incomplete. It shows no proof of the cases with empty sets. Please ensure that it meets all task requirements and remove this message.

<lang julia>function powerset (x)

 result = {{}}
 for i in x, j = 1:length(result)
   push!(result, [result[j],i])
 end
 result

end</lang>

Output:

julia> show(powerset({1,2,3}))
{{},{1},{2},{1,2},{3},{1,3},{2,3},{1,2,3}}

K

This example is incomplete. It shows no proof of the cases with empty sets. Please ensure that it meets all task requirements and remove this message.

  ps:{x@&:'+2_vs!_2^#x}

</lang>Usage:

   ps "ABC"
(""
 ,"C"
 ,"B"
 "BC"
 ,"A"
 "AC"
 "AB"
 "ABC")

Logo

This example is incomplete. It shows no proof of the cases with empty sets. Please ensure that it meets all task requirements and remove this message.

<lang logo>to powerset :set

 if empty? :set [output [[]]]
 localmake "rest powerset butfirst :set
 output sentence  map [sentence first :set ?] :rest  :rest

end

show powerset [1 2 3] [[1 2 3] [1 2] [1 3] [1] [2 3] [2] [3] []]</lang>

Logtalk

This example is incomplete. It shows no proof of the cases with empty sets. Please ensure that it meets all task requirements and remove this message.

<lang logtalk>:- object(set).

   :- public(powerset/2).

   powerset(Set, PowerSet) :-
       reverse(Set, RSet),
       powerset_1(RSet, [[]], PowerSet).

   powerset_1([], PowerSet, PowerSet).
   powerset_1([X| Xs], Yss0, Yss) :-
       powerset_2(Yss0, X, Yss1),
       powerset_1(Xs, Yss1, Yss).

   powerset_2([], _, []).
   powerset_2([Zs| Zss], X, [Zs, [X| Zs]| Yss]) :-
       powerset_2(Zss, X, Yss).

   reverse(List, Reversed) :-
       reverse(List, [], Reversed).

   reverse([], Reversed, Reversed).
   reverse([Head| Tail], List, Reversed) :-
       reverse(Tail, [Head| List], Reversed).

- end_object.</lang>

Usage example: <lang logtalk>| ?- set::powerset([1, 2, 3, 4], PowerSet).

PowerSet = [[],[1],[2],[1,2],[3],[1,3],[2,3],[1,2,3],[4],[1,4],[2,4],[1,2,4],[3,4],[1,3,4],[2,3,4],[1,2,3,4]] yes</lang>

Lua

This example is incomplete. It shows no proof of the cases with empty sets. Please ensure that it meets all task requirements and remove this message.

--returns the powerset of s, out of order. function powerset(s, start)

 start = start or 1
 if(start > #s) then return {{}} end
 local ret = powerset(s, start + 1)
 for i = 1, #ret do
   ret[#ret + 1] = {s[start], unpack(ret[i])}
 end
 return ret

end

--non-recurse implementation function powerset(s)

  local t = {{}}
  for i = 1, #s do
     for j = 1, #t do
        t[#t+1] = {s[i],unpack(t[j])}
     end
  end
  return t

end

--alternative, copied from the Python implementation function powerset2(s)

 local ret = {{}}
 for i = 1, #s do
   local k = #ret
   for j = 1, k do
     ret[k + j] = {s[i], unpack(ret[j])}
   end
 end
 return ret

end </lang>

M4

This example is incomplete. It shows no proof of the cases with empty sets. Please ensure that it meets all task requirements and remove this message.

<lang M4>define(`for',

 `ifelse($#, 0, ``$0,
         eval($2 <= $3), 1,
         `pushdef(`$1', `$2')$4`'popdef(
            `$1')$0(`$1', incr($2), $3, `$4')')')dnl

define(`nth',

 `ifelse($1, 1, $2,
         `nth(decr($1), shift(shift($@)))')')dnl

define(`range',

 `for(`x', eval($1 + 2), eval($2 + 2),
      `nth(x, $@)`'ifelse(x, eval($2+2), `', `,')')')dnl

define(`powerpart',

 `{range(2, incr($1), $@)}`'ifelse(incr($1), $#, `',
    `for(`x', eval($1+2), $#,
       `,powerpart(incr($1), ifelse(
          eval(2 <= ($1 + 1)), 1,
          `range(2,incr($1), $@), ')`'nth(x, $@)`'ifelse(
             eval((x + 1) <= $#),1,`,range(incr(x), $#, $@)'))')')')dnl

define(`powerset',

 `{powerpart(0, substr(`$1', 1, eval(len(`$1') - 2)))}')dnl

dnl powerset(`{a,b,c}')</lang>

{{},{a},{a,b},{a,b,c},{a,c},{b},{b,c},{c}}

Maple

This example is incomplete. It shows no proof of the cases with empty sets. Please ensure that it meets all task requirements and remove this message.

<lang Maple>with(combinat): powerset({1,2,3,4});</lang>

Output:

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

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

Mathematica

This example is incomplete. It shows no proof of the cases with empty sets. Please ensure that it meets all task requirements and remove this message.

Built-in function that either gives all possible subsets, subsets with at most n elements, subsets with exactly n elements or subsets containing between n and m elements. Example of all subsets:

<lang Mathematica>Subsets[{a, b, c}]</lang>gives:

{{}, {a}, {b}, {c}, {a, b}, {a, c}, {b, c}, {a, b, c}}

MATLAB

Sets are not an explicit data type in MATLAB, but cell arrays can be used for the same purpose. In fact, cell arrays have the benefit of containing any kind of data structure. So, this powerset function will work on a set of any type of data structure, without the need to overload any operators.<lang MATLAB>function pset = powerset(theSet)

   pset = cell(size(theSet)); %Preallocate memory

   %Generate all numbers from 0 to 2^(num elements of the set)-1
   for i = ( 0:(2^numel(theSet))-1 )
      
       %Convert i into binary, convert each digit in binary to a boolean
       %and store that array of booleans
       indicies = logical(bitget( i,(1:numel(theSet)) )); 
       
       %Use the array of booleans to extract the members of the original
       %set, and store the set containing these members in the powerset
       pset(i+1) = {theSet(indicies)};
      
   end

end</lang>

Sample Usage: Powerset of the set of the empty set.<lang MATLAB>powerset({{}})

ans =

    {}    {1x1 cell} %This is the same as { {},{{}} }</lang>

Powerset of { {1,2},3 }. <lang MATLAB>powerset({{1,2},3})

ans =

   {1x0 cell}    {1x1 cell}    {1x1 cell}    {1x2 cell} %This is the same as { {},Template:1,2,{3},{{1,2},3} }</lang>

Maxima

This example is incomplete. It shows no proof of the cases with empty sets. Please ensure that it meets all task requirements and remove this message.

<lang maxima>powerset({1, 2, 3, 4});

/* {{}, {1}, {1, 2}, {1, 2, 3}, {1, 2, 3, 4}, {1, 2, 4}, {1, 3}, {1, 3, 4},

  {1, 4}, {2}, {2, 3}, {2, 3, 4}, {2, 4}, {3}, {3, 4}, {4}} */</lang>

Nimrod

This example is incomplete. It shows no proof of the cases with empty sets. Please ensure that it meets all task requirements and remove this message.

<lang nimrod>import sets, hashes

proc hash(x): THash =

 var h = 0
 for i in x: h = h !& hash(i)
 result = !$h

proc powerset[T](inset: TSet[T]): auto =

 result = toSet([initSet[T]()])

 for i in inset:
   var tmp = result
   for j in result:
     var k = j
     k.incl(i)
     tmp.incl(k)
   result = tmp

echo powerset(toSet([1,2,3,4]))</lang>

Objective-C

This example is incomplete. It shows no proof of the cases with empty sets. Please ensure that it meets all task requirements and remove this message.

<lang objc>#import <Foundation/Foundation.h>

+ (NSArray *)powerSetForArray:(NSArray *)array { UInt32 subsetCount = 1 << array.count; NSMutableArray *subsets = [NSMutableArray arrayWithCapacity:subsetCount]; for(int subsetIndex = 0; subsetIndex < subsetCount; subsetIndex++) { NSMutableArray *subset = [[NSMutableArray alloc] init]; for (int itemIndex = 0; itemIndex < array.count; itemIndex++) { if((subsetIndex >> itemIndex) & 0x1) { [subset addObject:array[itemIndex]]; } } [subsets addObject:subset]; } return subsets; }</lang>

OCaml

This example is incomplete. It shows no proof of the cases with empty sets. Please ensure that it meets all task requirements and remove this message.

The standard library already implements a proper Set datatype. As the base type is unspecified, the powerset must be parameterized as a module. Also, the library is lacking a map operation, which we have to implement first.<lang ocaml>module PowerSet(S: Set.S) =

struct

 include Set.Make (S)

 let map f s =
   let work x r = add (f x) r in
   fold work s empty
 ;;

 let powerset s = 
   let base = singleton (S.empty) in
   let work x r = union r (map (S.add x) r) in 
   S.fold work s base
 ;;

end;; (* PowerSet *)</lang>version for lists:<lang ocaml>let subsets xs = List.fold_right (fun x rest -> rest @ List.map (fun ys -> x::ys) rest) xs [[]]</lang>

OPL

This example is incomplete. It shows no proof of the cases with empty sets. Please ensure that it meets all task requirements and remove this message.

<lang OPL>{string} s={"A","B","C","D"};

range r=1.. ftoi(pow(2,card(s))); {string} s2 [k in r] = {i | i in s: ((k div (ftoi(pow(2,(ord(s,i))))) mod 2) == 1)};

execute {

writeln(s2);

}</lang>

which gives:

[{} {"A"} {"B"} {"A" "B"} {"C"} {"A" "C"} {"B" "C"} {"A" "B" "C"} {"D"} {"A"
        "D"} {"B" "D"} {"A" "B" "D"} {"C" "D"} {"A" "C" "D"} {"B" "C" "D"}
        {"A" "B" "C" "D"}]

Oz

This example is incomplete. It shows no proof of the cases with empty sets. Please ensure that it meets all task requirements and remove this message.

Oz has a library for finite set constraints. Creating a power set is a trivial application of that:<lang oz>declare

 %% Given a set as a list, returns its powerset (again as a list)
 fun {Powerset Set}
    proc {Describe Root}
       %% Describe sets by lower bound (nil) and upper bound (Set)
       Root = {FS.var.bounds nil Set}
       %% enumerate all possible sets
       {FS.distribute naive [Root]}
    end
    AllSets = {SearchAll Describe}
 in
    %% convert to list representation
    {Map AllSets FS.reflect.lowerBoundList}
 end

in

 {Inspect {Powerset [1 2 3 4]}}</lang>A more convential implementation without finite set constaints:

<lang oz>fun {Powerset2 Set}

  case Set of nil then [nil]
  [] H|T thens
     Acc = {Powerset2 T}
  in
     {Append Acc {Map Acc fun {$ A} H|A end}}
  end

end</lang>

PARI/GP

This example is incomplete. It shows no proof of the cases with empty sets. Please ensure that it meets all task requirements and remove this message.

<lang parigp>vector(1<<#S,i,vecextract(S,i-1))</lang>

Perl

This example is incomplete. It shows no proof of the cases with empty sets. Please ensure that it meets all task requirements and remove this message.

Perl does not have a built-in set data-type. However, you can...

Use a third-party module

The CPAN module Set::Object provides a set implementation for sets of arbitrary objects, for which a powerset function could be defined and used like so:
<lang perl>use Set::Object qw(set);
sub powerset {
```
   my $p = Set::Object->new( set() );
   foreach my $i (shift->elements) {
       $p->insert( map { set($_->elements, $i) } $p->elements );
   }
   return $p;
```
}
my $set = set(1, 2, 3); my $powerset = powerset($set);
print $powerset->as_string, "\n";</lang>
Output:
```
Set::Object(Set::Object() Set::Object(1 2 3) Set::Object(1 2) Set::Object(1 3) Set::Object(1) Set::Object(2 3) Set::Object(2) Set::Object(3))
```
Use a simple custom hash-based set type

It's also easy to define a custom type for sets of strings or numbers, using a hash as the underlying representation (like the task description suggests):<lang perl>package Set {
```
   sub new       { bless { map {$_ => undef} @_[1..$#_] }, shift; }
   sub elements  { sort keys %{shift()} }
   sub as_string { 'Set(' . join(' ', sort keys %{shift()}) . ')' }
   # ...more set methods could be defined here...
```
}</lang>(Note: For a ready-to-use module that uses this approach, and comes with all the standard set methods that you would expect, see the CPAN module Set::Tiny)
The limitation of this approach is that only primitive strings/numbers are allowed as hash keys in Perl, so a Set of Set's cannot be represented, and the return value of our powerset function will thus have to be a list of sets rather than being a Set object itself.
We could implement the function as an imperative foreach loop similar to the Set::Object based solution above, but using list folding (with the help of Perl's List::Util core module) seems a little more elegant in this case:<lang perl>use List::Util qw(reduce);
sub powerset {
```
   @{( reduce { [@$a, map { Set->new($_->elements, $b) } @$a ] }
              [Set->new()], shift->elements )};
```
}
my $set = Set->new(1, 2, 3); my @subsets = powerset($set);
print $_->as_string, "\n" for @subsets;</lang>

Output:
```
Set()
Set(1)
Set(2)
Set(1 2)
Set(3)
Set(1 3)
Set(2 3)

Set(1 2 3)
```
Use arrays
If you don't actually need a proper set data-type that guarantees uniqueness of its elements, the simplest approach is to use arrays to store "sets" of items, in which case the implementation of the powerset function becomes quite short.
Recursive solution:<lang perl>sub powerset {
```
   @_ ? map { $_, [$_[0], @$_] } powerset(@_[1..$#_]) : [];
```
}</lang>List folding solution:<lang perl>use List::Util qw(reduce);
sub powerset {
```
   @{( reduce { [@$a, map([@$_, $b], @$a)] } [[]], @_ )}
```
}</lang>

Usage & output:
<lang perl>my @set = (1, 2, 3);
my @powerset = powerset(@set);
sub set_to_string {
```
   "{" . join(", ", map { ref $_ ? set_to_string(@$_) : $_ } @_) . "}"
```
}
print set_to_string(@powerset), "\n";</lang>
{{}, {1}, {2}, {1, 2}, {3}, {1, 3}, {2, 3}, {1, 2, 3}}

Perl 6

This example is incomplete. It shows no proof of the cases with empty sets. Please ensure that it meets all task requirements and remove this message.

Works with: rakudo version 2014-02-25

<lang perl6>sub powerset(Set $s) { $s.combinations.map(*.Set).Set } say powerset set <a b c d>;</lang>

Output:

set(set(), set(a), set(b), set(c), set(d), set(a, b), set(a, c), set(a, d), set(b, c), set(b, d), set(c, d), set(a, b, c), set(a, b, d), set(a, c, d), set(b, c, d), set(a, b, c, d))

If you don't care about the actual Set type, the .combinations method by itself may be good enough for you:

<lang perl6>.say for <a b c d>.combinations</lang>

Output:

a b c d a b a c a d b c b d c d a b c a b d a c d b c d

a b c d

PHP

This example is incomplete. It shows no proof of the cases with empty sets. Please ensure that it meets all task requirements and remove this message.

<lang PHP><?php function get_subset($binary, $arr) {

 // based on true/false values in $binary array, include/exclude
 // values from $arr
 $subset = array();
 foreach (range(0, count($arr)-1) as $i) {
   if ($binary[$i]) {
     $subset[] = $arr[count($arr) - $i - 1];
   } 
 }
 return $subset;

}

function print_array($arr) {

 if (count($arr) > 0) {
   echo join(" ", $arr);
 } else {
   echo "(empty)";
 }
 echo '
';

}

function print_power_sets($arr) {

 echo "POWER SET of [" . join(", ", $arr) . "]
";
 foreach (power_set($arr) as $subset) {
   print_array($subset);
 }

}

function power_set($arr) {

 $binary = array();
 foreach (range(1, count($arr)) as $i) {
   $binary[] = false;
 }
 $n = count($arr);
 $powerset = array();
 
 while (count($binary) <= count($arr)) {
   $powerset[] = get_subset($binary, $arr);
   $i = 0;
   while (true) {
     if ($binary[$i]) {
       $binary[$i] = false;
       $i += 1;
     } else {
       $binary[$i] = true;
       break;
     }
   }
   $binary[$i] = true;
 }
 
 return $powerset;

}

print_power_sets(array()); print_power_sets(array('singleton')); print_power_sets(array('dog', 'c', 'b', 'a'));

?></lang>

Output in browser:

POWER SET of []
POWER SET of [singleton]
(empty)
singleton
POWER SET of [dog, c, b, a]
(empty)
a
b
a b
c
a c
b c
a b c
dog
a dog
b dog
a b dog
c dog
a c dog
b c dog

a b c dog

PicoLisp

This example is incomplete. It shows no proof of the cases with empty sets. Please ensure that it meets all task requirements and remove this message.

<lang PicoLisp>(de powerset (Lst)

  (ifn Lst
     (cons)
     (let L (powerset (cdr Lst))
        (conc
           (mapcar '((X) (cons (car Lst) X)) L)
           L ) ) ) )</lang>

PL/I

This example is incomplete. It shows no proof of the cases with empty sets. Please ensure that it meets all task requirements and remove this message.

<lang pli>*process source attributes xref or(!);

/*--------------------------------------------------------------------
* 06.01.2014 Walter Pachl  translated from REXX
*-------------------------------------------------------------------*/
powerset: Proc Options(main);
Dcl (hbound,index,left,substr) Builtin;
Dcl sysprint Print;
Dcl s(4) Char(5) Var Init('one','two','three','four');
Dcl ps   Char(1000) Var;
Dcl (n,chunk,p) Bin Fixed(31);
n=hbound(s);                      /* number of items in the list.   */
ps='{} ';                         /* start with a null power set.   */
Do chunk=1 To n;                  /* loop through the ...     .     */
  ps=ps!!combn(chunk);            /* a CHUNK at a time.             */
  End;
Do While(ps>);
  p=index(ps,' ');
  Put Edit(left(ps,p-1))(Skip,a);
  ps=substr(ps,p+1);
  End;

combn: Proc(y) Returns(Char(1000) Var);
/*--------------------------------------------------------------------
* returns the list of subsets with y elements of set s
*-------------------------------------------------------------------*/
Dcl (y,base,bbase,ym,p,j,d,u) Bin Fixed(31);
Dcl (z,l) Char(1000) Var Init();
Dcl a(20) Bin Fixed(31) Init((20)0);
Dcl i Bin Fixed(31);
base=hbound(s)+1;
bbase=base-y;
ym=y-1;
Do p=1 To y;
  a(p)=p;
  End;
Do j=1 By 1;
  l=;
  Do d=1 To y;
    u=a(d);
    l=l!!','!!s(u);
    End;
  z=z!!'{'!!substr(l,2)!!'} ';
  a(y)=a(y)+1;
  If a(y)=base Then
    If combu(ym) Then
      Leave;
  End;
/* Put Edit('combn',y,z)(Skip,a,f(2),x(1),a); */
Return(z);

combu: Proc(d) Recursive Returns(Bin Fixed(31));
Dcl (d,u) Bin Fixed(31);
If d=0 Then
  Return(1);
p=a(d);
Do u=d To y;
  a(u)=p+1;
  If a(u)=bbase+u Then
    Return(combu(u-1));
  p=a(u);
  End;
Return(0);
End;
End;

End;</lang>

Output:

{}
{one}
{two}
{three}
{four}
{one,two}
{one,three}
{one,four}
{two,three}
{two,four}
{three,four}
{one,two,three}
{one,two,four}
{one,three,four}
{two,three,four}

{one,two,three,four}

Prolog

This example is incomplete. It shows no proof of the cases with empty sets. Please ensure that it meets all task requirements and remove this message.

Logical (cut-free) Definition

The predicate powerset(X,Y) defined here can be read as "Y is the powerset of X", it being understood that lists are used to represent sets.

The predicate subseq(X,Y) is true if and only if the list X is a subsequence of the list Y.

The definitions here are elementary, logical (cut-free), and efficient (within the class of comparably generic implementations).

<lang Prolog>powerset(X,Y) :- bagof( S, subseq(S,X), Y).

subseq( [], []). subseq( [], [_|_]). subseq( [X|Xs], [X|Ys] ) :- subseq(Xs, Ys).

subseq( [X|Xs], [_|Ys] ) :- append(_, [X|Zs], Ys), subseq(Xs, Zs).</lang>

Output:

?- powerset([1,2,3], X).
X = [[], [1], [1, 2], [1, 2, 3], [1, 3], [2], [2, 3], [3]].
% Symbolic:
?- powerset( [X,Y], S).
S = [[], [X], [X, Y], [Y]].
% In reverse:
?- powerset( [X,Y], [[], [1], [1, 2], [2]] ).
X = 1,

Y = 2.

Single-Functor Definition

<lang Prolog>power_set( [], [[]]). power_set( [X|Xs], PS) :-

 power_set(Xs, PS1),
 maplist( append([X]), PS1, PS2 ), % i.e. prepend X to each PS1
 append(PS1, PS2, PS).</lang>

Output:

?- power_set([1,2,3,4,5,6,7,8], X), length(X,N), writeln(N).
256

Constraint Handling Rules

Library: chr

CHR is a programming language created by Professor Thom Frühwirth. Works with SWI-Prolog and module chr written by Tom Schrijvers and Jan Wielemaker.<lang Prolog>:- use_module(library(chr)).

- chr_constraint chr_power_set/2, chr_power_set/1, clean/0.

clean @ clean \ chr_power_set(_) <=> true. clean @ clean <=> true.

only_one @ chr_power_set(A) \ chr_power_set(A) <=> true.

creation @ chr_power_set([H | T], A) <=>

          append(A, [H], B),

chr_power_set(T, A),

          chr_power_set(T, B),

chr_power_set(B).

empty_element @ chr_power_set([], _) <=> chr_power_set([]).</lang>

Example of output:

 ?- chr_power_set([1,2,3,4], []), findall(L, find_chr_constraint(chr_power_set(L)), LL), clean.
LL = [[1],[1,2],[1,2,3],[1,2,3,4],[1,2,4],[1,3],[1,3,4],[1,4],[2],[2,3],[2,3,4],[2,4],[3],[3,4],[4],[]] .

PureBasic

This example is incomplete. It shows no proof of the cases with empty sets. Please ensure that it meets all task requirements and remove this message.

This code is for console mode.<lang PureBasic>If OpenConsole()

 Define argc=CountProgramParameters()
 If argc>=(SizeOf(Integer)*8) Or argc<1
   PrintN("Set out of range.")
   End 1
 Else
   Define i, j, text$
   Define.q bset=1<<argc
   Print("{")
   For i=0 To bset-1   ; check all binary combinations
     If Not i: text$=  "{"
     Else    : text$=", {"
     EndIf
     k=0
     For j=0 To argc-1  ; step through each bit   
       If i&(1<<j)
         If k: text$+", ": EndIf         ; pad the output 
         text$+ProgramParameter(j): k+1  ; append each matching bit 
       EndIf
     Next j
     Print(text$+"}")
   Next i
   PrintN("}")
 EndIf

EndIf</lang>

Sample output:

C:\Users\PureBasic_User\Desktop>"Power Set.exe" 1 2 3 4
{{}, {1}, {2}, {1, 2}, {3}, {1, 3}, {2, 3}, {1, 2, 3}, {4}, {1, 4},
{2, 4}, {1, 2, 4}, {3, 4}, {1, 3, 4}, {2, 3, 4}, {1, 2, 3, 4}}

Python

This example is incomplete. It shows no proof of the cases with empty sets. Please ensure that it meets all task requirements and remove this message.

<lang python>def list_powerset(lst):

   # the power set of the empty set has one element, the empty set
   result = [[]]
   for x in lst:
       # for every additional element in our set
       # the power set consists of the subsets that don't
       # contain this element (just take the previous power set)
       # plus the subsets that do contain the element (use list
       # comprehension to add [x] onto everything in the
       # previous power set)
       result.extend([subset + [x] for subset in result])
   return result

the above function in one statement

def list_powerset2(lst):

   return reduce(lambda result, x: result + [subset + [x] for subset in result],
                 lst, [[]])

def powerset(s):

   return frozenset(map(frozenset, list_powerset(list(s))))</lang>list_powerset computes the power set of a list of distinct elements. powerset simply converts the input and output from lists to sets. We use the frozenset type here for immutable sets, because unlike mutable sets, it can be put into other sets.

Example:


>>> list_powerset([1,2,3])
[[], [1], [2], [1, 2], [3], [1, 3], [2, 3], [1, 2, 3]]
>>> powerset(frozenset([1,2,3]))
frozenset([frozenset([3]), frozenset([1, 2]), frozenset([]), frozenset([2, 3]), frozenset([1]), frozenset([1, 3]), frozenset([1, 2, 3]), frozenset([2])])

Further Explanation

If you take out the requirement to produce sets and produce list versions of each powerset element, then add a print to trace the execution, you get this simplified version of the program above where it is easier to trace the inner workings<lang python>def powersetlist(s):

   r = [[]]
   for e in s:
       print "r: %-55r e: %r" % (r,e)
       r += [x+[e] for x in r]
   return r

s= [0,1,2,3]

print "\npowersetlist(%r) =\n %r" % (s, powersetlist(s))</lang>

Sample output:

r: [[]]                                                    e: 0
r: [[], [0]]                                               e: 1
r: [[], [0], [1], [0, 1]]                                  e: 2
r: [[], [0], [1], [0, 1], [2], [0, 2], [1, 2], [0, 1, 2]]  e: 3

powersetlist([0, 1, 2, 3]) =
 [[], [0], [1], [0, 1], [2], [0, 2], [1, 2], [0, 1, 2], [3], [0, 3], [1, 3], [0, 1, 3], [2, 3], [0, 2, 3], [1, 2, 3], [0, 1, 2, 3]]

Binary Count method

If you list the members of the set and include them according to if the corresponding bit position of a binary count is true then you generate the powerset. (Note that only frozensets can be members of a set in the second function)<lang python>def powersequence(val):

    Generate a 'powerset' for sequence types that are indexable by integers.
       Uses a binary count to enumerate the members and returns a list

       Examples:
           >>> powersequence('STR')   # String
           [, 'S', 'T', 'ST', 'R', 'SR', 'TR', 'STR']
           >>> powersequence([0,1,2]) # List
           [[], [0], [1], [0, 1], [2], [0, 2], [1, 2], [0, 1, 2]]
           >>> powersequence((3,4,5)) # Tuple
           [(), (3,), (4,), (3, 4), (5,), (3, 5), (4, 5), (3, 4, 5)]
           >>> 
   
   vtype = type(val); vlen = len(val); vrange = range(vlen)
   return [ reduce( lambda x,y: x+y, (val[i:i+1] for i in vrange if 2**i & n), vtype())
            for n in range(2**vlen) ]

def powerset(s):

    Generate the powerset of s

       Example:
           >>> powerset(set([6,7,8]))
           set([frozenset([7]), frozenset([8, 6, 7]), frozenset([6]), frozenset([6, 7]), frozenset([]), frozenset([8]), frozenset([8, 7]), frozenset([8, 6])])
   
   return set( frozenset(x) for x in powersequence(list(s)) )</lang>

Recursive Alternative

This is an (inefficient) recursive version that almost reflects the recursive definition of a power set as explained in http://en.wikipedia.org/wiki/Power_set#Algorithms. It does not create a sorted output.<lang python>def p(l):

   if not l: return [[]]
   return p(l[1:]) + [[l[0]] + x for x in p(l[1:])]</lang>

Python: Standard documentation

Pythons documentation has a method that produces the groupings, but not as sets:<lang python>>>> from pprint import pprint as pp >>> from itertools import chain, combinations >>> >>> def powerset(iterable):

   "powerset([1,2,3]) --> () (1,) (2,) (3,) (1,2) (1,3) (2,3) (1,2,3)"
   s = list(iterable)
   return chain.from_iterable(combinations(s, r) for r in range(len(s)+1))

>>> pp(set(powerset({1,2,3,4}))) {(),

(1,),
(1, 2),
(1, 2, 3),
(1, 2, 3, 4),
(1, 2, 4),
(1, 3),
(1, 3, 4),
(1, 4),
(2,),
(2, 3),
(2, 3, 4),
(2, 4),
(3,),
(3, 4),
(4,)}

>>> </lang>

Qi

This example is incomplete. It shows no proof of the cases with empty sets. Please ensure that it meets all task requirements and remove this message.

Translation of: Scheme

<lang qi>(define powerset

 [] -> [[]]
 [A|As] -> (append (map (cons A) (powerset As))
                   (powerset As)))</lang>

R

This example is incomplete. It shows no proof of the cases with empty sets. Please ensure that it meets all task requirements and remove this message.

Non-recursive version

The conceptual basis for this algorithm is the following:<lang>for each element in the set: for each subset constructed so far: new subset = (subset + element)</lang>This method is much faster than a recursive method, though the speed is still O(2^n).<lang R>powerset = function(set){ ps = list() ps1 = numeric() #Start with the empty set. for(element in set){ #For each element in the set, take all subsets temp = vector(mode="list",length=length(ps)) #currently in "ps" and create new subsets (in "temp") for(subset in 1:length(ps)){ #by adding "element" to each of them. tempsubset = c(pssubset,element) } ps=c(ps,temp) #Add the additional subsets ("temp") to "ps". } return(ps) }

powerset(1:4)</lang>The list "temp" is a compromise between the speed costs of doing arithmetic and of creating new lists (since R lists are immutable, appending to a list means actually creating a new list object). Thus, "temp" collects new subsets that are later added to the power set. This improves the speed by 4x compared to extending the list "ps" at every step.

Recursive version

Library: sets

The sets package includes a recursive method to calculate the power set. However, this method takes ~100 times longer than the non-recursive method above.<lang R>library(sets)</lang> An example with a vector.<lang R>v <- (1:3)^2 sv <- as.set(v) 2^sv</lang>

{{}, {1}, {4}, {9}, {1, 4}, {1, 9}, {4, 9}, {1, 4, 9}}

An example with a list.<lang R>l <- list(a=1, b="qwerty", c=list(d=TRUE, e=1:3)) sl <- as.set(l) 2^sl</lang>

{{}, {1}, {"qwerty"}, {<<list(2)>>}, {1, <<list(2)>>}, {"qwerty",
 1}, {"qwerty", <<list(2)>>}, {"qwerty", 1, <<list(2)>>}}

Racket

This example is incomplete. It shows no proof of the cases with empty sets. Please ensure that it meets all task requirements and remove this message.

<lang racket>;;; Direct translation of 'functional' ruby method

(define (powerset s)

 (for/fold ([outer-set (set(set))]) ([element s])
   (set-union outer-set 
              (list->set (set-map outer-set
                                  (λ(inner-set) (set-add inner-set element)))))))</lang>

Rascal

This example is incomplete. It shows no proof of the cases with empty sets. Please ensure that it meets all task requirements and remove this message.

<lang rascal>import Set;

public set[set[&T]] PowerSet(set[&T] s) = power(s);</lang>

An example output:

rascal>PowerSet({1,2,3,4})
set[set[int]]: {
 {4,3},
 {4,2,1},
 {4,3,1},
 {4,2},
 {4,3,2},
 {4,1},
 {4,3,2,1},
 {4},
 {3},
 {2,1},
 {3,1},
 {2},
 {3,2},
 {1},
 {3,2,1},
 {}
}

REXX

This example is incomplete. It shows no proof of the cases with empty sets. Please ensure that it meets all task requirements and remove this message.

<lang rexx>/*REXX program to display a power set, items may be anything (no blanks)*/

parse arg S /*let user specify the set. */ if S= then S='one two three four' /*None specified? Use default*/ N=words(S) /*number of items in the list.*/ ps='{}' /*start with a null power set.*/

             do chunk=1  for N           /*traipse through the items.  */
             ps=ps combN(N,chunk)        /*N items, a CHUNK at a time. */
             end    /*chunk*/

w=words(ps)

             do k=1  for w               /*show combinations, one/line.*/
             say right(k,length(w)) word(ps,k)
             end    /*k*/

exit /*stick a fork in it, we done.*/ /*─────────────────────────────────────$COMBN subroutine────────────────*/ combN: procedure expose $ S; parse arg x,y; $= !.=0; base=x+1; bbase=base-y; ym=y-1; do p=1 for y; !.p=p; end

               do j=1; L=
                          do d=1  for y;  _=!.d;  L=L','word(S,_);  end
               $=$ '{'strip(L,'L',",")'}'
               !.y=!.y+1;   if !.y==base  then if .combU(ym)  then leave
               end   /*j*/

return strip($) /*return with partial powerset*/

.combU: procedure expose !. y bbase; parse arg d; if d==0 then return 1 p=!.d; do u=d to y; !.u=p+1

         if !.u==bbase+u  then return .combU(u-1)
         p=!.u
         end   /*u*/

return 0</lang>

when using the default input:

 1 {}
 2 {one}
 3 {two}
 4 {three}
 5 {four}
 6 {one,two}
 7 {one,three}
 8 {one,four}
 9 {two,three}
10 {two,four}
11 {three,four}
12 {one,two,three}
13 {one,two,four}
14 {one,three,four}
15 {two,three,four}
16 {one,two,three,four}

Ruby

This example is incomplete. It shows no proof of the cases with empty sets. Please ensure that it meets all task requirements and remove this message.

<lang ruby># Based on http://johncarrino.net/blog/2006/08/11/powerset-in-ruby/

See the link if you want a shorter version. This was intended to show the reader how the method works.

class Array

 # Adds a power_set method to every array, i.e.: [1, 2].power_set
 def power_set
   
   # Injects into a blank array of arrays.
   # acc is what we're injecting into
   # you is each element of the array
   inject([[]]) do |acc, you|
     ret = []             # Set up a new array to add into
     acc.each do |i|      # For each array in the injected array,
       ret << i           # Add itself into the new array
       ret << i + [you]   # Merge the array with a new array of the current element
     end
     ret       # Return the array we're looking at to inject more.
   end
   
 end
 
 # A more functional and even clearer variant.
 def func_power_set
   inject([[]]) { |ps,item|    # for each item in the Array
     ps +                      # take the powerset up to now and add
     ps.map { |e| e + [item] } # it again, with the item appended to each element
   }
 end

end

A direct translation of the "power array" version above

require 'set' class Set

 def powerset 
   inject(Set[Set[]]) do |ps, item| 
     ps.union ps.map {|e| e.union (Set.new [item])}
   end
 end

end

p [1,2,3,4].power_set p %w(one two three).func_power_set

p Set[1,2,3].powerset</lang>

Output:


[[], [4], [3], [3, 4], [2], [2, 4], [2, 3], [2, 3, 4], [1], [1, 4], [1, 3], [1, 3, 4], [1, 2], [1, 2, 4], [1, 2, 3], [1, 2, 3, 4]]
[[], ["one"], ["two"], ["one", "two"], ["three"], ["one", "three"], ["two", "three"], ["one", "two", "three"]]

<Set: {#<Set: {}>, #<Set: {1}>, #<Set: {2}>, #<Set: {1, 2}>, #<Set: {3}>, #<Set: {1, 3}>, #<Set: {2, 3}>, #<Set: {1, 2, 3}>}>

SAS

This example is incomplete. It shows no proof of the cases with empty sets. Please ensure that it meets all task requirements and remove this message.

<lang SAS>options mprint mlogic symbolgen source source2;

%macro SubSets (FieldCount = ); data _NULL_; Fields = &FieldCount; SubSets = 2**Fields; call symput ("NumSubSets", SubSets); run;

%put &NumSubSets;

data inital; %do j = 1 %to &FieldCount; F&j. = 1; %end; run;

data SubSets; set inital; RowCount =_n_; call symput("SetCount",RowCount); run;

%put SetCount ;

%do %while (&SetCount < &NumSubSets);

data loop; %do j=1 %to &FieldCount; if rand('GAUSSIAN') > rand('GAUSSIAN') then F&j. = 1; %end;

data SubSets_ ; set SubSets loop; run;

proc sort data=SubSets_ nodupkey; by F1 - F&FieldCount.; run;

data Subsets; set SubSets_; RowCount =_n_; run;

proc sql noprint; select max(RowCount) into :SetCount from SubSets; quit; run;

%end; %Mend SubSets;</lang>You can then call the macro as:

<lang SAS>%SubSets(FieldCount = 5);</lang>The output will be the dataset SUBSETS and will have a 5 columns F1, F2, F3, F4, F5 and 32 columns, one with each combination of 1 and missing values.

Output:

Obs F1 F2 F3 F4 F5 RowCount 1 . . . . . 1 2 . . . . 1 2 3 . . . 1 . 3 4 . . . 1 1 4 5 . . 1 . . 5 6 . . 1 . 1 6 7 . . 1 1 . 7 8 . . 1 1 1 8 9 . 1 . . . 9 10 . 1 . . 1 10 11 . 1 . 1 . 11 12 . 1 . 1 1 12 13 . 1 1 . . 13 14 . 1 1 . 1 14 15 . 1 1 1 . 15 16 . 1 1 1 1 16 17 1 . . . . 17 18 1 . . . 1 18 19 1 . . 1 . 19 20 1 . . 1 1 20 21 1 . 1 . . 21 22 1 . 1 . 1 22 23 1 . 1 1 . 23 24 1 . 1 1 1 24 25 1 1 . . . 25 26 1 1 . . 1 26 27 1 1 . 1 . 27 28 1 1 . 1 1 28 29 1 1 1 . . 29 30 1 1 1 . 1 30 31 1 1 1 1 . 31

32 1 1 1 1 1 32

Scala

Testing the cases:

{1,2,3,4}
${\mathcal {P}}$ ( $\varnothing$ ) = { $\varnothing$ }
${\mathcal {P}}$ ({ $\varnothing$ }) = { $\varnothing$ , { $\varnothing$ } }

<lang scala>import scala.compat.Platform.currentTime

object Powerset extends App {

 def powerset[A](s: Set[A]) = s.foldLeft(Set(Set.empty[A])) { case (ss, el) => ss ++ ss.map(_ + el)}

 assert(powerset(Set(1, 2, 3, 4)) == Set(Set.empty, Set(1), Set(2), Set(3), Set(4), Set(1, 2), Set(1, 3), Set(1, 4),
   Set(2, 3), Set(2, 4), Set(3, 4), Set(1, 2, 3), Set(1, 3, 4), Set(1, 2, 4), Set(2, 3, 4), Set(1, 2, 3, 4)))
 assert(powerset(Set()) == Set(Set()))
 assert(powerset(Set(Set())) == Set(Set(), Set(Set())))

 println(s"Successfully completed without errors. [total ${currentTime - executionStart} ms]")

}</lang>Another option that produces a iterator of the sets:<lang scala>def powerset[A](s: Set[A]) = (0 to s.size).map(s.toSeq.combinations(_)).reduce(_ ++ _).map(_.toSet)</lang>

Scheme

This example is incomplete. It shows no proof of the cases with empty sets. Please ensure that it meets all task requirements and remove this message.

Translation of: Common Lisp

<lang scheme>(define (power-set set)

 (if (null? set)
     '(())
     (let ((rest (power-set (cdr set))))
       (append (map (lambda (element) (cons (car set) element))
                    rest)
               rest))))

(display (power-set (list 1 2 3))) (newline)

(display (power-set (list "A" "C" "E")))

(newline)</lang>

Output:

((1 2 3) (1 2) (1 3) (1) (2 3) (2) (3) ())
((A C E) (A C) (A E) (A) (C E) (C) (E) ())

Call/cc generation:<lang lisp>(define (power-set lst)

 (define (iter yield)
   (let recur ((a '()) (b lst))
     (if (null? b) (set! yield

(call-with-current-continuation (lambda (resume) (set! iter resume) (yield a)))) (begin (recur (append a (list (car b))) (cdr b)) (recur a (cdr b)))))

   ;; signal end of generation
   (yield 'end-of-seq))

 (lambda () (call-with-current-continuation iter)))

(define x (power-set '(1 2 3))) (let loop ((a (x)))

 (if (eq? a 'end-of-seq) #f
   (begin
     (display a)
     (newline)
     (loop (x)))))</lang>output<lang>(1 2)

(1 3) (1) (2 3) (2) (3) ()</lang>Iterative:<lang scheme>(define (power_set_iter set)

 (let loop ((res '(())) (s set))
   (if (empty? s)
       res
       (loop (append (map (lambda (i) (cons (car s) i)) res) res) (cdr s)))))

</lang>

Output:

'((e d c b a)

 (e d c b)
 (e d c a)
 (e d c)
 (e d b a)
 (e d b)
 (e d a)
 (e d)
 (e c b a)
 (e c b)
 (e c a)
 (e c)
 (e b a)
 (e b)
 (e a)
 (e)
 (d c b a)
 (d c b)
 (d c a)
 (d c)
 (d b a)
 (d b)
 (d a)
 (d)
 (c b a)
 (c b)
 (c a)
 (c)
 (b a)
 (b)
 (a)
 ())

Seed7

This example is incomplete. It shows no proof of the cases with empty sets. Please ensure that it meets all task requirements and remove this message.

<lang seed7>$ include "seed7_05.s7i";

const func array bitset: powerSet (in bitset: baseSet) is func

 result
   var array bitset: pwrSet is [] (bitset.value);
 local
   var integer: element is 0;
   var integer: index is 0;
   var bitset: aSet is bitset.value;
 begin
   for element range baseSet do
     for key index range pwrSet do
       aSet := pwrSet[index];
       if element not in aSet then
         incl(aSet, element);
         pwrSet &:= aSet;
       end if;
     end for;
   end for;
 end func;

const proc: main is func

 local
   var bitset: aSet is bitset.value;
 begin
   for aSet range powerSet({1, 2, 3, 4}) do
     writeln(aSet);
   end for;

end func;</lang>

Output:

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

{1, 2, 3, 4}

Smalltalk

This example is incomplete. It shows no proof of the cases with empty sets. Please ensure that it meets all task requirements and remove this message.

Works with: GNU Smalltalk

Code from Bonzini's blog<lang smalltalk>Collection extend [

   power [
       ^(0 to: (1 bitShift: self size) - 1) readStream collect: [ :each || i |
           i := 0.
           self select: [ :elem | (each bitAt: (i := i + 1)) = 1 ] ]
   ]

].</lang><lang smalltalk>#(1 2 4) power do: [ :each |

   each asArray printNl ].

( 'A' 'C' 'E' ) power do: [ :each |

   each asArray printNl ].</lang>

Standard ML

This example is incomplete. It shows no proof of the cases with empty sets. Please ensure that it meets all task requirements and remove this message.

Version for lists:<lang sml>fun subsets xs = foldr (fn (x, rest) => rest @ map (fn ys => x::ys) rest) [[]] xs</lang>

Tcl

This example is incomplete. It shows no proof of the cases with empty sets. Please ensure that it meets all task requirements and remove this message.

<lang tcl>proc subsets {l} {

   set res [list [list]]
   foreach e $l {
       foreach subset $res {lappend res [lappend subset $e]}
   }
   return $res

}

puts [subsets {a b c d}]</lang>

Output:

{} a b {a b} c {a c} {b c} {a b c} d {a d} {b d} {a b d} {c d} {a c d} {b c d} {a b c d}

Binary Count Method

<lang tcl>proc powersetb set {

  set res {}
  for {set i 0} {$i < 2**[llength $set]} {incr i} {
     set pos -1
     set pset {}
     foreach el $set {
         if {$i & 1<<[incr pos]} {lappend pset $el}
     }
     lappend res $pset
  }
  return $res

}</lang>

TXR

This example is incomplete. It shows no proof of the cases with empty sets. Please ensure that it meets all task requirements and remove this message.

Translation of: Common Lisp

The power set function can be written concisely like this:<lang txr>(defun power-set (s)

(reduce-right
  (op append (mapcar (op cons @@1) @2) @2)
  s '(())))</lang>A complete program which takes command line arguments and prints the power set in comma-separated brace notation:<lang txr>@(do (defun power-set (s)
      (reduce-right  
        (op append (mapcar (op cons @@1) @2) @2)
        s '(()))))

@(bind pset @(power-set *args*)) @(output) @ (repeat) {@(rep)@pset, @(last)@pset@(empty)@(end)} @ (end)

@(end)</lang>

$ txr rosetta/power-set.txr  1 2 3
{1, 2, 3}
{1, 2}
{1, 3}
{1}
{2, 3}
{2}
{3}
{}

What is not obvious is that the above power-set function generalizes to strings and vectors.<lang txr>@(do (defun power-set (s)

      (reduce-right 
        (op append (mapcar (op cons @@1) @2) @2) 
        s '(())))
    (prinl (power-set "abc"))
    (prinl (power-set ""))
    (prinl (power-set #(1 2 3))))</lang>txr power-set-generic.txr
((#\a #\b #\c) (#\a #\b) (#\a #\c) (#\a) (#\b #\c) (#\b) (#\c) nil)
((nil) nil)
((1 2 3) (1 2) (1 3) (1) (2 3) (2) (3) nil)

UnixPipes

This example is incomplete. It shows no proof of the cases with empty sets. Please ensure that it meets all task requirements and remove this message.

<lang ksh>| cat A

a b c

| cat A |\

  xargs -n 1 ksh -c 'echo \{`cat A`\}' |\
  xargs |\
  sed -e 's; ;,;g' \
      -e 's;^;echo ;g' \
      -e 's;\},;}\\ ;g' |\
  ksh |unfold `wc -l A` |\
  xargs -n1 -I{} ksh -c 'echo {} |\
       unfold 1 |sort -u |xargs' |sort -u

a a b a b c a c b b c c</lang>

UNIX Shell

This example is incomplete. It shows no proof of the cases with empty sets. Please ensure that it meets all task requirements and remove this message.

From here <lang bash>p() { [ $# -eq 0 ] && echo || (shift; p "$@") | while read r ; do echo -e "$1 $r\n$r"; done }</lang> Usage <lang bash>|p `cat` | sort | uniq A C E ^D</lang>

Ursala

This example is incomplete. It shows no proof of the cases with empty sets. Please ensure that it meets all task requirements and remove this message.

Sets are a built in type constructor in Ursala, represented as lexically sorted lists with duplicates removed. The powerset function is a standard library function, but could be defined as shown below. <lang Ursala>powerset = ~&NiC+ ~&i&& ~&at^?\~&aNC ~&ahPfatPRXlNrCDrT</lang> test program: <lang Ursala>#cast %sSS

test = powerset {'a','b','c','d'}</lang>

Output:

{
  {},
  {'a'},
  {'a','b'},
  {'a','b','c'},
  {'a','b','c','d'},
  {'a','b','d'},
  {'a','c'},
  {'a','c','d'},
  {'a','d'},
  {'b'},
  {'b','c'},
  {'b','c','d'},
  {'b','d'},
  {'c'},
  {'c','d'},
  {'d'}}

V

This example is incomplete. It shows no proof of the cases with empty sets. Please ensure that it meets all task requirements and remove this message.

V has a built in called powerlist <lang v>[A C E] powerlist =[[A C E] [A C] [A E] [A] [C E] [C] [E] []]</lang> Its implementation in std.v is (like joy) <lang v>[powerlist

  [null?]
  [unitlist]
  [uncons]
  [dup swapd [cons] map popd swoncat]
   linrec].</lang>

zkl

Using a combinations function, build the power set from combinations of 1,2,... items. <lang zkl>fcn pwerSet(list){

 (0).pump(list.len(),List, Utils.Helpers.pickNFrom.fp1(list),
    T(Void.Write,Void.Write) ) .append(list)

}</lang>

Output:

foreach n in (5){
   ps:=pwerSet((1).pump(n,List)); ps.println(" Size = ",ps.len());
}

L(L()) Size = 1
L(L(),L(1)) Size = 2
L(L(),L(1),L(2),L(1,2)) Size = 4
L(L(),L(1),L(2),L(3),L(1,2),L(1,3),L(2,3),L(1,2,3)) Size = 8
L(L(),L(1),L(2),L(3),L(4),L(1,2),L(1,3),L(1,4),L(2,3),L(2,4),
  L(3,4),L(1,2,3),L(1,2,4),L(1,3,4),L(2,3,4),L(1,2,3,4)) Size = 16