Power set: Difference between revisions

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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 [[wp:Power_set|power set]] (or powerset) of S, written P(S), or 2S, is the set of all subsets of S.

Given a set S, the [[wp:Power_set|power set]] (or powerset) of S, written P(S), or 2S, 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 2S 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}}.

'''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 2S of S.

For a set which contains n elements, the corresponding power set has 2n elements, including the edge cases of [[wp:Empty_set|empty set]].The power set of the empty set is the set which contains itself (20 = 1):

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}}.

<math>\mathcal{P}</math>(<math>\varnothing</math>) = { <math>\varnothing</math> }

~~For~~ a set which contains ~~n elements,~~ the ~~corresponding power~~ set has ~~2n~~ ~~elements~~, ~~including~~ the ~~edge~~ ~~cases~~ of ~~[[wp:Empty_set|~~empty set~~]].~~

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):

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

<math>\mathcal{P}</math>(<math>\varnothing</math>) = { <math>\varnothing</math> }

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):

<math>\mathcal{P}</math>({<math>\varnothing</math>}) = { <math>\varnothing</math>, { <math>\varnothing</math> } }

=={{header|Ada}}==

{{incomplete|Ada|It shows no proof of the cases with empty sets.}}This solution prints the power set of words read from the command line.<lang ada>with Ada.Text_IO, Ada.Command_Line;

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

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end loop;

Print_All_Subsets(Set); -- do the work

end Power_Set;</lang>

end Power_Set;</lang>{{out}}<pre>>./power_set cat dog mouse

<pre>>./power_set cat dog mouse

{ cat, dog, mouse }

{ cat, dog }

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{ 2 }

{ }</pre>

=={{header|ALGOL 68}}==

{{incomplete|ALGOL 68|It shows no proof of the cases with empty sets.}}{{works with|ALGOL 68|Revision 1 - no extensions to language used}}{{works with|ALGOL 68G|Any - tested with release [http://sourceforge.net/projects/algol68/files/algol68g/algol68g-1.18.0/algol68g-1.18.0-9h.tiny.el5.centos.fc11.i386.rpm/download 1.18.0-9h.tiny]}}{{wont work with|ELLA ALGOL 68|Any (with appropriate job cards) - tested with release [http://sourceforge.net/projects/algol68/files/algol68toc/algol68toc-1.8.8d/algol68toc-1.8-8d.fc9.i386.rpm/download 1.8-8d] - due to extensive use of FORMATted transput}}

{{works with|ALGOL 68|Revision 1 - no extensions to language used}}

{{works with|ALGOL 68G|Any - tested with release [http://sourceforge.net/projects/algol68/files/algol68g/algol68g-1.18.0/algol68g-1.18.0-9h.tiny.el5.centos.fc11.i386.rpm/download 1.18.0-9h.tiny]}}

{{wont work with|ELLA ALGOL 68|Any (with appropriate job cards) - tested with release [http://sourceforge.net/projects/algol68/files/algol68toc/algol68toc-1.8.8d/algol68toc-1.8-8d.fc9.i386.rpm/download 1.8-8d] - due to extensive use of FORMATted transput}}

Requires: ALGOL 68g mk14.1+

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

<lang algol68>MODE MEMBER = INT;

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

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printf(($"set["d"] = "$,member, repr set, set[member],$l$))

OD

)</lang>

)</lang>{{out}}<pre>set[1] = ();

Output:

<pre>

set[1] = ();

set[2] = (1);

set[3] = (2);

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set[6] = (1, 4);

set[7] = (2, 4);

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

set[8] = (1, 2, 4);</pre>

</pre>

=={{header|AutoHotkey}}==

ahk [http://www.autohotkey.com/forum/viewtopic.php?t=44657&postdays=0&postorder=asc&start=147 discussion]

{{incomplete|AutoHotkey|It shows no proof of the cases with empty sets.}}ahk [http://www.autohotkey.com/forum/viewtopic.php?t=44657&postdays=0&postorder=asc&start=147 discussion]<lang autohotkey>a = 1,a,-- ; elements separated by commas

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

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

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}

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

=={{header|BBC BASIC}}==

The elements of a set are represented as the bits in an integer (hence the maximum size of set is 32).

{{incomplete|BBC BASIC|It shows no proof of the cases with empty sets.}}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"

<lang bbcbasic> DIM list$(3) : list$() = "1", "2", "3", "4"

PRINT FNpowerset(list$())

END

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s$ += "},"

NEXT i%

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

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

{{},{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}}

'''Output:'''

<pre>

{{},{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}}

</pre>

=={{header|Bracmat}}==

<lang bracmat>( ( powerset

{{incomplete|Bracmat|It shows no proof of the cases with empty sets.}}<lang bracmat>( ( powerset

= done todo first

. !arg:(?done.?todo)

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)

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

);</lang>

);</lang>{{out}}<pre> 1 2 3 4

Output:

<pre> 1 2 3 4

, 1 2 3

, 1 2 4

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, 4

,</pre>

=={{header|Burlesque}}==

{{incomplete|Burlesque|It shows no proof of the cases with empty sets.}}<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>

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>

=={{header|C}}==

<lang c>#include <stdio.h>

{{incomplete|C|It shows no proof of the cases with empty sets.}}<lang c>#include <stdio.h>

struct node {

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powerset(argv + 1, argc - 1, 0);

return 0;

}</lang>

}</lang>{{out}}<pre>% ./a.out 1 2 3

<pre>

% ./a.out 1 2 3

[ ]

[ 3 ]

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[ 3 1 ]

[ 2 1 ]

[ 3 2 1 ]

[ 3 2 1 ]</pre>

</pre>

=={{header|C++}}==

{{incomplete|C++|It shows no proof of the cases with empty sets.}}

=== Non-recursive version ===

<lang cpp>#include <iostream>

#include <set>

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std::cout << " }\n";

}

}</lang>

}</lang>{{out}}<pre>

Output:

<pre>

{ }

{ 2 }

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=== Recursive version ===

<lang cpp>#include <iostream>

#include <set>

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{

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

}</lang>

}

</lang>

=={{header|C sharp|C#}}==

{{incomplete|C sharp|It shows no proof of the cases with empty sets.}}<lang csharp>public IEnumerable<IEnumerable<T>> GetPowerSet<T>(List<T> list)

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

{

return from m in Enumerable.Range(0, 1 << list.Count)

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result.Select(subset =>

string.Join(",", subset.Select(clr => clr.ToString()).ToArray())).ToArray()));

}</lang>

}

</lang>

Output:

<lang>

Red

Green

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Green,Blue,Yellow

Red,Green,Blue,Yellow

</lang>

An alternative implementation for an arbitrary number of elements:

<lang csharp> public IEnumerable<IEnumerable<T>> GetPowerSet<T>(IEnumerable<T> input) {

public IEnumerable<IEnumerable<T>> GetPowerSet<T>(IEnumerable<T> input) {

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

as IEnumerable<IEnumerable<T>>;

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return input.Aggregate(seed, (a, b) =>

a.Concat(a.Select(x => x.Concat(new List<T>() { b }))));

}

}</lang>

</lang>

=={{header|Clojure}}==

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

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

~~<lang Clojure>~~#{#{} #{1} #{2} #{1 2} #{3} #{1 3} #{2 3} #{1 2 3}}~~</lang>~~

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

=={{header|CoffeeScript}}==

{{incomplete|CoffeeScript|It shows no proof of the cases with empty sets.}}<lang coffeescript>print_power_set = (arr) ->

print_power_set = (arr) ->

console.log "POWER SET of #{arr}"

for subset in power_set(arr)

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print_power_set []

print_power_set [4, 2, 1]

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

print_power_set ['dog', 'c', 'b', 'a']</lang>{{out}}<pre>> coffee power_set.coffee

</lang>

output

<lang>

> coffee power_set.coffee

POWER SET of

[]

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

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

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

[ 'dog', 'c', 'b', 'a' ]</pre>

</lang>

=={{header|ColdFusion}}==

{{incomplete|ColdFusion|It shows no proof of the cases with empty sets.}}Port from the [[#JavaScript|JavaScript]] version, compatible with ColdFusion 8+ or Railo 3+<lang javascript>public array function powerset(required array data)

Port from the [[#JavaScript|JavaScript]] version, compatible with ColdFusion 8+ or Railo 3+

<lang javascript>public array function powerset(required array data)

{

var ps = [""];

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}

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

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

["","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"]

Outputs:

=={{header|Common Lisp}}==

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

{{incomplete|Common Lisp|It shows no proof of the cases with empty sets.}}<lang lisp>(defun power-set (s)

(reduce #'(lambda (item ps)

(append (mapcar #'(lambda (e) (cons item e))

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s

:from-end t

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

:initial-value '(())))</lang>{{out}}

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)

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(append (mapcar #'(lambda (elt) (cons (car l) elt)) prev)

prev))))</lang>

Imperative-style using LOOP:

<lang lisp>(defun powerset (xs)

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

((1) (1 3) (1 2 3) (1 2) (2) (2 3) (3) NIL)

=={{header|D}}==

Version using just arrays (it assumes the input to contain distinct items):

{{incomplete|D|It shows no proof of the cases with empty sets.}}Version using just arrays (it assumes the input to contain distinct items):<lang d>T[][] powerSet(T)(in T[] s) pure nothrow @safe {

<lang d>T[][] powerSet(T)(in T[] s) pure nothrow @safe {

auto r = new typeof(return)(1, 0);

foreach (e; s) {

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

}</lang>

}</lang>{{out}}

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

===Lazy Version===

Compile with -version=power_set2_main to run the main.

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

}

}</lang>

}</lang>Same output.

Same output.

A [[Power_set/D|set implementation]] and its power set function.

=={{header|Déjà Vu}}==

{{incomplete|Déjà Vu|It shows no proof of the cases with empty sets.}}In Déjà Vu, sets are dictionaries with all values <code>true</code> and the default set to <code>false</code>.<lang dejavu>powerset s:

In Déjà Vu, sets are dictionaries with all values <code>true</code> and the default set to <code>false</code>.

<lang dejavu>powerset s:

local :out [ set{ } ]

for value in keys s:

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out

!. powerset set{ 1 2 3 4 }</lang>{{out}}<pre>[ 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 } ]</pre>

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

=={{header|E}}==

{{incomplete|E|It shows no proof of the cases with empty sets.}}[[Category:E examples needing attention]]<lang e>pragma.enable("accumulator")

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

def powerset(s) {

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})

}

}</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.

}</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. [[Category:E examples needing attention]]

=={{header|Erlang}}==

{{incomplete|Erlang|It shows no proof of the cases with empty sets.}}

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

<pre>

<pre>For [1 2 3]:

For [1 2 3]:

[ ] | 0 0 0 | 0

[ 3] | 0 0 1 | 1

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[1 2 ] | 1 1 0 | 6

[1 2 3] | 1 1 1 | 7

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯</pre><lang erlang>powerset(Lst) ->

</pre>

<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>{{out|Which outputs}}<pre>[[], [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]]</pre>Alternate shorter and more efficient version:<lang erlang>powerset([]) -> [[]];

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

Which outputs:

Alternate shorter and more efficient version:

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

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

[ [H|X] || X <- PT ] ++ PT.</lang>

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

or even more efficient version:

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

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

powerset(H, PT, PT).

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powerset(_, [], Acc) -> Acc;

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

=={{header|F Sharp|F#}}==

almost exact copy of OCaml version

{{incomplete|F Sharp|It shows no proof of the cases with empty sets.}}almost exact copy of OCaml version

let subsets xs = List.foldBack (fun x rest -> rest @ List.map (fun ys -> x::ys) rest) xs [[]]

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 =

</lang>

alternatively with list comprehension

let rec pow =

function

| [] -> [[]]

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

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

</lang>

=={{header|Factor}}==

We use hash sets, denoted by <code>HS{ }</code> brackets, for our sets. <code>members</code> converts from a set to a sequence, and <code><hash-set></code> converts back.

{{incomplete|Factor|It shows no proof of the cases with empty sets.}}We use hash sets, denoted by <code>HS{ }</code> brackets, for our sets. <code>members</code> converts from a set to a sequence, and <code><hash-set></code> converts back.<lang factor>USING: kernel prettyprint sequences arrays sets hash-sets ;

<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>

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

Usage:

<lang factor>( scratchpad ) HS{ 1 2 3 4 } powerset .

HS{

HS{ 1 2 3 4 }

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HS{ 1 3 4 }

HS{ 2 3 4 }

}</~~lang~~>

}</pre>

=={{header|Forth}}==

{{incomplete|Forth|It shows no proof of the cases with empty sets.}}{{works with|4tH|3.61.0}}.{{trans|C}}<lang forth>: ?print dup 1 and if over args type space then ;

{{works with|4tH|3.61.0}}.

<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 ;

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

powerset</lang>

powerset</lang>{{out}}<pre>$ 4th cxq powerset.4th 1 2 3 4

Output:

<pre>

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

( )

( 1 )

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( 1 3 4 )

( 2 3 4 )

( 1 2 3 4 )

( 1 2 3 4 )</pre>

</pre>

=={{header|Frink}}==

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.

{{incomplete|Frink|It shows no proof of the cases with empty sets.}}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[]

a.subsets[]</lang>

</lang>

=={{header|FunL}}==

FunL uses Scala type <code>scala.collection.immutable.Set</code> as it's set type, which has a built-in method <code>subsets</code> returning an (Scala) iterator over subsets.

{{incomplete|FunL|It shows no proof of the cases with empty sets.}}FunL uses Scala type <code>scala.collection.immutable.Set</code> as it's set type, which has a built-in method <code>subsets</code> 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

<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 ) =

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

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

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

{{}, {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}}

<pre>

{{}, {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}}

</pre>

=={{header|GAP}}==

<lang gap># Built-in

{{incomplete|GAP|It shows no proof of the cases with empty sets.}}<lang gap># Built-in

Combinations([1, 2, 3]);

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

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# [ [ ], [ 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>

=={{header|Go}}==

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.

{{incomplete|Go|It shows no proof of the cases with empty sets.}}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

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.

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

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fmt.Println(ps)

fmt.Println("length =", len(ps))

}</lang>

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

<pre>

{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

length = 16</pre>

</pre>

=={{header|Groovy}}==

Builds on the [[Combinations#Groovy|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'''.

{{incomplete|Groovy|It shows no proof of the cases with empty sets.}}Builds on the [[Combinations#Groovy|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

<lang groovy>def comb

comb = { m, List list ->

def n = list.size()

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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

}</lang>

Test program:

<lang groovy>def vocalists = [ "C", "S", "N", "Y" ] as LinkedHashSet

println "${vocalists}"

println powerSet(vocalists)</lang>

Line 985:

Line 839:

=={{header|Haskell}}==

<lang Haskell>import Data.Set

{{incomplete|Haskell|It shows no proof of the cases with empty sets.}}<lang Haskell>import Data.Set

import Control.Monad

Line 992:

Line 846:

listPowerset :: [a] -> [[a]]

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

listPowerset = filterM (const [True, False])</lang>

<tt>listPowerset</tt> describes the result as all possible (using the list monad) filterings (using <tt>filterM</tt>) of the input list, regardless (using <tt>const</tt>) of each item's value. <tt>powerset</tt> 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>

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

or

<lang Haskell>powerset = foldr (\x acc -> acc ++ map (x:) acc) [[]]</lang>

Examples:

Line 1,014:

Line 865:

'''Alternate solution'''

A method using only set operations and set mapping is also possible. Ideally, <code>Set</code> 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.

A method using only set operations and set mapping is also possible. Ideally, <code>Set</code> 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

<lang Haskell>import qualified Data.Set as Set

type Set=Set.Set

unionAll :: (Ord a) => Set (Set a) -> Set a

Line 1,030:

Line 880:

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]]

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]]</lang>

=={{header|Icon}} and {{header|Unicon}}==

{{incomplete|Icon|It shows no proof of the cases with empty sets.}}{{incomplete|Unicon|It shows no proof of the cases with empty sets.}}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.

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)

The following version returns a set containing the powerset:

procedure power_set (s)

result := set ()

if *s = 0

Line 1,057:

Line 899:

}

return result

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

end

</lang>

To test the above procedure:

procedure main ()

every s := !power_set (set(1,2,3,4)) do { # requires '!' to generate items in the result set

writes ("[ ")

Line 1,069:

Line 905:

write ("]")

}

end

end</lang>

</lang>

Output:

Line 1,091:

Line 926:

[ 2 1 ]

</pre>

===Generator===

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

An alternative version, which generates each item in the power set in turn:

procedure power_set (s)

if *s = 0

then suspend set ()

Line 1,115:

Line 945:

write ("]")

}

end

end</lang>

</lang>

=={{header|J}}==

{{incomplete|J|It shows no proof of the cases with empty sets.}}There are a [http://www.jsoftware.com/jwiki/Essays/Power_Set number of ways] to generate a power set in J. Here's one:

There are a [http://www.jsoftware.com/jwiki/Essays/Power_Set number of ways] to generate a power set in J. Here's one:

For example:

Line 1,131:

Line 958:

AE

AC

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

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.

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.

=={{header|Java}}==

{{incomplete|Java|It shows no proof of the cases with empty sets.}}{{works with|Java|1.5+}}

===Recursion===

[[Category: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.

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)

<lang java5>public static ArrayList<String> getpowerset(int a[],int n,ArrayList<String> ps)

{

if(n<0)

Line 1,174:

Line 993:

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.

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) {

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

Line 1,200:

Line 1,016:

}

return ps;

}</lang>

}

</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 <tt>n</tt> elements in the original set, each combination can be represented by a binary string of length <tt>n</tt>. To get all possible combinations, all you need is a counter from 0 to 2n - 1. If the kth bit in the binary string is 1, the kth element of the original set is in this combination.

This implementation works on idea that each element in the original set can either be in the power set or not in it. With <tt>n</tt> elements in the original set, each combination can be represented by a binary string of length <tt>n</tt>. To get all possible combinations, all you need is a counter from 0 to 2n - 1. If the kth bit in the binary string is 1, the kth element of the original set is in this combination.<lang java5>public static <T extends Comparable<? super T>> LinkedList<LinkedList<T>> BinPowSet(

<lang java5>public static <T extends Comparable<? super T>> LinkedList<LinkedList<T>> BinPowSet(

LinkedList<T> A){

LinkedList<LinkedList<T>> ans= new LinkedList<LinkedList<T>>();

Line 1,221:

Line 1,034:

return ans;

}</lang>

=={{header|JavaScript}}==

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

{{incomplete|JavaScript|It shows no proof of the cases with empty sets.}}Uses a JSON stringifier from http://www.json.org/js.html{{works with|SpiderMonkey}}<lang javascript>function powerset(ary) {

<lang javascript>function powerset(ary) {

var ps = [[]];

for (var i=0; i < ary.length; i++) {

Line 1,239:

Line 1,048:

load('json2.js');

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

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

[[],[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]]

Outputs:

=={{header|Julia}}==

<lang julia>function powerset (x)

{{incomplete|Julia|It shows no proof of the cases with empty sets.}}<lang julia>function powerset (x)

result = {{}}

for i in x, j = 1:length(result)

Line 1,251:

Line 1,057:

end

result

end</lang>

end</lang>{{Out}}

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

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

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

{{},{1},{2},{1,2},{3},{1,3},{2,3},{1,2,3}}</pre>

=={{header|K}}==

{{incomplete|K|It shows no proof of the cases with empty sets.}}<lang K>

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

</lang>

</lang>Usage:<pre>

Usage:

ps "ABC"

(""

Line 1,270:

Line 1,072:

"AC"

"AB"

"ABC")

"ABC")</pre>

</lang>

=={{header|Logo}}==

<lang logo>to powerset :set

{{incomplete|Logo|It shows no proof of the cases with empty sets.}}<lang logo>to powerset :set

if empty? :set [output [[]]]

localmake "rest powerset butfirst :set

Line 1,284:

Line 1,084:

=={{header|Logtalk}}==

<lang logtalk>:- object(set).

{{incomplete|Logtalk|It shows no proof of the cases with empty sets.}}<lang logtalk>:- object(set).

:- public(powerset/2).

Line 1,314:

Line 1,114:

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>

=={{header|Lua}}==

{{incomplete|Lua|It shows no proof of the cases with empty sets.}}<lang lua>

--returns the powerset of s, out of order.

function powerset(s, start)

Line 1,351:

Line 1,150:

end

</lang>

=={{header|M4}}==

<lang M4>define(`for',

{{incomplete|M4|It shows no proof of the cases with empty sets.}}<lang M4>define(`for',

`ifelse($#, 0, ``$0'',

eval($2 <= $3), 1,

Line 1,375:

Line 1,173:

dnl

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

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

Output:

<pre>

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

</pre>

=={{header|Maple}}==

{{incomplete|Maple|It shows no proof of the cases with empty sets.}}<lang Maple>with(combinat):

with(combinat):

powerset({1,2,3,4});

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

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

</lang>

Output:

<pre>

{{}, {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}}

</pre>

=={{header|Mathematica}}==

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:

{{incomplete|Mathematica|It shows no proof of the cases with empty sets.}}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>

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

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

gives:

Subsets[list, {n, Infinity}] gives all the subsets that have n elements or more.

Subsets[list, n] gives all the subsets that have at most n elements.

Subsets[list, {n}] gives all the subsets that have exactly n elements.

Subsets[list, {m,n}] gives all the subsets that have between m and n elements.

=={{header|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)

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

Line 1,432:

Line 1,206:

Sample Usage:

Powerset of the set of the empty set.

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

<lang MATLAB>powerset({{}})

ans =

Line 1,445:

Line 1,218:

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

=={{header|Maxima}}==

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

{{incomplete|Maxima|It shows no proof of the cases with empty sets.}}<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>

=={{header|Nimrod}}==

<lang nimrod>import sets, hashes

{{incomplete|Nimrod|It shows no proof of the cases with empty sets.}}<lang nimrod>import sets, hashes

proc hash(x): THash =

Line 1,471:

Line 1,242:

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

=={{header|Objective-C}}==

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

{{incomplete|Objective-C|It shows no proof of the cases with empty sets.}}<lang objc>#import <Foundation/Foundation.h>

+ (NSArray *)powerSetForArray:(NSArray *)array {

Line 1,489:

Line 1,259:

return subsets;

}</lang>

=={{header|OCaml}}==

{{incomplete|OCaml|It shows no proof of the cases with empty sets.}}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) =

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

Line 1,510:

Line 1,276:

;;

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>

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>

=={{header|OPL}}==

{{incomplete|OPL|It shows no proof of the cases with empty sets.}}<lang OPL>{string} s={"A","B","C","D"};

{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)};

Line 1,526:

Line 1,285:

{

writeln(s2);

}</lang>{{out|which gives}}

}

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

</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"}]

</lang>

=={{header|Oz}}==

Oz has a library for finite set constraints. Creating a power set is a trivial application of that:

{{incomplete|Oz|It shows no proof of the cases with empty sets.}}Oz has a library for finite set constraints. Creating a power set is a trivial application of that:<lang oz>declare

<lang oz>declare

%% Given a set as a list, returns its powerset (again as a list)

fun {Powerset Set}

Line 1,560:

Line 1,305:

end

in

{Inspect {Powerset [1 2 3 4]}}</lang>

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

A more convential implementation without finite set constaints:

<lang oz>fun {Powerset2 Set}

case Set of nil then [nil]

Line 1,573:

Line 1,316:

=={{header|PARI/GP}}==

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

{{incomplete|PARI/GP|It shows no proof of the cases with empty sets.}}<lang parigp>vector(1<<#S,i,vecextract(S,i-1))</lang>

=={{header|Perl}}==

{{incomplete|Perl|It shows no proof of the cases with empty sets.}}Perl does not have a built-in set data-type. However, you can...

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

<ul>

<li>'''''Use a third-party module'''''

Line 1,599:

Line 1,341:

Output:

<pre>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))</pre>

<pre>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))</pre></li><li>'''''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 {

</li>

<li>'''''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 [https://metacpan.org/pod/Set::Tiny Set::Tiny])''

}</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 [https://metacpan.org/pod/Set::Tiny 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 <code>Set::Object</code> based solution above, but using list folding (with the help of Perl's <code>List::Util</code> core module) seems a little more elegant in this case:

We could implement the function as an imperative foreach loop similar to the <code>Set::Object</code> based solution above, but using list folding (with the help of Perl's <code>List::Util</code> core module) seems a little more elegant in this case:<lang perl>use List::Util qw(reduce);

<lang perl>use List::Util qw(reduce);

sub powerset {

Line 1,629:

Line 1,362:

my @subsets = powerset($set);

print $_->as_string, "\n" for @subsets;</lang>

print $_->as_string, "\n" for @subsets;</lang>{{out}}<pre>Set()

Output:

<pre>

Set()

Set(1)

Set(2)

Line 1,640:

Line 1,369:

Set(1 3)

Set(2 3)

Set(1 2 3)</pre></li><li>'''''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.

Set(1 2 3)

</pre>

Recursive solution:<lang perl>sub powerset {

</li>

<li>'''''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);

}</lang>

List folding solution:

<lang perl>use List::Util qw(reduce);

sub powerset {

@{( reduce { [@$a, map([@$_, $b], @$a)] } [[]], @_ )}

}</lang>{{out|Usage & output}}<lang perl>my @set = (1, 2, 3);

}</lang>

Usage & output:

<lang perl>my @set = (1, 2, 3);

my @powerset = powerset(@set);

Line 1,671:

Line 1,385:

print set_to_string(@powerset), "\n";</lang>

{{}, {1}, {2}, {1, 2}, {3}, {1, 3}, {2, 3}, {1, 2, 3}}</li></ul>

<pre>

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

</pre>

</li>

</ul>

=={{header|Perl 6}}==

{{incomplete|Perl 6|It shows no proof of the cases with empty sets.}}{{works with|rakudo|2014-02-25}}<lang perl6>sub powerset(Set $s) { $s.combinations.map(*.Set).Set }

say powerset set <a b c d>;</lang>{{out}}

<lang perl6>sub powerset(Set $s) { $s.combinations.map(*.Set).Set }

say powerset set <a b c d>;</lang>

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

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

<lang perl6>.say for <a b c d>.combinations</lang>{{out}}<pre>

<pre>

a

b

Line 1,704:

Line 1,407:

b c d

a b c d</pre>

=={{header|PHP}}==

{{incomplete|PHP|It shows no proof of the cases with empty sets.}}

<?php

<lang PHP><?php

function get_subset($binary, $arr) {

// based on true/false values in $binary array, include/exclude

Line 1,765:

Line 1,467:

print_power_sets(array('singleton'));

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

?></lang>{{out|Output in browser}}<pre>POWER SET of []

?>

</lang>

Output in browser:

<lang>

POWER SET of []

POWER SET of [singleton]

(empty)

Line 1,789:

Line 1,487:

a c dog

b c dog

a b c dog

a b c dog</pre>

</lang>

=={{header|PicoLisp}}==

<lang PicoLisp>(de powerset (Lst)

{{incomplete|PicoLisp|It shows no proof of the cases with empty sets.}}<lang PicoLisp>(de powerset (Lst)

(ifn Lst

(cons)

Line 1,800:

Line 1,496:

(mapcar '((X) (cons (car Lst) X)) L)

L ) ) ) )</lang>

=={{header|PL/I}}==

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

{{incomplete|PL/I|It shows no proof of the cases with empty sets.}}<lang pli>*process source attributes xref or(!);

/*--------------------------------------------------------------------

* 06.01.2014 Walter Pachl translated from REXX

Line 1,868:

Line 1,563:

End;</lang>

{{out}}<pre>{}

'''output'''

<pre>{}

{one}

{two}

Line 1,885:

Line 1,579:

{two,three,four}

{one,two,three,four}</pre>

=={{header|Prolog}}==

{{incomplete|Prolog|It shows no proof of the cases with empty sets.}}

===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).

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).

Line 1,900:

Line 1,590:

subseq( [], [_|_]).

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

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

subseq( [X|Xs], [_|Ys] ) :- append(_, [X|Zs], Ys), subseq(Xs, Zs).</lang>{{out}}<pre>?- powerset([1,2,3], X).

</lang>

Output :

<pre>?- powerset([1,2,3], X).

X = [[], [1], [1, 2], [1, 2, 3], [1, 3], [2], [2, 3], [3]].

Line 1,914:

Line 1,601:

X = 1,

Y = 2.</pre>

===Single-Functor Definition===

<lang Prolog>power_set( [], [[]]).

Line 1,921:

Line 1,607:

maplist( append([X]), PS1, PS2 ), % i.e. prepend X to each PS1

append(PS1, PS2, PS).</lang>

{{out}}<pre>?- power_set([1,2,3,4,5,6,7,8], X), length(X,N), writeln(N).

Output :

256</pre>

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

256

</pre>

===Constraint Handling Rules===

CHR is a programming language created by '''Professor Thom Frühwirth'''.

{{libheader|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)).

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.

Line 1,937:

Line 1,619:

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

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

Line 1,945:

Line 1,626:

chr_power_set(B).

empty_element @ chr_power_set([], _) <=> chr_power_set([]).</lang>{{out|Example of output}}<pre> ?- 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],[]] .</pre>

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

</lang>

Example of output :

<pre> ?- 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],[]] .

</pre>

=={{header|PureBasic}}==

This code is for console mode.

{{incomplete|PureBasic|It shows no proof of the cases with empty sets.}}This code is for console mode.<lang PureBasic>If OpenConsole()

<lang PureBasic>If OpenConsole()

Define argc=CountProgramParameters()

If argc>=(SizeOf(Integer)*8) Or argc<1

Line 1,979:

Line 1,653:

PrintN("}")

EndIf

EndIf</lang>

EndIf</lang>{{out|Sample output}}

Sample output

<pre>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}}</pre>

=={{header|Python}}==

{{incomplete|Python|It shows no proof of the cases with empty sets.}}

<lang python>def list_powerset(lst):

# the power set of the empty set has one element, the empty set

Line 2,006:

Line 1,678:

def powerset(s):

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

return frozenset(map(frozenset, list_powerset(list(s))))</lang>

{{out|Example}}<pre>

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

Example:

<pre>

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

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

Line 2,018:

Line 1,686:

</pre>

==== 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

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):

<lang python>def powersetlist(s):

r = [[]]

for e in s:

Line 2,027:

Line 1,694:

s= [0,1,2,3]

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

print "\npowersetlist(%r) =\n %r" % (s, powersetlist(s))</lang>{{out|Sample output}}

r: [[]] e: 0

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

Sample output:

~~<pre>~~r: [[]] e: 0

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

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

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

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

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

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]]

</pre>

=== 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)

(Note that only frozensets can be members of a set in the second function)<lang python>def powersequence(val):

<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

Line 2,067:

Line 1,730:

'''

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.

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):

def p(l):

if not l: return [[]]

return p(l[1:]) + [[l[0]] + x for x in p(l[1:])]

return p(l[1:]) + [[l[0]] + x for x in p(l[1:])]</lang>

</lang>

===Python: Standard documentation===

Pythons [http://docs.python.org/3/library/itertools.html?highlight=powerset#itertools-recipes documentation] has a method that produces the groupings, but not as sets:

Pythons [http://docs.python.org/3/library/itertools.html?highlight=powerset#itertools-recipes documentation] has a method that produces the groupings, but not as sets:<lang python>>>> from pprint import pprint as pp

<lang python>>>> from pprint import pprint as pp

>>> from itertools import chain, combinations

>>>

Line 2,106:

Line 1,761:

(4,)}

>>> </lang>

=={{header|Qi}}==

{{incomplete|Qi|It shows no proof of the cases with empty sets.}}{{trans|Scheme}}<lang qi>(define powerset

(define powerset

[] -> [[]]

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

(powerset As)))

(powerset As)))</lang>

</lang>

=={{header|R}}==

===Non-recursive version===

The conceptual basis for this algorithm is the following:

The conceptual basis for this algorithm is the following:<lang>for each element in the set:

<lang>for each element in the set:

for each subset constructed so far:

new subset = (subset + element)

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){

</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()

ps[[1]] = numeric() #Start with the empty set.

Line 2,139:

Line 1,784:

}

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.

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===

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.

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

<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.

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

<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)>>}}

=={{header|Racket}}==

{{incomplete|Racket|It shows no proof of the cases with empty sets.}}<lang racket>;;; Direct translation of 'functional' ruby method

;;; 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)))))))

(λ(inner-set) (set-add inner-set element)))))))</lang>

</lang>

=={{header|Rascal}}==

{{incomplete|Rascal|It shows no proof of the cases with empty sets.}}

import Set;

<lang rascal>import Set;

public set[set[&T]] PowerSet(set[&T] s) = power(s);

public set[set[&T]] PowerSet(set[&T] s) = power(s);</lang>{{out|An example output}}

rascal>PowerSet({1,2,3,4})

</lang>

set[set[int]]: {

An example output:

rascal>PowerSet({1,2,3,4})

set[set[int]]: {

{4,3},

{4,2,1},

Line 2,198:

Line 1,827:

{3,2,1},

{}

}

</lang>

=={{header|REXX}}==

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

{{incomplete|REXX|It shows no proof of the cases with empty sets.}}<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*/

Line 2,231:

Line 1,858:

end /*u*/

return 0</lang>

~~'''output'''~~ when using the default input:

<pre> 1 {}

1 {}

2 {one}

3 {two}

Line 2,248:

Line 1,874:

14 {one,three,four}

15 {two,three,four}

16 {one,two,three,four}

16 {one,two,three,four}</pre>

</pre>

=={{header|Ruby}}==

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

{{incomplete|Ruby|It shows no proof of the cases with empty sets.}}<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

Line 2,294:

Line 1,918:

p %w(one two three).func_power_set

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

p Set[1,2,3].powerset</lang>{{out}}<pre>

<pre>

[[], [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}>}>

</pre>

=={{header|SAS}}==

{{incomplete|SAS|It shows no proof of the cases with empty sets.}}<lang SAS>options mprint mlogic symbolgen source source2;

options mprint mlogic symbolgen source source2;

%macro SubSets (FieldCount = );

Line 2,356:

Line 1,976:

%end;

%Mend SubSets;

%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.{{out}}<pre>

</lang>

You can then call the macro as:

%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:

<pre>

Obs F1 F2 F3 F4 F5 RowCount

1 . . . . . 1

Line 2,400:

Line 2,010:

30 1 1 1 . 1 30

31 1 1 1 1 . 31

32 1 1 1 1 1 32

32 1 1 1 1 1 32</pre>

</pre>

=={{header|Scala}}==

[[Category:Scala Implementations]]~~<lang scala>import scala.compat.Platform.currentTime~~

[[Category:Scala Implementations]]

Testing the cases:

#{1,2,3,4}

#<math>\mathcal{P}</math>(<math>\varnothing</math>) = { <math>\varnothing</math> }

#<math>\mathcal{P}</math>({<math>\varnothing</math>}) = { <math>\varnothing</math>, { <math>\varnothing</math> } }

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

object Powerset extends App {

Line 2,411:

Line 2,024:

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>

}</lang>

Another option that produces lazy sequence of the sets:

<lang scala>def powerset[A](s: Set[A]) = (0 to s.size).map(s.toSeq.combinations(_)).reduce(_ ++ _).map(_.toSet)</lang>

=={{header|Scheme}}==

{{incomplete|Scheme|It shows no proof of the cases with empty sets.}}{{trans|Common Lisp}}

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

(if (null? set)

Line 2,432:

Line 2,043:

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

(newline)</lang>

(newline)</lang>{{out}}

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)

Line 2,465:

Line 2,074:

(2)

(3)

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

()</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>

</lang>{{out}} '((e d c b a)

Output:<lang output>

'((e d c b a)

(e d c b)

(e d c a)

Line 2,508:

Line 2,111:

(a)

())

</lang>

=={{header|Seed7}}==

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

{{incomplete|Seed7|It shows no proof of the cases with empty sets.}}<lang seed7>$ include "seed7_05.s7i";

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

Line 2,539:

Line 2,140:

writeln(aSet);

end for;

end func;</lang>

end func;</lang>{{out}}<pre>{}

Output:

<pre>

{}

{1}

{2}

Line 2,558:

Line 2,155:

{1, 3, 4}

{2, 3, 4}

{1, 2, 3, 4}

{1, 2, 3, 4}</pre>

</pre>

=={{header|Smalltalk}}==

{{incomplete|Smalltalk|It shows no proof of the cases with empty sets.}}

Code from [http://smalltalk.gnu.org/blog/bonzinip/fun-generators Bonzini's blog]

Code from [http://smalltalk.gnu.org/blog/bonzinip/fun-generators Bonzini's blog]<lang smalltalk>Collection extend [

<lang smalltalk>Collection extend [

power [

^(0 to: (1 bitShift: self size) - 1) readStream collect: [ :each || i |

Line 2,571:

Line 2,165:

self select: [ :elem | (each bitAt: (i := i + 1)) = 1 ] ]

]

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

].</lang>

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

each asArray printNl ].

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

each asArray printNl ].</lang>

=={{header|Standard ML}}==

{{incomplete|Standard ML|It shows no proof of the cases with empty sets.}}Version for lists:<lang sml>fun subsets xs = foldr (fn (x, rest) => rest @ map (fn ys => x::ys) rest) [[]] xs</lang>

version for lists:

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

=={{header|Tcl}}==

<lang tcl>proc subsets {l} {

{{incomplete|Tcl|It shows no proof of the cases with empty sets.}}<lang tcl>proc subsets {l} {

set res [list [list]]

foreach e $l {

Line 2,592:

Line 2,180:

return $res

}

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

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

{} 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}

Output:

===Binary Count Method===

<lang tcl>proc powersetb set {

Line 2,608:

Line 2,195:

return $res

}</lang>

=={{header|TXR}}==

{{incomplete|TXR|It shows no proof of the cases with empty sets.}}

{{trans|Common Lisp}}The power set function can be written concisely like this:<lang txr>(defun power-set (s)

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)

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)

Line 2,631:

Line 2,209:

{@(rep)@pset, @(last)@pset@(empty)@(end)}

@ (end)

@(end)</lang>

@(end)</lang><pre>$ txr rosetta/power-set.txr 1 2 3

<pre>$ txr rosetta/power-set.txr 1 2 3

{1, 2, 3}

{1, 2}

Line 2,641:

Line 2,217:

{2}

{3}

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

{}</pre>

What is not obvious is that the above <code>power-set</code> function generalizes to strings and vectors.

<lang txr>@(do (defun power-set (s)

(reduce-right

(op append (mapcar (op cons @@1) @2) @2)

Line 2,651:

Line 2,223:

(prinl (power-set "abc"))

(prinl (power-set ""))

(prinl (power-set #(1 2 3))))</lang>

(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)

<pre>txr power-set-generic.txr

((~~#\a~~ ~~#\b~~ ~~#\c~~) (~~#\a~~ ~~#\b~~) (~~#\a~~ ~~#\c~~) (~~#\a~~) (~~#\b~~ ~~#\c~~) (~~#\b~~) (~~#\c~~) nil)

((1 2 3) (1 2) (1 3) (1) (2 3) (2) (3) nil)

((nil) nil)

((1 2 3) (1 2) (1 3) (1) (2 3) (2) (3) nil)</pre>

=={{header|UnixPipes}}==

{{incomplete|UnixPipes|It shows no proof of the cases with empty sets.}}<lang ksh>| cat A

| cat A

a

b

Line 2,681:

Line 2,249:

b

b c

c</lang>

c

</lang>

=={{header|UNIX Shell}}==

{{incomplete|UNIX Shell|It shows no proof of the cases with empty sets.}}

From [http://www.catonmat.net/blog/set-operations-in-unix-shell/ here]

<lang bash>p() { [ $# -eq 0 ] && echo || (shift; p "$@") | while read r ; do echo -e "$1 $r\n$r"; done }</lang>

Line 2,695:

Line 2,263:

=={{header|Ursala}}==

{{incomplete|Ursala|It shows no proof of the cases with empty sets.}}

Sets are a built in type constructor in Ursala, represented as

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.

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:

Line 2,705:

Line 2,270:

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

output:

{

<pre>{

{},

{'a'},

Line 2,722:

Line 2,287:

{'c'},

{'c','d'},

{'d'}}~~</pre>~~

{'d'}}

=={{header|V}}==

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)

its implementation in std.v is (like joy)

<lang v>[powerlist

[null?]

Line 2,735:

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[uncons]

[dup swapd [cons] map popd swoncat]

linrec].

linrec].</lang>

</lang>

=={{header|zkl}}==

Using a combinations function, build the power set from combinations of 1,2,... items.

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foreach n in (5){

ps:=pwerSet((1).pump(n,List)); ps.println(" Size = ",ps.len());

}</pre>

}

L(L()) Size = 1

</pre>

L(L(),L(1)) Size = 2

<pre>

L(L()) Size = 1

L(L(),L(1),L(2),L(1,2)) Size = 4

L(L(),L(1)) Size = 2

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(1,2)) ~~Size =~~ 4

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(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

</pre>