Programming Comparisons: Power Set

Problem: give the smallest possible complete subroutine/function/method definition in a language of your choice which, when called with a list, array, whatever, returns the power set of that list. It may be in any order, but must not contain duplicates.
e.g.: p([1, 2, 3]) returns: [[], [1], [2], [3], [2, 3], [1, 2], [1, 3], [1, 2, 3]]

haskell: 46 (86 bz2)

{- p [1,2,3] -}
p=foldr(\h t->t++map(h:)t)[[]]

perl: 53 (93 bz2)

# p(1, 2, 3)
sub p{my$y=pop;$y?map{[$y,@$_],$_}&p:[]}

haskell*: 64 (102 bz2)

{- p [0..] -}
p l=[]:do{(n,x)<-zip[0..]l;map(x:)$take (2^n)$p l}

ruby: 69 (105 bz2)

# r([1, 2, 3])
def r(l)l==[]?[l]:(a=l.pop;s=r l;s+s.map{|i|[a]+i})end

python: 73 (99 bz2)

# p([1, 2, 3])
p=lambda l:l and p(l[1:])+[l[:1]+y for y in p(l[1:])]or[l]

ocaml: 77 (117 bz2)

(* p [1; 2; 3];; *)
let rec p=function a::b->p b@List.map((@)[a])(p b)|l->[l]

coffeescript: 85 (117 bz2)

# p [1, 2, 3]
p=(l)->return[[]]if!l.length;s=p l[1..];s.concat([l[0],x...]for x in s)

prolog: 89 (112 bz2) <% '%' %> p([1,2,3],X). s([],L).s([A|B],L):-member(A,L),s(B,L).p(L,X):-setof(Z,s(Z,L),X).

lisp: 88 (109 bz2)

; (p '(1 2 3))
(defun p(l)(if
l(mapcan #'(lambda(x)`(,x(,(car l).,x)))(p(cdr l)))`(,l)))

javascript: 152 (158 bz2)

// p([1, 2, 3])
function p(l){if(!l.length)return[[]];var a=[];var s=p(l.slice(1));for(var
i=0;i<s.length;i++)a.push(s[i],[l[0]].concat(s[i]));return a}

php: 157 (150 bz2)

# p(array(1, 2, 3))
function p($l){if(!$l)return
array($l);$a=array();foreach(p(array_slice($l,1))as$i){$a[]=$i;$a[]=array_merge(array($l[0]),$i);}return$a;}

gnu smalltalk: 163 (170 bz2)

"(Set from: #(1 2 3)) p"
Set extend[p[|a s|^self isEmpty
ifTrue:[Set with:self]ifFalse:[a:=self remove:self anyOne. s:=self p.
s+(s collect:[:i|i+(Set with:a)])]]]

*: The second haskell entry works for infinite lists!