Code golf: find all anagrams

Question

A word is an anagram if the letters in that word can be re-arranged to form a different word.

Task:

• The shortest source code by character count to find

Answer 1

C++, 542 chars

#include <iostream>
#include <map>
#include <vector>
#include <boost/algorithm/string.hpp>
#define ci const_iterator
int main(){using namespace std;typedef string s;typedef vector<s> vs;vs l;
copy(istream_iterator<s>(cin),istream_iterator<s>(),back_inserter(l));map<s, vs> r;
for (vs::ci i=l.begin(),e=l.end();i!=e;++i){s a=boost::to_lower_copy(*i);
sort(a.begin(),a.end());r[a].push_back(*i);}for (map<s,vs>::ci i=r.begin(),e=r.end();
i!=e;++i)if(i->second.size()>1)*copy(i->second.begin(),i->second.end(),
ostream_iterator<s>(cout," "))="\n";}

Answer 2

Python, O(n^2)

import sys;
words=sys.stdin.readlines()
def s(x):return sorted(x.lower());
print '\n'.join([''.join([a.replace('\n',' ') for a in words if(s(a)==s(w))]) for w in words])

Answer 3

Python, 167 characters, includes I/O

import sys
d={}
for l in sys.stdin.readlines():
 l=l[:-1]
 k=''.join(sorted(l)).lower()
 d[k]=d.pop(k,[])+[l]
for k in d:
 if len(d[k])>1: print(' '.join(d[k]))

Without the input code (i.e. if we assume the wordlist already in a list w), it's only 134 characters:

d={}
for l in w:
 l=l[:-1]
 k=''.join(lower(sorted(l)))
 d[k]=d.pop(k,[])+[l]
for k in d:
 if len(d[k])>1: print(' '.join(d[k]))

Answer 4

AWK - 119

{split(toupper($1),a,"");asort(a);s="";for(i=1;a[i];)s=a[i++]s;x[s]=x[s]$1" "}
END{for(i in x)if(x[i]~/ .* /)print x[i]}

AWK does not have a join function like Python, or it could have been shorter...

It assumes uppercase and lowercase as different.

Answer 5

Perl, 59 characters

chop,$_{join'',sort split//,lc}.="$_ "for<>;/ ./&&say for%_

Note that this requires Perl 5.10 (for the say function).

Answer 6

Ruby, 94 characters

h={};(h[$_.upcase.bytes.sort]||=[])<<$_ while gets&&chomp;h.each{|k,v|puts v.join' 'if v.at 1}