Write a function f(m, G) that accepts as its arguments a mapping m, and a set/list of distinct, non-negative integers G.

m should map pairs of integers in G to new integers in G. (G, m) is guaranteed to form a finite abelian group, but any element of G may be the identity.

There is an important theorem that says:

[Each finite abelian group] is isomorphic to a direct product of cyclic groups of prime power order.

f must return a list of prime powers [p1, ... pn] in ascending order such that G is isomorphic to Z_p1 times ... times Z_pn


  • f((a, b) → (a+b) mod 4, [0, 1, 2, 3]) should return [4], as the parameters describe the group Z4.

  • f((a, b) → a xor b, [0, 1, 2, 3]) should return [2, 2], as the parameters describe a group isomorphic to Z2 × Z2.

  • f((a, b) → a, [9]) should return [], as the parameters describe the trivial group; i.e., the product of zero cyclic groups.

  • Define m as follows:

    (a, b) → (a mod 3 + b mod 3) mod 3
           + ((floor(a / 3) + floor(b / 3)) mod 3) * 3
           + ((floor(a / 9) + floor(b / 9)) mod 9) * 9

    Then f(m, [0, 1, ..., 80]) should return [3, 3, 9], as this group is isomorphic to Z3 × Z3 × Z9


  • m may either be a function (or function pointer to some function) Int × Int → Int, or a dictionary mapping pairs in G × G to new elements of G.

  • f may take its parameters in the opposite order, i.e. you may also implement f(G, m).

  • Your implementation should theoretically work for arbitrarily large inputs, but need not actually be efficient.

  • There is no limitation on using built-ins of any kind.

  • Standard rules apply. Shortest code in bytes wins.


For your score to appear on the board, it should be in this format:

# Language, Bytes

var QUESTION_ID=67252,OVERRIDE_USER=3852;function answersUrl(e){return"http://api.stackexchange.com/2.2/questions/"+QUESTION_ID+"/answers?page="+e+"&pagesize=100&order=desc&sort=creation&site=codegolf&filter="+ANSWER_FILTER}function commentUrl(e,s){return"http://api.stackexchange.com/2.2/answers/"+s.join(";")+"/comments?page="+e+"&pagesize=100&order=desc&sort=creation&site=codegolf&filter="+COMMENT_FILTER}function getAnswers(){jQuery.ajax({url:answersUrl(answer_page++),method:"get",dataType:"jsonp",crossDomain:!0,success:function(e){answers.push.apply(answers,e.items),answers_hash=[],answer_ids=[],e.items.forEach(function(e){e.comments=[];var s=+e.share_link.match(/\d+/);answer_ids.push(s),answers_hash[s]=e}),e.has_more||(more_answers=!1),comment_page=1,getComments()}})}function getComments(){jQuery.ajax({url:commentUrl(comment_page++,answer_ids),method:"get",dataType:"jsonp",crossDomain:!0,success:function(e){e.items.forEach(function(e){e.owner.user_id===OVERRIDE_USER&&answers_hash[e.post_id].comments.push(e)}),e.has_more?getComments():more_answers?getAnswers():process()}})}function getAuthorName(e){return e.owner.display_name}function process(){var e=[];answers.forEach(function(s){var r=s.body;s.comments.forEach(function(e){OVERRIDE_REG.test(e.body)&&(r="<h1>"+e.body.replace(OVERRIDE_REG,"")+"</h1>")});var a=r.match(SCORE_REG);a&&e.push({user:getAuthorName(s),size:+a[2],language:a[1],link:s.share_link})}),e.sort(function(e,s){var r=e.size,a=s.size;return r-a});var s={},r=1,a=null,n=1;e.forEach(function(e){e.size!=a&&(n=r),a=e.size,++r;var t=jQuery("#answer-template").html();t=t.replace("{{PLACE}}",n+".").replace("{{NAME}}",e.user).replace("{{LANGUAGE}}",e.language).replace("{{SIZE}}",e.size).replace("{{LINK}}",e.link),t=jQuery(t),jQuery("#answers").append(t);var o=e.language;/<a/.test(o)&&(o=jQuery(o).text()),s[o]=s[o]||{lang:e.language,user:e.user,size:e.size,link:e.link}});var t=[];for(var o in s)s.hasOwnProperty(o)&&t.push(s[o]);t.sort(function(e,s){return e.lang>s.lang?1:e.lang<s.lang?-1:0});for(var c=0;c<t.length;++c){var i=jQuery("#language-template").html(),o=t[c];i=i.replace("{{LANGUAGE}}",o.lang).replace("{{NAME}}",o.user).replace("{{SIZE}}",o.size).replace("{{LINK}}",o.link),i=jQuery(i),jQuery("#languages").append(i)}}var ANSWER_FILTER="!t)IWYnsLAZle2tQ3KqrVveCRJfxcRLe",COMMENT_FILTER="!)Q2B_A2kjfAiU78X(md6BoYk",answers=[],answers_hash,answer_ids,answer_page=1,more_answers=!0,comment_page;getAnswers();var SCORE_REG=/<h\d>\s*([^\n,]*[^\s,]),.*?(\d+)(?=[^\n\d<>]*(?:<(?:s>[^\n<>]*<\/s>|[^\n<>]+>)[^\n\d<>]*)*<\/h\d>)/,OVERRIDE_REG=/^Override\s*header:\s*/i;
body{text-align:left!important}#answer-list,#language-list{padding:10px;width:290px;float:left}table thead{font-weight:700}table td{padding:5px}
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script> <link rel="stylesheet" type="text/css" href="//cdn.sstatic.net/codegolf/all.css?v=83c949450c8b"> <div id="answer-list"> <h2>Leaderboard</h2> <table class="answer-list"> <thead> <tr><td></td><td>Author</td><td>Language</td><td>Size</td></tr></thead> <tbody id="answers"> </tbody> </table> </div><div id="language-list"> <h2>Winners by Language</h2> <table class="language-list"> <thead> <tr><td>Language</td><td>User</td><td>Score</td></tr></thead> <tbody id="languages"> </tbody> </table> </div><table style="display: none"> <tbody id="answer-template"> <tr><td>{{PLACE}}</td><td>{{NAME}}</td><td>{{LANGUAGE}}</td><td>{{SIZE}}</td><td><a href="{{LINK}}">Link</a></td></tr></tbody> </table> <table style="display: none"> <tbody id="language-template"> <tr><td>{{LANGUAGE}}</td><td>{{NAME}}</td><td>{{SIZE}}</td><td><a href="{{LINK}}">Link</a></td></tr></tbody> </table>

  • \$\begingroup\$ If m is allowed to be a dictionary, could you provide the test cases as dictionaries as well? \$\endgroup\$ Dec 21, 2015 at 17:20
  • \$\begingroup\$ I considered it, but they're pretty big, especially the last case (thousands of key-value pairs), and I can't think of a very good format for them. It's probably easier for answerers to copy the function definitions, and then construct the dictionaries themselves (with something like for a in G: for b in G: d[(a, b)] = m(a, b)). \$\endgroup\$
    – Lynn
    Dec 21, 2015 at 17:25
  • \$\begingroup\$ I think it is correct. I can't make sense of your paste well enough to verify what is going on, but this should prove it: bpaste.net/show/5821182a9b48 \$\endgroup\$
    – Lynn
    Dec 21, 2015 at 20:55
  • \$\begingroup\$ To help wrap your head around it: it operates on ternary numbers with trits in the format AABC, treating them as triples (A, B, C), with pairwise addition modulo (9, 3, 3). \$\endgroup\$
    – Lynn
    Dec 21, 2015 at 20:58
  • \$\begingroup\$ Oh, I just realized my (very stupid) mistake. Thank you for your snippet! \$\endgroup\$
    – flawr
    Dec 21, 2015 at 21:27

2 Answers 2


Matlab, 326 bytes

With some group theory the idea is quite simple: Here the TL;DR Calculate all possible orders of elements of the group. Then find the biggest subgroup of a certain prime power order and "factorize" it out of the group, rinse, repeat.

function r=c(h,l)

                            %factorize group order
P=unique(f);                %prime factors
for k=1:numel(P);
    E(k)=sum(f==P(k));    %exponents of unique factors

                            %calculate the order O of each element
for k=2:N+1;


    O(l==L & O<0)=k-1


O=unique(O);               % (optional, just for speedupt)
                           % for each prime,find the highest power that
                           % divides any of the orders of the element, and
                           % each time substract that from the remaining
                           % exponent in the prime factorization of the
                           % group order
for p=1:nnz(P);                          % loop over primes
    while E(p)>1;                        % loop over remaining exponent
        for e=E(p):-1:1;                 % find the highest exponent
            if any(B)
                R=[R,P(p)^e];            % if found, add to list
    if E(p)==1;

Example inputs:

L = 0:3;
L = 0:80;
h=@(a,b)mod(mod(a,3)+mod(b,3),3)+mod(floor(a/3)+floor(b/3),3)*3+ mod(floor(a/9)+floor(b/9),9)*9; 

Golfed version:

function r=c(h,l);N=numel(L);f=factor(N);P=unique(f);for k=1:numel(P);E(k)=sum(f==P(k));end;O=L*0-1;l=L;for k=2:N+1;l=h(l,L);O(l==L&O<0)=k-1;end;R=[];for p=1:nnz(P);while E(p)>1;for e=E(p):-1:1;B=mod(O,P(p)^e)==0; if any(B);R=[R,P(p)^e]; O(B)=O(B)/(P(p)^e);E(p)=E(p)-e;break;end;end;end;if E(p)==1;R=[R,P(p)];end;end;r=sort(R)

GAP, 159 111 bytes

GAP allows us to simply construct a group by a multiplication table and compute its abelian invariants:

ai:=    # the golfed version states the function w/o assigning it
  local t;
  # t is inlined in the golfed version
  return AbelianInvariants(GroupByMultiplicationTable(t));

The old version

The finitely presented group with generators G and relations a*b=m(a,b) (for all a, b from G) is the group (G,m) we started with. We can create it and compute its abelian invariants with GAP:

ai:=    # the golfed version states the function w/o assigning it
  local F,n,rels;
                  F.(i)*F.(j)/F.(Position(G,m(G[i],G[j]))) ) ));
  # rels is inlined in the golfed version
  return AbelianInvariants(F/rels);


m1:=function(a,b) return (a+b) mod 4; end;
# I don't feel like implementing xor
m3:=function(a,b) return 9; end;
  return (a+b) mod 3 # no need for inner mod
         + ((QuoInt(a,3)+QuoInt(b,3)) mod 3) * 3
         + ((QuoInt(a,9)+QuoInt(b,9)) mod 9) * 9;

Now we can do:

gap> ai(m1,[0..3]);
[ 4 ]

Actually, we are not restricted to using lists of integers. Using the correct domain, we can just use the general plus:

ai(\+,List(Integers mod 4));
[ 4 ]

So I can essentially do the second example using that its group is isomorphic to the additive group of the 2 dimensional vector space over the field with 2 elements:

gap> ai(\+,List(GF(2)^2));
[ 2, 2 ]

And the remaining examples:

gap> ai(m3,[9]);
[  ]
gap> ai(m4,[0..80]);
[ 3, 3, 9 ]

Additional remarks

In the old version, m did not need to define a group composition for G. If m(a,b)=m(a,c), that just says that b=c. So we could do ai(m1,[0..5]) and ai(m3,[5..15]). The new version will fail horrible in these cases, as will both versions if m returns values that are not in G.

If (G,m) is not abelian, we get a description of the abelianized version of it, that is its biggest abelian factor group:

gap> ai(\*,List(SymmetricGroup(4)));
[ 2 ]

This is what AbelianInvariants is usually used for, we would normally just call AbelianInvariants(SymmetricGroup(4)).

The golfed version

function(m,G)return AbelianInvariants(GroupByMultiplicationTable(List(G,a->List(G,b->Position(G,m(a,b))))));end

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.