# Finding the spanning subgraphs of a complete bipartite graph

Let $K_{(m,n)}$ be the complete bipartite graph with $m$ and $n$ being the number of vertices in each partition. Is there an efficient way to list down or construct all its spanning subgraphs up to isomorphism?

I tried finding the spanning subgraphs for small $m$ and $n$ and what I am doing is I start by distributing edges. The number of edges is greater than or equal to $0$ and less than or equal to $mn$. There is only one spanning subgraph with $0$ and $mn$ edges. There is only one spanning subgraph with $1$ edge also. For $2$ edges there are two if $m+n\geq 4$ and only one otherwise. Then I proceed until I have used up all the possible number of edges. This is quite tedious.

Can anyone suggest an alternative method? I figured some programs might be able to enumerate all the spanning subgraphs fast so there must be a way on how they do it.

Any help or idea is appreciated. Thank you!

UPDATE:

I'm also interested on the number of these graphs. How many spanning subgraphs does $K_{(m,n)}$ have? I've read somewhere that I should use Polya's Enumeration Theorem. I'm not yet familiar on how to use that. I'd be really thankful if someone could give a hint and point me to the right direction. Thanks!

Introductory remark. The following discussion uses two different definition of spanning subgraphs of $$K_{n,m}$$, one being subgraphs with the same vertex set and the second subgraphs with the same vertex set where there are no isolated vertices. The first of these is equivalent to coloring the graph with two colors. Call these two model $$Q$$ and model $$P$$ respectively.

The goal here is to enumerate spanning subgraphs of $$K_{n,m}$$ where we will treat the simple case where even if $$n=m$$ there are no flips above a central vertical axis i.e. no reflections. We can do much better than NAUTY as we are only counting these graphs as opposed to enumerating them. We use the Polya Enumeration Theorem (PET) to obtain the count of all non-isomorphic subgraphs of $$K_{n,m}$$ (model $$Q$$) and the principle of inclusion-exclusion (PIE) to extract the count of the spanning subgraphs (model $$P$$).

We will consult NAUTY just the same to get sample data to match against in our mathematical analysis. The following Perl script was used.

#! /usr/bin/perl -w
#

sub decode_graph {
my ($str) = @_; sub R { my (@args) = map { sprintf "%06b",$_;
} @_;
join '', @args;
}

my (@ents) = map {
ord($_) - 63 } split //,$str;

my $n = shift @ents; my @adj_data = split //, R(@ents); my$adj = []; my $pos = 0; for(my$ind2 = 1; $ind2 <$n; $ind2++){ for(my$ind1 = 0; $ind1 <$ind2; $ind1++){$adj->[$ind1]->[$ind2] = $adj_data[$pos];
$adj->[$ind2]->[$ind1] =$adj_data[$pos];$pos++;
}
}

return $adj; } MAIN: { my$mx = shift || 3;

die "out of range for GENBG: $mx" if 2*$mx < 2 || 2*$mx > 32; for(my$comp_a = 1; $comp_a <=$mx; $comp_a++){ for(my$comp_b = 1; $comp_b <=$mx; $comp_b++){ my$vcount = $comp_a +$comp_b;

my $cmd = sprintf "./genbg %d %d",$comp_a, $comp_b; my$count = 0;

open GENBG, "$cmd 2>/dev/null|"; while(my$bp = <GENBG>){
chomp $bp; my$adj = decode_graph $bp; for($v = 0; $v <$vcount; $v++){ my$deg = 0;
for(my $w = 0;$w < $vcount;$w++){
my $ent =$adj->[$v]->[$w];
$deg++ if defined($ent) && $ent == 1; } last if$deg == 0;
}

$count++ if$v == $vcount; } close GENBG; printf " " if$comp_b > 1;
printf "%06d", $count; } printf "\n"; }; }  This gave the following table: $ ./scripts/bipartite.pl 6
000001 000001 000001 000001 000001 000001
000001 000003 000005 000008 000011 000015
000001 000005 000017 000042 000091 000180
000001 000008 000042 000179 000633 002001
000001 000011 000091 000633 003835 020755
000001 000015 000180 002001 020755 200082


Now for the mathematics. We use the Polya Enumeration Theorem as conjectured by the OP. To do this we need the cycle index of the action on the edges of the group that permutes the vertices in partition $$A$$ of size $$n$$ according to the symmetric group $$S_n$$ and the vertices in partition $$B$$ of size $$m$$ according to $$S_m.$$

These cycle indices are easy to compute and we do not need to iterate over all $$n!\times m!$$ pairs of permutations (acting on $$A$$ and $$B$$) but instead it is sufficient to iterate over pairs of terms from the cycle indices $$Z(S_n)$$ and $$Z(S_m)$$ of the symmetric groups $$S_n$$ and $$S_m$$ according to their multiplicities to obtain the cycle index $$Z(Q_{n,m})$$ of the combined action on $$A$$ and $$B$$. The number of terms here is the much better count of the number of partitions of $$n$$ and $$m$$ (upper bound).

The classic approach to the calculation of these cycle indices is based on the simple observation that for two cycles, one of length $$l_1$$ from a permutation $$\alpha$$ of $$A$$ and another of length $$l_2$$ from a column permutation $$\beta$$ of $$B$$ their contribution to the disjoint cycle decomposition product for $$(\alpha,\beta)$$ in the cycle index $$Z(Q_{n,m})$$ is by inspection

$$a_{\mathrm{lcm}(l_1, l_2)}^{l_1 l_2 / \mathrm{lcm}(l_1, l_2)} = a_{\mathrm{lcm}(l_1, l_2)}^{\gcd(l_1, l_2)}.$$

Once we have the cycle indices we evaluate $$Z(Q_{n,m})(1+z)$$ which is the standard substitution to produce the OGF. If we are only looking to obtain the count, we may use $$Z(Q_{n,m})_{\Large a_1=a_2=a_3=\cdots=2}.$$

Here is an example: $$Z(Q_{3,4}) = {\frac {{a_{{1}}}^{12}}{144}}+1/24\,{a_{{2}}}^{3}{a_{{1}}}^{6}+1/18\,{a_{{3 }}}^{3}{a_{{1}}}^{3}+1/12\,{a_{{2}}}^{6}+1/6\,{a_{{4}}}^{3} \\+1/48\,{a_{{2}}} ^{4}{a_{{1}}}^{4}+1/8\,{a_{{1}}}^{2}{a_{{2}}}^{5}+1/6\,a_{{1}}a_{{2}}a_{{3} }a_{{6}}+1/8\,{a_{{3}}}^{4} \\+1/12\,{a_{{3}}}^{2}a_{{6}}+1/24\,{a_{{6}}}^{2}+ 1/12\,a_{{12}}.$$ The substituted cycle index becomes $$Z(Q_{3,4})(1+z) = {z}^{12}+{z}^{11}+3\,{z}^{10}+6\,{z}^{9}+11\,{z}^{8} \\ +13\,{z}^{7}+17\,{z}^{6 }+13\,{z}^{5}+11\,{z}^{4}+6\,{z}^{3}+3\,{z}^{2}+z+1.$$

At this point we have just about everything we need, the only problem as is evident from the substituted cycle index (indexed by edge count) is that we are counting all subgraphs including those that obviously cannot span $$K_{3,4}.$$ Observe however that $$Z(Q_{3,4})$$ includes the count from all graphs $$K_{a,b}$$ where $$1\le a \le 3$$ and $$1\le b \le 4$$ and this observation holds for the $$Z(Q_{a,b})$$ as well. The way to identify a spanning subgraph of $$K_{3,4}$$ is that every vertex in the vertex set has degree at least one, which means these are just the graphs that cannot possibly be counted by $$Z(Q_{a,b})$$ with $$(a,b)\ne (3,4)$$ because of the missing vertices. Therefore we apply PIE to the poset where the nodes corresponding to the $$(a,b)$$ is the set of graphs counted by the corresponding substituted cycle index. We have by inspection that this poset is isomorphic to the divisor poset of $$2^{n-1}\times 3^{m-1}$$ so that we may use the ordinary Möbius function from number theory as our Möbius function for the PIE computation. (We are not including the empty graph in the sets at the nodes from the poset.)

This is implemented in the following Maple program.

with(numtheory);

pet_cycleind_symm :=
proc(n)
option remember;

if n=0 then return 1; fi;

end;

pet_varinto_cind :=
proc(poly, ind)
local subs1, subsl, polyvars, indvars, v, pot;

polyvars := indets(poly);
indvars := indets(ind);

subsl := [];

for v in indvars do
pot := op(1, v);

subs1 :=
[seq(polyvars[k]=polyvars[k]^pot,
k=1..nops(polyvars))];

subsl := [op(subsl), v=subs(subs1, poly)];
od;

subs(subsl, ind);
end;

pet_cycleind_knm :=
proc(n, m)
option remember;
local cind, sind1, sind2, t1, t2, q,
v1, v2, len, len1, len2;

cind := 0;

if n=1 then
sind1 := [a[1]];
else
sind1 := pet_cycleind_symm(n);
fi;

if m=1 then
sind2 := [a[1]];
else
sind2 := pet_cycleind_symm(m);
fi;

for t1 in sind1 do
for t2 in sind2 do
q := 1;

for v1 in indets(t1) do
len1 := op(1, v1);

for v2 in indets(t2) do
len2 := op(1, v2);

len := lcm(len1, len2);
q := q *
a[len]^((len1*len2/len) *
degree(t1, v1)*degree(t2, v2));
od;
od;

cind := cind +
lcoeff(t1)*lcoeff(t2)*q;
od;
od;

cind;
end;

v_pre_pie :=
proc(n, m)
option remember;
local cind;

cind := pet_cycleind_knm(n, m);
subs([seq(a[v]=2, v=1..n*m)], cind);
end;

v :=
proc(n, m)
local q, a, b, res;

q := 2^(n-1)*3^(m-1);

res := 0;
for a to n do
for b to m do
res := res +
mobius(q/2^(a-1)/3^(b-1))*
(v_pre_pie(a, b)-1);
od;
od;

res;
end;

print_table :=
proc(mx)
local n, m;

for n to mx do
for m to mx do
if m>1 then printf(" ") fi;
printf("%06d", v(n, m));
od;

printf("\n");
od;
end;


The above Maple code produces the following table:

> print_table(6);
000001 000001 000001 000001 000001 000001
000001 000003 000005 000008 000011 000015
000001 000005 000017 000042 000091 000180
000001 000008 000042 000179 000633 002001
000001 000011 000091 000633 003835 020755
000001 000015 000180 002001 020755 200082


It can calculate values that are completely out of reach for NAUTY like the sequence of non-isomorphic spanning subgraphs of $$K_{n,n}$$ which is $$1, 3, 17, 179, 3835, 200082, 29610804, 13702979132, \\ 20677458750966, 103609939177198046, 1745061194503344181714, \\ 99860890306900024150675406,\ldots$$ which points us to OEIS A054976 where we find confirmation of the above calculation and a slightly different interpretation of the problem statement.

The function v in the Maple program implements model $$P$$ and the function v_pre_pie implements model $$Q.$$

For bicolored versions of $$K_{n,n}$$ model $$Q$$ gives the sequence $$2, 7, 36, 317, 5624, 251610, 33642660, 14685630688, \\ 21467043671008, 105735224248507784, 1764356230257807614296, \\ 100455994644460412263071692,\ldots$$ which points us to OEIS A002724, where the calculation is confirmed.

This MSE Meta Link has many more PET computations by various users.

Thanks go to the authors of the NAUTY package.

• Excellent! Thanks for taking the time to answer. I'm studying it right now. Quick question about 'the sequence of non-isomorphic spanning subgraphs of $K_{(n,n)}$. If I take $n=2$ I found that there are $6$ spanning subgraphs of $K_{(2,2)}$ namely: the empty graph of four vertices, $P_2\cup \{a\}\cup\{b\}, P_2\cup P_2, P_3\cup \{a\}, P_4, K_{(2,2)}$. Thank you! – chowching Feb 25 '15 at 6:12
• We already discussed this at your other MSE question. There seven bicolored graphs on $K_{2,2}$ but a bicolored graph is not the same as a spanning subgraph. Please consult the discussion on the other question. E.g. the empty graph does not span $K_{2,2}.$ This Wikipedia entry is the definition of a spanning subgraph. The above program also counts bicolored graphs. Note that v_pre_pi(2,2)=7 and v(2,2)=3. – Marko Riedel Feb 25 '15 at 18:50
• I see what you mean now, I have added a comment. – Marko Riedel Feb 25 '15 at 19:05
• I'm not impressed to have yet another impressing anwer from you. – Jorge Feb 26 '15 at 1:31

The following data augment those in the first answer, namely we treat the problem of computing $$Z(Q_{n})$$ for $$K_{n,n}$$ where we include reflections. The algorithm here is slightly different from Harary and Palmer. Clearly all permutations from the first answer represented in $$Z(Q_{n,n})$$ contribute as well. Now for the reflections, which are simple, fortunately. These (the vertex permutations) can be obtained from the permutations in $$S_n$$ by placing vertices from component $$A$$ alternating with component $$B$$ on a cycle with twice the length of a cycle contained in a permutation $$\gamma$$ of $$S_n.$$ The induced action on the set of edges is the contribution to $$Z(Q_n).$$ In the following we have used an algorithmic approach rather than the formula from Harary and Palmer, namely we construct a canonical representative of each permutation shape from the cycle index $$Z(S_n)$$ that doubles the length of every cycle and alternates elements from the two components. We let that act on the set of edges and factor the result.

This gives the following cycle indices, the first of which matches the result given by Harary and Palmer. For $$n=3$$, $$Z(Q_3) = {\frac {{a_{{1}}}^{9}}{72}}+1/6\,{a_{{1}}}^{3}{a_{{2}}}^{3} \\+1/8\,a_{{1}}{a_{{2}}}^{4}+1/4\,a_{{1}}{a_{{4}}}^{2}+1/9\,{ a_{{3}}}^{3}+1/3\,a_{{3}}a_{{6}} ,$$ for $$n=4$$, $$Z(Q_4) = {\frac {{a_{{1}}}^{16}}{1152}}+{\frac {{a_{{1}}}^{8}{a_{{2} }}^{4}}{96}}+{\frac {5\,{a_{{1}}}^{4}{a_{{2}}}^{6}}{96}}\\+{ \frac {{a_{{1}}}^{4}{a_{{3}}}^{4}}{72}}+{\frac {17\,{a_{{2} }}^{8}}{384}}+1/12\,{a_{{1}}}^{2}a_{{2}}{a_{{3}}}^{2}a_{{6} }+1/8\,{a_{{1}}}^{2}a_{{2}}{a_{{4}}}^{3}\\+1/18\,a_{{1}}{a_{{ 3}}}^{5}+1/6\,a_{{1}}a_{{3}}{a_{{6}}}^{2}+1/24\,{a_{{2}}}^{ 2}{a_{{6}}}^{2}+{\frac {19\,{a_{{4}}}^{4}}{96}}+1/12\,a_{{4 }}a_{{12}}+1/8\,{a_{{8}}}^{2} ,$$ and for $$n=5$$, $$Z(Q_5) = {\frac {{a_{{1}}}^{25}}{28800}}+{\frac {{a_{{1}}}^{15}{a_{{ 2}}}^{5}}{1440}}+{\frac {{a_{{1}}}^{9}{a_{{2}}}^{8}}{288}}+ {\frac {{a_{{1}}}^{10}{a_{{3}}}^{5}}{720}}+{\frac {{a_{{1}} }^{5}{a_{{2}}}^{10}}{192}}\\+{\frac {{a_{{1}}}^{3}{a_{{2}}}^{ 11}}{96}}+{\frac {a_{{1}}{a_{{2}}}^{12}}{128}}+{\frac {{a_{ {1}}}^{6}{a_{{2}}}^{2}{a_{{3}}}^{3}a_{{6}}}{72}}+{\frac {{a _{{1}}}^{4}{a_{{3}}}^{7}}{72}}\\+{\frac {{a_{{1}}}^{5}{a_{{4} }}^{5}}{480}}+1/24\,{a_{{1}}}^{3}{a_{{2}}}^{3}{a_{{4}}}^{4} +{\frac {{a_{{2}}}^{5}{a_{{3}}}^{5}}{720}}+1/48\,{a_{{1}}}^ {3}a_{{2}}{a_{{4}}}^{5}\\+1/48\,{a_{{1}}}^{2}{a_{{2}}}^{4}a_{ {3}}{a_{{6}}}^{2}+{\frac {{a_{{2}}}^{5}{a_{{3}}}^{3}a_{{6}} }{72}}+1/32\,a_{{1}}{a_{{2}}}^{2}{a_{{4}}}^{5}\\+1/48\,{a_{{2 }}}^{5}a_{{3}}{a_{{6}}}^{2}+1/36\,{a_{{2}}}^{2}{a_{{3}}}^{5 }a_{{6}}+1/12\,{a_{{1}}}^{2}a_{{2}}a_{{3}}{a_{{6}}}^{3}\\+{ \frac {3\,a_{{1}}{a_{{4}}}^{6}}{32}}+{\frac {{a_{{2}}}^{2}{ a_{{3}}}^{3}{a_{{6}}}^{2}}{72}}+1/24\,{a_{{1}}}^{2}a_{{3}}{ a_{{4}}}^{2}a_{{12}}+1/24\,a_{{2}}a_{{3}}{a_{{4}}}^{2}a_{{ 12}}\\+{\frac {13\,{a_{{5}}}^{5}}{600}}+1/8\,a_{{1}}{a_{{8}}} ^{3}+1/12\,a_{{3}}a_{{4}}a_{{6}}a_{{12}}+{\frac {{a_{{5}}}^ {3}a_{{10}}}{60}}+1/30\,{a_{{5}}}^{2}a_{{15}}\\+1/8\,a_{{5}}{ a_{{10}}}^{2}+1/20\,a_{{5}}a_{{20}}+1/30\,a_{{10}}a_{{15}} .$$

These cycle indices can be calculated for large $$n$$ but the pattern should be clear. The count of these graphs gives the sequence

$$2, 6, 26, 192, 3014, 127757, 16853750, \\ 7343780765, 10733574184956, 52867617324773592,\\882178116079222400788, 50227997322550920824045262, \ldots$$

which points us to OEIS A007139 where we find confirmation of this calculation.

This was the Maple code used to obtain these values.

with(numtheory);

pet_cycleind_symm :=
proc(n)
local p, s;
option remember;

if n=0 then return 1; fi;

end;

pet_autom2cycles :=
proc(src, aut)
local numa, numsubs;
local marks, pos, cycs, cpos, clen;

numsubs := [seq(src[k]=k, k=1..nops(src))];
numa := subs(numsubs, aut);

marks := Array([seq(true, pos=1..nops(aut))]);

cycs := []; pos := 1;

while pos <= nops(aut) do
if marks[pos] then
clen := 0; cpos := pos;

while marks[cpos] do
marks[cpos] := false;
cpos := numa[cpos];
clen := clen+1;
od;

cycs := [op(cycs), clen];
fi;

pos := pos+1;
od;

return mul(a[cycs[k]], k=1..nops(cycs));
end;

pet_varinto_cind :=
proc(poly, ind)
local subs1, subs2, polyvars, indvars, v, pot, res;

res := ind;

polyvars := indets(poly);
indvars := indets(ind);

for v in indvars do
pot := op(1, v);

subs1 :=
[seq(polyvars[k]=polyvars[k]^pot,
k=1..nops(polyvars))];

subs2 := [v=subs(subs1, poly)];

res := subs(subs2, res);
od;

res;
end;

pet_flatten_term :=
proc(varp)
local terml, d, cf, v;

terml := [];

cf := varp;
for v in indets(varp) do
d := degree(varp, v);
terml := [op(terml), seq(v, k=1..d)];
cf := cf/v^d;
od;

[cf, terml];
end;

pet_flat2rep_cyc :=
proc(f)
local p, q, res, cyc, t, len;

q := 1; res := [];

for t in f do
len := op(1, t);
cyc := [seq(p, p=q+1..q+len-1), q];
res := [op(res), cyc];
q := q+len;
od;

res;
end;

pet_cycs2table :=
proc(cycs)
local pairs, cyc, p, ent;

pairs := [];
for cyc in cycs do
pairs :=
[op(pairs),
seq([cyc[p], cyc[1 + (p mod nops(cyc))]],
p = 1 .. nops(cyc))];
od;

map(ent->ent[2], sort(pairs, (a,b)->a[1] < b[1]));
end;

pet_cycleind_knn :=
proc(n)
option remember;
local cindA, cindB, sind, t1, t2, p, q, cyc1, cyc2,
flat, len, len1, len2, v1, v2,
edges, edgeperm, rep, cycsA, cycsB, cycsrc, cyc;

if n=1 then
sind := [a[1]];
else
sind := pet_cycleind_symm(n);
fi;

cindA := 0;

for t1 in sind do
for t2 in sind do
q := 1;

for v1 in indets(t1) do
len1 := op(1, v1);

for v2 in indets(t2) do
len2 := op(1, v2);

len := lcm(len1, len2);
q := q *
a[len]^((len1*len2/len) *
degree(t1, v1)*degree(t2, v2));
od;
od;

cindA := cindA +
lcoeff(t1)*lcoeff(t2)*q;
od;
od;

edges := [seq(seq({p, q+n}, q=1..n), p=1..n)];

cindB := 0;

for t1 in sind do
flat := pet_flatten_term(t1);

cycsA := pet_flat2rep_cyc(flat[2]);
cycsB := [];
for cycsrc in cycsA do
cyc := [];
for q in cycsrc do
cyc := [op(cyc), q, q+n];
od;

cycsB := [op(cycsB), cyc];
od;

rep := pet_cycs2table(cycsB);

edgeperm :=
subs([seq(q=rep[q], q=1..2*n)], edges);

cindB := cindB + flat[1]*
pet_autom2cycles(edges, edgeperm);
od;

(cindA+cindB)/2;
end;

Q :=
proc(n)
option remember;
local cind;

cind := pet_cycleind_knn(n);
subs([seq(a[p]=2, p=1..n*n)], cind);
end;


Remarks, per request, Dec 2020. Here are some observations concerning reflections. We need the factorization into cycles of the vertices in order to compute the action on the edges. Now a reflection when factored into cycles must alternate between left and right vertices. That means left and right vertices are interleaved on those cycles. Hence we can obtain the factorizations of all reflections by taking a permutation from $$S_n$$ (which gives the left vertices) and doubling all its cycles in length, inserting a permutation of the right vertices, for a total of $$n! \times n!$$ reflections. The first factorial comes from the permutations of the left vertices which can be interleaved with the right ones in $$n!$$ ways, yielding the second factorial. Now consider the factorization into cycles of the action on the edges. Left and right vertices are disjoint, so no matter the order in which we interleave the right vertices, we get the same shape of factorizations into cycles (just permute the right vertices, which maintains the cycle structure of the edges). Therefore we can take one representative for each term in $$Z(S_n)$$, interleave it with some permutation of the right, and factor that (i.e. factor the action on the edges) to get the contribution to $$Z(Q_n)$$, which must be multiplied by $$n!$$ for the number of reflections covered by this term. This is what the algorithm does. The $$n!$$ is canceled when we divide by the total number of permutations in the group.

• I have trouble understanding the algorithm for calculating the cycle index when reflections are concerned. I obtain the same results when I consider each of the $n!^2$ pairs of permutations, but it's of course infeasible for larger $n$. Could you describe your approach in more detail? From what I understand, "permutation shape" is just its cycle notation. But two different pairs of permutations with the same cycle notations can contribute different terms to the cycle index when the vertices are additionally reflected. I don't understand how the alternation of components works here. – Adam S. Dec 15 '20 at 23:28
• I have added some commentary. Please note that the 2017 answer is the reference here. – Marko Riedel Dec 16 '20 at 2:53
• Wow, brilliant. Thank you! My mistake was to treat the two permutations and the reflection completely separately, which made the cycle factorization difficult. Embedding the reflection into one of the permutations makes things so much easier and indeed reduces the computational cost from $n!^2$ just to the number of different permutation shapes in $S_n$. This will help me speed up my little project: github.com/adamsol/burnside-counter. – Adam S. Dec 16 '20 at 17:30
• It is now tempting to try to simplify also the case without reflections, so that we don't have to iterate over pairs of permutation shapes at all. But it seems that this case is paradoxically more complicated and it's not possible. – Adam S. Dec 16 '20 at 17:38
• Good that I could help. I might do some additional simplifications later today. – Marko Riedel Dec 16 '20 at 17:40

The question is equivalent to finding all non-isomorphic bipartite graphs with bi-partitions of sizes $n$ and $m$ respectively and is hard to solve in general. In particular there is no simple characterization of such graphs.

If you need the sequence for the number of such graphs then you can take a look at this oeis entry http://oeis.org/A033995.

If you need to compute such graphs (up to a reasonable number) then your best choice is the tool genbg which you can find in McKay's nauty package. In some sense genbg is currently the most efficient way to construct such graphs and hence answers your question.

• Yeah I need to find all those non-isomorphic bipartite graphs with sizes $m$ and $n$ in each partition. The sequence above is for the number of bipartite graphs of $n$ vertices. But that number is greater than the number of graphs that I am trying to find. For example if I have four vertices, there are $7$ bipartite graphs but if I'm only taking all the spanning subgraphs of $K_{(2,2)}$ then there are just $6$ of them. The other bipartite graph is a claw which is bipartite with sizes 1 and 3 in each partition. – chowching Feb 17 '15 at 12:43
• @chowching You should be more specific about what you're looking for. Note that in general there is no closed formula for the number of bipartite graphs with bipartitions of cardinality n and m. – Jernej Feb 17 '15 at 13:00
• Thanks for the reply. I'm studying Polya Enumeration Theorem now to see if I can apply it to find how many such graphs are there given a certain $n$ and $m$. – chowching Feb 17 '15 at 13:00
• @chowching You can but the formula will most likely require using computer computations. – Jernej Feb 17 '15 at 13:01
• @chowching I suggest that for a start you look at Harary's and Palmer's book called "graphical enumeration" – Jernej Feb 17 '15 at 13:02

Adding another answer after receiving a pointer from a reader of this thread. We treat the case of spanning subgraphs of $$K_{n,n}$$ including reflections. The algorithm has been greatly simplified. There are two types of symmetries, namely shuffles within the two sets of vertices and reflections that map one to the other. The first type has been simplified as follows: we continue to iterate over pairs of permutation shapes from $$S_n$$ but the contribution to the cycle index $$Z(Q_n)$$ is now computed by iterating over pairs of cycle types (same length) rather than single cycles. The second type (reflections) has been replaced in its entirety. We continue to produce contributions by building factorized permutations consisting of cycles that alternate between the two classes of vertices (a precondition of this problem) but whereas in the first version we would construct a representative of the factored permutation, apply it to the edges, and factor the result into cycles, the new version does not construct representatives and does no factorization and builds the terms for the cycle index from first principles, namely that we enumerate the cycles of edges (two-sets of vertices) that are obtained from a vertex permutation by choosing pairs of vertices, counting these, and computing the lengths of the cycles on which they reside. These optimizations make for a faster and much more compact program. This was the code:

pet_cycleind_symm :=
proc(n)
option remember;

if n=0 then return 1; fi;

end;

pet_varinto_cind :=
proc(poly, ind)
local subs1, subs2, polyvars, indvars, v, pot, res;

res := ind;

polyvars := indets(poly);
indvars := indets(ind);

for v in indvars do
pot := op(1, v);

subs1 :=
[seq(polyvars[k]=polyvars[k]^pot,
k=1..nops(polyvars))];

subs2 := [v=subs(subs1, poly)];

res := subs(subs2, res);
od;

res;
end;

pet_flatten_termA :=
proc(varp)
local terml, d, cf, v;

terml := [];

cf := varp;
for v in indets(varp) do
d := degree(varp, v);
terml := [op(terml), [op(1,v), d]];
cf := cf/v^d;
od;

[cf, terml];
end;

pet_cycleind_knn :=
proc(n)
option remember;
local cindA, cindB, sind, t1, t2, term, res,
flat, flat1, flat2, len, l1, l2, cycs, uidx, vidx,
u, v, inst1;

if n=1 then
sind := [a[1]];
else
sind := pet_cycleind_symm(n);
fi;

cindA := 0;

for t1 in sind do
flat1 := pet_flatten_termA(t1);

for t2 in sind do
flat2 := pet_flatten_termA(t2);

res := 1;

for u in flat1[2] do
l1 := op(1, u);

for v in flat2[2] do
l2 := op(1, v);

len := lcm(l1, l2);
res := res *
a[len]^(op(2, u)*op(2, v)
*l1*l2/len);
od;
od;

cindA := cindA + flat1[1]*flat2[1]*res;
od;
od;

cindB := 0;

for term in sind do
flat := pet_flatten_termA(term);

cycs := map(cyc -> [2*cyc[1], cyc[2]], flat[2]);
res := 1;

# edges on different cycles of different sizes
for uidx to nops(cycs) do
u := cycs[uidx];

for vidx from uidx+1 to nops(cycs) do
v := cycs[vidx];

l1 := op(1, u); l2 := op(1, v);
res := res *
a[lcm(l1, l2)]^((l1*l2/2/lcm(l1, l2))*
op(2, u)*op(2, v));
od;
od;

# edges on different cycles of the same size
for uidx to nops(cycs) do
u := cycs[uidx];

l1 := op(1, u); inst1 := op(2, u);
# a[l1]^(1/2*inst1*(inst1-1)*l1*l1/2/l1)
res := res *
a[l1]^(1/2*inst1*(inst1-1)*l1/2);
od;

# edges on identical cycles of some size
for uidx to nops(cycs) do
u := cycs[uidx];

l1 := op(1, u); inst1 := op(2, u);
if type(l1/2, even) then
# a[l1]^((l1/2)^2/l1);
res := res *
(a[l1]^(l1/4))^inst1;
else
# a[l1/2]^(l1/2/(l1/2))*a[l1]^(((l1/2)^2-l1/2)/l1)
res := res *
(a[l1/2]*a[l1]^(l1/4-1/2))^inst1;
fi;
od;

cindB := cindB + flat[1]*res;
od;

(cindA+cindB)/2;
end;



This will produce the following data for the non-isomorphic spanning subgraphs of $$K_{6,6}$$ including reflections classified according to the number of edges:

$${z}^{36}+{z}^{35}+2\,{z}^{34}+4\,{z}^{33}+10\,{z}^{32} +20\,{z}^{31}+50\,{z}^{30}+99\,{z}^{29}+227\,{z}^{28} \\ +458\,{z}^{27}+934\,{z}^{26}+ 1711\,{z}^{25}+3057\,{z}^{24}+4889\,{z}^{23}+7400\,{z}^{22} \\ +10071\,{z}^{21}+12755\,{z}^{20}+14524\,{z}^{19} +15331\,{z}^{18}+14524\,{z}^{17}+12755\,{z}^{16} \\+10071\,{z}^{15}+7400\,{z}^{14}+4889\,{z}^{13}+ 3057\,{z}^{12}+1711\,{z}^{11}+934\,{z}^{10} \\ +458\,{z}^{9}+227\,{z}^{8}+99\,{z}^{7}+50\,{z}^{6}+20\,{z}^{5} +10\,{z}^{4}+4\,{z}^{3}+2\,{z}^{2}+z+1.$$

i.e. the value four for three edges counts, first, a set of three edges, not connected, second, two edges sharing a vertex and a single edge, third, three edges emanating from the same vertex and fourth, a path of three edges.

**Addendum.** We may use Burnside as shown below if we are not interested in classifying according to the number of edges and require only the total count. This gives:

Q :=
proc(n)
option remember;
local cind;

cind := pet_cycleind_knn(n);
subs([seq(a[p]=2, p=1..n*n)], cind);
end;


The routine that substitutes polynomials into cycle indices is no longer needed in this case. Observe that further simplification at the expense of readability is possible by not computing the cycle index at all and setting all $$a[p]$$ to two during the computation of what used to be the cycle index.

Remark, Dec 2020. We can in fact optimize away all flattening operations and use the fact that Maple monomials i.e. the terms of the cycle index are perfectly sufficient to implement a multiset data structure where the variables represent cycles of some length and the degrees, the number of instances, as is of course the standard semantics of cycle indices. This is shown below. (Duplicate code omitted.)

pet_cycleind_knn :=
proc(n)
option remember;
local cindA, cindB, sind, t1, t2, term,
res, len, len1, len2, cycs, uidx, vidx, interlv,
u, v, inst, q;

if n=1 then
sind := [a[1]];
else
sind := pet_cycleind_symm(n);
fi;

cindA := 0;

for t1 in sind do
for t2 in sind do
q := 1;

for u in indets(t1) do
len1 := op(1, u);

for v in indets(t2) do
len2 := op(1, v);

len := lcm(len1, len2);
q := q *
a[len]^((len1*len2/len) *
degree(t1, u)*degree(t2, v));
od;
od;

cindA := cindA +
lcoeff(t1)*lcoeff(t2)*q;
od;
od;

cindB := 0;

interlv := subs({seq(a[q]=a[2*q], q=1..n)}, sind);
for term in interlv do
termvars := indets(term); res := 1;

# edges on different cycles of different sizes
for uidx to nops(termvars) do
u := op(uidx, termvars);
len1 := op(1, u);

for vidx from uidx+1 to nops(termvars) do
v := op(vidx, termvars);
len2 := op(1, v);

res := res *
a[lcm(len1, len2)]
^((len1*len2/2/lcm(len1, len2))*
degree(term, u)*degree(term, v));
od;
od;

# edges on different cycles of the same size
for u in termvars do
len := op(1, u); inst := degree(term, u);
# a[len]^(1/2*inst*(inst-1)*len*len/2/len)
res := res *
a[len]^(1/4*inst*(inst-1)*len);
od;

# edges on identical cycles of some size
for u in termvars do
len := op(1, u); inst := degree(term, u);
if type(len/2, even) then
# a[len]^((len/2)^2/len);
res := res *
(a[len]^(len/4))^inst;
else
# a[len/2]^(len/2/(len/2))
# *a[len]^(((len/2)^2-len/2)/len)
res := res *
(a[len/2]*a[len]^(len/4-1/2))^inst;
fi;
od;

cindB := cindB + lcoeff(term)*res;
od;

(cindA+cindB)/2;
end;

Q :=
proc(n)
option remember;
local cind;

cind := pet_cycleind_knn(n);
subs([seq(a[p]=2, p=1..n*n)], cind);
end;