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!
Best Answer
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; expand(1/n*add(a[l]*pet_cycleind_symm(n-l), l=1..n)); 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.