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How many distinct non-isomorphic bipartite graphs with parts of size $m$ and $n$ exist? (Two bipartite graphs are distinct if there is no way to just rearrange the vertices within a part set of one of them to become the other.) It's not necessary that every vertex has an edge. Is there a closed form formula in terms of $m$ and $n$? If not, is there a recursive way to count it? If there is a good recursive algorithm to count it, what would its pseudocode look like?

Thanks!

Anon
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  • https://arxiv.org/pdf/1304.0139.pdf – HEKTO Jun 13 '17 at 23:30
  • I don't think species are needed though, it falls to simple polya. Although yeah, species are awesome :p – Asinomás Jun 13 '17 at 23:31
  • How would I use Polya's theorem in this problem? Would I have to examine each of the $m!n!$ possible symmetries and count how many graphs are fixed by each one? That seems too complicated to find an effective way to count the number of such graphs. – Anon Jun 14 '17 at 01:08
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    This problem appeared at the following MSE link. Complexity is not factorial but rather the number of terms in the cycle index of the symmetric group (partition function). – Marko Riedel Jun 14 '17 at 01:34
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    Maybe you want http://oeis.org/A028657 – Gerry Myerson Jun 14 '17 at 01:55
  • @MarkoRiedel Thanks for the link! It really helped. On another note, is there anoter way to solve this without Burnside's lemma and Polya's theorem using recursion (its fine if the algorithm is a bit slower)? I don't know how to use symbolic manipulation in programming. – Anon Jun 14 '17 at 05:23
  • No recursion yet, but a thank you nonetheless for pointing me to this problem that I worked on more than two years ago. I posted another answer just now, greatly simplifying the algorithm and obtaining a more compact, faster program. You might want to consult this new answer, and thanks again. – Marko Riedel Jun 14 '17 at 23:15
  • When you say that the complexity is the partition function, do you mean that its $O(f(mn))$ where $f(x)$ counts the number of partitions of an integer? Also, in using Polya's theorem, is the permutation group that's being considered the permutations of the $mn$ edges? Thanks! – Anon Jun 16 '17 at 01:12

3 Answers3

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On your question whether or not there exists a closed-form formula for the number of bipartite graphs with parts of size $m$ and $n$ (denoted by $|B_u(m,n)|$ below), my coauthor and I proved the following formulas for $m = 2$ and $m = 3$ in an upcoming paper. $|B_u(2,n)|$ corresponds to the integer sequence A002623, i.e., 1, 3, 7, 13, 22, 34, 50, 70, 95,..., and $|B_u(3,n)|$ corresponds to the integer sequence A002727, i.e., 1, 4, 13, 36, 87, 190, 386, 734, 1324, 2284,... in Sloane's classification of integer sequences. Generalizing these closed-form formulas for $m = 4,5,6,...$ remains an open problem to the best of my knowledge.

$$|B_u(2,n)|\!=\! \frac{2n^{3}+15n^{2} + 34n + 22.5 + 1.5\left ( -1 \right )^{n}}{24}, n=0,1,2,...$$

$\,$

$\!\!\!\!\!\!\!|B_{u}\left( 3,n \right)|\! = \left\{\begin{matrix} \frac{1}{6}\left [ \binom{n+7}{7}\! + \frac{3\left ( n+4 \right )\left ( 2n^{4}+32n^{3}+172n^{2} + 352n + 15\left ( -1 \right )^{n} +225 \right )}{960} +\! \frac{2(n^{3}+12n^{2}+45n+54)}{54} \right ] &\!\!\!\!\!\!\text{if}\, n \bmod\!\! \text{ } 3 = 0, \\ \\ \frac{1}{6}\left [ \binom{n+7}{7}\! + \frac{3\left ( n+4 \right )\left ( 2n^{4}+32n^{3}+172n^{2} + 352n + 15\left ( -1 \right )^{n} +225 \right )}{960} +\! \frac{2(n^{3}+12n^{2}+45n+50)}{54} \right ] &\!\!\!\!\!\!\!\text{ if}\, n \bmod\!\! \text{ } 3 = 1, \\ \\ \frac{1}{6}\left [ \binom{n+7}{7}\! + \frac{3\left ( n+4 \right )\left ( 2n^{4}+32n^{3}+172n^{2} + 352n + 15\left ( -1 \right )^{n} +225 \right )}{960} +\! \frac{2(n^{3}+12n^{2}+39n+28)}{54} \right ] &\!\!\!\!\!\!\!\!\! \text{if}\, n \bmod\!\! \text{ } 3 = 2. \!\!\!\! \end{matrix}\right. $

Ref: Abdullah Atmaca and A. Yavuz Oruc. "On The Size Of Two Families Of Unlabeled Bipartite Graphs." AKCE International Journal of Graphs and Combinatorics. To appear.

AYO
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In a more recent article by A. Atmaca and A. Yavuz Oruc, the following more general result for the number of unlabeled graphs with $n$ left vertices and $r$ right vertices (denoted by $|B_u(n,r)|$) has been proven:

$$\frac{{r+2^n-1\choose r}}{n!}\le |B_u(n,r)| \le 2\frac{{r+2^n-1\choose r}}{n!}, n < r. $$

Given that $|B_u(n,r)|= |B_u(r,n)|$, the following inequality holds as well when $n > r:$

$$\frac{{n+2^r-1\choose n}}{r!}\le |B_u(n,r)| \le 2\frac{{n+2^r-1\choose n}}{r!}, n > r. $$

Note that the upper bound is twice as large as the lower bound. Tightening the constant factors, i.e., 1 and 2 on the lefthand and righthand side of the inequality remains open.

AYO
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  • When $r=n$ are isomorphisms allowed to swap the two partite sets? – bof Jul 14 '18 at 10:00
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    In these formulas, it is assumed that left and right sets of vertices are not swappable whether or not $n = r$ as in Harrison, Michael A. "On the number of classes of binary matrices." IEEE Transactions on Computers 100.12 (1973): 1048-1052. For example, if $n = r = 2$ with left set of vertices denoted $X = {x_1,x_2}$ and right set of vertices denoted $Y = {y_1,y_2}$, the graph that connects $x_1$ to $y_1$ and $y_2$ and the graph that connects $y_1$ to $x_1$ and $x_2$ are considered non-isomorphic. – AYO Jul 14 '18 at 15:07
  • It should also be added that the lower bounds in the two inequalities coincide and also hold when $n = r$ under the same assumption as in my earlier comment. – AYO Jul 14 '18 at 15:13
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On your question whether or not there exists a recursive formula for the number of bipartite graphs with parts of size m and n (denoted by $|B_u(m,n)|$ below), and based on the notations and formula used in the other answers, a recursive formula for $m=2$ is $$|B_u (2,n+1)| = 2 |B_u (2,n-1)| - |B_u (2,n-3)| + 2n +3$$ for $n>2$, and $|B_u (2,0)|=1$, $|B_u (2,1)|=3$, $|B_u (2,2)|=7$.

Ad Huikeshoven
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    As it’s currently written, your answer is unclear. Please [edit] to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center. – CrSb0001 Sep 11 '23 at 16:13