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Apologies if this is a basic question, I'm a bit of a rookie when it comes to maths.

If I have two sets of nodes $U$ and $V$, is there a way to calculate how many bipartite graphs could be formed if:

  • All the nodes in $U$ must have a degree $\geq$ 1
  • All the nodes in $V$ must have a degree of 0 or 1

Thank you!

David Poxon
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    Even if you are a rookie in maths, do not doubt your own ability to at least probe questions. For example, you can start by assuming that both the sides are of very small size, say one or two, and explicitly drawing all bipartite graphs. This will give you at least an idea as to how to extend a counting logic from a smaller to a bigger number of nodes. – Sarvesh Ravichandran Iyer Jul 16 '18 at 07:01
  • Are you asking for simultaneous conditions? Or is each bullet point a separate case? – qualcuno Jul 16 '18 at 07:13
  • @GuidoA. the conditions are simultaneous. – David Poxon Jul 16 '18 at 07:19

3 Answers3

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Each $v \in V$ is assigned either a vertex in $U$ or is left disconnected. Thus, each bipartite graph where the vertices in $V$ have degree at most one is equivalent to a function $f:V \to U \cup \{*\}$ where we interpret $f(v) = *$ as $v$ having no neighbour. Now, as an extra condition, we need each vertex in $U$ to be connected to one in $V$. Thus, if $V' = V \setminus f^{-1}(\{*\})$, the mapping $f\restriction_{V'}$ should be surjective. This also chacterizes our graphs, since each solution can be written is this form.

Therefore, we can count these in the following manner: first, we select a subset $W$ of vertices of $V$ that will be disconnected so that at least $|U|$ vertices of $V$ are left. Afterwards, each possible option to connect the vertices on $V \setminus W$ corresponds to a surjective function from $V\setminus W$ to $U$. More concretely, it will suffice to count the following set

$$ \coprod_{\quad W \subseteq V \\ |W| \leq |V| - |U|} \{f:V\setminus W \to U : f \text{ is surjective}\} $$

whose cardinal is

$$ \sum_{\quad W \subseteq V \\ |W| \leq |V| - |U|} S(|V \setminus V|,|U|)|U|! $$

where $S(i,j)$ denotes the Stirling number of second kind. A bit more explicitly, if $|V| = m$, $|U| = n$, we have

$$ \sum_{i = 0}^{m-n}{m\choose i}S(m-i,n)n! $$

possible options (here we use that sets of same size give an equal amount of surjective functions since these only depend on the size of the domain and codomain). Making a change of variable as $k = m - i$ and using the symmetry of the binomial coefficients, we get:

$$ \sum_{k = m}^{n}{m\choose k}S(k,n)n! $$

qualcuno
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Let $|V|=n\ge|U|=m$ We want to count the number of functions from subsets of $|V|$ onto $|U|.$ Let $f(k,m)$ be the number of functions from a set of $k$ elements onto a set of $m$ elements. Since we have $\binom{n}{k}$ ways to choose $k$ elements of |$V|,$ the number we seek is $$\sum_{k=m}^n\binom{n}{k}f(k,m)$$.

It remains to calculate $f(k,m).$ These number are given by the Stirling numbers of the second kind, as explained here.

saulspatz
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The answer to this question depends quite a bit on what it means for two graphs to be the same. The other answers attempt to address what happens when the labels in $U$ and $V$ matter. The question is just as fun, if a bit more simplistic, when we only consider the count up to isomorphism.

  • Our first observation is that two such graphs are isomorphic iff the vertices in $U$ have the same degrees up to a re-ordering.
  • Our second observation is that any total number of edges $n$ with $|U|\leq n\leq |V|$ is allowed, and no other number of edges satisfies the necessary constraints.
  • Lastly, for any fixed total number of edges $n$, the number of ways of assigning vertex degrees to the vertices in $|U|$ in a way which satisfies the constraints is the number of partitions of $n$, denoted $p(n)$.

Putting it all together, we have $$\boxed{\sum_{n=|U|}^{|V|}p(n)}$$ graphs up to isomorphism.

Hans Musgrave
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