I would like to calculate the expected maximum tree size in a randomly generated pseudoforest of $N$ labelled nodes where self-loops are not permitted. Empty and single-node trees are also not permitted.

For example, if we have $4$ labelled nodes, we can generate $3$ pseudoforests with a largest tree size of $2$, and $78$ pseudoforests with a maximum tree size of $4$.

There are a total of $(n-1)^n$ possible pseudoforests, thus for $N = 4$ there are $81$.

The expected maximum tree size for $N = 4$ would therefore be: $$ E(x) = \sum_{i=1}^{n}i\cdot p(i) = 2 \cdot \frac{3}{81} + 4\cdot\frac{78}{81} = 3.925... $$

Some observations:

  1. There will never be a pseudoforest where the maximum tree size is $n-1$.
  2. The number of pseudoforests of $N$ nodes containing only one connected tree (therefore maximum tree size of $N$) can be calculated using sequences $A000435$ or $A001864 / n$.

    For $N = 4$, this gives us $78$, ie. when $i = n$ in the summation.

  3. The minimum tree size is $2$ if $N$ is even, and $3$ if $N$ is odd.
  4. The sum of the numerators of $p(i)$ is equal to $(n-1)^n$

When $N = 5$, the summation is: $$ 3 \cdot \frac{80}{1024} + 5\cdot\frac{944}{1024} = 4.84375 $$

When $N = 6$, the summation is: $$ 2 \cdot \frac{15}{15625} + 3\cdot\frac{640}{15625} + 4\cdot\frac{1170}{15625} + 6\cdot\frac{13800}{15625} = 5.72352 $$
How can I calculate the numerators of $p(i)$ when $i < n$?


We observe that all fractions in $E(n)$ have the same denominator, so let's directly calculate the total numerator $S(n)$; for instance $S(5) = 3 \cdot 80 + 5 \cdot 944$.

Let $T(n)$ be the number of pseudoforests with exactly one connected component, or the number of "good graphs" in the notation of this question.

Let $\pi(n)$ be the set of the integer partitions of $n$.

The numerator can then be found with the following algorithm:

$\texttt{result} \leftarrow 0$
$\quad\quad\quad\quad \texttt{continue}$
$\quad\quad\quad\quad\texttt{result +=} \max(\tau) \cdot C(\tau) \cdot \prod_{t\in \tau}{T(t)}$,

where $C(\tau)$ is the number of ways $n$ labelled items can be split according to $\tau$.

Let $t_1,\dots,t_r$ be the elements of $\tau$, let $s_1,\dots,s_s$ be the elements of $\tau$ without repeats and let $m_1,\dots,m_s$ be their respective multiplicities. Then we have:

$$C(\tau) = \frac{{n \choose t_1} \cdot {n - t_1 \choose t2} \cdot \dots \cdot {t_r \choose t_r}}{\prod_{i = 1}^{s}{m_i!}}\text{.}$$

Essentially, $C(\tau)$ counts the number of ways of splitting $n$ labelled nodes into connected components of the sizes given by $\tau$, while $\prod{T(n)}$ counts the number of ways of connecting the labelled nodes inside each connected component.

  • $\begingroup$ I think I'm very close to this now. If I'm understanding correctly, $C(\tau)$ is ${n \choose t_1} \cdot {n-t_1 \choose t_2} \cdot {n-t_2 \choose t_3} \dots {n-t_{r-1} \choose t_r}$ divided by the product of the multiplicities of the elements of $\tau$? For example a partition of [2,2,3] results in a $2!1!$ denominator? $\endgroup$ – rtheunissen Jan 10 '15 at 1:14
  • $\begingroup$ Almost. $C(\tau)$ is ${n \choose t_1} \cdot {n-t_1 \choose t_2} \cdot {n-t_1-t_2 \choose t_3} \cdot \dots \cdot {n-t_1-\dots-t_{r-1} \choose t_r}$ divided by their multiplicities. That formula has a few typos, which I'm going to fix now. $\endgroup$ – Jacopo Notarstefano Jan 10 '15 at 1:27
  • $\begingroup$ To help your intuition: the numerator assigns the nodes according to the elements of the partition, but we are counting twice some assignments; in particular we count $k!$ times the assignments where $k$ elements of the partition of the same size split the same subset of nodes. Here's similar argument (fourth bullet point): math.stackexchange.com/a/393606/4471 $\endgroup$ – Jacopo Notarstefano Jan 10 '15 at 1:38
  • $\begingroup$ Uh, I just noticed that that product telescopes, which results in a much simpler formula for $C(\tau)$! $\endgroup$ – Jacopo Notarstefano Jan 10 '15 at 1:45

I am sending some preliminary, as yet not fine-tuned and optimized obervations to generate activity on this question. I am sure these admit improvements (the time and space complexity of the algorithm is poor).

If I have decoded the problem statement correctly we are enumerating labeled endofunctions with no fixed points as appeared at this MSE link so that these pseudo-forests are directed.

The species $\mathcal{Q}$ under consideration is $$\mathcal{Q} = \mathfrak{P}(\mathfrak{C}_{=2}(\mathcal{T}) + \mathfrak{C}_{=3}(\mathcal{T}) + \mathfrak{C}_{=4}(\mathcal{T}) + \cdots).$$ where $\mathcal{T}$ represents labeled rooted trees with EGF $T(z)$ and functional equation $T(z) = z\exp T(z),$ the labeled tree function.

It follows that the distribution of the maxima of the tree sizes (where unicyclic components are counted by the number of nodes) on $n$ nodes can be found by considering the species $$\mathcal{Q}_{\le n} = \mathfrak{P}(\mathfrak{C}_{=2}(\mathcal{T}_{\le n}) + \mathfrak{C}_{=3}(\mathcal{T}_{\le n}) + \mathfrak{C}_{=4}(\mathcal{T}_{\le n}) + \cdots + \mathfrak{C}_{=n}(\mathcal{T}_{\le n}))$$ and marking the cycles with a variable $\mathcal{V}_q$ indicating the number of nodes in the component using the generating function $$Q(z) = \exp \left(\frac{T_{\le n}(z)^2}{2} + \frac{T_{\le n}(z)^3}{3} + \frac{T_{\le n}(z)^4}{4} + \cdots + \frac{T_{\le n}(z)^n}{n}\right).$$

This will produce the following distributions: $${u}^{2},\\ 8\,{u}^{3},\\ 78\,{u}^{4}+3\,{u}^{2},\\ 944\,{u}^{5}+80\,{u}^{3},\\ 13800\,{u}^{6}+1170\,{u}^{4}+640\,{u}^{3}+15\,{u}^{2},\\ 237432\,{u}^{7}+19824\,{u}^{5}+21840\,{u}^{4}+840\,{u}^{3},\\ 4708144\,{u}^{8}+386400\,{u}^{6}+422912\,{u}^{5}+229320\,{u}^{4} \\+17920\,{u}^{3}+105\,{u}^{2},\\ 105822432\,{u}^{9}+8547552\,{u}^{7}+9273600\,{u}^{6}+9634464\,{u}^{5}\\ +786240\,{u}^{4}+153440\,{u}^{3},\ldots$$

This gives the following for the expected maximum tree size: $$2., 3., 3.925925926, 4.843750000, 5.723520000, 6.612311385, \\ 7.471092584, 8.342072010, 9.189007167, 10.04727275, \\ 10.88589292, 11.73525388, 12.56739638, 13.40959924,\ldots$$ which would seem to indicate that most of these consist of one connected component.

The following Maple code was used to compute these values. It took $58$ seconds to compute the distribution for $n=35,$ which represents $$399725722782532944388077717044552088857010024925364224$$ pseudoforests ($34^{35}$).

gf_le :=
    option remember;
    local Tle, term, res, p;

    Tle := add(q^(q-1)*z^q/q!, q=1..n);

    res := 0;

    for term in expand(add(Tle^q/q, q=2..n)) do
        p := degree(term, z);

        res := res + v[p]*coeff(term, z, p)*z^p;


gf :=
    option remember;
    local res, cft, vs, term;

    res := 0; cft := n!*coeftayl(gf_le(n), z=0, n);

    if not type(cft, `+`) then
        cft := [cft];

    for term in cft do
        vs := indets(term);
        res := res +
        lcoeff(term)*u^max(map(t->op(1, t), vs));

  • $\begingroup$ Thank you Marko this looks very good although as I am not very confident with analytic combinatronics and Maple, a lot of it doesn't make immediate sense to me. How would you explain your solution (and how to generate it) to someone with basic knowledge of polynomials, coefficients, sums etc. but not necessarily Maple? $\endgroup$ – rtheunissen Jan 9 '15 at 11:10
  • $\begingroup$ This looks difficult. On principle I include Maple code, C or Perl when I believe it substantially assists in understanding the main text. This is the case here and the answer makes considerable usage of multivariate polynomials, without which there would be a lot of extra code, not all of it necessarily aiding understanding. My second answer may be easier to port. I don't know if it makes sense to attack this without a CAS. Maybe someone will post a SAGE implementation or one in Mathematica. Unfortunately there is less of an incentive to do so now given Jacopo's good answer. $\endgroup$ – Marko Riedel Jan 9 '15 at 22:19
  • $\begingroup$ Thank you so much for your answer, if anything my mind has now been opened to analytic combinatronics and multivariate polynomials. $\endgroup$ – rtheunissen Jan 10 '15 at 4:10

I implemented @Jacopo Notarstefano's method, which is a good answer and has better space complexity than my initial draft. Time complexity is middling. It can compute the distribution for $n=60$ in $291$ seconds or about five minutes according to my tests.

Here are some of the coefficients for $n=60$: $$2: 29215606371473169285018060091249259296875\\ 3: 6771490865749881440878442829560908615652159...\\ 180618395077599800000000\\ 4: 1005563324091428408820112074277646586848640...\\ 7341577965559630663675312982983593750\\ 10: 312487203633572495389224364764544078483826...\\ 833508949993777620300122108361941948220628320222797082880\\ 12: 198506674917987164954385699985381605466521...\\ 98813653741864939041891599818516344945718291826495396755200$$

This is the Maple code.


conn :=
    option remember;

    n*(n-1)^n + n!*
                binomial(p+q-2, q-2), p=0..n),
            q=2..k), k=2..n);


pf_dist :=
    option remember;
    local res, lambda, cf, cfc, scm, q;

    res := 0;

    lambda := firstpart(n);
    while type(lambda, `list`) do
        cf := n!/mul(q!, q in lambda);
        cfc := mul(conn(q), q in lambda);
        scm := mul(q[2]!, q in convert(lambda, `multiset`));

        res := res + cf*cfc*u^max(lambda)/scm;

        lambda := nextpart(lambda);


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