Approximating the Cumulant Generating Function of Triangles in the Erdös–Rényi Random Graph

Abstract

We study the pressure of the “edge-triangle model”, which is equivalent to the cumulant generating function of triangles in the Erdös–Rényi random graph. The investigation involves a population dynamics method on finite graphs of increasing volume, as well as a discretization of the graphon variational problem arising in the infinite volume limit. As a result, we locate a curve in the parameter space where a one-step replica symmetry breaking transition occurs. Sampling a large graph in the broken symmetry phase is well described by a graphon with a structure very close to the one of an equi-bipartite graph.

Appendix: \(n=3\)

In order to show how the cloning algorithm works, we explicitly compute the cumulant generating function of triangles in the simplest case, that is the graph of size $n=3$. Since the probability of the unique triangle is $p^3$, we have

$$\begin{aligned} \mu _{3,p}(\alpha ) = \frac{1}{3} \ln \left\langle \exp \frac{\alpha }{3} T_3(X) \right\rangle ^{ER}_3 = \frac{1}{3} \ln \left[ \mathrm{e}^\frac{\alpha }{3}p^3 +1-p^3 \right] . \end{aligned}$$

(46)

We compute again (46) by applying the dynamics described in Sect. 3 to a family of M clones. The three steps required to construct the edges of the graph $G_3$ are represented in Fig. 11.

Fig. 11

Graphical representation of the evolution step, according to the tilted dynamics $P_\alpha (\cdot , \cdot )$ of cloning scheme. The levels of the tree display the values of the elements of the adjacency matrix $x_{1,2}, x_{1,3}, x_{2,3}$ during the evolution leading to $G_3$. M is the initial size of the population. In all levels the expected sizes of the sub-populations with a given configuration of values $x_{1,2}, x_{1,3}, x_{2,3}$ are reported

The leafs of the tree represent the occupation variables of the three possible edges, $x_{1,2}, x_{1,3}, x_{2,3}$. In the first step the edge connecting two vertices, say 1 and 2, of each clone is added with probability p. Thus, in the clone population about pM graphs have the edge (1, 2) and $(1-p)M$ graphs have not this edge, see the first level in Fig.11 . In the second step the edge (1, 3) is added, still with probability p, leading to four possible values for the pair $(x_{1,2},x_{1,3})$. Thus, the expected numbers of types are $Mp^2, Mp(1-p), Mp(1-p), M(1-p)^2$, see level 2 in Fig. 11. Let us observe that in the first two steps, since $\Delta T(x,y)=0$, the original and tilted transition probabilities coincide: $P_\alpha (x,y)=P(x,y)$, see (30). The situation changes in the last step. Indeed, the configuration of edges $x=11$ (which means that $x_{1,2}=1$ and $x_{1,3}=1$) may evolve to $y=111$, with $\Delta T(x,y)=1$, or to $y=110$, in which case $\Delta T(x,y)=0$. For all the other configurations x we have $\Delta T(x,y)=0$. Thus, see the definition (30) of the tilted probability:

$$\begin{aligned} P_\alpha (11,111)= \frac{\mathrm{e}^\frac{\alpha }{3}p}{ \mathrm{e}^\frac{\alpha }{3}p+1-p}, \quad P_\alpha (11,110)= \frac{1-p}{ \mathrm{e}^\frac{\alpha }{3}p+1-p}, \end{aligned}$$

(47)

being $P(11,111)=p$, $P(11,110)=1-p$ and $k_\alpha (11)= p\mathrm{e}^\frac{\alpha }{3}+1-p$. Then, the probability of the paths connecting the root $\phi$ to the leafs 111, respectively 110, will be, :

$$\begin{aligned} {\mathbb {P}}(\phi ,111)= \frac{\mathrm{e}^\frac{\alpha }{3}p^3}{ \mathrm{e}^\frac{\alpha }{3}p+1-p},\quad {\mathbb {P}}(\phi ,110)= \frac{p^2(1-p)}{ \mathrm{e}^\frac{\alpha }{3}p+1-p}. \end{aligned}$$

(48)

The average in (31) can be computed as the sum over the paths from the root to the leafs (equivalently, as the sum over the leafs). Recalling that $k_\alpha (11)= p\mathrm{e}^\frac{\alpha }{3}+1-p$ and observing also that $k_\alpha (x)=1$ if $x\ne 11$, from (31) we have:

$$\begin{aligned} \mu _{3,p}(\alpha )= & {} \frac{1}{3} \ln \left[ {\mathbb {P}}(\phi ,111) k_\alpha (11)+ {\mathbb {P}}(\phi ,110) k_\alpha (11)+ \sum _{x\notin \{111, 110\}}{\mathbb {P}}(\phi ,x ) \right] \end{aligned}$$

(49)

$$\begin{aligned}= & {} \frac{1}{3} \ln \left[ \frac{p^3\mathrm{e}^\frac{\alpha }{3}}{ p\mathrm{e}^\frac{\alpha }{3}+1-p} (p\mathrm{e}^\frac{\alpha }{3}+1-p) + \frac{p^2(1-p)}{ p\mathrm{e}^\frac{\alpha }{3}+1-p} (p\mathrm{e}^\frac{\alpha }{3}+1-p) + (1-p^2) \right] \nonumber \\ \end{aligned}$$

(50)

$$\begin{aligned}= & {} \frac{1}{3} \ln \left[ p^3\mathrm{e}^\frac{\alpha }{3} +1-p^3 \right] , \end{aligned}$$

(51)

The cloning algorithm simulates (51) by producing a final population of expected size

$$\begin{aligned} M_3=M\,{\mathbb {P}}(\phi ,111) k_\alpha (11)+ M\, {\mathbb {P}}(\phi ,110) k_\alpha (11)+\sum _{x\notin \{111, 110 \}} M\, {\mathbb {P}}(\phi , x). \end{aligned}$$

Then, the cumulant generating function is computed from the size of the final population $M_3$ as

$$\begin{aligned} \mu _{3,p}(\alpha ) = \frac{1}{3} \ln \left[ \frac{M_3}{M} \right] . \end{aligned}$$

(52)