# The Giant Component and Explosive Percolation

Last time we left off with a tantalizing conjecture: a random graph with edge probability $p = 5/n$ is almost surely a connected graph. We arrived at that conjecture from some ad-hoc data analysis, so let’s go back and treat it with some more rigorous mathematical techniques. As we do, we’ll discover some very interesting “threshold theorems” that essentially say a random graph will either certainly have a property, or it will certainly not have it.

The phase transition we empirically observed from last time.

## Big components

Recalling the basic definition: an Erdős-Rényi (ER) random graph with $n$ vertices and edge probability $p$ is a probability distribution over all graphs on $n$ vertices. Generatively, you draw from an ER distribution by flipping a $p$-biased coin for each pair of vertices, and adding the edge if you flip heads. We call the random event of drawing a graph from this distribution a “random graph” even though it’s not a graph, and we denote an ER random graph by $G(n,p)$. When $p = 1/2$, the distribution $G(n,1/2)$ is the uniform distribution over all graphs on $n$ vertices.

Now let’s get to some theorems. The main tools we’ll use are called the first and second moment method. Let’s illustrate them by example.

### The first moment method

Say we want to know what values of $p$ are likely to produce graphs with isolated vertices (vertices with no neighbors), and which are not. Of course, the value of $p$ will depend on $n \to \infty$ in general, but we can already see by example that if $p = 1/2$ then the probability of a fixed vertex being isolated is $2^{-n} \to 0$. We can use the union bound (sum this value over all vertices) to show that the probability of any vertex being isolated is at most $n2^{-n}$ which also tends to zero very quickly. This is not the first moment method, I’m just making the point that all of our results will be interpreted asymptotically as $n \to \infty$.

So now we can ask: what is the expected number of isolated vertices? If I call $X$ the random variable that counts the expected number of isolated vertices, then I’m asking about $\mathbb{E}[X]$. Really what I’m doing is interpreting $X$ as a random variable depending on $n, p(n)$, and asking about the evolution of $\mathbb{E}[X]$ as $n \to \infty$.

Now the first moment method states, somewhat obviously, that if the expectation tends to zero then the value of $X$ itself also tends to zero. Indeed, this follows from Markov’s inequality, which states that the probability that $X \geq a$ is bounded by $\mathbb{E}[X]/a$. In symbols,

$\displaystyle \Pr[X \geq a] \leq \frac{\mathbb{E}[X]}{a}$.

In our case $X$ is counting something (it’s integer valued), so asking whether $X > 0$ is equivalent to asking whether $X \geq 1$. The upper bound on the probability of $X$ being strictly positive is then just $\mathbb{E}[X]$.

So let’s find out when the expected number of isolated vertices goes to zero. We’ll use the wondrous linearity of expectation to split $X$ into a sum of counts for each vertex. That is, if $X_i$ is 1 when vertex $i$ is isolated and 0 otherwise (this is called an indicator variable), then $X = \sum_{i=1}^n X_i$ and linearity of expectation gives

$\displaystyle \mathbb{E}[X] = \mathbb{E}[\sum_{i=1}^n X_i] = \sum_{i=1}^n \mathbb{E}[X_i]$

Now the expectation of an indicator random variable is just the probability that the event occurs (it’s trivial to check). It’s easy to compute the probability that a vertex is isolated: it’s $(1-p)^n$. So the sum above works out to be $n(1-p)^n$. It should really be $n(1-p)^{n-1}$ but the extra factor of $(1-p)$ doesn’t change anything. The question is what’s the “smallest” way to set $p$ as a function of $n$ in order to make the above thing go to zero? Using the fact that $(1-x) < e^{-x}$ for all $x > 0$, we get

$n(1-p)^n < ne^{-pn}$

And setting $p = (\log n) / n$ simplifies the right hand side to $ne^{- \log n} = n / n = 1$. This is almost what we want, so let’s set $p$ to be anything that grows asymptotically faster than $(\log n) / n$. The notation for this is $\omega((\log n) / n)$. Then using some slick asymptotic notation we can prove that the RHS of the inequality above goes to zero, and so the LHS must as well. Back to the big picture: we just showed that the expectation of $X$ (the expected number of isolated vertices) goes to zero, and so by the first moment method the value of $X$ (the actual number of isolated vertices) has to go to zero with probability tending to 1.

Some quick interpretations: when $p = (\log n) / n$ each vertex has $\log n$ neighbors in expectation. Moreover, having no isolated vertices is just a little bit short of the entire graph being connected (our ultimate goal is to figure out exactly when this happens). But already we can see that our conjecture from the beginning is probably false: we aren’t able to use this same method to show that when $p = c/n$ for some constant $c$ rules out isolated vertices as $n \to \infty$. We just got lucky in our data analysis that 5 is about the natural log of 100 (which is 4.6).

### The second moment method

Now what about the other side of the coin? If $p$ is asymptotically less than $(\log n) / n$ do we necessarily get isolated vertices? That would really put our conjecture to rest. In this case the answer is yes, but it might not be in general. Let’s discuss.

We said that in general if $\mathbb{E}[X] \to 0$ then the value of $X$ has to go to zero too (that’s the first moment method). The flip side of this is: if $\mathbb{E}[X] \to \infty$ does necessarily the value of $X$ also tend to infinity? The answer is not always yes. Here is a gruesome example I originally heard from a book: say $X$ is the number of people that will die in the next decade due to an asteroid hitting the earth. The probability that the event happens is quite small, but if it does happen then the number of people that will die is quite large. It is perfectly reasonable for this to drag up the expectation (as the world population grows every decade), but at least we hope a growing population doesn’t by itself increase the value of $X$.

Mathematics is on our side here. We’re asking under what conditions on $\mathbb{E}[X]$ does the following implication hold: $\mathbb{E}[X] \to \infty$ implies $\Pr[X > 0] \to 1$.

With the first moment method we used Markov’s inequality (a statement about expectation, also called the first moment). With the second moment method we’ll use a statement about the second moment (variances), and the most common is Chebyshev’s inequality. Chebyshev’s inequality states that the probability $X$ deviates from its expectation by more than $c$ is bounded by $\textup{Var}[X] / c^2$. In symbols, for all $c > 0$ we have

$\displaystyle \Pr[|X - \mathbb{E}[X]| \geq c] \leq \frac{\textup{Var}[X]}{c^2}$

Now the opposite of $X > 0$, written in terms of deviation from expectation, is $|X - \mathbb{E}[X]| \geq \mathbb{E}[X]$. In words, in order for any number $a$ to be zero, it has to have a distance of at least $b$ from any number $b$. It’s such a stupidly simple statement it’s almost confusing. So then we’re saying that

$\displaystyle \Pr[X = 0] \leq \frac{\textup{Var}[X]}{\mathbb{E}[X]^2}$.

In order to make this probability go to zero, it’s enough to have $\textup{Var}[X] = o(\mathbb{E}[X]^2)$. Again, the little-o means “grows asymptotically slower than.” So the numerator of the fraction on the RHS will grow asymptotically slower than the denominator, meaning the whole fraction tends to zero. This condition and its implication are together called the “second moment method.”

Great! So we just need to compute $\textup{Var}[X]$ and check what conditions on $p$ make it fit the theorem. Recall that $\textup{Var}[X] = \mathbb{E}[X^2] - \mathbb{E}[X]^2$, and we want to upper bound this in terms of $\mathbb{E}[X]^2$. Let’s compute $\mathbb{E}[X]^2$ first.

$\displaystyle \mathbb{E}[X]^2 = n^2(1-p)^{2n}$

Now the variance.

$\displaystyle \textup{Var}[X] = \mathbb{E}[X^2] - n^2(1-p)^{2n}$

Expanding $X$ as a sum of indicator variables $X_i$ for each vertex, we can split the square into a sum over pairs. Note that $X_i^2 = X_i$ since they are 0-1 valued indicator variables, and $X_iX_j$ is the indicator variable for both events happening simultaneously.

\displaystyle \begin{aligned} \mathbb{E}[X^2] &= \mathbb{E}[\sum_{i,j} X_{i,j}] \\ &=\mathbb{E} \left [ \sum_i X_i^2 + \sum_{i \neq j} X_iX_j \right ] \\ &= \sum_i \mathbb{E}[X_i^2] + \sum_{i \neq j} \mathbb{E}[X_iX_j] \end{aligned}

By what we said about indicators, the last line is just

$\displaystyle \sum_i \Pr[i \textup{ is isolated}] + \sum_{i \neq j} \Pr[i,j \textup{ are both isolated}]$

And we can compute each of these pieces quite easily. They are (asymptotically ignoring some constants):

$\displaystyle n(1-p)^n + n^2(1-p)(1-p)^{2n-4}$

Now combining the two terms together (subtracting off the square of the expectation),

\displaystyle \begin{aligned} \textup{Var}[X] &\leq n(1-p)^n + n^2(1-p)^{-3}(1-p)^{2n} - n^2(1-p)^{2n} \\ &= n(1-p)^n + n^2(1-p)^{2n} \left ( (1-p)^{-3} - 1 \right ) \end{aligned}

Now we divide by $\mathbb{E}[X]^2$ to get $n^{-1}(1-p)^{-n} + (1-p)^{-3} - 1$. Since we’re trying to see if $p = (\log n) / n$ is a sharp threshold, the natural choice is to let $p = o((\log n) / n)$. Indeed, using the $1-x < e^{-x}$ upper bound and plugging in the little-o bounds the whole quantity by

$\displaystyle \frac{1}{n}e^{o(\log n)} + o(n^{1/n}) - 1 = o(1)$

i.e., the whole thing tends to zero, as desired.

## Other thresholds

So we just showed that the property of having no isolated vertices in a random graph has a sharp threshold at $p = (\log n) / n$. Meaning at any larger probability the graph is almost surely devoid of isolated vertices, and at any lower probability the graph almost surely has some isolated vertices.

This might seem like a miracle theorem, but there turns out to be similar theorems for lots of properties. Most of them you can also prove using basically the same method we’ve been using here. I’ll list some below. Also note they are all sharp, two-sided thresholds in the same way that the isolated vertex boundary is.

• The existence of a component of size $\omega(\log (n))$ has a threshold of $1/n$.
• $p = c/n$ for any $c > 0$ is a threshold for the existence of a giant component of linear size $\Theta(n)$. Moreover, above this threshold no other components will have size $\omega(\log n)$.
• In addition to $(\log n) / n$ being a threshold for having no isolated vertices, it is also a threshold for connectivity.
• $p = (\log n + \log \log n + c(n)) / n$ is a sharp threshold for the existence of Hamiltonian cycles in the following sense: if $c(n) = \omega(1)$ then there will be a Hamilton cycle almost surely, if $c(n) \to -\infty$ there will be no Hamiltonian cycle almost surely, and if $c(n) \to c$ the probability of a Hamiltonian cycle is $e^{-e^{-c}}$. This was proved by Kolmos and Szemeredi in 1983. Moreover, there is an efficient algorithm to find Hamiltonian cycles in these random graphs when they exist with high probability.

## Explosive Percolation

So now we know that as the probability of an edge increases, at some point the graph will spontaneously become connected; at some time that is roughly $\log(n)$ before, the so-called “giant component” will emerge and quickly engulf the entire graph.

Here’s a different perspective on this situation originally set forth by Achlioptas, D’Souza, and Spencer in 2009. It has since become called an “Achlioptas process.”

The idea is that you are watching a random graph grow. Rather than think about random graphs as having a probability above or below some threshold, you can think of it as the number of edges growing (so the thresholds will all be multiplied by $n$). Then you can imagine that you start with an empty graph, and at every time step someone is adding a new random edge to your graph. Fine, eventually you’ll get so many edges that a giant component emerges and you can measure when that happens.

But now imagine that instead of being given a single random new edge, you are given a choice. Say God presents you with two random edges, and you must pick which to add to your graph. Obviously you will eventually still get a giant component, but the question is how long can you prevent it from occurring? That is, how far back can we push the threshold for connectedness by cleverly selecting the new edge?

What Achlioptas and company conjectured was that you can push it back (some), but that when you push it back as far as it can go, the threshold becomes discontinuous. That is, they believed there was a constant $\delta \geq 1/2$ such that the size of the largest component jumps from $o(n)$ to $\delta n$ in $o(n)$ steps.

This turned out to be false, and Riordan and Warnke proved it. Nevertheless, the idea has been interpreted in an interesting light. People have claimed it is a useful model of disaster in the following sense. If you imagine that an edge between two vertices is a “crisis” relating two entities. Then in every step God presents you with two crises and you only have the resources to fix one. The idea is that when the entire graph is connected, you have this one big disaster where all the problems are interacting with each other. The percolation process describes how long you can “survive” while avoiding the big disaster.

There are critiques of this interpretation, though, mainly about how simplistic it is. In particular, an Achlioptas process models a crisis as an exogenous force when in reality problems are usually endogenous. You don’t expect a meteor to hit the Earth, but you do expect humans to have an impact on the environment. Also, not everybody in the network is trying to avoid errors. Some companies thrive in economic downturns by managing your toxic assets, for example. So one could reasonably argue that Achlioptas processes aren’t complex enough to model the realistic types of disasters we face.

Either way, I find it fantastic that something like a random graph (which for decades was securely in pure combinatorics away from applications) is spurring such discussion.

Next time, we’ll take one more dive into the theory of Erdős-Rényi random graphs to prove a very “meta” theorem about sharp thresholds. Then we’ll turn our attention to other models of random graphs, hopefully more realistic ones 🙂

Until then!

# The Erdős-Rényi Random Graph

During the 1950’s the famous mathematician Paul Erdős and Alfred Rényi put forth the concept of a random graph and in the subsequent years of study transformed the world of combinatorics. The random graph is the perfect example of a good mathematical definition: it’s simple, has surprisingly intricate structure, and yields many applications.

In this post we’ll explore basic facts about random graphs, slowly detail a proof on their applications to graph theory, and explore their more interesting properties computationally (a prelude to proofs about their structure). We assume the reader is familiar with the definition of a graph, which we’ve written about at length for non-mathematical audiences, and has some familiarity with undergraduate-level probability and combinatorics for the more mathematical sections of the post. We’ll do our best to remind the reader of these prerequisites as we go, and we welcome any clarification questions in the comment section.

## The Erdős-Rényi Model

The definition of a random graph is simple enough that we need not defer it to the technical section of the article.

Definition: Given a positive integer $n$ and a probability value $0 \leq p \leq 1$, define the graph $G(n,p)$ to be the undirected graph on $n$ vertices whose edges are chosen as follows. For all pairs of vertices $v,w$ there is an edge $(v,w)$ with probability $p$.

Of course, there is no single random graph. What we’ve described here is a process for constructing a graph. We create a set of $n$ vertices, and for each possible pair of vertices we flip a coin (often a biased coin) to determine if we should add an edge connecting them. Indeed, every graph can be made by this process if one is sufficiently lucky (or unlucky), but it’s very unlikely that we will have no edges at all if $p$ is large enough. So $G(n,p)$ is really a probability distribution over the set of all possible graphs on $n$ vertices. We commit a horrendous abuse of notation by saying $G$ or $G(n,p)$ is a random graph instead of saying that $G$ is sampled from the distribution. The reader will get used to it in time.

## Why Do We Care?

Random graphs of all sorts (not just Erdős’s model) find applications in two very different worlds. The first is pure combinatorics, and the second is in the analysis of networks of all kinds.

In combinatorics we often wonder if graphs exist with certain properties. For instance, in graph theory we have the notion of graph colorability: can we color the vertices of a graph with $k$ colors so that none of its edges are monochromatic? (See this blog’s primer on graph coloring for more) Indeed, coloring is known to be a very difficult problem on general graphs. The problem of determining whether a graph can be colored with a fixed number of colors has no known efficient algorithm; it is NP-complete. Even worse, much of our intuition about graphs fails for graph coloring. We would expect that sparse-looking graphs can be colored with fewer colors than dense graphs. One naive way to measure sparsity of a graph is to measure the length of its shortest cycle (recall that a cycle is a path which starts and ends at the same vertex). This measurement is called the girth of a graph. But Paul Erdős proved using random graphs, as we will momentarily, that for any choice of integers $g,k$ there are graphs of girth $\geq g$ which cannot be colored with fewer than $k$ colors. Preventing short cycles in graphs doesn’t make coloring easier.

The role that random graphs play in this picture is to give us ways to ensure the existence of graphs with certain properties, even if we don’t know how to construct an example of such a graph. Indeed, for every theorem proved using random graphs, there is a theorem (or open problem) concerning how to algorithmically construct those graphs which are known to exist.

Pure combinatorics may not seem very useful to the real world, but models of random graphs (even those beyond the Erdős-Rényi model) are quite relevant. Here is a simple example. One can take a Facebook user $u$ and form a graph of that users network of immediate friends $N(u)$ (excluding $u$ itself), where vertices are people and two people are connected by an edge if they are mutual friends; call this the user’s friendship neighborhood. It turns out that the characteristics of the average Facebook user’s friendship neighborhood resembles a random graph. So understanding random graphs helps us understand the structure of small networks of friends. If we’re particularly insightful, we can do quite useful things like identify anomalies, such as duplicitous accounts, which deviate quite far from the expected model. They can also help discover trends or identify characteristics that can allow for more accurate ad targeting. For more details on how such an idea is translated into mathematics and code, see Modularity (we plan to talk about modularity on this blog in the near future; lots of great linear algebra there!).

Random graphs, when they exhibit observed phenomena, have important philosophical consequences. From a bird’s-eye view, there are two camps of scientists. The first are those who care about leveraging empirically observed phenomena to solve problems. Many statisticians fit into this realm: they do wonderful magic with data fit to certain distributions, but they often don’t know and don’t care whether the data they use truly has their assumed properties. The other camp is those who want to discover generative models for the data with theoretical principles. This is more like theoretical physics, where we invent an arguably computational notion of gravity whose consequences explain our observations.

For applied purposes, the Erdős-Rényi random graph model is in the second camp. In particular, if something fits in the Erdős-Rényi model, then it’s both highly structured (as we will make clear in the sequel) and “essentially random.” Bringing back the example of Facebook, this says that most people in the average user’s immediate friendship neighborhood are essentially the same and essentially random in their friendships among the friends of $u$. This is not quite correct, but it’s close enough to motivate our investigation of random graph models. Indeed, even Paul Erdős in his landmark paper mentioned that equiprobability among all vertices in a graph is unrealistic. See this survey for a more thorough (and advanced!) overview, and we promise to cover models which better represent social phenomena in the future.

So lets go ahead and look at a technical proof using random graphs from combinatorics, and then write some programs to generate random graphs.

## Girth and Chromatic Number, and Counting Triangles

As a fair warning, this proof has a lot of moving parts. Skip to the next section if you’re eager to see some programs.

Say we have a $k$ and a $g$, and we wonder whether a graph can exist which simultaneously has no cycles of length less than $g$ (the girth) and needs at least $k$ colors to color. The following theorem settles this question affirmatively.  A bit of terminology: the chromatic number of a graph $G$, denoted $\chi(G)$, is the smallest number of colors needed to properly color $G$.

Theorem: For any natural numbers $k,g$, there exist graphs of chromatic number at least $k$ and girth at least $g$.

Proof. Taking our cue from random graphs, let’s see what the probability is that a random graph $G(n,p)$ on $n$ vertices will have our desired properties. Or easier, what’s the chance that it will not have the right properties? This is essentially a fancy counting argument, but it’s nicer if we phrase it in the language of probability theory.

The proof has a few twists and turns for those uninitiated to the probabilistic method of proof. First, we will look at an arbitrary $G(n,p)$ (where $n,p$ are variable) and ask two questions about it: what is the expected number of short cycles, and what is the expected “independence number” (which we will see is related to coloring). We’ll then pick a value of $p$, depending crucially on $n$, which makes both of these expectations small. Next, we’ll use the fact that if the probability that $G(n,p)$ doesn’t have our properties is strictly less than 1, then there has to be some instance in our probability space which has those properties (if no instance had the property, then the probability would be one!). Though we will not know what the graphs look like, their existence is enough to prove the theorem.

So let’s start with cycles. If we’re given a desired girth of $g$, the expected number of cycles of length $\leq g$ in $G(n,p)$ can be bounded by $(np)^{g+1}/(np-1)$. To see this, the two main points are how to count the number of ways to pick $k$ vertices in order to form a cycle, and a typical fact about sums of powers. Indeed, we can think of a cycle of length $k$ as a way to seat a choice of $k$ people around a circular table. There are $\binom{n}{k}$ possible groups of people, and $(k-1)!$ ways to seat one group. If we fix $k$ and let $n$ grow, then the product $\binom{n}{k}(k-1)!$ will eventually be smaller than $n^k$ (actually, this happens almost immediately in almost all cases). For each such choice of an ordering, the probability of the needed edges occurring to form a cycle is $p^j$, since all edges must occur independently of each other with probability $p$.

So the probability that we get a cycle of $j$ vertices is

$\displaystyle \binom{n}{j}(j-1)!p^j$

And by the reasoning above we can bound this by $n^jp^j$. Summing over all numbers $j = 3, \dots, g$ (we are secretly using the union bound), we bound the expected number of cycles of length $\leq g$ from above:

$\displaystyle \sum_{j=3}^g n^j p^j < \sum_{j=0}^g n^j p^j = \frac{(np)^{g+1}}{np - 1}$

Since we want relatively few cycles to occur, we want it to be the case that the last quantity, $(np)^{j+1}/(np-1)$, goes to zero as $n$ goes to infinity. One trick is to pick $p$ depending on $n$. If $p = n^l$, our upper bound becomes $n^{(l+1)(g+1)} / (n^{1+l} - 1)$, and if we want the quantity to tend to zero it must be that $(l+1)(g+1) < 1$. Solving this we get that $-1 < l < \frac{1}{g+1} - 1 < 0$. Pick such an $l$ (it doesn’t matter which), and keep this in mind: for our choice of $p$, the expected number of cycles goes to zero as $n$ tends to infinity.

On the other hand, we want to make sure that such a graph has high chromatic number. To do this we’ll look at a related property: the size of the largest independent set. An independent set of a graph $G = (V,E)$ is a set of vertices $S \subset V$ so that there are no edges in $E$ between vertices of $S$. We call $\alpha(G)$ the size of the largest independent set. The values $\alpha(G)$ and $\chi(G)$ are related, because any time you have an independent set you can color all the vertices with a single color. In particular, this proves the inequality $\chi(G) \alpha(G) \geq n$, the number of vertices of $G$, or equivalently $\chi(G) \geq n / \alpha(G)$. So if we want to ensure $\chi(G)$ is large, it suffices to show $\alpha(G)$ is small (rigorously, $\alpha(G) \leq n / k$ implies $\chi(G) \geq k$).

The expected number of independent sets (again using the union bound) is at most the product of the number of possible independent sets and the probability of one of these having no edges. We let $r$ be arbitrary and look at independent sets of size $r$ Since there are $\binom{n}{r}$ sets and each has a probability $(1-p)^{\binom{r}{2}}$ of being independent, we get the probability that there is an independent set of size $r$ is bounded by

$\displaystyle \textup{P}(\alpha(G) \geq r) \leq \binom{n}{r}(1-p)^{\binom{r}{2}}$

We use the fact that $1-x < e^{-x}$ for all $x$ to translate the $(1-p)$ part. Combining this with the usual $\binom{n}{r} \leq n^r$ we get the probability of having an independent set of size $r$ at most

$\displaystyle \textup{P}(\alpha(G) \geq r) \leq \displaystyle n^re^{-pr(r-1)/2}$

Now again we want to pick $r$ so that this quantity goes to zero asymptotically, and it’s not hard to see that $r = \frac{3}{p}\log(n)$ is good enough. With a little arithmetic we get the probability is at most $n^{(1-a)/2}$, where $a > 1$.

So now we have two statements: the expected number of short cycles goes to zero, and the probability that there is an independent set of size at least $r$ goes to zero. If we pick a large enough $n$, then the expected number of short cycles is less than $n/5$, and using Markov’s inequality we see that the probability that there are more than $n/2$ cycles of length at most $g$ is strictly less than 1/2. At the same time, if we pick a large enough $n$ then $\alpha(G) \geq r$ with probability strictly less than 1/2. Combining these two (once more with the union bound), we get

$\textup{P}(\textup{at least } n/2 \textup{ cycles of length } \leq g \textup{ and } \alpha(G) \geq r) < 1$

Now we can conclude that for all sufficiently large $n$ there has to be a graph on at least $n$ vertices which has neither of these two properties! Pick one and call it $G$. Now $G$ still has cycles of length $\leq g$, but we can fix that by removing a vertex from each short cycle (it doesn’t matter which). Call this new graph $G'$. How does this operation affect independent sets, i.e. what is $\alpha(G')$? Well removing vertices can only decrease the size of the largest independent set. So by our earlier inequality, and calling $n'$ the number of vertices of $G'$, we can make a statement about the chromatic number:

$\displaystyle \chi(G') \geq \frac{n'}{\alpha(G')} \geq \frac{n/2}{\log(n) 3/p} = \frac{n/2}{3n^l \log(n)} = \frac{n^{1-l}}{6 \log(n)}$

Since $-1 < l < 0$ the numerator grows asymptotically faster than the denominator, and so for sufficiently large $n$ the chromatic number will be larger than any $k$ we wish. Hence we have found a graph with girth at least $g$ and chromatic number at least $k$.

$\square$

## Connected Components

The statistical properties of a random graph are often quite easy to reason about. For instance, the degree of each vertex in $G(n,p)$ is $np$ in expectation. Local properties like this are easy, but global properties are a priori very mysterious. One natural question we can ask in this vein is: when is $G(n,p)$ connected? We would very much expect the answer to depend on how $p$ changes in relation to $n$. For instance, $p$ might look like $p(n) = 1/n^2$ or $\log(n) / n$ or something similar. We could ask the following question:

As $n$ tends to infinity, what limiting proportion of random graphs $G(n,p)$ are connected?

Certainly for some $p(n)$ which are egregiously small (for example, $p(n) = 0$), the answer will be that no graphs are connected. On the other extreme, if $p(n) = 1$ then all graphs will be connected (and complete graphs, at that!). So our goal is to study the transition phase between when the graphs are disconnected and when they are connected. A priori this boundary could be a gradual slope, where the proportion grows from zero to one, or it could be a sharp jump. Next time, we’ll formally state and prove the truth, but for now let’s see if we can figure out which answer to expect by writing an exploratory program.

We wrote the code for this post in Python, and as usual it is all available for download on this blog’s Github page.

We start with a very basic Node class to represent each vertex in a graph, and a function to generate random graphs

import random
class Node:
def __init__(self, index):
self.index = index
self.neighbors = []

def __repr__(self):
return repr(self.index)

def randomGraph(n,p):
vertices = [Node(i) for i in range(n)]
edges = [(i,j) for i in xrange(n) for j in xrange(i) if random.random() < p]

for (i,j) in edges:
vertices[i].neighbors.append(vertices[j])
vertices[j].neighbors.append(vertices[i])

return vertices


The randomGraph function simply creates a list of edges chosen uniformly at random from all possible edges, and then constructs the corresponding graph. Next we have a familiar sight: the depth-first search. We use it to compute the graph components one by one (until all vertices have been found in some run of a DFS).

def dfsComponent(node, visited):
for v in node.neighbors:
if v not in visited:
dfsComponent(v, visited)

def connectedComponents(vertices):
components = []
cumulativeVisited = set()

for v in vertices:
if v not in cumulativeVisited:
componentVisited = set([v])
dfsComponent(v, componentVisited)

components.append(componentVisited)
cumulativeVisited |= componentVisited

return components


The dfsComponent function simply searches in breadth-first fashion, adding every vertex it finds to the “visited” set.  The connectedComponents function keeps track of the list of components found so far, as well as the cumulative set of all vertices found in any run of bfsComponent. Hence, as we iterate through the vertices we can ignore vertices we’ve found in previous runs of bfsComponent. The “x |= y” notation is python shorthand for updating x via a union with y.

Finally, we can make a graph of the largest component of (independently generated) random graphs as the probability of an edge varies.

def sizeOfLargestComponent(vertices):
return max(len(c) for c in connectedComponents(vertices))

def graphLargestComponentSize(n, theRange):
return [(p, sizeOfLargestComponent(randomGraph(n, p))) for p in theRange]


Running this code and plotting it for $p$ varying from zero to 0.5 gives the following graph.

The blue line is the size of the largest component, and the red line gives a moving average estimate of the data.  As we can see, there is a very sharp jump peaking at $p=0.1$ at which point the whole graph is connected. It would appear there is a relatively quick “phase transition” happening here. Let’s zoom in closer on the interesting part.

It looks like the transition begins around 0.02, and finishes at around 0.1. Interesting… Let’s change the parameters a bit, and increase the size of the graph. Here’s the same chart (in the same range of $p$ values) for a graph with a hundred vertices.

Now the phase transition appears to have shifted to about $(0.01, 0.05)$, which is about multiplying the endpoints of the phase transition interval above by 1/2. The plot thickens… Once more, let’s move up to a graph on 500 vertices.

Again it’s too hard to see, so let’s zoom in.

This one looks like the transition starts at 0.002 and ends at 0.01. This is a 5-fold decrease from the previous one, and we increased the number of vertices by 5. Could this be a pattern? Here’s a conjecture to formalize it:

Conjecture: The random graph $G(n,p)$ enters a phase transition at $p=1/n$ and becomes connected almost surely at $p=5/n$.

This is not quite rigorous enough to be a true conjecture, but it sums up our intuition that we’ve learned so far. Just to back this up even further, here’s an animation showing the progression of the phase transition as $n = 20 \dots 500$ in steps of twenty. Note that the $p$ range is changing to maintain our conjectured window.

Looks pretty good. Next time we’ll see some formal mathematics validating our intuition (albeit reformulated in a nicer way), and we’ll continue to investigate other random graph models.

Until then!