2.1. Notation

Recall that

\begin{equation*} d_1 = pn_2\qquad\text{and}\qquad d_2 = pn_1\end{equation*}

are the degrees of vertices in a p-biregular graph, and thus, the number of edges in any p-biregular graph $H \in {{\mathcal{R}}(n_1,n_2,p)}$ is

\begin{equation*} M\,{:\!=}\,pN\qquad\text{where}\qquad N=n_1n_2.\end{equation*}

Throughout the proofs we also use shorthand notation

(12) \begin{equation} q = 1 - p, \quad \hat p = \min \left\{ p, q \right\}, \quad \hat n = \min \left\{ n_1,n_2 \right\},\end{equation}

and $[n] \,{:\!=}\, \left\{ 1, \dots, n \right\}$ . All logarithms appearing in this paper are natural.

By $\Gamma_{G}(v)$ we denote the set of neighbours of a vertex v in a graph G.

2.2. Switchings

The switching technique is used to compare the size of two classes of graphs, say ${\mathcal{R}}$ and ${\mathcal{R}}'$ , by defining an auxiliary bipartite graph $B\,{:\!=}\,B({\mathcal{R}},{\mathcal{R}}')$ , in which two graphs $H \in {\mathcal{R}}$ , $H' \in {\mathcal{R}}'$ are connected by an edge whenever H can be transformed into H ^′ by some operation (a forward switching) that deletes and/or creates some edges of H. By counting the number of edges of $B({\mathcal{R}},{\mathcal{R}}')$ in two ways, we see that

(13) \begin{equation} \sum_{H\in{\mathcal{R}}}\deg_B(H) = \sum_{H'\in{\mathcal{R}}'}\deg_B(H'),\end{equation}

which easily implies that

(14) \begin{equation}\frac{\min_{H' \in {\mathcal{R}}'} \deg_B(H')}{\max_{H \in {\mathcal{R}}} \deg_B(H)} \le \frac{|{\mathcal{R}}|}{|{\mathcal{R}}'|} \le \frac{\max_{H' \in {\mathcal{R}}'} \deg_B(H')}{\min_{H \in {\mathcal{R}}} \deg_B(H)}.\end{equation}

The reverse operation mapping $H' \in {\mathcal{R}}'$ to its neighbours in graph B, is called a backward switching. Usually, one defines the forward switching in such a way that the backward switching can be easily described.

All switchings used in this paper follow the same pattern. For a fixed graph $G \subseteq K$ (possibly empty), where $K \,{:\!=}\, {K_{n_1,n_2}}$ , the families ${\mathcal{R}}, {\mathcal{R}}'$ will be subsets of

(15) \begin{equation} {\mathcal{R}}_G \,{:\!=}\, \{ H \in {\mathcal{R}}(n_1,n_2,p)\ \, :\, \ G \subseteq H \}.\end{equation}

Every $H \in {\mathcal{R}}_G$ will be interpreted as a blue-red colouring of the edges of $K\setminus G$ : those in $H\setminus G$ will be coloured blue and those in $K \setminus H$ — red. Given $H\in{\mathcal{R}}$ , consider a subset S of the edges of $K\setminus G$ in which for every vertex $v \in V_1 \cup V_2$ the blue degree equals the red degree, i.e., $\deg_{(H\setminus G)\cap S}(v)=\deg_{(K\setminus H)\cap S}(v)$ . Then switching the colours within S produces another graph $H' \in {\mathcal{R}}_G$ . Formally, E(H ^′) is the symmetric difference $E(H) \triangle S$ . In each application of the switching technique, we will restrict the choices of S to make sure that $H' \in {\mathcal{R}}'$ .

As an elementary illustration of this technique, which nevertheless turns out to be useful in Section 6, we prove here the following result. A cycle in $K\setminus G$ is alternating if it is a union of a blue matching and a red matching. Note that the definition depends on H. We will omit mentioning this dependence, as H will always be clear from the context. Given $e \in K\setminus G$ , set

(16) \begin{equation} {\mathcal{R}}_{G,e} \,{:\!=}\, \left\{ H \in {\mathcal{R}}_G\ \, :\, \ e \in H \right\}\qquad\text{and }\qquad {\mathcal{R}}_{G,\neg e} \,{:\!=}\, \left\{ H \in {\mathcal{R}}_G\ \, :\, \ e \not\in H \right\}.\end{equation}

Proposition 4. Let a graph $G\subseteq K$ be such that ${\mathcal{R}}_G\neq\emptyset$ and let $e\in K\setminus G$ . Assume that for some number $D > 0$ and every $H\in {\mathcal{R}}_{G}$ the edge e is contained in an alternating cycle of length at most 2D. Then ${\mathcal{R}}_{G,\neg e}\neq\emptyset$ , ${\mathcal{R}}_{G, e}\neq\emptyset$ , and

\begin{equation*} \frac{1}{N^D - 1} \le \frac{|{\mathcal{R}}_{G,\neg e}|}{|{\mathcal{R}}_{G,e}|} \le N^D - 1. \end{equation*}

Proof. Let $B = B({\mathcal{R}}_{G,e}, {\mathcal{R}}_{G,\neg e})$ be the switching graph corresponding to the following forward switching: choose an alternating cycle S of length at most 2D containing e and switch the colours of edges within S. Note that the backward switching does precisely the same. Note that by the assumption, the minimum degree $\delta(B)$ is at least 1.

Since ${\mathcal{R}}_G={\mathcal{R}}_{G, e}\cup{\mathcal{R}}_{G, \neg e} \neq \emptyset$ , one of the classes ${\mathcal{R}}_{G, e}$ and ${\mathcal{R}}_{G, \neg e}$ is non-empty and, in view of $\delta(B) \ge 1$ , the other one is non-empty as well. For $\ell = 2, \dots, D$ , the number of cycles of length $2\ell$ containing e is, crudely, at most $n_1^{\ell - 1} n_2^{\ell - 1} = N^{\ell - 1}$ , hence the maximum degree is

\begin{equation*} \Delta(B)\le N + N^2 + \cdots + N^{D - 1} \le N^D - 1. \end{equation*}

Thus, by (14), we obtain the claimed bounds.

2.3. Probabilistic inequalities

We first state a few basic concentration inequalities that we apply in our proofs. For a sum X of independent Bernoulli (not necessarily identically distributed) random variables, writing $\mu = \operatorname{\mathbb{E}} X$ , we have (see Theorem 2.8, (2.5), (2.6), and (2.11) in [Reference Janson, Łuczak and Rucinski5]) that

(17) \begin{equation} \mathbb{P}\left(X \ge \mu + t\right) \le \exp \left\{ - \frac{t^2}{2\left( \mu + t/3 \right)} \right\}, \qquad t \ge 0, \end{equation}

(18) \begin{equation} \mathbb{P}\left(X \le \mu - t\right) \le \exp \left\{ - \frac{t^2}{2\mu} \right\}, \qquad t \ge 0. \end{equation}

Let $\Gamma$ be a set of size $|\Gamma|=g$ and let $A \subseteq \Gamma$ , $|A|=a \ge1$ . For an integer $r \in [0,g]$ , choose uniformly at random a subset $R \subseteq \Gamma$ of size $|R|=r$ . The random variable $Y = |A \cap R|$ has then the hypergeometric distribution $\operatorname{Hyp}(g, a, r)$ with expectation $\mu \,{:\!=}\, \operatorname{\mathbb{E}} Y = ar/g$ . By Theorem 2.10 in [Reference Janson, Łuczak and Rucinski5], inequalities (17) and (18) hold for Y, too.

Moreover, by Remark 2.6 in [Reference Janson, Łuczak and Rucinski5], inequalities (17) and (18) also hold for a random variable Z which has Poisson distribution $\operatorname{Po}(\mu)$ with expectation $\mu$ . In this case, we also have the following simple fact. For $k\ge0$ , set $q_k=\mathbb{P}\left(Z=k\right)$ . Then $q_k/q_{k-1}=\mu/k$ , and hence $k = \lfloor\mu\rfloor$ maximises $q_k$ (we say that such k is a mode of Z). Since $\operatorname{Var} Z = \mu$ , by Chebyshev’s inequality, $\mathbb{P}\left(|Z-\mu|<\sqrt{2\mu}\right)\ge1/2$ . Moreover, the interval $(\mu - \sqrt{2 \mu}, \mu + \sqrt{2 \mu})$ contains at most $\lceil \sqrt{8\mu}\:\rceil$ integers, hence it follows that

(19) \begin{equation} q_{\lfloor \mu \rfloor} \ge \frac{1/2}{\lceil \sqrt{8\mu}\: \rceil}.\end{equation}

2.4. Asymptotic enumeration of dense bipartite graphs

To estimate co-degrees of $\mathbb{R}(n,n,p)$ we will use the following asymptotic formula by Canfield, Greenhill and McKay [Reference Canfield, Greenhill and McKay1]. We reformulate it slightly for our convenience.

Given two vectors ${\mathbf{d}}_1 = (d_{1,v}, v \in V_1)$ and ${\mathbf{d}}_2 = (d_{2,v}, v \in V_2)$ of positive integers such that $\sum_{v \in V_1} d_{1, v} = \sum_{v \in V_2} d_{2,v}$ , let ${{\mathcal{R}}({\mathbf{d}}_1,{\mathbf{d}}_2)}$ be the class of bipartite graphs on $(V_1, V_2)$ with vertex degrees $\deg(v) = d_{i,v}, v \in V_i, i = 1,2$ . Let $|V_i| = n_i$ and write $\bar d_i = {n_i}^{-1}\sum_{v \in V_i} d_{i,v}$ , $D_i = \sum_{v \in V_i} (d_{i,v} - \bar d_i)^2$ , $p = \bar d_1/n_2 = \bar d_2/n_1$ , and $q = 1 - p$ .

Theorem 5. ([Reference Canfield, Greenhill and McKay1]) Given any positive constants a, b, C such that $a+b < 1/2$ , there exists a constant $\epsilon > 0$ so that the following holds. Consider the set of degree sequences ${\mathbf{d}}_1,{\mathbf{d}}_2$ satisfying

1. i. $\max_{v \in V_1}|d_{1,v} - \bar d_1| \le Cn_2^{1/2 + \epsilon}$ , $\max_{v \in V_2} |d_{2,v} - \bar d_2| \le Cn_1^{1/2 + \epsilon}$

2. ii. $\max \{n_1, n_2\} \le C(pq)^2(\min \left\{ n_1, n_2 \right\})^{1 + \epsilon}$

3. iii. $ \frac{(1 - 2p)^2}{4pq} \left( 1 + \frac{5n_1}{6n_2} + \frac{5n_2}{6n_1} \right) \le a \log \max \left\{ n_1, n_2 \right\}$ .

If $\min\{n_1, n_2\} \to \infty$ , then uniformly for all such ${\mathbf{d}}_1,{\mathbf{d}}_2$

(20) \begin{align} |{\mathcal{R}}({\mathbf{d}}_1,{\mathbf{d}}_2)| &= \binom{n_1n_2}{p n_1n_2}^{-1} \prod_{v \in V_1}\binom {n_2}{d_{1,v}} \prod_{v \in V_2}\binom {n_1}{d_{2,v}} \times \nonumber\\&\quad \times \exp \left[ -\frac{1}{2}\left( 1- \frac{D_1}{pqn_1n_2} \right)\left(1- \frac{D_2}{pqn_1n_2} \right) + O\left((\max\left\{ n_1,n_2 \right\})^{-b}\right) \right],\end{align}

where the constants implicit in the error term may depend on a,b,C.

Note that condition (ii) of Theorem 5 implies the corresponding condition in [Reference Canfield, Greenhill and McKay1] after adjusting $\epsilon$ . Also, the uniformity of the bound is not explicitly stated in [Reference Canfield, Greenhill and McKay1], but, given $n_1, n_2$ , one should take ${\mathbf{d}}_1, {\mathbf{d}}_2$ with the worst error and apply the result in [Reference Canfield, Greenhill and McKay1].

3.1. The set-up

Recall that $K = {K_{n_1,n_2}}$ has $N = n_1n_2$ edges. Consider a sequence of graphs $\mathbb{G}(t) \subseteq K$ for $t = 0, \dots , N$ , where $\mathbb{G}(0)$ is empty and, for $t < N$ , $\mathbb{G}(t + 1)$ is obtained from $\mathbb{G}(t)$ by adding an edge $\varepsilon_{t + 1}$ chosen from $K \setminus \mathbb{G}(t)$ uniformly at random, that is, for every graph $G \subseteq K$ of size t and every edge $e \in K \setminus G$

(21) \begin{equation}\mathbb{P}\left({\varepsilon_{t+1} = e}\,|\,{\mathbb{G}(t) = G}\right) = \frac{1}{N - t}.\end{equation}

Of course, $(\varepsilon_1, \dots, \varepsilon_N)$ is just a uniformly random ordering of the edges of K.

Our approach to proving Theorem 2 is to represent the random regular graph $\mathbb{R}(n_1,n_2,p)$ as the outcome of a random process which behaves similarly to $(\mathbb{G}(t))_t$ . Recalling that a p-biregular graph has $M = pN$ edges, let

\begin{equation*}(\eta_1, \dots, \eta_M)\end{equation*}

be a uniformly random ordering of the edges of $\mathbb{R}(n_1,n_2,p)$ . By taking the initial segments, we obtain a sequence of random graphs

\begin{equation*} \mathbb{R}(t) = \{\eta_1, \dots, \eta_t\}, \qquad t = 0, \dots, M.\end{equation*}

For convenience, we shorten

\begin{equation*} \mathbb{R}\,{:\!=}\, \mathbb{R}(M) = \mathbb{R}(n_1,n_2,p)\;.\end{equation*}

Let us mention here that for a fixed $H \in {{\mathcal{R}}(n_1,n_2,p)}$ , conditioning on $\mathbb{R} = H$ , the edge set of the random subgraph $\mathbb{R}(t)$ is a uniformly random t-element subset of the edge set of H. This observation often leads to a hypergeometric distribution and will be utilised several times in our proofs.

We say that a graph G with t edges is admissible, if the family ${\mathcal{R}}_G$ (see definition (15)) is non-empty, or, equivalently,

\begin{equation*} \mathbb{P}\left(\mathbb{R}(t) = G\right) > 0.\end{equation*}

For an admissible graph G with t edges and any edge $ e \in K \setminus G$ , let

(22) \begin{equation}p_{t+1}(e,G) \,{:\!=}\, \mathbb{P}\left({\eta_{t+1} = e}\,|\,{\mathbb{R}(t) = G}\right).\end{equation}

The conditional space underlying (22) can be described as first extending G uniformly at random to an element of ${\mathcal{R}}_G$ and then randomly permuting the new $M-t$ edges.

The main idea behind the proof of Theorem 2 is that the conditional probabilities in (22) behave similarly to those in (21). Observe that $p_{t+1}(e,\mathbb{R}(t)) = \mathbb{P}\left({\eta_{t+1} = e}\,|\,{\mathbb{R}(t)}\right)$ . Given a real number $\chi \ge 0$ and $t \in \{0, \dots, M - 1\}$ , we define an $\mathbb{R}(t)$ -measurable event

\begin{equation*} {\mathcal{A}}(t,\chi) \,{:\!=}\, \left\{ p_{t+1}(e,\mathbb{R}(t)) \ge \frac{1 - \chi}{N - t} \text{ for every } e \in K \setminus \mathbb{R}(t) \right\}.\end{equation*}

In the crucial lemma below, we are going to show that for suitably chosen $\gamma_0, \gamma_1, \dots$ , a.a.s. the events ${\mathcal{A}}(t,\gamma_t)$ occur simultaneously for all $t = 0,\dots,t_0-1$ where $t_0$ is quite close to M. Postponing the choice of $\gamma_t$ and $t_0$ , we define an event

(23) \begin{equation} {\mathcal{A}} \,{:\!=}\, \bigcap_{t = 0}^{ t_0 - 1}{\mathcal{A}}(t, \gamma_t).\end{equation}

Intuitively, the event ${\mathcal{A}}$ asserts that up to time $t_0$ the process $\mathbb{R}(t)$ stays ‘almost uniform’, which will enable us to embed ${\mathbb{G}(n_1,n_2,m)}$ into $\mathbb{R}(t_0)$ .

To define time $t_0$ , it is convenient to parametrise the time by the proportion of edges of $\mathbb{R}$ ‘not yet revealed’ after t steps. For this, we define by

(24) \begin{equation} \tau = \tau(t) \,{:\!=}\, 1 - \frac{t}{M} \in [0,1]\qquad\text{and so}\qquad t = (1 - \tau)M.\end{equation}

Given a constant $C > 0$ and, we define (recalling the notation in (12)) the ‘final’ value $\tau_0$ of $\tau$ as

(25) \begin{equation} \tau_0 \,{:\!=}\, \begin{cases} \dfrac{3 \cdot 3240^2(C + 4)\log N}{p \hat n}, \quad & p \le 0.49, \\ \\[-9pt] 700(3(C + 4))^{1/4} \left( q^{3/2}\mathbb{I} + \left(\frac{\log N}{\hat n}\right)^{1/4}\right), \quad & p > 0.49. \end{cases}\end{equation}

(Some of the constants appearing here and below are sharp or almost sharp, but others have room to spare as we round them up to the nearest ‘nice’ number.)

Consider the following assumptions on p (which we will later show to follow from the assumptions of Theorem 2):

(26) \begin{equation} \hat p \ge \frac{3 \cdot 3240^2(C+4)\log N}{\hat n},\end{equation}

(27) \begin{equation} \hat p \cdot \mathbb{I} \le \frac{49}{51}\cdot\frac1{340^2 (C + 4)^{1/6}}, \end{equation}

and

(28) \begin{equation} q \ge 680\left( \frac{3(C+4)\log N}{\hat{n}}\right)^{1/4}. \end{equation}

At the end of this subsection we show that these three assumptions imply

(29) \begin{equation} \tau_0 \le 1,\end{equation}

so that

\begin{equation*} t_0 \,{:\!=}\, \lfloor (1 - \tau_0)M \rfloor\end{equation*}

is a non-negative integer. Further, for $t = 0, \dots, M - 1$ , define

(30) \begin{equation} \gamma_t \,{:\!=}\, 1080 \hat p^2 \mathbb{I} + \begin{cases} 3240 \sqrt{\dfrac{2(C + 3)\log N}{\tau p \hat n}}, \quad & p \le 0.49, \\ \\[-9pt] 25,000 \sqrt{\dfrac{(C + 3)\log N}{ \tau^2q^2 \hat n}} , \quad & p > 0.49. \end{cases}\end{equation}

Taking (29) for granted, we now state our crucial lemma, which is proved in Section 4.

Lemma 6. For every constant $C > 0$ , if assumptions (26), (27) and (28) hold, then

(31) \begin{equation} \mathbb{P}\left({\mathcal{A}}\right) = 1- O(N^{-C}),\end{equation}

where the constant implicit in the O-term in (31) may also depend on C.

It remains to show (29). When $p \le 0.49$ , inequality (29) is equivalent to (26). For $p > 0.49$ , we have $q \le 51\hat p/49$ , which together with assumptions (27) and (28) implies that

\begin{equation*} \tau_0 \le 700 \left( 3(C+4) \right)^{1/4} \left( \frac{51}{49}\hat p \cdot \mathbb{I} \right)^{3/2} + 700 \cdot \frac{q}{680} \le \frac{700 \cdot 3^{1/4}}{340^3} + \frac{700 \cdot 0.51}{680} < 1.\end{equation*}

3.2. Proof of Theorem 2

From Lemma 6 we are going to deduce Theorem 2 using a coupling argument similar to the one which was employed by Dudek, Frieze, Ruciński and Šileikis [Reference Dudek, Frieze, Ruciński and Šileikis2], but with an extra tweak (inspired by Kim and Vu [Reference Kim and Vu6]) of letting the probabilities $\gamma_t$ of Bernoulli random variables depend on t, which reduces the error $\gamma$ in (5).

It is easy to check that the assumptions of Lemma 6 follow from the assumptions of Theorem 2. Indeed, (28) coincides with assumption (4), while (26) and (27) follow from the assumption $\gamma \le 1$ (see (5)) for sufficiently large $C^*$ .

Our aim is to couple $(\mathbb{G}(t))_t$ and $(\mathbb{R}(t))_t$ on the event ${\mathcal{A}}$ defined in (23). For this we will define a graph process $\mathbb{R}'(t) \,{:\!=}\, \{\eta^{\prime}_1, \dots, \eta^{\prime}_t\}, t = 0, \dots, t_0$ so that for every admissible graph G with $t \in [0, M -1]$ edges and every $e \in K \setminus G$

(32) \begin{equation} \mathbb{P}\left({\eta^{\prime}_{t+1} = e}\,|\,{\mathbb{R}'(t) = G}\right) = p_{t+1}(e, G),\end{equation}

where $p_{t+1}(e, G)$ was defined in (22). Note that $\mathbb{R}'(0)$ is an empty graph. Since the distribution of the process $(\mathbb{R}(t))_t$ is determined by the conditional probabilities (22), in view of (32), the distribution of $\mathbb{R}'(t_0)$ is the same as that of $\mathbb{R}(t_0)$ and therefore we will identify $\mathbb{R}'(t_0)$ with $\mathbb{R}(t_0)$ . As the second step, we will show that a.a.s. ${\mathbb{G}(n_1,n_2,m)}$ can be sampled from $\mathbb{R}'(t_0) = \mathbb{R}(t_0)$ .

Proceeding with the definition, set $\mathbb{R}'(0)$ to be the empty graph and define graphs $\mathbb{R}'(t)$ , $t = 1, \dots, t_0$ , inductively, as follows. Hence, let us further fix $t \in [0, t_0-1]$ and suppose that

\begin{equation*} \mathbb{R}'(t)=R_t\quad\mbox{and}\quad \mathbb{G}(t)=G_t\end{equation*}

have been already chosen. Our immediate goal is to select a random pair of edges $\varepsilon_{t+1}$ and $\eta_{t+1}^{\prime}$ , according to, resp., (21) and (32), in such a way that the event $\varepsilon_{t+1}\in \mathbb{R}'(t+1)$ is quite likely.

To this end, draw $\varepsilon_{t + 1}$ uniformly at random from $K \setminus G_t$ and, independently, generate a Bernoulli random variable $\xi_{t+1}$ with the probability of success $1 - \gamma_t$ (which is in [0,1] by (146)). If event ${\mathcal{A}}(t, \gamma_t)$ has occurred, that is, if

(33) \begin{equation}p_{t+1}(e, R_t) \ge \frac{1 - \gamma_t}{N - t} \quad \text{ for every } \quad e \in K\setminus R_t,\end{equation}

then draw a random edge $\zeta_{t+1} \in K \setminus R_t$ according to the distribution

\begin{equation*}\mathbb{P}\left({\zeta_{t+1} = e}\,|\,{\mathbb{R}'(t) = R_t}\right) \,{:\!=}\, \frac{p_{t+1}(e, R_t) - (1 - \gamma_t)/(N - t) }{\gamma_t} \ge 0,\end{equation*}

where the inequality holds by (33). Observe also that

\begin{equation*}\sum_{e\in K \setminus R_t}\mathbb{P}\left({\zeta_{t+1} = e}\,|\,{\mathbb{R}'(t) = R_t}\right) = 1,\end{equation*}

so $\zeta_{t+1}$ has a properly defined distribution. Finally, fix an arbitrary bijection

\begin{equation*} f_{R_t, G_t}\ \, :\, \ R_t \setminus G_t \to G_t \setminus R_t\end{equation*}

between the sets of edges and define

\begin{equation*}\eta^{\prime}_{t+1} = \begin{cases}\varepsilon_{t+1}, &\text{ if } \xi_{t + 1} = 1, \varepsilon_{t+1} \in K \setminus R_t,\\ \\[-9pt] f_{R_t, G_t}(\varepsilon_{t+1}), &\text{ if } \xi_{t + 1} = 1, \varepsilon_{t+1} \in R_t,\\ \\[-9pt] \zeta_{t+1}, &\text{ if } \xi_{t + 1} = 0. \\\end{cases}\end{equation*}

On the other hand, if event ${\mathcal{A}}(t, \gamma_t)$ has failed, then $\eta^{\prime}_{t+1}$ is sampled directly (without defining $\zeta_{t+1}$ ) according to the distribution (32). With this definition of $(\mathbb{R}'(t))_{t= 0}^{t_0}$ , it is easy to check that for $\eta^{\prime}_{t+1}$ defined above, (32) indeed holds, so from now on we drop the prime $^{\prime}$ and identify $\mathbb{R}'(t)$ with $\mathbb{R}(t)$ , which is a subset of $\mathbb{R}(n_1,n_2,p)$ .

Most importantly, we conclude that, for $t = 0, \cdots, t_0 - 1$

(34) \begin{equation}{\mathcal{A}}(t, \gamma_t) \cap \{\xi_{t+1} = 1\} \quad \implies \quad \varepsilon_{t+1} \in \mathbb{R}(t+1).\end{equation}

In view of this, define

\begin{equation*} S \,{:\!=}\, \left\{ t \in [t_0]\ \, :\ \, \xi_t = 1 \right\}\end{equation*}

and recall that $m=\lceil (1-\gamma)M\rceil$ . If $|S| \ge m$ , define ${\mathbb{G}(n_1,n_2,m)}$ as, say, the edges indexed by the smallest m elements of S (note that since the vectors $(\xi_i)$ and $(\varepsilon_i)$ are independent, after conditioning on S, these m edges are uniformly distributed), and if $|S| < m$ , then define ${\mathbb{G}(n_1,n_2,m)}$ as, say, the graph with edges $\{\varepsilon_1, \dots, \varepsilon_m\}$ . Recalling the definition (23) of the event ${\mathcal{A}}$ , by (34) we observe that ${\mathcal{A}}$ implies the inclusion $\left\{ \varepsilon_t\ \, :\, \ t \in S \right\} \subseteq \mathbb{R}(t_0) \subset \mathbb{R}(M) = \mathbb{R}(n_1,n_2,p)$ . On the other hand $|S| \ge m$ implies ${\mathbb{G}(n_1,n_2,m)} \subseteq \left\{ \varepsilon_t\ \, :\, \ t \in S \right\}$ , so

\begin{equation*}\mathbb{P}\left({\mathbb{G}(n_1,n_2,m)} \subseteq \mathbb{R}\right) \ge \mathbb{P}\left(\{|S| \ge m\}\cap {\mathcal{A}} \right).\end{equation*}

Since, by Lemma 6, event ${\mathcal{A}}$ holds with probability $1 - O(N^{-C})$ , to complete the proof of (6) it suffices to show that also

\begin{equation*} \mathbb{P}\left(|S| \ge m\right) = 1 - O\left(N^{-C}\right) .\end{equation*}

For this we need the following claim whose technical proof is deferred to Section 8.

Claim 7. We have

(35) \begin{equation} \operatorname{\mathbb{E}} |S| \ge t_0 - \theta M,\end{equation}

where

\begin{equation*} \theta \,{:\!=}\, 1080 \hat p^2 \mathbb{I} + \begin{cases} 6480\sqrt{\dfrac{2(C + 3)\log N}{p\hat n}}, &\quad p \le 0.49, \\ \\[-9pt] 6250\sqrt{\dfrac{(C + 3)\log N}{q^2\hat n } }\log \frac{\hat n}{\log N}, &\quad p > 0.49, \end{cases}\end{equation*}

and, with $\gamma$ as in (5),

(36) \begin{equation} \gamma \ge \tau_0 + \theta + 2/M + \sqrt{\frac{2C \log N}{M}}.\end{equation}

Recalling that $t_0 = \lfloor (1 - \tau_0)M \rfloor$ and $m = \lceil(1-\gamma)M\rceil$ , we have

(37) \begin{equation} \begin{split} t_0 - \theta M - m &\ge (1-\tau_0)M - 1 - \theta M - (1-\gamma)M - 1 \\[3pt] &= (\gamma - \tau_0 - \theta)M - 2 \overset{\mbox{(36)}}{\ge} \sqrt{2C M \log N}. \end{split} \end{equation}

Since $|S|$ is a sum of independent Bernoulli random variables,

\begin{align*} \mathbb{P}\left(|S| < m\right) &= \mathbb{P}\left(|S| < \operatorname{\mathbb{E}}|S|- (\operatorname{\mathbb{E}}|S| - m) \right) \overset{\mbox{(35)}}{\le} \mathbb{P}\left(|S| < \operatorname{\mathbb{E}}|S| -(t_0 - \theta M - m)\right) \\ \fbox{(18), $\operatorname{\mathbb{E}} |S| \le M$}\quad &\overset{\mbox{}}{\le} \exp \left( -\frac{(t_0 - \theta M - m)^2}{2M} \right) \overset{\mbox{(38)}}{\le} \exp \left( - C \log N \right) = N^{-C},\end{align*}

which as mentioned above, implies (6).

Finally, we prove (7) by coupling $\mathbb{G}(n_1, n_2, p')$ with $\mathbb{G}(n_1,n_2, m)$ so that the former is a subset of the latter with probability $1 - O(N^{-C})$ . Denote by $X \,{:\!=}\, e(\mathbb{G}(n_1,n_2,p'))$ the number of edges of $\mathbb{G}(n_1,n_2, p')$ . Whenever $X \le m$ , choose X edges of $\mathbb{G}(n_1,n_2,m)$ at random and declare them to be a copy of $\mathbb{G}(n_1,n_2,p')$ . Otherwise sample $\mathbb{G}(n_1,n_2,p')$ independently from $\mathbb{G}(n_1,n_2,m)$ . Hence it remains to show that $\mathbb{P}\left(X > m\right) = O(N^{-C})$ .

Recalling that $m = \lceil{{(1-\gamma)}}\rceil pN$ and $p' = (1 - 2\gamma)p$ , we have $m/N \ge (1 - \gamma)p = p' + \gamma p$ . Since $X \sim \operatorname{Bin}(N,p')$ , Chernoff’s bound (17) implies that

(38) \begin{equation} \mathbb{P}\left(X > m\right) \le \mathbb{P}\left(X > p'N + \gamma p N\right) \le \exp\left\{ -\frac{(\gamma pN)^2}{2(p'N + \gamma pN/3)} \right\} \le e^{-\gamma^2pN/2}. \end{equation}

Choosing $C^*$ sufficiently large, the definition (5) implies $\gamma \ge \sqrt{\frac{2C\log N}{p N}}$ with a lot of room (to see this easier in the case $p > 0.49$ , note that (4) implies $\log(\hat n/\log N) \ge 1$ , say). So (38) implies $\mathbb{P}\left(X > m\right) \le N^{-C}$ .

4.1. Preparations

Recall our notation $K = {K_{n_1,n_2}}$ , and ${\mathcal{R}}_G$ defined in (15). Fix an integer $t \in [0, t_0)$ . For a graph G with t edges and $e, f \in K\setminus G$ , define

\begin{equation*} {\mathcal{R}}_{G,e,\neg f} \,{:\!=}\, \left\{ H \in {\mathcal{R}}_G\ \, :\, \ e \in H, f \notin H\right\}.\end{equation*}

By skipping a few technical steps, we will see that the essence of the proof of Lemma 6 is to show that the ratio ${|{\mathcal{R}}_{G,e,\neg f}|} / {|{\mathcal{R}}_{G,f,\neg e}|}$ is approximately 1 for any pair of edges e,f, where G is a ‘typical’ instance of $\mathbb{R}(t)$ .

Recalling our generic definition of switchings from Subsection 2.2, let us treat the graphs in ${\mathcal{R}} \,{:\!=}\, {\mathcal{R}}_{G,e,\neg f}$ and ${\mathcal{R}}' \,{:\!=}\, {\mathcal{R}}_{G,f,\neg e}$ as blue-red edge colourings of the graph $K \setminus G$ . Recall that a path or a cycle in $K \setminus G$ is alternating if no two consecutive edges have the same colour.

When edges e, f are disjoint, we define the switching graph $B = B({\mathcal{R}},{\mathcal{R}}')$ by putting an edge between $H \in {\mathcal{R}}$ and $H' \in {\mathcal{R}}'$ whenever there is an alternating 6-cycle containing e and f in H such that switching the colours in the cycle gives H $^{\prime}$ (see Figure 2). If e and f share a vertex, instead of 6-cycles we use alternating 4-cycles containing e and f.

Figure 2. Switching between H and H $^{\prime}$ when e and f are disjoint: solid edges are in $H \setminus G$ (or $H' \setminus G$ ) and the dashed ones in $K \setminus H$ (or $K \setminus H'$ ).

It is easy to describe the vertex degrees in B. For distinct $u,v\in V_i$ , let $\theta_{G,H}(u,v)$ be the number of (alternating) paths uxv such that ux is blue and xv is red. Note that $\theta_{G,H}(u,v) = |\Gamma_{H\setminus G}(u)\cap\Gamma_{K\setminus H}(v)|$ . Then, setting $f = u_1u_2$ and $e = v_1v_2$ , where $u_i,v_i\in V_i$ , $i = 1,2$ ,

(39) \begin{equation} \deg_B(H) = \prod_{i: u_i \neq v_i}\theta_{G,H}(u_i,v_i), \quad H \in {\mathcal{R}}\end{equation}

and

(40) \begin{equation} \deg_B(H') = \prod_{i: u_i \neq v_i}\theta_{G,H'}(v_i,u_i), \quad H' \in {\mathcal{R}}'.\end{equation}

Note that equation (13) is equivalent to

(41) \begin{equation} \frac{|{\mathcal{R}}|}{|{\mathcal{R}}'|} = \frac{\frac1{|{\mathcal{R}}'|}\sum_{H'\in{\mathcal{R}}'}\deg_B(H')}{\frac1{|{\mathcal{R}}|}\sum_{H\in{\mathcal{R}}}\deg_B(H)}.\end{equation}

In view of (39) and (40), the denominator of the RHS above is the (conditional) expectation of the random variable $\prod_{i : u_i \neq v_i} \theta_{G,\mathbb{R}}(u_i,v_i)$ , given that $\mathbb{R}$ contains G and e, but not f (and similarly for the numerator).

To get an idea of how large that expectation could be, let us focus on one factor, say, $\theta_{G,\mathbb{R}}(u_1,v_1)$ , assuming $u_1 \neq v_1$ . Clearly, the red degree of $v_1$ equals $|\Gamma_{K\setminus\mathbb{R}}(v_1)| = n_2 - d_2 = qn_2$ . Since $|\Gamma_{\mathbb{R}}(v_1)| = pn_2$ , viewing $\mathbb{R} \setminus G$ as a $\tau$ -dense subgraph of $\mathbb{R}$ (see (24)) we expect that the blue neighbourhood $|\Gamma_{\mathbb{R}\setminus G}(u_1)|$ is approximately $\tau p n_2$ . It is reasonable to expect that for typical G and $\mathbb{R}$ the red and blue neighbourhoods intersect proportionally, that is, on a set of size about $q \cdot \tau p \cdot n_2 = \tau qd_1$ .

Inspired by this heuristic, we say that, for $\delta > 0$ , an admissible graph G with t edges is $\delta$ -typical if

(42) \begin{equation} \max_{u_1u_2, v_1v_2 \in K \setminus G} \mathbb{P}\left({\max_{i \in [2] : u_i \neq v_i}\left|\frac{\theta_{G,\mathbb{R}}(u_i,v_i)}{\tau qd_i } - 1\right| > \delta }\,|\,{G \subseteq \mathbb{R}, v_1v_2 \in \mathbb{R}, u_1u_2 \notin \mathbb{R}}\right) \le \tau^2 \delta,\end{equation}

where the outer maximum is taken over distinct pairs of edges. (The bound $\tau^2 \delta$ has hardly any intuition, but it is simple and sufficient for our purposes.)

Lemma 6 is a relatively easy consequence of the upcoming Lemma 8, which states that for a suitably chosen function $\delta(t)$ it is very likely that the initial segments of $\mathbb{R}$ are $\delta(t)$ -typical.

For each $t = 0, \dots, t_0 - 1$ , define

(43) \begin{equation} \delta(t) \,{:\!=}\, 120\hat p^2 \mathbb{I} + 360 \sqrt{\frac{(C + 3)\lambda(t)}{6\tau pq\hat n }} ,\end{equation}

where

(44) \begin{equation} \lambda(t) \,{:\!=}\,\begin{cases} 6 \log N, \quad &p \le 0.49, \\ \\[-9pt] 6\log N + \dfrac{64\log N}{\tau p q}, \quad &p > 0.49.\end{cases}\end{equation}

For future reference, note that

(45) \begin{equation}\delta(t) \le \gamma_t/9.\end{equation}

Indeed, recalling (30), for $p\le0.49$ ,

\begin{equation*} 9\delta(t) \le 1080\hat p^2 \mathbb{I} + 3240 \sqrt{\frac{2(C + 3)\log N}{\tau p\hat n }} =\gamma_t,\end{equation*}

while, for $p > 0.49$ , noting that $\lambda(t)\le 70\log N/(\tau pq)$ , we have

\begin{equation*}9\delta(t) \le 1080\hat p^2 \mathbb{I} + 3240 \sqrt{\frac{70(C + 3)\log N}{0.49^2 \cdot 6\tau^2q^2\hat n}} \le \gamma_t.\end{equation*}

Lemma 8. For every constant $C > 0$ , under the conditions of Lemma 6,

\begin{align*} \mathbb{P}\left(\mathbb{R}(t) \text{ is } \delta(t)\text{-typical for all } t < t_0\right) = 1 - O(N^{-C}).\\[-25pt]\end{align*}

Note that the LHS of (42) is a function of graph G, say f(G). Hence, Lemma 8 asserts that, with high probability, $f(\mathbb{R}(t))$ is small for all $t < t_0$ . The main idea of the proof of Lemma 8, which we defer to Subsection 4.3, is to bound $f(\mathbb{R}(t))$ by a ratio of two simpler functions (again conditional probabilities) of $\mathbb{R}(t)$ . Lemmas 9 and 10 below bound each of these conditional probabilities separately.

For any $t = 0, \dots, M$ and distinct $u_i, v_i \in V_i$ , $i = 1,2$ , set

(46) \begin{equation}\theta_t(u_i,v_i) \,{:\!=}\, \theta_{\mathbb{R}(t),\mathbb{R}}(u_i,v_i),\end{equation}

where $\theta_{G,H}(u_i,v_i) = |\Gamma_{H\setminus G}(u_i)\cap\Gamma_{K\setminus H}(v_i)|$ was introduced earlier in this subsection.

Lemma 9. For every constant $C>0$ , under the conditions of Lemma 6, for $i \in [2]$ we have that with probability $1 - O(N^{-C})$ ,

(47) \begin{equation} \mathbb{P}\left({ \max_{u,v \in V_i, u \neq v} \left|\frac{\theta_t(u,v)}{\tau qd_{i} } - 1\right| > \delta(t)}\,|\,{\mathbb{R}(t)}\right) \le e^{ - 2\lambda(t) }, \qquad \text{for all } t < t_0. \end{equation}

Lemma 10. For every constant $C > 0$ , under the conditions of Lemma 6, with probability $1 - O(N^{-C})$

(48) \begin{equation} \min_{e, f \in K \setminus \mathbb{R}(t), e \neq f}\mathbb{P}\left({e \in \mathbb{R}, f \notin \mathbb{R}}\,|\,{\mathbb{R}(t)}\right) \ge e^{ - \lambda(t) }, \qquad \text{for all } t < t_0.\end{equation}

4.2. Proof of Lemma 6

In view of Lemma 8, it suffices to show that if $t < t_0$ , and a t-edge graph G is $\delta(t)$ -typical, then

\begin{equation*} \min_{e \in K \setminus G}p_{t+1}(e,G) \ge \frac{1 - \gamma_t}{N - t}.\end{equation*}

Fix $f \in K \setminus G$ which maximises $p_{t+1}(f,G)$ . Since the average of $p_{t+1}(e,G)$ over $e \in K \setminus G$ is exactly $\frac{1}{N - t}$ , we have $p_{t+1}(f,G) \ge \frac{1}{N - t}$ and therefore it is enough to prove that, for every $e \in K \setminus (G \cup \{f\})$ ,

\begin{equation*}\frac{p_{t+1}(e,G)}{p_{t+1}(f,G)} \ge 1 - \gamma_t.\end{equation*}

Recall the definitions of ${\mathcal{R}}_G$ in (15), ${\mathcal{R}}_{G,e}, {\mathcal{R}}_{G, \neg e}$ in (16), and ${\mathcal{R}} = {\mathcal{R}}_{G,e,\neg f}$ and ${\mathcal{R}}' = {\mathcal{R}}_{G, f, \neg e}$ from Subsection 4.1 and observe that ${\mathcal{R}} = {\mathcal{R}}_{G\cup \{e\},\neg f}$ too. In view of the remark immediately following (22),

\begin{equation*} p_{t+1}(e,G) = \mathbb{P}\left({e \in \mathbb{R}}\,|\,{\mathbb{R}(t) = G}\right) \cdot \mathbb{P}\left({\eta_{t+1} = e}\,|\,{\mathbb{R}(t) = G, e \in \mathbb{R}}\right) = \frac{|{\mathcal{R}}_{G,e}|}{|{\mathcal{R}}_G|} \cdot \frac{1}{M-t}.\end{equation*}

Therefore,

\begin{equation*} 1 \ge \frac{p_{t+1}(e,G)}{p_{t+1}(f,G)} = \frac{|{\mathcal{R}}_{G,e}|}{|{\mathcal{R}}_{G,f}|} = \frac{|{\mathcal{R}}_{G \cup \{e,f\}}| + |{\mathcal{R}}_{G,e,\neg f}|}{|{\mathcal{R}}_{G \cup \{e, f\}}| + |{\mathcal{R}}_{G,f,\neg e}| } \ge \frac{|{\mathcal{R}}|}{|{\mathcal{R}}'|}.\end{equation*}

Write $e = v_1v_2$ and $f = u_1u_2$ and for simplicity assume that both $u_1 \neq v_1$ and $u_2 \neq v_2$ (otherwise the proof goes mutatis mutandis and is, in fact, a bit simpler). By (39), (40), and (41),

\begin{equation*} \frac{|{\mathcal{R}}|}{|{\mathcal{R}}'|} = \frac{\operatorname{\mathbb{E}}_{\text{top}}}{\operatorname{\mathbb{E}}_{\text{bottom}}},\end{equation*}

where

\begin{equation*} \operatorname{\mathbb{E}}_{\text{top}} \,{:\!=}\, \frac1{|{\mathcal{R}}'|}\sum_{H'\in{\mathcal{R}}}\deg_B(H')= \operatorname{\mathbb{E}}\left[{\theta_{G,\mathbb{R}}(v_1,u_1)\theta_{G,\mathbb{R}}(v_2,u_2)}\,|\,{G \subseteq \mathbb{R}, f \in \mathbb{R}, e \notin \mathbb{R}}\right]\end{equation*}

and

\begin{equation*} \operatorname{\mathbb{E}}_{\text{bottom}} \,{:\!=}\, \frac1{|{\mathcal{R}}|}\sum_{H\in{\mathcal{R}}}\deg_B(H)= \operatorname{\mathbb{E}}\left[{\theta_{G,\mathbb{R}}(u_1,v_1)\theta_{G,\mathbb{R}}(u_2,v_2)}\,|\,{G \subseteq \mathbb{R}, e \in \mathbb{R}, f \notin \mathbb{R}}\right].\end{equation*}

Since G is $\delta$ -typical with $\delta = \delta(t)$ , denoting the LHS of (42) as $p_*$ , we have $p_* \le \tau^2 \delta$ , so

(49) \begin{align} \notag \operatorname{\mathbb{E}}_{\text{top}} &\ge (1 - \delta) \tau qd_1 \cdot (1 - \delta)\tau qd_2 \cdot (1 - p_*) + 0 \cdot p_* \\[3pt] &\ge (1 - \delta)^2 \tau^2p^2q^2n_1n_2(1 - \tau^2\delta ) \ge (1 - \delta)^3 \tau^2p^2q^2N.\end{align}

Moreover, since, deterministically,

\begin{equation*} \theta_{G,\mathbb{R}}(u_1, v_1)\theta_{G, \mathbb{R}}(u_2, v_2) \le \min\{p,q\} n_2\cdot \min\{p,q\}n_1 \le 4p^2q^2N,\end{equation*}

using (42) again, we infer that

(50) \begin{align} \operatorname{\mathbb{E}}_{\text{bottom}} \notag &\le (1 + \delta) \tau qd_1 \cdot (1 + \delta)\tau qd_2 \cdot (1 - p_*) + 4p^2q^2N \cdot p_*\\[3pt] &\le (1 + \delta)^2\tau^2p^2q^2N + 4p^2q^2N \cdot \tau^2\delta \le (1 + 6\delta + \delta^2)\tau^2p^2q^2N .\end{align}

Finally, combining the bounds on $\operatorname{\mathbb{E}}_{\text{top}}$ and $\operatorname{\mathbb{E}}_{\text{bottom}}$ and using (45), we conclude that

(51) \begin{equation} \frac{p_{t+1}(e,G)}{p_{t+1}(f,G)} \ge \frac{(1 - \delta)^3}{(1 + 6 \delta + \delta^2)} \ge 1 - 9\delta \ge 1 - \gamma_t,\end{equation}

and the proof of Lemma 6 is complete.

4.3. Proof of Lemma 8

By Lemmas 9 and 10 the events (47) (for each $i \in [2]$ ) and (48) hold simultaneously with probability $1 - O(N^{-C})$ . Hence it suffices to fix an arbitrary integer $t < t_0$ and a realisation of $\mathbb{R}(t)$ , say $\mathbb{R}(t) = G$ satisfying inequalities (47) (for each $i \in [2]$ ) and (48) and to prove that G is $\delta(t)$ -typical. Fix any such G and define events

\begin{equation*} {\mathcal{E}} \,{:\!=}\, \bigcup_{i \in [2]} \left\{ \max_{u,v \in V_i, u \neq v}\left|\frac{\theta_{G,\mathbb{R}}(u,v)}{\tau qd_i } - 1\right| > \delta(t) \right\} \quad\text{and}\quad {\mathcal{F}}_{e,f} \,{:\!=}\, \{ e\in \mathbb{R}, f \notin \mathbb{R} \}.\end{equation*}

Since, conditioning on the event $\mathbb{R}(t) = G$ , we have $\theta_{G,\mathbb{R}}(u_i,v_i) = \theta_t(u_i,v_i)$ , by the choice of G we have

(52) \begin{equation} \mathbb{P}\left({{\mathcal{E}}}\,|\,{\mathbb{R}(t) = G}\right) \le 2e^{-2\lambda(t)}\end{equation}

and

(53) \begin{equation} \min_{e,f\in K \setminus G}\mathbb{P}\left({{\mathcal{F}}_{e,f}}\,|\,{\mathbb{R}(t) = G}\right) \ge e^{-\lambda(t)}.\end{equation}

Note that the probability on the LHS of (42) does not change if we replace $G \subseteq \mathbb{R}$ by $\mathbb{R}(t) = G$ , since conditioning on either event makes $\mathbb{R}$ uniformly distributed over ${\mathcal{R}}_G$ (i.e., biregular graphs containing G) and the random variables $\theta_{G,\mathbb{R}}(u_i,v_i)$ do not depend on the random ordering of the edges of $\mathbb{R}$ . Hence it remains to prove that (52) and (53) imply, for any distinct edges $e = v_1v_2,$ $f = u_1u_2 \in K \setminus G$ , that

(54) \begin{equation} \mathbb{P}\left({\max_{i \in [2] : u_i \neq v_i}\left|\frac{\theta_{G,\mathbb{R}}(u_i,v_i)}{\tau qd_i } - 1\right| > \delta(t) }\,|\,{\mathbb{R}(t) = G, e \in \mathbb{R}, f \notin \mathbb{R}}\right) \le \tau^2 \delta(t).\end{equation}

The probability on the LHS of (54) is at most $\mathbb{P}\left({{\mathcal{E}}}\,|\,{\mathbb{R}(t) = G, {\mathcal{F}}_{e,f}}\right)$ . Inequalities (52) and (53) imply that

\begin{equation*} \mathbb{P}\left({{\mathcal{E}}}\,|\,{\mathbb{R}(t) = G, {\mathcal{F}}_{e,f}}\right) = \frac{\mathbb{P}\left({{\mathcal{E}}\cap {\mathcal{F}}_{e,f}}\,|\,{\mathbb{R}(t) = G}\right)}{\mathbb{P}\left({{\mathcal{F}}_{e,f}}\,|\,{\mathbb{R}(t) = G}\right)} \le \frac{\mathbb{P}\left({{\mathcal{E}}}\,|\,{\mathbb{R}(t) = G}\right)}{\mathbb{P}\left({{\mathcal{F}}_{e,f}}\,|\,{\mathbb{R}(t) = G}\right)} \le 2 e^{ - \lambda(t)}. \end{equation*}

Finally, even a quick glance at the definitions of $\delta(t)$ and $\tau_0$ (see (43) and (25)) reveals that $\min \{ \tau_0,\delta(t) \} \ge 1/\hat n$ . Thus, we get, with a huge margin,

\begin{align*} 2e^{-\lambda(t)}\le 2/N^6 \le 1/N^2 \le 1/\hat n^4 \le \tau_0^2\delta(t) \le \tau^2\delta(t).\\[-30pt]\end{align*}

5.1. Co-degrees in the random biregular graph

Recall that $\Gamma_F(v)$ is the set of neighbours of a vertex v in a graph F. We define the co-degree of two distinct vertices u, v as

(55) \begin{equation} \operatorname{cod}_F(u,v) \,{:\!=}\, |\Gamma_F(u) \cap \Gamma_F(v)|.\end{equation}

A few times we will use the following simple observation: for $F \subseteq K$ and distinct $u,v\in V_1$ ,

(56) \begin{equation} \begin{split} \operatorname{cod}_{F}(u, v) &= |\Gamma_{F}(u)| + |\Gamma_{F}(v)| - |\Gamma_{F}(u) \cup \Gamma_{F}(v)| \\[3pt] &= \deg_F(u) + \deg_F(v) - \left( n_2 - \operatorname{cod}_{K \setminus F}(u,v) \right), \end{split}\end{equation}

where, recall, $K = {K_{n_1,n_2}}$ is the complete bipartite graph.

Due to symmetry, we prove the following concentration result for pairs of vertices on one side of the bipartition only.

Lemma 11. Suppose that $\hat p\hat n \to \infty$ and let $\lambda = \lambda(n_1,n_2)$ be such that $\lambda \to \infty$ . Then, for any distinct $u_1, v_1 \in V_1$ ,

\begin{equation*} \mathbb{P}\left( |\operatorname{cod}_{\mathbb{R}}(u_1,v_1) - p^2n_{2}| > 20\left( \hat p^3n_{2}\mathbb{I} + \frac{\hat pn_{2}}{\hat n} + \sqrt{\lambda \hat p^2 n_2} \right) + \lambda \right) = O\left( \sqrt N e^{ - \lambda } \right),\end{equation*}

where $\mathbb{I}$ is defined in (3).

Proof. We first claim that it is sufficient to assume $p \le 1/2$ and prove that for any distinct $u_1, v_1 \in V_1$

(57) \begin{equation} \mathbb{P}\left( |\operatorname{cod}_{\mathbb{R}}(u_1,v_1) - p^2n_2| > 20\left(p^3n_2\mathbb{I} + \frac{pn_2}{\hat n} + \sqrt{\lambda p^2n_2 } \right) + \lambda \right) = O\left( \sqrt N e^{ - \lambda } \right).\end{equation}

To see this, note that by (56), recalling $q = 1 - p$ ,

\begin{equation*} \operatorname{cod}_{\mathbb{R}}(u_1, v_1) - p^2n_2= 2pn_2 - (n_2 - \operatorname{cod}_{K \setminus \mathbb{R}}(u_1,v_1))- p^2n_2=\operatorname{cod}_{K \setminus \mathbb{R}}(u_1,v_1) - q^2n_2. \end{equation*}

Further, $K \setminus \mathbb{R} = K \setminus \mathbb{R}(n_1,n_2,p)$ has the same distribution as $\mathbb{R}(n_1,n_2,q)$ . So, if $p > 1/2$ , the lemma follows by applying (57) with q instead of p.

The crude proof idea comes from our anticipation that $\operatorname{cod}_\mathbb{R}(u_1,v_1)$ behaves similarly to $\operatorname{cod}_{{\mathbb{G}(n_1,n_2,p)}}(u_1,v_1)$ , which is distributed as $\operatorname{Bin}(n_2, p^2)$ or, approximately, as $\operatorname{Po}(p^2n_2)$ . We will show that each tail of $\operatorname{cod}_\mathbb{R}(u_1,u_2)$ is comparable to the tail of $\operatorname{Po}(\mu)$ with expectation $\mu$ fairly close to $p^2n_2$ and then apply the Chernoff bounds (this is packaged in Claim 12 below). Further we consider the cases $\mathbb{I} = 1$ and $\mathbb{I} = 0$ and apply Claim 12 analogously to the proof of Theorem 2.1 in [Reference Krivelevich, Sudakov, Vu and Wormald7]: in the case $\mathbb{I} = 1$ we use switchings and in the case $\mathbb{I} = 0$ we use asymptotic enumeration (Theorem 5).

Fix distinct $u_1, v_1 \in V_1$ and set $X(u_1,v_1): = \operatorname{cod}_{\mathbb{R}}(u_1,v_1)$ . Further, recall that ${{\mathcal{R}}(n_1,n_2,p)}$ is the class of subgraphs of ${K_{n_1,n_2}}$ such that every $v \in V_i$ has degree $d_i$ , for $i = 1, 2$ , and let

\begin{equation*} {\mathcal{R}}_k(u_1,v_1) = \{H \in {{\mathcal{R}}(n_1,n_2,p)}\ \, :\ \, \operatorname{cod}_H(u_1,v_1) = k \}, \qquad k = 0, \dots, d_1. \end{equation*}

Claim 12. Fix two vertices $u_1,v_1\in V_1$ . Suppose that positive numbers $r_k, k = 0, \dots, d_1,$ are such that

\begin{equation*} |{\mathcal{R}}_k(u_1,v_1)| \sim r_k, \text{ uniformly in } \textit{k} \text{ as } (n_1, n_2) \to \infty, \end{equation*}

and there exist numbers $0 \le \mu_- \le \mu_+ \le N$ such that

\begin{equation*} \frac{r_k}{r_{k-1}} \le \frac{\mu_+}{k}, \text{ for } k \in [\mu_+, d_1] \quad \text{ and } \quad \frac{r_{k-1}}{r_k} \le \frac{k}{\mu_-}, \text{ for } k \in [1, \mu_-]. \end{equation*}

If $x \ge \sqrt{2\mu_+ \lambda} + \lambda$ , then

(58) \begin{equation} \mathbb{P}\left(X(u_1,v_1) \ge \mu_+ + x\right) + \mathbb{P}\left(X(u_1,v_1) \le \mu_- - x\right) = O(\sqrt{N} e^{-\lambda}). \end{equation}

Proof. Note that if $Z_+ \sim \operatorname{Po}(\mu_+)$ , then $\mu_+/k = \mathbb{P}\left(Z_+ = k\right)/\mathbb{P}\left(Z_+ = k - 1\right)$ . Setting $m = \lfloor \mu_+ \rfloor$ , for any integer $i\in\{m,\dots,d_1\}$ , and abbreviating $X\,{:\!=}\,X(u_1,v_1)$ and ${\mathcal{R}}_j\,{:\!=}\,{\mathcal{R}}_j(u_1,v_1)$ , we have

\begin{align*} \mathbb{P}\left(X \ge i\right) &= \frac{\sum_{j \ge i}|{\mathcal{R}}_j| }{|{{\mathcal{R}}(n_1,n_2,p)}|}\le \sum_{j \ge i} \frac{|{\mathcal{R}}_j|}{|{\mathcal{R}}_{m}|} \sim \sum_{j \ge i} \frac{r_j}{r_m} = \sum_{j \ge i} \prod_{k = m + 1}^{j} \frac{r_k}{r_{k-1}} \le \sum_{j \ge i} \prod_{k = m + 1}^{j} \frac{\mu_+}{k}\\ &= \sum_{j \ge i} \prod_{k = m + 1}^{j}\frac{\mathbb{P}\left(Z_+ = k\right)}{\mathbb{P}\left(Z_+ = k-1\right)} = \sum_{j \ge i} \frac{\mathbb{P}\left(Z_+ = j\right)}{\mathbb{P}\left(Z_+ = m\right)} = \frac{\mathbb{P}\left(Z_+ \ge i\right)}{\mathbb{P}\left(Z_+ = m\right)}. \end{align*}

Since $\mu_+ \le N$ , by (19) we have $\mathbb{P}\left(Z_+ = m\right) = \Omega(1/\sqrt{\mu_+}) = \Omega(1/\sqrt{N})$ , and conclude that

(59) \begin{equation} \mathbb{P}\left(X \ge i\right) = O\left( \sqrt{N} \cdot \mathbb{P}\left(Z_+ \ge i\right) \right), \qquad i \ge \mu_+. \end{equation}

Similarly, if $Z_- \sim \operatorname{Po}(\mu_-)$ , then for $i\le m \,{:\!=}\, \lfloor \mu_- \rfloor$ , using $k/\mu_- = \mathbb{P}\left(Z_- = k-1\right)/\mathbb{P}\left(Z_- = k\right)$ ,

\begin{equation*} \mathbb{P}\left(X \le i\right)\le \sum_{j \le i} \frac{|{\mathcal{R}}_j|}{|{\mathcal{R}}_{m}|} \sim \sum_{j \le i}\prod_{k = j + 1}^{ m} \frac{r_{k-1}}{r_{k}} \le \sum_{j \le i}\prod_{k = j + 1}^{ m} \frac{\mathbb{P}\left(Z_- = k-1\right)}{\mathbb{P}\left(Z_- = k\right)} = \frac{\mathbb{P}\left(Z_- \le i\right)}{\mathbb{P}\left(Z_- = m\right)}, \end{equation*}

and therefore, again using (19),

(60) \begin{equation} \mathbb{P}\left(X \le i\right) = O\left( \sqrt{N} \cdot \mathbb{P}\left(Z_- \le i\right) \right), \qquad i \le \mu_-. \end{equation}

Since the RHS of (58) does not depend on x, we can assume the equality: $x = \sqrt{2\mu_+\lambda} + \lambda$ . Inequalities (17) and (18) imply

\begin{equation*} \begin{split} \mathbb{P}\left(Z_+ \ge \mu_+ + x\right) &+ \mathbb{P}\left(Z_- \le \mu_- - x\right) \le \exp \left( - \frac{x^2}{2(\mu_+ + x/3)} \right) + \exp \left( -\frac{x^2}{2\mu_-} \right) \\[3pt] \fbox{$\mu_- \le \mu_+, x > 0$}\quad &\le 2 \exp \left( -\frac{x^2}{2(\mu_+ + x/3)} \right) \\[3pt] \fbox{$x = \sqrt{2 \mu_+ \lambda} + \lambda$}\quad &= 2 \exp \left( -\frac{\lambda(2\mu_+ + 2\sqrt{2\mu_+ \lambda} + \lambda)}{2\mu_+ + \sqrt{2\mu_+ \lambda}/3 + \lambda/3} \right) \le 2e^{-\lambda}. \end{split}\end{equation*}

Combining this with (59) and (60) yields (58).

It remains to prove (57). We consider separately the cases $\mathbb{I} = 1$ and $\mathbb{I} = 0$ .

Case $\mathbb{I} = 1$ . If $p \ge 1/5$ , then deterministically $\operatorname{cod}_\mathbb{R}(u_1,u_2) \le pn_2\le 5p^2n_2 \le p^2 n_2 + 20 p^3 n_2$ and $\operatorname{cod}_\mathbb{R}(u_1, u_2) - p^2 n_2 \ge - p^2 n_2 \ge -5p^3 n_2$ , and hence (57) holds trivially. So we further assume $p < 1/5$ .

Setting $r_k = |{\mathcal{R}}_k|$ , we are going to prove bounds on $r_k/r_{k-1}$ , $k\ge 1$ , using a switching between ${\mathcal{R}} = {\mathcal{R}}_k$ and ${\mathcal{R}}' = {\mathcal{R}}_{k-1}$ . To this end, recall our terminology from Subsection 2.2 (with $G = \emptyset$ ): any graph $H \subseteq K$ is interpreted as a colouring of the edges of K, with the edges in H blue and the rest red. We define a forward switching as follows: pick a common blue neighbour $w_2 \in \Gamma_H(u_1) \cap \Gamma_H(v_1)$ and find two alternating 4-cycles $u_1w_2x_1x_2$ and $v_1w_2y_1y_2$ so that $x_1 \neq y_1$ , $x_2 \neq y_2$ . Moreover, we restrict the choice of the cycles to those for which $v_1x_2$ and $u_1y_2$ are red; this is to make sure that swapping of the colours on each of the two cycles (but keeping the colour of $v_1x_2,u_1y_2$ red) indeed decreases $\operatorname{cod}_{\mathbb{R}}(u_1,v_1)$ by one, thus mapping $H \in {\mathcal{R}}_k$ to $H' \in {\mathcal{R}}_{k-1}$ (see Figure 3).

Figure 3. Switching between $H \in {\mathcal{R}}_k(u_1,v_1)$ and $H' \in {\mathcal{R}}_{k-1}(u_1,v_1)$ : solid edges are in H and H $^{\prime}$ and the dashed ones in $K \setminus H$ and $K \setminus H'$ , respectively.

Let $B_k \,{:\!=}\, B({\mathcal{R}}_k, {\mathcal{R}}_{k-1})$ be the auxiliary graph corresponding to the described switching. Since the number of choices of $w_2$ is k for any $H \in {\mathcal{R}}_k$ , we will show upper and lower bounds on $\deg_{B_k}(H)/k$ by fixing $w_2$ and proving upper and lower bounds on the possible choices of $(x_1, x_2, y_1, y_2)$ .

For the upper bound, since there are $n_2 - 2d_1 + k$ common red neighbours of $u_1$ and $v_1$ (one can use (56) for this), the number of choices of distinct $x_2, y_2$ is exactly $(n_2 - 2d_1 + k)_2$ , while the number of choices of $x_1$ and $y_1$ , as blue neighbours of, resp., $x_2$ and $y_2$ , is at most $d_2^2$ . Ignoring the requirements that $x_1\neq y_1$ and that both $x_1w_2$ and $y_1w_2$ should be red), we have shown

(61) \begin{equation} \deg_{B_k}(H)/k \le (n_2 - 2d_1 + k)_2d_2^2.\end{equation}

For the lower bound, we subtract from the upper bound (61) the number of choices of $(x_1, x_2, y_1, y_2)$ for which either $x_1 = y_1$ or at least one of $x_1w_2$ and $y_1w_2$ is blue (the assumption $x_2 \neq y_2$ was taken into account in the upper bound). The number of choices with $x_1 = y_1$ is at most $n_1d_1^2$ , and those with, say, $x_1w_2$ blue, is at most $d_2d_1 \cdot (n_2 - 2d_1 + k)d_2 $ . Indeed, there are $\deg_H(w_2) \le d_2$ candidates for $x_1$ and, then, at most $d_1$ candidates for $x_2$ , $\operatorname{cod}_{K \setminus H}(u_1,u_2) = n_2 - 2d_1 + k$ candidates for $y_2$ , and then $d_2$ choices of $y_1$ . Therefore

\begin{align*} \deg_{B_k}(H)/k &\ge (n_2 - 2d_1 + k)_2d_2^2 - n_1d_1^2 - 2(n_2 - 2d_1 + k)d_1d_2^2 \\[3pt]&= (n_2 - 2d_1 + k)_2 d_2^2 \left( 1 - \frac{n_1d_1^2}{(n_2 - 2d_1 + k)_2 d_2^2} - \frac{2d_1}{n_2 - 2d_1 + k - 1}\right) \\[3pt]\fbox{$d_1 = n_2p, d_2 = n_1p, k \ge 1$}\quad& \ge (n_2 - 2d_1 + k)_2 d_2^2 \left( 1 - \frac{n_2^2}{(n_2 - 2pn_2)^2 n_1} - \frac{2pn_2}{n_2 - 2pn_2}\right) \\[3pt]&= (n_2 - 2d_1 + k)_2 d_2^2 \left( 1 - \frac{1}{(1 - 2p)^2n_1} - \frac{2p}{1 - 2p}\right) \\[3pt]\fbox{$p\le1/5$}\quad &\ge (n_2 - 2d_1 + k)_2 d_2^2 \left( 1 - \left( \frac{25}{9pn_1} + \frac{10}{3} \right) p \right) \\[3pt]\fbox{$\hat p \hat n \to \infty$}\quad &\ge (n_2 - 2d_1 + k)_2 d_2^2 \left( 1 - 4p \right).\end{align*}

A moment of thought (and a glance at Figure 3) reveals that the backward switching corresponds to choosing a common red neighbour $w_2$ of $u_1$ and $v_1$ , choosing alternating cycles $u_1w_2x_1x_2$ and $v_1w_2y_1y_2$ such that $x_1 \neq y_1$ and the edges $v_1x_2$ and $u_1y_2$ are red (note that these assumptions imply $x_2 \neq y_2$ ), and swapping the colours along the cycles.

Since for every $H' \in {\mathcal{R}}_{k-1}$ the number of choices of $x_2 \in \Gamma_{H'}(u_1) \setminus \Gamma_{H'}(v_1)$ and $ y_2 \in \Gamma_{H'}(v_1) \setminus \Gamma_{H'}(u_1)$ is exactly $(d_1 - k + 1)^2$ , we will bound $\deg_{B_k}(H')/(d_1 - k + 1)^2$ from above and below by fixing $x_2, y_2$ and estimating the number of possible triplets $(w_2, x_1,y_1)$ . For the upper bound, we can choose $w_2$ in exactly $\operatorname{cod}_{K \setminus H'}(u_1,u_2) = n_2 - 2d_1 + k - 1$ ways, and a pair $x_1, y_1$ of distinct blue neighbours of $w_2$ in at most $(d_2)_2$ ways (we ignore the requirement that $x_1x_2$ and $y_1y_2$ are red). Thus,

(62) \begin{equation} \deg_{B_k}(H')/(d_1 - k + 1)^2 \le (n_2 - 2d_1 + k - 1)(d_2)_2. \end{equation}

For the lower bound, given $x_2, y_2$ , we need to subtract from the upper bound (62) the number of choices of $w_2, x_1, y_1$ for which $x_1x_2$ or $y_1y_2$ is blue. In the former case, ignoring other constraints, there are at most $d_2$ choices of a blue neighbour $x_1$ of $x_2$ , at most $d_1$ choices of a blue neighbour $w_2$ of $x_1$ , and at most $d_2$ choices of a blue neighbour $y_1$ of $w_2$ . By symmetry, the total number of bad choices is at most $2d_1d_2^2$ , whence

\begin{align*} \deg_{B_k}(H')/(d_1 - k + 1)^2 &\ge (n_2 - 2d_1 + k - 1)(d_2)_2 - 2d_1d_2^2 \\[3pt] \fbox{$k \ge 1$}\quad &\ge (n_2 - 2d_1 + k - 1)d_2^2\left( 1 - \frac{1}{d_2} - \frac{2d_1}{n_2 - 2d_1} \right) \\[3pt] \fbox{$d_1 = n_2p, p \le 1/5$}\quad &\ge (n_2 - 2d_1 + k - 1)d_2^2\left( 1- d_2^{-1} - 4p \right).\end{align*}

Since, by our assumptions, $d_2 \ge \hat p\hat n \to \infty$ and $p \le 1/5$ , it follows that the graphs $B_k$ for $k = 1, \dots, d_1$ have minimum degrees at least 1. In particular, this means that starting with any p-biregular graph, by applying a certain number of switchings, we can obtain a p-biregular graph with arbitrary co-degree. Since we implicitly assume ${{\mathcal{R}}(n_1,n_2,p)}$ is non-empty, this implies that all classes ${\mathcal{R}}_k, k = 0, \dots, d_1,$ are non-empty, i.e., the numbers $r_k = |{\mathcal{R}}_k|$ are all positive, satisfying one of the conditions of Claim 12.

Using (14) and bounds on the degrees in $B_k$ , for $k = 1, \dots, d_1,$ we have

(63) \begin{align} \notag \frac{r_k}{r_{k-1}} \le \frac{\max_{H' \in {\mathcal{R}}' }\deg_{B_k}(H')}{\min_{H \in {\mathcal{R}}} \deg_{B_k}(H)} &\le \frac{(d_1 - k + 1)^2}{k(n_2 - 2d_1 + k)}\cdot\frac{1}{1 - 4p} \\ &\le \frac{(d_1 - k + 1)^2}{k(n_2 - 2d_1 + k)} \cdot \left( 1 + 20p \right),\end{align}

(where the last inequality holds because $p \le 1/5$ implies $\frac{1}{1 - 4p} = 1 + \frac{4p}{1 - 4p} \le 1 + 20p$ ), and

(64) \begin{equation} \frac{r_k}{r_{k-1}} \ge \frac{\min_{H' \in {\mathcal{R}}' }\deg_{B_k}(H')}{\max_{H \in {\mathcal{R}}} \deg_{B_k}(H)} \ge \frac{(d_1 - k + 1)^2}{k(n_2 - 2d_1 + k)} \left( 1 - d_2^{-1} -4p \right). \end{equation}

Let $\mu$ be a real number such that

(65) \begin{equation}\frac{(d_1 - \mu + 1)^2}{\mu(n_2 - 2d_1 + \mu)} = 1.\end{equation}

After solving for $\mu$ , we have

(66) \begin{equation} \mu = \frac{(d_1 + 1)^2}{n_2 + 2} \in [\:p^2n_2, p^2n_2(1 + 3d_1^{-1})\:]. \end{equation}

Define

(67) \begin{equation} \mu_+ \,{:\!=}\, \mu(1 + 20p), \quad \text{and} \quad \mu_- \,{:\!=}\, \mu(1 - d_2^{-1} -4p). \end{equation}

From (63), (65), and (67), for $\mu \le k \le d_1$ , we have

\begin{equation*} \frac{r_k}{r_{k-1}} \le \frac{(d_1 - \mu + 1)^2\left( 1 + 20p \right)}{\mu (n_2 - 2d_1 + \mu)} \cdot \frac{\mu}{k} = \frac{\mu_+}{k}. \end{equation*}

On the other hand, from (64), (65), and (67) that, for $1\le k \le \mu$ , we have

\begin{equation*} \frac{r_{k-1}}{r_k} \le \frac{\mu (n_2 - 2d_1 + \mu)}{(d_1 - \mu + 1)^2\left( 1 - 1/d_2 -4p \right)} \cdot \frac{k}{\mu} = \frac{k}{\mu_-}. \end{equation*}

Note that, since $\min\{d_1,d_2\} = p\hat n \to \infty$ and $p\le1/5$ , it follows by (66), and (67) that

(68) \begin{equation} p^2n_2(1 - (p\hat n)^{-1} - 4p) \le \mu_-\le \mu_+\le p^2n_2(1 + 15(p\hat n)^{-1} + 20p)\le 6p^2n_2 .\end{equation}

Therefore,

(69) \begin{equation} \mu_+ \le p^2n_2 + 20\left(p^3n_2 + \frac{pn_2}{\hat n} \right) \quad\text{and}\quad \mu_- \ge p^2n_2 - 20\left(p^3n_2 + \frac{pn_2}{\hat n} \right).\end{equation}

Consequently, setting

(70) \begin{equation} x \,{:\!=}\, 20 \sqrt{\lambda p^2 n_2} + \lambda,\end{equation}

we have, by (69)

\begin{equation*} \mathbb{P}\left(|X - p^2n_2| \ge 20\left(p^3n_2 + \frac{pn_2}{\hat n}\right) + x\right) \le \mathbb{P}\left(X \ge \mu_+ + x\right) + \mathbb{P}\left(X\le \mu_- - x\right). \end{equation*}

Noting that the last inequality in (68) implies $x \ge \sqrt{\mu_+n_2} + \lambda $ , using Claim 12 we obtain (57), completing the proof in the case $\mathbb{I} = 1$ .

Case $\mathbb{I} = 0$ . For the switching argument used in the previous case it was crucial that p was small, as otherwise several estimates would be negative and so meaningless. Therefore, it cannot be used now. Fortunately, in this case we are in a position to apply an asymptotic enumeration approach based on Theorem 5 and analogous to the proof of Theorem 2.1 in [Reference Krivelevich, Sudakov, Vu and Wormald7].

Recall the notation ${{\mathcal{R}}({\mathbf{d}}_1,{\mathbf{d}}_2)}$ defined before Theorem 5. Writing $A = \Gamma_H(u_1)$ and $B = \Gamma_H(v_1)$ , we note that every graph in ${\mathcal{R}}_k$ induces an ordered partition of $V_2$ into four sets $A \cap B$ , $A \setminus B$ , $B \setminus A$ and $V_2 \setminus (A \cup B)$ , of sizes, respectively, $k, d_1 - k, d_1 - k$ , and $n_2 - 2d_1 + k$ . There are exactly

\begin{equation*}\Pi(n_2,d_1,d_2,k)\,{:\!=}\,\frac{n_2!}{k!(d_1 - k)!^2(n_2 - 2d_2 + k)!} \end{equation*}

such partitions. After removing vertices $u_1$ and $v_1$ , we obtain a graph $H^* \in {{\mathcal{R}}({\mathbf{d}}_1,{\mathbf{d}}_2)}$ on $(V_1\setminus\{u_1,v_1\}, V_2)$ , with ${\mathbf{d}}_1$ having all its $n_1 - 2$ entries equal to $d_1$ , and entries of ${\mathbf{d}}_2$ determined by the partition: entries equal to $d_2 - 2$ on $A \cap B$ , to $d_2 - 1$ on $A \triangle B$ and the remaining ones equal to $d_2$ . Since $H^*$ together with the ordered partition uniquely determines H, we have

\begin{equation*} |{\mathcal{R}}_k| = \Pi(n_2,d_1,d_2,k)\times |{\mathcal{R}}({\mathbf{d}}_1,{\mathbf{d}}_2)|.\end{equation*}

Let us check that the three assumptions (i)–(iii) of Theorem 5 are satisfied. When doing so, we should remember that we have $n_1 - 2$ instead of $n_1$ , but that the parameter $p = d_1/n_2$ remains intact. With foresight, choose $a = 0.35$ and, say, $b = 0.1$ , and, for convenience $C = 1$ , which determine, via Theorem 5, an $\epsilon > 0$ .

Note that $\mathbb{I} = 0$ implies

(71) \begin{equation} \hat n \ge \frac{4\max \left\{ n_1,n_2 \right\}} {\log N} \quad \text{and} \quad \frac{n_1}{n_2} + \frac{n_2}{n_1} \le \frac{p \log N}{2} \le p \log \max \left\{ n_1, n_2 \right\}.\end{equation}

As the degrees on one side are all equal and on the other side they differ from each other by at most 2, assumption (i) holds true with a big margin. Using $q \ge 1/2$ and the first inequality in (71), we get

\begin{equation*} (pq)^2\min\{n_1 - 2,n_2\}^{1 + \epsilon} \ge \frac{p^2\hat n^{1 + \epsilon}}{4}(1 + o(1)) = \Omega \left( \frac{ \left(\max \left\{ n_1, n_2 \right\}\right)^{1 + \epsilon}}{(\log N)^{3 + \epsilon}} \right) \gg \max \left\{ n_1, n_2 \right\},\end{equation*}

which implies assumption (ii) for large $\hat n$ . Finally, using elementary inequalities $(1 - 2p)^2 \le q$ (since $p \le 1/2$ ), $1 \le \left( n_1/n_2 + n_2/n_1 \right)/2$ , and the second inequality in (71), we obtain

\begin{align*} \frac{(1 - 2p)^2}{4pq} &\left(1 + \frac{5(n_1 - 2)}{6n_2} + \frac{5n_2}{6(n_1 - 1)}\right) \le \frac{1}{4p}\left( \frac{1}{2} + \frac{5}{6} \right)\left(\frac{n_1}{n_2} + \frac{n_2}{n_1}\right) (1 + o(1)) \\ &\le \frac{1}{3}\log \max \left\{ n_1, n_2 \right\} (1 + o(1)) \ll 0.35 \cdot \log \max \left\{ n_1 - 2, n_2 \right\},\end{align*}

which for large $\hat n$ implies assumption (iii) with $a = 0.35$ .

Note that in (20) we have $D_1 = 0$ , while $D_2 \le 4n_2 \ll pq n_1n_2$ , by the assumption $\hat p\hat n\to\infty$ . Thus, uniformly over k, the exponent in (20) is $-1/2 + o(1)$ and, by Theorem 5,

\begin{equation*} |{\mathcal{R}}_k| \sim \frac {e^{-1/2}n_2!\binom {n_2}{d_1}^{n_1 - 2} \binom {n_1 - 2}{d_{2} - 2}^{k}\binom {n_1 - 2}{d_{2} - 1}^{2d_1 - 2k}\binom {n_1 - 2}{d_{2}}^{n_1 - 2d_1 + k}} {k!(d_1 - k)!^2(n_2 - 2d_1 + k)!\binom{(n_1 - 2)n_2}{(n_1 - 2)d_1}} =: r_k.\end{equation*}

Straightforward calculations yield

\begin{equation*}\frac{r_k}{r_{k-1}} =\frac {(d_1 - k + 1)^2(n_1 - 2 - d_2 + 1)(d_2 - 1)} {k(n_2 - 2d_1 + k)\cdot d_2(n_1 - d_2)} \le \frac{(d_1 - k + 1)^2}{k(n_2 - 2d_1 + k)}\end{equation*}

and, because $d_2/n_1 = p\le1/2$ ,

\begin{equation*}\frac{r_k}{r_{k-1}} \ge \frac{(d_1 - k + 1)^2}{k(n_2 - 2d_1 + k)}\left(1 - 2d_2^{-1} \right).\end{equation*}

(Compare with (63) and (64) to note the absence of $\Theta(p)$ error terms.) With $\mu$ as in (66), we redefine

(72) \begin{equation}\mu_+ \,{:\!=}\, \mu\quad\text{and}\quad\mu_- \,{:\!=}\, \mu (1 - 2d_2^{-1}).\end{equation}

From (66) and (72) it follows that

(73) \begin{equation} \mu_{+} = \mu \le p^2n_2 (1 + 3d_1^{-1}) \le p^2n_2 + \frac{3pn_2}{n_1}\le p^2n_2 + \frac{20pn_2}{\hat n}\end{equation}

and

(74) \begin{equation} \mu_{-} = p^2n_2 (1 - 2d_2^{-1}) \ge p^2n_2 - 2p \ge p^2n_2 - \frac{20pn_2}{\hat n}.\end{equation}

From (73) it follows $\mu_+ \le 6p^2n_2$ , which implies that x defined in (70) satisfies $x \ge \sqrt{2\mu_+\lambda} + \lambda$ . Consequently, using (73), (74) and Claim 12 we infer

\begin{equation*} \mathbb{P}\left(|X - p^2n_2| \ge \frac{20pn_2}{\hat n} + x\right) \le \mathbb{P}\left(X \ge \mu_+ + x\right) + \mathbb{P}\left(X\le \mu_- - x\right) = O\left( \sqrt{N} e^{-\lambda} \right).\end{equation*}

We have obtained (57) in the case $\mathbb{I} = 0$ .

5.2. Degrees and co-degrees in $\mathbb{R}(t)$

For convenience, having fixed $H\in{{\mathcal{R}}(n_1,n_2,p)}$ , we will denote by $\mathbb{P}_H$ and $\operatorname{\mathbb{E}}_H$ the conditional probability and expectation with respect to the event $\mathbb{R} = H$ . In the conditional space defined by such an event, $(\eta_1, \dots, \eta_M)$ is just a uniformly random permutation of the edges of H and therefore $\mathbb{R}(t)$ is a uniformly random t-subset of edges of H.

We first show that the degrees of vertices in the process $\mathbb{R}(t)$ grow proportionally almost until the end. Recall that $d_1 = pn_2,\; d_2 = pn_1$ , while $\tau=\tau(t)$ and $\tau_0$ are defined, respectively, in (24) and (25).

Lemma 13. If $\lambda = \lambda(n_1,n_2) \le \tau_0 p \hat n$ , then for $i = 1, 2$ , with probability $1 - O(N^2e^{-9\lambda/4})$

(75) \begin{equation} \forall t < t_0 \quad \forall v \in V_i \quad |\deg_{\mathbb{R}(t)}(v) - (1 - \tau) d_i| \le 3 \sqrt{\lambda \tau d_i}.\end{equation}

Proof. Fix $t < t_0$ and $v\in V_i$ and set $X_t(v) \,{:\!=}\, \deg_{\mathbb{R} \setminus \mathbb{R}(t)}(v)$ . For any $H \in {{\mathcal{R}}(n_1,n_2,p)}$ , conditioning on $\mathbb{R} = H$ , the random variable $X_t(v) \sim \operatorname{Hyp}(M, d_i, M - t)$ has a hypergeometric distribution. Since the distribution does not depend on H, $X_t(v)$ has the same distribution unconditionally. In view of (24), $\operatorname{\mathbb{E}} X_t(v) = \tau d_i$ , therefore combining (17) and (18), for all $x > 0$ ,

(76) \begin{align}\mathbb{P}\left(|X_t(v) - \tau d_i| \ge x\right)\le 2 \exp \left\{ - \frac{x^2}{2\left(\tau d_i + x/3 \right)} \right\}= 2 \exp \left\{ - \frac{x^2}{2\tau d_i\left(1 + x/(3\tau d_i) \right)} \right\}.\end{align}

Let $x \,{:\!=}\, 3\sqrt{\lambda\tau d_i}$ . Since $\tau \ge \tau_0$ , by the assumption $\lambda\le \tau_0 p\hat n $ ,

(77) \begin{equation} x/(3\tau d_i) = \sqrt{\lambda/(\tau d_i)} \le \sqrt{\tau_0 \hat n p/(\tau_0 d_i)} \le 1.\end{equation}

Consequently, taking the union bound over all $(1 - \tau_0)Mn_i \le N^2$ choices of t and v and using (76) and (77), we conclude that (75) fails with probability at most

\begin{equation*}2N^2\exp \left\{ - \frac{x^2}{2\tau d_i\left(1 + x/(3\tau d_i) \right)} \right\}\le 2N^2\exp \left\{ -\frac{9 \tau d_i \lambda}{4\tau d_i} \right\} = 2N^2e^{-9\lambda/4}.\end{equation*}

In the proof of Lemma 10, to have a pseudorandom-like property of $\mathbb{R} \setminus \mathbb{R}(t)$ for $p > 0.49$ , we will need a bound on the upper tail of co-degrees in $\mathbb{R}(t)$ . Recall the definition of co-degree from (55).

Lemma 14. Assume that $n_1 \ge n_2$ and $p > 0.49$ . If $\lambda = \lambda(n_1,n_2) \ge \log N$ , then, with probability $1 - O(N^3e^{-\lambda/3 })$ ,

(78) \begin{equation} \forall t \le M \quad \forall u_1 \neq v_1 \quad\operatorname{cod}_{\mathbb{R}(t)}(u_1,v_1) \le(1 - \tau)^2p^2 n_{2} + 20 q^3n_2\mathbb{I} + 15\sqrt{\lambda n_{2}}. \end{equation}

Proof. If $\lambda \ge n_2$ , inequality (78) holds trivially, so let us assume $\lambda < n_2$ . We can condition on $\mathbb{R} = H$ satisfying

(79) \begin{align} \notag \forall u_1,v_1 \in V_1 \quad\operatorname{cod}_{H}(u_1,v_1) & \le p^2n_2+ 20\left(\hat p^3n_2\mathbb{I} + n_2\hat p/\hat n + \sqrt{\hat p^2\lambda n_2}\right) + \lambda\\[3pt] \notag \fbox{$\hat p \le \min \{q, 0.5\}, n_2 = \hat n$}\quad&\le p^2n_2+ 20\left(q^3n_2\mathbb{I} + 0.5 + 0.5\sqrt{\lambda n_2}\right) + \lambda\\[3pt] \fbox{$1 \ll \lambda < n_2$}\quad &\le p^2n_2 + 20q^3n_2 \mathbb{I} + 12\sqrt{\lambda n_2}.\end{align}

Indeed, the probability of the opposite event can be bounded, by Lemma 11 and the union bound, by $O\left(n_1^2\sqrt Ne^{-\lambda}\right) \ll N^3e^{-\lambda}$ . Taking this bound into account, it thus suffices to show that if H satisfies (79), then for any distinct $u_1, v_1\in V_1$ and $t \le M$

(80) \begin{equation} \mathbb{P}_H\left({\operatorname{cod}_{\mathbb{R}(t)}(u_1,v_1) > (1 - \tau)^2p^2 n_{2} + 20 q^3n_2\mathbb{I} + 15\sqrt{\lambda n_{2}}}\right) \le 2e^{-\lambda/3}, \end{equation}

since then the proof is completed by a union bound over $O(N^3)$ choices of $t, u_1, v_1$ .

Fix $t \le M, u_1 \neq v_1$ and let $Y \,{:\!=}\, \left| \Gamma_{\mathbb{R}(t)}(u_1) \cap \Gamma_{H}(v_1)\right|$ and $\operatorname{cod}_H \,{:\!=}\, \operatorname{cod}_H(u_1, v_1)$ . The distribution of Y conditioned on $\mathbb{R} = H$ is hypergeometric $\operatorname{Hyp}(M, \operatorname{cod}_H, t)$ , and hence, by (79),

\begin{equation*} \mu_Y \,{:\!=}\, \operatorname{\mathbb{E}}_H{Y} = \frac{t\operatorname{cod}_H}{M} = (1 - \tau) \operatorname{cod}_H \le (1 - \tau)p^2n_2 + 20q^3n_2 \mathbb{I} + 12 \sqrt{\lambda n_2}. \end{equation*}

Using (17), a trivial bound $\mu_Y \le n_2$ , and our assumption $\lambda < n_2$ ,

\begin{equation*} \mathbb{P}_H\left({ Y \ge \mu_Y + \sqrt{\lambda n_2}}\right) \le \exp\left\{-\frac{\lambda n_2}{2(\mu_Y + \sqrt{\lambda n_2}/3)}\right\} \le e^{-3\lambda/8}\le e^{-\lambda/3}. \end{equation*}

Since also trivially $Y \le \min\{d_1, t\}$ , we have shown that given $\mathbb{R} = H$ , with probability at least $1 - e^{-\lambda/3}$ ,

(81) \begin{equation} Y \le y_0 \,{:\!=}\, \min \{\mu_Y + \sqrt{\lambda n_2}, d_1, t\} \le (1 - \tau)p^2n_2 + 20q^3n_2 \mathbb{I} + 13 \sqrt{\lambda n_2}.\end{equation}

Let ${\mathcal{E}}_{(80)}$ be the event that the inequality in (80) holds. By the law of total probability,

\begin{align*} \mathbb{P}_H\left({{\mathcal{E}}_{(80)}}\right) &= \sum_y \mathbb{P}_H\left({{\mathcal{E}}_{(80)}}\,|\,{Y = y}\right)\mathbb{P}_H\left({Y = y}\right) \\[3pt] \fbox{(81)}\quad &\le \sum_{y \le y_0}\mathbb{P}_H\left({{\mathcal{E}}_{(80)}}\,|\,{Y = y}\right)\mathbb{P}_H\left({Y = y}\right) + e^{-\lambda/3}.\end{align*}

Hence, to prove (80), it suffices to show that, for any $ y = 0, \dots, y_0$ ,

(82) \begin{equation}\mathbb{P}_H\left({{\mathcal{E}}_{(80)}}\,|\,{Y = y}\right) \le e^{-\lambda/3}.\end{equation}

Fix an integer $y \in [0, y_0]$ and a set $S\subseteq V_2$ of size $|S| = y \le y_0$ . Under an additional conditioning $\Gamma_{\mathbb{R}(t)}(u_1)\cap \Gamma_H(v_1) = S$ , set $\mathbb{R}(t)$ is the union of a fixed y-element set $\{u_1w\,\ :\,\ w \in S\}$ and a random $(t - y)$ -element subset of $E(H) \setminus \{u_1w\,\ :\,\ w \in \Gamma_H(u_1)\cap \Gamma_H(v_1)\}$ . Thus, in this conditional space, $X \,{:\!=}\, \operatorname{cod}_{\mathbb{R}(t)}(u_1,v_1)$ counts how many of these $t-y$ random edges fall into the set $\{v_1w\,\ :\,\ w \in S\}$ and therefore $X \sim \operatorname{Hyp}(M - \operatorname{cod}_H, y, t - y)$ . Moreover, the distribution of X is the same for all S of size y, so X has the same distribution when conditioned on $Y = y$ . In particular,

\begin{align*} \mu_X \,{:\!=}\, \operatorname{\mathbb{E}}_H\left({X}\,|\,{Y = y}\right) &= \frac{y(t-y)}{M-\operatorname{cod}_H} \\[3pt] \fbox{$y \ge 0$}\quad &\le \frac{yt}{M} \left(1 + \frac{\operatorname{cod}_H}{M - \operatorname{cod}_H}\right) \\[3pt] \fbox{$M = pn_1n_2, \operatorname{cod}_H \le pn_2$}\quad &\le (1-\tau) y \left(1 + \frac{1}{n_1 - 1}\right) \\[3pt] \fbox{$y \le y_0$ and (81)}\quad &\le (1-\tau)^2p^2n_2 + 20q^3n_2 \mathbb{I} + 13 \sqrt{\lambda n_2} + \frac{n_2}{n_1 - 1}. \\[3pt] \fbox{$n_1 \ge n_2$}\quad &\le (1-\tau)^2p^2n_2 + 20q^3n_2 \mathbb{I} + 14 \sqrt{\lambda n_2}. \end{align*}

Using again (17) as well as inequalities $\mu_X \le n_2$ and $\lambda \le n_2$ , we infer that

\begin{align*}\mathbb{P}_H\left({X \ge \mu_X + \sqrt{\lambda n_2}}\,|\,{Y = y}\right) \le \exp\left\{-\frac{\lambda n_2}{2(\mu_X + \sqrt{\lambda n_2}/3)}\right\} \le e^{-\lambda/3} ,\end{align*}

which, together with the above upper bound on $\mu_X$ , implies (82). This completes the proof of Lemma 14.

5.3. Proof of Lemma 9

Recall the definitions of $\tau=\tau(t)$ in (24), $\delta(t)$ in (43), and $\lambda(t)$ in (44). In this proof we utilise some technical bounds on $\tau\,{:\!=}\,\tau(t)$ and $\gamma_t$ proved in Section 8 (see Proposition 23). In particular, the bound (146) on $\gamma_t$ , together with (45), implies that

(83) \begin{equation} \delta(t) \le 1/9\le 1.\end{equation}

(We do not even need to remember what $\gamma_t$ is to see that.)

We now derive an upper bound on $\lambda(t)$ . By (147), with a huge margin,

\begin{equation*} 6 \log N \le \frac{\tau \hat p \hat n}{16(C + 3)} \le \frac{\tau pq\hat n }{8(C+3)}.\end{equation*}

On the other hand, for $p > 0.49$ , squaring and rearranging the inequality (145) implies

\begin{equation*} \frac{64\log N}{\tau p q} \le \frac{\tau pq\hat n }{8(C+3)}.\end{equation*}

Summing up, for any p,

(84) \begin{equation}\lambda(t)\le \frac{\tau pq \hat n}{4(C + 3)}.\end{equation}

Without loss of generality, we assume that $i = 1$ . Let ${\mathcal{F}}_t$ be the family of graphs H satisfying, for any distinct $u_1, v_1 \in V_1$ ,

(85) \begin{equation} |d_1 - \operatorname{cod}_H(u_1, v_1) - pqn_2| \le 20\hat p n_{2} \left(\hat p^2\mathbb{I} + \frac{1}{\hat n} + \sqrt{\frac{(C+3)\lambda(t)}{n_{2}}} + \frac{(C+3)\lambda(t)}{20\hat pn_2}\right).\end{equation}

Lemma 11 with $\lambda = (C+3)\lambda(t)$ and a union bound over the $O(N^2)$ choices of $u_1, v_1$ imply

(86) \begin{equation} \mathbb{P}\left(\mathbb{R} \in {\mathcal{F}}_t\right) = 1 - O(N^{2.5} e^{-(C+3)\lambda(t)}) \overset{{\lambda(t) \ge 6\log n}}{\ge} 1 - O(N^{-(C+1)}e^{-2\lambda(t)}).\end{equation}

Writing $\delta_* = \sqrt{\frac{(C+3)\lambda(t)}{\tau p q \hat n}}$ and noting that (84) implies $\delta_* \le 1/2$ , the sum of the last three terms in the parentheses in (85) is at most

\begin{equation*} 2 \cdot \sqrt{\frac{(C+3)\lambda(t)}{\hat n}} + \frac{(C+3)\lambda(t)}{20\hat p \hat n} {\le} 2\sqrt{pq}\delta_* + \frac{pq\delta_*^2}{20\hat p} {\le} (1 + 1/40) \delta_*\le \frac{3\delta_*}{\sqrt{6}}.\end{equation*}

In addition, the factor in front of the brackets is at most $40pq n_2$ . Thus, (85) implies

(87) \begin{equation} |d_1 - \operatorname{cod}_H(u_1, v_1) - pqn_2| \le 40 pqn_2 \left( \hat p^2 \mathbb{I} + \frac{3\delta_*}{\sqrt{6}} \right) = pqn_2 \cdot \delta(t)/3.\end{equation}

We claim that for $t = 0, \dots, t_0 - 1$

(88) \begin{equation} \max_{H \in {\mathcal{F}}_t}\mathbb{P}_H\left({\max_{u_1 \ne v_1} \left|\frac{\theta_t(u_1,v_1)}{\tau pqn_2 } - 1\right| \ge \delta(t)}\right) \le 2N^2e^{-(C+3)\lambda(t)}, \end{equation}

and deferring its proof to the end we first show how (86) and (88) imply the lemma.

Inequalities (86) and (88) imply

(89) \begin{align} \notag \mathbb{P}\left(\max_{u_1 \ne v_1} \left|\frac{\theta_t(u_1,v_1)}{\tau pqn_2 } - 1\right| \ge \delta(t)\right) &\le 2N^2e^{-(C+3)\lambda(t)} + \mathbb{P}\left(\mathbb{R} \notin {\mathcal{F}}_t\right) \\[3pt] &= O(N^{-(C+1)}e^{-2\lambda(t))}).\end{align}

Consider random variables, for $t = 0, \dots, t_0 - 1$ ,

\begin{equation*} Y_t \,{:\!=}\, \mathbb{P}\left({\max_{u_1 \ne v_1} \left|\frac{\theta_t(u_1,v_1)}{\tau pqn_2 } - 1\right| > \delta(t)}\,|\,{\mathbb{R}(t)}\right). \end{equation*}

Using Markov’s inequality and (89), we infer that

\begin{align*} \mathbb{P}\left(Y_t > e^{-2\lambda(t)}\right) &\le e^{2\lambda(t)}\operatorname{\mathbb{E}} Y_t = e^{2\lambda(t)}\mathbb{P}\left(\max_{u_1 \ne v_1} \left|\frac{\theta_t(u_1,v_1)}{\tau pqn_2 } - 1\right| \ge \delta(t)\right) = O( N^{-(C+1)}), \end{align*}

which, taking the union bound over the O(N) choices of t, implies that (47) holds with the desired probability, completing the proof of lemma.

Returning to the proof of (88), fix $t < t_0, H \in {\mathcal{F}}_t$ and two distinct vertices $u_1,v_1\in V_1$ . Conditioning on $\mathbb{R} = H$ , note that, recalling (46), random variable $X \,{:\!=}\, \theta_t(u_1,v_1) = |\Gamma_{H \setminus \mathbb{R}(t)}(u_i)\cap\Gamma_{K\setminus H}(v_i)|$ counts elements in the intersection of two subsets of E(H): a fixed set $ \left\{ (u_i,w)\ \, :\ \, w \in \Gamma_H(u_i) \cap \Gamma_{K \setminus H}(v_i) \right\}$ of size $d_1 - \operatorname{cod}_H(u_1, v_1)$ and a random set $H \setminus \mathbb{R}(t)$ of size $M - t$ . Hence $X \sim \operatorname{Hyp}(M, d_1 - \operatorname{cod}_H(u_1,v_1), M -t)$ has the hypergeometric distribution with expectation

\begin{equation*} \mu_X \,{:\!=}\, \operatorname{\mathbb{E}} X = (d_1 - \operatorname{cod}_H(u_1,v_1))(M - t)/M = \tau (d_1 - \operatorname{cod}_H(u_1,v_1)).\end{equation*}

Note that by (87),

(90) \begin{equation}|\mu_X - \tau p q n_2| \le \tau pqn_2 \cdot \delta(t)/3.\end{equation}

Let $\lambda^* \,{:\!=}\, (3C + 9)\lambda(t)$ . From (90), (83), and (84) it follows that

(91) \begin{equation} \mu_X \ge \frac{26}{27}\tau pqn_{2} \ge \lambda^*.\end{equation}

Note that (43) implies

(92) \begin{equation} \delta(t)^2\ge7200\frac{\lambda^*}{\tau pqn_2}\ge 10\frac{\lambda^*}{\tau pqn_2}. \end{equation}

By (17), (18), and (91),

\begin{equation*} \begin{split} \mathbb{P}_H\left({|X - \mu_X| \ge \sqrt{\mu_X \lambda^*} }\right) &\le 2 \exp \left\{ - \frac{\lambda^*}{2\left(1 + \frac{1}{3}\sqrt{\lambda^*/\mu_X } \right)} \right\} \\[3pt] &\le 2 e^{-3\lambda^*/8}\le 2 e^{-\lambda^*/3} = 2e^{-(C + 3)\lambda(t)}. \end{split}\end{equation*}

Thus, with probability $1 - 2e^{-(C+3)\lambda(t)}$

\begin{align*} |\theta_t(u_1,v_1) - \tau p q n_2| &\le |X - \mu_X| + |\mu_X - \tau p q n_2| \\[4pt] &\le \sqrt{\mu_X \lambda^*} + |\mu_X - \tau p q n_2| \\[4pt] \fbox{(90), (92)}\quad &\le \sqrt{\left( 1 + \delta(t)/3 \right)\tau p q n_2 \cdot \delta^2(t)\tau p q n_2 / 10} + (\delta(t)/3) \tau p q n_2\\[4pt] \fbox{(83)}\quad &\le \delta(t)\tau pqn_2 (\sqrt{(1 + 1/27)/10} + 1/3) \\[4pt] &\le \delta(t)\tau pqn_2.\end{align*}

Hence, applying also the union bound over all $n_1^2 \le N^2$ choices of $u_1,v_1$ , we infer (88).

6.1. Jumbled graphs

Let $K\,{:\!=}\, {K_{n_1,n_2}}$ be the complete bipartite graph with partition classes $V_1$ and $V_2$ , where $|V_i| = n_i$ . Given a bipartite graph $F \subseteq K$ and two subsets $A \subseteq V_1$ , $B \subseteq V_2$ , denote by $e_F(A,B)$ the number of edges of F between A and B. Recall that $N = n_1n_2$ and $M = pN$ .

Given real numbers $\pi,\delta\in(0,1)$ , we say that a graph $F \subseteq K$ is $(\pi, \delta)$ -jumbled if for every $A \subseteq V_1$ and $B \subseteq V_2$

\begin{equation*} |e_F(A,B) - \pi|A||B|| \le \delta \sqrt{N|A||B|}.\end{equation*}

The following result of Thomason [14, Theorem 2], which quantifies a variant of jumbledness in terms of the degrees and co-degrees of a graph, will turn out to be crucial for us.

Theorem 15 ([Reference Thomason14]). Let $F \subseteq K$ be a bipartite graph and $\rho \in (0,1)$ and $\mu \ge 0$ be given. If

(93) \begin{equation} \min_{v \in V_1} \deg_F(v) \ge \rho n_2\quad\text{and}\quad\max_{u,v\in V_1: u \neq v} \operatorname{cod}_F(u,v) \le \rho^2n_2 + \mu, \end{equation}

then, for all $A \subseteq V_1$ and $B \subseteq V_2$ ,

\begin{equation*} |e_F(A,B) - \rho|A||B|| \le \sqrt{(\rho n_2 + \mu |A|)|A||B|} + |B|\mathbb{I}_{\left\{ |A|\rho < 1 \right\}}. \end{equation*}

Recall that $K\setminus \mathbb{R}(t)$ has precisely $N - t = N - (1-\tau)M = (\tau p + q)N$ edges. The following technical result states that, under the conditions of Lemma 6, with high probability, $K\setminus \mathbb{R}(t)$ is jumbled for parameters which are tailored for Lemma 20.

Lemma 17. Let $\alpha = \min \{ \tau p, q \}$ and $\pi \,{:\!=}\, \tau p + q$ . For every constant $C > 0$ and $p > 0.49$ , if assumptions (27) and (28) hold, then, with probability $1 - O(N^{-C })$ , for all $t < t_0$

\begin{equation*} K \setminus \mathbb{R}(t)\quad \text{is} \quad(\pi, \alpha/16)\text{-jumbled}\end{equation*}

and for any $e \in K \setminus \mathbb{R}(t)$

\begin{equation*} K \setminus (\mathbb{R}(t) \cup \{ e \}) \quad \text{is} \quad(\pi, \alpha/16)\text{-jumbled}.\end{equation*}

Proof. W.l.o.g., we assume that $n_1\ge n_2$ . Let $\lambda \,{:\!=}\, 3(C + 4)\log N$ , and

(94) \begin{equation} \delta \,{:\!=}\, 20 \left( \lambda/n_2 \right)^{1/4} + 10q^{3/2}\mathbb{I}. \end{equation}

The plan is to show that

(95) \begin{equation} \delta \le \frac{\alpha}{16}\end{equation}

and that with the correct probability, $K \setminus \mathbb{R}(t)$ and $K \setminus \mathbb{R}(t) \cup \{e\}$ are in fact $(\pi, \delta)$ -jumbled.

We start with the proof of (95). Notice that, by (28),

(96) \begin{equation}\sqrt{\lambda/{n_2}}\le\left(\lambda/{n_2}\right)^{1/4}\le\frac q{680}\le\frac\pi{680}\le\frac1{680}.\end{equation}

Since $\hat p > \tfrac{49}{51}q$ , condition (27) implies that $1/320 \ge q^{1/2} \mathbb{I}$ . After multiplying both sides by 10q, we get

\begin{equation*} \frac q{32} \ge 10\cdot q^{3/2} \mathbb{I},\end{equation*}

which together with the second inequality in (96) implies

\begin{equation*} \frac{q}{16} = \frac{q}{32} + \frac{q}{32} \ge 20 \left( \lambda/{n_2} \right)^{1/4} + 10\cdot q^{3/2} \mathbb{I} = \delta.\end{equation*}

On the other hand, using $p > 0.49$ , $\tau \ge \tau_0$ and the definition (25) of $\tau_0$ , we infer that

\begin{align*} \frac{\tau p}{16} &\ge \frac{0.49 \cdot \tau_0}{16} = \frac{0.49 \cdot 700\cdot (3(C + 4))^{1/4}}{16} \left( \left(\frac{\log N}{n_2}\right)^{1/4} + q^{3/2} \mathbb{I}\right) \\ &\ge 20 \left( \lambda/{n_2} \right)^{1/4} + 10q^{3/2} \mathbb{I} = \delta.\end{align*}

Hence $\alpha/16 = \min \left\{ \tau p/16, q/16 \right\} \ge \delta$ , implying (95).

We now prove the jumbledness, first focusing on $K \setminus \mathbb{R}(t)$ and then indicating the tiny change in calculation for $K \setminus (\mathbb{R}(t) \cup \{e\})$ . Fixing an arbitrary $t < t_0$ , we will first show that, with probability $1 - O(N^{-C -1})$ , conditions (93) of Theorem 15 are satisfied by $F = K \setminus \mathbb{R}(t)$ and $F = K \setminus (\mathbb{R}(t) \cup \{e\})$ for suitably chosen $\rho$ and $\mu$ . Then we will apply Theorem 15 to deduce that $K \setminus \mathbb{R}(t)$ is $(\pi, \delta)$ -jumbled. Lemma 17 will follow by applying the union bound over all (at most $t_0\le M\le N$ ) choices of t.

By (147), $\lambda \le \tau_0 p \hat n$ with a big room to spare. Note that

(97) \begin{equation} (1-\tau)p = 1 - \pi,\end{equation}

which implies that $|\deg_{K \setminus \mathbb{R}(t)}(v) - \pi n_2|=|\deg_{\mathbb{R}(t)}(v) - (1-\tau)d_1|$ . Hence, by Lemma 13, with probability $1 - O(N^2e^{-\lambda})$ ,

(98) \begin{equation} \max_{v \in V_1} |\deg_{K \setminus \mathbb{R}(t)}(v) - \pi n_2| \le 3\sqrt{\tfrac{4}{9}\lambda\tau d_1} \le 2\sqrt{\lambda n_2} \le 3\sqrt{\lambda n_2}.\end{equation}

Moreover, recalling that $n_1\ge n_2$ and, again using (97), Lemma 14 implies that, with probability $1 - O(N^3e^{-\lambda/3})$ ,

(99) \begin{equation} \max_{u, v \in V_1, u \ne v} \operatorname{cod}_{\mathbb{R}(t)}(u,v) \le (1 - \pi)^2 n_2 + 20q^3n_2\mathbb{I} + 15 \sqrt{\lambda n_2}. \end{equation}

Since $\lambda = 3(C + 4)\log N$ , the intersection of events (98) and (99) holds with probability $1 - O(N^3e^{-\lambda/3}) = 1 - O(N^{-C - 1})$ .

Note that for distinct $u, v \in V_1$ , by (56) and (98)

(100) \begin{equation} \operatorname{cod}_{K \setminus \mathbb{R}(t)}(u, v) \le \operatorname{cod}_{\mathbb{R}(t)}(u,v) + (2 \pi - 1)n_2 + 6\sqrt{\lambda n_2}, \end{equation}

which, by (99), implies that

(101) \begin{equation} \max_{u, v \in V_1, u \ne v} \operatorname{cod}_{K \setminus \mathbb{R}(t)}(u,v) \le \pi^2 n_2 + 20q^3n_2\mathbb{I} + (15 + 6) \sqrt{\lambda n_2}. \end{equation}

Set

\begin{equation*} \rho \,{:\!=}\, \pi - 3\sqrt{\lambda/n_2} \quad \text{and} \quad \mu \,{:\!=}\, 20q^3n_2\mathbb{I} + (15 + 12) n_2\sqrt{\lambda/ n_2},\end{equation*}

and note that by the inequality $\pi\le1$ and by (96), we have $0 < \rho < 1$ . Furthermore, $\rho^2\ge\pi^2-6\sqrt{\lambda/ n_2}$ . Hence, (98) and (101) imply the assumptions (93) for $F = K \setminus \mathbb{R}(t)$ with the above $\rho$ and $\mu$ . Consequently, by Theorem 15 (using $a \le n_1$ and $\rho\le\pi$ ),

\begin{equation*} |e_{K \setminus \mathbb{R}(t)}(A,B) - \rho ab| \le \sqrt{ \left( \pi n_2 + 20q^3n_1n_2\mathbb{I} + (15 + 12)n_1n_2\sqrt{\lambda /n_2}\right)ab} + b. \end{equation*}

Further, since $N=n_1n_2$ , $n_1\ge n_2 \ge b$ and $a,\lambda \ge 1$ , we have, with a big margin,

\begin{equation*} \pi \le1 \le n_1 \sqrt{\lambda/n_2}\quad\text{and}\quad b \le \sqrt{b n_2} \le \sqrt{Nab}(\lambda/n_2)^{1/4}.\end{equation*}

It follows, applying the inequality $\sqrt{x + y} \le \sqrt x+ \sqrt y$ as well as (94), that

(102) \begin{equation} |e_{K \setminus \mathbb{R}(t)}(A,B) - \rho ab| \le \left( (\sqrt{15 + 13} + 1)(\lambda/n_2)^{1/4} + \sqrt{20} q^{3/2}\mathbb{I}\right) \sqrt{Nab} \le \frac{\delta}{2}\sqrt{Nab}. \end{equation}

Moreover, note that using $ab \le n_1n_2 = N$ and the first inequality in (96),

(103) \begin{equation}(\pi - \rho)ab = 3\sqrt{\lambda/n_2} \cdot ab \le 3\left(\lambda/n_2\right)^{1/4} \cdot \sqrt{Nab} \le \frac{\delta}{2}\sqrt{Nab},\end{equation}

Hence, (102) and (103) imply

\begin{equation*} |e_{K \setminus \mathbb{R}(t)}(A,B) - \pi ab| \le |e_{K \setminus \mathbb{R}(t)}(A,B) - \rho ab| + (\pi - \rho)ab \le \delta \sqrt{Nab},\end{equation*}

meaning that $K \setminus \mathbb{R}(t)$ is $(\pi, \delta)$ -jumbled.

If above we replace $K \setminus \mathbb{R}(t)$ by $K \setminus (\mathbb{R}(t) \cup \left\{ e \right\})$ , the upper bound in (98) and the bound in (101) still hold trivially, while the lower bound $\pi n_2 - 3 \sqrt{\lambda n_2}$ in (98) remains correct, since $\deg_{K \setminus (\mathbb{R}(t) \cup \{e\})}(v) \ge \deg_{K \setminus \mathbb{R}(t)}(v) - 1$ and we have plenty of room in (98). Hence Theorem 15 applies with the same $\rho$ and $\mu$ , implying that $K \setminus (\mathbb{R}(t) \cup \{e\})$ is also $(\pi, \delta)$ -jumbled.

6.2. A technical inequality for blue-red graphs

We find it convenient to introduce relative counterparts of basic graph quantities. Below $i\in\{1,2\}$ . As before, let $K\,{:\!=}\, {K_{n_1,n_2}}$ be the complete bipartite graph with partition classes $V_1$ and $V_2$ , where $|V_i| = n_i$ . The relative size of a subset of vertices $S\subseteq V_i$ is

\begin{equation*} s({S}) \,{:\!=}\, \frac{|S|}{n_i}.\end{equation*}

Further, for $X\subseteq V_1$ and $Y\subseteq V_{2}$ and a subgraph $F \subseteq K$ , we define the relative edge count

\begin{equation*} \varepsilon_F(X,Y) = \varepsilon_F(Y,X) = \frac{e_F(X,Y)}{n_1n_2}.\end{equation*}

Moreover, for $v\in V_i$ and $Y\subseteq V_{3-i}$ , we define the relative degree

\begin{equation*} d_F(v,Y) = \frac{e_F(\left\{ v \right\}, Y)}{n_{3-i}}.\end{equation*}

If $Y = V_{3-i}$ , we shorten $d_F(v,Y)$ to $d_F(v)$ .

In this notation, a graph F is $(\pi,\delta)$ -jumbled if for every $X \subseteq V_1$ , $Y \subseteq V_2$

(104) \begin{equation} |\varepsilon_F(X,Y) - \pi s(X)s(Y)| \le \delta \sqrt{s(X)s(Y)}.\end{equation}

Now, let the edges of a graph F be 2-coloured by blue and red, and let B and R be the subgraphs of F induced by the edges of colour, resp., blue and red. We then call $F = B \cup R$ a blue-red graph. Note that

\begin{equation*} \varepsilon_{B\cup R}(X,Y) = \varepsilon_F(X,Y)= \varepsilon_{R}(X,Y) + \varepsilon_{B}(X,Y).\end{equation*}

We say that a blue-red graph F is $(r,b,\delta)$ -regular, if

(105) \begin{equation} b - \delta \le {d_{B}} (v) \le b + \delta, \quad\text{and}\quad r - \delta \le d_{R} (v) \le r + \delta \quad \text{ for every } v\in V_1\cup V_2.\end{equation}

If F is at the same time $(r+b, \delta)$ -jumbled and $(r,b,\delta)$ -regular, as in the technical lemma below, we will sometimes loosely refer to such a graph as regularly jumbled.

Finally, for every $S\subseteq V_i$ , $i = 1,2$ , set $\overline{S}\,{:\!=}\,V_i\setminus S$ .

Lemma 18. Let $r, b \in (0,1)$ be real numbers and define $\alpha \,{:\!=}\, \min \{r,b\}$ . Let $\nu<\alpha/16$ and $\delta \leq \alpha/16$ be positive reals and let $F \subseteq K$ be a $(r, b, \delta)$ -regular, $(b + r, \delta)$ -jumbled bipartite blue-red graph. If sets $X \subseteq V_i$ , $Y \subseteq V_{3-i}$ satisfy

(106) \begin{equation} \varepsilon_{B}(X,\overline{Y}) + \varepsilon_{R}(\overline{X},Y) \le \nu,\end{equation}

and

(107) \begin{equation} \min \{bs(X),rs(Y)\} \leq \frac{rb}{r + b},\end{equation}

then

(108) \begin{equation} \max \{bs(X), rs(Y)\} \le \frac{\nu}{1 - 7\delta/\alpha}.\end{equation}

Proof. Since

\begin{equation*} \varepsilon_{R}(X,Y) = \frac{e_R(X,Y)}{n_1n_2} = \frac{\sum_{v\in X}e_R(\left\{ v \right\}, Y)}{n_1n_2} = \frac1{n_1}\sum_{v\in X}d_R(v,Y),\end{equation*}

from (105) we have

(109) \begin{equation}\varepsilon_{R}(X,Y) + \varepsilon_{R}(X,\overline{Y}) \leq (r + \delta)s(X)\end{equation}

and, similarly,

(110) \begin{equation} \varepsilon_{B}(X,Y) + \varepsilon_{B}(\overline{X},Y) \leq (b + \delta)s(Y).\end{equation}

By summing (106), (109) and (110), we infer that

(111) \begin{equation} \varepsilon_F(X,Y) + \varepsilon_F(X,\overline{Y}) + \varepsilon_F(\overline{X},Y)\leq \nu + (r + \delta)s(X) + (b + \delta)s(Y). \end{equation}

On the other hand, by (105),

\begin{equation*} \varepsilon_F(X,\overline{Y}) = \varepsilon_F(X, V_2) - \varepsilon_F(X,Y) \ge (b + r - 2\delta)s(X) - \varepsilon_F(X,Y)\end{equation*}

and

\begin{equation*} \varepsilon_F(\overline{X},Y) = \varepsilon_F(V_1, Y) - \varepsilon_F(X,Y) \ge (b + r - 2\delta)s(Y) - \varepsilon_F(X,Y).\end{equation*}

Hence, by (104) with $\pi = b + r$ ,

\begin{align*} \varepsilon_F(X,Y) + \varepsilon_F(X,\overline{Y}) + \varepsilon_F(\overline{X},Y) &\geq (b + r - 2\delta) (s(X) + s(Y)) - \varepsilon_F(X, Y) \\[3pt] & \ge (b + r - 2\delta) (s(X) + s(Y)) - (b + r)s(X)s(Y) - \delta \sqrt{s(X)s(Y)}. \end{align*}

Comparing with (111), we obtain the inequality

\begin{equation*} b s(X) + r s(Y) - \frac{(b + r)bs(X) rs(Y)}{br} \le \nu + \delta\left(3s(X) + 3s(Y) + \sqrt{s(X)s(Y)}\right).\\\end{equation*}

Denoting $x \,{:\!=}\, bs(X)$ and $y \,{:\!=}\, rs(Y)$ and $h \,{:\!=}\, rb/(b + r)$ , this becomes

(112) \begin{equation} x + y - \frac{x y}h \le \nu + \delta \left( \frac{3x}b + \frac{3y}r +\sqrt{\frac{xy}{br}} \right) =: \psi.\end{equation}

Trivially, by the definitions of s(X) and $\alpha$ , and by our assumptions on $\nu$ and $\delta$ , we have

\begin{equation*} \psi \le \nu + 7\delta < \frac12\alpha\le \frac{1}{1/b + 1/r} = h. \end{equation*}

Since our goal—inequality (108)—now reads as

\begin{equation*} \max \left\{ x, y \right\} \le \frac{\nu}{1 - 7\delta /\alpha},\end{equation*}

to complete the proof it is enough to assume, without loss of generality, that $\max \{x, y\} = y$ and show, equivalently, that

(113) \begin{equation} y \le \nu + 7\delta y/\alpha.\end{equation}

By (107) we have $x = \min\{x,y\} \le h$ . Note that $x = h$ cannot hold, since then the LHS of (112) would equal h, contradicting the fact that $\psi < h$ . Hence, we have $x < h$ , which, together with $\psi < h$ and (112), implies that

\begin{align*} y &\le \frac{\psi - x}{1 - x/h} = h - \frac{h - \psi}{1 - x/h} \le h - (h - \psi) = \psi = \nu + \delta \left( \frac{3x}b + \frac{3y}r + \sqrt{\frac{xy}{br}} \right) \le \nu + \frac{7\delta y}{\alpha},\end{align*}

and (113) is proved.

6.3. Alternating walks and cycles

A cycle in a blue-red bipartite graph is said to be alternating if it is a union of a red matching and a blue matching, that is, every other edge is blue and the remaining edges are red. The ultimate goal of this subsection is to show that for every edge in a regularly jumbled blue-red bipartite graph, there is an alternating cycle of bounded length containing that edge. We are going to achieve it by utilising walks.

Given $x,y\in V_1\cup V_2$ , an alternating walk from x to y in a blue-red graph F is a sequence of (not necessarily distinct) vertices $(v_1 = x,\dots,v_s = y)$ such that for each $i = 1,\dots,s-1$ , $v_iv_{i+1}\in F$ , every other edge is blue and the remaining edges are red. There is no restriction on the colour of the initial edge $v_1v_2$ . The length of a walk is defined as the number of edges, or $s-1$ .

If the vertices $v_1,\dots,v_s$ are all distinct, an alternating walk is called an alternating path. Note also that if F is bipartite, $x\in V_1,\;y\in V_2$ , and the edge xy is, say, blue, then every alternating path from x to y which begins (and thus ends) with a red edge together with xy forms an alternating cycle containing xy.

The first result of this section asserts that regularly jumbled blue-red bipartite graphs have a short alternating walk between any pair of vertices.

Proof. For $w \in V_1 \cup V_2$ and an integer $k \ge 1$ , define $R^w_k$ and $B^w_k$ as the sets of vertices $v\in V(F)$ such that there is an alternating walk from v to w of length $\ell \le k$ , $\ell\equiv k \pmod{2}$ , starting with, respectively, a red edge and a blue edge. (Note that these definitions concern walks ending with w.)

Clearly, for every $k\ge3$ , $B^w_{k-2}\subseteq B^w_k$ and $R^w_{k-2}\subseteq R^w_k$ . Observe also that for any $k \geq 2$ , by definition the sets $R^w_{k-1}$ and $B^w_k$ are contained in opposite sides of the bipartition $(V_1,V_2)$ and, moreover,

(114) \begin{equation} \varepsilon_{B}\left(\overline{B^w_k}, R^w_{k-1}\right) = 0.\end{equation}

By symmetry, $B_{k-1}^w$ and $R_k^w$ are contained in opposite sides of $(V_1,V_2)$ and

(115) \begin{equation} \varepsilon_{R}\left(\overline{R^w_k}, B^w_{k-1}\right) = 0.\end{equation}

Set $\nu = rb/16$ and note that, since $r,b < 1$ , we have $\nu < \alpha/16$ . There exists an integer $t\leq T \,{:\!=}\, \lceil 1/\nu \rceil = \lceil 16/rb \rceil$ such that

(116) \begin{equation}s\left({R^w_{2t + 1}\setminus R^w_{2t - 1}}\right)\leq \nu,\end{equation}

since otherwise $1 \ge s({R^w_{2T + 1}}) \ge \sum_{i = 1}^{T} s({R^w_{2i+1} \setminus R^w_{2i-1}}) > \nu T \ge 1$ , which is a contradiction.

By (114) and (116),

(117) \begin{equation} \varepsilon_{B}\left(R^w_{2t+1}, \overline{B^w_{2t}}\right) = \varepsilon_{B}\left(R^w_{2t - 1},\overline{B^w_{2t}}\right)+ \varepsilon_{B}\left(R^w_{2t + 1}\setminus R^w_{2t - 1}, \overline{B^w_{2t}}\right)\leq s\left({R^w_{2t + 1}\setminus R^w_{2t - 1}}\right)\leq \nu.\end{equation}

Combining (115) for $k = 2t + 1$ and (117), we get

(118) \begin{equation} \varepsilon_{B}\left(R^w_{2t + 1}, \overline{B^w_{2t}}\right)+ \varepsilon_{R}\left(\overline{R^w_{2t + 1}}, B^w_{2t}\right) \leq \nu.\end{equation}

Set $X = R^w_{2t + 1}, Y = B^w_{2t}$ , for convenience. We claim that

(119) \begin{equation} s(X) > r/(r + b)\quad\text{and}\quad s(Y) > b/(r + b). \end{equation}

Assuming the contrary, we have

\begin{equation*} \min \left\{ bs(X), rs(Y) \right\} \le br/(r + b), \end{equation*}

which, together with (118), constitute the assumptions of Lemma 18. Applying it, we get

(120) \begin{equation} \max \left\{ bs(X), rs(Y) \right\} \le \frac{\nu}{1 - 7\delta/\alpha} = \frac{rb}{16(1 - 7\delta/\alpha)} \le \frac{rb}{16(1 - 7/16)} = \frac{rb}{9}, \end{equation}

where the second inequality follows by our assumption $\delta \le \alpha/16$ .

On the other hand, since X contains the set $R^w_1 = \Gamma_R(w)$ of red neighbours of w, by $(r, b, \delta)$ -regularity of F we have $s(X) \ge r - \delta \ge \frac{15}{16}r$ , a contradiction with (120). Hence, we have shown (119). Since $X = R^w_{2t + 1}\subseteq R^w_{2T + 1}$ and $Y = B^w_{2t}\subseteq B^w_{2T}$ , we also have

(121) \begin{equation}s\left({R^w_{2T + 1}}\right) > \frac{r}{r + b}, \quad s\left({B^w_{2T}}\right) > \frac{b}{r + b}.\end{equation}

Since we chose w arbitrarily, (121) holds for $w \in \left\{ x,y \right\}$ , implying

(122) \begin{equation} s\left({R^y_{2T + 1}}\right) > \frac{r}{r + b}, \quad s\left({B^x_{2T}}\right) > \frac{b}{r + b}.\end{equation}

Let us assume, without loss of generality, that $x \in V_1$ and $y \in V_2$ . Then, $B^x_{2T}, R^y_{2T + 1}\subseteq V_1$ and, in particular, for every vertex $v\in B^x_{2T}$ there is a walk (of even length at most 2T) from v to x starting with a blue and thus ending with a red edge. By (122), $s({B^x_{2T}}) + s({R^y_{2T + 1}}) > 1$ , so there exists $v \in B^x_{2T} \cap R^y_{2T + 1}$ . This means that there is an alternating walk of length at most $2T + (2T + 1)= 4\lceil 16/rb \rceil + 1 = L$ from x to y (through v) that starts with a red edge.

By an analogous reasoning with the roles of the colours red and blue swapped, $R^x_{2T}\cap B^y_{2T + 1}\neq \emptyset$ , and thus there is an alternating walk of length at most $2T + (2T + 1) = L $ from x to y which starts with a blue edge.

The following result is an easy consequence of Lemma 19.

6.4. Proof of Lemma 10

Let $\alpha = \min \left\{ \tau p, q \right\}$ . Applying Lemma 13 with $\lambda = (C + 1)\log N$ (note that the condition $\lambda \le \tau_0p\hat n$ follows generously from (147)) we have that, with probability $1 - O(N^{-C})$ , for every $t < t_0$

(123) \begin{equation} \forall v_i \in V_i \quad \tau d_i (1 - \delta) \le d_i - \deg_{\mathbb{R}(t)}(v_i) \le \tau d_i (1 + \delta), \qquad i \in \left\{ 1,2 \right\},\end{equation}

where

\begin{equation*} \delta = 3\sqrt{(C+1) \log N/(\tau p\hat n )}\end{equation*}

Whenever $p > 0.49$ , by Lemma 17, with probability $1 - O(N^{-C})$ for every $t < t_0$ we have that