2. Overview of the proof of Theorem 1.9

As mentioned earlier, to prove Theorem 1.9 we will employ a variant of a standard approach for attacking $0$ -statements of Ramsey problems. For attacking the $0$ -statement of Conjecture 1.8, this standard approach is as follows:

• • For $G = G_{n,p}$ , assume $G \to (H_1, H_2)$ ;

• • Use structural properties of $G$ (resulting from this assumption) to show that $G$ contains at least one of a sufficiently small collection of non-isomorphic graphs $\mathcal{F}$ ;

• • Show that there exists a constant $b \gt 0$ such that for $p \leq bn^{-1/m_2(H_1,H_2)}$ we have that $G$ contains no graph in $\mathcal{F}$ a.a.s.;

• • Conclude, by contradiction, that $G \not \to (H_1, H_2)$ a.a.s.

The variant of this approach we will use is due to Marciniszyn, Skokan, Spöhel and Steger [Reference Marciniszyn, Skokan, Spöhel and Steger14], who proved Conjecture 1.3 for cliques. In [Reference Marciniszyn, Skokan, Spöhel and Steger14], for $r \gt \ell \geq 3$ , they employ an algorithm called Asym-Edge-Col which either produces a valid edge-colouring for $K_r$ and $K_{\ell }$ of $G$ (showing that $G \not \to (K_r, K_{\ell })$ ) or encounters an error. Instead of assuming $G \to (K_r, K_{\ell })$ , they assume algorithm Asym-Edge-Col encounters an error, and proceed with the standard approach from there. One of the advantages of this approach is that it provides an algorithm for constructing a valid edge-colouring for $K_r$ and $K_{\ell }$ , rather than just proving the existence of such a colouring.

2.1 On Conjecture 1.8

As mentioned earlier, we provide all but one step, Conjecture 1.8, of this approach. Let us consider how Conjecture 1.8 relates to previous work on the $0$ -statement of Conjecture 1.3. Firstly, Conjecture 1.8 was implicitly proven for pairs of cliques in [Reference Marciniszyn, Skokan, Spöhel and Steger14] and pairs of a clique and a cycle in [Reference Liebenau, Mattos, Mendonça and Skokan12]. More specifically, when $H_1$ and $H_2$ are both cliques (except when $H_1 = H_2 = K_3$ )Footnote ⁷, the authors of [Reference Marciniszyn, Skokan, Spöhel and Steger14] prove a slightly more general version of Conjecture 1.8 (Lemma 8 in [Reference Marciniszyn, Skokan, Spöhel and Steger14]) where $\hat{\mathcal{A}}(H_1, H_2, \varepsilon )$ is replaced with the set

\begin{equation*}\mathcal {A}(H_1,H_2) \,{:\!=}\, \{A \in \mathcal {C}(H_1, H_2) \,{:}\, m(A) \leq m_2(H_1, H_2) + 0.01 \land A \ \text {is }2\text {-connected}\}.\end{equation*}

Note that the proof of Lemma 8 in [Reference Marciniszyn, Skokan, Spöhel and Steger14] shows that $\mathcal{A}(H_1,H_2) \neq \emptyset$ for certain pairs of cliques $H_1$ and $H_2$ . When $H_1$ is a clique, $H_2$ is a cycle and $H_1 \neq H_2$ (that is, excluding again the case when $H_1 = H_2 = K_3$ ), the proof of Lemma 3.3 in [Reference Liebenau, Mattos, Mendonça and Skokan12] implies that there exists a constant $\varepsilon \gt 0$ such that $\hat{\mathcal{A}}(H_1, H_2, \varepsilon ) = \emptyset$ .

For reference, we note here the places in our proof of Theorem 1.9 where we specifically need Conjecture 1.8 to hold:

• • the proof of Lemma 5.3;

• • the proofs of Claims 6.6 and 7.6;

• • the definition of $\gamma = \gamma (H_1, H_2)$ in Section 6.

3. Organisation of paper

The paper is organised as follows. In Section 4, we collect together notation, density measures and several useful results we will need. In Section 5, we give our algorithm Asym-Edge-Col for producing a valid edge-colouring for $H_1$ and $H_2$ of $G = G_{n,p}$ provided it does not encounter an error (and Conjecture 1.8 holds for $H_1$ and $H_2$ ). In Sections 6-6.3, we prove that Asym-Edge-Col does not encounter an error a.a.s. (Lemma 5.5) in the case when $m_2(H_1) \gt m_2(H_2)$ . In Section 7, we prove Lemma 5.5 in the case when $m_2(H_1) = m_2(H_2)$ . In Section 8, we prove Theorem 1.10, before providing some concluding remarks in Section 9.

4. Notation, density measures and useful results

As far as possible we keep to the notation used in [Reference Marciniszyn, Skokan, Spöhel and Steger14]. Also, we repeat several definitions used earlier for ease of reference.

Let $G = (V,E)$ be a graph. We denote the number of vertices in $G$ by $v(G) = v_G \,{:\!=}\, |V(G)|$ and the number of edges in $G$ by $e(G) = e_G \,{:\!=}\, |E(G)|$ . Moreover, for graphs $H_1$ and $H_2$ we let $v_1 \,{:\!=}\, |V(H_1)|$ , $e_1 \,{:\!=}\, |E(H_1)|$ , $v_2 \,{:\!=}\, |V(H_2)|$ and $e_2 \,{:\!=}\, |E(H_2)|$ .

Let $H$ be a graph. The most well-known density measure is

\begin{align*} d(H) \,{:\!=}\, \begin{cases} e_H/v_H & \quad \text{if } v(H) \geq 1,\\ \\[-9pt] 0 & \quad \text{otherwise}. \end{cases} \end{align*}

Taking the maximum value of $d$ over all subgraphs $J \subseteq H$ , we have the following measure

\begin{equation*}m(H) \,{:\!=}\, \max \{d(J)\,{:}\, J \subseteq H\}.\end{equation*}

(We say that a graph $H$ is balanced with respect to $d$ , or just balanced, if we have $d(H) = m(H)$ . Moreover, we say $H$ is strictly balanced if for every proper subgraph $J \subsetneq H$ , we have $d(J) \lt m(H)$ .)

In [Reference Rödl and Ruciński19], Rödl and Ruciński introduced the following so-called $2$ -density measure.

\begin{align*} d_2(H) \,{:\!=}\, \begin{cases} (e_H - 1)/(v_H - 2) & \quad \text{if $H$ is non-empty with} \ v(H) \geq 3,\\ \\[-9pt] 1/2 & \quad \text{if} \ H \cong K_2,\\ \\[-9pt] 0 & \quad \text{otherwise}. \end{cases} \end{align*}

As with $d$ , we have an associated measure based on maximising $d_2$ over subgraphs of $H$ :

\begin{equation*}m_2(H) \,{:\!=}\, \max \!\left \{d_2(J)\,{:}\, J \subseteq H\right \}.\end{equation*}

Analogously to the notion of balancedness, we say that a graph $H$ is $2$ -balanced if $d_2(H) = m_2(H)$ , and strictly $2$ -balanced if for all proper subgraphs $J \subsetneq H$ , we have $d_2(J) \lt m_2(H)$ .

Regarding asymmetric Ramsey properties, in [Reference Kohayakawa and Kreuter10], Kohayakawa and Kreuter introduced the following generalisation of $d_2$ . Let $H_1$ and $H_2$ be any graphs, and define

\begin{align*} d_2(H_1,H_2) \,{:\!=}\, \begin{cases} \frac{e_1}{v_1 - 2 + \frac{1}{m_2(H_2)}} & \quad \text{if } H_2 \text{ is non-empty and} \ v_1 \geq 2,\\ \\[-9pt] 0 & \quad \text{otherwise}. \end{cases} \end{align*}

Similarly to before, we have the following measure based on maximising $d_2$ over all subgraphs $J \subseteq H_1$ .

\begin{equation*}m_2(H_1, H_2) \,{:\!=}\, \max \!\left \{d_2(J,H_2)\,{:}\, J \subseteq H_1\right \}.\end{equation*}

We say that $H_1$ is balanced with respect to $d_2(\cdot, H_2)$ if we have $d_2(H_1, H_2) = m_2(H_1, H_2)$ and strictly balanced with respect to $d_2(\cdot, H_2)$ if for all proper subgraphs $J \subsetneq H_1$ we have $d_2(J, H_2) \lt m_2(H_1, H_2)$ .

Observe that $m_2(\cdot, \cdot )$ is not symmetric in both arguments. Recall Proposition 1.7.

Proposition 1.7. Suppose that $H_1$ and $H_2$ are non-empty graphs with $m_2(H_1) \geq m_2(H_2)$ . Then we have

\begin{equation*}m_2(H_1) \geq m_2(H_1, H_2) \geq m_2(H_2).\end{equation*}

Moreover,

\begin{equation*}m_2(H_1) \gt m_2(H_1, H_2) \gt m_2(H_2) \ \text {whenever} \ m_2(H_1) \gt m_2(H_2).\end{equation*}

Note that if $m_2(H_1) = m_2(H_2)$ and $H_1$ and $H_2$ are non-empty graphs, then $H_1$ cannot be strictly balanced with respect to $d_2(\cdot, H_2)$ unless $H_1 \cong K_2$ . Indeed, otherwise, by Proposition 1.7 we would then have that

\begin{equation*}m_2(H_2) = m_2(H_1, H_2) \gt d_2(K_2, H_2) = m_2(H_2).\end{equation*}

The following fact will be useful in the proofs of Lemmas 4.2 and 6.8.

Fact 4.1. For $a,c,C \in \mathbb{R}$ and $b,d \gt 0$ , we have

\begin{equation*}(i) \ \ \frac {a}{b} \leq C \ \land \ \frac {c}{d} \leq C \implies \frac {a+c}{b+d} \leq C \ \text {and} \ \ (ii) \ \ \frac {a}{b} \geq C \ \land\ \frac {c}{d} \ \geq C \implies \frac {a+c}{b+d} \geq C\end{equation*}

and similarly, if also $b \gt d$ ,

\begin{equation*}(iii) \ \ \frac {a}{b} \leq C \ \land \ \frac {c}{d} \geq C \implies \frac {a-c}{b-d} \leq C \ \text {and} \ \ (iv) \ \ \frac {a}{b} \geq C \ \land\ \frac {c}{d} \ \leq C \implies \frac {a-c}{b-d} \geq C.\end{equation*}

The following result will be very useful for us, creating an important connection between types of balancedness and $2$ -connectivity.

Proof. By [Reference Nenadov and Steger16, Lemma 3.3], strictly $2$ -balanced graphs are $2$ -connected, hence $(ii)$ holds and $H_2$ is $2$ -connected in case $(i)$ . We now use a very similar method to that of the proof of Lemma 3.3 to show that $H_1$ is $2$ -connected in case $(i)$ . Since $H_1$ is strictly balanced with respect to $d_2(\cdot, H_2)$ , we have that $H_1$ is connected. Indeed, assume not. Let $H_1$ have $k \geq 2$ components and denote the number of vertices and edges in each component by $u_1, \ldots, u_k$ and $d_1, \ldots, d_k$ , respectively. Then since $H_1$ is strictly balanced with respect to $d_2(\cdot, H_2)$ , we must have that

\begin{equation*}\frac {\sum _{i=1}^k d_i}{\sum _{i=1}^k u_i - 2 + \frac {1}{m_2(H_2)}} \gt \frac {d_1}{u_1 - 2 + \frac {1}{m_2(H_2)}}\end{equation*}

and

\begin{equation*}\frac {\sum _{i=1}^k d_i}{\sum _{i=1}^k u_i - 2 + \frac {1}{m_2(H_2)}} \gt \frac {\sum _{i=2}^k d_i}{\sum _{i=2}^k u_i - 2 + \frac {1}{m_2(H_2)}}.\end{equation*}

Since $m_2(H_2) \gt 1$ , by Fact 4.1(i) we get that

\begin{equation*}\frac {\sum _{i=1}^k d_i}{\sum _{i=1}^k u_i - 2 + \frac {1}{m_2(H_2)}} \geq \frac {\sum _{i=1}^k d_i}{\sum _{i=1}^k u_i - 4 + \frac {2}{m_2(H_2)}} \gt \frac {\sum _{i=1}^k d_i}{\sum _{i=1}^k u_i - 2 + \frac {1}{m_2(H_2)}},\end{equation*}

a contradiction.

Assume $H_1$ is not $2$ -connected. Then there exists a cut vertexFootnote ⁹ $v \in V(H_1)$ . Further, using Proposition 1.7 alongside that $H_1$ is strictly balanced with respect to $d_2(\cdot, H_2)$ and $m_2(H_1) \gt m_2(H_2) \gt 1$ , we can show that $H_1$ does not contain any vertex of degree 1. Indeed, otherwise $\frac{e_1 - 1}{v_1 - 3 + \frac{1}{m_2(H_2)}} \gt \frac{e_1}{v_1 - 2 + \frac{1}{m_2(H_2)}} = m_2(H_1, H_2)$ , contradicting that $H_1$ is strictly balanced with respect to $d_2(\cdot, H_2)$ . Thus there exist subgraphs $J_1$ and $J_2$ of $H_1$ such that $|E(J_1)|, |E(J_2)| \geq 1$ , $J_1 \cup J_2 = H_1$ and $V(J_1) \cap V(J_2) = \{v\}$ . Using Fact 4.1(i) and that $H_1$ is strictly balanced with respect to $d_2(\cdot, H_2)$ , we have that

\begin{align*} e_1 = e_{J_1} + e_{J_2} & \lt\, m_2(H_1, H_2)\!\left (v_{J_1} - 2 + \frac{1}{m_2(H_2)} + v_{J_2} - 2 + \frac{1}{m_2(H_2)}\right ) \\ & =\, m_2(H_1,H_2)\!\left (v_1 - 3 + \frac{2}{m_2(H_2)}\right ). \end{align*}

However, since $m_2(H_2) \gt 1$ we also have that

\begin{equation*}\frac {e_1}{v_1 - 3 + \frac {2}{m_2(H_2)}} \gt \frac {e_1}{v_1 - 2 + \frac {1}{m_2(H_2)}} = m_2(H_1, H_2),\end{equation*}

contradicting the inequality above. Hence $H_1$ is $2$ -connected.

5. Algorithm for computing valid edge-colourings: Asym-Edge-Col

To prove Theorem 1.9, we can clearly assume $H_1$ and $H_2$ are non-empty graphs satisfying the criteria of Conjecture 1.8 and that Conjecture 1.8 itself holds. Suppose $G = G_{n,p}$ and $p \leq bn^{-1/m_2(H_1, H_2)}$ where $b$ will be a small constant defined later. As noted earlier, to prove Conjecture 1.3 we can show that a.a.s. $G$ has a valid edge-colouring for $H_1$ and $H_2$ . We construct our valid edge-colouring using an algorithm Asym-Edge-Col (see Figure 3). In order to state the algorithm succinctly, we need to define a considerable amount of notation, almost all of which we keep very similar to that in [Reference Marciniszyn, Skokan, Spöhel and Steger14].

Figure 3.

The implementation of algorithm Asym-Edge-Col.

Recall Definition 1.5. In particular, recall that for any graph $G$ we define the families

\begin{equation*}\mathcal {R}_G \,{:\!=}\, \{R \subseteq G \,{:}\, R \cong H_1\} \ \text {and}\ \mathcal {L}_G \,{:\!=}\, \{L \subseteq G \,{:}\, L \cong H_2\}\end{equation*}

of all copies of $H_1$ and $H_2$ in $G$ , respectively. Also, recall the family

\begin{equation*}\mathcal {L}^*_G \,{:\!=}\, \{L \in \mathcal {L}_G \,{:}\, \forall e \in E(L) \ \exists R \in \mathcal {R}_G \ \text {s.t.} \ E(L) \cap E(R) = \{e\}\}.\end{equation*}

We highlight here that if $E(L) \cap E(R) = \{e\}$ for some $L \in \mathcal{L}_G$ and $R \in \mathcal{R}_G$ then it is still possible that $|V(L) \cap V(R)| \gt 2$ .

Recall Definition 1.6. Intuitively, the graphs in $\hat{\mathcal{A}}$ are the building blocks of the graphs $\hat{G}$ which may remain after the edge deletion process in Asym-Edge-Col (described later).

For any graph $G$ , define

\begin{equation*}\mathcal {S}_G \,{:\!=}\, \{S \subseteq G\,{:}\, S \cong A \in \hat {\mathcal {A}} \land \nexists S' \supset S \ \text {with} \ S' \subseteq G, \ S' \cong A' \in \hat {\mathcal {A}}\},\end{equation*}

that is, the family $\mathcal{S}_G$ contains all maximal subgraphs of $G$ isomorphic to a member of $\hat{\mathcal{A}}$ . Hence, there are no two members $S_1, S_2 \in \mathcal{S}_G$ such that $S_1 \subsetneq S_2$ . For any edge $e \in E(G)$ , let

\begin{equation*}\mathcal {S}_G(e) \,{:\!=}\, \{S \in \mathcal {S}_G \,{:}\, e \in E(S)\}.\end{equation*}

Definition 5.1. We call $G$ an $\hat{\mathcal{A}}$ -graph if, for all $e \in E(G)$ , we have

\begin{align*} |\mathcal{S}_G(e)| = 1. \end{align*}

In particular, an $\hat{\mathcal{A}}$ -graph is an edge-disjoint union of graphs from $\hat{\mathcal{A}}$ . In an $\hat{\mathcal{A}}$ -graph $G$ , a copy of $H_1$ or $H_2$ can be a subgraph of $G$ in two particular ways: either it is a subgraph of an $S \in \mathcal{S}_G$ or it is a subgraph with edges in at least two different graphs from $\mathcal{S}_G$ . The former we call trivial copies of $H_1$ and $H_2$ , and we define

\begin{equation*}\mathcal {T}_G \,{:\!=}\, \left \{T \subseteq G \,{:}\, (T \cong H_1 \vee T \cong H_2) \land \left |\bigcup _{\substack {e \in E(T)}}\mathcal {S}_G(e)\right | \geq 2\right \}\end{equation*}

to be the family of all non-trivial copies of $H_1$ and $H_2$ in $G$ .

Our next lemma asserts that $(H_1, H_2)$ -sparse $\hat{\mathcal{A}}$ -graphs are easily colourable, provided Conjecture 1.8 holds.

Note that we did not use that $\hat{\mathcal{A}}$ is finite, as given by Conjecture 1.8, in our proof of Lemma 5.3, only that ‘every graph in $\hat{\mathcal{A}}$ has a valid edge-colouring for $H_1$ and $H_2$ ’. The finiteness of $\hat{\mathcal{A}}$ will be essential later for the proofs of Claims 6.6 and 7.6.

Now let us describe the algorithm Asym-Edge-Col which if successful outputs a valid edge-colouring of $G$ . In Asym-Edge-Col, edges are removed from and then inserted back into a working copy $G'=(V, E')$ of $G$ . Each edge is removed in the first while-loop only when it is not the unique intersection of the edge sets of some copy of $H_1$ and some copy of $H_2$ in $G'$ (line 6). It is then ‘pushed’Footnote ¹⁰ onto a stack $s$ such that when we reinsert edges (in reverse order) in the second while-loop we can colour them to construct a valid edge-colouring for $H_1$ and $H_2$ of $G$ ; if at any point $G'$ is an $(H_1, H_2)$ -sparse $\hat{\mathcal{A}}$ -graph, then we combine the colouring of these edges with a valid edge-colouring for $H_1$ and $H_2$ of $G'$ provided by A-Colour. We also keep track of the copies of $H_2$ in $G$ and push abstract representations of some of them (or all of them if $G'$ is never an $(H_1, H_2)$ -sparse $\hat{\mathcal{A}}$ -graph during Asym-Edge-Col) onto $s$ (lines 8 and 15) to be used later in the colour swapping stage of the second while-loop (lines 30-32).

Let us consider algorithm Asym-Edge-Col in detail. In line 5, we check whether $G'$ is an $(H_1, H_2)$ -sparse $\hat{\mathcal{A}}$ -graph or not. If not, then we enter the first while-loop. In line 6, we choose an edge $e$ which is not the unique intersection of the edge sets of some copy of $H_1$ and some copy of $H_2$ in $G'$ (if such an edge $e$ exists). Then in lines 7-12 we push each copy of $H_2$ in $G'$ that contains $e$ onto $s$ before pushing $e$ onto $s$ as well. Now, if every edge $e \in E'$ is the unique intersection of the edge sets of some copy of $H_1$ and some copy of $H_2$ in $G'$ , then we push onto $s$ a copy $L$ of $H_2$ in $G'$ which contains an edge that is not the unique intersection of the edge set of $L$ and the edge set of some copy of $H_1$ in $G'$ . If no such copies $L$ of $H_2$ exist, then the algorithm has an error in line 18. If Asym-Edge-Col does not run into an error, then we enter the second while-loop with input $G'$ . Observe that $G'$ is either the empty graph on vertex set $V$ or some $(H_1, H_2)$ -sparse $\hat{\mathcal{A}}$ -graph. By Lemma 5.3, $G'$ has a valid edge-colouring for $H_1$ and $H_2$ . The second while-loop successively removes edges (line 25) and copies of $L$ (line 29) from $s$ in the reverse order in which they were added onto $s$ , with the edges added back into $E'$ . Each time an edge is added back it is coloured blue, and if a monochromatic blue copy $L$ of $H_2$ is constructed, we make one of the edges of $L$ red (lines 30-32). This colouring process is then repeated until we have a valid edge-colouring for $H_1$ and $H_2$ of $G$ .

The following lemma confirms that our colouring process in the second while-loop produces a valid edge-colouring for $H_1$ and $H_2$ of $G$ .

To prove Theorem 1.9, it now suffices to prove the following lemma.

We split our proof of Lemma 5.5 into two cases: (1) when $m_2(H_1) \gt m_2(H_2)$ and (2) when $m_2(H_1) = m_2(H_2)$ . Notice that this accords with our definition of $\hat{\mathcal{A}}$ .

6. Case 1: $m_2(H_1) \gt m_2(H_2)$ .

We will prove Case 1 of Lemma 5.5 using an auxiliary algorithm Grow (see Figure 4). If Asym-Edge-Col has an error, then Grow computes a subgraph $F \subseteq G$ which is either too large in size or too dense to appear in $G_{n,p}$ a.a.s. (with $p$ as in Lemma 5.5). Indeed, letting $\mathcal{F}$ be the class of all graphs that can possibly be returned by Grow, we will show that the expected number of copies of graphs from $\mathcal{F}$ contained in $G_{n,p}$ is $o(1)$ , which with Markov’s inequality implies that $G_{n,p}$ a.a.s. contains no graph from $\mathcal{F}$ . This in turn implies Lemma 5.5 by contradiction. Note that algorithm Grow is only used for proving Lemma 5.5 and hence does not add anything on to the run-time of Asym-Edge-Col.

Figure 4.

The implementation of algorithm Grow.

To state Grow we require the following definitions. Let

(1) \begin{equation} \gamma = \gamma (H_1,H_2) \,{:\!=}\, \frac{1}{m_2(H_1, H_2)} - \frac{1}{m_2(H_1, H_2) + \varepsilon (H_1, H_2)} \gt 0, \end{equation}

where $\varepsilon (H_1, H_2)$ is the constant in Conjecture 1.8. Recall that for any graph $F$ , we have

\begin{equation*}\lambda (F) \,{:\!=}\, v(F) - \frac {e(F)}{m_2(H_1,H_2)}\end{equation*}

and that this definition is motivated by the fact that the expected number of copies of $F$ in $G_{n,p}$ with $p = bn^{-1/m_2(H_1,H_2)}$ has order of magnitude

\begin{equation*}n^{v(F)}p^{e(F)} = b^{e(F)}n^{\lambda (F)}.\end{equation*}

Also, recall that

\begin{equation*}\mathcal {T}_G \,{:\!=}\, \left \{T \subseteq G \,{:}\, (T \cong H_1 \vee T \cong H_2) \land \left |\bigcup _{\substack {e \in E(T)}}\mathcal {S}_G(e)\right | \geq 2\right \}.\end{equation*}

For any graph $F$ and edge $e \in E(F)$ , we say that $e$ is eligible for extension in Grow if it satisfies

\begin{equation*}\nexists L \in \mathcal {L}^*_F \ \text {s.t.} \ e \in E(L),\end{equation*}

and observe that $F$ is in $\mathcal{C}^*$ (see Definition 1.5) if and only if it contains no edge that is eligible for extension in Grow.

Algorithm Grow has as input the graph $G' \subseteq G$ that Asym-Edge-Col got stuck on. Let us consider the properties of $G'$ when Asym-Edge-Col got stuck. Because the condition in line 6 of Asym-Edge-Col fails, $G'$ is in the family $\mathcal{C}$ , where we recall

\begin{equation*}\mathcal {C} = \mathcal {C}(H_1, H_2) \,{:\!=}\, \{G = (V,E) \,{:}\, \forall e \in E\ \exists (L,R) \in \mathcal {L}_G \times \mathcal {R}_G\ \text {s.t.}\ E(L) \cap E(R) = \{e\}\}.\end{equation*}

In particular, every edge of $G'$ is contained in a copy $L \in \mathcal{L}_{G^{\prime}}$ of $H_2$ , and, because the condition in line 14 fails, we can assume in addition that $L$ belongs to $\mathcal{L}^*_{G^{\prime}}$ . Hence, $G'$ is actually in the family $\mathcal{C}^* = \mathcal{C}^*(H_1, H_2)$ where we recall

\begin{equation*}\mathcal {C}^* = \mathcal {C}^*(H_1, H_2) \,{:\!=}\, \{G = (V,E)\,{:}\, \forall e \in E \ \exists L \in \mathcal {L}^*_{G} \ \text {s.t.} \ e \in E(L)\}.\end{equation*}

Lastly, $G'$ is not $(H_1, H_2)$ -sparse or not an $\hat{\mathcal{A}}$ -graph because Asym-Edge-Col ended with an error.

We now outline algorithm Grow. Firstly, Grow checks whether either of two special cases occur (lines 2-9). The first case corresponds to when $G'$ is an $\hat{\mathcal{A}}$ -graph, which is not $(H_1, H_2)$ -sparse as it is a graph that Asym-Edge-Col got stuck on. The second case happens if there are $2$ graphs in $\mathcal{S}_{G^{\prime}}(e)$ that overlap in (at least) the edge $e$ . The outputs of these two special cases are graphs which the while-loop of Grow could get stuck on. Indeed, if neither of these cases happen, then in line 10 we can choose an edge $e$ that does not belong to any graph in $\mathcal{S}_G$ . Crucially, algorithm Grow chooses a graph $R \in \mathcal{R}_{G^{\prime}}$ which contains such an edge $e$ and makes it the seed $F_0$ for a growing procedure (line 11). This choice of $F_0$ will allow us to conclude later that there always exists an edge eligible for extension in Grow (see the proof of Claim 6.1), that is, the while-loop of Grow operates as desired and doesn’t get stuck.

In each iteration $i$ of the while-loop, the growing procedure extends $F_i$ to $F_{i+1}$ in one of two ways. The first (lines 14-15) is by attaching a copy of $H_1$ in $G'$ that intersects $F_i$ in at least two vertices but is not contained in $F_i$ . The second is more involved and begins with calling a function Eligible-Edge which maps $F_i$ to an edge $e \in E(F_i)$ which is eligible for extension in Grow (we will show that such an edge always exists). Importantly, Eligible-Edge selects this edge $e$ to be unique up to isomorphism of $F_i$ , that is, for any two isomorphic graphs $F$ and $F'$ , there exists an isomorphism $\phi$ with $\phi (F) = F'$ such that

In particular, our choice of $e$ depends only on $F_i$ and not on the surrounding graph $G'$ or any previous graph $F_j$ with $j \lt i$ (indeed, there may be many ways that Grow could construct a graph isomorphic to $F_i$ ). One could implement Eligible-Edge by having an enormous table of representatives for all isomorphism classes of graphs with up to $n$ vertices. Since we do not care about complexity here, and only want to show the existence of certain structures in $G'$ , the time Eligible-Edge would take to be implemented is unimportant. What is important is that Eligible-Edge does not itself increase the number of graphs $F$ that Grow can output.

Once we have our edge $e \in E(F_i)$ eligible for extension in Grow, we apply a procedure called Extend-L which attaches a graph $L \in \mathcal{L}^*_{G^{\prime}}$ that contains $e$ to $F_i$ (line 18). We then attach to each new edge $e' \in E(L)\setminus E(F_i)$ a graph $R_{e^{\prime}} \in \mathcal{R}_{G^{\prime}}$ such that $E(L) \cap E(R_{e^{\prime}}) = \{e'\}$ (lines 4-6 of Extend-L). (We will show later that such a graph $L$ and graphs $R_{e^{\prime}}$ exist and that $E(L)\setminus E(F_i)$ is non-empty.) The algorithm comes to an end when either $i \geq \ln\!(n)$ or $\lambda (\tilde{F}) \leq -\gamma$ for some subgraph $\tilde{F} \subseteq F_i$ . In the former, the algorithm returns $F_i$ (line 23); in the latter, the algorithm returns a subgraph $\tilde{F} \subseteq F_i$ that minimises $\lambda (\tilde{F})$ (line 25). For each graph $F$ , the function Minimising Subgraph( $F$ ) returns such a minimising subgraph that is unique up to isomorphism. Once again, this is to ensure that Minimising Subgraph( $F$ ) does not itself artificially increase the number of graphs that Grow can output. As with function Eligible-Edge, one could implement Minimising Subgraph using an enormous look-up table.

We will now argue that Grow terminates without error, that is, Eligible-Edge always finds an edge eligible for extension in Grow and all ‘any’-assignments in Grow and Extend-L are always successful.

Proof. Our proof is very similar to the proof of Claim 13 in [Reference Marciniszyn, Skokan, Spöhel and Steger14].

We first show that the special cases in lines 2-9 always function as desired. The first case occurs if and only if $G'$ is an $\hat{\mathcal{A}}$ -graph. By assumption, $G'$ is not $(H_1, H_2)$ -sparse, hence the family $\mathcal{T}_{G^{\prime}}$ is not empty. Hence the assignment in line 3 is successful. Clearly, the assignment in line 7 is always successful due to the if-condition in line 6.

One can also easily see that the assignments in lines 10 and 11 are successful. Indeed, neither of the two special cases occur so we must have an edge $e \in E$ that is not contained in any $S \in \mathcal{S}_{G^{\prime}}$ . Also, there must exist a member of $\mathcal{R}_{G^{\prime}}$ that contains $e$ because $G'$ is a member of $\mathcal{C}^* \subseteq \mathcal{C}$ .

Next, we show that the call to Eligible-Edge in line 17 is always successful. Recall (1) on page 16. Indeed, suppose for a contradiction that no edge in $F_i$ is eligible for extension in Grow for some $i \geq 0$ . Then every edge $e \in E(F_i)$ is in some $L \in \mathcal{L}^*_{F_i}$ , by definition. Hence $F \in \mathcal{C}^*$ . Recall that $H_1$ and $H_2$ satisfy the criteria of Conjecture 1.8. Hence $H_2$ is strictly $2$ -balanced, $H_1$ is strictly balanced with respect to $d_2(\cdot, H_2)$ ) and $m_2(H_1) \geq m_2(H_2) \gt 1$ . Then, by Lemma 4.2, $H_1$ and $H_2$ are $2$ -connected, hence $F_i$ is $2$ -connected by construction. However, our choice of $F_0$ in line 11 guarantees that $F_i$ is not in $\hat{\mathcal{A}}$ . Indeed, the edge $e$ selected in line 10 satisfying $|\mathcal{S}_{G^{\prime}}(e)| = 0$ is an edge of $F_0$ and $F_0 \subseteq F_i \subseteq G'$ . Thus, by the definition of $\hat{\mathcal{A}}$ and that $m_2(H_1) \gt m_2(H_2)$ , we have that $m(F_i) \gt m_2(H_1, H_2) + \varepsilon$ . Thus, there exists a non-empty graph $\tilde{F} \subseteq F_i$ with $d(\tilde{F}) = m(F_i)$ such that

\begin{align*} \lambda (\tilde{F}) & = v(\tilde{F}) - \frac{e(\tilde{F})}{m_2(H_1, H_2)} \\ & = e(\tilde{F})\left (\frac{1}{m(F_i)} - \frac{1}{m_2(H_1, H_2)}\right ) \\ & \lt e(\tilde{F})\left (\frac{1}{m_2(H_1, H_2) + \varepsilon } - \frac{1}{m_2(H_1, H_2)}\right ) \\ & = -\gamma e(\tilde{F}) \leq -\gamma . \end{align*}

Thus Grow terminates in line 13 without calling Eligible-Edge, and so every call to Eligible-Edge is successful and returns an edge $e$ . Since $G' \in \mathcal{C^*}$ , the call to Extend-L $(F_i, e, G')$ is also successful and thus there exist suitable graphs $L \in \mathcal{L}^*_{G^{\prime}}$ with $e \in E(L)$ and $R_{e^{\prime}}$ for each $e' \in E(L) \setminus E(F_i)$ .

It remains to show that for every iteration $i$ of the while-loop, we have $e(F_{i+1}) \gt e(F_i)$ . Since a copy $R$ of $H_1$ found in line 14 is a copy of $H_1$ in $G'$ but not in $F_i$ (and $H_1$ is connected), we must have that $F_{i+1} = F_i \cup R$ contains at least one more edge than $F_i$ .

So assume lines 17 and 18 are called in iteration $i$ and let $e$ be the edge chosen in line 17 and $L$ the subgraph selected in line 2 of Extend-L $(F_i, e, G')$ . By the definition of $\mathcal{L}^*_{G^{\prime}}$ , for each $e' \in E(L)$ there exists $R_{e^{\prime}} \in \mathcal{R}_{G^{\prime}}$ such that $E(L) \cap E(R_{e^{\prime}}) = \{e'\}$ . If $|E(L) \setminus E(F_i)| \gt 0$ , then $e(F_{i+1}) \geq e(F_i \cup L) \gt e(F_i)$ . Otherwise, $L \subseteq F_i$ . But since $e$ is eligible for extension in Grow, we must have $L \notin \mathcal{L}^*_{F_i}$ . Thus there exists $e' \in L$ such that $R_{e^{\prime}} \in \mathcal{R}_{G^{\prime}}\setminus \mathcal{R}_{F_i}$ and $|V(R_{e^{\prime}}) \cap V(F_i)| \geq 2$ , contradicting that lines 17 and 18 are called in iteration $i$ .

6.1 Proof of Lemma 5.5

We consider the evolution of $F_i$ now in more detail. We call iteration $i$ of the while-loop in algorithm Grow non-degenerate (See Figure 5) if all of the following hold:

• • The condition in line 14 evaluates to false (and Extend-L is called);

• • In line 3 of Extend-L, we have $V(F) \cap V(L) = e$ ;

• • In every execution of line 6 of Extend-L, we have $V(F') \cap V(R_{e^{\prime}}) = e'$ .

Figure 5.

A graph $F_2$ resulting from two non-degenerate iterations for $H_1 = K_4$ and $H_2 = C_4$ . The two central copies of $H_2$ are shaded.

Otherwise, we call iteration $i$ degenerate. Note that, in non-degenerate iterations, there are only a constant number of graphs $F_{i+1}$ that can result from any given $F_i$ since Eligible-Edge determines the exact position where to attach the copy $L$ of $H_2$ , $V(F_i) \cap V(L) = e$ and for every execution of line 6 of Extend-L we have $V(F') \cap V(R_{e^{\prime}}) = e'$ (recall that the edge $e$ found by Eligible-Edge( $F_i$ ) is unique up to isomorphism of $F_i$ ).

Claim 6.2. If iteration $i$ of the while-loop in procedure Grow is non-degenerate, we have

\begin{equation*}\lambda (F_{i+1}) = \lambda (F_i).\end{equation*}

Proof. In a non-degenerate iteration we add $v_2 - 2$ vertices and $e_2 - 1$ edges for the copy of $H_2$ and then $(e_2 - 1)(v_1 - 2)$ new vertices and $(e_2 - 1)(e_1 - 1)$ new edges to complete the copies of $H_1$ . This gives

\begin{align*} \lambda (F_{i+1}) - \lambda (F_i) & = v_2 - 2 + (e_2 - 1)(v_1 - 2) - \frac{(e_2 - 1)e_1}{m_2(H_1,H_2)} \\[2pt] & = v_2 - 2 + (e_2 - 1)(v_1 - 2) - (e_2 - 1)\left (v_1 - 2 + \frac{1}{m_2(H_2)}\right ) \\[2pt] & = 0, \end{align*}

where we have used in the penultimate equality that $H_1$ is (strictly) balanced with respect to $d_2(\cdot, H_2)$ and in the final inequality that $H_2$ is (strictly) $2$ -balanced.

When we have a degenerate iteration $i$ , the structure of $F_{i+1}$ may vary considerably and also depend on the structure of $G'$ . Indeed, if $F_i$ is extended by a copy $R$ of $H_1$ in line 15, then $R$ could intersect $F_i$ in a multitude of ways. Moreover, there may be several copies of $H_1$ that satisfy the condition in line 14. The same is true for graphs added in lines 3 and 6 of Extend-L. Thus, degenerate iterations cause us difficulties since they enlarge the family of graphs algorithm Grow can return. However, we will show that at most a constant number of degenerate iterations can happen before algorithm Grow terminates, allowing us to bound from above sufficiently well the number of non-isomorphic graphs Grow can return. Pivotal in proving this is the following claim.

Claim 6.3. There exists a constant $\kappa = \kappa (H_1,H_2) \gt 0$ such that if iteration $i$ of the while-loop in procedure Grow is degenerate then we have

\begin{equation*}\lambda (F_{i+1}) \leq \lambda (F_i) - \kappa .\end{equation*}

We prove Claim 6.3 in Section 6.2. Together, Claims 6.2 and 6.3 yield the following claim.

Claim 6.4. There exists a constant $q_1 = q_1(H_1,H_2)$ such that algorithm Grow performs at most $q_1$ degenerate iterations before it terminates, regardless of the input instance $G'$ .

Proof. By Claim 6.2, the value of the function $\lambda$ remains the same in every non-degenerate iteration of the while-loop of algorithm Grow. However, Claim 6.3 yields a constant $\kappa$ , which depends solely on $H_1$ and $H_2$ , such that

\begin{equation*}\lambda (F_{i+1}) \leq \lambda (F_i) - \kappa \end{equation*}

for every degenerate iteration $i$ .

Hence, after at most

\begin{equation*}q_1 \,{:\!=}\, \frac {\lambda (F_0) + \gamma }{\kappa }\end{equation*}

degenerate iterations, we have $\lambda (F_i) \leq -\gamma$ , and algorithm Grow terminates.

For $0 \leq d \leq t \lt \lceil \ln\!(n) \rceil$ , let $\mathcal{F}(t,d)$ denote a family of representatives for the isomorphism classes of all graphs $F_t$ that algorithm Grow can possibly generate after exactly $t$ iterations of the while-loop with exactly $d$ of those $t$ iterations being degenerate. Let $f(t,d) \,{:\!=}\, |\mathcal{F}(t,d)|$ .

Claim 6.5. There exist constants $C_0 = C_0(H_1, H_2)$ and $A = A(H_1,H_2)$ such that

\begin{equation*}f(t,d)\ \leq \lceil \ln\!(n)\rceil ^{(C_0 +1)d}\cdot \ A^{t-d}\end{equation*}

for $n$ sufficiently large.

Proof. By Claim 6.1, in every iteration $i$ of the while-loop of Grow, we add new edges onto $F_i$ . These new edges span a graph on at most

\begin{equation*}K \,{:\!=}\, v_2 + (e_2 - 1)(v_1 - 2)\end{equation*}

vertices. Thus $v(F_t) \leq v_1 + Kt$ . Let $\mathcal{G}_K$ denote the set of all graphs on at most $K$ vertices. In iteration $i$ of the while-loop, $F_{i+1}$ is uniquely defined if one specifies the graph $G \in \mathcal{G}_K$ with edges $E(F_{i+1})\setminus E(F_{i})$ , the number $y$ of vertices in which $G$ intersects $F_i$ , and two ordered lists of vertices from $G$ and $F_i$ respectively of length $y$ , which specify the mapping of the intersection vertices from $G$ onto $F_i$ . Thus, the number of ways that $F_i$ can be extended to $F_{i+1}$ is bounded from above by

\begin{equation*}\sum _{G \in \mathcal {G}_K}\sum _{y = 2}^{v(G)}v(G)^yv(F_i)^y \leq |\mathcal {G}_K|\cdot K \cdot K^K(v_1 + Kt)^K \leq \lceil \ln\!(n) \rceil ^{C_0},\end{equation*}

where $C_0$ depends only on $v_1$ , $v_2$ and $e_2$ , and $n$ is sufficiently large. The last inequality follows from the fact that $t \lt \ln\!(n)$ as otherwise the while-loop would have already ended.

Recall that, since Eligible-Edge determines the exact position where to attach the copy of $H_2$ , in non-degenerate iterations $i$ there are at most

\begin{align*} 2e_2(2e_1)^{e_2-1} \,{=\!:}\, A \end{align*}

ways to extend $F_i$ to $F_{i+1}$ , where the coefficients of 2 correspond with the orientations of the edge of the copy of $H_2$ we attach to $F_i$ and the edges of the copies of $H_1$ we attach to said copy of $H_2$ . Hence, for $0 \leq d \leq t \lt \lceil \ln\!(n) \rceil$ ,

\begin{align*} f(t,d) \leq \binom{t}{d}(\lceil \ln\!(n) \rceil ^{C_0})^d \cdot A^{t-d} \leq \lceil \ln\!(n) \rceil ^{(C_0 +1)d} \cdot A^{t-d}, \end{align*}

where the binomial coefficient corresponds to the choice of when in the $t$ iterations the $d$ degenerate iterations happen.

A reader of [Reference Marciniszyn, Skokan, Spöhel and Steger14] may observe that Claim 6.5 is not analogous to Claim 17 in [Reference Marciniszyn, Skokan, Spöhel and Steger14]. Indeed, since we have a constant number of non-degenerate iterations, instead of a unique non-degenerate iteration as in [Reference Marciniszyn, Skokan, Spöhel and Steger14], we truncated the proof of Claim 17 in order to have the appropriate bound to prove the following claim. Let $\mathcal{F} = \mathcal{F}(H_1, H_2, n)$ be a family of representatives for the isomorphism classes of all graphs that can be outputted by Grow (whether Grow enters the while-loop or not). Note that the proof of the following claim requires Conjecture 1.8 to be true; in particular, we need that $\hat{\mathcal{A}}(H_1, H_2, \varepsilon )$ is finite when $m_2(H_1) \gt m_2(H_2)$ .

Claim 6.6. There exists a constant $b = b(H_1,H_2) \gt 0$ such that for all $p \leq bn^{-1/m_2(H_1,H_2)}$ , $G_{n,p}$ does not contain any graph from $\mathcal{F}(H_1,H_2,n)$ a.a.s.

Proof. We first consider the two special cases in lines 2-9 of Grow. Let $\mathcal{F}_0 = \mathcal{F}_0(H_1,H_2) \subseteq \mathcal{F}$ denote the class of graphs that can be outputted by Grow if one of these two cases happens. We can see that any $F \in \mathcal{F}_0$ is either of the form

\begin{equation*}F = \bigcup _{\substack {e \in E(T)}}\mathcal {S}_{G^{\prime}}(e),\end{equation*}

for some graph $T \in \mathcal{T}_{G^{\prime}}$ , or of the form

\begin{equation*}F = S_1 \cup S_2,\end{equation*}

for some edge-intersecting $S_1, S_2 \in \mathcal{S}_{G^{\prime}}$ . Whichever of these forms $F$ has, since every element of $\mathcal{S}_{G^{\prime}}$ is $2$ -connected and in $\mathcal{C}^*$ , and $T$ is $2$ -connectedFootnote ¹², we have that $F$ is $2$ -connected and in $\mathcal{C}^*$ . On the other hand, $F \subseteq G'$ is not in $\mathcal{S}_{G^{\prime}}$ and thus not isomorphic to a graph in $\hat{\mathcal{A}}$ . Indeed, otherwise the graphs $S$ forming $F$ would not be in $\mathcal{S}_{G^{\prime}}$ due to the maximality condition in the definition of $\mathcal{S}_{G^{\prime}}$ . It follows that $m(F) \gt m_2(H_1, H_2) + \varepsilon (H_1, H_2)$ . Since we assumed Conjecture 1.8 holds, the family $\mathcal{F}_0$ is finite. Hence Markov’s inequality yields that $G_{n,p}$ contains no graph from $\mathcal{F}_0$ a.a.s.

Let $\tilde{\mathcal{F}} = \tilde{\mathcal{F}}(H_1, H_2, n)$ denote a family of representatives for the isomorphism classes of all graphs that can be the output of Grow with parameters $n$ and $\gamma (H_1, H_2)$ on any input instance $G^{\prime}$ for which it enters the while-loop. Observe that $\mathcal{F} = \mathcal{F}_0 \cup \tilde{\mathcal{F}}$ . Let $\mathcal{F}_1$ and $\mathcal{F}_2$ denote the classes of graphs that algorithm Grow can output in lines 23 and 25, respectively. For each $F \in \mathcal{F}_1$ , we have that $e(F) \geq \ln\!(n)$ , as $F$ was generated in $\lceil \ln\!(n) \rceil$ iterations, each of which introduces at least one new edge by Claim 6.1. Moreover, Claims 6.2 and 6.3 imply that $\lambda (F_i)$ is non-increasing. Thus, we have that $\lambda (F) \leq \lambda (F_0)$ for all $F \in \mathcal{F}_1$ . For all $F \in \mathcal{F}_2$ , we have that $\lambda (F) \leq -\gamma$ due to the condition in line 13 of Grow. Let $A \,{:\!=}\, A(H_1, H_2)$ be the constant found in the proof of Claim 6.5. Since we have chosen $F_0 \cong H_1$ as the seed of the growing procedure, it follows that for

\begin{equation*}b \,{:\!=}\, (Ae)^{-\lambda (F_0)-\gamma } \leq 1,\end{equation*}

the expected number of copies of graphs from $\tilde{\mathcal{F}}$ in $G_{n,p}$ with $p \leq bn^{-1/m_2(H_1,H_2)}$ is bounded by

(2) \begin{align} \sum _{F \in \tilde{\mathcal{F}}}n^{v(F)}p^{e(F)} & \leq \sum _{F \in \tilde{\mathcal{F}}}b^{e(F)}n^{\lambda (F)} \nonumber\\[2pt] & \leq \sum _{F \in \mathcal{F}_1}(eA)^{(-\lambda (F_0) - \gamma )\ln\!(n)}n^{\lambda (F_0)} + \sum _{F \in \mathcal{F}_2}b^{e(F)}n^{-\gamma }\nonumber \\[2pt] & = \sum _{F \in \mathcal{F}_1}A^{(-\lambda (F_0) - \gamma )\ln\!(n)}n^{-\gamma } + \sum _{F \in \mathcal{F}_2}b^{e(F)}n^{-\gamma }. \end{align}

Observe that, since $m_2(F_2) \geq 1$ , we have that

(3) \begin{equation} \lambda (F_0) = v_1 - \frac{e_1}{m_2(F_1,F_2)} = 2 - \frac{1}{m_2(F_2)} \geq 1 \end{equation}

By Claims 6.1, 6.4 and 6.5, and (3), we have that

(4) \begin{align} \sum _{F \in \mathcal{F}_1}A^{(-\lambda (F_0) - \gamma )\ln\!(n)}n^{-\gamma } & \leq \sum _{d = 0}^{\min \{t, q_1\}}f(\lceil \ln\!(n) \rceil,d)A^{(-\lambda (F_0) - \gamma )\ln\!(n)}n^{-\gamma } \nonumber\\[2pt] & \leq (q_1 + 1)\lceil \ln\!(n) \rceil ^{(C_0 + 1)q_1} \cdot A^{\lceil \ln\!(n) \rceil }A^{(-\lambda (F_0) - \gamma )\ln\!(n)}n^{-\gamma }\nonumber \\[2pt] & \leq (\!\ln\!(n))^{2(C_0 + 1)q_1}n^{-\gamma }. \end{align}

Observe that, by Claim 6.1, if some graph $F \in \mathcal{F}_2$ is the output of Grow after precisely $t$ iterations of the while-loop then $e(F) \geq t$ . Since $b \lt 1$ , this implies

(5) \begin{equation} b^{e(F)} \leq b^t, \end{equation}

for such a graph $F$ . Using (5) and Claims 6.1, 6.4 and 6.5, we have that

(6) \begin{align} \sum _{F \in \mathcal{F}_2}b^{e(F)}n^{-\gamma } & \leq \sum _{t = 0}^{\lceil \ln\!(n) \rceil }\sum _{d = 0}^{\min \{t, q_1\}}f(t,d)b^tn^{-\gamma } \nonumber \\[2pt] & \leq \sum _{t = 0}^{\lceil \ln\!(n) \rceil }\sum _{d = 0}^{\min \{t, q_1\}}\lceil \ln\!(n) \rceil ^{(C_0 +1)d} \cdot A^{t-d}(Ae)^{(-\lambda (F_0)-\gamma )t}n^{-\gamma }\nonumber \\[2pt] & \leq (\lceil \ln\!(n) \rceil + 1)(q_1 + 1)\lceil \ln\!(n) \rceil ^{(C_0 +1)q_1} n^{-\gamma }\nonumber \\[2pt] & \leq (\!\ln\!(n))^{2(C_0 + 1)q_1}n^{-\gamma }. \end{align}

Thus, by (2), (4) and (6), we have that $\sum _{F \in \tilde{\mathcal{F}}}n^{v(F)}p^{e(F)} = o(1)$ . Consequently, Markov’s inequality implies that $G_{n,p}$ a.a.s. contains no graph from $\tilde{\mathcal{F}}$ .

Combined with the earlier observation that $G_{n,p}$ a.a.s. contains no graph from $\mathcal{F}_0$ , we have that $G_{n,p}$ a.a.s. contains no graph from $\mathcal{F} = \mathcal{F}_0 \cup \tilde{\mathcal{F}}$ .

Proof of Lemma 5.5 : Case 1. Suppose that the call to Asym-Edge-Col $(G)$ gets stuck for some graph $G$ , and consider $G' \subseteq G$ at this moment. Then Grow $(G')$ returns a copy of a graph $F \in \mathcal{F}(H_1, H_2, n)$ that is contained in $G' \subseteq G$ . Provided Claim 6.3 holds, by Claim 6.6 this event a.a.s. does not occur in $G = G_{n,p}$ with $p$ as claimed. Thus Asym-Edge-Col does not get stuck a.a.s. and, by Lemma 5.4, finds a valid colouring for $H_1$ and $H_2$ of $G_{n,p}$ with $p \leq bn^{-1/m_2(H_1,H_2)}$ a.a.s.

6.2 Proof of Claim 6.3

Our strategy for proving Claim 6.3 revolves around comparing our degenerate iteration $i$ of the while-loop of algorithm Grow with any non-degenerate iteration which could have occurred instead. In accordance with this strategy, we have the following technical lemma which will be crucial in proving Claim 6.3.Footnote ¹³ The lemma will play the same role as Lemma 21 does in [Reference Marciniszyn, Skokan, Spöhel and Steger14], but is considerably different. In order to state our technical lemma, we define the following families of graphs.

Definition 6.7. Let $F$ , $H_1$ and $H_2$ be graphs and $\hat{e} \in E(F)$ . We define $\mathcal{H}(F, \hat{e}, H_1, H_2)$ to be the family of graphs constructed from $F$ in the following way: Attach a copy $H_{\hat{e}}$ of $H_2$ to $F$ such that $E(H_{\hat{e}}) \cap E(F) = \{\hat{e}\}$ and $V(H_{\hat{e}}) \cap V(F) = \hat{e}$ . Then, for each edge $f \in E(H_{\hat{e}})\setminus \{\hat{e}\}$ , attach a copy $H_f$ of $H_1$ to $F \cup H_{\hat{e}}$ such that $E(F \cup H_{\hat{e}}) \cap E(H_f) = \{f\}$ and $(V(F)\setminus \hat{e}) \cap V(H_f) = \emptyset$ .

Notice that, during construction of a graph $J \in \mathcal{H}(F, \hat{e}, H_1, H_2)$ , the edge of $H_{\hat{e}}$ intersecting at $\hat{e}$ and the edge of each copy $H_f$ of $H_1$ intersecting at an edge $f \in E(H_{\hat{e}})\setminus \{\hat{e}\}$ are not stipulated. That is, we may end up with different graphs after the construction process if we choose different edges of $H_{\hat{e}}$ to intersect $F$ at $\hat{e}$ and different edges of the copies $H_f$ of $H_1$ to intersect the edges in $E(H_{\hat{e}})\setminus \{\hat{e}\}$ . Observe that although $E(F \cup H_{\hat{e}}) \cap E(H_f) = \{f\}$ and $(V(F)\setminus \hat{e}) \cap V(H_f) = \emptyset$ for each $f \in E(H_{\hat{e}}) - \{\hat{e}\}$ , the construction may result in one or more graphs $H_f$ intersecting $H_{\hat{e}}$ in more than two vertices, including possibly in vertices of $\hat{e}$ (e.g. $H_{f_3}$ in Figure 6). Also, the graphs $H_f$ may intersect with each other in vertices and/or edges (e.g. $H_{f_1}$ and $H_{f_2}$ in Figure 6).

Figure 6.

A graph $J \in \mathcal{H}(F, \hat{e}, C_5, C_6)\setminus \mathcal{H}^*(F, \hat{e}, C_5, C_6)$ .

Borrowing notation and language from [Reference Marciniszyn, Skokan, Spöhel and Steger14], for any $J \in \mathcal{H}(F, \hat{e}, H_1, H_2)$ we call $V_J \,{:\!=}\, V(H_{\hat{e}})\setminus \hat{e}$ the inner vertices of $J$ and $E_J \,{:\!=}\, E(H_{\hat{e}})\setminus \{\hat{e}\}$ the inner edges of $J$ . Let $H_{\hat{e}}^{J}$ be the inner graph on vertex set $V_J \ \dot{\cup } \ \hat{e}$ and edge set $E_J$ , and observe that this graph $H_{\hat{e}}^{J}$ is isomorphic to a copy of $H_2$ minus some edge. Further, for each copy $H_f$ of $H_1$ , we define $U_J(\,f) \,{:\!=}\, V(H_f)\setminus f$ and $D_J(\,f) \,{:\!=}\, E(H_f) \setminus \{f\}$ and call

\begin{equation*}U_J \,{:\!=}\, \bigcup _{\substack {{f \in E_J}}} U_J(\,f),\end{equation*}

the set of outer vertices of $J$ and

\begin{equation*}D_J \,{:\!=}\, \bigcup _{\substack {{f \in E_J}}} D_J(\,f),\end{equation*}

the set of outer edges of $J$ . Observe that the sets $U_J(\,f)$ may overlap with each other and, as noted earlier, with $V(H_{\hat{e}}^{J})$ . However, the sets $D_J(\,f)$ may overlap only with each other. Further, define $\mathcal{H}^*(F, \hat{e}, H_1, H_2) \subseteq \mathcal{H}(F, \hat{e}, H_1, H_2)$ such that for any $J^* \in \mathcal{H}^*(F, \hat{e}, H_1, H_2)$ we have $U_{J^*}(\,f_1) \cap U_{J^*}(\,f_2) = \emptyset$ and $D_{J^*}(\,f_1) \cap D_{J^*}(\,f_2) = \emptyset$ for all $f_1, f_2 \in E_{J^*}$ , $f_1 \neq f_2$ , and $U_{J^*}(\,f) \cap V\left(H_{\hat{e}}^{J^*}\right) = \emptyset$ for all $f \in E_{J^*}$ ; that is, the copies of $H_1$ are, in some sense, pairwise disjoint. Note that each $J^* \in \mathcal{H}^*(F, \hat{e}, H_1, H_2)$ corresponds with a non-degenerate iteration $i$ of the while-loop of algorithm Grow when $F = F_i$ , $J^* = F_{i+1}$ and $\hat{e}$ is the edge chosen by Eligible-Edge( $F_i$ ). This observation will be very helpful several times later. For any $J \in \mathcal{H}(F, \hat{e}, H_1, H_2)$ , define

\begin{equation*}v^{+}(J) \,{:\!=}\, |V(J)\setminus V(F)| = v(J) - v(F),\end{equation*}

and

\begin{equation*}e^{+}(J) \,{:\!=}\, |E(J)\setminus E(F)| = e(J) - e(F),\end{equation*}

and call $\frac{e^{+}(J)}{v^{+}(J)}$ the $F$ -external density of $J$ . The following lemma relates the $F$ -external density of any $J^* \in \mathcal{H}^*(F, \hat{e}, H_1, H_2)$ to that of any $J \in \mathcal{H}(F, \hat{e}, H_1, H_2)\setminus \mathcal{H}^*(F, \hat{e}, H_1, H_2)$ .

Lemma 6.8. Let $F$ be a graph and $\hat{e} \in E(F)$ . Then for any $J \in \mathcal{H}(F, \hat{e}, H_1, H_2)\setminus \mathcal{H}^*(F, \hat{e}, H_1, H_2)$ and any $J^* \in \mathcal{H}^*(F, \hat{e}, H_1, H_2)$ , we have

\begin{equation*}\frac {e^{+}(J)}{v^{+}(J)} \gt \frac {e^{+}(J^*)}{v^{+}(J^*)}.\end{equation*}

We prove Lemma 6.8 in Section 6.3.

Claim 6.3 will follow from the next two claims. We say that algorithm Grow encounters a degeneracy of type 1 in iteration $i$ of the while-loop if line 14 returns true, that is, $\exists R \in \mathcal{R}_{G^{\prime}}\setminus \mathcal{R}_{F_i} \,{:}\, |V(R) \cap V(F_i)| \geq 2$ . Note that the following claim requires that $m_2(H_1) \gt m_2(H_2)$ .

Claim 6.9. There exists a constant $\kappa _1 = \kappa _1(H_1, H_2) \gt 0$ such that if procedure Grow encounters a degeneracy of type 1 in iteration $i$ of the while-loop, we have

\begin{equation*}\lambda (F_{i+1}) \leq \lambda (F_i) - \kappa _1.\end{equation*}

Proof. Let $F \,{:\!=}\, F_i$ be the graph before the operation in line 15 is carried out (that is, before $F_{i+1} \gets F_i \cup R$ ), let $R$ be the copy of $H_1$ merged with $F$ in line 15 and let $F' \,{:\!=}\, F_{i+1}$ be the output from line 15. We aim to show there exists a constant $\kappa _1 = \kappa _1(H_1, H_2) \gt 0$ such that

\begin{equation*} \lambda (F) - \lambda (F') = v(F) - v(F') - \frac {e(F) - e(F')}{m_2(H_1, H_2)} \geq \kappa _1. \end{equation*}

Choose any edge $\hat{e} \in E(F)$ (the edge $\hat{e}$ need not be in the intersection of $R$ and $F$ ). Choose any $F^* \in \mathcal{H}^*(F, \hat{e}, H_1, H_2)$ . Our strategy is to compare our degenerate outcome $F'$ with $F^*$ . As noted earlier, $F^*$ corresponds to a non-degenerate iteration of the while-loop of algorithm Grow (if $\hat{e}$ was the edge chosen by Eligible-Edge). Then Claim 6.2 gives us that $\lambda (F) = \lambda (F^*)$ . Then

\begin{align*} \lambda (F) - \lambda (F') & = \lambda (F^*) - \lambda (F') = v(F^*) - v(F') - \frac{e(F^*) - e(F')}{m_2(H_1, H_2)}. \end{align*}

Hence we aim to show that there exists $\kappa _1 = \kappa _1(H_1, H_2) \gt 0$ such that

(7) \begin{equation} v(F^*) - v(F') - \frac{e(F^*) - e(F')}{m_2(H_1, H_2)} \geq \kappa _1. \end{equation}

Define $R'$ to be the graph with vertex set $V' \,{:\!=}\, V(R) \cap V(F)$ and edge set $E' \,{:\!=}\, E(R) \cap E(F)$ , and let $v' \,{:\!=}\, |V'|$ and $e' \,{:\!=}\, |E'|$ . Observe that $R' \subsetneq R$ . Since $F^*$ corresponds with a non-degenerate iteration of the while-loop of algorithm Grow, $H_2$ is (strictly) $2$ -balanced and $H_1$ is (strictly) balanced with respect to $d_2(\cdot, H_2)$ , we have

(8) \begin{align} v(F^*) - v(F') - \frac{e(F^*) - e(F')}{m_2(H_1, H_2)} & = (e_2 - 1)(v_1 - 2) + (v_2 - 2) - (v_1 - v')\nonumber \\[3pt] & \quad - \frac{(e_2 - 1)e_1 - (e_1 - e')}{m_2(H_1, H_2)}\nonumber \\[3pt] & = (e_2 - 1)(v_1 - 2) + (v_2 - 2) \nonumber \\ & \quad - (e_2 - 1)\left (v_1 - 2 + \frac{1}{m_2(H_2)}\right ) \nonumber \\[3pt] & \quad + \frac{e_1 - e'}{m_2(H_1, H_2)} - (v_1 - v')\nonumber \\[3pt] & = \frac{e_1 - e'}{m_2(H_1, H_2)} - (v_1 - v')\nonumber \\[3pt] & = v' - 2 + \frac{1}{m_2(H_2)} - \frac{e'}{m_2(H_1, H_2)}. \end{align}

Also, since Grow encountered a degeneracy of type 1, we must have $v' \geq 2$ . Hence, if $e' = 0$ and $v' \geq 2$ , then

\begin{equation*}v' - 2 + \frac {1}{m_2(H_2)} - \frac {e'}{m_2(H_1, H_2)} \geq \frac {1}{m_2(H_2)} \gt 0.\end{equation*}

If $e' \geq 1$ , then since $R$ is a copy of $H_1$ , $H_1$ is strictly balanced with respect to $d_2(\cdot, H_2)$ and $R' \subsetneq R$ with $|E(R')| = e' \geq 1$ , we have that $0 \lt d_2(R', H_2) \lt m_2(H_1, H_2)$ , and so

(9) \begin{equation} -\frac{1}{m_2(H_1, H_2)} \gt -\frac{1}{d_2(R', H_2)}. \end{equation}

Then by (8) and (9), we have that

\begin{align*} v(F^*) - v(F') - \frac {e(F^*) - e(F')}{m_2(H_1, H_2)} &= v' - 2 + \frac {1}{m_2(H_2)} - \frac {e'}{m_2(H_1, H_2)}\\ & \gt v' - 2 + \frac {1}{m_2(H_2)} - \frac {e'}{d_2(R', H_2)} = 0, \end{align*}

using the definition of $d_2(R', H_2)$ . Thus (7) holds for

\begin{equation*}\kappa _1 = \min _{R' \subsetneq R} \!\left \{ \frac {1}{m_2(H_2)}, \ v' - 2 + \frac {1}{m_2(H_2)} - \frac {e'}{m_2(H_1, H_2)}\right \}.\end{equation*}

We say that algorithm Grow encounters a degeneracy of type 2 in iteration $i$ of the while-loop if, when we call Extend-L $(F_i, e, G')$ , the graph $L$ found in line 2 overlaps with $F_i$ in more than 2 vertices, or if there exists some edge $e' \in E(L)\setminus E(F_i)$ such that the graph $R_{e^{\prime}}$ found in line 5 overlaps in more than 2 vertices with $F'$ . The following result corresponds to Claim 22 in [Reference Marciniszyn, Skokan, Spöhel and Steger14]. As in the proof of Claim 22 in [Reference Marciniszyn, Skokan, Spöhel and Steger14], we transform $F'$ into the output of a non-degenerate iteration $F^*$ in three steps. However, we swap the order of the latter two steps in our proof. More precisely, we transform $F'$ into a graph $F^2 \in \mathcal{H}(F_i, e, H_1, H_2)$ in the first two steps, then transform $F^2$ into a graph $F^3 \,{:\!=}\, F^* \in \mathcal{H}^*(F_i, e, H_1, H_2)$ . In this last step we require Lemma 6.8.

Claim 6.10. There exists a constant $\kappa _2 = \kappa _2(H_1, H_2) \gt 0$ such that if procedure Grow encounters a degeneracy of type 2 in iteration $i$ of the while-loop, we have

\begin{equation*}\lambda (F_{i+1}) \leq \lambda (F_i) - \kappa _2.\end{equation*}

Proof. Let $F \,{:\!=}\, F_i$ be the graph passed to Extend-L and let $F' \,{:\!=}\, F_{i+1}$ be its output. We aim to show that there exists a constant $\kappa _2 = \kappa _2(H_1, H_2) \gt 0$ such that

(10) \begin{equation} \lambda (F) - \lambda (F') = v(F) - v(F') - \frac{e(F) - e(F')}{m_2(H_1, H_2)} \geq \kappa _2. \end{equation}

Recall that $F'$ would be one of a constant number of graphs if iteration $i$ was non-degenerate. Our strategy is to transform $F'$ into the output of such a non-degenerate iteration $F^*$ in three steps

\begin{equation*}F' \,{=\!:}\, F^0 \overset {(i)}{\to } F^1 \overset {(ii)}{\to } F^2 \overset {(iii)}{\to } F^3 \,{:\!=}\, F^*,\end{equation*}

with each step carefully resolving a different facet of a degeneracy of type 2. By Claim 6.2, we have $\lambda (F) = \lambda (F^*)$ , hence we have that

\begin{align*} \lambda (F) - \lambda (F') & = \lambda (F^*) - \lambda (F') = \sum _{j=1}^3\left (\lambda (F^j) - \lambda (F^{j-1})\right ) \\ & = \sum _{j=1}^3\left (v(F^j) - v(F^{j-1}) - \frac{e(F^j) - e(F^{j-1})}{m_2(H_1, H_2)}\right ). \end{align*}

We shall show that there exists $\kappa _2 = \kappa _2(H_1, H_2) \gt 0$ such that

(11) \begin{equation} \left (v(F^j) - v(F^{j-1}) - \frac{e(F^j) - e(F^{j-1})}{m_2(H_1, H_2)}\right ) \geq \kappa _2, \end{equation}

for each $j \in \{1,2,3\}$ , whenever $F^j$ and $F^{j-1}$ are not isomorphic. In each step we will look at a different structural property of $F'$ that may result from a degeneracy of type 2. We do not know the exact structure of $F'$ , and so, for each $j$ , step $j$ may not necessarily modify $F^{j-1}$ . However, since $F'$ is not isomorphic to $F^*$ , as $F'$ resulted from a degeneracy of type 2, we know that for at least one $j$ that $F^j$ is not isomorphic to $F^{j-1}$ . This will allow us to conclude (10) from (11).

We will now analyse the graph that Extend-L attaches to $F$ when a degeneracy of type 2 occurs. First of all, Extend-L attaches a graph $L \cong H_2$ to $F$ such that $L \in \mathcal{L}^*_{G^{\prime}}$ . Let $x$ be the number of new vertices that are added onto $F$ when $L$ is attached, that is, $x = |V(L)\setminus (V(F) \cap V(L))|$ . Since $L$ overlaps with the edge $e$ determined by Eligible-Edge in line 17 of Grow, we must have that $x \leq v_2 - 2$ . Further, as $L \in \mathcal{L}_{G^{\prime}}^*$ , every edge of $L$ is covered by a copy of $H_1$ . Thus, since the condition in line 14 of Grow came out as false in iteration $i$ , we must have that

(12) \begin{equation} \text{for all}\ u,v \in V(F) \cap V(L),\ \text{if}\ uv \in E(L)\ \text{then}\ uv \in E(F). \end{equation}

(By Claim 6.1, (12) implies that $x \geq 1$ since $F$ must be extended by at least one edge.)

Let $L' \subseteq L$ denote the subgraph of $L$ obtained by removing every edge in $E(F) \cap E(L)$ . Observe that $|V(L')| = |V(L)| = v_2$ and $|E(L')| \geq 1$ (see the remark above). Extend-L attaches to each edge $e' \in E(L')$ a copy $R_{e^{\prime}}$ of $H_1$ in line 6 such that $E(L') \cap E(R_{e^{\prime}}) = \{e'\}$ . As the condition in line 14 of Grow came out as false, each graph $R_{e^{\prime}}$ intersects $F$ in at most one vertex and, hence, zero edges. Let

\begin{equation*}L_R' \,{:\!=}\, L' \cup \bigcup _{e' \in E(L')} R_{e^{\prime}}.\end{equation*}

Then $F'$ is the same as $F \cup L_R'$ , and since every graph $R_{e^{\prime}}$ contains at most one vertex of $F$ , we have that $E(F') = E(F)\ \dot{\cup }\ E(L_R')$ . Therefore,

\begin{equation*}e(F') - e(F) = e(L_R^{\prime}).\end{equation*}

Observe that $|V(F) \cap V(L')| = v_2 - x$ and so

\begin{align*} v(F') - v(F) & = v(L_R^{\prime}) - |V(F) \cap V(L_R^{\prime})| \\[3pt] & = v(L_R^{\prime}) - (v_2 - x) - |V(F) \cap (V(L_R^{\prime})\setminus V(L^{\prime}))|. \end{align*}

Transformation (i): $F^0 \to F^1$ . If $|V(F) \cap (V(L_R^{\prime})\setminus V(L'))| \geq 1$ , then we apply transformation (i), mapping $F^0$ to $F^1$ : For each vertex $v \in V(F) \cap (V(L_R^{\prime})\setminus V(L'))$ , transformation (i) introduces a new vertex $v'$ . Every edge incident to $v$ in $E(F)$ remains connected to $v$ and all those edges incident to $v$ in $E(L_R^{\prime})$ are redirected to $v'$ . In $L_R^{\prime}$ we replace the vertices in $V(F) \cap (V(L_R^{\prime})\setminus V(L'))$ with the new vertices. So now we have $|V(F) \cap (V(L_R^{\prime})\setminus V(L'))| = 0$ . Since $E(F) \cap E(L_R^{\prime}) = \emptyset$ , the output of this transformation is uniquely defined. Moreover, the structure of $L_R^{\prime}$ is completely unchanged. Hence, since $|V(F) \cap (V(L_R^{\prime})\setminus V(L'))| \geq 1$ , and $|E(F')| = |E(F)\ \dot{\cup }\ E(L_R^{\prime})|$ remained the same after transformation (i), we have that

\begin{equation*} v(F^1) - v(F^{0}) - \frac {e(F^1) - e(F^{0})}{m_2(H_1, H_2)} = |V(F) \cap (V(L_R^{\prime})\setminus V(L'))| \geq 1. \end{equation*}

Transformation (ii): $F^1 \to F^2$ . Recall the definition of $\mathcal{H}(F, e, H_1, H_2)$ . If $x \leq v_2 - 3$ , then we apply transformation (ii), mapping $F^1$ to $F^2$ by replacing $L_{R}^{\prime}$ with a graph $L_{R}^{\prime\prime}$ such that $F \cup L_{R}^{\prime\prime} \in \mathcal{H}(F, e, H_1, H_2)$ .

If $x = v_2 - 2$ , observe that already $F \cup L_{R}^{\prime} \in \mathcal{H}(F, e, H_1, H_2)$ and we continue to transformation (iii). So assume $x \leq v_2 - 3$ . Consider the proper subgraph $L_F \,{:\!=}\, L[V(F) \cap V(L)] \subsetneq L$ obtained by removing all $x$ vertices in $V(L) \setminus V(F)$ and their incident edges from $L$ . Observe that $v(L_F) = v_2 - x \geq 3$ and also that $L_F \subseteq F$ by (12). Assign labels to $V(L_F)$ so that $V(L_F) = \{y,z,w_1, \ldots, w_{v_2 - (x+2)}\}$ where $e = \{y,z\}$ and $w_1, \ldots, w_{v_2 - (x+2)}$ are arbitrarily assigned. At the start of transformation (ii), we create $v_2 - (x+2)$ new vertices $w_1^{\prime}, \ldots, w^{\prime}_{v_2 - (x+2)}$ and also new edges such that $\left\{y,z,w_1^{\prime}, \ldots, w^{\prime}_{v_2 - (x+2)}\right\}$ induces a copy $\hat{L}_F$ of $L_F$ , and for all $i,j \in \{1, \ldots, v_2 - (x+2)\}$ , $i \neq j$ ,

\begin{align*} & \text{if} \ w_iw_j \in E(L_F)\ \text{then} \ w^{\prime}_iw^{\prime}_j \in E(\hat{L}_F); \\[3pt] & \text{if} \ w_iy \in E(L_F)\ \text{then} \ w^{\prime}_iy \in E(\hat{L}_F); \\[3pt] & \text{if} \ w_iz \in E(L_F)\ \text{then} \ w^{\prime}_iz \in E(\hat{L}_F); \\[3pt] & \text{and} \ e = yz \in E(\hat{L}_F). \end{align*}

We also transform $L_R^{\prime}$ . For each edge in $E(L_{R}^{\prime})$ incident to a vertex $w_i$ in $L_F$ , redirect the edge to $w^{\prime}_i$ , and remove $w_1, \ldots, w_{v_2 - (x+2)}$ from $V(L_R^{\prime})$ . Hence the structure of $L_R^{\prime}$ remains the same except for the vertices $w_1, \ldots, w_{v_2 - (x+2)}$ that we removed. Define $L'' \,{:\!=}\, L_R^{\prime} \cup \hat{L}_F$ and observe that $V(L'') \cap V(F) = e$ .

Continuing transformation (ii), for each $e' \in E(\hat{L}_{F})\setminus \{e\}$ , attach a copy $R_{e^{\prime}}$ of $H_1$ to $L''$ such that $E(R_{e^{\prime}}) \cap E(L'' \cup F) = \{e'\}$ and $V(R_{e^{\prime}}) \cap V(L'' \cup F) = e'$ . That is, all these new copies $R_{e^{\prime}}$ of $H_1$ are, in some sense, pairwise disjoint. Observe that $E(L'') \cap E(F) = \{e\}$ and define

\begin{equation*}L_R^{\prime\prime} \,{:\!=}\, L^{\prime\prime} \cup \bigcup _{e' \in E(\hat {L}_F)\setminus \{e\}} R_{e^{\prime}}.\end{equation*}

Then $F \cup L_R^{\prime\prime} \in \mathcal{H}(F, e, H_1, H_2)$ . (See Figure 7 for an example of transformation (ii).) Let $F^2 \,{:\!=}\, F \cup L_R^{\prime\prime}$ . Then,

\begin{align*} \ & v(F^2) - v(F^1) - \frac{e(F^2) - e(F^1)}{m_2(H_1, H_2)} \\ = \ & (e(\hat{L}_F) - 1)(v_1 - 2) + v(\hat{L}_F) - 2 - \frac{(e(\hat{L}_F) - 1)e_1}{m_2(H_1, H_2)} \\ = \ & v(\hat{L}_F) - 2 - \frac{(e(\hat{L}_F) - 1)}{m_2(H_2)} \\ = \ & \frac{(v(\hat{L}_F) - 2)\left (m_2(H_2) - \frac{e(\hat{L}_F) - 1}{v(\hat{L}_F) - 2}\right )}{m_2(H_2)} \\ \geq \ & \delta _1 \end{align*}

for some $\delta _1 = \delta _1(H_1, H_2) \gt 0$ , where the second equality follows from $H_1$ being (strictly) balanced with respect to $d_2(\cdot, H_2)$ , the third equality follows from $v(\hat{L}_F) = v(L_F) \geq 3$ and the last inequality follows from $\hat{L}_F$ being a copy of $L_F \subsetneq L \cong H_2$ and $H_2$ being strictly $2$ -balanced.

Figure 7.

An example of transformation (ii) where $H_1 = K_3$ and $H_2 = C_8$ . Observe that edges $aw_1$ and $bw_1$ are replaced by edges $aw^{\prime}_1$ and $bw^{\prime}_1$ .

Transformation (iii): $F^2 \to F^3$ . Recall that for any $J \in \mathcal{H}(F, \hat{e}, H_1, H_2)$ , we define $v^{+}(J) \,{:\!=}\, |V(J)\setminus V(F)| = v(J) - v(F)$ and $e^{+}(J) \,{:\!=}\, |E(J)\setminus E(F)| = e(J) - e(F)$ . Remove the edge $e$ from $E(L'')$ (and $E(L^{\prime\prime}_R)$ ) to give $E(L'') \cap E(F) = \emptyset$ . Then

\begin{equation*}e^+(F \cup L_R^{\prime\prime}) = e(L_R^{\prime\prime}),\end{equation*}

and

\begin{equation*}v^+(F \cup L_R^{\prime\prime}) = v(L_R^{\prime\prime}) - 2.\end{equation*}

If $F^2 = F \cup L_R^{\prime\prime} \in \mathcal{H}^*(F, e, H_1, H_2)$ , then transformation (iii) sets $F^3 \,{:\!=}\, F^2$ . Otherwise we have that $F \cup L_R^{\prime\prime} \in \mathcal{H}(F, e, H_1, H_2)\setminus \mathcal{H}^*(F, e, H_1, H_2)$ . Let $F^3 \,{:\!=}\, J^*$ where $J^*$ is any member of $\mathcal{H}^*(F, e, H_1, H_2)$ and recall that, indeed, $J^*$ is a possible output of a non-degenerate iteration of the while-loop of Grow.

Then, in transformation (iii), we replace $F \cup L_R^{\prime\prime}$ with the graph $J^*$ . Since $H_2$ is (strictly) $2$ -balanced and $H_1$ is (strictly) balanced with respect to $d_2(\cdot, H_2)$ , we have that

(13) \begin{equation} m_2(H_1, H_2) = \frac{e_1}{v_1 - 2 + \frac{1}{m_2(H_2)}} = \frac{e_1(e_2 - 1)}{(v_1 - 2)(e_2 - 1) + v_2 - 2} = \frac{e^{+}(J^*)}{v^{+}(J^*)}. \end{equation}

Using (13) and Lemma 6.8, and that $H_2$ is (strictly) $2$ -balanced and $H_1$ is (strictly) balanced with respect to $d_2(\cdot, H_2)$ , we have that

\begin{align*} \ & v(F^3) - v(F^2) - \frac{e(F^3) - e(F^2)}{m_2(H_1, H_2)} \\ = \ & v(J^*) - v(F \cup L_R^{\prime\prime}) - \frac{e(J^*) - e(F \cup L_R^{\prime\prime})}{m_2(H_1, H_2)} \\ = \ & v^+(J^*) - v^+(F \cup L_R^{\prime\prime}) - \frac{e^+(J^*) - e^+(F \cup L_R^{\prime\prime})}{m_2(H_1, H_2)} \\ \overset{L.6.8}{\gt } \ & v^+(J^*) - v^+(F \cup L_R^{\prime\prime}) - \frac{e^+(J^*) - e^+(J^*)\left (\frac{v^+(F \cup L_R^{\prime\prime})}{v^+(J^*)}\right )}{m_2(H_1, H_2)} \\ = \ & \left (v^+(J^*) - v^+(F \cup L_R^{\prime\prime})\right ) \left (1 - \frac{e^+(J^*)}{v^+(J^*)m_2(H_1, H_2)}\right ) \\ = \ & 0. \end{align*}

Since $v^+(J^*)$ , $v^+(F \cup L_R^{\prime\prime})$ , $e^+(J^*)$ , $e^+(F \cup L_R^{\prime\prime})$ and $m_2(H_1, H_2)$ only rely on $H_1$ and $H_2$ , there exists $\delta _2 = \delta _2(H_1, H_2) \gt 0$ such that

\begin{equation*}v(F^3) - v(F^2) - \frac {e(F^3) - e(F^2)}{m_2(H_1, H_2)} \geq \delta _2.\end{equation*}

Taking

\begin{equation*}\kappa _2 \,{:\!=}\, \min \{1, \delta _1, \delta _2\},\end{equation*}

we see that (11) holds.

As stated earlier, Claim 6.3 follows from Claims 6.9 and 6.10. All that remains to prove Case 1 of Lemma 5.5 is to prove Lemma 6.8.