Proof of Theorem 1.4.Let $\beta _{2.3}, \gamma _{2.3}\gt 0$ and $C_{2.3}$ be given by Theorem 2.3 on input $\alpha _0=1/256$ . Then let $\gamma _{2.2}\gt 0$ and $C_{2.2}$ be given by Theorem 2.2 on input $\beta = \min \{ \alpha _0/4, \beta _{2.3} \}$ . Moreover, let $C_{2.4}$ be given by Theorem 2.4. Define $C = \max \{ C_{2.3},C_{2.4} \}$ and $\gamma =\min \{ \gamma _{2.3},\gamma _{2.2} \}$ .

Let $G$ be any $n$ -vertex graph and $p \ge C \log n/n$ , and define $m= \min \{ \delta (G), \lfloor n/3 \rfloor \}$ . If $m \le n/256$ , then we get from Theorem 2.4 that a.a.s. $G \cup G(n,p)$ contains at least $m$ pairwise vertex-disjoint triangles, as $C \ge C_{2.4}$ . Otherwise, $m \gt n/256$ and we can choose $\alpha \in (\alpha _0,1/3]$ such that $(\alpha -\gamma )n \le m \le \alpha n$ . If $G$ is $(\alpha,\beta )$ -stable, then $G$ is also $(\alpha,\beta _{2.3})$ -stable and, by Theorem 2.3, there are a.a.s. at least $\min \{ \delta (G),\lfloor \alpha n\rfloor \} \ge m$ pairwise vertex-disjoint triangles in $G \cup G(n,p)$ , as $\alpha _0 \lt \alpha \le 1/3$ and $C \ge C_{2.3}$ . Otherwise, $G$ is not $(\alpha,\beta )$ -stable and, by Theorem 2.2, a.a.s. $G \cup G(n,p)$ contains at least $ \min \{ \alpha n, \lfloor n/3 \rfloor \} \ge m$ pairwise vertex-disjoint triangles, as $p = \omega (1/n)$ .

Proof of Theorem 2.3.Let $0 \lt \alpha _0 \le 1/3$ and choose $d=1/2$ . Let $\varepsilon \gt 0$ and $C_{4.1}\gt 0$ be given by Lemma 4.1 on input $d$ . We can assume $\varepsilon \lt 1/5$ and then choose $0\lt \beta \lt \alpha _0\, \varepsilon/36$ and $0\lt \gamma \lt \beta/11$ . With $C_{2.4}$ given by Theorem 2.4, let $C \ge 2 C_{2.4} + 1 + 2 C_{4.1}/\alpha _0$ . Finally, let $\alpha _0 \le \alpha \le 1/3$ .

Given $n$ , let $p\ge C \log n/n$ . With our choice of $C$ , we can reveal $G(n,p)$ in three rounds $G_1 \sim G(n,2 C_{2.4} \log n/n)$ , $G_2 \sim G(n,\log n/n)$ , and $G_3 \sim G(n,\tfrac{2 C_{4.1}}{\alpha _0} \log n/n)$ . We will only know later in which subset we will use $G_1$ and $G_3$ , but we have that a.a.s. there is an edge of $G_2$ between any two not necessarily disjoint sets of size $\beta n$ . Indeed, fixed two such sets, the probability that there is no edge of $G_2$ is at most $(1-\log n/n)^{(\beta n)^2} \le \exp (-\beta ^2 n \log n)$ , and we conclude by an union bound over the at most $2^{2n}$ choices for the two sets. Now let $G$ be an $n$ -vertex graph with minimum degree $\delta (G) \ge \left ( \alpha -\gamma \right ) n$ that is $(\alpha,\beta )$ -stable and define $m_0= \max \{ n/3 -\delta (G), n/3-\lfloor \alpha n \rfloor \}$ . Our goal is to a.a.s. find pairwise vertex-disjoint triangles in $G \cup G(n,p)$ such that at most $3 m_0$ vertices are left uncovered.

To aid with calculations we let $\kappa = n/3 - \lceil \alpha n \rceil$ and observe that $\kappa \in \{ 0,-1/3,-2/3 \}$ if $\alpha =1/3$ and that $\kappa \gt 0$ if $\alpha \lt 1/3$ and $n$ large enough. Also note that $m_0-\kappa$ is an integer and that $3 \kappa = \lfloor (1-\alpha ) n \rfloor - 2 \lceil \alpha n \rceil$ . With this we set $w = \max \{ 3 \kappa, 0\}$ . As $G$ is $(\alpha, \beta )$ -stable we get a partition of $V(G)$ into sets $A$ and $B$ satisfying the conditions of Definition 2.1.

The sets $A_1$ and $B_1$ partition $V(G) \setminus (V(\mathcal{T}_1) \cup W)$ and, after proving Claim 5.1, we will cover all but $w$ vertices from $A_1 \cup B_1$ with additional triangles. Hence, if we manage to find these triangles, we have covered all but $|W|+w \le 3 m_0$ vertices, as desired. We remark for later that $|W| \le 3m_0-w \le 4 \gamma n$ .

Proof of Claim 5.1.We have either $|B| \gt \lfloor (1-\alpha ) n \rfloor$ or $|A| \ge \lceil \alpha n \rceil$ . First suppose that we are in the first case, where $|B|=\lfloor (1-\alpha )n \rfloor +m$ for some $1 \le m \le \beta n$ (and $|A|=\lceil \alpha n \rceil - m$ ), and note that

\begin{equation*} |B|-2|A|= n-3 \lceil \alpha n\rceil +3m =3m+3\kappa \gt 0 . \end{equation*}

If $1 \le m \le m_0 - \kappa$ , then $0 \lt 3m \le 3 m_0 - 3 \kappa$ and we let $W$ be any set with $\min \{3m,3m+3 \kappa \} \le 3m_0-w$ vertices from $B$ . Then with the choice of $\mathcal{T}_1=\emptyset$ , we have that the sets $A_1=A$ and $B_1=B \setminus W$ partition $V(G) \setminus W$ , and $|A_1|=\lceil \alpha n \rceil -m$ and $|B_1|=|B|-\min \{ 3m,3m+3 \kappa \}=2|A_1|+ w$ .

If on the other hand, $m_0 \lt m + \kappa$ , then

\begin{equation*} \delta (G[B]) \ge \delta (G) - |A| \ge (n/3-m_0)-(\lceil \alpha n \rceil -m) = m-m_0+\kappa \gt 0 \end{equation*}

and we observe that $m-m_0+\kappa$ is an integer. Moreover, $m-m_0+\kappa \le m \le |B|/256$ , where we use $m_0-\kappa \ge 0$ , $m \le \beta n \le \alpha _0 \varepsilon n/36 \le n/(3 \cdot 5 \cdot 36)$ and $|B| = \lfloor (1-\alpha )n \rfloor + m \ge n/2 +m$ . Thus, by Theorem 2.4 and as $2 C_{2.4} \log n/n \ge C_{2.4} \log |B|/ |B|$ , we a.a.s. find $m-m_0+\kappa$ pairwise vertex-disjoint triangles in $(G \cup G_1)[B]$ . Denote by $\mathcal{T}_1$ the collection of these $m-m_0+\kappa$ triangles. Let $W$ be any set of $3m_0-w$ vertices from $B$ not covered by any triangle in $\mathcal{T}_1$ . Then the sets $A_1=A$ and $B_1=B \setminus (V(\mathcal{T}_1) \cup W)$ partition $V(G) \setminus (V(\mathcal{T}_1) \cup W)$ , and we have $|A_1|=\lceil \alpha n \rceil -m$ and

\begin{equation*} |B_1|=|B|-3(m-m_0+\kappa )-(3 m_0 - w)=2|A_1| + w. \end{equation*}

It remains to consider the second case, where $|A|= \lceil \alpha n \rceil +m$ for some $0 \le m \le \beta n$ . First, we greedily pick $m$ pairwise vertex-disjoint triangles in $G \cup G_2$ each with two vertices in $A$ and one vertex in $B$ . Indeed during the process, there is always a vertex $v$ in $B$ , not yet contained in a triangle, with at least $\deg (v,A)-2m \ge (\alpha/4-2\beta )n \ge \beta n$ uncovered neighbours in $A$ in the graph $G$ . By the property assumed in $G_2$ , we can then find an edge within these neighbours of $v$ to get a triangle. Denote by $\mathcal{T}_1$ the collection of these $m$ triangles.

If $\kappa \ge 0$ , then, with the choice of $W=\emptyset$ , we have that $A_1=A \setminus V(\mathcal{T}_1)$ and $B_1=B \setminus V(\mathcal{T}_1)$ partition $V(G) \setminus V(\mathcal{T}_1)$ and

\begin{equation*} |B_1|=|B|-|\mathcal {T}_1|= \lfloor (1-\alpha )n \rfloor - 2m = 2(\lceil \alpha n\rceil -m)+ 3 \kappa =2|A_1|+w. \end{equation*}

If $\kappa \lt 0$ , we additionally pick a set $W$ of vertices not covered by triangles from $\mathcal{T}_1$ , such that $|W|=1$ , $|W \cap A|=1$ , $|W \cap B|=0$ if $\kappa =-2/3$ , and $|W|=2$ , $|W \cap A|=|W\cap B|=1$ if $\kappa =-1/3$ . Then, the sets $A_1=A \setminus (V(\mathcal{T}_1) \cup W)$ and $B_1=B \setminus (V(\mathcal{T}_1) \cup W)$ partition $V(G) \setminus (V(\mathcal{T}_1) \cup W)$ , and $|B_1|=2|A_1|+w$ . Indeed, if $\kappa =-2/3$ , we have $|A_1|=|A|-2|\mathcal{T}_1|-1=\lceil \alpha n\rceil - m-1$ and

\begin{equation*} |B_1|=|B|-|\mathcal {T}_1|= \lfloor (1-\alpha )n \rfloor - 2m = 2 (\lceil \alpha n\rceil -m) + 3 \kappa = 2|A_1| \end{equation*}

and if $\kappa =-1/3$ we have $|A_1|=|A|-2|\mathcal{T}_1|-1=\lceil \alpha n\rceil - m-1$ and

\begin{equation*} |B_1|=|B|-|\mathcal {T}_1|-1= \lfloor (1-\alpha )n \rfloor - 2m -1 = 2 (\lceil \alpha n\rceil -m) + 3 \kappa -1= 2|A_1|\, . \end{equation*}

Observe, that in both the first and the second case $|B_1|=2|A_1|+w$ and $|W| \le 3m_0-w$ . Moreover, as we remove at most $3m_0-w \le 4 \gamma n \le \alpha n/20$ vertices from each $A$ and $B$ , the minimum degree between $A_1$ and $B_1$ is at least $\alpha n/4 - \alpha n/20 = \alpha n/5$ . The other conditions on the degrees between $A_1$ and $B_1$ are clearly satisfied, because for all but at most $\beta n$ vertices from each set there are still at most $\beta n$ non-neighbours in the other set. The bounds $|\mathcal{T}| \le \beta n$ and $|A_1| \le \lceil \alpha n \rceil$ also hold in all cases.

We want to cover all but $w$ vertices in $A_1 \cup B_1$ and we start from those vertices in $A_1$ and $B_1$ that do not have a high degree to the other part. We will always cover them with triangles with one vertex in $A_1$ and two vertices in $B_1$ to ensure that the relation between the number of vertices remaining in $A_1$ and $B_1$ does not change. Let

\begin{equation*} \tilde {A_1} = \{v \in A_1\;:\; \deg (v,B_1) \le |B_1|-9\beta n \} \text { and } \tilde {B_1} = \{v \in B_1\;:\; \deg (v,A_1) \le |A_1|-9\beta n \}, \end{equation*}

and observe that $|\tilde{A_1}|,|\tilde{B_1}| \le \beta n$ .

We claim that a.a.s. we can greedily pick pairwise vertex-disjoint triangles in $(G \cup G_2)[A_1 \cup B_1]$ that cover all vertices of $\tilde{A_1}$ , with each triangle having one vertex in $\tilde{A_1}$ and two vertices in $B_1 \setminus \tilde{B_1}$ . Indeed, at each step during the process, an uncovered vertex $v$ in $\tilde{A_1}$ has at least $\deg (v,B_1)- |\tilde{B_1}| - 2 |\tilde{A_1}| \ge (\alpha/5-3 \beta )n \ge \beta n$ uncovered neighbours in $B_1 \setminus \tilde{B_1}$ in the graph $G$ . We then find an edge of $G_2$ within these neighbours of $v$ and build a triangle. Denote by $\mathcal{T}_2$ the collection of these triangles and note that $|\mathcal{T}_2| \le \beta n$ .

Observe that at this point $2 |\mathcal{T}_2| \le 2\beta n$ vertices of $B_1 \setminus \tilde{B_1}$ have already been covered. We claim that a.a.s. we can greedly pick pairwise vertex-disjoint triangles in $(G \cup G_2)[(A_1 \cup B_1) \setminus V(\mathcal{T}_2)]$ that cover all vertices of $\tilde{B_1}$ , where each triangle has one vertex in $A_1 \setminus \tilde{A_1}$ , one vertex in $\tilde{B_1}$ , and one vertex in $B_1 \setminus \tilde{B_1}$ . Indeed, at each step during the process, an uncovered vertex $v$ in $\tilde{B_1}$ has at least $\deg (v,A_1) - |\tilde{A_1}| - |\tilde{B_1}| \ge (\alpha/5-2 \beta )n \ge \beta n$ uncovered neighbours in $A_1 \setminus \tilde{A_1}$ in the graph $G$ and at least

\begin{align*} \delta (G)& - 3|\mathcal{T}_1| - |W| - \deg (v,A_1) - 2 |\mathcal{T}_2| - 2|\tilde{B_1}| \\ &\ge (\alpha - \gamma )n - 3\beta n - 4 \gamma n - (\lceil \alpha n \rceil -9 \beta n) - 4 \beta n \ge \beta n \end{align*}

uncovered neighbours in $B_1 \setminus \tilde{B_1}$ in the graph $G$ . We then find an edge of $G_2$ between these two neighbourhood sets to get a triangle. Denote by $\mathcal{T}_3$ the collection of these triangles and note that $|\mathcal{T}_3| \le \beta n$ .

The sets $A_2=A_1 \setminus V(\mathcal{T}_2 \cup \mathcal{T}_3)$ and $B_2=B_1 \setminus V(\mathcal{T}_2 \cup \mathcal{T}_3)$ give a partition of the remaining vertices in $V(G) \setminus (V(\mathcal{T}_1) \cup V(\mathcal{T}_2) \cup V(\mathcal{T}_3) \cup W)$ . We have

\begin{equation*}|A_2|\ge |A| - 2|\mathcal {T}_1| - |\mathcal {T}_2| - |\mathcal {T}_3| - |W| \ge \alpha n - 5 \beta n - 4 \gamma n \ge \alpha n/2\, \end{equation*}

and $|B_2|=2|A_2| + w$ . Moreover, the degree from $A_2$ to $B_2$ is at least $|B_1|-9 \beta n - 2|\mathcal{T}_2 \cup \mathcal{T}_3| = |B_2| - 9 \beta n$ and the degree from $B_2$ to $A_2$ is at least $|A_1|- 9 \beta n - |\mathcal{T}_2 \cup \mathcal{T}_3| = |A_2| - 9 \beta n$ . We partition $B_2$ arbitrarily into three subsets $B'_{\!\!1}$ , ${B''}_{\!\!\!2}$ , and $W'$ of size $|B'_{\!\!1}|=|{B''}_{\!\!\!2}|=|A_2|$ and $|W'|=|B_2|-2|A_2|=w$ . Then the degree from $A_2$ to $B'_{\!\!1}$ and the degree from $A_2$ to ${B''}_{\!\!\!2}$ are at least $|B'_{\!\!1}| - 9 \beta n$ .

We claim that the pair $(A_2,B'_{\!\!1})$ is $(\varepsilon,1/2)$ -super-regular, with $\varepsilon$ being chosen as stated at the beginning of the proof. Indeed for all $X \subseteq A_2$ and $Y \subseteq B'_{\!\!1}$ with $|X| \ge \varepsilon |A_2|$ and $|Y| \ge \varepsilon |B'_{\!\!1}|$ , we have

\begin{equation*}e(X,Y) \ge |X| (|Y| - 9 \beta n) \ge \frac {1}{2} |X| |Y| \end{equation*}

and $\deg (a,B'_{\!\!1}) \ge |B'_{\!\!1}| - 9 \beta n \ge |B'_{\!\!1}|/2$ for all $a \in A_2$ , and $\deg (b,A_2) \ge |A_2| - 9 \beta n \ge |A_2|/2$ for all $b \in B'_{\!\!1}$ , where for all inequalities we use $\beta \le \alpha/36$ . For the same reason, the pair $(A_2,{B''}_{\!\!\!2})$ is $(\varepsilon,1/2)$ -super-regular as well.

Now, as $\tfrac{2C_{4.1}}{\alpha _0} \log n/n \ge C_{4.1} \log |A_2|/|A_2|$ , we can apply Lemma 4.1 to $A_2$ , $B'_{\!\!1}$ and ${B''}_{\!\!\!2}$ , with $d=1/2$ and $\varepsilon$ , and a.a.s. get a triangle factor $\mathcal{T}_4$ in $(G \cup G_3) [V']$ , where $V'=V(G) \setminus (V(\mathcal{T}_1) \cup V(\mathcal{T}_2) \cup V(\mathcal{T}_3) \cup W \cup W')$ . Then $\mathcal{T}_1 \cup \mathcal{T}_2 \cup \mathcal{T}_3 \cup \mathcal{T}_4$ contains at least

\begin{equation*}(n - |W| - |W'|)/3 \ge n/3- m_0 \ge \min \{ \delta (G), \lfloor \alpha n \rfloor \}\end{equation*}

pairwise vertex-disjoint triangles covering $V(G) \setminus (W \cup W')$ .

Proof of Theorem 2.2.We start by defining necessary constants. Given $0\lt \beta \lt 1/12$ , let $d\gt 0$ be obtained from Lemma 4.4, and set $\gamma =d/2$ and $t_0=11/d$ . Next, we take any $0\lt \delta ' \lt 160^{-2} d^2$ and use Lemma 4.2 on input $d/2$ and $\delta '$ to obtain $\delta _0, \delta, \varepsilon '$ with $\delta ' \ge \delta _0\gt \delta \gt \varepsilon '\gt 0$ and $C_1$ . Additionally, we assume that $C_1$ is large enough and $\varepsilon '$ is small enough for Lemma 4.3 to hold with input $d/2$ . Finally, let $C_2$ be given by Lemma 3.2 on input $d/2$ . We let $0\lt \varepsilon \le \varepsilon '/2$ . In summary, the dependencies between our constants are as follows:

\begin{equation*} \varepsilon \ll \varepsilon ' \lt \delta \lt \delta _0 \le \delta ' \ll d \ll \beta \lt \frac {1}{12} \quad \text {and} \quad \frac {1}{t_0}, \gamma \ll d. \end{equation*}

We apply Lemma 3.5 with $\varepsilon$ and $t_0$ to obtain $T$ . We take $C$ large enough such that, for $p \ge C/n$ , the random graph $G(n,p)$ contains the union $G_1 \cup G_2 \cup G_3$ , where $G_1 \sim G(n,2C_1T/n)$ , $G_2 \sim G(n,4C_2T/(dn))$ , and $G_3 \sim G(n,96 T^2/(d^2n))$ .

Now, for any $\alpha$ with $4 \beta \le \alpha \le 1/3$ , let $G$ be an $n$ -vertex graph on the vertex set $V$ with minimum degree $\delta (G) \ge (\alpha -\gamma )n$ that is not $(\alpha,\beta )$ -stable. With the regularity lemma (Lemma 3.5) applied to $G$ , we get $G'$ , $t_0 \lt t+1 \le T$ and a partition $V_0,\dots,V_t$ of $V(G)$ such that (P1)–(P4) hold. Define $n_0=|V_1|=|V_2|=\dots =|V_t|$ and observe that $(1-\varepsilon )n/t \le n_0 \le n/t$ . We denote by $R$ the $(\varepsilon,d)$ -reduced graph for $G$ , that is, the graph on the vertex set $[t]$ with edges $ij$ corresponding to $\varepsilon$ -regular pairs $(V_i,V_j)$ of density at least $d$ in $G'$ . We observe that the minimum degree of $R$ satisfies $\delta (R) \ge (\alpha -2d) t$ because, otherwise, thre would be vertices with degree at most $(\alpha -2d)t(n/t)+\varepsilon n \lt (\alpha - \gamma )n - (d+\varepsilon )n$ in $G'$ , contradicting (P2).

The purpose of $G_1$ will become clear later, but we describe some useful properties of $G_2$ and $G_3$ now. Let $U$ and $W$ be any two clusters that give an edge in $R$ , $V$ any cluster, and $U' \subseteq U$ , $W'\subseteq W$ , $V'\subseteq V$ three pairwise disjoint subsets each of size $dn_0/2$ . Then, with $G_2$ and as $4C_2T/(dn) \ge 2C_2/(dn_0)$ , by Lemma 3.2 we have that with probability at least $1-2^{-4(dn_0/2)/(d/2)}=1-2^{-4n_0}$

(1) \begin{align} \text{there is a triangle in $G \cup G_2$ with one vertex in each set $U'$, $W'$, $V'$.} \end{align}

With a union bound over the at most $t^32^{3n_0}$ choices for $U$ , $W$ , $V$ and $U'$ , $W'$ , $V'$ , we conclude that a.a.s. (1) holds for all choices as above.

With $G_3$ we a.a.s. have that

(2) \begin{align} \text{ any set $A$ of size at least $dn_0/2$ contains an edge of $G_3$.} \end{align}

In fact, given any set $A$ of size at least $dn_0/2$ , the expected number of edges of $G_3$ in $A$ is

\begin{equation*} \binom {|A|}{2} \cdot \frac {96 T^2}{d^2 n} \ge \frac {1}{3} \cdot \frac {d^2 n_0^2}{4} \cdot \frac {96 T^2}{d^2 n} =8T^2 \frac {n_0^2}{n} \ge 2n\,, \end{equation*}

where we used that $n/n_0 \le t/(1-\varepsilon ) \le 2T$ . Therefore, the probability that the set $A$ does not contain an edge of $G_3$ is at most $(1-\frac{96 T^2}{d^2 n})^{\binom{|A|}{2}} \le \exp \left (-\binom{|A|}{2} \cdot \frac{96 T^2}{d^2 n}\right ) \le \exp (-2n)$ and (2) follows from a union bound over the at most $2^n$ choices for $A$ .

Now let $M_1$ be a largest matching in $R$ . Since $G$ is not $(\alpha,\beta )$ -stable, using Lemma 4.4, we conclude that $|M_1| \ge (\alpha + 2d)t$ . At this point, for the sake of clarity, we split our proof into two cases – $0 \lt \alpha \lt 1/3 - d/3$ and $1/3 - d/3 \le \alpha \le 1/3$ – although some steps will be the same. The first case is indeed much easier, as we do not need to cover all the graph with triangles, while in the second case we are looking for a spanning structure and we want to find $\lfloor n/3 \rfloor$ pairwise vertex-disjoint triangles.

Case $0 \lt \alpha \lt 1/3 - d/3$

As $M_1$ is a largest matching in $R$ , the set $V(R) \setminus V(M_1)$ is independent and only one endpoint of each edge of $M_1$ can be adjacent to more than one vertex from $V(R) \setminus V(M_1)$ . Therefore, we can greedily pick a second matching $M_2$ such that each edge of $M_2$ contains a vertex of $V(R) \setminus V(M_1)$ and a vertex of $V(M_1)$ , and $M_2$ covers at least $\min \{|V(R) \setminus V(M_1)|,\delta (R) \}$ vertices of $V(R) \setminus V(M_1)$ . The two matchings $M_1$ and $M_2$ together cover a subset $V(M_1 \cup M_2) \subseteq V(R)$ of

\begin{equation*} 2|M_1| + \min \{|V(R) \setminus V(M_1)|,\delta (R) \} \ge \min \{t,(3\alpha +2d)t\} \ge (3 \alpha + d)t \end{equation*}

vertices, and we can extract a collection of $|M_2|$ vertex-disjoint cherries and a disjoint matching that cover such vertices. This gives a subgraph $R'\subseteq R$ consisting of cherries and a matching such that for all edges $ij \in E(R')$ the pair $(V_i,V_j)$ is $\varepsilon$ -regular of density at least $d$ in $G'$ , and therefore in $G$ as well. We denote by $\mathcal{J} \subseteq [t]$ the indices of the clusters $V_i$ of the cherries and the matching edges in $R'$ and we observe from above that $|\mathcal{J}| \ge (3 \alpha + d)t$ . We add to $V_0$ all the vertices of $G$ that are in the clusters $V_j$ for $j \notin \mathcal{J}$ .

Then we make all pairs associated with the edges of $R'$ super-regular. Given a pair $(A,B)$ , by Lemma 3.3, all but at most $\varepsilon n_0$ vertices of $A$ (resp. $B$ ) have degree at least $(d-\varepsilon )n_0$ to $B$ (resp. $A$ ). For every such pair we remove these vertices from $A$ and $B$ , and remove additional vertices to ensure all clusters have the same size. As $R'$ only contains vertex-disjoint cherries and a disjoint matching, we can achieve that by removing a total of at most $2 \varepsilon n_0$ vertices from each cluster. We add all the removed vertices to $V_0$ . Observe that afterwards all the pairs $(A,B)$ associated with the edges of $R'$ are $(2\varepsilon,d-3\varepsilon )$ -super-regular, because every vertex $a \in A$ has degree at least $(d-\varepsilon )n_0 - 2\varepsilon n_0 \ge (d-3\varepsilon )|B|$ into $B$ , and every vertex $b \in B$ has degree at least $(d-3\varepsilon )|A|$ into $A$ .

Recall that for a later application of Lemma 4.2, we need that for each cherry the sizes of the leaf-clusters are smaller than the size of the centre-cluster. Thus, for each cherry $ijk$ of $R'$ , with $j$ being the centre, we additionally remove $\delta |V_j| \le \delta n_0$ vertices from the leaves $V_i$ and $V_k$ , and add them to $V_0$ . We have $|V_i|=|V_k|=(1-\delta )|V_j|$ that implies $|V_i|=|V_k| \ge (1-\delta _0)|V_j|$ , as $\delta _0 \gt \delta$ . We have that all edges of $R'$ still give $(2\varepsilon,d-3\varepsilon -\delta )$ -super-regular pairs. Moreover,

\begin{equation*} \left | \bigcup _{j \in \mathcal {J}} V_j \right | \ge (1-2\varepsilon -\delta ) n_0 |\mathcal {J}| \ge (1-\varepsilon )(1-2\varepsilon -\delta )(3\alpha +d) n \ge 3 \alpha n\, . \end{equation*}

We can assume (by moving only a few additional vertices to $V_0$ that do not harm the bounds above) that for all cherries and matching edges in $R'$ the number of vertices in the clusters together is divisible by three.

For each such super-regular cherry $ijk$ of $R'$ , after revealing $G_1[V_i \cup V_j \cup V_k]$ we find by Lemma 4.2 a.a.s. a triangle factor covering all the vertices in $V_i \cup V_j \cup V_k$ . Similarly for any matching edge $ij$ of $R'$ , after revealing $G_1[V_i \cup V_j]$ we find by Lemma 4.3 a.a.s. a triangle factor covering all the vertices in $V_i \cup V_j$ . Note that we apply Lemmas 4.2 and 4.3 only constantly many times and thus a.a.s. we get a triangle factor in all such applications. Let $\mathcal{T}$ be the union of all such triangle factors. Then $\mathcal{T}$ covers $\left |\bigcup _{j \in \mathcal{J}} V_j \right | \ge 3 \alpha n$ vertices and gives at least $\alpha n=\min \{\alpha n, \lfloor n/3 \rfloor \}$ pairwise vertex-disjoint triangles in $G \cup G(n,p)$ .

Case $1/3 - d/3 \le \alpha \le 1/3$

As discussed in the overview, here we cannot directly apply Lemmas 4.2 and 4.3 as in the case $0 \lt \alpha \lt 1/3-d/3$ , but we need additional steps. However, even with a lower minimum degree, we will cover all vertices of $G$ and find $\lfloor n/3 \rfloor$ pairwise vertex-disjoint triangles. Recall that $M_1$ is a largest matching and that $|M_1| \ge (\alpha +2d)t$ . Then the set $V(R) \setminus V(M_1)$ is independent, has size

\begin{equation*} |V(R) \setminus V(M_1)| = t - 2|M_1| \le (1-2\alpha -4d)t \le (\alpha -3d)t \end{equation*}

and only one endpoint of each edge of $M_1$ can be adjacent to more than one vertex from $V(R) \setminus V(M_1)$ . Given that $\delta (R) \ge (\alpha -2d)t$ , we can greedily pick a second matching $M_2$ such that each edge of $M_2$ contains a vertex of $V(R) \setminus V(M_1)$ and a vertex of $V(M_1)$ , and $M_2$ covers the remaining vertices $V(R) \setminus V(M_1)$ completely. Therefore, the two matchings $M_1$ and $M_2$ together cover the vertex set $V(R)$ and we can extract a collection of $\ell = |M_2| \le (\alpha -3d)t$ vertex-disjoint cherries and a disjoint matching that cover $V(R)$ .

This gives a spanning subgraph $R'\subseteq R$ on vertex set $[t]$ containing $\ell \le (\alpha -3d)t$ cherries and a matching of size $(t-3\ell )/2 \ge (1-3 \alpha + 9d)t/2 \ge 9dt/2$ such that for all edges $ij \in E(R')$ the pair $(V_i,V_j)$ is $\varepsilon$ -regular of density at least $d$ in $G'$ , and therefore in $G$ as well. We denote by $\mathcal{I} \subseteq [t]$ the indices of the clusters $V_i$ that are not the centre of a cherry in $R'$ . As above, with Lemma 3.3, we can make the pairs associated with the edges of $R'$ $(2\varepsilon,d-3\varepsilon )$ -super-regular, while keeping the clusters all of the same size. For this we have to remove at most $t 2 \varepsilon n_0 \le t 2 \varepsilon n/t = 2 \varepsilon n$ vertices, which we add to $V_0$ . Next, as for a later application of Lemma 4.2 we need that for each cherry the sizes of the leaf-clusters are smaller than the size of the centre-cluster, we remove for each $i \in \mathcal{I}$ additionally $\delta |V_i| \le \delta n_0 \le \delta n/t$ vertices from $V_i$ and add them to $V_0$ . Note that we remove vertices from the clusters of matching edges as well, although this is not necessary. We then get $|V_0| \le \varepsilon n + 2 \varepsilon n + t \delta n/t \le 2 \delta n$ . We can assume (by moving only a few additional vertices to $V_0$ that do not harm the bounds above) that for all cherries and matching edges in $R'$ the number of vertices in the clusters together is divisible by three. By removing $n \pmod 3 \in \{0,1,2\}$ vertices from $V_0$ , we also have $|V_0| \equiv 0 \pmod 3$ ; note that this only happens when $n$ is not divisible by $3$ and we can discard these vertices.

Covering $V_0$ with triangles. We now want to cover the exceptional vertices in $V_0$ by triangles. It would be easy to do this greedily by just using (2), but it might happen that afterwards in many of the cherries the number of vertices is not divisible by three or that the centre cluster gets too small. To avoid both these issues, we will cover $V_0$ while using the same number of vertices from clusters that are together in a cherry or matching edge. For this we will always cover three vertices at a time and combine (1) with (2) to find additional triangles. Observe that $|V_0 \cup \bigcup _{i \not \in \mathcal{I} } V_i| \le 2 \delta n + \ell n/t \le (\alpha -\gamma )n - 2dn$ and, therefore, any $v \in V_0$ has at least $2 dn$ neighbours in $\bigcup _{i \in \mathcal{I}} V_i$ .

Assume we have already covered $V' \subseteq V_0$ vertices of $V_0$ using at most $5|V'|$ triangles in total. Let $W'$ be the set of vertices from $\bigcup _{i \in \mathcal{I}} V_i$ used for the triangles covering $V'$ and note that $|W'| \le 30 \delta n$ . Then let $\mathcal{I}' \subseteq \mathcal{I}$ be the set of indices of clusters $V_i$ with $i \in \mathcal{I}$ which intersect $W'$ in at least $\sqrt{\delta }n_0$ vertices and note that $|\mathcal{I}'| \le |W'|/(\sqrt{\delta }n_0) \le 30 \sqrt{\delta } t/ (1-\varepsilon ) \le 40 \sqrt{\delta } t \le dt/4$ . Moreover, notice that as for each $v \in V_0$ we have $\deg _G(v,\bigcup _{i \in \mathcal{I}} V_i) \ge 2dn$ , there are at least $dt$ indices $i \in \mathcal{I}$ such that $v$ has at least $dn_0$ neighbours in $V_i$ . In particular, as $|\mathcal{I}'| \le dt/4$ and $t \ge 10/d$ , there are at least $dt-|\mathcal{I}'| \ge 3dt/4 \ge 7$ indices $i \in \mathcal{I} \setminus \mathcal{I}'$ such that $v$ has at least $dn_0$ neighbours in $V_i$ . Therefore, we can pick three vertices $v_1,v_2,v_3 \in V_0 \setminus V'$ and three indices $i_1,i_2,i_3$ in $\mathcal{I} \setminus \mathcal{I}'$ such that $v_j$ has $dn_0$ neighbours in $V_{i_j}$ for $j=1,2,3$ and the clusters $V_{i_1},V_{i_2},V_{i_3}$ belong to pairwise different cherries or matching edges. For $j=1,2,3$ with (2), we find an edge $e_j$ in $G_3[N(v_j,V_{i_j}) \setminus W']$ and we cover the three vertices with triangles. It is easy to show that we can find at most $10$ additional triangles with the help of (1) and (2), in such a way that, overall, for each cherry and matching edge, we use the same number of vertices from each of their clusters; in particular, the number of vertices used from each cherry and matching edge is divisible by three. The clusters $V_{i_1},V_{i_2},V_{i_3}$ can belong to three cherries, two cherries and one matching edge, one cherry and two matching edges, or three matching edges. We give details in the case where they are all leaves of (different) cherries, and we refer to Figure 1 for the other three cases. With (1) we find four triangles: two with a vertex in each of the other cluster of the cherry containing $V_{i_1}$ and the third vertex in one of the other clusters of the cherry containing $V_{i_2}$ , and other two triangles with one vertex in each of the other cluster of the cherry containing $V_{i_3}$ and the third vertex in the remaining cluster of the cherry containing $V_{i_2}$ . When a $V_{i_j}$ belongs to a matching edge of $R'$ , we first find with (2) two triangles inside this matching edge each with one vertex in the cluster $V_{i_j}$ and the other two vertices in the other cluster of the matching edge, then we proceed as before (see Figure 1). Note that we cover three vertices of $V_0$ using at most $13$ triangles, and thus to cover $V' \subseteq V_0$ we use at most $13 |V'|/3 \le 5 |V'|$ , as claimed above. Therefore we can repeat this procedure until $V'=V_0$ .

Figure 1. Embeddings of triangles for absorbing $V_0$ while using the same number of vertices from each cluster within a cherry or a matching edge. Each red triangle covers a vertex of $V_0$ . Each blue triangle stands for two triangles with end-points in the same clusters; we only draw one for simplicity.

Let $\mathcal{T}_1$ be the set of triangles we found above to cover $V_0$ and keep the divisibility condition. We now update the regularity partition by deleting $V(\mathcal{T}_1)$ from each $V_i$ for $i \in [t]$ and note that for all cherries and matchings from $R'$ the number of vertices in the clusters together is divisible by three. We recall that so far we removed at most $(2 \varepsilon + \delta + 2\sqrt{\delta }) n_0$ vertices from each cluster, where the first (resp. second, third) term bounds the number of vertices removed for making each pair super-regular (resp. for a later application of Lemma 4.2, for covering $V_0$ ).

Balancing the partition. Now the matching edges in $R'$ are already ready for an application of Lemma 4.3, and we will not modify the corresponding clusters anymore. However, before an application of Lemma 4.2 to the cherries in $R'$ , we need to ensure that the ratio between their size and the size of the centre-cluster satisfies the hypotheses of the lemma. This is what we are going to do now. For $i \not \in \mathcal{I}$ we denote by $W_i$ and $U_i$ the leaf clusters of the cherry centred at $V_i$ . Before covering the vertices of $V_0$ , we had $|W_i| = |U_i| \le (1-\delta ) |V_i|$ for $i \not \in \mathcal{I}$ , which still holds as we removed the same number of vertices from each cluster of a cherry.

However we still need to guarantee the other inequality $|W_i|,|U_i| \ge (1-\delta _0) |V_i|$ . For that, we find $2m$ triangles with two vertices in $V_i$ , of which one half has the third vertex in $U_i$ and the other half in $W_i$ , where $m$ is the smallest integer such that

(3) \begin{equation} U_i|-m \ge (1-\delta _0)(|V_i|-4m). \end{equation}

Then after removing these $2m$ triangles, we will have precisely $(1-\delta _0)|V_i| \le |U_i| = |W_i|$ . Observe that the inequality (3) implies that $m \ge \frac{(1-\delta _0)|V_i|-|U_i|}{4(1-\delta _0)-1}$ and, as we chose the smallest such $m$ , we get $m \le \lceil \frac{(1-\delta _0)|V_i|-|U_i|}{4(1-\delta _0)-1} \rceil$ . Moreover, as $\delta \lt \delta _0$ , $|V_i| \le n_0$ and $|U_i| \ge (1-2\varepsilon -\delta -2 \sqrt{\delta } ) n_0$ , we have $\frac{(1-\delta _0)|V_i|-|U_i|}{4(1-\delta _0)-1} \lt \frac{(1-\delta _0)-(1-2\varepsilon -\delta -2 \sqrt{\delta } )}{2}n_0 \lt 2 \sqrt{\delta }n_0$ . Therefore, for $n$ (and thus $n_0$ ) large enough, $m \le 2 \sqrt{\delta }n_0$ . We can find these at most $4 \sqrt{\delta } n_0$ triangles, by iteratively picking them with (2) and removing the corresponding vertices from $U_i$ , $W_i$ , and $V_i$ . Indeed, for any $v \in W_i \cup U_i$ we have degree into $V_i$ at least $(d-3 \varepsilon - \delta -10 \sqrt{\delta })n_0 \ge dn_0/2$ , as we started from $(2\varepsilon,d-3\varepsilon )$ -super-regular pairs and $\delta \lt \delta ' \lt 160^{-2} d^2$ .

Note that afterwards we still have $|U_i| \le (1-\delta )|V_i|$ as for large enough $n$ and with $\delta \lt \delta _0$ we have $m \le \lceil \frac{(1-\delta _0)|V_i|-|U_i|}{4(1-\delta _0)-1} \rceil \le \frac{(1-\delta )|V_i|-|U_i|}{4(1-\delta )-1}$ . Therefore, we have $(1-\delta _0)|V_i| \le |U_i| = |W_i| \le (1-\delta ) |V_i|$ . Moreover with $d-3 \varepsilon -\delta -10 \sqrt{\delta } \ge d/2$ and $2\varepsilon \le \varepsilon '$ , we get that the pairs $(U_i,V_i)$ and $(W_i,V_i)$ are $(\varepsilon ',d/2)$ -super-regular. Let $\mathcal{T}_2$ be the set of triangles we removed during this phase.

Completing the triangles. Now for any $i \not \in \mathcal{I}$ , after revealing $G_1[V_i \cup W_i \cup U_i]$ , we a.a.s. find a triangle factor covering the vertices of $U_i$ , $W_i$ , and $V_i$ by Lemma 4.2. Similarly for any matching edge $ij$ of $R'$ observe that $(V_i,V_j)$ is a $(\varepsilon ',d/2)$ -super-regular pair. Then after revealing $G_1[V_i \cup V_j]$ , we a.a.s. find a triangle factor covering the vertices of $V_i$ and $V_j$ by Lemma 4.3. Note that we apply Lemma 4.2 and Lemma 4.3 only constantly many times and thus a.a.s. we get a triangle factor in all such applications. Let $\mathcal{T}_3$ be the union of the triangle factors we obtain for each $i \not \in \mathcal{I}$ and each matching edge $ij$ from $R'$ . Then $\mathcal{T}_1 \cup \mathcal{T}_2 \cup \mathcal{T}_3$ gives $\lfloor n/3 \rfloor$ pairwise vertex-disjoint triangles in $G \cup G(n,p)$ .

Proof of Theorem 2.4.Let $1 \le m \le n/256$ and let $G$ be an $n$ -vertex graph on vertex set $V$ with minimum degree $\delta (G) \ge m$ . We let $C$ be large enough such that with $p \ge C \log n/n$ we can expose $G(n,p)$ in four rounds as $\bigcup _{i=1}^4 G_i$ with $G_i \sim G(n,C_i \log n/n)$ for $i=1,\dots,4$ such that the following hold. We let $C_1=1$ and observe that, by a union bound, a.a.s. for any set of vertices $U$ of size at least $n/256$ there is at least one edge in $G_1[U]$ . Next, we let $C_2$ be large enough such that for a set of vertices $U$ of size at least $n/2$ there are a.a.s. at least $\log ^3 n$ pairwise vertex-disjoint triangles in $G_2[U]$ [Reference Janson, Łuczak and Ruciński19, Theorem 3.29]. Finally, let $C_3=1$ (this is sufficient for our application of Proposition 7.1 because we do not need the $\log n$ term) and $C_4$ be such that we can apply Proposition 7.2 with $C_4/2$ . We expose $G_1$ already now and assume that the described property holds, while we leave $G_2$ , $G_3$ , and $G_4$ until we need them.

To apply one of the two propositions to a large subgraph $G'$ of $G$ , we need $\Delta (G') \le v(G')/32$ . For this let $V'$ be the set of vertices from $G$ of degree at least $n/64$ . If $|V'|\ge m$ , then we let $V''$ be any subset of $V'$ of size $m$ and we greedily find $m$ pairwise vertex-disjoint triangles in $G \cup G_1$ , each containing exactly one vertex from $V''$ . Indeed, as long as we have less than $m$ triangles, there is a vertex $v \in V''$ not yet contained in a triangle. Then there is a set $U \subseteq N_G(v) \setminus V''$ of at least $n/64-3m \ge n/256$ vertices not covered by triangles, and we can find an edge within $G_1[U]$ that gives us a triangle containing $v$ and two vertices from $U$ .

Otherwise, $|V'|\lt m$ and we remove $V'$ from $G$ to obtain $G'=G[V \setminus V']$ . Note that we have $v(G') = n-|V'| \ge n/2$ , minimum degree $\delta (G') \ge m'= m-|V'|$ , and maximum degree $\Delta (G') \lt n/64 \le v(G')/32$ . If $m' \lt (\log n)^3$ , then we a.a.s. find $m'$ pairwise vertex-disjoint triangles within $G_2[V(G')]$ . If $(\log v(G'))^3 \le (\log n)^3 \le m' \le \sqrt{v(G')}$ , then by Proposition 7.1 and as $\log n/n= \omega (1/v(G'))$ there are a.a.s. at least $m'$ pairwise vertex-disjoint triangles in $G' \cup G_3[V(G')]$ . Finally, if $\sqrt{v(G')} \le m' \le n/256 \le v(G')/32$ , then by Proposition 7.2 and as $C_4 \log n/n \ge \frac{C_4}{2} \log v(G')/ v(G')$ , there are a.a.s. at least $m'$ pairwise vertex-disjoint triangles in $G' \cup G_4[V(G')]$ .

Now, that we found $m'$ pairwise vertex-disjoint triangles, we can greedily add triangles by using the $m-m'$ vertices from $V'$ and an edge in their neighbourhood until we have $m$ triangles. Analogous to above, as long as we have less than $m$ triangles, for each available vertex $v \in V'$ , there is a set $U \subseteq N_G(v) \setminus V'$ of at least $n/256$ vertices not covered by triangles, and we find an edge within $G_1[U]$ .

Proof of Lemma 7.3.Given $0\lt \gamma \lt 1/2$ and an integer $s$ we let $\varepsilon \gt 0$ such that $\varepsilon \le 1/(6s)$ and $\varepsilon \lt 1/2- s \gamma$ . Moreover, we let $n$ be large enough for our calculations and, for simplicity, assume that $\varepsilon \sqrt{n}$ is an integer. Then let $2/\varepsilon \le m \le \sqrt{n}$ and $G$ be an $n$ -vertex graph on vertex set $V$ with $\delta (G) \ge m$ and $\Delta (G) \le \gamma n$ .

Let $\mathcal{K}$ be a family of vertex-disjoint stars in $G$ with $\varepsilon m \le g_K \le \varepsilon \sqrt{n}$ for all $K \in \mathcal{K}$ , that maximizes the sum

(4) \begin{equation} \sum _{K \in \mathcal{K}} g_K^2 \end{equation}

among all such families. Note that each star in $\mathcal{K}$ has at least $2$ leaves because $g_K \ge \varepsilon m$ and $m \ge 2/\varepsilon$ .

If the sum in (4) is bigger than $s \varepsilon ^2 n m$ we are done. So we assume the family $\mathcal{K}$ satisfies