Triangles in $K_s$-saturated graphs with minimum degree $t$

TL;DR

This paper determines the minimum number of triangles in certain saturated graphs and explores how the count of larger cliques varies with minimum degree, revealing new extremal graph structures.

Contribution

It establishes exact triangle counts in $K_4$-saturated graphs with minimum degree 4 and constructs graphs with prescribed clique counts for larger $s$ and $r$, showing diverse behaviors.

Findings

Minimum triangles in $K_4$-saturated graphs with degree 4 is exactly $2n-4$.

Unique extremal graph achieving this minimum.

Number of $K_r$ in $K_s$-saturated graphs can be controlled independently of minimum degree.

Abstract

For $n \geq 15$, we prove that the minimum number of triangles in an $n$-vertex $K_4$-saturated graph with minimum degree 4 is exactly $2n-4$, and that there is a unique extremal graph. This is a triangle version of a result of Alon, Erd\H{o}s, Holzman, and Krivelevich from 1996. Additionally, we show that for any $s > r \geq 3$ and $t \geq 2 (s-2)+1$, there is a $K_s$-saturated $n$-vertex graph with minimum degree $t$ that has $\binom{ s-2}{r-1}2^{r-1} n + c_{s,r,t}$ copies of $K_r$. This shows that unlike the number of edges, the number of $K_r$'s ($r >2$) in a $K_s$-saturated graph is not forced to grow with the minimum degree, except for possibly in lower order terms.

Equations

$$\textup{sat}(n,F)=\min\{|E(G)|:\mbox{$G$ has $n$ vertices and is $F$-saturated}\}.$$

$$\text{sat}(n,K_s)=(s-2)(n-s+2)+\left(\genfrac{}{}{0ex}{}{s-2}{2}\right).$$

$$\textup{sat}_{t}(n,F)=\min\{|E(G)|:\mbox{$G$ has $n$ vertices, is $F$-saturated, and has $\delta(G)=t$ }\}.$$

$${\text{sat}}_t(n,K_s)\ge tn-c_t$$

$${\text{sat}}_t(n,H,F)$$

$$\text{sat}(n,K_3,K_4)=n-2$$

$$\left(\genfrac{}{}{0ex}{}{s-2}{r}\right)+(n-s+2)\left(\genfrac{}{}{0ex}{}{s-2}{r-1}\right)$$

$${\text{sat}}_t(n,K_3,K_4)\le 2n+2t-12.$$

$${\text{sat}}_t(n,K_3,K_4)=2n+2t-12$$

$${\text{sat}}_t(n,K_r,K_s)\le \left(\genfrac{}{}{0ex}{}{s-2}{r-1}\right)2^{r-1}n+C_{s,r,t}$$

$${\text{sat}}_t(n,K_3,K_5)\le 9n+C_t$$

$$\left(\genfrac{}{}{0ex}{}{s-2}{2}\right)(n-2)\le {\text{sat}}_t(n,K_3,K_s).$$

$$t>\min \{|A|,|B|,{\delta }_{S(A,B)}(A),{\delta }_{S(A,B)}(B)\}\text{mbox}andn>|A|+|B|+N+M,$$

$${\delta }_G(Y\cup A)$$

$$|X|k_{r-1}(A)=(n-|A|-|B|-M)k_{r-1}(A).$$

$$\{i,i+1,i+2,\dots ,i+s-4\}\cup ({\mathcal{B}}_1\backslash \{b_i\}).$$

$$i_{\ell }\le a\le i_{\ell }+s-4$$

$$1$$

$$(s-\beta -2)+\beta =s-2$$

$$A_1=\{a_{147},a_{258},a_{369}\},,A_2=\{a_{159},a_{267},a_{369}\},\text{mbox}and{A}_3=\{a_{169},a_{249},a_{357}\}.$$

$$B_i=\{b_{i,1},b_{i,2},b_{i,3}\}.$$

$$N(x)=\{x_1,x_2,x_3,x_4\}\text{mbox}andN[x]=\{x\}\cup N(x).$$

$$V_S\sim V_T$$

$$\bigcup _{T\in {\mathcal{T}}_{S,i}}T$$

$$k_3(G)\ge 2n-4.$$

$$Y=V_{123}\cup V_{124}\cup V_{134}\cup V_{234}\cup V_{1234}.$$

$$Y^{\prime }:=V_{123}\cup V_{124}\cup V_{134}\cup V_{234}.$$

$$k_3(G)\ge k_3(G_1^{\prime })+2(n-9)=14+2(n-9)=2n-4.$$

$$V_S:y{\alpha }_1{\beta }_1,y{\alpha }_2{\beta }_2,\dots ,y{\alpha }_k{\beta }_k.$$

$$k_3(G)\ge 3n-18.$$

Full text

1 Introduction

Let $F$ be a graph. A graph $G$ is $F$-free if $G$ does not contain $F$ as a subgraph. A graph $G$ is $F$-saturated if $G$ is $F$-free, and adding a new edge to $G$ creates a copy of $F$. The minimum number of edges in an $F$-saturated graph with $n$ vertices is called the saturation number of $F$. Write $\textup{sat}(n,F)$ for this minimum, so

[TABLE]

An $n$-vertex $F$-saturated graph with $\textup{sat}(n,F)$ edges is called an extremal graph.

1.1 History and Previous Results

One of the most important results on graph saturation is that for any graph $F$ with at least one edge, there is a constant $C_{F}$ such that $\textup{sat}(n,F)\leq C_{F}n$. This was proved by Kászonyi and Tuza in 1986 [14], and shows that saturation numbers are linear in $n$. Since then, the study of saturation has become an established branch of extremal graph theory. Saturation numbers of hypergraphs and of random graphs have been studied as well [4, 16, 20, 19]. The survey paper of Faudree, Faudree, and Schmitt [9] contains many results and references.

For complete graphs, Erdős, Hajnal, and Moon [7] proved that for $s\geq 3$,

[TABLE]

Furthermore, there is a unique extremal graph which is, up to isomorphism, $K_{s-2}+\overline{K_{n-s+2}}$ (the join of a clique with $s-2$ vertices and an independent set with $n-s+2$ vertices). This graph has minimum degree $s-2$, and the minimum degree of a $K_{s}$-saturated graph is at least $s-2$ (since nonadjacent vertices must have a $K_{s-2}$ in their common neighborhood). A natural question is to ask for the minimum number of edges in an $F$-saturated graph $G$ with $\delta(G)=t$ where $t>s-2$. Given a graph $F$ and an integer $t$, let

[TABLE]

Duffus and Hanson [6] proved that $\textup{sat}_{2}(n,K_{3})=2n-5$ for $n\geq 5$, and characterized the extremal graphs. They also showed that $\textup{sat}_{3}(n,K_{3})=3n-15$ for $n\geq 10$. For larger $t$ and $s$, the bounds are not exact. In 2014, Day [5], resolving a conjecture of Bollobás [13] from 1996, showed that

[TABLE]

for any $s\geq 3$, $t\geq s-2$. Here $c_{t}$ is a constant depending only on $t$. Further discussion on saturated graphs with degree constraints can be found in [9].

In 2014, Alon and Shikhelman [2] introduced a very important generalization of Turán numbers which has since been extensively studied ([8, 10, 12, 17] to name a few). It is connected to the widely studied Turán problem for Berge hypergraphs ([11, 18], for instance). Motivated by this generalization, Kritschgau et. al. [15] defined an analogous generalization of saturation numbers. For graphs $H$ and $F$, let $\textup{sat}(n,H,F)$ be the minimum number of copies of $H$ in an $F$-saturated graph with $n$ vertices. Observe that $\textup{sat}(n,K_{2},F)=\textup{sat}(n,F)$. In this paper, we introduce the function

[TABLE]

which is defined to be the minimum number of copies of $H$ in an $n$-vertex $F$-saturated graph with minimum degree $t$. This function generalizes both $\textup{sat}_{t}(n,F)$ and $\textup{sat}(n,H,F)$. The question we put forth is the following.

Question 1.1

Let $H$ and $F$ be graphs. What is the minimum number of copies of $H$ in an $F$-saturated $n$-vertex graph with minimum degree $t$?

Before stating our results, let us recall a result from [15] which we take as a starting point for our work. In [15], the formula

[TABLE]

was proved, and it was shown that $K_{2}+\overline{K_{n-2}}$ is the unique extremal graph. Bounds on $\textup{sat}(n,K_{r},K_{s})$ for all $s>r\geq 3$ were also proved in [15]. However, an asymptotic formula for $\textup{sat}(n,K_{r},K_{s})$ is known only in the case when $(r,s)=(3,4)$.

Motivated by the fact that (2) gives a formula for $\textup{sat}(n,K_{3},K_{4})$, we will study the function $\textup{sat}_{t}(n,K_{3},K_{4})$ in detail. Our aim is to determine whether or not a statement similar to (1) holds when counting $K_{3}$’s in a $K_{4}$-saturated graph with minimum degree at least $t$. First let us state some results that answer Question 1.1 in certain cases. The first proposition we give is easy to prove using the fact that in a $K_{s}$-saturated graph, any two nonadjacent vertices must have a $K_{s-2}$ in their common neighborhood.

Proposition 1.2

Let $s>r\geq 2$ be integers. If $G$ is an $n$-vertex $K_{s}$-saturated graph with $\delta(G)=s-2$, then $G$ is isomorphic to $K_{s-2}+\overline{K_{n-s+2}}$ and consequently, has

[TABLE]

copies of $K_{r}$.

In the special case $(r,s)=(3,4)$ and $\delta(G)=2$, we have exactly $n-2$ triangles. A close look at the proof of (2) in [15] shows that if $G$ is not isomorphic to $K_{2}+\overline{K_{n-2}}$, then $G$ has at least $n$ triangles. Now if $G$ is $K_{4}$-saturated and is not isomorphic to $K_{2}+\overline{K_{n-2}}$, then one can prove that $G$ has minimum degree at least 3. Thus, we see a small jump in the number of triangles when $G$ is no longer allowed to have a vertex of degree 2. It turns out that when the minimum degree is 3, we do not just get 2 additional triangles, but we must get and additional $n-5$ triangles. Before formalizing this as a proposition, we need to introduce a new graph.

Let $a_{1}a_{2}a_{3}a_{4}a_{5}a_{1}$ be a cycle of length five and let $b$ be a vertex joined to each $a_{i}$. Call this graph $W$. Let $m_{1},m_{3},m_{4}$ be positive integers with $m_{1}+m_{3}+m_{4}=n-s+1$. In $W$, replace $b$ with a clique of size $s-3$, and for $i\in\{1,3,4\}$, replace $a_{i}$ with an independent set of size $m_{i}$. Vertices in this new graph are adjacent if and only if the vertices they replaced are adjacent vertices in $W$. Write $W_{s}(m_{1},1,m_{3},m_{4},1)$ for this graph, which has appeared in the literature (see [3]).

Proposition 1.3

Let $s\geq 3$. If $G$ is an $n$-vertex $K_{s}$-saturated graph with $\delta(G)=s-1$, then $G$ is isomorphic to either $(K_{s-1}-e)+\overline{K_{n-s+1}}$, or $W_{s}(m_{1},1,m_{3},m_{4},1)$ for some $m_{1},m_{3},m_{4}$ with $m_{1}+m_{3}+m_{4}=n-s+1$.

It follows from Proposition 1.3 that if $G$ is an $n$-vertex $K_{4}$-saturated graph with minimum degree 3, then $G$ has at least $2n-7$ triangles, and equality holds only if $G$ is isomorphic to $W_{4}(m_{1},1,m_{3},1,1)$ for some $m_{1}+m_{3}=n-4$. For a proof of Proposition 1.3, see [15].

1.2 New Results

In light of Day’s Theorem (1) and that

• •

$|V(G)|=n$, $G$ is $K_{4}$-saturated, and $\delta(G)=2$    $\Rightarrow$    $G$ has at least $n-2$ triangles,

• •

$|V(G)|=n$, $G$ is $K_{4}$-saturated, and $\delta(G)=3$    $\Rightarrow$    $G$ has at least $2n-7$ triangles,

one may be tempted to conjecture that in general, $\delta(G)=t$ forces at least $(t-1)n-O(1)$ triangles in any $n$-vertex $K_{4}$-saturated graph. This would then give a version of Day’s result for triangles.

It turns out, rather surprisingly, that for any $t\geq 4$ and $n\geq t+5$, there is an $n$-vertex $K_{4}$-saturated graph that has minimum degree $t$ and only $2n+2t-12$ triangles. We call this graph $H_{t}(n)$ and it is defined in Section 2. Our main theorem determines $\textup{sat}_{t}(n,K_{3},K_{4})$, and shows $H_{t}(n)$ is the unique extremal graph.

Theorem 1.4

Let $n\geq 14$ and $G$ be an $n$-vertex $K_{4}$-saturated graph with $\delta(G)=4$. Then $G$ contains at least $2n-4$ triangles. Furthermore, if $G$ contains exactly $2n-4$ triangles, then $G$ is isomorphic to $H_{t}(n)$.

For $t\geq 4$, the graph $H_{t}(n)$ implies the following upper bound on $\textup{sat}_{t}(n,K_{3},K_{4})$.

Theorem 1.5

For integers $t\geq 4$ and $n\geq 2t$,

[TABLE]

We conjecture that the upper bound in Theorem 1.5 is best possible, and that $H_{t}(n)$ is the unique extremal graph for all sufficiently large $n$.

Conjecture 1.6

For any integer $t\geq 4$, there is an integer $n_{t}$ such that for all $n\geq n_{t}$,

[TABLE]

and $H_{t}(n)$ is the unique extremal graph.

Theorem 1.4 shows that Conjecture 1.6 is correct when $t=4$. The next theorem gives an upper bound for arbitrary $s>r\geq 3$.

Theorem 1.7

Let $s>r\geq 3$ and $t\geq 2(s-2)+1$ be integers. For $n\geq 2(s-2)+2t$,

[TABLE]

where $C_{s,r,t}$ is a constant depending only on $r$, $s$, and $t$.

It would be interesting to determine if there is a constant $c_{r,s}$, depending only on $r$ and $s$, such that for all $n\geq n(r,s,t)$ and $t\geq s-2$, there is an $n$-vertex $K_{s}$-saturated graph with minimum degree $t$ having at most $c_{r,s}n+o(n)$ copies of $K_{r}$. Theorem 1.5 shows that such a constant exists in the case when $(r,s)=(3,4)$. Theorem 1.7 covers all $s>r\geq 3$, but assumes $t\geq 2(s-2)+1$, and perhaps the coefficient of $n$ could be improved (as in Theorem 1.5 when $(r,s)=(3,4)$). We can improve this upper bound in the case $(r,s)=(3,5)$ as the following result shows.

Theorem 1.8

For any $t\geq 8$ and $n\geq t+30$,

[TABLE]

where $C_{t}$ is a constant dependent only on $t$.

Finally, a simple argument in the case $r=3$ gives a lower bound on $\textup{sat}_{t}(n,K_{3},K_{s})$.

Proposition 1.9

Let $s>3$ and $t\geq 6\binom{s-2}{2}$ be integers. For $n\geq 2s-2$,

[TABLE]

Proposition 1.9 implies $\textup{sat}_{t}(n,K_{3},K_{5})\geq 3(n-2)$ for $t\geq 18$. We suspect the upper bound of Theorem 1.8 could give the correct coefficient of $n$.

In Section 2, we define the graphs that prove Theorems 1.5, 1.7, and Proposition 1.8. In Section 3, we prove Theorem 1.4. In Section 4, we prove Proposition 1.9.

2 Proof of Theorems 1.5,

1.7, and 1.8

A natural question one might ask is whether or not every edge $e$ in a $K_{s}$-saturated graph is in a triangle. It turns out that this is not necessarily true. In fact, $K_{s}$-saturated graphs that have an edge in no triangle contain certain subgraphs–what we call $K_{s}$-support structures–which pave the way for constructions proving Theorems 1.5 and 1.7. For this and the following sections, we will use $k_{r}(A)$ to denote the number of copies of $K_{r}$ in $A$.

Definition 2.1

Let $S(A,B)$ be a graph with vertex set $A\cup B$, where $A$ and $B$ are disjoint sets. We call $S(A,B)$ a pre-$K_{s}$-support-structure provided:

• •

$A$* and $B$ are both $K_{s-1}$-free,*

• •

$N(a)\cap B$* has a copy of $K_{s-2}$ for every vertex $a\in A$,*

• •

$N(b)\cap A$* has a copy of $K_{s-2}$ for every vertex $b\in B$, and*

• •

$S(A,B)$* itself is $K_{s}$-free.*

If $S(A,B)$ has the added condition that the subgraph induced by $A$ and the subgraph induced by $B$ are $K_{s-1}$-saturated, and adding any missing edge between $A$ and $B$ would create a $K_{s}$, then we call $S(A,B)$ a $K_{s}$-support-structure.

Lemma 2.2

Given a pre-$K_{s}$-support-structure $S(A,B)$, there is a $K_{s}$-support-structure $S^{\prime}(A,B)$ where $S^{\prime}(A,B)$ is a supergraph of $S(A,B)$.

Proof. We add edges one at a time to $S(A,B)$ arbitrarily, adding each missing edge $e$ only if it does not create a $K_{s-1}$ in $A$, a $K_{s-1}$ in $B$, or a $K_{s}$ in $S(A,B)$. We continue this process until there are no longer any edges to add. Certainly after at most $\binom{|A\cup B|}{2}$ steps this process will end. Notice that with each edge added, the new graph is still a pre-$K_{s}$-support-structure, and in fact the final graph $S^{\prime}(A,B)$ is a $K_{s}$-support-structure.

Lemma 2.3

Let $S(A,B)$ be a $K_{s}$-support-structure, and set $N=t-\min\{|B|,\delta_{S(A,B)}(A)\}$ and $M=t-\min\{|A|,\delta_{S(A,B)}(B)\}$. Given integers

[TABLE]

there exists a $K_{s}$-saturated supergraph $G$ of $S(A,B)$ with order n and minimum degree t containing $nk_{r-1}(A)+C_{|A|,|B|,t}$ copies of $K_{r}$.

Proof. Define $G$ as the graph obtained from $S(A,B)$ by taking two independent sets $X=\{x_{1},x_{2},\dots,x_{n-|A|-|B|-M}\}$ and $Y=\{y_{1},y_{2},\dots,y_{M}\}$, and joining $X$ to $A\cup Y$, and joining $Y$ to $B\cup X$. Note that since $S(A,B)$ is a $K_{s}$-support-structure, $G$ is $K_{s}$-saturated. Next we show that the minimum degree of G is $t$. We have

[TABLE]

By symmetry, $\delta_{G}(X\cup B)\geq t$. Putting these two inequalities together, we see that $\delta(G)=\min\{\delta_{G}(Y\cup A),\delta_{G}(X\cup B)\}=t$.

We now count the number of $K_{r}$’s in $G$. Note that by $G$’s construction, the number of vertices not in $X$ is dependent only on $|A|$, $|B|$ and $t$. Thus, the number of copies of $K_{r}$ that do not use a vertex in $X$ is also dependent only on these parameters. Suppose $K$ is a $K_{r}$ that does contain a vertex in $X$. Then $K$ can only contain one vertex in $X$, since $X$ is an independent set. Also, $K$ cannot contain any vertex in $Y$ because $r\geq 3$ and $Y$ is an independent set. Thus, all the vertices in $K$ that are not in $X$ must be in $A$. We conclude that the number of copies of $K_{r}$ containing a vertex in $X$ is

[TABLE]

We now define three graphs by defining pre-$K_{s}$-support structures, and then applying Lemmas 2.2 and 2.3.

The graph $H_{t}(n)$: Let $t\geq 4$ and $n>2t$ be integers. Let $A=\{a_{1},a_{2},a_{3},a_{4}\}$ and $B=\{b_{123},b_{124},b_{134},b_{234}\}$, and construct the graph $S(A,B)$ with vertex set $A\cup B$, and the following edges:

• •

$a_{1}a_{2}$ and $a_{3}a_{4}$,

• •

$b_{123}b_{234}$, $b_{234}b_{124}$, $b_{124}b_{134}$, and $b_{134}b_{123}$,

• •

$a_{r}b_{ijk}$ if and only if $r\in\{i,j,k\}$.

It is easy to verify that $S(A,B)$ is a $K_{s}$-support-structure. Employing Lemmas 2.2 and 2.3, we obtain a $K_{4}$-saturated graph, which we call $H_{t}(n)$, that has $n$-vertices, minimum degree $t$, and containing $2n+2t-12$ triangles. Thus, $H_{t}(n)$ proves Theorem 1.5.

Remark: The graph $H_{4}(n)$ appears briefly in [1], but the focus in that paper is on minimizing the number of edges in a $K_{4}$-saturated graph with minimum degree 4. Up to isomorphism, there is a unique graph that minimizes the number of edges in a $K_{4}$-saturated graph with minimum degree 4, but this graph does not minimize the number of triangles.

Next we construct the graph that proves Theorem 1.7.

The graph $F_{s,t}(n)$: Let $s>r\geq 3$, $t\geq 2(s-2)+1$, and $n\geq 2(s-2)+2t$. Let $A=\{a_{1},a_{2},\dots,a_{2(s-2)}\}$ and $B=\{b_{1},b_{2},\dots,b_{2(s-2)}\}$, and construct the graph $S(A,B)$ with vertex set $A\cup B$, and the following edges:

• •

$a_{i}a_{j}$ if and only if $i\not\equiv j+(s-2)\pmod{2(s-2)}$,

• •

$b_{i}b_{j}$ if and only if $i\not\equiv j+(s-2)\pmod{2(s-2)}$,

• •

$a_{i}b_{i\pmod{2(s-2)}}$, $a_{i}b_{i+1\pmod{2(s-2)}}$, …, and $a_{i}b_{i+(s-2)-1\pmod{2(s-2)}}$.

Notice that the vertices in $A$ and the vertices in $B$ form two complete $(s-2)$-partite graphs with 2 vertices in each part.

We now show that $S(A,B)$ is a pre-$K_{s}$-support-structure. It is easy to verify that $A$ and $B$ are $K_{s-1}$-free, and that each $a\in A$ and $b\in B$ has a copy of $K_{s-2}$ in $N(a)\cap B$ and $N(a)\cap B$, respectively. Next we prove that $S(A,B)$ is $K_{s}$-free. It is enough to show that $a_{1}$ does not lie in a $K_{s}$ since $S(A,B)$ is vertex transitive. The neighborhood of $a_{1}$ is the union of the two sets

$\mathcal{A}_{1}=\{a_{2},a_{3},\dots,a_{2(s-2)}\}\backslash\{a_{s-1}\}$   and   $\mathcal{B}_{1}=\{b_{1},b_{2},\dots,b_{s-2}\}$.

Let $H_{1}$ be the subgraph of $S(A,B)$ induced by $\mathcal{A}_{1}\cup\mathcal{B}_{1}$. We will show $H_{1}$ is $K_{s-1}$-free. It will be convenient to relabel the vertices in $\mathcal{A}_{1}$ as

$a_{s}=1$, $a_{s+1}=2$, $a_{s+2}=3,\dots,a_{2(s-2)}=s-3$,

and

$a_{2}=s-2$, $a_{3}=s-1,a_{4}=s,\dots,a_{s-2}=2s-6$.

With this relabeling, in the graph $H_{1}$ the vertex $b_{i}$ ($1\leq i\leq s-2$) is adjacent to all vertices in

[TABLE]

Now suppose $K$ is a clique in $H_{1}$. Let $b_{i_{1}},b_{i_{2}},\dots,b_{i_{\beta}}$ be the vertices in $K$ that are in $\mathcal{B}_{1}$, where $1\leq i_{1}<i_{2}<\dots<i_{\beta}\leq s-2$. By (3), if $a\in\{1,2,3,\dots,2s-6\}$ is a vertex in $\mathcal{A}_{1}\cap K$, then

[TABLE]

for $\ell=1,2,\dots,\beta$. In particular, $i_{\beta}\leq a\leq i_{1}+s-4$. Adding $-i_{\beta}+1$ throughout leads to

[TABLE]

Therefore, there are at most $s-\beta-2$ vertices in $\mathcal{A}_{1}$ that are in $K$, so $K$ contains at most

[TABLE]

vertices. We conclude that the neighborhood of $a_{1}$ is $K_{s-1}$-free which implies $S(A,B)$ is $K_{s}$-free.

Adding edges to $S(A,B)$ using the process described in Lemma 2.2, we obtain a $K_{s}$-support-structure $S^{\prime}(A,B)$. Notice that since $A$ and $B$ were already $K_{r-1}$-saturated, the only edges added were between $A$ and $B$, so the subgraphs induced by $A$ and $B$ have not changed. Applying Lemma 2.3 to $S^{\prime}(A,B)$, we obtain a $K_{s}$-saturated graph $F_{s,t}(n)$ with $n$ vertices, minimum degree $t$, and containing at most $nk_{r-1}(A)+C_{r,s,t}=\binom{s-2}{r-1}2^{r-1}n+C_{r,s,t}$ copies of $K_{r}$.

A computation, using the assumption that $n\geq 2(s-2)+2t$, shows that all vertices in $F_{s,t}(n)$ have degree at least $t$ (vertices in $X$ will have degree $t$). The graph $F_{s,t}(n)$ proves Theorem 1.7.

The graph $R_{t}(n)$: Let $t>9$, and $n>2t+12$. Let $A$ be the disjoint union of the three $K_{3}$’s

[TABLE]

The subscripts appearing in a given triangle form a class of parallel lines in an affine plane of order 3.

Next, let $B$ be the disjoint union of triangles $B_{1}$ through $B_{9}$, where each $B_{i}$ is given by

[TABLE]

Construct the graph $S(A,B)$ by connecting each vertex $a_{ijk}\in A_{s}$ to all vertices in the union $B_{i}\cup B_{j}\cup B_{k}\cup\{b_{t,s}:1\leq t\leq 9\}$.

We now show that $S(A,B)$ is a pre-$K_{5}$-support structure. It is easy to verify that $A$ and $B$ are $K_{4}$-free, and that each $a\in A$ and $b\in B$ has a copy of $K_{3}$ in $N(a)\cap B$ and $N(a)\cap B$, respectively. Also, since $A$ and $B$ are $K_{4}$-free, a possible $K_{5}$ in $S(A,B)$ must have at least two vertices in $A$, say $a_{i_{1}j_{1}k_{1}},a_{i_{2}j_{2}k_{2}}\in A_{s}$. However, by $S(A,B)$’s construction, their common neighborhood in $B$ contains at most 1 vertex. Hence $S(A,B)$ is $K_{5}$-free.

Adding edges to $S(A,B)$ using the process described in Lemma 2.2, we obtain a $K_{s}$-support-structure $S^{\prime}(A,B)$. Importantly, since by $S(A,B)$’s construction every pair of non-adjacent vertices in $A$ must have a $K_{3}$ in their common neighborhood in $B$, no edges were added to $A$, so the subgraph induced by $A$ has not changed. Applying Lemma 2.3 to $S^{\prime}(A,B)$, we obtain a $K_{5}$-saturated graph $R_{t}(n)$ with $n$ vertices, minimum degree $t$, and containing at most $nk_{3}(A)+C_{t}=9n+C_{t}$ copies of $K_{3}$.

Remark: The construction of $R_{t}(n)$ can be generalized to create a $K_{s}$-support-structure $R_{s,t}(n)$ for any $s\geq 3$, but for brevity’s sake we have chosen to omit this. Notice that for $s=5$, $R_{5,t}(n)$’s $9n+C_{t}$ copies of $K_{3}$ is less than $F_{5,t}(n)$’s $12n+C_{t}$ copies of $K_{3}$. This turns out to be an exception; for $s=6$, $R_{6,t}(n)$ and $F_{6,t}(n)$ both have $24n+C_{t}$ copies of $K_{3}$, and for $s\geq 7$, $R_{7,t}(n)$ contains more copies of $K_{3}$ than $F_{7,t}(n)$. In general, $R_{s,t}(n)$ will have fewer copies of $K_{r}$ than $F_{s,t}(n)$ for $s\in\{4,5,\dots,2^{r-1}+1\}$, but when $s>2^{r-1}+2$, $R_{s,t}(n)$ contains more copies of $K_{r}$ than $F_{s,t}(n)$. This suggests that a better construction may exist.

3 Proof of Theorem 1.4

In this section, we prove Theorem 1.4. The proof will be broken up into several lemmas. We write $k_{3}(G)$ for the number of triangles in $G$. For $x\in V(G)$, $N(x)$ is the neighborhood of $x$. Let us set up some specialized notation that will be used in the rest of this section. This same notation was used in [1].

Setup and Notation: For the remainder of Section 3, let $G$ be an $n$-vertex graph that is $K_{4}$-saturated, and has $\delta(G)=4$. Let $x\in V(G)$ be a vertex of degree 4, and let

[TABLE]

Let $Y=V(G)\backslash N[x]$. If $y\in Y$, then $N(y)\cap N(x)$ must contain an edge so that every vertex in $Y$ is adjacent to at least two vertices in $N(x)$. For $S\subseteq\{1,2,3,4\}$, let $V_{S}$ be the vertices $y\in Y$ such that $y$ is a neighbor of $x_{i}$ if and only if $i\in S$. For notational convenience, we will omit braces and commas so if, say $S=\{1,3,4\}$, then we write $V_{134}$ rather than $V_{\{1,3,4\}}$. Given sets $S,T\subseteq\{1,2,3,4\}$, write

[TABLE]

if all vertices in $V_{S}$ are adjacent to all vertices in $V_{T}$, and $V_{S}\nsim V_{T}$ if no vertex in $V_{S}$ is adjacent to a vertex in $V_{T}$.

An important observation is that if $V_{S}\neq\emptyset$, then there must be a pair $\{i,j\}\subseteq S$ such that $x_{i}$ is adjacent to $x_{j}$. Also, if $\{x_{i}:i\in S\}$ forms an independent set in $N(x)$, then $V_{S}=\emptyset$. Because of this, we ignore all $V_{S}$ for which $\{x_{i}:i\in S\}$ is an independent set in $N(x)$.

Given a subset $S\subset\{1,2,3,4\}$ with $i\notin S$ for some $i\in\{1,2,3,4\}$, let $\mathcal{T}_{S,i}$ be the set of all $T\subseteq\{1,2,3,4\}$ for which $i\in T$, and for any $j,k\in S$ for which $x_{j}$ and $x_{k}$ are adjacent vertices in $N(x)$, we have $|T\cap\{j,k\}|\leq 1$.

Lemma 3.1** (Rules Lemma)**

Let $S\subset\{1,2,3,4\}$ with $i\notin S$ for some $i\in\{1,2,3,4\}$. If $V_{S}\neq\emptyset$, then $y$ is adjacent to some vertex in

[TABLE]

and in particular, this union is not empty.

Proof. Suppose $y\in V_{S}$. Since $i\notin S$, the vertex $y$ is not adjacent to $x_{i}$ where $x_{i}\in N(x)$. There must be an edge $\alpha\beta$ with both $\alpha$ and $\beta$ in the intersection $N(y)\cap N(x_{i})$. If $\alpha,\beta\in N(x)$, then $\{x,x_{i},\alpha,\beta\}$ is a $K_{4}$ which is a contradiction. Thus, we may assume that $\alpha\in Y$, say $\alpha\in V_{T}$. Because $\alpha$ is adjacent to $x_{i}$, we have $i\in T$. If there is a $j,k\in S$ for which $x_{j}x_{k}$ is an edge in $N(x)$ and $j,k\in T$, then $\{y,\alpha,x_{j},x_{k}\}$ is a $K_{4}$. We conclude that $T$ must be some member of $\mathcal{T}_{S,i}$.

It will be convenient to represent Lemma 3.1 with an arrow diagram and we refer to these as Rules. Assuming the set up and conclusion of Lemma 3.1 where $\mathcal{T}_{S,i}=\{T_{1},T_{2},\dots,T_{\ell}\}$, then we illustrate using the figure below.

V_{S}$$V_{T_{\ell}}$$V_{T_{2}}$$V_{T_{1}}$$\vdots

Rule for $V_{S}$ and choice of $i\notin S$

We also use the notation $V_{S}\rightarrow V_{T_{1}}\cup V_{T_{2}}\cup\dots\cup V_{T_{\ell}}$. Note that the assertion of Lemma 3.1 is not that $T_{j}\neq\emptyset$ for every $T_{j}\in\mathcal{T}_{S,i}$, but only that at least one of these sets is not empty.

The first application of Lemma 3.1 will be in the proof of Lemma 3.4. First we prove two easier lemmas.

Lemma 3.2

The number of edges in $N(x)$ is at least 2.

Proof of Lemma 3.2. See Lemma 3.2 of [15] (the proof is not difficult).

Lemma 3.3

If $n\geq 14$ and the number of edges in $N(x)$ is 2, then

[TABLE]

Furthermore, equality holds if and only if $G$ is isomorphic to $H_{4}(n)$.

Proof of Lemma 3.3. There are two possibilities for the edges in $N(x)$.

Case 1: The two edges in $N(x)$ form a path.

Suppose the edges in $N(x)$ are $x_{1}x_{2}$ and $x_{2}x_{3}$. Every vertex in $Y$ must be adjacent to $x_{2}$ since $N(x)\cap N(y)$ must contain an edge for all $y\in Y$, and every edge in $N(x)$ has $x_{2}$ as an endpoint. Therefore, $x_{2}$ is adjacent to all vertices in $G$ except for $x_{4}$. The intersection $N(x_{3})\cap N(x_{4})$ must contain an edge, say $y_{1}y_{2}$. The edge $y_{1}y_{2}$ must have both endpoints in $Y$ since $x$ is the only neighbor of $x_{4}$ in $N[x]$, but then $\{y_{1},y_{2},x_{2},x_{3}\}$ is a $K_{4}$ in $G$ ($y_{1}$ and $y_{2}$ are both adjacent to $x_{2}$). This is a contradiction.

Case 2: The two edges in $N(x)$ share no endpoints.

Assume the edges in $N(x)$ are $x_{1}x_{2}$ and $x_{3}x_{4}$. Then $V_{12}$ and $V_{34}$ must be empty (see [1]). Since $N(y)\cap N(x)$ must contain an edge for each $y\in Y$ and $V_{12}=V_{34}=\emptyset$, we can partition $Y$ as follows:

[TABLE]

Note that every vertex in $Y$ is adjacent to either both $x_{1}$ and $x_{2}$, or to both $x_{3}$ and $x_{4}$.

Suppose $y\in V_{1234}$. We claim that the degree of $y$ is exactly 4. If $y$ is adjacent to some $z\in Y\backslash\{y\}$, then we may assume, without loss of generality, that $z$ is adjacent to both $x_{1}$ and $x_{2}$. However, this implies $\{y,z,x_{1},x_{2}\}$ is a $K_{4}$ in $G$ which is a contradiction. Consequently, $d(y)=4$ whenever $y\in V_{1234}$, and $y$ lies in exactly two triangles: $yx_{1}x_{2}$ and $yx_{3}x_{4}$.

We now focus on

[TABLE]

The intersection $N(x_{1})\cap N(x_{3})$ must contain an edge $\alpha\beta$. This edge has both endpoints in $Y$ since the only common neighbor of $x_{1}$ and $x_{3}$ in $N[x]$ is $x$. Because $\alpha$ and $\beta$ are adjacent vertices in $Y$, neither can be in $V_{1234}$. Thus, $\alpha,\beta\in V_{123}\cup V_{134}$. If $\alpha$ and $\beta$ are both in $V_{123}$, then $\{\alpha,\beta,x_{1},x_{2}\}$ is a $K_{4}$. A similar argument shows $\alpha$ and $\beta$ cannot both be in $V_{134}$. We may assume that $\alpha\in V_{123}$ and $\beta\in V_{134}$, which shows $V_{123}\neq\emptyset$ and $V_{134}\neq\emptyset$. By symmetry, $V_{124}$ and $V_{234}$ are not empty.

It is proved in [1] that $G[Y^{\prime}]$ is a complete bipartite graph with parts

$V_{123}\cup V_{124}$ and $V_{134}\cup V_{234}$.

This determines the graph $G$ up to the number of vertices in the parts $V_{123}$, $V_{124}$, $V_{134}$, and $V_{234}$ (which are all not empty), and $V_{1234}$ (which may be empty).

Let $y_{123}\in V_{123}$, and define $y_{124}$, $y_{134}$, and $y_{234}$ similarly. The subgraph $G_{1}^{\prime}$ of $G$ induced by $N[x]\cup\{y_{123},y_{124},y_{134},y_{234}\}$ has 9 vertices and 14 triangles; it is exactly one of the graphs appearing in the proof of Theorem 8 in [1]. These triangles are

$xx_{1}x_{2}$, $xx_{3}x_{4}$, $x_{1}x_{2}y_{123}$, $x_{1}x_{2}y_{124}$, $x_{3}x_{4}y_{134}$, $x_{3}x_{4}y_{234}$ $y_{123}y_{134}x_{1}$, $y_{123}y_{134}x_{3}$,

$y_{134}y_{124}x_{1}$, $y_{134}y_{124}x_{4}$, $y_{124}y_{234}x_{2}$, $y_{124}y_{234}x_{4}$, $y_{123}y_{234}x_{2}$, $y_{123}y_{234}x_{3}$.

We now estimate the number of triangles in $G$ by adding back the remaining $n-9$ vertices in $V(G)\backslash V(G_{1}^{\prime})$ one by one. At each step, we count the number of triangles that contain the new added vertex and vertices in $V(G_{1}^{\prime})$. By counting triangles in this way, we never count the same triangle more than once.

If a vertex $y$ is added to $V_{1234}$, then we obtain exactly two new triangles: $yx_{1}x_{2}$ and $yx_{3}x_{4}$. We can add any number of vertices to $V_{1234}$ and the resulting graph is $K_{4}$-saturated. Now suppose a vertex $y$ is added to $V_{123}$ (the other cases are the same by symmetry). This creates at least five new triangles: $yx_{1}x_{2}$, $yy_{134}x_{1}$, $yy_{134}x_{3}$, $yy_{234}x_{2}$, and $yy_{234}x_{3}$. The conclusion is that

[TABLE]

We have $k_{3}(G)=2n-4$ if and only if all of the $n-9$ vertices $V(G)\backslash V(G_{1}^{\prime})$ are contained in $V_{1234}$. This is precisely the graph $H_{4}(n)$.

Method and Notation For Triangle Counting: The counting method used in the last three paragraphs of the proof of Lemma 3.3 will be used multiple times in the proof of Lemmas 3.4 and 3.5. We find a small subgraph $G_{i}^{\prime}$ of $G$, count the number of triangles in $G_{i}^{\prime}$, and then count the number of triangles created when a vertex $y$ is added to $G_{i}^{\prime}$. The crucial point is that when a vertex $y$ is added, we are only counting triangles that contain $y$ and vertices in $G_{i}^{\prime}$. This means that we never count the same triangle more than once. In an effort to be concise, we will always write $y$ for the added vertex, and then list the triangles in the following way. If $y$ is added to $V_{S}$ and this creates triangles $y\alpha_{1}\beta_{1}$, $y\alpha_{2}\beta_{2},\dots,y\alpha_{k}\beta_{k}$, we will write

[TABLE]

We preface this with the statement “When $y$ is added to $V_{S}$,” and then list the triangles containing $y$ using the notation shown above.

Lemma 3.4

Suppose $n\geq 12$. If the number of edges in $N(x)$ is 3, then

[TABLE]

Proof of Lemma 3.4. Since $N(x)$ must be triangle free, the three edges in $N(x)$ either form a star or a path of length three.

Suppose first the edges in $N(x)$ are $x_{1}x_{2}$, $x_{1}x_{3}$, and $x_{1}x_{4}$. Since every edge in $N(x)$ has $x_{1}$ as an endpoint and every vertex $y\in Y$ must be joined to an edge in $N(x)$, vertex $x_{1}$ is adjacent to all vertices in $G$. Let $H$ be the subgraph of $G$ induced by $N(x_{1})$. Then $H$ is a $K_{3}$-saturated $(n-1)$-vertex graph with $\delta(H)=3$. Since $n\geq 11$, a result of Duffus and Hanson [6] implies that $H$ has at least $3(n-1)-15=3n-18$ edges. Each of these edges forms a triangle with $x$ and so $k_{3}(G)\geq 3n-18$.

Assume now the edges in $N(x)$ are $x_{1}x_{2}$, $x_{2}x_{3}$, and $x_{3}x_{4}$. First we show that $V_{12}$ and $V_{34}$ are empty. We prove this in full, but then we will make use of the results proved in [1] concerning adjacencies between a $V_{S}$ and a $V_{T}$. If $y\in V_{12}$, then $y$ is not adjacent to $x_{4}$, and so there must be an edge $\alpha\beta$ in the intersection $N(y)\cap N(x_{4})$. Since $y$ and $x_{4}$ have no common neighbors in $N[x]$, the endpoints of $\alpha\beta$ lie in $Y$. If $\alpha$ or $\beta$ is adjacent to both $x_{1}$ and $x_{2}$, then either $\{x_{1},x_{2},y,\alpha\}$ or $\{x_{1},x_{2},y,\beta\}$ is a $K_{4}$. The sets $N(\alpha)\cap N(x)$ and $N(\beta)\cap N(x)$ must contain at least one of the edges in $N(x)$, other than $x_{1}x_{2}$. All of the edges in $N(x)$, other than $x_{1}x_{2}$, have $x_{3}$ as an endpoint and so $\alpha$ and $\beta$ are both adjacent to $x_{3}$, but then $\{\alpha,\beta,x_{3},x_{4}\}$ is a $K_{4}$. The conclusion is that $V_{12}=\emptyset$. By symmetry, $V_{34}=\emptyset$.

As in the proof of Lemma 3.3, if $y\in V_{1234}$, then $d(y)=4$ and $N(y)=\{x_{1},x_{2},x_{3},x_{4}\}$. Let $Y^{\prime}=Y\backslash V_{1234}$. Because $V_{12}=V_{34}=\emptyset$, we have

[TABLE]

If $y$ and $z$ are adjacent vertices with $y,z\in V_{23}\cup V_{123}\cup V_{234}$, then $\{y,z,x_{2},x_{3}\}$ is a $K_{4}$. Thus,

[TABLE]

Similar arguments show that

[TABLE]

Summarizing, $V_{S}\sim V_{T}$ if the intersection $S\cap T$ does not contain one of the pairs $\{1,2\}$, $\{2,3\}$, or $\{3,4\}$. Also, $V_{S}\nsim V_{T}$ if $S\cap T$ contains at least one of the pairs $\{1,2\}$, $\{2,3\}$, or $\{3,4\}$. We represent this using a graph. A solid edge between $V_{S}$ and $V_{T}$ indicates $V_{S}\sim V_{T}$, a dashed edge indicates $V_{S}\nsim V_{T}$, and no edge between $V_{S}$ and $V_{T}$ indicates that it is possible for a vertex in $V_{S}$ to be adjacent or not adjacent to vertex in $V_{T}$ (this last possibility does not occur here, but will occur in the proof of Lemma 3.5).

V_{123}$$V_{124}$$V_{134}$$V_{234}$$V_{23}

Adjacencies among the $V_{S}$ and $V_{T}$

From Lemma 3.1, we have that two Rules for this graph are

V_{123}$$V_{124}$$V_{134}$$V_{234}$$V_{23}$$V_{123}$$V_{124}$$V_{134}$$V_{234}$$V_{23}

Rule 1 (on the left) and Rule 2 (on the right)

By symmetry (which is highlighted by the way we have chosen to show Rules 1 and 2), we have $V_{234}\rightarrow V_{124}$ and $V_{134}\rightarrow V_{23}\cup V_{123}\cup V_{124}$. We also refer to the assertion $V_{234}\rightarrow V_{124}$ as Rule 1, and the assertion $V_{134}\rightarrow V_{23}\cup V_{123}\cup V_{124}$ as Rule 2.

We take a moment to carefully show how Lemma 3.1 gives Rule 2. Choosing $S=\{1,2,4\}$ and $i=3$, we find all $T$ for which $3\in T$, and $|T\cap\{1,2\}|\leq 1$. These two conditions fail for $\{1,2,3\}$, but hold for $\{2,3\}$, $\{1,3,4\}$, and $\{2,3,4\}$. Thus, $V_{124}\rightarrow V_{23}\cup V_{134}\cup V_{234}$.

The structure of $G$ is now determined up to the sizes of the parts $V_{1234}$, $V_{23}$, $V_{123}$, $V_{124}$, $V_{134}$, and $V_{234}$.

If all of the vertices in $Y$ belong to $V_{1234}$, then $k_{3}(G)\geq 3+3(n-5)=3n-12$ and we are done. Let $Y^{\prime}:=Y\backslash V_{1234}$ and assume $Y^{\prime}\neq\emptyset$.

Case 1: $V_{23}\neq\emptyset$.

Let $y_{23}\in V_{23}$. Since $\delta(G)\geq 4$, $y_{23}$ has at least two other neighbors aside from $x_{2}$ and $x_{3}$. Let $z_{1}$, $z_{2}$ be two other neighbors of $y_{23}$. These two neighbors must be in the union $V_{124}\cup V_{134}$.

First suppose $z_{1}:=z_{124}\in V_{124}$ and $z_{2}:=z_{134}\in V_{134}$. Let $G_{2}^{\prime}$ be the subgraph of $G$ induced by $N[x]\cup\{y_{23},z_{124},z_{134}\}$. This subgraph has 8 vertices and 11 triangles (see the Appendix for the list of triangles). When $y$ is added to $V_{S}$,

$V_{23}~{}:~{}yx_{2}x_{3},yz_{124}x_{2},yz_{134}x_{3}$, $yz_{124}z_{134}$

$V_{123}~{}:~{}yx_{1}x_{2},yx_{2}x_{3},yx_{1}z_{134},yx_{3}z_{134}$ ($V_{234}$ is the same by symmetry)

$V_{124}~{}:~{}yx_{1}x_{2},yx_{2}y_{23},yz_{134}y_{23}$ ($V_{134}$ is the same by symmetry)

The conclusion is that

[TABLE]

Now suppose that $z_{1}:=z_{124}$ and $z_{2}:=z_{124}^{\prime}$ are in $V_{124}$, and that $V_{134}=\emptyset$. If $V_{123}\neq\emptyset$, then by Rule 2, $V_{123}\rightarrow V_{134}$ implying $V_{134}\neq\emptyset$, a contradiction. Therefore, $V_{123}=\emptyset$.

We consider two subcases.

Subcase 1: $V_{234}\neq\emptyset$.

Let $z_{234}\in V_{234}$ and $G_{3}^{\prime}$ be the subgraph of $G$ induced by $N[x]\cup\{y_{23},z_{124},z_{124}^{\prime},z_{234}\}$. This subgraph has 9 vertices and 14 triangles. When $y$ is added to $V_{S}$,

$V_{23}~{}:~{}yx_{2}x_{3},yz_{124}x_{2},yz_{124}^{\prime}x_{2}$       $V_{124}~{}:~{}yx_{1}x_{2},yy_{23}x_{2},yz_{234}x_{2},yz_{234}x_{4}$

$V_{234}~{}:~{}yx_{2}x_{3},yx_{3}x_{4},yz_{124}x_{2},yz_{124}^{\prime}x_{2}$

We conclude that $k_{3}(G)\geq k_{3}(G_{3}^{\prime})+3(n-9)=14+3(n-9)=3n-13$.

Subcase 2: $V_{234}=\emptyset$.

The neighborhood $x_{1}x_{4}$ must contain an edge, say $\alpha\beta$. This edge cannot use the vertex $x$ since $x_{1}$ is not adjacent to $x_{3}$, and $x_{4}$ is not adjacent to $x_{2}$. Furthermore, neither endpoint is in $V_{23}$ because $x_{1}$ is not adjacent to any vertex in $V_{23}$. We conclude that the endpoints of $\alpha$ and $\beta$ must both be in $V_{124}$, but then $\{\alpha,\beta,x_{1},x_{2}\}$ is a $K_{4}$.

Case 2: $V_{23}=\emptyset$.

Here we will consider two subcases.

Subcase 1: $V_{123}\neq\emptyset$

Let $y_{123}\in V_{123}$. By Rule 1, $V_{123}\rightarrow V_{134}$ so there is a $y_{134}\in V_{134}$, and $y_{123}$ is adjacent to $y_{134}$.

If $V_{234}\neq\emptyset$, say $y_{234}\in V_{234}$, then by Rule 1, $V_{234}\rightarrow V_{124}$. Let $y_{124}\in V_{124}$ be a neighbor of $y_{234}$. Let $G_{4}^{\prime}$ be the subgraph of $G$ induced by $N[x]\cup\{y_{123},y_{134},y_{234},y_{124}\}$. This graph has 9 vertices and 15 triangles. When $y$ is added to $V_{S}$,

$V_{123}~{}:~{}yx_{1}x_{2},yx_{2}x_{3},yy_{134}x_{1},yy_{134}x_{3}$ ($V_{234}$ is the same by symmetry)

$V_{134}~{}:~{}yx_{3}x_{4},yy_{123}x_{1},yy_{123}x_{3},yy_{124}x_{4}$ ($V_{124}$ is the same by symmetry)

Thus,

[TABLE]

Now suppose $V_{234}=\emptyset$. Let $G_{5}^{\prime}$ be the subgraph induced by $N[x]\cup\{y_{123},y_{134}\}$. The graph $G_{5}^{\prime}$ has 7 vertices and 8 triangles. When $y$ is added to $V_{S}$,

$V_{123}~{}:~{}yx_{1}x_{2},yx_{2}x_{3},yy_{134}x_{1},yy_{134}x_{3}$       $V_{134}~{}:~{}yx_{3}x_{4},yy_{123}x_{1},yy_{123}x_{3}$

$V_{124}~{}:~{}yx_{1}x_{2},yy_{134}x_{1},yy_{134}x_{4}$

Thus, $k_{3}(G)\geq k_{3}(G^{\prime})+3(n-7)=8+3(n-7)=3n-13$.

Subcase 2: $V_{123}=\emptyset$.

By symmetry, we may assume $V_{234}=\emptyset$ (otherwise we are back in Subcase 1 with $V_{234}$ replacing $V_{123}$). Then all of the vertices of $G$ not in $N[x]$ must be in $V_{124}\cup V_{134}\cup V_{1234}$. Let $y_{124}\in V_{124}$. By Rule 1, $V_{124}\rightarrow V_{23}\cup V_{234}\cup V_{134}$, but $V_{23}=V_{234}=\emptyset$. Thus, there is a $y_{134}\in V_{134}$. If we take $G_{6}^{\prime}$ to be the subgraph of $G$ induced by $N[x]\cup\{y_{124},y_{134}\}$, then $G_{6}^{\prime}$ has 6 vertices and 7 triangles. When $y$ is added to $V_{S}$,

$V_{124}~{}:~{}yx_{1}x_{2},yy_{134}x_{1},yy_{134}x_{4}$ ($V_{134}$ is the same by symmetry)

Therefore, $k_{3}(G)\geq 3(n-7)+7=3n-14$.

Lemma 3.5

If $n\geq 15$ and the number of edges in $N(x)$ is 4, then

[TABLE]

Proof of Lemma 3.5. Since $G$ is $K_{4}$-free, $N(x)$ must be triangle-free and so the four edges in $N(x)$ form a $C_{4}$. Assume $x_{1}x_{2}$, $x_{2}x_{3}$, $x_{3}x_{4}$, and $x_{4}x_{1}$ are the four edges in $N(x)$. The set $Y$ can be partitioned into the disjoint union

[TABLE]

The first step is to deal with vertices in $V_{1234}$. If $y\in V_{1234}$, then $y$ cannot have a neighbor in $Y$, otherwise we obtain a $K_{4}$. Thus, $N(y)=\{x_{1},x_{2},x_{3},x_{4}\}$ for all $y\in V_{1234}$. Such a vertex lies in four triangles: $yx_{1}x_{2}$, $yx_{2}x_{3}$, $yx_{3}x_{4}$, and $yx_{4}x_{1}$. We will therefore assume that $V_{1234}=\emptyset$ (if $G^{\prime}$ is the subgraph of $G$ obtained by removing the vertices in $Y_{1234}$ and we can prove the result for $G^{\prime}$, then the result easily follows for $G$).

For the rest of the proof, we focus on

[TABLE]

As stated in [1], the following relationships hold among these 8 sets. Recall a solid/dashed edge indicates that all vertices of $V_{S}$ are adjacent/not adjacent to all vertices of $V_{T}$.

V_{412}$$V_{123}$$V_{234}$$V_{341}$$V_{41}$$V_{12}$$V_{23}$$V_{34}

The only missing edges in this figure are those with endpoints in $\{V_{12},V_{23},V_{34},V_{41}\}$. At this stage, we do not have enough information to determine adjacencies between these sets.

Next, we apply Lemma 3.1 to obtain Rules 3 and 4.

V_{412}$$V_{41}$$V_{12}$$V_{341}$$V_{123}$$V_{34}$$V_{23}$$V_{234}$$V_{412}$$V_{41}$$V_{12}$$V_{341}$$V_{123}$$V_{34}$$V_{23}$$V_{234}

Rule 3 (on the left) and Rule 4 (on the right)

Again, the pictures are drawn to highlight the symmetry, and we use Rule 3 and Rule 4 accordingly (for example, the assertion $V_{412}\rightarrow V_{23}\cup V_{234}\cup V_{34}$ is also called Rule 4). Rule 3 is obtained by applying Lemma 3.1 with $S=\{1,2\}$ and $i=3$, which gives $V_{12}\rightarrow V_{23}\cup V_{234}$, and then again with $S=\{1,2\}$ with $i=4$, which gives $V_{12}\rightarrow V_{41}\cup V_{341}$.

The proof of Lemma 3.5 from this point forward will be divided into cases.

Case 1: $V_{12}\cup V_{23}\cup V_{34}\cup V_{14}=\emptyset$.

Since $G$ has more than 5 vertices, we can assume there is some $y_{123}\in V_{123}$. By Rule 4, $V_{123}\rightarrow V_{41}\cup V_{341}\cup V_{34}$, but $V_{41}=V_{34}=\emptyset$, so $y_{123}$ has a neighbor $y_{134}\in V_{134}$. Hence, we know that $V_{123}\neq\emptyset$ and $V_{134}\neq\emptyset$.

Subcase 1: $V_{234}\cup V_{124}=\emptyset$.

Let $G_{7}^{\prime}$ be the subgraph of $G$ induced by $N[x]\cup\{y_{123},y_{134}\}$. Then $G_{7}^{\prime}$ has 7 vertices and 10 triangles. When $y$ is added to $V_{S}$,

$V_{123}~{}:~{}yx_{1}x_{2},yx_{2}x_{3},yy_{134}x_{1},yy_{134}x_{3}$    ($V_{341}$ is the same by symmetry)

Thus, $k_{3}(G)\geq 10+4(n-7)=4n-18\geq 2n-2$.

Subcase 2: All of $V_{123}$, $V_{134}$, $V_{234}$, and $V_{124}$ are not empty.

Before dealing with Subcase 2, we note that Subcases 1 and 2 do cover all possibilities. This is because if $V_{234}\neq\emptyset$ (or $V_{124}\neq\emptyset$), then Rule 4 and the fact that $V_{12}\cup V_{23}\cup V_{34}\cup V_{41}=\emptyset$ implies $V_{124}\neq\emptyset$ (or $V_{234}\neq\emptyset$). Therefore, it cannot be the case that exactly one of $V_{123}$, $V_{134}$, $V_{234}$, $V_{124}$ is empty.

Assume $y_{234}\in V_{234}$ and $y_{412}\in V_{412}$. Let $G_{8}^{\prime}$ be the subgraph of $G$ induced by $N[x]\cup\{y_{123},y_{234},y_{134},y_{412}\}$. This graph has 9 vertices and 16 triangles. When $y$ is added to $V_{S}$,

$V_{123}~{}:~{}yx_{1}x_{2},yx_{2}x_{3},yy_{134}x_{1},yy_{134}x_{3}$    ($V_{234}$, $V_{134}$, $V_{412}$ are the same)

Thus, $k_{3}(G)\geq 16+4(n-9)=4n-20\geq 2n-2$.

Case 2: $V_{12}\cup V_{23}\cup V_{34}\cup V_{14}\neq\emptyset$.

For Case 2 we need an additional rule which is proved in [1].

Rule 5: If $F$ is the subgraph of $G$ induced by $V_{12}\cup V_{23}\cup V_{34}\cup V_{14}$, then $F$ is a 4-partite graph that is $K_{4}$-saturated with respect to the parts.

In particular, Rule 5 implies that if one of the four parts $V_{12},V_{23},V_{34},V_{14}$ is empty, then the remaining parts induce a complete 1-partite, 2-partite, or 3-partite graph depending on the number of nonempty parts.

Subcase 1: $V_{12}\neq\emptyset$ and $V_{23}=V_{34}=V_{14}=\emptyset$.

Let $y_{12}\in V_{12}$. By Rule 3, $V_{12}\rightarrow V_{41}\cup V_{341}$ and $V_{12}\rightarrow V_{23}\cup V_{234}$, but $V_{41}=V_{23}=\emptyset$ so there must be vertices $y_{234}\in V_{234}$ and $y_{134}\in V_{134}$ that are both adjacent to $y_{12}$. Let $G_{9}^{\prime}$ be the subgraph of $G$ induced by $N[x]\cup\{y_{12},y_{234},y_{134}\}$. Then $G^{\prime}$ has 11 triangles and 8 vertices. When $y$ is added to $V_{S}$,

$V_{12}~{}:~{}yx_{1}x_{2},yy_{234}x_{2},yy_{134}x_{1}$            $V_{123}~{}:~{}yx_{1}x_{2},yx_{2}x_{3},yy_{134}x_{1}$ ($V_{412}$ is the same)

$V_{234}~{}:~{}yx_{2}x_{3},yx_{3}x_{4},yy_{12}x_{2}$ ($V_{341}$ is the same)

We conclude that

[TABLE]

where we have used the assumption $n\geq 15$ for the last inequality.

Subcase 2: $V_{12}\neq\emptyset$ and $V_{23}\neq\emptyset$, $V_{34}=V_{14}=\emptyset$.

Let $y_{12}\in V_{12}$ and $y_{23}\in V_{23}$. By Rule 5, any vertex in $V_{12}$ will be adjacent to all vertices in $V_{23}$. By Rule 3, $V_{12}\rightarrow V_{41}\cup V_{341}$ so, since $V_{41}=\emptyset$, there is a vertex $y_{134}\in V_{134}$. This vertex must be adjacent to both $y_{12}$ and $y_{23}$. Let $G_{10}^{\prime}$ be the subgraph of $G$ induced by $N[x]\cup\{y_{12},y_{23},y_{134}\}$. Then $G_{10}^{\prime}$ has 8 vertices and 12 triangles. When $y$ is added to $V_{S}$,

$V_{12}$ :  $yx_{1}x_{2}$ $yy_{23}x_{2}$, $yy_{134}x_{1}$ ($V_{23}$ is the same),

$V_{123}$: $yx_{1}x_{2}$, $yx_{2}x_{3}$, $yy_{134}x_{1}$,$yy_{134}x_{4}$           $V_{134}$ :$yx_{1}x_{4}$, $yx_{3}x_{4}$, $yy_{12}x_{1}$,

$V_{234}$: $yx_{2}x_{3}$, $yx_{3}x_{4}$, $yy_{12}x_{2}$ ($V_{412}$ is the same).

Therefore, $k_{3}(G)\geq 12+3(n-8)=3n-12\geq 2n-2$.

Subcase 3: $V_{12}\neq\emptyset$ and $V_{34}\neq\emptyset$, $V_{23}=V_{14}=\emptyset$.

Let $y_{12}\in V_{12}$ and $y_{34}\in V_{34}$. By Rule 3, $V_{12}\rightarrow V_{23}\cup V_{234}$, $V_{12}\rightarrow V_{41}\cup V_{341}$, but $V_{23}=V_{14}=\emptyset$. Thus, there are $y_{134}\in V_{134}$ and $y_{234}\in V_{234}$ with both $y_{134}$ and $y_{234}$ adjacent to $y_{12}$. Also by Rule 3, $V_{34}\rightarrow V_{23}\cup V_{123}$ and $V_{34}\rightarrow\cup V_{41}\cup V_{412}$. This implies there are vertices $y_{123}\in V_{123}$ and $y_{412}\in V_{412}$ where both of these vertices are adjacent to $y_{34}$. Let $G_{11}^{\prime}$ be the subgraph of $G$ induced by $N[x]\cup\{y_{12},y_{34},y_{134},y_{234},y_{123},y_{412}\}$. Then $G^{\prime}$ has 11 vertices and has 22 triangles. When $y$ is added to $V_{S}$,

$V_{12}$: $yx_{1}x_{2}$ $yy_{134}x_{1}$, $yy_{234}x_{2}$ ($V_{34}$ is the same),

$V_{123}$: $yx_{1}x_{2}$, $yx_{2}x_{3}$, $yy_{34}x_{3}$, $yy_{134}x_{1}$, $yy_{134}x_{3}$ ($V_{234}$, $V_{134}$, and $V_{124}$ are the same).

Thus, $k_{3}(G)\geq k_{3}(G_{11}^{\prime})+3(n-11)=3n-11\geq 2n-2$.

Subcase 4: $V_{12}\neq\emptyset$ , $V_{23}\neq\emptyset$, $V_{34}\neq\emptyset$, $V_{14}=\emptyset$.

Let $y_{12}\in V_{12}$, $y_{23}\in V_{23}$, and $y_{34}\in V_{34}$. Note that $V_{12}\cup V_{23}\cup V_{34}$ is a complete 3-partite graph by Rule 5. By Rule 3, $y_{34}\rightarrow V_{14}\cup V_{124}$, but $V_{14}=\emptyset$. Therefore, there is a $y_{124}\in V_{124}$ and this vertex is adjacent to both $y_{34}$ and $y_{23}$. Let $G_{12}^{\prime}$ be the subgraph of $G$ induced by $N[x]\cup\{y_{12},y_{23},y_{34},y_{124}\}$. Then $G_{12}^{\prime}$ has 9 vertices and has 15 triangles. When $y$ is added to $V_{S}$,

$V_{12}$ :  $yx_{1}x_{2}$ $yy_{23}x_{2}$, $yy_{23}y_{34}$     $V_{23}$ :  $yx_{2}x_{3}$, $yy_{34}x_{3}$, $yy_{124}x_{2}$, $yy_{34}y_{12}$, $yy_{34}y_{124}$

$V_{34}$ :  $yx_{3}x_{4}$, $yy_{23}x_{3}$, $yy_{124}x_{4}$, $yy_{23}y_{12},yy_{23}y_{124}$       $V_{123}$ : $yx_{1}x_{2}$ $yx_{2}x_{3}$, $yy_{34}x_{3}$

$V_{234}$ :  $yx_{2}x_{3}$, $yx_{3}x_{4}$, $yy_{12}y_{2}$,       $V_{124}$ : $yx_{1}x_{2}$, $yx_{1}x_{4}$, $yy_{23}x_{2}$, $yy_{34}x_{4}$

$V_{341}$: $yx_{3}x_{4}$, $yx_{1}x_{4}$, $yy_{12}x_{1}$.

Hence, $k_{3}(G)\geq k_{3}(G_{12}^{\prime})+3(n-9)=3n-12$.

Subcase 5: Each of $V_{12}$, $V_{23}$, $V_{34}$, $V_{14}$ is not empty.

For this subcase, we will count triangles in $G$ in a different way than the previous subcases. Let

$X=\{x_{1},x_{2},x_{3},x_{4}\}$, $S=V_{12}\cup V_{23}\cup V_{34}\cup V_{41}$, and $T=V_{123}\cup V_{234}\cup V_{341}\cup V_{412}$.

Claim 3.6

The number of triangles that contain at least one vertex in $X$ is at least $2n-6$.

Proof of Claim 3.6. There are four triangles that contain $x$, $2|T|$ triangles that contain two vertices in $X$ and one in $T$, and $|S|$ triangles that contain two vertices in $X$ and one in $S$. By Rule 3, a vertex $y_{12}\in V_{12}$ lies in a triangle of the form $y_{12}zx_{2}$ where $z$ is some vertex in $V_{23}\cup V_{234}$. Similar statements hold for vertices in $V_{23}$, $V_{34}$, and $V_{41}$. Altogether, we have $4+2|T|+2|S|=4+2(n-5)=2n-6$ triangles.

If $|S|=4$, then we let $V_{ij}=\{y_{ij}\}$ for $(i,j)\in\{(1,2),(2,3),(3,4),(4,1)\}$. Since $G$ is $K_{4}$-free, Rule 5 implies that $S$ induces a $K_{4}-e$. The two distinct cases, up to symmetry, are the missing edge $e$ is $y_{12}y_{23}$ or $y_{12}y_{34}$.

Suppose first $e=y_{12}y_{34}$. For any $y_{123}\in V_{123}$, we have that $y_{123}y_{34}y_{14}$ is a triangle. Likewise, $y_{234}\in V_{234}$ implies $y_{234}y_{12}y_{14}$ is a triangle, $y_{341}\in V_{341}$ implies $y_{341}y_{23}y_{12}$ is a triangle, and $y_{412}\in V_{234}$ implies $y_{412}y_{34}y_{12}$ is a triangle. The two triangles $y_{12}y_{23}y_{34}$ and $y_{12}y_{41}y_{34}$ have no vertex in $X$. The number of triangles containing no vertex in $X$ is at least

[TABLE]

Thus, by Claim 3.6, $G$ contains at least $3n-13\geq 2n-2$ triangles.

Now suppose $e=y_{12}y_{23}$. By Rule 3, $y_{12}$ has a neighbor in $V_{23}\cup V_{234}=\{y_{23}\}\cup V_{234}$, but $y_{12}$ is not adjacent to $y_{23}$ so $V_{234}\neq\emptyset$. Similarly, $y_{23}$ has a neighbor in $V_{12}\cup V_{412}$ but $y_{23}$ is not adjacent to $y_{12}$ so $V_{412}\neq\emptyset$. For any $y_{234}\in V_{234}$, $y_{234}y_{12}y_{41}$ is a triangle. Likewise, $y_{412}\in V_{412}$ implies $y_{412}y_{23}y_{34}$ is a triangle, and $y_{123}\in V_{123}$ implies $y_{123}y_{34}y_{41}$ is a triangle. Therefore, there are at least

[TABLE]

triangles in $G$ with no vertex in $X$. Since $V_{234}$ and $V_{412}$ are not empty,

[TABLE]

Hence, $\gamma\geq 2+(n-9)-(n-11)=4$. Combining this with Claim 3.6 gives

[TABLE]

The final possibility is if at least one of the parts $V_{12}$, $V_{23}$, $V_{34}$, $V_{41}$ contains 2 or more vertices. Assume $|V_{12}|>1$ and let $y_{12}^{1},y_{12}^{2}$ be distinct vertices in $V_{12}$. Let $y_{23}\in V_{23}$, $y_{34}\in V_{34}$, and $y_{41}\in V_{41}$. The set $\{y_{12}^{1},y_{23},y_{34},y_{41}\}$ cannot form a $K_{4}$ and so by Rule 5, this set of four vertices induces a $K_{4}-e$. There are two triangles using these vertices. Similarly, there are two triangles using the vertices $\{y_{12}^{2},y_{23},y_{34},y_{41}\}$ and regardless of which pair of vertices is not adjacent in this set, one of the triangles must contain $y_{12}^{2}$. Thus, we have 3 distinct triangles contained in $S$. By Claim 3.6,

[TABLE]

This completes the proof of Lemma 3.5.

Combining Lemmas 3.2, 3.3, 3.4, and 3.5 implies Theorem 1.4.

4 Proof of Proposition 1.9

Let $s>3$ and $t\geq 6\binom{s-2}{2}$ be integers. Suppose $G$ is an $n$-vertex $K_{s}$-saturated graph with minimum degree $t$ where $n\geq 2s-2$. We must show that $G$ has at least $\binom{s-2}{2}(n-2)$ triangles.

First assume every edge of $G$ lies in a triangle. If $t(e)$ is the number of triangles that contain the edge $e$, then the number of triangles in $G$ is

[TABLE]

Now assume there is an edge $xy$ in $G$ that does not lie in any triangle. Let $A=N(x)$ and $B=N(y)$. Because $xy$ is not in any triangle, $A\cap B=\emptyset$. If $a\in A$, then $N(y)\cap N(a)$ must contain a copy of $K_{s-2}$. The number of triangles that contain $a$ and an edge from this $K_{s-2}$ is $\binom{s-2}{2}$. The same argument applies to a vertex $b\in B$. Let $C=V(G)\backslash(\{x,y\}\cup A\cup B)$. If $c\in C$, then both $N(c)\cap N(x)$ and $N(c)\cap N(y)$ must contain a copy of $K_{s-2}$. These two copies of $K_{s-2}$ cannot share any edges because $A\cap B=\emptyset$. Thus, $G$ must contain at least

[TABLE]

triangles. This completes the proof of Proposition 4.

5 Appendix

For $G_{1}^{\prime}$, the edges in $N(x)$ are $x_{1}x_{2}$ and $x_{3}x_{4}$.

14 Triangles in $G_{1}^{\prime}$: $xx_{1}x_{2}$, $xx_{3}x_{4}$, $y_{123}x_{1}x_{2}$, $y_{124}x_{1}x_{2}$, $y_{134}x_{3}x_{4}$, $y_{234}x_{3}x_{4}$, $y_{123}y_{134}x_{1}$, $y_{123}y_{134}x_{3}$, $y_{123}y_{234}x_{2}$, $y_{123}y_{234}x_{3}$, $y_{124}y_{134}x_{1}$, $y_{124}y_{134}x_{4}$, $y_{124}y_{234}x_{2}$, $y_{124}y_{234}x_{4}$

For $G_{2}^{\prime}$ through $G_{6}^{\prime}$, the edges in $N(x)$ are $x_{1}x_{2}$, $x_{2}x_{3}$, and $x_{3}x_{4}$.

V_{123}$$V_{124}$$V_{134}$$V_{234}$$V_{23}

11 Triangles in $G_{2}^{\prime}$: $xx_{1}x_{2}$, $xx_{2}x_{3}$, $xx_{3}x_{4}$, $y_{23}x_{2}x_{3}$, $z_{124}x_{1}x_{2}$, $z_{134}x_{3}x_{4}$, $y_{23}z_{124}x_{2}$,

$y_{23}z_{134}x_{3}$, $z_{124}z_{134}x_{1}$, $z_{124}z_{134}x_{4}$, $y_{23}z_{124}z_{134}$

14 Triangles in $G_{3}^{\prime}$: $xx_{1}x_{2}$, $xx_{2}x_{3}$, $xx_{3}x_{4}$, $y_{23}x_{2}x_{3}$, $z_{124}x_{1}x_{2}$, $z_{124}^{\prime}x_{1}x_{2}$, $z_{234}x_{2}x_{3}$,

$z_{234}x_{3}x_{4}$, $y_{23}z_{124}x_{2}$, $y_{23}z_{124}^{\prime}x_{2}$, $z_{124}z_{234}x_{2}$, $z_{124}z_{234}x_{4}$, $z_{124}^{\prime}z_{234}x_{2}$, $z_{124}^{\prime}z_{234}x_{4}$

15 Triangles in $G_{4}^{\prime}$: $xx_{1}x_{2}$, $xx_{2}x_{3}$, $xx_{3}x_{4}$, $y_{123}x_{1}x_{2}$, $y_{123}x_{2}x_{3}$, $y_{124}x_{1}x_{2}$, $y_{234}x_{2}x_{3}$, $y_{234}x_{3}x_{4}$, $y_{134}x_{3}x_{4}$, $y_{123}x_{1}y_{134}$, $y_{123}x_{3}y_{134}$, $y_{124}x_{1}y_{134}$, $y_{124}x_{4}y_{134}$, $y_{124}x_{2}y_{234},y_{124}x_{4}y_{234}$

8 Triangles in $G_{5}^{\prime}$: $xx_{1}x_{2}$, $xx_{2}x_{3}$, $xx_{3}x_{4}$, $y_{123}x_{1}x_{2}$, $y_{123}x_{2}x_{3}$, $y_{134}x_{3}x_{4}$, $y_{123}y_{134}x_{1}$, $y_{123}y_{134}x_{3}$

7 Triangles in $G_{6}^{\prime}$: $xx_{1}x_{2}$, $xx_{2}x_{3}$, $xx_{3}x_{4}$, $y_{124}x_{1}x_{2}$, $y_{134}x_{3}x_{4}$, $y_{124}y_{134}x_{1}$, $y_{124}y_{134}x_{4}$

For $G_{7}^{\prime}$ through $G_{12}^{\prime}$, the edges in $N(x)$ are $x_{1}x_{2}$, $x_{2}x_{3}$, $x_{3}x_{4}$, and $x_{4}x_{1}$.

V_{412}$$V_{123}$$V_{234}$$V_{341}$$V_{41}$$V_{12}$$V_{23}$$V_{34}

10 Triangles in $G_{7}^{\prime}$: $xx_{1}x_{2}$, $xx_{2}x_{3}$, $xx_{3}x_{4}$, $xx_{4}x_{1}$, $y_{123}x_{1}x_{2}$, $y_{123}x_{2}x_{3}$, $y_{134}x_{3}x_{4}$, $y_{134}x_{4}x_{1}$, $y_{123}y_{134}x_{1}$, $y_{123}y_{134}x_{3}$

16 Triangles in $G_{8}^{\prime}$: $xx_{1}x_{2}$, $xx_{2}x_{3}$, $xx_{3}x_{4}$, $xx_{4}x_{1}$, $y_{123}x_{1}x_{2}$, $y_{123}x_{2}x_{3}$, $y_{412}x_{1}x_{2}$, $y_{412}x_{4}x_{1}$, $y_{234}x_{2}x_{3}$, $y_{234}x_{3}x_{4}$, $y_{134}x_{3}x_{4}$, $y_{134}x_{4}x_{1}$, $y_{123}y_{134}x_{1}$, $y_{123}y_{134}x_{3}$, $y_{412}y_{234}x_{2}$, $y_{412}y_{234}x_{4}$

11 Triangles in $G_{9}^{\prime}$: $xx_{1}x_{2}$, $xx_{2}x_{3}$, $xx_{3}x_{4}$, $xx_{4}x_{1}$, $y_{12}x_{1}x_{2}$, $y_{134}x_{3}x_{4}$, $y_{134}x_{4}x_{1}$, $y_{234}x_{2}x_{3}$, $y_{234}x_{3}x_{4}$, $y_{12}y_{134}x_{1}$, $y_{12}y_{234}x_{2}$

12 Triangles in $G_{10}^{\prime}$: $xx_{1}x_{2}$, $xx_{2}x_{3}$, $xx_{3}x_{4}$, $xx_{4}x_{1}$, $y_{12}x_{1}x_{2}$, $y_{23}x_{2}x_{3}$, $y_{134}x_{3}x_{4}$, $y_{134}x_{4}x_{1}$, $y_{12}y_{134}x_{1}$, $y_{23}y_{134}x_{1}$, $y_{12}y_{23}x_{2}$, $y_{12}y_{23}y_{134}$

22 Triangles in $G_{11}^{\prime}$: $xx_{1}x_{2}$, $xx_{2}x_{3}$, $xx_{3}x_{4}$, $xx_{4}x_{1}$, $y_{12}x_{1}x_{2}$, $y_{34}x_{3}x_{4}$, $y_{123}x_{1}x_{2}$, $y_{123}x_{2}x_{3}$, $y_{412}x_{4}x_{1}$, $y_{412}x_{1}x_{2}$, $y_{234}x_{2}x_{3}$, $y_{234}x_{3}x_{4}$, $y_{341}x_{3}x_{4}$, $y_{341}x_{4}x_{1}$, $y_{12}y_{134}x_{1}$, $y_{12}y_{234}x_{2}$, $y_{34}y_{124}x_{4}$, $y_{34}y_{123}x_{3}$, $y_{134}y_{123}x_{1}$, $y_{134}y_{123}x_{3}$, $y_{234}y_{412}x_{2}$, $y_{234}y_{412}x_{4}$

15 Triangles in $G_{12}^{\prime}$: $xx_{1}x_{2}$, $xx_{2}x_{3}$, $xx_{3}x_{4}$, $xx_{4}x_{1}$, $y_{12}x_{1}x_{2}$, $y_{23}x_{2}x_{3}$, $y_{34}x_{3}x_{4}$, $y_{124}x_{1}x_{2}$, $y_{124}x_{4}x_{1}$, $y_{12}y_{23}x_{2}$, $y_{23}y_{34}x_{3}$, $y_{124}y_{23}x_{2}$, $y_{124}y_{34}x_{4}$, $y_{12}y_{23}y_{34}$, $y_{124}y_{23}y_{34}$