Proof of claim. Suppose the claim is false. Let $D\,:\!=\,\max \{ d(v,V_i) \colon v \in V(G), i \in [r]\}\geq n/(t-1)$ . Without loss of generality, we may assume that there exists $x_1 \in V_1$ such that $d(x_1,V_2) =D$ . Since $\Delta (G) \lt n$ , there exists $x_2 \in V_2 \setminus N(x_1)$ . Furthermore, since $\Delta (G) \lt n$ and $r =2t-1$ , we can greedily find $x_3, \ldots, x_{t-1}$ such that $S = \{x_1,\ldots, x_{t-1}\}$ is a crossing independent set.Footnote ² Without loss of generality, assume that $x_i \in V_i$ for $i \in [t-1]$ .

For each $i\in [ t]$ , let $W_i$ consist of all the vertices of $V_{r+1-i}$ that are not adjacent to any vertex in $S$ . Set $n_i \,:\!=\, |W_i|$ and without loss of generality, assume that $n_1 \ge n_2 \ge \ldots \ge n_{t}$ . Let $\ell \,:\!=\, \max \{ i \in [t] \colon n_i \gt 0 \}$ . Note that

\begin{align*} n_1+ \ldots + n_{\ell } & = n_1 + \ldots +n_{t} = \left | \bigcup _{t \le i \le r} V_{i} \setminus \bigcup _{j \in [t-1]}N(x_j) \right | \\ & \ge tn - ((t-1)\Delta (G)-d(x_1,V_2)) \gt n+D . \end{align*}

By averaging, we have

(2.1) \begin{align} n_1+ \ldots + n_{\ell -1} \gt (\ell -1) (n+D)/ \ell . \end{align}

Notice that the $\ell$ -partite subgraph of $G$ induced by $W_1,\ldots, W_{\ell }$ must be complete (otherwise one can extend $S$ into a crossing independent set of size $t+1$ , a contradiction). Hence, there exists $y \in W_{\ell }$ such that $\bigcup _{i \in [\ell -1]} W_i \subseteq N(y)$ . By the definition of $D$ , we deduce that

\begin{align*} n_1 + \ldots + n_{\ell -1} \le \sum _{i \in [\ell -1]} d(y,V_{r+1-i}) \le (\ell -1)D. \end{align*}

Together with (2.1), this implies that $D \gt n/(\ell -1)$ and so

\begin{align*} \Delta (G) \ge d(y) \ge n_1 + \ldots + n_{\ell -1} \gt (\ell -1) (n+D)/ \ell \ge n, \end{align*}

a contradiction.

Given a crossing independent set $S$ of $G$ , let $\sigma (S) \,:\!=\, \sum _{x \in S} d(x,V_S)$ where $V_S$ is the union of all $V_i$ that contain a vertex from $S$ . As mentioned in Footnote 2, we can greedily construct a crossing independent set of size $t$ . Let $S$ be a crossing independent set of size $t$ with $\sigma (S)$ maximal. Without loss of generality, $S \cap \bigcup _{i \in [t-1]} V_i= \emptyset$ .

Consider any $(t-1)$ -set $S^{\prime}\subset S$ . By Claim 2.2, there exists a vertex $y \in V_1$ that is not adjacent to any vertex in $S^{\prime}$ . Note that $S^{\prime}\cup \{y\}$ is a crossing independent set of size $t$ . Hence, $ \sigma (S) \ge \sigma (S^{\prime}\cup \{y\} ) \ge \sigma (S^{\prime}) + \sum _{x\in S^{\prime}} d(x,V_1)$ . By summing over all $(t-1)$ -sets $S^{\prime}\subset S$ , we obtain that

\begin{align*} t \sigma (S) & \ge \sum _{S^{\prime} \subset S \colon |S^{\prime}|=t-1} \sigma (S^{\prime}) + (t-1) \sum _{x \in S}d(x,V_1) = (t-2)\sigma (S) + (t-1) \sum _{x \in S}d(x,V_1)\\ & \ge (t-2)\sigma (S) + (t-1)n, \end{align*}

where the last inequality follows as every vertex in $V_1$ must be adjacent to at least one vertex in $S$ (else there exists a crossing independent set of size $t+1$ , a contradiction). Thus, $ \sigma (S) \ge (t-1) n/2$ .

As $S \cap \bigcup _{i \in [t-1]} V_i= \emptyset$ , then $\bigcup _{i \in [t-1]} V_i \subseteq \bigcup _{x \in S} N(x)$ or else there exists a crossing independent set of size $t+1$ , a contradiction. Therefore,

\begin{align*} t \Delta (G) \ge \sum _{x \in S} d(x) \ge \sigma (S) + \left |\bigcup _{i \in [t-1]} V_i \right | \ge \frac{(t-1)n}2 + (t-1) n \ge t n, \end{align*}

implying that $\Delta (G) \ge n$ , a contradiction.

Proof. Let $G$ be a $K_{t+1}$ -free $r$ -partite graph with $n$ vertices in each part. If $\delta (G)\gt \frac{3t-4}{3t-1} (rn)$ , then $\chi (G)\le t$ by Theorem 3.1. Thus $G\in \mathcal{G}(n, r, t)$ and $\delta (G)\le \delta (n, r, t)$ .

Proof. (1.2) covers the case when $a =0$ so we assume that $a \in [t-1]$ . By Corollary 3.2, it suffices to show that $\delta (n,r,t) \ge \frac{3t-4}{3t-1} rn$ . By (3.1), we have

\begin{align*} \delta (n,r,t) &\ge \left ( r - \left \lceil \frac{r}{t}\right \rceil \right ) n = \left ( r - \frac{r+a}{t} \right ) n = \left ( \frac{r}{t(3t-1)} + \frac{3t-4}{3t-1}r -\frac{a}{t} \right ) n \ge \frac{3t-4}{3t-1} rn, \end{align*}

where we use the fact that $r \ge a(3t-1)$ in the last inequality.

Proof. When $r=mt-1$ , the desired bound follows from (3.1). We thus assume $r\le mt-2$ . Let $\ell \,:\!=\, \lceil (r-1)n/(mt-2) \rceil \le n$ . Let $K\,:\!=\,K_r(n)$ and let $V_1, \ldots, V_{r}$ denote its parts. For $i \in [t-1]$ , let $B_i \,:\!=\, \{(i-1)m+1, \ldots, im\}$ . So $B_1, \ldots, B_{t-1}$ form an equipartition of $[m(t-1)]$ . For $i \in [t-1]$ , let $W_i \subset \bigcup _{j \in B_i} V_j$ be such that $|W_i \cap V_j| = \ell$ for $j \in B_i$ . Let $W_t \,:\!=\, V(K) \setminus \bigcup _{i \in [t-1]}W_i$ . Then

\begin{equation*} |W_1| = \cdots = |W_{t-1}|= \ell m \quad \text {and} \quad |W_t|= rn - m (t-1) \ell . \end{equation*}

Let $G^{\prime}$ be the complete $t$ -partite graph with parts $W_1, \ldots, W_{t}$ . We set $G \,:\!=\, K \cap G^{\prime}$ ; that is, $G$ is the graph on $V(K)$ such that, for $x \in V_i \cap W_j$ and $x^{\prime} \in V_{i^{\prime}} \cap W_{j^{\prime}}$ , we have $xx^{\prime} \in E(G)$ if and only if $i \ne i^{\prime}$ and $j \ne j^{\prime}$ . Clearly $\chi (G) \le \chi (G^{\prime}) \le t$ . If $x \in V_i$ with $i \notin [m(t-1)]$ , then as $x$ is only non-adjacent to the vertices in $W_t$ , we have that $d(x) = r n - |W_t|$ . If $x \in V_i$ with $i \in [m(t-1)]$ , then

\begin{align*} d(x) = \begin{cases} rn - (n + |W_1| - \ell ) & \text{ if } x \notin W_t,\\[4pt] rn - (n + |W_t| - ( n- \ell ) ) & \text{ if }x \in W_t. \end{cases} \end{align*}

By our choice of $\ell$ ,

\begin{align*} \delta (G) & = (r-1) n - \max \{ (m-1) \ell, (r-1)n - (m(t-1)-1) \ell \}\\[3pt] & = (r-1) n - (m-1) \ell = (r-1)n - (m-1) \left \lceil \frac{(r-1)n}{mt-2} \right \rceil, \end{align*}

as required.

Proof. Let $t^{\prime}\,:\!=\, t-a+m$ and $r^{\prime} \,:\!=\, m(t^{\prime}-1)$ . By Proposition 4.1, there exists a graph $G^{\prime} \in \mathcal{G}(n,r^{\prime},t^{\prime})$ such that

\begin{align*} \delta (G^{\prime}) = (r^{\prime}-1)n - (m-1) \left \lceil \frac{(r^{\prime}-1)n}{mt^{\prime}-2} \right \rceil \le (r^{\prime}-1)n - (m-2)n, \end{align*}

where the last inequality is due to fact that

\begin{align*} (m-1)(r^{\prime}-1) - (mt^{\prime}-2)(m-2) = m(t^{\prime}-m+2)-3 = (t-a+2)m-3 \gt 0. \end{align*}

We now construct a graph $G \in \mathcal{G}(n,r,t)$ from $G^{\prime}$ as follows. Let $W_{a-m+1} \,:\!=\, V(G^{\prime})$ . Let $W_1, \ldots, W_{a-m}$ be vertex sets each of size $(m-1)n$ such that $W_1, \ldots, W_{a-m+1}$ are disjoint. Let $G$ be the resulting graph on $\bigcup _{i \in [a-m+1]} W_i$ obtained from $G^{\prime}$ by adding edges $xx^{\prime}$ for all $x \in W_i$ and $x^{\prime} \in W_{i^{\prime}}$ with $i \ne i^{\prime}$ . Note that $\chi (G) = \chi (G^{\prime}) + a-m \le t$ . Since each $W_i$ with $i \in [a-m]$ can be partitioned into $m-1$ vertex classes of size $n$ , we deduce that $G \in \mathcal{G}(n,r,t)$ .

For $x \in V(G) \setminus V(G^{\prime})$ , $x$ is non-adjacent to the vertices in the class $W_i$ it lies in and so $d_G(x) = r n -(m-1) n$ . For $x \in V(G^{\prime})$ , we have $d_G(x) = (a-m)(m-1) n + d_{G^{\prime}}(x)$ and so the vertex $y \in V(G^{\prime})$ with smallest degree $d_G(y)$ satisfies

\begin{align*} d_G(y) = (a-m)(m-1) n + \delta (G^{\prime}) = (r-1)n - (m-1) \left \lceil \frac{(r^{\prime}-1)n}{mt^{\prime}-2} \right \rceil \le (r-1)n - (m-2)n. \end{align*}

Hence,

\begin{align*} \delta (G) = (r-1)n - (m-1) \left \lceil \frac{(r^{\prime}-1)n}{mt^{\prime}-2} \right \rceil, \end{align*}

as required.

Proof. By (3.1) and Corollary 3.2, it suffices to prove that

(4.2) \begin{align} \delta (n,r,t) \geq \frac{3t-4}{3t-1} rn. \end{align}

Proposition 4.1 and (4.1) together imply that

\begin{equation*} \delta (n,r,t)\gt (r-m)n + \frac {(a-1)(m-1)}{mt-2}n - (m-1). \end{equation*}

On the other hand, $\frac{3t-4}{3t-1} r = r-m +\frac{a}t-\frac{r}{t(3t-1)}.$ Thus, to prove (4.2), it suffices to have

\begin{align*} \frac{(a-1)(m-1)}{mt-2}n - (m-1)\ge \frac{an}t-\frac{rn}{t(3t-1)}. \end{align*}

Indeed, this is equivalent to our assumption

\begin{equation*} \frac {r}{t(3t-1)(m-1)}-\frac {a}{t(m-1)} + \frac {a-1}{mt-2} \geq \frac {1}{n}. \end{equation*}

Proof. Let $\Delta ^* \,:\!=\, (m-1) (r-1)n/(mt-2)$ . Since $\delta (n,r,t)$ is an integer, it suffices to show that $\delta (n,r,t) \le (r-1)n - \Delta ^*$ . Suppose to the contrary that $\delta (n,r, t) \gt (r-1)n - \Delta ^*$ . Let $G \in \mathcal{G}(n,r,t)$ be such that $\delta (G) =\delta (n,r, t)$ . As $G \in \mathcal{G}(n,r,t)$ , $V(G)$ can be partitioned into $r$ independent sets $V_1,\ldots, V_r$ each of size $n$ ; as $\chi (G) \le t$ , $V(G)$ can be partitioned into $t$ colour classes $W_1, \ldots, W_{t}$ . For every $x\in V_i\cap W_j$ , $x$ is non-adjacent to those vertices in $V_i \cup W_j$ and so $d(x)\leq rn -|V_i|- |W_j|+|V_i\cap W_j| =(r-1)n - |W_j|+|V_i\cap W_j|$ .

For $i\in [r]$ , let $C(i)\,:\!=\,\{j\in [t]: |V_i \cap W_j|\ne 0\}$ be the set of colours present in $V_i$ . For $j\in [t]$ , let $\text{supp}(j)\,:\!=\,\{i\in [r]: |V_i \cap W_j|\ne 0\}$ be the set of parts that colour $j$ is present. Let

\begin{align*} \Delta _j \,:\!=\, |W_j| - \min _{i\in \text{supp}(j)} |V_i \cap W_j|. \end{align*}

Thus $\delta (G)\leq \min _{j\in [t]} \{(r-1)n - \Delta _j \}$ . Hence we have for all $j \in [t]$ ,

(5.1) \begin{align} \Delta _j \lt \Delta ^* = \frac{(m-1)(r-1)n}{mt-2}. \end{align}

Proof of claim. Note that

\begin{align*} \frac{m \Delta ^*}{m-1} = \frac{m (r-1)n}{mt-2} \geq \frac{m (m-1)t n }{mt-2} \gt (m-1)n. \end{align*}

If $|\text{supp}(j)| \le m-1$ , then $|W_j|\le (m-1)n \lt \frac{m}{m-1} \Delta ^*$ as desired. Hence, we may assume that $|\text{supp}(j)| \ge m$ . Using the definition of $\Delta _j$ and (5.1), we obtain that

\begin{align*} \frac{m-1}{m} |W_j| \le \frac{|\text{supp}(j)|-1}{|\text{supp}(j)|} |W_j| \le \Delta _j \lt \Delta ^*. \end{align*}

Hence the claim follows.

Suppose that $|C(i)| =1$ for all $i \in [r]$ . Every $V_i$ is a subset of some $W_j$ and consequently, there exists $j \in [t]$ with $|W_j| \ge \lceil r/t \rceil n \ge m n$ . It follows that

\begin{equation*} \Delta _j \ge |W_j|-n \ge (m-1)n \ge (m-1)\frac { r-1}{mt-2}n = \Delta ^*, \end{equation*}

a contradiction.

Without loss of generality, we assume that $|C(1)|=s\ge 2$ . For every $j \in C(1)$ , we know that $|W_j| \le \Delta _j + |V_1 \cap W_j|$ from the definition of $\Delta _j$ . Hence,

\begin{align*} \sum _{j \in C(1)} |W_j| \le \sum _{j \in C(1)} \left( \Delta _j + |V_1 \cap W_j|\right) \stackrel{(5.1)}{\lt } s \Delta ^* + n. \end{align*}

Together with Claim 5.2, this gives

\begin{align*} rn = \sum _{j \in [t]} |W_j| & \lt s \Delta ^* + n + (t-s) \frac{m}{m-1} \Delta ^* = \frac{mt-s}{m-1}\Delta ^* + n \le \frac{mt-2}{m-1}\Delta ^* + n = r n, \end{align*}

a contradiction.

Proof of Theorem 1.2. The $m=2$ case of the theorem is precisely Lemma 2.1. For $m\geq 3$ , we may apply Corollary 3.3 (with $a\,:\!=\,1$ ) to conclude that $f(n,r,t+1)=\delta (n,r,t)$ . Then Proposition 5.1 implies $\delta (n,r,t)\leq (r-m)n=(r-\lceil r/t \rceil )n$ ; together with the lower bound in (1.3) this completes the proof.

Proof of Theorem 1.3. Under Condition (i), we first apply Corollary 3.3 to obtain that $f(n,r,t+1)=\delta (n,r,t)$ . Then Propositions 4.1 and 5.1 give the desired lower and upper bounds, respectively. Under Condition (ii), we apply Corollary 4.3 instead of Corollary 3.3.

Proof of Corollary 1.4. The case when $r \equiv 0 \pmod{3}$ follows from (1.2). The case when $r \equiv 2 \pmod{3}$ follows immediately from Theorem 1.2. If $r \equiv 1 \pmod{3}$ then $r=3m-2$ for some $m \geq 4$ . Thus Condition (i) of Theorem 1.3 holds provided that $m \geq 6$ . If $r=10$ and $n \geq 60$ or $r=13$ and $n \geq 22$ , then it is easy to check that Condition (ii) of Theorem 1.3 holds. Thus, Theorem 1.3 yields the corollary in this case.