Prague Dimension of Random Graphs

Abstract

The Prague dimension of graphs was introduced by Nešetřil, Pultr and Rödl in the 1970s. Proving a conjecture of Füredi and Kantor, we show that the Prague dimension of the binomial random graph is typically of order $n/\log n$ for constant edge-probabilities. The main new proof ingredient is a Pippenger–Spencer type edge-coloring result for random hypergraphs with large uniformities, i.e., edges of size $O(\log n)$.

Appendices

A Lower bounds: proof of Lemma 20

Proof of Lemma 20

Writing ${\mathcal {S}}$ for the event that the largest clique of $G_{n,p}$ has size at most $s=\lceil 2\log _{1/p} n\rceil$, it well-known that ${\mathcal {S}}$ holds whp (by a straightforward first moment argument). Writing ${\mathcal {E}}$ for the event that $G_{n,p}$ contains $(1\pm \epsilon )\left( {\begin{array}{c}n\\ 2\end{array}}\right) p$ edges for $\epsilon :=n^{-1/2}$, say, it is easy to see that ${\mathcal {E}}$ holds whp (using Chebychev’s inequality). Furthermore, recalling $\varphi (p)={(1-p)\log (1-p)/(p\log p)}$, the probability that $G_{n,p}$ equals any fixed spanning subgraph $G \subseteq K_n$ with $e(G) = (1\pm \epsilon )\left( {\begin{array}{c}n\\ 2\end{array}}\right) p$ edges is routinely seen to be at most

$$\begin{aligned} \begin{aligned} \Pi&:= \max _{m \in (1\pm \epsilon )\left( {\begin{array}{c}n\\ 2\end{array}}\right) p} p^m (1-p)^{\left( {\begin{array}{c}n\\ 2\end{array}}\right) -m} \\&= \max _{m \in (1\pm \epsilon )\left( {\begin{array}{c}n\\ 2\end{array}}\right) p} \exp \biggl ( -\genfrac(){0.0pt}1{n}{2}p \biggl [\frac{m}{\genfrac(){0.0pt}1{n}{2}p} + \biggl (1-\frac{m}{\genfrac(){0.0pt}1{n}{2}}\biggr )\frac{\log (1-p)}{p\log p}\biggr ] \! \cdot \! \log (1/p)\biggr )\\&\le \exp \biggl ( -\genfrac(){0.0pt}1{n}{2}p \biggl [1-\epsilon + \biggl (1-\frac{\epsilon p}{1-p}\biggr )\varphi (p)\biggr ] \! \cdot \! \log (1/p)\biggr )\\&\le \exp \Bigl (-(1-o(1)) \! \cdot \! \genfrac(){0.0pt}1{n}{2}p \bigl (1+\varphi (p)\bigr ) \! \cdot \! \log (1/p)\Bigr ), \end{aligned} \end{aligned}$$

(77)

where we used $\varphi (p) \ge 0$ as well as $\epsilon =o(1)$ and $\epsilon p/(1-p)=o(1)$ for the last inequality.

For the clique covering number $\text {cc}(G_{n,p})$, the crux is that there are at most

$$\begin{aligned} \left( {\begin{array}{c}n+s\\ s\end{array}}\right) ^T \, \le \, o(n^{sT}) \end{aligned}$$

many collections $\{C_1, \ldots , C_t\}$ with $t \le T$ that are a clique covering for some graph $G \subseteq K_n$ with largest clique of size at most s. Hence, since each clique covering uniquely determines the entire edge-set and thus the underlying spanning subgraph $G \subseteq K_n$, it follows by a union bound argument that

$$\begin{aligned} {{\mathbb {P}}{}}(\text {cc}(G_{n,p}) \le T) \; \le \; {{\mathbb {P}}{}}(\lnot {\mathcal {S}}\text { or } \lnot {\mathcal {E}}) \, + \, o(n^{sT}) \cdot \Pi . \end{aligned}$$

(78)

Note that ${{\mathbb {P}}{}}(\lnot {\mathcal {S}}\text { or } \lnot {\mathcal {E}})=o(1)$ and $s\log n \sim \left( {\begin{array}{c}s\\ 2\end{array}}\right) \cdot \log (1/p)$. In view of inequality (77), for any $\epsilon \in (0,1)$ it follows that (78) is at most o(1) when $T \le (1-\epsilon ) \cdot (1+\varphi (p)) \left( {\begin{array}{c}n\\ 2\end{array}}\right) p/\left( {\begin{array}{c}s\\ 2\end{array}}\right)$, establishing (75).

Turning to the thickness $\text {cc}_{\Delta }(G_{n,p})$, we associate each clique covering ${\mathcal {C}}$ of some graph $G \subseteq K_n$ with an auxiliary bipartite graph ${\mathcal {B}}$ on vertex-set ${[n] \cup {\mathcal {C}}}$, where ${v \in [n]}$ and ${C_i \in {\mathcal {C}}}$ are connected by an edge whenever ${v \in V(C_i)}$. If the thickness of ${\mathcal {C}}$ is at most T, then in ${\mathcal {B}}$ the degree of each $v \in [n]$ is at most $\lfloor T\rfloor$, which also gives ${|{\mathcal {C}}| \le n \lfloor T\rfloor }$. Since the structure of the auxiliary bipartite graph ${\mathcal {B}}$ uniquely determines ${\mathcal {C}}$ (as the neighbors of $C_i$ in ${\mathcal {B}}$ determine the clique vertex-set $V(C_i)$), it follows that there are at most

$$\begin{aligned} \left( {\begin{array}{c}n\lfloor T\rfloor + \lfloor T\rfloor \\ \lfloor T\rfloor \end{array}}\right) ^n \, \le \, O\bigl ((6n)^{nT}\bigr ) \end{aligned}$$

many collections ${\mathcal {C}}$ with thickness at most T that are a clique covering of some graph $G \subseteq K_n$. Since each such ${\mathcal {C}}$ uniquely determines the underlying spanning subgraph $G \subseteq K_n$, we obtain similarly to (78) that

$$\begin{aligned} {{\mathbb {P}}{}}(\text {cc}_{\Delta }(G_{n,p}) \le T) \; \le \; {{\mathbb {P}}{}}(\lnot {\mathcal {E}}) \, + \, O\bigl ((6n)^{nT}\bigr ) \cdot \Pi . \end{aligned}$$

(79)

Note that ${{\mathbb {P}}{}}(\lnot {\mathcal {E}})=o(1)$ and $n \log (6n) \sim \left( {\begin{array}{c}n\\ 2\end{array}}\right) \log (1/p) \cdot (s-1)/n$. In view of inequality (77), for any $\epsilon \in (0,1)$ it follows that (79) is at most o(1) when $T \le (1-\epsilon ) \cdot (1+\varphi (p)) np/(s-1)$, completing the proof of (76). $\square$

B Variant of Theorem 2: proof of Corollary 10

Proof of Corollary 10

Choosing $\xi =\xi (\delta ) \in (0,1/16]$ such that $(1+\delta )(1+\xi )/(1-4\xi )^2 \le 1+2\delta$, set ${m_0:= \lfloor (1+\xi )m\rfloor }$, ${m_1:= \lfloor m_0/(1-4\xi )^2\rfloor }$, and ${c:= (1+\delta )r m_1/n}$. Let ${\mathcal {H}}_{i}^*$ be chosen uniformly at random from all $\left( {\begin{array}{c}|E({\mathcal {H}})|\\ i\end{array}}\right)$ subhypergraphs of ${\mathcal {H}}$ with exactly i edges. Since ${\mathcal {H}}_q$ conditioned on having exactly i edges has the same distribution as ${\mathcal {H}}_i^*$, by the law of total probability and monotonicity it follows that

$$\begin{aligned} \begin{aligned} {{\mathbb {P}}{}}(\chi '({\mathcal {H}}_q) \ge c)&\, \le \, {{\mathbb {P}}{}}(|E({\mathcal {H}}_q)| > m_0) + \sum _{0 \le i \le m_0}{{\mathbb {P}}{}}(\chi '({\mathcal {H}}_i^*) \ge c){{\mathbb {P}}{}}(|E({\mathcal {H}}_q)| = i) \\&\, \le \, n^{-\omega (r)} + {{\mathbb {P}}{}}(\chi '({\mathcal {H}}_{m_0}^*) \ge c), \end{aligned} \end{aligned}$$

(80)

where we used standard Chernoff bounds (such as [21, Theorem 2.1]) and ${{\mathbb {E}}{}}|E({\mathcal {H}}_q)| = |E({\mathcal {H}})|q = m \ge n^{1+\sigma } \gg r \log n$. Sequentially choosing the random edges $e_1, \ldots , e_{m_1} \in E({\mathcal {H}})$ of ${\mathcal {H}}_{m_1}$ as defined in Theorem 2, note that $e_{i+1} \in {E({\mathcal {H}}) {\setminus } \{e_1, \ldots , e_i\}}$ holds with probability at least ${1-m_1/e({\mathcal {H}}) > 1-4\xi }$, as $m_1 < 4 m \le 4\xi e({\mathcal {H}})$. Since we can equivalently construct the edge-set $\{f_1, \ldots , f_{m_0}\}$ of ${\mathcal {H}}^*_{m_0}$ by sequentially choosing $f_{i+1} \in {E({\mathcal {H}}) {\setminus } \{f_1, \ldots , f_i\}}$ uniformly at random, a natural coupling of ${\mathcal {H}}_{m_1}$ and ${\mathcal {H}}^*_{m_0}$ thus satisfies

$$\begin{aligned} \begin{aligned} {{\mathbb {P}}{}}({\mathcal {H}}_{m_0}^* \subseteq {\mathcal {H}}_{m_1}) \, \ge \, {{\mathbb {P}}{}}({\text {Bin}}(m_1,1-4\xi ) \ge m_0) \, \ge \, 1- n^{-\omega (r)}, \end{aligned} \end{aligned}$$

where we used standard Chernoff bounds and that $m_1 (1-4\xi ) > m_0/(1-\xi )$ for $n \ge n_0(\xi )$. Hence

$$\begin{aligned} {{\mathbb {P}}{}}(\chi '({\mathcal {H}}_{m_0}^*) \ge c) \, \le \, {{\mathbb {P}}{}}(\chi '({\mathcal {H}}_{m_1}) \ge c) + n^{-\omega (r)} \, \le \, n^{-\omega (r)} , \end{aligned}$$

(81)

where we invoked Theorem 2 with m set to $m_1$ (which applies since ${n^{1+\sigma } \le m} \le m_1< {4 \xi e({\mathcal {H}}) < n^r}$). This completes the proof by combining (80) and (81) with $c \le (1+2\delta )rm/n$. $\square$

C Heuristics: random greedy edge coloring algorithm

In this appendix we give, for the greedy coloring algorithm from Sect. 3, two heuristic explanations for the trajectories $|Q_{e}(i)| \approx {\hat{q}}(t)$ and $|Y_{v,c}(i)| \approx {\hat{y}}(t)$ that these random variables follow, where ${t = t(i,m) = i /m}$.

For our first pseudo-random heuristic, we write $E_i=\{e_1, \ldots , e_i\}$ for the multi-set of edges appearing during the first i steps of the algorithm. Ignoring that edges can appear multiple times, our pseudo-random ansatz is that the edges in $E_i$ and their assigned colors are approximately independent with

$$\begin{aligned} {{\mathbb {P}}{}}\bigl (\text {e in} E_i\, \text {and colored}\,c\bigr ) \approx \frac{|E_i|}{|E({\mathcal {H}})|} \cdot \frac{1}{q} \approx \frac{i}{nD/r} \cdot \frac{1}{rm/n} = \frac{t}{D} =: p(t,D) = p, \end{aligned}$$

where independence only holds with respect to colorings that are proper, i.e., possible in the algorithm. Using this heuristic ansatz, we now consider the event ${\mathcal {E}}_{v,c}$ that no edge $f \in E_i$ with $v \in f$ is colored c. Exploiting that no two distinct edges containing v can receive the same color in the algorithm (since this coloring would not be proper), our pseudo-random ansatz and the degree assumption (2) then suggests that

$$\begin{aligned} {{\mathbb {P}}{}}(\lnot {\mathcal {E}}_{v,c}) = \sum _{f \in E({\mathcal {H}}): v \in f} \hspace{-0.25em} {{\mathbb {P}}{}}(f \hbox {in}\, E_i\, \text {and colored}~c) \approx D \cdot p = t. \end{aligned}$$

Since for every pair of vertices there are only at most $n^{-\sigma }D$ edges containing both (by the codegree assumption), for $\ell = o(\log n)$ distinct vertices $v_{1},\ldots ,v_{\ell }$ our pseudo-random ansatz also loosely suggests that

$$\begin{aligned} {{\mathbb {P}}{}}\Bigl ( \bigcap _{i \in [\ell ]}{\mathcal {E}}_{v_i,c}\Bigr ) \approx \prod _{i \in [\ell ]}{{\mathbb {P}}{}}({\mathcal {E}}_{v_i,c}) + O\bigl (\ell ^2 \cdot n^{-\sigma }D \cdot p\bigr )\approx (1-t)^\ell . \end{aligned}$$

Recalling (46) from Sect. 3, using linearity of expectation we then anticipate $|Q_{e}(i)| \approx {\hat{q}}(t)$ based on

$$\begin{aligned} {{\mathbb {E}}{}}|Q_{e}(i)| = \sum _{c \in [q]}{{\mathbb {P}}{}}\bigl (c \in Q_{e}(i)\bigr ) = \sum _{c \in [q]}{{\mathbb {P}}{}}\Bigl ( \bigcap _{v \in e}{\mathcal {E}}_{v,c}\Bigr ) \approx q \cdot (1-t)^r= {\hat{q}}(t). \end{aligned}$$

Mimicking this reasoning, recalling (47) we similarly anticipate $|Y_{v,c}(i)| \approx {\hat{y}}(t)$ based on

$$\begin{aligned} {{\mathbb {E}}{}}|Y_{v,c}(i)| = \sum _{f \in E({\mathcal {H}}): v \in f} \hspace{-0.25em} {{\mathbb {P}}{}}\bigl (c \in Q_{f \setminus \{v\}}(i)\bigr ) \approx D \cdot (1-t)^{r-1} = {\hat{y}}(t). \end{aligned}$$

In our second expected one-step changes heuristic we assume for simplicity that there are deterministic approximations $|Q_{e}(i)| \approx f(t) q$ and $|Y_{v,c}(i)| \approx g(t) D$. Using these approximations and $q \approx rm/n$, the calculations leading to (55)–(56) and (58)–(59) in Sect. 3.2.1 then suggest that

$$\begin{aligned} {{\mathbb {E}}{}}\bigl (|Q_{e}(i+1)|-|Q_{e}(i)| \, \big | \, {\mathcal {F}}_i\bigr )&\approx - \frac{f(t) q \cdot r\cdot g(t)D}{nD/r\cdot f(t) q} \approx -\frac{rg(t)q}{m} , \end{aligned}$$

(82)

$$\begin{aligned} {{\mathbb {E}}{}}\bigl (|Y_{v,c}(i+1)|-|Y_{v,c}(i)| \, \big | \, {\mathcal {F}}_i\bigr )&\approx - \frac{g(t)D \cdot (r-1) \cdot g(t)D}{nD/r\cdot f(t) q} \approx -\frac{(r-1)g^2(t)D}{f(t) m} , \end{aligned}$$

(83)

where ${\mathcal {F}}_i$ denotes the natural filtration of the algorithm after i steps. Since the left-hand sides of (82)–(83) are approximately equal to $[f(t+1/m)-f(t)]q \approx f'(t)q/m$ and $g'(t)D/m$, respectively, we anticipate

$$\begin{aligned} f'(t) = -rg(t) \quad \text { and } \quad g'(t) = -(r-1) g^2(t)/f(t). \end{aligned}$$

(84)

Noting $|Q_{e}(0)| = q$ and $|Y_{v,c}(0)| \approx D$, we also anticipate $f(0)=g(0)=1$. The solutions ${f(t)=(1-t)^r}$ and ${g(t)=(1-t)^{r-1}}$ then make $|Q_{e}(i)| \approx f(t) q = {\hat{q}}(t)$ and $|Y_{v,c}(i)| \approx g(t) D = {\hat{y}}(t)$ plausible.