Cliques and Chromatic Number in Multiregime Random Graphs

Abstract

In this paper, we study cliques and chromatic number in the random subgraph G_n of the complete graph K_n on n vertices, where each edge is independently open with a probability p_n. Associating G_n with the probability measure ℙ_n, we say that the sequence {ℙ_n} is multiregime if the edge probability sequence {p_n} is not convergent. Using a recursive method we obtain uniform bounds on the maximum clique size and chromatic number for such multiregime random graphs.

Appendices

Appendix

Appendix: Proof of Theorems 3 and 4

Proof of Theorem 3

We use Theorem 1 with appropriate choices of η and γ.

• (i) Here α₁ = θ₁ < 2, α₂ = 0 and W_n = log n log(1 p_n) = 1 θ₁ . For 0 < ξ < 2−θ₁ θ₁ , we set 𝜖 = ξθ₁ 2 , η = θ₁ 2 + ξθ₁ and γ = ξθ₁ 2 so that Eq. 1.3 is satisfied and θ_clq = ξ. Using 𝜖(2 + 2𝜖)W_n logn ≥ 2𝜖W_n logn = ξ logn in Eq. 1.5, we then get Eq. 1.12.

• (ii) Here α₁ = α₂ = 0 and W_n = log n log(1 p). We let 0 < ξ < 1 and set 𝜖 = ξ 2, η = ξ and γ = ξ 2 and so that Eq. 1.3 is satisfied and θ_clq = ξ. As before, we use 𝜖(2 + 2𝜖)W_n logn ≥ 2𝜖W_n logn = ξ(log n)² log(1 p) and get Eq. 1.13 from Eq. 1.5.

• (iii) Here α₁ = 0 and α₂ = θ₂ < 1 and letting 0 < ξ < 1 we set 𝜖 = ξ 4, η = ξ 2 −θ₂ξ 2 and γ > 0 smaller than η. With these choices (1.3) is satisfied and moreover θ_clq = θ₂ + 2(η − γ) > θ₂. To evaluate W_n, use the log estimates (2.1) and (2.1) to get that $\frac {\frac {1}{n^{\theta _{2}}}}{1-n^{-\theta _{2}}} > \log \left (\frac {1}{p_{n}}\right ) = -\log \left (1-\frac {1}{n^{\theta _{2}}}\right ) > \frac {1}{n^{\theta _{2}}}$ and so

  $$ n^{\theta_{2}}\log{n}\left( 1-\frac{1}{n^{\theta_{2}}}\right) \leq W_{n} = \frac{\log{n}}{\log\left( \frac{1}{p_{n}}\right)} \leq n^{\theta_{2}}\log{n} $$

  (A.1)

  for all n large. Moreover

  $$ (1-\eta - \alpha_{2})W_{n} = (1-\theta_{2})\left( 1-\frac{\xi}{2}\right)W_{n} \geq (1-\theta_{2})(1-\xi)n^{\theta_{2}} \log{n} $$

  for all n large and $(2+2\epsilon )W_{n} +1 \leq (2+\xi )n^{\theta _{2}}\log {n}$ and $\epsilon (2+2\epsilon )W_{n}\log {n} \geq \frac {\xi }{4} n^{\theta _{2}} (\log {n})^{2}$ for all n large. Plugging the above into Eq. 1.5 we get Eq. 1.14.

Proof of Theorem 4

We use Theorem 2 with p_n = 1 − r_n and appropriate choices of η and γ.

• (i) Here p_n = 1 − r_n with α₁ = 0 and α₂ = θ₂ < 1 2. Thus η₀ := 1 2(1−θ₂) − θ₂ < 1 − θ₂ and for 0 < ξ < 1 to be determined later, we set

  $$ \epsilon = \frac{\xi}{6}, c = (1-\theta_{2})\left( 1-\frac{\xi^{3}}{6}\right) , \eta = \eta_{0} \left( 1+\frac{\xi^{2}}{6}\right) \text{ and } \gamma = \frac{\eta_{0} \xi^{2}}{12}. $$

  (A.2)

  We need to ensure that conditions (1.8) and (1.9) hold with α₁ = 0 and α₂ = θ₂. By definition c < 1 − θ₂ and η₀ < 1 − θ₂ and so max(η, c) < 1 − θ₂ provided ξ > 0 is small. We choose ξ > 0 smaller if necessary so that

  $$ c(\theta_{2} + \eta) = \frac{1}{2} - \frac{\xi^{3}}{12} + \frac{\eta_{0}\xi^{2}}{6}(1-\theta_{2})\left( 1-\frac{\xi^{3}}{6}\right) \geq \frac{1}{2} + \frac{\eta_{0}\xi^{2}}{12}(1-\theta_{2})\left( 1-\frac{\xi^{3}}{6}\right) $$

  and so θ₂ < 1 2 < c(θ₂ + η) < 1. To ensure the third condition in Eq. 1.8, we have

  $$ 2c(\theta_{2} + \eta-\gamma) = 1-\frac{\xi^{3}}{6} + \frac{\eta_{0}\xi^{2}}{6}(1-\theta_{2})\left( 1 - \frac{\xi^{3}}{6}\right) \!\geq\! 1 + \frac{\eta_{0}\xi^{2}}{12}(1-\theta_{2})\left( 1 - \frac{\xi^{3}}{6}\right) $$

  provided ξ > 0 is small. Fixing such a ξ we get 2c(θ₂ + η − γ) > 1 > 2θ₂, since θ₂ < 1 2. Thus Eq. 1.8 holds.

  To ensure (1.9), we write θ_chr = 1 1−θ₂ + η₀ξ² 6 − θ₂ (1−θ₂)(1−ξ³ 6) and choose ξ > 0 small so that (1 −ξ³ 6)^− 1 ≤ 1 + ξ³ 4 and so θ_chr ≥ 1 + η₀ξ² 6 − θ₂ (1−θ₂) ξ³ 4 ≥ 1 + η₀ξ² 12 . For future use we choose ξ > 0 smaller if necessary so that

  $$ c\theta_{chr} \geq (1-\theta_{2})\left( 1-\frac{\xi^{3}}{6}\right)\left( 1 + \frac{\eta_{0} \xi^{2}}{12}\right) \geq (1-\theta_{2})\left( 1+\frac{\eta_{0}\xi^{2}}{24}\right) > 1-\theta_{2}. $$

  (A.3)

  Thus the bounds in Eq. 1.11 is true. We now evaluate the upper and lower bounds in Eq. 1.11. From Eq. A.1 and the fact that 0 < ξ < 1, we get $(2+2\epsilon )W_{n} + 1 \leq \frac {2n^{\theta _{2}}\log {n}}{1-\xi }.$ Similarly

  $$ \frac{1+\epsilon}{c(1-\eta - \theta_{2})} = \frac{2\left( 1+\frac{\xi}{6}\right)}{\left( 1-\frac{\xi^{3}}{6}\right)\left( 1-2\theta_{2} - \frac{\eta_{0}\xi^{2}}{3}(1-\theta_{2})\right)} \leq \frac{2(1+\xi)}{1-2\theta_{2}}, $$

  provided ξ > 0 is small and these estimates obtain the bounds for χ(.) in Eq. 1.15.

  To evaluate the exponents in Eq. 1.11, we use Eq. A.1 to get that $\epsilon (2+\epsilon )W_{n} \log {n} \geq \frac {\xi }{4} n^{\theta _{2}}(\log {n})^{2}$ for all n large. Similarly from Eq. A.3 we get cθ_chr > 1 − θ₂ and this obtains (1.15).

• (ii) Here p_n = 1 − r_n = 1 − p and so α₁ = α₂ = 0. Letting ξ be small such that

  $$ \epsilon = \frac{\xi}{6}, \eta = \frac{1}{2} + 2\xi^{2} < 1, \gamma = \xi^{2} \text{ and } c=1-\xi^{3} $$

  (A.4)

  we get that the conditions in Eq. 1.8 are true. Also θ_chr = 1 + 2ξ² > 1 and so Eq. 1.9 is also true. Thus the bounds in Eq. 1.11 hold. Recalling that W_n = log n log(1 p_n) we have that (2 + 2𝜖)W_n + 1 = (2 + ξ 3) log n log(1 1−p) + 1 ≤ 2 (1−ξ) log n log(1 1−p) for all n large. Similarly, the scaling factor in the upper bound in Eq. 1.11 is 1+𝜖c(1−η−α₂) = 1+ξ6 (1−ξ³) (1 2 − 2ξ²) ≤ (1 + ξ) if ξ > 0 is small. The exponents in Eq. 1.11 evaluate to cθ_chr = (1 − ξ³)(1 + 2ξ²) ≥ 1 + ξ² for all ξ > 0 small and 𝜖(2 + 2𝜖)W_n ≥ 2𝜖W_n = ξ 3 log n log(1 1−p). This obtains (1.16).

• (iii) Here $p_{n} = 1-r_{n} = \frac {1}{n^{\theta _{1}}}$ and so α₁ = θ₁ < 1 and α₂ = 0. Let ξ be small such that

  $$ \epsilon = \frac{\xi^{2}}{6}, \eta = \frac{1+\theta_{1}}{2} + 2\theta_{1}\xi^{2} < 1, \gamma = \theta_{1}\xi^{2} \text{ and } c=1-\xi^{3}. $$

  (A.5)

  Recalling condition (1.8), we have max(η, c) < 1 = 1 − α₂,0 < cη = c(α₂ + η) < 1 and

  $$ \begin{array}{@{}rcl@{}} 2c(\alpha_{2} + \eta-\gamma) &=& 2(1-\xi^{3})(\eta-\gamma) = 2(\eta-\gamma) - 2\xi^{3}(\eta-\gamma)\\ &=& 1+\theta_{1} +2\theta_{1} \xi^{2} - 2\xi^{3}(\eta-\gamma). \end{array} $$

  (A.6)

  which is greater than one if ξ > 0 small. Thus Eq. 1.8 is true. Also θ_chr = 1 + θ₁ + 2θ₁ξ² − θ₁ 1−ξ³ and for all ξ > 0 small, we have 11−ξ³ ≤ 1 + ξ² and for such ξ, we have θ_chr ≥ 1 + θ₁ + 2θ₁ξ² − θ₁(1 + ξ²) = 1 + θ₁ξ² > 1. For future use we set ξ > 0 smaller if necessary so that

  $$ c\theta_{chr} \geq (1-\xi^{3}) (1+\theta_{1}\xi^{2}) \geq 1+\frac{\theta_{1}\xi^{2}}{2} > 1. $$

  (A.7)

  Thus Eq. 1.9 is also true and consequently, the bounds in Eq. 1.11 hold.

Recalling that W_n = log n log(1 p_n) = 1 θ₁ we have from Eq. A.5 that (2 + 2𝜖)W_n + 1 = (2 + ξ² 3) 1 θ₁ + 1 ≤ 2+θ₁ θ₁(1−ξ) for all n large, provided ξ > 0 small. Fixing such a ξ, the scaling factor in the upper bound in Eq. 1.11 is

$$ \frac{1+\epsilon}{c(1-\eta-\alpha_{2})} = \frac{1+\frac{\xi^{2}}{6}}{\left( 1-\xi^{3}\right)\left( 1-\frac{1+\theta_{1}}{2}-2\xi^{2}\right)} \leq (1+\xi)\frac{2}{1-\theta_{1}} $$

(A.8)

provided we set ξ > 0 smaller if necessary.

Finally, regarding the exponents in Eq. 1.17, we have from Eq. A.7 that cθ_chr > 1 and moreover 𝜖(2 + 2𝜖)W_n ≥ 2𝜖W_n = ξ² 3 1 θ₁ . This obtains (1.17).