2. Preliminaries

First, two technical notes: we use the standard notation of symbols $o, O, \Theta, \omega, \Omega, \ll, \gg$ , and $\sim$ to compare functions asymptotically (e.g. see pages 9-10 of [Reference Janson, Ruciński and Luczak7]). We also note that, throughout the paper, we assume $n$ to be arbitrarily large.

In this section we set up sequences of random variables, describe how the evolution of the $d$ -process depends on these, and deduce properties of certain approximating functions; such functions are used to estimate the number of vertices of given degrees throughout the process (much of this is also described in ref. [Reference Telcs, Wormald and Zhou13] with similar notation; the one major difference is that we use $i$ instead of $t$ for the number of time steps, and $t$ instead of $x$ for the corresponding time variable). Consider a sequence of graphs $G_0,G_1,\dots,G_{\lfloor dn/2 \rfloor }$ , where $G_0$ is the empty graph of $n$ vertices, and for $i \in [n]$ , let $G_i$ be formed by adding an edge uniformly at random to $G_{i-1}$ so that the maximum degree of $G_i$ is at most $d$ (in the unlikely event that there are no valid edges to add after $s$ steps for some $s \lt \lfloor dn/2 \rfloor$ , let $G_i = G_{s}$ for all $i \gt s$ ). Next, we define several sequences of random variables: For all $i, j, j_1, j_2$ such that $0 \leq i \leq \lfloor \frac{dn}{2} \rfloor$ , $0 \leq j \leq d$ , and $0 \leq j_1 \leq j_2 \leq d-1$ , define:

$Y^{(j)}_i \,:\!=\,$ the number of vertices in $G_i$ with degree $j$

$S^{(j)}_i \,:\!=\,$ the number of vertices in $G_i$ with degree at most $j$

$Z^{(j_1,j_2)}_i \,:\!=\,$ the number of edges $\{v_1,v_2\}$ in $G_i$ for which

\begin{equation*}\min \{\deg (v_1), \deg (v_2) \} = j_1 \ \text { and } \ \max \{\deg (v_1), \deg (v_2)\} = j_2\end{equation*}

$Z_i \,:\!=\, \displaystyle \sum _{0 \leq j_1 \leq j_2 \leq d-1} Z^{(j_1,j_2)}_i$ .

By definition and by edge-counting we have

\begin{equation*}\sum _{j=0}^{d} Y_i^{(j)} = n \qquad \text {and} \qquad \sum _{j=1}^{d} j Y_i^{(j)} = 2i.\end{equation*}

Combining the equations above gives us

(1) \begin{equation} \sum _{j=0}^{d-1} S_i^{(j)} = \sum _{j=0}^{d-1} (d-j) Y_i^{(j)} = dn-2i. \end{equation}

One can also quickly verify from (1) and by definition of $S^{(d-1)}_i$ , $Z^{(j_1,j_2)}_{i}$ , and $Z_i$ , that

(2) \begin{equation} d S^{(d-1)}_i \geq \max \{2 Z_i, dn - 2i\} \qquad \text{and} \qquad j Y^{(j)}_i \geq \sum _{k \leq j} Z^{(k,j)}_i + \sum _{k \geq j} Z^{(j,k)}_i. \end{equation}

Throughout the process we will keep track of the variables $S^{(j)}$ using martingale arguments. This is sufficient for us, as the $Y$ variables can be derived from the $S$ variables, and because none of the $Z$ variables will have any significant effect in any of our calculations, as we will see later.

Our next step is to estimate the expected on-step change of $S^{(j)}_i$ , known as the “trend hypothesis” in ref. [Reference Wormald14]. Note that $S^{(j)}_{i} - S^{(j)}_{i+1}$ equals the number of vertices of degree $j$ that are picked at the $i+1$ time step; hence, for all $j \in \{0\} \cup [d-1]$ :

(3) \begin{align} \mathbb{E}\left[S^{(j)}_{i+1} - S^{(j)}_{i} \mid G_i\right] &= \frac{-Y^{(j)}_i \left(S^{(d-1)}_i - 1\right) + \displaystyle \sum _{k \leq j} Z^{(k,j)}_i + \sum _{k \geq j} Z^{(j,k)}_i }{\binom{S^{(d-1)}_i}{2} - Z_i} \nonumber \\&= \frac{-2Y^{(j)}_i}{S^{(d-1)}_i}\left (1 + O\left (\frac{1}{dn-2i}\right )\right ) \qquad \text{by (2)} \end{align}

(4) \begin{align} &= \frac{-2 Y^{(j)}_i}{S^{(d-1)}_i}+ O\left (\frac{1}{dn-2i}\right ). \end{align}

For $j \in \{0\} \cup [d-1]$ we define approximating functions $y_j:[0,d/2) \to{\mathbb{R}}$ and $s_j:[0,d/2) \to{\mathbb{R}}$ : let $y_j(t)$ , $s_j(t)$ be functions such that $s_j = \sum _{k=0}^{j} y_k$ , $y_0(0) = 1$ and $y_k(0) = 0$ for all $k \in [d-1]$ (equivalent to $s_j(0) = 1$ for all $j$ ), and (assuming the “dummy functions” $y_{-1}(t) = s_{-1}(t) = 0$ ):

(5) \begin{equation} \frac{ds_j}{dt} = \frac{-2y_{j}}{s_{d-1}} = \frac{2(s_{j-1} - s_j)}{s_{d-1}} \ \qquad \ \frac{dy_j}{dt} = \frac{2(y_{j-1} - y_j)}{s_{d-1}}. \end{equation}

By (5) and the chain rule, for all $j \in [d-1]$ :

\begin{equation*}\frac {d y_j}{d y_0} - \frac {y_j}{y_0} = -\frac {y_{j-1}}{y_0}.\end{equation*}

Since the above equation is first-order linear, we have, for some constant $C_j$ :

\begin{equation*}y_j = y_0\left ({-}\int \frac {y_{j-1}}{y_0^2} d y_0 + C_j\right ).\end{equation*}

Using the above recursively with initial conditions, we have, for all $j \in [d-1]$ :

(6) \begin{equation} y_j = \frac{y_0 ({-}\ln (y_0))^j}{j!}. \end{equation}

To solve for an explicit formula relating $y_0$ and $t$ , note that, by (5):

\begin{equation*}\frac {d}{dt} \left ( \sum _{j=0}^{d-1} (d-j) y_j \right )= \frac {-2 \displaystyle \sum _{j=0}^{d-1}y_j}{s_{d-1}} = -2,\end{equation*}

hence, using initial conditions (note the resemblance to (1)):

(7) \begin{equation} \sum _{j=0}^{d-1} s_j = \sum _{j=0}^{d-1} (d-j) y_j = d - 2t. \end{equation}

Equations (6) and (7) together give us a complete description of the functions $y_j$ and $s_j$ . We will now prove some useful properties of these functions. To start, we can combine (6) and (7) to get

(8) \begin{equation} \sum _{j=0}^{d-1} \frac{y_0 ({-}\ln (y_0))^j (d-j)}{j!} = d - 2t. \end{equation}

Note that, by continuity of $y_0$ and by (8), $y_0 \gt 0$ over its domain. Next, by summing up (6) over $j \in \{0\} \cup [d-1]$ ((6) holds for $j=0$ also) one can see that $s_{d-1}$ is positive if $y_0 \leq 1$ . This, combined with $\frac{d y_0}{dt} = \frac{-2y_0}{s_{d-1}}$ , tells us that $y_0$ is decreasing and $s_{d-1}$ is positive. It follows from $y_0 \in (0,1]$ and (6) that each $y_j$ is positive for $t \not = 0$ . In turn, this implies that $0 \leq y_j \leq s_j \leq s_{d-1}$ for each $j$ . From this it follows that $ds_{d-1}$ is at least the left expression of (7), so $s_{d-1} \geq 1 - \frac{2t}{d}$ . We make a special note of the last couple of properties mentioned:

(9) \begin{equation} 0 \leq y_j \leq s_j \leq s_{d-1} \text{ for all } j \quad \text{ and } \quad s_{d-1}(i/n) \geq 1 - \frac{2i}{dn}. \end{equation}

Next, we want to understand the behaviour of each function when $t$ is close to $\frac{d}{2}$ , as this is the most critical point of the process. Consider (8) again. As $t \to \frac{d}{2}, y_0 \to 0$ , so $\frac{y_0({-} \ln (y_0))^{d-1}}{(d-1)!}$ will be the most dominant term on the left; hence,

\begin{equation*} t \to \frac {d}{2} \ \Longrightarrow \ y_0 \sim \frac {(d-1)! (d-2t)}{({-}\ln (d-2t))^{d-1}}.\end{equation*}

This, combined with (6) gives us, for all $j \in \{0 \} \cup [d-1]$ :

(10) \begin{equation} t \to \frac{d}{2} \ \Longrightarrow \ y_j(t) \sim s_j(t) \sim \frac{(d-1)!(d-2t)}{j!({-}\ln (d-2t))^{d-1-j}}. \end{equation}

For large enough $t$ (and hence, for a large enough step $i$ ), we can approximate the above expression:

(11) \begin{equation} i \geq \frac{dn}{2} - n^{1 - 1/(100d)} \ \Longrightarrow \ ny_j(i/n) \sim ns_j(i/n) = \Theta \left (\ln (n)^{-d+1+j}\left ( \frac{dn}{2}-i\right ) \right ). \end{equation}

One can now see that, near the end of the process, $s_j/s_{j-1} = \Theta (\ln (n))$ , as mentioned in the introduction.

Finally, we introduce two martingale inequalities from a result of Bohman [Reference Bohman2] which will be used in Section 4 in a slightly modified form. The original inequalities are as follows:

Lemma 2 (Lemma 6 from [Reference Bohman2]). Suppose $a, \eta,$ and $N$ are positive, $\eta \leq N/2$ , and $a \lt \eta m$ . If $0 = A_0, A_1,\dots, A_m$ is a submartingale such that $-\eta \leq A_{i+1} - A_{i} \leq N$ for all $i$ , then

\begin{equation*} \mathbb {P}[A_m \leq -a] \leq e^{-\frac {a^2}{3\eta N m}}. \end{equation*}

Lemma 3 (Lemma 7 from [Reference Bohman2]). Suppose $a, \eta,$ and $N$ are positive, $\eta \leq N/10$ , and $a \lt \eta m$ . If $0 = A_0, A_1,\dots, A_m$ is a supermartingale such that $-\eta \leq A_{i+1} - A_{i} \leq N$ for all $i$ , then

\begin{equation*} \mathbb {P}[A_m \geq a] \leq e^{-\frac {a^2}{3\eta N m}}. \end{equation*}

We present the following modification, which removes the requirement $a \lt \eta m$ and modifies one of the inequalities slightly:

Corollary 4. Suppose $a, \eta,$ and $N$ are positive, and $\eta \leq N/2$ . If $0 = A_0, A_1,\dots, A_m$ is a submartingale such that $-\eta \leq A_{i+1} - A_{i} \leq N$ for all $i$ , then

\begin{equation*} \mathbb {P}[A_m \leq -a] \leq e^{-\frac {a^2}{3\eta N m}}.\end{equation*}

Corollary 5. Suppose $a, \eta,$ and $N$ are positive, and $\eta \leq N/10$ . If $0 = A_0, A_1,\dots, A_m$ is a supermartingale such that $-\eta \leq A_{i+1} - A_{i} \leq N$ for all $i$ , then

\begin{equation*} \mathbb {P}[A_m \geq a] \leq e^{-\frac {a^2}{3\eta N m}} + e^{-\frac {a}{6N}}.\end{equation*}

Corollary 4 is nearly immediate from Lemma 2: first, one can extend the result to include $a = \eta m$ by using left-continuity (with respect to $a$ ) of both sides of the inequality; we hence assume $a \gt \eta m$ . Since $A_{i+1} - A_i \geq -\eta \gt -a/m$ , then $A_m = A_m - A_0 \gt -a$ . We now derive Corollary 5 from Lemma 3: assume $a \geq \eta m$ , and let $m^{\prime} \in \mathbb{Z}^+$ such that $a \lt \eta m^{\prime} \leq 2a$ . Extend the martingale by adding variables $A_{m+1},\dots,A_{m^{\prime}}$ which are all equal to $A_m$ . Apply Lemma 3 with $m$ replaced with $m^{\prime}$ , and use $\eta m^{\prime} \leq 2a$ to get

\begin{equation*}\mathbb {P}[A_{m} \geq a] = \mathbb {P}[A_{m^{\prime}} \geq a] \leq e^{-\frac {a}{6N}}.\end{equation*}

Combining the case $a \lt \eta m$ from Lemma 3 and the case $a \geq \eta m$ above gives Corollary 5.

3. First phase

Let $i_{trans} = \lfloor \frac{dn}{2} - n^{1 - 1/(100d)} \rfloor$ . The objective of this section is to prove the following Theorem:

Theorem 6. Define $E_{first}(i) \,:\!=\, n^{0.6} \left (\frac{dn}{dn-2i}\right )^{4d}$ . With high probability, for all $i \leq i_{trans}$ and all $j \in \{ 0 \} \cup [d-1]$ :

(12) \begin{equation} \left |S^{(j)}_{i} - n s_j\left (\frac{i}{n}\right )\right | \leq E_{first}(i). \end{equation}

Now we define two new random variables for each $j$ :

\begin{align*} S^{(j)+}_{i} &\,:\!=\, S^{(j)}_i - n s_j(i/n) - E_{first}(i)\\ S^{(j)-}_{i} &\,:\!=\, S^{(j)}_i - n s_j(i/n) + E_{first}(i). \end{align*}

Next, we introduce a stopping time $T$ , defined as the first step $i \leq i_{trans}$ for which (12) is not satisfied for some $j$ ; if (12) always holds, then let $T = \infty$ . Although this stopping time is not necessarily needed to prove Theorem 6, it does make some calculations easier, and moreover, a similar stopping time will be necessary for the following section; hence, this serves as a good warm-up. Let variable name $W$ be introduced to equip this stopping time to variable $S$ , i.e.

\begin{align*} &W^{(j)+}_{i} \,:\!=\, \begin{cases} S^{(j)+}_{i}, i \lt T \\[5pt] S^{(j)+}_{T}, i \geq T \end{cases} \quad W^{(j)-}_i \,:\!=\, \begin{cases} S^{(j)-}_i, i \lt T \\[5pt] S^{(j)-}_T, i \geq T. \end{cases} \end{align*}

Note that $W^{(j)+}_{i}$ corresponds to the upper boundary and $W^{(j)-}_i$ to the lower one in the sense that crossing the corresponding boundary will make the corresponding variable change signs; furthermore, the inequality of Theorem 6 holds if and only if $W^{(j)+}_{i_{trans}} \leq 0$ and $W^{(j)-}_{i_{trans}} \geq 0$ for each $j$ . We now state our martingale Lemma:

Proof. Here we just prove the first part of the Lemma; the second part follows from nearly identical calculations. Fix some arbitrary $i \leq i_{trans}$ ; we need to show that

\begin{equation*}\mathbb {E}\left[W^{(j)-}_{i+1} - W^{(j)-}_{i} \mid G_i\right] \geq 0.\end{equation*}

Also assume that $T \geq i+1$ , else $W^{(j)-}_{i+1} - W^{(j)-}_i = 0$ and we are done; it follows that $W^{(j)-}_i = S^{(j)-}_i, W^{(j)-}_{i+1} = S^{(j)-}_{i+1}$ , and (12) holds for the fixed $i$ . By (4) and (5) and using Taylor’s Theorem, we have, for some $\psi \in [i,i+1]$ :

\begin{align*} \mathbb{E}\left[S^{(j)-}_{i+1} - S^{(j)-}_i \mid G_i\right] &= \frac{-2 Y^{(j)}_i}{S^{(d-1)}_i} + O\left (\frac{1}{dn-2i}\right ) + \frac{2y_j(i/n)}{s_{d-1}(i/n)} -\frac{d^2}{d\mu ^2}\left (\frac{ns_j(\mu/n)}{2}\right )\Big |_{\mu = \psi } \\[5pt]& \quad + \left(E_{first}(i+1) - E_{first}(i)\right). \end{align*}

We split the above expression $\Big($ excluding $O\left (\frac{1}{dn-2i}\right )\Big)$ into three summands.

1. 1. Here we make use of the fact that $Y^{(j)}_i = S^{(j)}_i - S^{(j-1)}_{i}$ and $y_j(t) = s_j(t) - s_{j-1}(t)$ . We have (putting $s_{j-1}$ and $S_i^{({-}1)} = 0$ ):

   \begin{align*} \frac{-2 Y^{(j)}_i}{S^{(d-1)}_i} + \frac{2y_j(i/n)}{s_{d-1}(i/n)} &= \frac{-2 S^{(j)}_i + 2ns_j(i/n) + 2 S^{(j-1)}_i - 2ns_{j-1}(i/n)}{n s_{d-1}(i/n)} \\[4pt] & \ \ \ + 2Y^{(j)}_i \left (\frac{1}{n s_{d-1}(i/n)} - \frac{1}{S^{(d-1)}_i}\right ) \\[4pt]&\geq \frac{-4 E_{first}(i)}{ns_{d-1}(i/n)} + 2Y^{(j)}_i \left (\frac{1}{n s_{d-1}(i/n)} - \frac{1}{S^{(d-1)}_i}\right ) \qquad \text{by (12) and}\ i \lt T \\[4pt] & \geq \frac{-4 E_{first}(i)}{ns_{d-1}(i/n)} - \frac{2 Y^{(j)}_{i} E_{first}(i)}{S^{(d-1)}_i (n s_{d-1}(i/n))} \qquad \text{by (12) and}\ i \lt T\\[4pt] & \geq \frac{-6 E_{first}(i)}{n s_{d-1}(i/n)}. \end{align*}

2. 2.

   \begin{align*} -\frac{d^2}{d\mu ^2}\left (\frac{ns_j(\mu/n)}{2}\right )\Big |_{\mu = \psi } &= \frac{2}{n}\left (\frac{s_{d-1}(\psi/n)(y_{j-1}(\psi/n) - y_j(\psi/n))+y_j(\psi/n)y_{d-1}(\psi/n)}{(s_{d-1}(\psi/n))^3} \right ) \\[4pt]&= O\left (\frac{1}{dn-2i}\right ) \qquad \text{by (9)}. \end{align*}

3. 3. For some $\phi \in [i,i+1]$ :

   (13) \begin{align} E_{first}(i+1) - E_{first}(i) &= \frac{dE_{first}(\mu )}{d\mu } \Big |_{\mu = \phi } \nonumber \\[4pt]&= 8d^{4d+1} n^{4d+0.6} \left (dn-2\phi \right )^{-4d-1} \nonumber \\[4pt]&= (1 + o(1))\frac{8dE_{first}(i)}{dn - 2i}. \end{align}

Now we put the three bounds together:

\begin{align*} \mathbb{E}\left[Y_{i+1}^- - Y_i^- \mid G_i\right] &\geq \frac{7dE_{first}(i)}{dn - 2i} - \frac{6 E_{first}(i)}{n s_{d-1}(i/n)} + O\left (\frac{1}{dn-2i}\right ) \\[4pt]&\geq \frac{dE_{first}(i) + O(1)}{dn - 2i} \qquad \text{by (9)} \\[4pt]&\geq 0. \end{align*}

Next, we need a Lipschitz condition on each of our variables. Note that $S^{(j)}_{i+1} - S^{(j)}_{i}$ is either $-2, -1$ , or 0; also, one can quickly verify that $|s_j((i+1)/n) - s_j(i/n)| \leq \frac{2}{n}$ by (5) and (9), and $\big|E_{first}(i+1) - E_{first}(i)\big| = o(1)$ by (13). Hence, we have, for all $i \leq i_{trans}$ and all $j$ :

(14) \begin{equation} \max \left\{\left|W^{(j)+}_{i+1} - W^{(j)+}_{i}\right|, \left|W^{(j)-}_{i+1} - W^{(j)-}_{i}\right| \right\} \leq 5. \end{equation}

We conclude the proof of Theorem 6 by noting that, by Lemma 7 and (14), we can use the standard Hoeffding-Azuma inequality for martingales (e.g. Theorem 7.2.1 in [Reference Alon and Spencer1]) to show that $\mathbb{P}\left[W^{(j)+}_{i_{trans}} \gt 0\right]$ and $\mathbb{P}\left[W^{(j)-}_{i_{trans}} \lt 0\right]$ are both $o(1)$ . For example, for the variable $W^{(j)+}_{i}$ one would get

\begin{equation*} \mathbb {P}\left[W^{(j)+}_{i_{trans}} \gt 0\right] \leq \exp \left \{ - \frac {n^{1.2}}{50 i_{trans}} \right \} = o(1). \end{equation*}

4. Second phase

The second phase is where the more sophisticated tools will be used, including the use of critical intervals, self-correcting estimates, and a more general martingale inequality. Furthermore, this phase is broken up into $d-1$ sub-phases, in relation to when each of the $d-1$ sequences $S^{(j)}$ (for $j \leq d-2$ ) terminate at 0. First, a few definitions: for all $k \in \{0\} \cup [d-2]$ , define

\begin{equation*} i_{after}(k) \,:\!=\, \left \lfloor \frac {dn}{2} - \ln (n)^{d-1.01-k} \right \rfloor .\end{equation*}

These step values will govern the endpoints of the sub-phases: define for all $k \in \{0\} \cup [d-2]$ :

\begin{align*} I_k \,:\!=\, \left\{\begin{array}{l@{\quad}l} \!\left[i_{trans} + 1, i_{after}(0)\right], & k = 0 \\[4pt]\!\left[i_{after}(k-1) + 1, i_{after}(k)\right], & k \gt 0. \end{array}\right.\end{align*}

Next, for all $i,j,k$ such that $0 \leq j \lt d$ , $0 \leq k \lt d-1$ , and $i \in I_k$ , define error functions

\begin{equation*} E_{j,k}(i) = E_{j}(i) \,:\!=\, 2^k \ln (n)^{0.05} (n s_{j}(i/n))^{0.7}.\end{equation*}

Note that, by (11), we have

(15) \begin{equation} E_j(i) = \Theta \left ( \ln (n)^{-0.7d + 0.75 + 0.7j} \left (\frac{dn}{2} - i\right )^{0.7}\right ). \end{equation}

Finally, for any $r \in{\mathbb{R}}_+$ and $\ell \in [d-2]$ , define

\begin{equation*} i(r,\ell ) = \frac {dn}{2} - \left (\frac {\ell ! }{2(d-1)!}\right ) r(\ln n)^{d-1-\ell }.\end{equation*}

The following Theorem will be proved by induction over the $d-1$ sub-phases governed by the index $k$ :

In the end, it is only the second statement with $k = d-2$ that matters for proving Theorem 1. We make the connection here:

Proof of Theorem 1 from Theorem 8. First, note that $S^{(\ell )}_{\lfloor i(r_{\ell },\ell ) \rfloor } = 0$ is the same as $T_{\ell } \leq i(r_{\ell },\ell )$ , hence by Theorem 8:

\begin{equation*} \mathbb {P} \left ( \bigcap _{\ell = 0}^{d-2} \left (T_{\ell } \leq i(r_{\ell },\ell )\right )\right ) \to \exp \left \{-\sum _{\ell =0}^{d-2} r_{\ell } \right \}.\end{equation*}

Using the Principle of Inclusion-Exclusion plus a simple limiting argument, one can derive

\begin{equation*}\mathbb {P} \left ( \bigcap _{\ell = 0}^{d-2} \left (\frac {(d-1)!(dn-2T_{\ell })}{\ell !(\ln (n))^{d-1-\ell }} \leq r_{\ell }\right )\right ) = \mathbb {P} \left ( \bigcap _{\ell = 0}^{d-2} \left (T_{\ell } \geq i(r_{\ell },\ell )\right )\right ) \to \prod _{\ell =0}^{d-2} \left(1 - e^{-r_{\ell }}\right),\end{equation*}

hence the $d-1$ -dimensional random vector with entries $V_{n}^{(\ell )} = \frac{(d-1)!(dn-2T_{\ell })}{\ell !(\ln (n))^{d-1-\ell }}$ converges in distribution to the product of $d-1$ independent exponential variables of mean 1.

The rest of this section is for proving the first statement of Theorem 8 (for some fixed $k$ using induction), and Section 5 will be for proving the second statement (again, for some fixed $k$ using induction, assuming the first statement holds for the same $k$ ). Hence, for the rest of the paper we will fix some $k \in \{0\} \cup [d-2]$ .

First, we note that (16) holds w.h.p. for all $j \lt k$ by a simple argument: by induction on the second statement of Theorem 8, w.h.p. if $i \in I_k$ then $S_{i}^{(j)} = 0$ . By (11) and by definition of $E_j(i)$ , if $i \in I_k$ then $ns_{j}(i/n) \ll E_{j}(i)$ , completing the argument.

Next, we prove that (16) holds for $j = d-1$ if it holds for all other values of $j$ : by combining (1) and (7), we have

\begin{align*} \left |S^{(d-1)}_{i} - n s_{d-1}\left (\frac{i}{n}\right )\right | &= \left | \sum _{j=0}^{d-2}\left (S^{(j)}_{i} - n s_{j}\left (\frac{i}{n}\right )\right ) \right | \\[4pt]&\leq \sum _{j=0}^{d-2} 4E_j(i) \qquad \text{by (16) for } j \leq d-2 \\[4pt]&\lt 4 E_{d-1}(i) \qquad \text{by (15).} \end{align*}

Hence, for the rest of this section, we need to show the first statement of Theorem 8 for $j \in [k, d-2]$ . From now on we always assume $j$ to be in this range. We will also assume that, for all $\lambda \lt k$ , $S_{i}^{(\lambda )} = 0$ if $i \in I_k$ (which holds w.h.p. from above).

In this section we will make use of so-called critical intervals, ranges of possible values for $S^{(j)}_i$ in which we apply a martingale argument. The lower critical interval will be

\begin{equation*}[n s_j(i/n) - 4E_j(i), ns_j(i/n) - 3E_j(i)],\end{equation*}

and the upper critical interval will be

\begin{equation*}[n s_j(i/n) + 3E_j(i), ns_j(i/n) + 4E_j(i)].\end{equation*}

Our goal is to show that w.h.p. $S^{(j)}_i$ does not cross either critical interval; however, we first need to show that $S^{(j)}_i$ sits between the critical intervals at the beginning of the phase (this is the reason why $E_j(i)$ has the $2^k$ factor; it makes a sudden jump in size between phases to accommodate a new martingale process), which is the statement of our first Lemma of this section:

Lemma 9. W.h.p., for all $j \in [k,d-2]$ (putting $i_{after}({-}1) = i_{trans}$ for convenience of notation):

\begin{equation*} \left |S^{(j)}_{i_{after}(k-1) + 1} - n s_j \left (\frac {i_{after}(k-1)+1}{n}\right ) \right | \lt 3E_j(i_{after}(k-1)+1). \end{equation*}

Next, like in Section 3, we define two new random variables for each $j$ and $i \in I_k$ :

\begin{align*} S^{(j)+}_{i} &\,:\!=\, S^{(j)}_i - n s_j(i/n) - 4 E_j(i)\\[3pt] S^{(j)-}_{i} &\,:\!=\, S^{(j)}_i - n s_j(i/n) + 4 E_j(i). \end{align*}

We also re-introduce the stopping time $T$ , now defined as the first step $i \in I_k$ for which (16) is not satisfied for some $j$ ; if (16) always holds, then let $T = \infty$ . Let variable name $W$ be introduced to equip this stopping time to variable $S$ , i.e.

\begin{align*} W^{(j)+}_{i} \,:\!=\, \begin{cases} S^{(j)+}_{i}, i \lt T \\[5pt] S^{(j)+}_{T}, i \geq T \end{cases} \quad W^{(j)-}_i \,:\!=\, \begin{cases} S^{(j)-}_i, i \lt T \\[5pt] S^{(j)-}_T, i \geq T. \end{cases} \end{align*}

Note that $W^{(j)+}_{i}$ corresponds to the upper critical interval, and $W^{(j)-}_i$ to the lower one. Furthermore, the inequality of Theorem 8 holds if and only if $W^{(j)+}_{i_{after}(k)} \leq 0$ and $W^{(j)-}_{i_{after}(k)} \geq 0$ for each $j$ (here we must make use of our assumption that $S_{i}^{(\lambda )} = 0$ for all $\lambda \lt k$ ). The next Lemma states that, within their respective critical intervals, they are a supermartingale and submartingale respectively:

Proof. Here we just prove the first part of the Lemma; the second part follows from nearly identical calculations. By the same logic as in the proof of Lemma 7 we work with $S^{(j)-}$ instead of $W^{(j)-}$ and assume that (16) holds for all $j$ . We also have the same expected change as in Lemma 7, except with $E_{first}(i)$ replaced with $4E_j(i)$ :

\begin{align*} \mathbb{E}\left[S^{(j)-}_{i+1} - S^{(j)-}_i \mid G_i\right] &= \frac{-2 Y^{(j)}_i}{S^{(d-1)}_i} + O\left (\frac{1}{dn-2i}\right ) + \frac{2y_j(i/n)}{s_{d-1}(i/n)} -\frac{d^2}{d\mu ^2}\left (\frac{ns_j(\mu/n)}{2}\right )\Big |_{\mu = \psi } \\& \quad + 4(E_{j}(i+1) - E_j(i)). \end{align*}

We split the above expression $\Big($ excluding $O\left (\frac{1}{dn-2i}\right )\Big)$ into three summands, assuming $S^{(j)-}_i \leq E_j(i) \ \Longleftrightarrow \ S^{(j)}_i - ns_j(i/n) \leq - 3E_j(i)$ $\big($ for convenience, for the case $j = 0$ , we put $S_i^{(j-1)}, s_{j-1}$ , and $E_{j-1}$ all equal to 0 $\big)$ :

1. 1.

   \begin{align*} \frac{-2 Y^{(j)}_i}{S^{(d-1)}_i} + \frac{2y_j(i/n)}{s_{d-1}(i/n)} &= \frac{-2 S^{(j)}_i + 2ns_j(i/n) + 2 S^{(j-1)}_i - 2ns_{j-1}(i/n)}{n s_{d-1}(i/n)} \\[3pt] & \ \ \ + 2Y^{(j)}_i \left (\frac{1}{n s_{d-1}(i/n)} - \frac{1}{S^{(d-1)}_i}\right ) \\[3pt]&\geq \frac{6 E_j(i) - 8 E_{j-1}(i)}{ns_{d-1}(i/n)} - \frac{8 S^{(j)}_{i} E_{d-1}(i)}{S^{(d-1)}_i (n s_{d-1}(i/n))} \qquad \text{by (16)} \\[3pt]&\geq \frac{5.9 E_j(i)}{ns_{d-1}(i/n)} - \frac{9 S^{(j)}_{i} E_{d-1}(i)}{(n s_{d-1}(i/n))^2} \qquad \text{by (15), (16), (11), and } i \leq i_{after}(k) \\[3pt]&\geq \frac{5.9 E_j(i)}{ns_{d-1}(i/n)} - \frac{9 (n s_{j}(i/n)) E_{d-1}(i)}{(n s_{d-1}(i/n))^2} - \frac{36 E_{j}(i) E_{d-1}(i)}{(n s_{d-1}(i/n))^2} \qquad \text{by (16)} \\[3pt]&= \left (\frac{E_j(i)}{n s_{d-1}(i/n)}\right )\left (5.9 - 9\left (\frac{s_{j}(i/n)}{s_{d-1}(i/n)}\right )^{0.3} - \frac{36*2^k \ln (n)^{0.05}}{(n s_{d-1}(i/n))^{0.3}}\right ) \\[3pt]&\geq \frac{5.8 E_j(i)}{n s_{d-1}(i/n)} \qquad \text{by } i \leq i_{after}(k)\ \text{and}\ (11). \end{align*}

2. 2. Just as in the proof of Lemma 7:

   \begin{equation*} -\frac {d^2}{d\mu ^2}\left (\frac {ns_j(\mu/n)}{2}\right )\Big |_{\mu = \psi } = O\left (\frac {1}{dn-2i}\right ). \end{equation*}

3. 3.

   (17) \begin{align} 4(E_j(i+1) - E_j(i)) &= 4 \frac{dE_j(\mu )}{d\mu }\Big |_{\mu = \phi } \qquad \text{for some}\ \phi \in [i, i+1] \nonumber \\[4pt]&= (4)(2^k) \ln (n)^{0.05} \left (\frac{0.7}{(ns_j(\phi/n))^{0.3}}\right )\left (\frac{-2y_j(\phi/n)}{s_{d-1}(\phi/n)}\right ) \qquad \text{by (5)} \nonumber \\[4pt]& = (4 + o(1))(2^k) \ln (n)^{0.05} \left (\frac{0.7}{(ns_j(\phi/n))^{0.3}}\right )\left (\frac{-2s_j(\phi/n)}{s_{d-1}(\phi/n)}\right ) \ \ \ \ \text{by (10)} \nonumber \\[4pt]&= \frac{-(5.6 + o(1)) E_j(\phi )}{n s_{d-1}(\phi/n)} = \frac{-(5.6 + o(1))E_j(i)}{n s_{d-1}(i/n)}. \end{align}

Now we put the above bounds together (using (11), (15), and $i \leq i_{after}(k) \leq i_{after}(j)$ ):

\begin{align*} \mathbb{E}\left[S_{i+1}^{(j)-} - S_{i}^{(j)-} \mid G_i\right] &\geq \frac{0.01E_j(i)}{n s_{d-1}(i/n)} + O\left (\frac{1}{dn-2i}\right ) \\&\geq 0. \end{align*}

We introduce the next Lemma to get sufficiently small bounds on the one-step changes in each time step (this is known as the “bounded hypothesis” from [Reference Wormald14]):

Lemma 11. For all $i \in I_k$ and all $j \in [k,d-2]$ ,

\begin{equation*} -3 \lt W^{(j) \xi }_{i+1} - W^{(j)\xi }_i \lt \ln (n)^{-d+1.06+j}\end{equation*}

where “ $\xi$ ” can be either “ $+$ ” or “ $-$ ”.

Proof. Like in the proofs of Lemma 7 and 10, we assume that $W^{\xi } = S^{(j)\xi }$ ( $\xi$ is $+$ or $-$ ), else $W^{\xi }_{i+1} - W^{\xi }_{i} = 0$ . Again, we have $-2 \leq S^{(j)}_{i+1} - S^{(j)}_{i} \leq 0$ . Secondly, we have

\begin{align*} & |-n s_j((i+1)/n)+ns_j(i/n) - C E_j(i+1)+C E_j(i)| \\[3pt] & \leq |-n s_j((i+1)/n)+ns_j(i/n)| + |- C E_j(i+1)+C E_j(i)| \\[3pt] &= O \left (\frac{y_j(i/n)}{s_{d-1}(i/n)} + \frac{E_j(i)}{n s_{d-1}(i/n)} \right ) \qquad \text{by (5), (11), and (17)} \\[3pt] &= o\left(\ln (n)^{-d+ 1.06+j}\right) \qquad \text{by (11), (15), and } i \leq i_{after}(k) \leq i_{after}(j). \end{align*}

Combining the inequalities completes the proof.

To put this all together to prove the first part of Theorem 8, we introduce a series of events: first, let $\mathcal{E}^{(j)+}$ denote the event that $W^{(j)+}_{i_{after}(k)} \gt 0$ and $\mathcal{E}^{(j)-}$ denote the event that $W^{(j)-}_{i_{after}(k)} \lt 0$ . Let $\mathcal{E} = \left (\bigcup _{j \geq k}\mathcal{E}^{(j)+}\right ) \cup \left (\bigcup _{j \geq k}\mathcal{E}^{(j)-}\right )$ ; we seek to bound $\mathbb{P}[\mathcal{E}]$ , since $\mathcal{E}$ is the event that (16) doesn’t hold for some $i \in I_k$ . Next, for all $\ell \in I_k$ , let $\mathcal{H}^{(j)+}_{\ell }$ be the event that $W^{(j)+}_{\ell - 1} \lt -E_j(\ell - 1)$ and $W^{(j)+}_{\ell } \geq -E_j(\ell )$ , and let

\begin{equation*} \mathcal {E}^{(j)+}_{\ell } \,:\!=\, \mathcal {H}^{(j)+}_{\ell } \cap \left\{W^{(j)+}_i \geq -E_j(i) \text { for all } i \geq \ell \right\} \cap \left\{W^{(j)+}_{i_{after(k)}} \gt 0\right\}.\end{equation*}

$\Big($ see Fig. 1 for a visual representation of event $\mathcal{E}^{(j)+}_{\ell }\Big)$

Figure 1.

Visual representation of event $\mathcal{E}^{(j)+}_{\ell }$ .

Similarly, for all $\ell \in I_k$ , let $\mathcal{H}^{(j)-}_{\ell }$ be the event that $W^{(j)-}_{\ell - 1} \gt E_j(\ell - 1)$ and $W^{(j)-}_{\ell } \leq E_j(\ell )$ , and let

\begin{equation*}\mathcal {E}^{(j)-}_{\ell } \,:\!=\, \mathcal {H}^{(j)-}_{\ell } \cap \left\{W^{(j)-}_i \leq E_j(i) \text { for all } i \geq \ell \right\} \cap \left\{W^{(j)-}_{i_{after}(k)} \lt 0\right\}.\end{equation*}

Finally, note that, by Lemma 9, with high probability we must have

\begin{equation*}W^{(j)+}_{i_{after}(k-1)+1} \lt -E_j(i_{after}(k-1)+1) \ \text { and } \ W^{(j)-}_{i_{after}(k-1)+1} \gt E_j(i_{after}(k-1)+1).\end{equation*}

Furthermore, assuming these two inequalities hold (and, once again, assuming that $S^{\lambda }_i = 0$ if $\lambda \lt k$ ), then if $W^{(j)+}_{i_{after}(k)} \gt 0$ for some $j$ , one of the events $\mathcal{E}^{(j)+}_{\ell }$ must happen; likewise, if $W^{(j)-}_{i_{after}(k)} \lt 0$ for some $j$ , one of the events $\mathcal{E}^{(j)-}_{\ell }$ must happen; hence, $\mathcal{E}^{(j)+} = \displaystyle \bigcup _{\ell } \mathcal{E}^{(j)+}_{\ell }$ and $\mathcal{E}^{(j)-} = \displaystyle \bigcup _{\ell } \mathcal{E}^{(j)-}_{\ell }$ .

We are now ready to prove the first statement of Theorem 8 in full.

Proof of the first part of Theorem 8 with fixed $k$ . First, we fix an arbitrary $j$ (in $[k,d-2]$ ). We prove that $\mathbb{P}\big[\mathcal{E}^{(j)-}\big] = \exp \big\{-\Omega \big(\ln (n)^{0.036}\big)\big\}$ ; the proof for bounding $\mathbb{P}\big[\mathcal{E}^{(j)+}\big]$ is nearly identical. We will use Corollary 5 to bound $\mathbb{P}\big[\mathcal{E}^{(j)-}_{\ell }\big]$ for each fixed $\ell$ . Given a fixed $\ell$ , we define a modified stopping time

\begin{equation*}T_{mod} \,:\!=\, \min _{i \in [\ell,i_{after}(k)]} \left\{W^{(j)-}_i \gt E_j(i) \text { or } i = T \right\}\end{equation*}

(letting $T_{mod} = \infty$ if the condition doesn’t hold for any $i$ in the range). Let variable $W^{\ell }_i$ be the variable $W^{(j)-}_{i}$ defined just on $i \in [\ell, i_{after}(k)]$ equipped with this stopping time (we drop the “ $(j)-$ ” here for convenience); i.e.

\begin{equation*} W^{\ell }_i \,:\!=\, \begin {cases} W^{(j)-}_i, i \lt T_{mod} \\[4pt] W^{(j)-}_{T_{mod}}, i \geq T_{mod}. \end {cases} \end{equation*}

Note that $\big(W^{\ell }_i\big)_i$ (over $i \in [\ell,i_{after}(k)]$ ) is a submartingale by Lemma 10, since our new stopping time negates the need for the condition $W^{(j)-}_i \leq E_j(i)$ ; also, $\big(W^{\ell }_i\big)_i$ satisfies Lemma 11. Since we want an upper bound for $\mathbb{P}\big[\mathcal{E}^{(j)-}_{\ell }\big]$ , we can condition on event $\mathcal{H}^{(j)-}_{\ell }$ , as $\mathcal{H}^{(j)-}_{\ell } \supseteq \mathcal{E}^{(j)-}_{\ell }$ . Now let

\begin{align*} A_i &= -W_{\ell +i}^{\ell } + W^{\ell }_{\ell },\\ \eta &= \ln (n)^{-d+1.06+j}, \\ N &= 3, \\ m &= i_{after}(k) - \ell,\\ a &= 0.9 E_j(\ell ). \end{align*}

Note that the conditions of Corollary 5 are satisfied: $0 = A_0$ and $\eta \lt N/10$ are obvious, Lemma 11 gives us $-\eta \leq A_{i+1} - A_{i} \leq N$ , and $(A_i)_i$ is a supermartingale since $\big(W^{\ell }_i\big)_i$ is a submartingale. We therefore implement Corollary 5, using $m \leq \frac{dn}{2} - \ell \leq d n s_{d-1}(\ell/n)$ (by (9)), (11), and (15):

(18) \begin{equation} \mathbb{P}[A_m \geq a] \leq e^{-\frac{a^2}{3\eta N m}}+ e^{-\frac{a}{6N}} = e^{-\Omega \left( \ln (n)^{0.04}(n s_j(\ell/n))^{0.4}\right)} + e^{-\Omega \left(\ln (n)^{0.05}(n s_j(\ell/n))^{0.7}\right)}. \end{equation}

To bound $\mathbb{P}\big[\mathcal{E}^{(j)-}_{\ell }\big]$ , we show that $\mathcal{E}^{(j)-}_{\ell } \subseteq \{ A_m \geq a\}$ and apply (18) while conditioning on $\mathcal{H}^{(j)-}_{\ell }$ . Given $\mathcal{H}^{(j)-}_{\ell }$ happens, we have $W^{\ell }_{\ell } = W^{(j)-}_{\ell } \gt 0.9 E_j(\ell ) = a$ by (15), Lemma 11, and $i \leq i_{after}(j)$ . Therefore $\mathcal{E}^{(j)-}_{\ell } = \mathcal{H}^{(j)-}_{\ell } \cap \left\{ W^{\ell }_{i_{after}(k)} \lt 0 \right\} \subseteq \{A_m \geq a\}$ , hence

\begin{equation*} \mathbb {P}\left [\mathcal {E}^{(j)-}_{\ell }\right ] = e^{-\Omega \left( \ln (n)^{0.04}(n s_j(\ell/n))^{0.4}\right)} + e^{-\Omega \left(\ln (n)^{0.05}(n s_j(\ell/n))^{0.7}\right)}.\end{equation*}

We now take a union bound to bound $\mathbb{P}\big[\mathcal{E}^{(j)-}\big]$ (using (11) where appropriate):

\begin{align*} \mathbb{P}\big[\mathcal{E}^{(j)-}\big] &\leq \sum _{\ell = i_{after}(k-1)+1}^{i_{after}(k)} \mathbb{P}\left [\mathcal{E}^{(j)-}_{\ell }\right ] \\[3pt] = & \sum _{\ell =i_{trans}}^{i_{after}(k)} \left (\exp \left \{-\Omega \left(\ln (n)^{0.04}(n s_j(\ell/n))^{0.4}\right)\right \} + \exp \left \{-\Omega \left(\ln (n)^{0.05}(n s_j(\ell/n))^{0.7}\right)\right \} \right )\\[3pt] = & \sum _{\ell =i_{trans}}^{i_{after}(j)} \left (\exp \left \{- \Omega \left (\frac{(dn-2\ell )^{0.4}}{\ln (n)^{0.4d-0.44-0.4j}}\right )\right \} + \exp \left \{- \Omega \left (\frac{(dn-2\ell )^{0.7}}{\ln (n)^{0.7d-0.75-0.7j}}\right )\right \}\right ) \\[3pt] = & \sum _{p =\lfloor \ln (n)^{d-1.01-j}\rfloor }^{\lceil n^{1 - 1/(100d)}\rceil } \left ( \exp \left \{- \Omega \left (\frac{p^{0.4}}{\ln (n)^{0.4d-0.44-0.4j}}\right )\right \} + \exp \left \{- \Omega \left (\frac{p^{0.7}}{\ln (n)^{0.7d-0.75-0.7j}}\right )\right \} \right )\\[3pt] = & \ln (n)^{d-1.01-j}\sum _{q=1}^{\infty } \left (\exp \left \{-\Omega \left(q^{0.4} \ln (n)^{0.036}\right)\right \} + \exp \left \{-\Omega \left(q^{0.7} \ln (n)^{0.043}\right)\right \}\right ) \\[3pt]=& \exp \left\{-\Omega \left(\ln (n)^{0.036}\right)\right\}. \end{align*}

We give a note for the aspects of the proof of bounding $\mathbb{P}[\mathcal{E}^{(j)+}]$ that are different from the above: use the variable $W^{(j)+}_i$ instead of $W^{(j)-}_i$ , events $\mathcal{E}^{(j)+}_{\ell }$ instead of $\mathcal{E}^{(j)-}_{\ell }$ , and $\mathcal{H}^{(j)+}_{\ell }$ instead of $\mathcal{H}^{(j)-}_{\ell }$ . Define $T_{mod}$ instead as

\begin{equation*} T_{mod} \,:\!=\, \min _{i \in [\ell,i_{after}(k)]} \left\{W^{(j)+}_i \lt -E_j(i) \text { or } i = T \right\}.\end{equation*}

Finally, use Corollary 4 instead of Corollary 5 (which will be slightly easier to implement).

5. Final phase

We continue our proof by induction of Theorem 8 with our fixed index $k$ ; now we prove the second part. We assume the first part of Theorem 8 to hold, as well as the second part of the Theorem for lesser $k$ ; for example, we have $S^{(k-1)}_{i_{after}(k-1)} = 0$ w.h.p. In this section we focus on the $d$ -process for a narrow domain of $i$ . Let

\begin{equation*} i_{before}(k) \,:\!=\, \left \lfloor \frac {dn}{2} - \ln (n)^{d-0.8-k} \right \rfloor .\end{equation*}

We will consider the $d$ -process starting at step $i_{before}(k)$ assuming that (16) holds at $i = i_{before}(k)$ ; we do not need the first part of Theorem 8 in this section otherwise. We do not use martingale arguments here, but rather we show that the distribution of the sequence of time steps at which a vertex of degree $k$ is chosen from the $d$ -process is similar to a uniform distribution over all possible such sequences. Theorem 8, (10), and (15) tell us that w.h.p. we will have $\sim \frac{2(d-1)!}{k!}\ln (n)^{0.2}$ vertices of degree at most $k$ (or degree equal to $k$ ; they are the same here) left when there are $\lfloor \ln (n)^{d-0.8-k} \rfloor$ steps left; hence, the average distance between steps at which we remove vertices of degree $k$ is $\frac{k!}{2(d-1)!} \ln (n)^{d-1-k}$ . When there are this many steps left times $r$ , we expect $r$ such vertices to remain, and for the probability that there are no vertices of degree $k$ to be $e^{-r}$ . Most of this section will build towards proving the following Theorem:

Theorem 12. Let $L(n)$ be an integer-valued function so that $L(n) = \Theta (\ln (n)^{0.2})$ and let $J(n) = \lfloor \frac{dn}{2}\rfloor - i_{before}(k) \sim \ln (n)^{d-0.8-k}$ . Let $H$ be any graph with $i_{before}(k)$ edges which satisfies ( 16 ) at $i = i_{before}(k)$ , has no vertices of degree at most $k-1$ , and has $L(n)$ vertices of degree $k$ . Also, let $r \in{\mathbb{R}}^+$ be arbitrary. Then

\begin{equation*}\mathbb{P} \left [S^{(k)}_ {\lfloor \frac {dn}{2} - \frac {r J(n)}{L(n)} \rfloor } = 0 \ \bigg | \ G_{i_{before}(k)} = H\right ] \to e^{-r}.\end{equation*}

First, we note that, given that (16) holds for $i = i_{before}(k)$ and by (1), that w.h.p. ${dn - 2i_{before}(k) - S^{(d-1)}_{i_{before}(k)}} = O\left (\frac{dn-2i_{before}}{\ln (n)}\right )$ $\Big($ consider $S^{(d-2)}_{i_{before(k)}}\Big)$ ; hence, for all $i \in \left[i_{before}(k),i_{after}(k)\right]$ :

(19) \begin{equation} S^{(d-1)}_{i} = dn - 2i + O\left (\frac{dn-2i_{before}}{\ln (n)}\right ) = (dn - 2i) \left (1 + O\left (\frac{1}{\ln (n)^{0.79}}\right ) \right ). \end{equation}

Let $t_{start} = i_{before}(k)$ and $t_{end} = \lfloor dn/2 - r J(n)/L(n) \rfloor$ . Consider the $d$ -process between $t_{start}$ and $t_{end}$ , given that $G_{t_{start}} = H$ . At each step two vertices are chosen; now assume that the pair at each step is ordered uniformly at random, so that a sequence of $2(t_{end} - t_{start})$ vertices is generated. We also generate a binary sequence simultaneously, each digit corresponding to a vertex: after a pair of vertices is picked for the vertex sequence, for each of the two vertices (in the order that they are randomly shuffled) append a “1” to the binary sequence if the corresponding vertex had degree $k$ just before it was picked, and append a “0” otherwise. Let $\mathcal{P}: \{0,1\}^{2(t_{end}-t_{start})} \to [0,1]$ be the corresponding probability function that arises from this process (note that, if $\gamma$ is a string with more than $L(n)$ “1”’s, then $\mathcal{P}(\gamma )$ = 0). Note that $\mathcal{P}$ depends on the graph $H$ . We compare this to a second probability function $\mathcal{Q}: \{0,1\}^{2(t_{end}-t_{start})} \to [0,1]$ , which is defined by picking a binary string with $L(n)$ 1’s and $2 J(n) - L(n)$ 0’s uniformly at random, then taking the first $2(t_{end} - t_{start})$ digits.

For any binary sequence $\gamma$ with $\ell$ digits, and $I \subset [\ell ]$ , let $\gamma _{I}$ be the subsequence with indices from $I$ ; for example, $\gamma _{[a]}$ would be the first $a$ digits of $\gamma$ , and $\gamma _{ \{a\}}$ would just be the $a$ -th digit (for notation’s sake, let “ $\gamma _{[0]}$ ” be the empty string). Also let $\|\gamma \|$ denote the number of 1’s in $\gamma$ . We now present the following Lemma:

Lemma 13. Let $\alpha$ be an arbitrary $2(t_{end} - t_{start})$ length binary sequence with at most $L(n)$ 1’s, and let $\gamma$ be the random binary sequence according to either $\mathcal{P}$ or $\mathcal{Q}$ . Let $i \in [2(t_{end} - t_{start})]$ . Then (letting $a_{\{0\}} = 1$ for sake of notation):

\begin{equation*} \frac {\mathbb {P}_{\mathcal {P}}[\gamma _{[i]} = \alpha _{[i]} \mid \gamma _{[i-1]} = \alpha _{[i-1]}]}{\mathbb {P}_{\mathcal {Q}}[\gamma _{[i]} = \alpha _{[i]} \mid \gamma _{[i-1]} = \alpha _{[i-1]}]} \begin {cases} = 1 + O\left (\frac {1}{J(n)\ln (n)^{0.39}}\right ) &\text { if } \alpha _{\{i\}} = 0 \text { and } \alpha _{\{i-1\}} = 0 \\[9pt] = 1 + O\left (\frac {\ln (n)^{0.4}}{J(n)}\right ) &\text { if } \alpha _{\{i\}} = 0 \text { and } \alpha _{\{i-1\}} = 1 \\[9pt] = 1 + O\left (\frac {1}{\ln (n)^{0.79}}\right ) &\text { if } \alpha _{\{i\}} = 1 \text { and } \alpha _{\{i-1\}} = 0 \\[9pt] \leq 1 + O\left (\frac {1}{\ln (n)^{0.79}}\right ) &\text { if } \alpha _{\{i\}} = 1 \text { and } \alpha _{\{i-1\}} = 1. \end {cases}\end{equation*}

Proof. First, we consider the cases where $\alpha _{\{i \}} = 1$ . We have

(20) \begin{equation} \mathbb{P}_{\mathcal{Q}}[\gamma _{\{i\}} = 1 \mid \gamma _{[i-1]} = \alpha _{[i-1]}] = \frac{L(n) - \| \alpha _{[i-1]} \|}{2J(n) - (i-1)}. \end{equation}

For the probability space $\mathcal{P}$ , we need to consider three subcases: we need to consider whether $i$ is even or odd, and if it is even, whether $\alpha _{ \{i-1\} }$ is 0 or 1, since each step of the $d$ -process outputs two digits of the binary string. Let’s say that $\tau$ corresponds to the last step in the $d$ -process before the $i$ -th binary digit is generated (recall that pairs of digits are generated together). Then if $i$ is odd:

(21) \begin{align} \mathbb{P}_{\mathcal{P}}[\gamma _{\{i\}} = 1 \mid \gamma _{[i-1]} = \alpha _{[i-1]}] \nonumber &= -\frac{1}{2}\mathbb{E}\left[S^{(k)}_{\tau +1} - S^{(k)}_{\tau } | G_{\tau }\right] \\[2pt]&= \frac{S^{(k)}_{\tau }}{S^{(d-1)}_{\tau }}\left (1 + O\left (\frac{1}{dn-2{\tau }}\right ) \right ) \nonumber \qquad \text{by (3)}\\[2pt]&= \frac{S^{(k)}_{\tau }}{2 \lfloor dn/2-\tau \rfloor }\left (1 + O\left (\frac{1}{\ln (n)^{0.79}}\right )\right ) \nonumber \qquad \text{by (19)} \\[2pt]&= \frac{L(n) - \| \alpha _{[i-1]} \|}{2J(n) - (i-1)}\left (1 + O\left (\frac{1}{\ln (n)^{0.79}}\right )\right ). \end{align}

If $i$ is even and $\alpha _{ \{i-1\} } = 1$ , then $S^{(k)}_{\tau } = L(n) - \| \alpha _{[i-1]} \| + 1$ . At step $\tau$ there are $S^{(k)}_{\tau } \left(S^{(d-1)}_{\tau } + O(1)\right)$ ordered pairs of vertices whose first vertex has degree $k$ , and at most $2\binom{S^{(k)}_{\tau }}{2}$ ordered pairs of vertices both with degree $k$ ; hence:

(22) \begin{align} \mathbb{P}_{\mathcal{P}}[\gamma _{\{i\}} = 1 \nonumber \mid \gamma _{[i-1]} = \alpha _{[i-1]}] &\leq \frac{S^{(k)}_{\tau } - 1}{S^{(d-1)}_{\tau } + O(1)}\\[4pt]&= \frac{S^{(k)}_{\tau } - 1}{2 \lfloor dn/2-\tau \rfloor }\left (1 + O\left (\frac{1}{\ln (n)^{0.79}}\right )\right ) \qquad \text{by (19)} \nonumber \\[4pt]&= \frac{L(n) - \| \alpha _{[i-1]} \|}{2J(n) - (i-1)}\left (1 + O\left (\frac{1}{\ln (n)^{0.79}}\right )\right ). \end{align}

Hence the final inequality of the Lemma holds by (21) and (22).

Next, consider the case where $i$ is even and $\alpha _{ \{i-1\} } = 0$ ; here, $S^{(k)}_{\tau } = L(n) - \| \alpha _{[i-1]} \|$ once again. At step $\tau$ there are $\left(S^{(d-1)}_{\tau } - S^{(k)}_{\tau }\right) \left(S^{(d-1)}_{\tau } + O(1)\right)$ ordered pairs of vertices whose first vertex has degree greater than $k$ , and $S^{(k)}_{\tau } \left(S^{(d-1)}_{\tau } - S^{(k)}_{\tau } + O(1)\right)$ ordered pairs of vertices for which the first vertex has degree greater $k$ and the second vertex has degree $k$ (one can “pick the second vertex first” to see this). Hence:

(23) \begin{align} \mathbb{P}_{\mathcal{P}}[\gamma _{\{i\}} = 1 \nonumber \mid \gamma _{[i-1]} = \alpha _{[i-1]}] &= \frac{S^{(k)}_{\tau } }{S^{(d-1)}_{\tau }} \left (1 + O \left (\frac{1}{S^{(d-1)}_{\tau }}\right )\right ) \qquad \text{since}\ S^{(d-1)}_{\tau } \gg S^{(k)}_{\tau } \text{ there} \\[4pt]&= \frac{S^{(k)}_{\tau } }{2 \lfloor dn/2-\tau \rfloor } \left (1 + O \left (\frac{1}{\ln (n)^{0.79}}\right )\right ) \nonumber \qquad \text{by (19)} \\[4pt]&= \frac{L(n) - \| \alpha _{[i-1]} \|}{2J(n) - (i-1)}\left (1 + O\left (\frac{1}{\ln (n)^{0.79}}\right )\right ), \end{align}

hence the third equality of the Lemma holds by (21) and (23).

Now consider $\alpha _{\{i\}} = 0$ . By modifying (20) to accommodate $\gamma _{\{i\}} = 0$ , we have

(24) \begin{equation} \mathbb{P}_{\mathcal{Q}}[\gamma _{\{i\}} = 0 \mid \gamma _{[i-1]} = \alpha _{[i-1]}] = 1 - \frac{L(n) - \| \alpha _{[i-1]} \|}{2J(n) - (i-1)}. \end{equation}

Similarly, by modifying (21) and (23), if $a_{\{i-1\}} = 0$ then

(25) \begin{equation} \mathbb{P}_{\mathcal{P}}[\gamma _{\{i\}} = 0 \mid \gamma _{[i-1]} = \alpha _{[i-1]}] = 1 - \frac{L(n) - \| \alpha _{[i-1]} \|}{2J(n) - (i-1)}\left (1 + O\left (\frac{1}{\ln (n)^{0.79}}\right )\right ). \end{equation}

By modifying (21) and (22), if $a_{\{i-1\}} = 1$ , then

(26) \begin{align} \mathbb{P}_{\mathcal{P}}[\gamma _{\{i\}} = 0 \mid \gamma _{[i-1]} = \alpha _{[i-1]}] & \geq 1 - \frac{L(n) - \| \alpha _{[i-1]} \|}{2J(n) - (i-1)}\left (1 + O\left (\frac{1}{\ln (n)^{0.79}}\right )\right ) \nonumber \\[4pt]&= 1 + O\left (\frac{L(n) - \| \alpha _{[i-1]} \|}{2J(n) - (i-1)}\right ). \end{align}

Since $L(n) = \Theta (\ln (n)^{0.2})$ and

\begin{equation*}2J(n) - (i-1) = 2 \lfloor dn/2 - \tau \rfloor = \Omega (dn/2 - t_{end}) = \Omega (J(n)/\ln (n)^{0.2}),\end{equation*}

then $\frac{L(n) - \| \alpha _{[i-1]} \|}{2J(n) - (i-1)} = O\left (\frac{\ln (n)^{0.4}}{J(n)}\right )$ . Therefore the ratio of (25) and (24) is $1 + O \left (\frac{1}{J(n) \ln (n)^{0.39}}\right )$ , verifying the first inequality of the Lemma, and the ratio of (26) and (24) is $1 + O \left (\frac{\ln (n)^{0.4}}{J(n)}\right )$ , verifying the second inequality of the Lemma.

Proof of Theorem 12. First, let $\alpha$ be an arbitrary string which satisfies the criteria in Lemma 13. By using the Lemma 13 recursively:

(27) \begin{align} \frac{\mathbb{P}_{\mathcal{P}}[\gamma = \alpha ]}{\mathbb{P}_{\mathcal{Q}}[\gamma = \alpha ]} &\leq \exp \left \{ O\left (J(n) \frac{1}{J(n) \ln (n)^{0.39}} + L(n) \left (\frac{1}{\ln (n)^{0.79}} + \frac{\ln (n)^{0.4}}{J(n)}\right )\right ) \right \}\nonumber \\[4pt]&= 1 + o(1), \end{align}

and if $\alpha$ is an arbitrary string with no two consecutive 1’s which satisfies the criteria in Lemma 13, then by similar logic,

(28) \begin{equation} \frac{\mathbb{P}_{\mathcal{P}}[\gamma = \alpha ]}{\mathbb{P}_{\mathcal{Q}}[\gamma = \alpha ]} = 1 + o(1). \end{equation}

Let $\mathcal{C}$ be the event that $\gamma$ has two consecutive 1’s; we consider $\mathbb{P}[ \, \mathcal{C} \mid G_{t_{start}} = H]$ . We consider probability space $\mathcal{Q}$ first. Recall that $\alpha$ is a string that has $\sim 2J(n) = \Omega (\ln (n)^{1.2})$ characters and at most $L(n) = \Theta (\ln (n)^{0.2})$ 1’s. Because $\mathcal{Q}$ is a truncation of a uniform distribution, the probability of having two consecutive 1’s will be $O\left (\frac{(L(n))^2}{J(n)}\right ) = O(\ln (n)^{-0.8})$ . Hence, by (27) we must have

(29) \begin{equation} \mathbb{P}_{\mathcal{P}}[ \, \mathcal{C} \mid G_{t_{start}} = H] = o(1) \ \text{ and } \ \mathbb{P}_{\mathcal{Q}}[ \, \mathcal{C} \mid G_{t_{start}} = H] = o(1). \end{equation}

We now combine (28) and (29) to prove Theorem 12 (for ease of notation, assume we are given $G_{t_{start}} = H$ ):

\begin{align*} \mathbb{P}\left [S^{(k)}_{\lfloor \frac{dn}{2} - \frac{r J(n)}{L(n)} \rfloor } = 0 \right ] &= \mathbb{P}_{\mathcal{P}}[\|\gamma \| = L(n)] \\[4pt]&= \mathbb{P}_{\mathcal{P}}[\|\gamma \| = L(n) \ | \ \mathcal{C}]\mathbb{P}_{\mathcal{P}}[\mathcal{C}] + \mathbb{P}_{\mathcal{P}}[\|\gamma \| = L(n) \ | \ \overline{\mathcal{C}}]\mathbb{P}_{\mathcal{P}}[\overline{\mathcal{C}}] \\[4pt]&= \mathbb{P}_{\mathcal{Q}}[||\gamma || = L(n)] + o(1) \qquad \text{by (28) and (29)}\\[4pt]&= \frac{\binom{2(t_{end}-t_{start})}{L(n)}}{\binom{2J(n)}{L(n)}} + o(1) \\[4pt]&= \frac{\binom{2J(n) - 2(\lfloor r J(n)/L(n) \rfloor )}{L(n)}}{\binom{2J(n)}{L(n)}} + o(1) \\[4pt]&= \left (1-\frac{r(1+ o(1))}{L(n)}\right )^{L(n)} + o(1) \\[4pt]& = e^{-r} + o(1). \end{align*}

We can now complete the proof of the second statement of Theorem 8 at value $k$ . Roughly speaking, we will use Theorem 12 with $L(n) \approx \frac{2(d-1)! \ln (n)^{0.2}}{k!}$ , so $\frac{dn}{2} - i(r_{k},k) \approx \frac{r_k J(n)}{L(n)}.$ First, note that $S^{(k)}_{i_{after}(k)} = 0$ (w.h.p.) comes automatically when the rest of the statement is proved (by putting $r_{\ell } = 0$ for $\ell \lt k$ and having $r_k \to 0$ ). Let $\mathcal{G}_{\ell }$ be the event that $S^{(\ell )}_{ \lfloor i(r_{\ell },\ell )\rfloor } = 0$ and $\mathcal{G} = \bigcap _{\ell \leq k} \mathcal{G}_{\ell }$ , let $\mathcal{F}$ be the event that (16) holds for $i = i_{before}(k)$ and $S^{(j)}_{i_{before(k)}} = 0$ for $j \lt k$ , and let $\mathcal{A} = \mathcal{F} \cap \bigcap _{\ell \lt k} \mathcal{G}_{\ell }$ . Also, let

\begin{equation*}\mathcal {I} = [ns_{k}(i_{before}(k)/n) - 4E_k(i_{before}(k)),ns_{k}(i_{before}(k)/n) + 4E_k(i_{before}(k))].\end{equation*}

Note that, by part 1 of Theorem 8, by induction on the second part Theorem 8, and since $i_{before}(k) \gt i_{after}(k-1)$ , $\mathcal{F}$ happens with probability $1 - o(1)$ . Therefore:

\begin{align*} \mathbb{P}[\mathcal{G}] &= \mathbb{P}\left [\mathcal{G}_{k} \cap \mathcal{A} \right ] + o(1)\\[5pt]&= \sum _{p \in \mathcal{I}} \mathbb{P}\left [\mathcal{G} \ \bigg | \ \mathcal{A} \cap \left(S^{(k)}_{i_{before}(k)} = p\right) \right ] \mathbb{P}\left [\mathcal{A} \cap \left(S^{(k)}_{i_{before}(k)} = p\right) \right ] + o(1). \end{align*}

We can now apply (10), (15), and Theorem 12 to get

\begin{equation*}\mathbb {P}\left [\mathcal {G} \ \bigg | \ \mathcal {A} \cap \left(S^{(k)}_{i_{before}(k)} = p\right) \right ] = e^{-r_{\ell }} + o(1)\end{equation*}

for $p \in \mathcal{I}$ . We note that all $o(1)$ functions in the sum can be made to be the same by carefully reviewing the proof of Theorem 12. Therefore:

\begin{align*} \mathbb{P}[\mathcal{G}] &= \sum _{p \in \mathcal{I}} (e^{-r_{\ell }} + o(1)) \mathbb{P}\left [\mathcal{A} \cap \left(S^{(k)}_{i_{before}(k)} = p\right) \right ] +o(1) \qquad \\&= e^{-r_{\ell }} \sum _{p \in \mathcal{I}}\mathbb{P}\left [\mathcal{A} \cap \left(S^{(k)}_{i_{before}(k)} = p\right) \right ] + o(1) \\&= e^{-r_{\ell }} \mathbb{P}\left [\bigcap _{\ell \lt k} \mathcal{G}_{\ell }\right ] + o(1) \qquad \text{by Theorem 8} \\&= \exp \left \{ \sum _{\ell = 0}^{k} e^{- r_{\ell }} \right \} + o(1) \qquad \text{by induction on Theorem 8,} \end{align*}

proving Theorem 8.