Time Complexity Analysis of Randomized Search Heuristics for the Dynamic Graph Coloring Problem

The claim essentially follows from the proofs of Theorems 3 and 5 in [28] where the authors investigate an equivalent problem on cycle graphs. Hence, we just sketch the idea here. Imagine we link two properly colored paths of length $n/2$ each with an edge $(u,v)$ which introduces a single conflict. The conflict splits the path into two paths that are properly colored and joined by a conflicting edge. Consider the length of the shortest properly colored path. As argued in [28], both RLS and (1+1) EA can either increase or decrease this length in fitness-neutral operations like recoloring one of the vertices involved in the conflict. If it has decreased to 1, the conflict has been propagated down to a leaf node where a single bit flip can get rid of it. Fischer and Wegener calculate bounds for the expected number of steps until this number reaches its minimum 1. This is achieved by estimating the number of so-called relevant steps, which either increase or decrease the length of the shortest properly colored path. The probability for a relevant step is $\Theta(1/n)$. The expected number of relevant steps is $\Theta(n^{2})$ since we have a fair random walk on states up to $n/2$. In summary, this results in a runtime bound of $\Theta(n^{3})$.

The proof uses and refines arguments from [29]. Let $e=\{r,v\}$ be the added edge with $r$ being the root of the $n$-vertex complete binary tree. If $c(r)\neq c(v)$ we are done and the coloring is already a proper 2-coloring. Hence, we assume that $c(r)=c(v)$ and there is exactly one conflict. This situation is a worst-case situation in vertex-coloring of complete binary trees, since many vertices must be recolored in the same mutation to produce an accepted candidate solution (see Figure 1 (left)). Let $\mathrm{OPT}$ be the set of the two possible proper colorings and let $A_{i}$, for $0\leq i\leq\log(n)-1$ be the set of colorings with one conflict such that the parent vertex of the conflicting edge has (graph) distance $i$ to the root. We have $\sum_{i=0}^{\log(n)-1}|A_{i}|=2n-2$ since we can choose the position of the conflicting edge among $n-1$ edges and there are two possible colors for its vertices. By the same argument, $|A_{0}|=4$ and $|A_{1}|=8$.

Starting with a coloring $x\in A_{0}$, the probability of reaching $\mathrm{OPT}$ in one mutation is at most $n^{-(n-1)/2}+n^{-(n+1)/2}=O(n^{-(n-1)/2})$ since all vertices on either side of the conflicting edge must be recolored in one mutation. The probability of reaching $A_{1}$ in one mutation is $\Omega(n^{-(n+1)/4})$ since a sufficient condition is to flip $v$ and all the vertices in one of $v$’s subtrees (see Figure 1 (left)). Since each subtree of $v$ has $(n-3)/4$ vertices, this means flipping $1+(n-3)/4=(n+1)/4$ many vertices. This probability is also $O(n^{-(n+1)/4})$ since the only other way to create some coloring in $A_{1}$ is to flip $r$, the sibling of $v$ (that we call $w$), and one of $w$’s subtrees. The probability to reach any solution in $\bigcup_{i=2}^{\log(n)-1}A_{i}$ is $O(n^{-(n+1)/4})$ as well since more than $(n+1)/4$ vertices would have to flip and there are $2n-2-|A_{0}|-|A_{1}|=2n-14$ solutions in $\bigcup_{i=2}^{\log(n)-1}A_{i}$. This implies the claimed lower bound as the probability to escape from $A_{0}$ in one mutation is $\Theta(n^{-(n+1)/4})$.

To show the claimed upper bound, we argue that in $\Theta(n^{(n+1)/4})$ expected time we do escape from $A_{0}$. If $\mathrm{OPT}$ is reached, we are done. Hence, we assume that $\bigcup_{i=1}^{\log(n)-1}A_{i}$ is reached. For each coloring in this set, there is a proper coloring within Hamming distance at most $(n-3)/4$ since, if $\{u,v\}$ denotes the conflicting edge with $u$ being the parent of $v$, it is sufficient to recolor the subtree at $v$ and this subtree has at most $(n-3)/4$ vertices (see Figure 1 (right)). Thus, the expected time to either reach $\mathrm{OPT}$ or to go back to $A_{0}$ is $O(n^{(n-3)/4})$. Since at least $(n+1)/4$ vertices would have to flip to go back to $A_{0}$ (and $|A_{0}|=O(1)$), the conditional probability to go back to $A_{0}$ is at most $O(1/n)$. If this happens, we repeat the above arguments; this clearly does not change the asymptotic runtime and we have shown an upper bound of $O(n^{(n+1)/4})$.

We first consider Kempe chains and distinguish between two cases: the Kempe chain involves colors $\Delta$ and $\Delta+1$ and the case that it involves colors $\Delta$ and a smaller color $c<\Delta$. Kempe chains involving two colors other than $\Delta$ cannot change the number of $\Delta$-colored vertices (albeit this may still happen in a subsequent Grundy local search, if $(\Delta+1)$-colored vertices are recolored with color $\Delta$). We start with the case of colors $\Delta$ and $\Delta+1$.

Assume that there exists a vertex $v$ that is being recolored from $\Delta+1$ to $\Delta$ in the Kempe chain (if no such vertex exists, the claim holds trivially). Since the coloring is a Grundy coloring, $v$ must have vertices of all colors in $\{1,\dots,\Delta\}$ in its neighborhood, and there can only be one vertex of each color (owing to the degree bound $\Delta$). Let $w$ be the $\Delta$-colored vertex and note that $w$ must have all colors from 1 to $\Delta-1$ in its neighborhood. Thus, $w$ must have neighbors $w_{1},\dots,w_{\Delta-1}$ such that $w_{i}$ is colored $i$. Since $w$ also has $v$ as its neighbor, $w$ cannot have any further neighbors apart from $v,w_{1},\dots,w_{\Delta-1}$; in particularly, $w$ cannot have any further $(\Delta+1)$-colored vertices as neighbors. Hence the subgraph $H_{\Delta(\Delta+1)}$ induced by vertices colored $\Delta$ or $\Delta+1$ contains $\{v,w\}$ as a connected component and a Kempe chain on this component will simply swap the colors of $v$ and $w$ without increasing the number of $\Delta$-colored vertices.

Now assume that the other color is $c<\Delta$. We show that in the subgraph $H_{c\Delta}$ induced by vertices colored $c$ or $\Delta$, the number of $c$-colored vertices in every connected component of $H_{c\Delta}$ is at most 1 larger than the number of $\Delta$-colored vertices. This implies the claim since a Kempe chain operation swaps colors $c$ and $\Delta$ in one connected component of $H_{c\Delta}$.

Every $\Delta$-colored vertex $w$ needs to have colors $\{1,\dots,\Delta-1\}$ in its neighborhood since the coloring is a Grundy coloring. Since the maximum degree is $\Delta$, $w$ can have at most two $c$-colored neighbors.

Consider a connected component of $H_{c\Delta}$ that contains a $c$-colored vertex $v$ (if no such vertex exists, the claim holds trivially). Imagine the breadth-first search (BFS) tree generated by running BFS in $H_{c\Delta}$ starting at $c$. Note that colors are alternating at different depths of the BFS tree, with $\Delta$-colored vertices at odd depths and $c$-colored vertices at even depths from the root. For odd depths $d$, all $\Delta$-colored vertices can only have 1 $c$-colored vertex at depth $d+1$ since they are already connected to one $c$-colored vertex at depth $d-1$. Hence there are at least as many $\Delta$-colored vertices at depth $d$ as $c$-colored vertices at depth $d+1$. Using this argument for all odd values of $d$ and noting that the root vertex $v$ is $c$-colored proves the claim.

The number of $(\Delta+1)$-colored vertices is non-increasing by design of the selection operator. Moreover, the number of $\Delta$-colored vertices can only increase if the number of $(\Delta+1)$-colored vertices decreases at the same time. If there are no $(\Delta+1)$-colored vertices, the number of $\Delta$-colored vertices is non-increasing. Hence we only need to consider the case where there is at least one $(\Delta+1)$-colored vertex.

The proof of Lemma 6 revealed that every Kempe chain or color elimination can only increase the number of $\Delta$-colored vertices by 1. Moreover, this can only happen if a Kempe chain affects a connected component $C$ of $H_{c\Delta}$, for a color $c<\Delta$, such that $C$ has one more $c$-colored vertex than $\Delta$-colored vertices. For this operation to be accepted by selection, the following Grundy local search must reduce the number of $(\Delta+1)$-colored vertices by at least 1.

The Grundy number of binary trees is at most $\Gamma\leq{\Delta+1}\leq 4$. By Lemma 8, the number of vertices colored 3 or 4 is at most $T$.

By design of our selection operator, the number of 4-colored vertices is non-increasing over time. For every 4-colored vertex $v$ there must be a Kempe chain operation recoloring a neighboring vertex whose color only appears once in the neighborhood of $v$. If there are $i$ 4-colored vertices, the probability of reducing this number is $\Omega(i/n)$ and the expected time for color 4 to disappear is $O(n)\cdot\sum_{i=1}^{T}1/i=O(n\log^{+}T)$.

Consider the connected components of the original graph. If an edge is added that runs within one connected component, it cannot create a conflict. This is because the connected component is properly 2-colored, with all vertices of the same color belonging to the same set of the bipartition. Since the graph is bipartite after edge insertions, the new edge must connect two vertices of different colors. Hence added edges can only create a conflict if they connect two different connected components that are colored inversely to each other.

Consider the subgraph induced by the added edges that are conflicting, and pick a connected component $C$ in this subgraph. Note that all vertices in $C$ have the same color $c\in\{1,2\}$ before Grundy local search is applied. Now Grundy local search will fix these conflicts by increasing the colors of vertices in $C$. We bound the value of the largest color $c_{\max}$ used and first consider the case where the largest color is $c_{\max}\geq 4$. For Grundy local search to assign a color $c_{\max}\geq 4$ to a vertex $v\in C$, all colors $1,\dots,c_{\max}-1$ must occur in the neighborhood of $v$ in the new graph. In particular, $C$ must contain vertices $v_{3},v_{4},\dots,v_{c_{\max}-1}$ respectively colored $3,\dots,c_{\max}-1$ that are neighbored to $v$. This implies that $c_{\max}-3$ edges incident to $v$, connecting $v$ to a smaller color, must have been added during the dynamic change. Applying the same argument to $v_{3},v_{4},\dots,v_{c_{\max}-1}$ yields that there must be at least $\sum_{j=1}^{c_{\max}-3}j=(c_{\max}-3)(c_{\max}-2)/2$ inserted edges in $C$. Thus $(c_{\max}-3)(c_{\max}-2)/2\leq T$, which implies $(c_{\max}-3)^{2}\leq 2T$ and this is equivalent to $c_{\max}\leq\sqrt{2T}+3$. Also $c_{\max}\leq\Gamma$ by definition of the Grundy number.

Now we can argue as in [30, Theorem 3]: the largest color can be eliminated from any bipartite graph in expected time $O(n\log n)$. (Note that these color eliminations can increase the number of vertices colored with large colors, so long as the number of the vertices with the largest color decreases.) Since at most $c_{\max}-2$ colors have to be eliminated, a bound of $O(c_{\max}n\log n)$ follows. Plugging in $c_{\max}=O(\min\{\sqrt{T},\Gamma\})$ completes the proof.

If no end point of an added edge is neighbored to an end point of another added edge, Grundy local search will only create colors up to 3. This is because Grundy local search will only increase the color of end points of added edges, and the condition implies that the colors of neighbors of all end points will remain fixed. Hence, Grundy local search will recolor vertices independently from each other. If $\Gamma\leq 4$, the largest color value is 4. Lemma 7 states that the number of vertices colored 3 or 4 cannot increase.

Following [30, Theorem 3], while there are $i$ vertices colored 4, a color elimination choosing such a vertex will lead to a smaller free color, reducing the number of 4-colored vertices. The expected time for this to happen is at most $n/i$, hence the total expected time to eliminate all color-4 vertices is at most $\sum_{i=1}^{T}n/i=O(n\log^{+}T)$. The same argument then applies to all 3-colored vertices.

Choose $T_{n}$ as the union of an isolated edge $\{u,v\}$ where $c(u)=2$ and $c(v)=1$ and a tree where the root $r$ has $N-1:={(n-3)/2}$ children and every child has exactly one leaf (cf. Figure 2). This graph was also used in [30] as an example where ILS with Kempe chains fail in a static setting. Since $n\equiv 1\bmod 4$, $N$ is an even number. Every feasible 2-coloring will color the root and the leaves in the same color and the root’s children in the remaining color. Assume the root and leaves are colored 2 as the other case is symmetric. Now add an edge $\{r,u\}$ to the graph. This creates a star of depth 2 (termed the depth-2 star in the following) where the root is the center and the root now has $N$ children.

This creates a conflict at $\{r,u\}$ that is being resolved by recoloring one of these vertices to color 3 in the next Grundy local search. With probability $1/2$, this is the root $r$.

From this situation, any Kempe chain affecting any vertex in $V\setminus\{r\}$ can swap the colors on an edge incident to a leaf. Let $X_{0},X_{1},\dots$ denote the random number of leaves colored 1, starting with $X_{0}=1$. We only consider steps in which this number is changed; note that the probability of such a change is $\Theta(1)$ as every Kempe chain on any vertex except for the root changes $X_{t}$ if an appropriate color value is chosen. There are $N:=(n-1)/2$ leaves and the number of 1-colored leaves performs a random walk biased towards $N/2$: $\text{Pr}\left(X_{t+1}=X_{t}+1\mid X_{t}\right)=(N-X_{t})/N$ and $\text{Pr}\left(X_{t+1}=X_{t}-1\mid X_{t}\right)=X_{t}/N$. This process is known as the Ehrenfest urn model: imagine two urns labelled 1 and 2 that together contain $N$ balls. In each step, we pick a ball uniformly at random and move it to the other urn. If $X_{t}$ denotes the number of balls in urn 1 at time $t$, we obtain the above transition probabilities.⁴⁴4This simple model was originally proposed to describe the process of substance exchange between two bordering containers of equal size which are separated by a permeable membrane. Consider $N$ particles spread across the containers and denote by $X_{t}$ the number of particles in the left container w. l. o. g. at time $t$. In each step one particle is chosen uniformly at random and swaps sides.

When $X_{t}\in\{0,N\}$ then a proper 2-coloring has been found. As long as $X_{t}\in\{2,\dots,N-2\}$, all Kempe chain moves involving the root will be rejected as the number of 3-colored vertices would increase. While $X_{t}\in\{1,N-1\}$ a Kempe chain move recoloring the root with the minority color will be accepted. This has probability $1/n\cdot 1/(N-1)=\Theta(1/N^{2})$ (as the color is chosen uniformly from $\{1,\dots,\deg(r)+1\}$) and then the following Grundy local search will produce a proper 2-coloring. Also considering possible transitions to neighbouring states $0$ or $N$, while $X_{t}\in\{1,N-1\}$ the conditional probability that a proper 2-coloring is found before moving to a state $X_{t}\in\{2,N-2\}$ is $\Theta(1/N)$.

For the Ehrenfest model it is known that the expected time to return to an initial state of $1$ is $1!(N-1)!/N!\cdot 2^{N}=2^{N}/N$ [40, equation (66)]. It is easy to show that this time remains in $\Theta(2^{N}/N)$ when considering $N-1$ as a symmetric target state, and when conditioning on traversing states $\{2,\dots,N-2\}$. A rigorous proof for this statement is given in the Lemma 12 stated after this proof.

Let $T_{a\to b}$ denote the first-passage time from a state $a$ to a state $b$ and $T_{a\to B}$ for a set $B$ denote the first-passage time from $a$ to any state in $B$. For the Ehrenfest model it is known that the expected time to return to a state of $1$ from a state of $1$ is $\text{E}\left(T_{1\to 1}\right)=1!(N-1)!/N!\cdot 2^{N}=2^{N}/N$ [40, equation (66)]. The Ehrenfest model starting in state 1 can return to state 1 either via state 0 or larger states. Since the former takes exactly 2 steps, the expected return time via larger states is also $\Theta(2^{N}/N)$ by the law of total expectation.