Fast diversified coherent core search on multi-layer graphs

Abstract

Mining dense subgraphs on multi-layer graphs is an interesting problem, which has witnessed lots of applications in practice. To overcome the limitations of the quasi-clique-based approach, we propose d-coherent core (d-CC), a new notion of dense subgraph on multi-layer graphs, which has several elegant properties. We formalize the diversified coherent core search (DCCS) problem, which finds kd-CCs that can cover the largest number of vertices. We propose a greedy algorithm with an approximation ratio of $1 - 1/e$ and two search algorithms with an approximation ratio of 1/4. Furthermore, we propose some optimization techniques to further speed up the algorithms. The experiments verify that the search algorithms are faster than the greedy algorithm and produce comparably good results as the greedy algorithm in practice. As opposed to the quasi-clique-based approach, our DCCS algorithms can fast detect larger dense subgraphs that cover most of the quasi-clique-based results.

Appendices

Missing proofs

Proof

(Property 1) Suppose $C^{d}_{L}({\mathcal {G}})$ is not unique. Let $C_1$ and $C_2$ be two distinct d-CCs of ${\mathcal {G}}$ w.r.t. L. Let $C = C_1 \cup C_2$. We have $C_1 \subset C$ and $C_2 \subset C$. On each layer $i \in L$, $G_i[C_1]$ is a subgraph of $G_i[C]$. Thus, for each vertex $v \in C$, we have $d_{G_i[C]}(v) \ge d_{G_i[C_1]}(v) \ge d$ for all $i \in L$. By the definition of d-CC, C is also a d-CC, so neither $C_1$ nor $C_2$ is maximum, which leads to a contradiction. Thus, $C^{d}_{L}({\mathcal {G}})$ is unique. $\square$

Proof

(Property 2) Let $d_1, d_2 \in {\mathbb {N}}$ and $d_1 > d_2$. For each vertex $v \in C^{d_1}_{L}({\mathcal {G}})$, we have $d_{G_l[C^{d_1}_{L}({\mathcal {G}})]}(v) \ge d_1 > d_2$ for every layer number $l \in L$. By the definition of d-CC, $C^{d_1}_{L}({\mathcal {G}}) \subseteq C^{d_2}_{L}({\mathcal {G}})$. Thus, the property holds. $\square$

Proof

(Property 3) For each vertex $v \in C^{d}_{L^{\prime }}({\mathcal {G}})$, we have $d_{G_l[C^{d}_{L^{\prime }}({\mathcal {G}})]}(v) \ge d$ for each layer number $l \in L$. Based on the definition of d-CC, we have $C^{d}_{L^{\prime }}({\mathcal {G}}) \subseteq C^{d}_{L}({\mathcal {G}})$. Hence, the property holds. $\square$

Proof

(Lemma 1) It is clear that $L_1 \subseteq L_1 \cup L_2$ and $L_2 \subseteq L_1 \cup L_2$. By Property 3, we have $C_{L_1 \cup L_2}^{d} ({\mathcal {G}}) \subseteq C_{L_1}^{d} ({\mathcal {G}})$ and $C_{L_1 \cup L_2}^{d} ({\mathcal {G}}) \subseteq C_{L_2}^{d} ({\mathcal {G}})$. Thus, $C_{L_1 \cup L_2}^{d} ({\mathcal {G}}) \subseteq C_{L_1}^{d} ({\mathcal {G}}) \cap C_{L_2}^{d}({\mathcal {G}})$. $\square$

Proof

(Lemma 2) Let $C^d_{L^{\prime }}({\mathcal {G}})$ be a descendant of $C^d_{L}({\mathcal {G}})$. We have $L \subseteq L^{\prime }$. By Property 3, we have $C^d_{L^{\prime }}({\mathcal {G}}) \subseteq C^d_{L}({\mathcal {G}})$. Thus, $\textsf {Cov}(({\mathcal {R}}- \{C^{*}({\mathcal {R}})\}) \cup \{C^d_{L^{\prime }}({\mathcal {G}})\}) \subseteq \textsf {Cov}(({\mathcal {R}}- \{C^{*}({\mathcal {R}})\}) \cup \{C^d_{L}({\mathcal {G}})\})$. Obviously, if we have $|\textsf {Cov}(({\mathcal {R}}- \{C^{*}({\mathcal {R}})\}) \cup \{C^d_{L}({\mathcal {G}})\})| < \left( 1 + \frac{1}{k}\right) |\textsf {Cov}({\mathcal {R}})|$, we must have $|\textsf {Cov}(({\mathcal {R}}- \{C^{*}({\mathcal {R}})\}) \cup \{C^d_{L^{\prime }}({\mathcal {G}})\})| < \left( 1 + \frac{1}{k}\right) |\textsf {Cov}({\mathcal {R}})|$. Thus, $C^d_{L^{\prime }}({\mathcal {G}})$ cannot satisfy Eq. (1), which means none of the descendants of $C^d_{L}({\mathcal {G}})$ satisfies Eq. (1). $\square$

Proof

(Lemma 3) For any subset $D \subseteq L_P$ such that $|D| = s - |L|$, since $I_{D} = \cap _{i \in D} C^{d}(G_i)$, we have $I_D \subseteq C^{d}(G_i)$ for each $i \in D$. Consequently, for each vertex $v \in I_D$, we have $v \in C^{d}(G_i)$ for each $i \in D$. That is, v must be contained in at least $s - |L|$d-cores on the layers in $L_P$, so $v \in I$. Therefore, we have $I_D \subseteq I$. $\square$

Proof

(Lemma 4) Let $C^d_{S}({\mathcal {G}})$ be a descendant of $C^d_{L}({\mathcal {G}})$ such that $|S| = s$. Let $D = S - L$ and $I_D = \cap _{i \in D} C^{d}(G_i)$. By Lemma 1, we have $C^d_{S}({\mathcal {G}}) \subseteq C^d_{L}({\mathcal {G}}) \cap I_D$. Since $I_D \subseteq I$ according to Lemma 3, we have $C^d_{S}({\mathcal {G}}) \subseteq C^d_{L}({\mathcal {G}}) \cap I$. For ease of presentation, let $C = C_{L}^d({\mathcal {G}}) \cap I$. We illustrate the relationships between $\textsf {Cov}({\mathcal {R}})$, $C^{*}({\mathcal {R}})$ and C in Fig. 40 with seven disjoint subsets A, B, D, E, F, G, and H. We have $|\textsf {Cov}({\mathcal {R}})| = |A| + |B| + |D| + |F| + |G| + |H|, |C^{*}({\mathcal {R}})| = |B| + |D| + |G| + |H|, |C| = |D| + |E| + |F| + |G|, |{\varDelta }({\mathcal {R}}, C^*({\mathcal {R}}))| = |D| + |H|$.

Since $|C| < \frac{1}{k}|\textsf {Cov}({\mathcal {R}})| + |{\varDelta }({\mathcal {R}}, C^*({\mathcal {R}}))|$, we have $|D| +|E| + |F| + |G| < \frac{1}{k} (|A| + |B| + |D| + |F| + |G| + |H|) + |D| + |H|$. Thus,

$$\begin{aligned} \begin{aligned}&|\textsf {Cov}(({\mathcal {R}}- \{C^{*}({\mathcal {R}})\}) \cup \{C\})| \\&\quad = |A| + |B| + |D| + |E| + |F| + |G| \\&\quad < \frac{1}{k} (|A| + |B| + |D| + |G| + |F| + |H|) + |A|\\&\quad + |B| + |D| + |H| \\&\quad \le \left( 1 + \frac{1}{k}\right) (|A| + |B| + |D| + |G| + |F| + |H|) \\&\quad = \left( 1 + \frac{1}{k}\right) |\textsf {Cov}({\mathcal {R}})|. \end{aligned} \end{aligned}$$

Since $C^d_{S}({\mathcal {G}}) \subseteq C$, we have $\textsf {Cov}(({\mathcal {R}}- \{C^{*}({\mathcal {R}})\}) \cup \{C^{d}_{S}({\mathcal {G}})\}) \subseteq \textsf {Cov}(({\mathcal {R}}- \{C^{*}({\mathcal {R}})\}) \cup \{C\})$. Then, we have $|\textsf {Cov}(({\mathcal {R}}- \{C^{*}({\mathcal {R}})\}) \cup \{C^{d}_{S}({\mathcal {G}})\})| \le |\textsf {Cov}(({\mathcal {R}}- \{C^{*}({\mathcal {R}})\}) \cup \{C\})| < \left( 1 + \frac{1}{k}\right) |\textsf {Cov}({\mathcal {R}})|$. Thus, the lemma thus holds. $\square$

Fig. 40

Relationships between $\textsf {Cov}({\mathcal {R}})$, $C^{*}({\mathcal {R}})$, and C

Proof

(Lemma 5) By the definitions of d-CC and d-core, we have $C^d(G_j) = C^d_{\{j\}}({\mathcal {G}})$. By Lemma 1, we have $C^{d}_{L \cup \{j\}}({\mathcal {G}}) \subseteq C_{L}^d({\mathcal {G}}) \cap C^d(G_j)$. Let $C = C_{L}^d({\mathcal {G}}) \cap C^d(G_j)$, in similar to the proof of Lemma 4, if $|C| < \frac{1}{k}|\textsf {Cov}({\mathcal {R}})| + |{\varDelta }({\mathcal {R}}, C^*({\mathcal {R}}))|$, we have $\textsf {Cov}(({\mathcal {R}}- \{C^{*}({\mathcal {R}})\}) \cup \{C^{d}_{L \cup \{j\} }({\mathcal {G}})\}) \subseteq \textsf {Cov}(({\mathcal {R}}- \{C^{*}({\mathcal {R}})\}) \cup \{C\})$. Then, we must have $|\textsf {Cov}(({\mathcal {R}}- \{C^{*}({\mathcal {R}})\}) \cup \{C^{d}_{L \cup \{j\} }({\mathcal {G}})\})|\le |\textsf {Cov}(({\mathcal {R}}- \{C^{*}({\mathcal {R}})\}) \cup \{C\})|< \left( 1 + \frac{1}{k}\right) |\textsf {Cov}({\mathcal {R}})|$. The lemma thus holds. $\square$

Proof

(Lemma 6) Since $L \subseteq S$, we have $L \cup \{j\} \subseteq S \cup \{ j\}$. According to Property 3, we have $C^d_{S \cup \{j\}}({\mathcal {G}}) \subseteq C^d_{L \cup \{j\}}({\mathcal {G}})$. Therefore, $\textsf {Cov}(({\mathcal {R}}- \{C^{*}({\mathcal {R}})\}) \cup \{C^d_{S \cup \{j\}}({\mathcal {G}})\}) \subseteq \textsf {Cov}(({\mathcal {R}}- \{C^{*}({\mathcal {R}})\}) \cup \{C^d_{L \cup \{j\}}({\mathcal {G}})\})$. Since $C^d_{L \cup \{j\}}({\mathcal {G}})$ does not satisfy Eq. (1), we must have $|\textsf {Cov}(({\mathcal {R}}- \{C^{*}({\mathcal {R}})\}) \cup \{C^d_{S \cup \{j\}}({\mathcal {G}})\})| < \left( 1 + \frac{1}{k}\right) |\textsf {Cov}({\mathcal {R}})|$. Thus, the lemma holds. $\square$

Proof

(Lemma 7) According to the usage of potential vertex sets, for any descendant $C^d_{L^{\prime }}({\mathcal {G}})$ of $C^d_L({\mathcal {G}})$ with $|L^{\prime }| = s$, we have $C^d_{L^{\prime }}({\mathcal {G}}) \subseteq U^d_{L}({\mathcal {G}})$. In similar to the proof of Lemma 2, the lemma holds. $\square$

Proof

(Lemma 8) Similar to the proof of Lemma 5, we have $|\textsf {Cov}(({\mathcal {R}}- \{C^{*}({\mathcal {R}})\}) \cup \{U^d_{L - \{ j\}}({\mathcal {G}})\})| < \left( 1 + \frac{1}{k}\right) |\textsf {Cov}({\mathcal {R}})|$. According to the usage of potential sets, for any descendant $C^d_{L^{\prime }}({\mathcal {G}})$ of $C^d_L({\mathcal {G}})$ with $|L^{\prime }| = s$, we have $C^d_{L^{\prime }}({\mathcal {G}}) \subseteq U^d_{L}({\mathcal {G}})$. Thus, we must have $|\textsf {Cov}(({\mathcal {R}}- \{C^{*}({\mathcal {R}})\}) \cup \{C^d_{L^{\prime }}({\mathcal {G}})\})| \le |\textsf {Cov}(({\mathcal {R}}- \{C^{*}({\mathcal {R}})\}) \cup \{U^d_{L - \{ j\}}({\mathcal {G}})\})| < \left( 1 + \frac{1}{k}\right) |\textsf {Cov}({\mathcal {R}})|$. Thus, the lemma holds. $\square$

Fig. 41

Relationships between $C^d_{L}({\mathcal {G}})$, $C^d_{S_1}({\mathcal {G}})$, $C^d_{S_2}({\mathcal {G}})$, and $U^d_{L}({\mathcal {G}})$

Proof

(Lemma 9) We illustrate the relationships between $C^d_{L}({\mathcal {G}})$, $C^d_{S_1}({\mathcal {G}})$, $C^d_{S_2}({\mathcal {G}})$, and $U^d_{L}({\mathcal {G}})$ in Fig. 41 with five disjoint subsets A, B, C, D, and E. We have $|C^d_{S_1}({\mathcal {G}})| = |A| + |B| + |C|, |C^d_{S_2}({\mathcal {G}})| = |A| + |C| + |D|, |U^{d}_{L}({\mathcal {G}})| = |A| + |B| + |C| + |D| + |E|, |C^d_{S_1}({\mathcal {G}}) \cap C^d_{S_2}({\mathcal {G}})| = |A|$.

Since $C^d_{S_1}({\mathcal {G}})$ can update ${\mathcal {R}}$, Lemma 5 implies that $|C^d_{S_1}({\mathcal {G}})| \ge \frac{1}{k}|\textsf {Cov}({\mathcal {R}})| + |{\varDelta }({\mathcal {R}}, C^*({\mathcal {R}}))|$. Let ${\mathcal {R}}^{\prime }$ be the resulting ${\mathcal {R}}$ after updating ${\mathcal {R}}$ with $C^d_{S_1}({\mathcal {G}})$. We have $|\textsf {Cov}({\mathcal {R}}^{\prime })| \ge \left( 1 + \frac{1}{k}\right) |\textsf {Cov}({\mathcal {R}})|$.

Suppose $C^d_{S_2}({\mathcal {G}})$ can update ${\mathcal {R}}^{\prime }$ again, then we have $|\textsf {Cov}(({\mathcal {R}}^{\prime } - \{C^{*}({\mathcal {R}}^{\prime })\}) \cup \{C^d_{S_2}({\mathcal {G}})\})| \ge \left( 1 + \frac{1}{k}\right) |\textsf {Cov}({\mathcal {R}}^{\prime })| \ge \left( \frac{1}{k} + \frac{1}{k^2}\right) |\textsf {Cov}({\mathcal {R}})|$. Since $A \cup C \subset C^d_{S_2}({\mathcal {G}})$, $\textsf {Cov}(({\mathcal {R}}^{\prime } - \{C^{*}({\mathcal {R}}^{\prime })\}) \cup \{C^d_{S_2}({\mathcal {G}})\}) = \textsf {Cov}({\mathcal {R}}^{\prime }) - {\varDelta }({\mathcal {R}}^{\prime }, C^{*}({\mathcal {R}}^{\prime })) + D \subseteq \textsf {Cov}({\mathcal {R}}^{\prime }) + D$.

Putting them together, we have $|\textsf {Cov}({\mathcal {R}}^{\prime })| + |D| \ge |\textsf {Cov}(({\mathcal {R}}^{\prime } - \{C^{*}({\mathcal {R}}^{\prime })\}) \cup \{C^d_{S_2}({\mathcal {G}})\})| \ge \left( 1 + \frac{1}{k}\right) |\textsf {Cov}({\mathcal {R}}^{\prime })|$. That means $|D| \ge \frac{1}{k} |\textsf {Cov}({\mathcal {R}}^{\prime })|$. Thus, for $U^d_{L}({\mathcal {G}})$, we have

$$\begin{aligned} |U^d_{L}({\mathcal {G}})|&= |A| + |B| + |C| + |D| + |E| \ge |C^d_{S_1}({\mathcal {G}})| + |D| \\&\ge \frac{1}{k}|\textsf {Cov}({\mathcal {R}})| + |{\varDelta }({\mathcal {R}}, C^*({\mathcal {R}}))| + \frac{1}{k} |\textsf {Cov}({\mathcal {R}}^{\prime })| \\&= \left( \frac{1}{k} + \frac{1}{k^2}\right) |\textsf {Cov}({\mathcal {R}})| + |{\varDelta }({\mathcal {R}}, C^*({\mathcal {R}}))|\\&\quad + \frac{1}{k}|\textsf {Cov}({\mathcal {R}})|\\&\ge \left( \frac{1}{k} + \frac{1}{k^2}\right) |\textsf {Cov}({\mathcal {R}})|\\&\quad + \left( 1 + \frac{1}{k}\right) |{\varDelta }({\mathcal {R}}, C^*({\mathcal {R}}))|. \end{aligned}$$

The last equation holds due to the pigeonhole principle. For each $C^{\prime } \in {\mathcal {R}}$, we must have $|{\varDelta }({\mathcal {R}}, C^{\prime })| \le \frac{1}{k}|\textsf {Cov}({\mathcal {R}})|$. Now, $|U^d_{L}({\mathcal {G}})|$ contradicts with Eq. (2). Thus, if $U^d_{L}({\mathcal {G}})$ satisfies Eq. (2), $C^d_{S_2}({\mathcal {G}})$ cannot update ${\mathcal {R}}$ any more. $\square$

Proof

(Lemma 10) We prove that if there not exists a candidate path in the index to w, w certainly does not exist in $C^{d}_{L}({\mathcal {G}})$.

First, for each vertex v in the lowest level, if $L \not \subseteq L(v)$, there must exist a layer number $j \in L$ such that $v \notin C^{d}(G_j)$. By Lemma 1, we must have $v \not \in C^{d}_{L}({\mathcal {G}})$. Thus, we can remove all such vertices from the graph and the index. After that, we consider each vertex w in the next level of the lowest level. At this time, all of w’s neighbors u in the lowest level such that $L \not \subseteq L(u)$ have already been removed from the graph. Thus, vertex w has the same neighbors as we build the index. If $L \not \subseteq L(w)$, there must exist a layer number $j^{\prime } \in L$ such that $w \notin C^{d}(G_{j^{\prime }})$. By Lemma 1, w cannot be contained in $C^{d}_{L}({\mathcal {G}})$. We can continue this process level by level. This implies that all the vertices that do not satisfy this condition cannot exist in $C^{d}_{L}({\mathcal {G}})$. $\square$

Proof

(Lemma 11) We have $C^{d}_{L}({\mathcal {G}}) \subseteq Y$ by the definition of Y. For each vertex v, if $v \in \bigcup _{h = 0}^{|L| - 1} I_{h}$, the support of v is less than |L|. Thus, v is unlikely to exist in a d-CC on at least |L| layers. Therefore, we must have $v \in \bigcup _{h = |L|}^{l({\mathcal {G}})} I_{h}$. Thus, the lemma holds. $\square$

Proof

(Lemma 12) We prove this lemma by simply contradiction. Suppose there exists a vertex $v \not \in C^{d}_{L}({\mathcal {G}})$ and v is not set to be discarded after the FastdCC procedure. We must have $d^{+}_{i}(v) \ge d$ for each $i \in L$. Otherwise, v must set to be discarded at line 2 of the ProcessUnd procedure or line 5 of the ProcessDis procedure. At this time, since v only connects to undetermined or existing vertices, we have $d_{G_i}(v) = d^{+}_{i}(v) \ge d$ for each $i \in L$. Therefore, we have $v \in C^{d}_{L}({\mathcal {G}})$, which leads to a contradiction. Thus, the lemma holds. $\square$

Proof

(Lemma 13) To analyze the time complexity of FastdCC procedure, we first analyze the cases when an edge can be accessed as follows:

• (1) At line 4 of the FastdCC procedure, when computing $d^{+}_{i}(v)$ and $d^{*}_{i}(v)$ of all $i \in L$ for each vertex v, each edge (u, v) on a layer $i \in L$ will be accessed exactly once.

• (2) At line 8 of the ProcessUnd procedure, when vertex u accesses a vertex v on a higher level, each edge (u, v) on a layer $i \in L^{\prime }$ will be accessed exactly once.

• (3) At line 3 of the ProcessDis procedure, when updating $d^{+}_{i}(u)$, the edge (u, v) on a layer $i \in L$ will be accessed. Note that the edge (u, v) on a layer $i \in L$ will be accessed only once. This is because, when updating $d^{+}_{i}(u)$, v has already been set to be discarded. The state of a discarded vertex will never change. Thus, v will never have chance to visit u any more. Meanwhile, since v is discarded, u also will not visit vertex v afterward.

• (4) At line 3 of the ProcessEst procedure, when updating $d^{*}_{i}(u)$, the edge (u, v) on a layer $i \in L$ will be accessed. In similar, at this time, v has already been set to be existing. The state of an existing vertex will also never change. Thus, the edge (u, v) on a layer $i \in L$ will also be accessed only once.

Thus, the total edge access time is $O(4\sum _{i \in L^{\prime }} |E_{i}({\mathcal {G}})|) = O(4m^{\prime })$.

Next, we analyze the time cost on comparing $d^{+}_{i}(v)$ and $d^{*}_{i}(v)$ on each vertex v w.r.t. d as follows.

• (1) At line 1 and line 3 of the ProcessUnd procedure, we compare $d^{+}_{i}(v)$ and $d^{*}_{i}(v)$ w.r.t. d for each vertex v which is set to undetermined at the first time. The time cost for vertex v is O(2l). Thus, the total time cost for all vertices is O(2nl).

• (2) At line 4 of the ProcessDis procedure which compares $d^{+}_{i}(v)$ w.r.t. d, since the comparison can be involved in the updating of $d^{+}_{i}(v)$ at line 3, the comparison times equals to the number of the edge access times. As we analyzed earlier, each edge on each level is accessed at most once, so the total time cost is $O(m^{\prime })$.

• (3) At line 4 of the ProcessEst procedure which compares $d^{*}_{i}(v)$ w.r.t. d, the comparison can also be involved in the updating of $d^{*}_{i}(v)$ at line 3. In similar, the comparison times equals to the number of the edge access times. The total time cost is $O(m^{\prime })$.

Meanwhile, the time cost to set the states of each vertex is at most O(4n) since there are only four states in the procedure. Putting them together, the time complexity of the FastdCC procedure is $O(2nl + 6m^{\prime } + 4n) = O(nl + m^{\prime })$. $\square$

Proof

(Theorem 1) Given a collection of sets ${\mathcal {F}}= \{C_1, C_2, \dots , C_n\}$ and $k \in {\mathbb {N}}$, the max-k-cover problem is to find a subset ${\mathcal {R}}\subseteq {\mathcal {F}}$ such that $|{\mathcal {R}}| = k$ and that $|\textsf {Cov}({\mathcal {R}})|$ is maximized. The max-k-cover problem has been proved to be NP-complete unless P $=$ NP [2].

It is easy to show that the DCCS problem is in NP. We prove the theorem by reduction from the max-k-cover problem in polynomial time. Given an instance $({\mathcal {F}}, k)$ of the max-k-cover problem, we first construct a multi-layer graph ${\mathcal {G}}$. The vertex set of ${\mathcal {G}}$ is $\bigcup _{i = 1}^n C_i$. There are n layers in ${\mathcal {G}}$. An edge (u, v) exists on layer i if and only if $u, v \in C_i$ and $u \ne v$. Then, we construct an instance of the DCCS problem $({\mathcal {G}}, d, s, k)$, where $d = 1$ and $s = 1$. The result of the DCCS problem instance $({\mathcal {G}}, d, s, k)$ is exactly the result of the max-k-cover problem instance $({\mathcal {F}}, k)$. The reduction can be done in polynomial time. Thus, the DCCS problem is NP-complete. $\square$

To prove Theorems 2 and 3, we first state the following claim. The correctness of the claim has been proved in [2].

Claim

Let ${\mathcal {F}}= \{C_1, C_2, \dots , C_n\}$ and $k \in {\mathbb {N}}$. Let ${\mathcal {R}}^{*}$ the subset of ${\mathcal {F}}$ such that $|{\mathcal {R}}^*| = k$ and $|\textsf {Cov}({\mathcal {R}}^*)|$ is maximized. Let ${\mathcal {R}}\subseteq {\mathcal {F}}$ be a set obtained in the following way. Initially, ${\mathcal {R}}= \emptyset$. We repeat taking an element C out of ${\mathcal {F}}$ randomly and updating ${\mathcal {R}}$ with C according to the two rules specified in Sect. 5.1 until ${\mathcal {F}}= \emptyset$. Finally, we have $|\textsf {Cov}({\mathcal {R}})| \ge \frac{1}{4} |\textsf {Cov}({\mathcal {R}}^{*})|$.

Proof

(Theorem 3) Note that the BU-DCCS algorithm uses the same procedure described in Claim A to update ${\mathcal {R}}$ except that some pruning techniques are applied as well. Therefore, we only need to show that the pruning techniques will not affect the approximation ratio stated in Claim A. Let C be a d-CC pruned by a pruning method and $D_C$ be the set of descendant candidate d-CCs of C in the search tree. For all $C^{\prime } \in D_C$, according to Lemmas 2, 4, or 5, $C^{\prime }$ must not update ${\mathcal {R}}$. By Claim A, candidate d-CCs can be taken in an arbitrary order without affecting the approximation ratio. Therefore, we can safely ignore all the d-CCs in $D_C$ without affecting the quality of ${\mathcal {R}}$. Finally, we have $|\textsf {Cov}({\mathcal {R}})| \ge \frac{1}{4} |\textsf {Cov}({\mathcal {R}}^*)|$. Thus, the theorem holds. $\square$

Proof

(Theorem 4) The TD-DCCS algorithm uses the same procedure described in Claim A to update ${\mathcal {R}}$ and applies some pruning techniques in addition. By the same arguments in the proof of Theorem 2, this theorem holds. $\square$

The Update procedure

We present the Update procedure in Fig. 42. The input includes the set ${\mathcal {R}}$ of temporary top-k diversified d-CCs, a newly generated d-CC C and $k \in {\mathbb {N}}$. For each d-CC $C^{\prime } \in {\mathcal {R}}$, we store both $C^{\prime }$ and the size $|{\varDelta }({\mathcal {R}}, C^{\prime })|$. To facilitate fast updating of ${\mathcal {R}}$, we build some auxiliary data structures. Specifically, we store ${\mathcal {R}}$ in two hash tables M and H. For each entry in M, the key of the entry is a vertex v, and the value of the entry is $M[v] = \{ C^{\prime } | C^{\prime } \in {\mathcal {R}}, v \in C^{\prime } \}$, that is, the set of d-CCs $C^{\prime } \in {\mathcal {R}}$ containing vertex v. For each entry in H, the key of the entry is an integer i, and the value of the entry H[i] is the set of d-CCs $C^{\prime } \in R$ such that $|{\varDelta }({\mathcal {R}}, C^{\prime })| = i$. Obviously, $C^{*}({\mathcal {R}})$ can be easily obtained from H by retrieving the entry of H indexed by the smallest key.

Given the temporary result set ${\mathcal {R}}$ and a new d-CC C, the procedure relies on three key operations to update ${\mathcal {R}}$, namely Size(${\mathcal {R}}$, C) that returns the size $|\textsf {Cov}( ({\mathcal {R}}- \{C^{*}({\mathcal {R}}) \} ) \cup \{ C \} )|$, Delete(${\mathcal {R}}$) that removes $C^{*}({\mathcal {R}})$ from ${\mathcal {R}}$, and Insert(${\mathcal {R}}$, C) that inserts C to ${\mathcal {R}}$. We describe these procedures as follows.

Fig. 42

The Update, Size, Delete, and Insert procedure

Operation${\mathsf {Size}({\mathcal {R}}, C)}$ Note that $\textsf {Cov}( ({\mathcal {R}}- \{C^{*}({\mathcal {R}}) \} ) \cup \{ C \}$ can be decomposed into three disjoint subsets $\textsf {Cov}({\mathcal {R}}- \{ C^{*}({\mathcal {R}}) \} )$, $C - \textsf {Cov}({\mathcal {R}})$ and $C \cap {\varDelta }({\mathcal {R}}, C^{*}({\mathcal {R}}))$. In the beginning, we can obtain $C^{*}({\mathcal {R}})$ and $|{\varDelta }({\mathcal {R}}, C^{*}({\mathcal {R}}) )|$ from H (line 1) and initialize the counter c to 0 (line 1). For each vertex $v \in C$, if v is not a key in M, we have $v \in C - \textsf {Cov}({\mathcal {R}})$, so we increase c by 1 (line 5). Otherwise, if $v \in C^{*}({\mathcal {R}})$ and M[v] only contains $C^{*}({\mathcal {R}})$, c is also increased by 1 (line 7) since $v \in C \cap {\varDelta }({\mathcal {R}}, C^{*}({\mathcal {R}}))$. Since $|\textsf {Cov}({\mathcal {R}}- \{ C^{*}({\mathcal {R}}) \} )|$ is equal to $\textsf {size}(M) - |{\varDelta }({\mathcal {R}}, C^{*}({\mathcal {R}}))|$, we accumulate $\textsf {size}(M) - |{\varDelta }({\mathcal {R}}, C^{*}({\mathcal {R}}))|$ to c (line 8) and return c as the result (line 9).

Operation${\mathsf {Delete}({\mathcal {R}})}$ First, we retrieve $C^{*}({\mathcal {R}})$ from H (line 1). For each vertex $v \in C^{*}({\mathcal {R}})$, $C^{*}({\mathcal {R}})$ is removed from M[v] (line 3). Note that, if M[v] contains a single element $C^{\prime }$ after removing $C^{*}({\mathcal {R}})$, v is a vertex only covered by $C^{\prime }$. Therefore, we move $C^{\prime }$ from $H[|{\varDelta }({\mathcal {R}}, C^{\prime })|]$ to $H[|{\varDelta }({\mathcal {R}}, C^{\prime })| + 1]$ (line 6) and increase $|{\varDelta }({\mathcal {R}}, C^{\prime })|$ by 1 (line 7). If M[v] is empty, v is not covered by ${\mathcal {R}}$, so v is removed from M (line 9).

Operation${\mathsf {Insert}({\mathcal {R}}, C)}$ First, we insert C to ${\mathcal {R}}$ (line 1) and then set $|{\varDelta }({\mathcal {R}}, C)|$ to 0 (line 2). For each vertex $v \in C$, if v is not a key in M, we insert an entry with key v and value C to hash table M (lines 5–6). At this moment, v is only covered by C, so $|{\varDelta }({\mathcal {R}}, C)|$ is increased by 1 (line 7). If v is a key in M, C can be directly inserted to M[v] (line 12). Note that, if M[v] contains a single element $C^{\prime }$ before insertion, v will not be covered only by $C^{\prime }$ after inserting C, so $C^{\prime }$ is moved in H from $H[|{\varDelta }({\mathcal {R}}, C^{\prime })|]$ to $H[|{\varDelta }({\mathcal {R}}, C^{\prime })| - 1]$ (line 11), and $|{\varDelta }({\mathcal {R}}, C^{\prime })|$ is decreased by 1 (line 12). After updating M, we obtain $|{\varDelta }({\mathcal {R}}, C)|$ and insert C to H accordingly (line 14).

Putting them altogether, we have the Update procedure. If $|{\mathcal {R}}| < k$, we directly insert C to ${\mathcal {R}}$ (line 2). If $|{\mathcal {R}}| \ge k$, the Size(${\mathcal {R}}$, C) procedure is invoked to check if C satisfies Rule 2 (line 5). If so, ${\mathcal {R}}$ is updated with C by invoking Delete(${\mathcal {R}}$) and Insert(${\mathcal {R}}$, C) (lines 6–7).

Complexity analysis First we analyze the time complexity of the Update procedure. Assume that an entry can be inserted to or deleted from a hash table in constant time. Thus, the time complexity of Size(${\mathcal {R}}$, C), Delete(${\mathcal {R}}$), and Insert(${\mathcal {R}}$, C) is O(|C|), $O(|C^{*}({\mathcal {R}})|)$ and O(|C|), respectively. Consequently, the time complexity of Update is obviously $O(\max \{|C|, |C^{*}({\mathcal {R}})|\})$.

The space cost for storing the result set ${\mathcal {R}}$ and maintaining the hash table M is $O(\sum _{C^{\prime } \in {\mathcal {R}}} |C^{\prime }|)$, and the space cost for storing $|{\varDelta }({\mathcal {R}}, C^{\prime })|$ and maintaining the hash table H is O(k). Thus, the space complexity of Update is $O(2\sum _{C^{\prime } \in {\mathcal {R}}} |C_j| + 2k) = O(\sum _{C^{\prime } \in {\mathcal {R}}} |C^{\prime }|)$.