Generalized core maintenance of dynamic bipartite graphs

Abstract

k-core is important in many graph mining applications, such as community detection and clique finding. As one generalized concept of k-core, (i, j)-core is more suited for bipartite graph analysis since it can identify the functions of two different types of vertices. Because (i, j)-cores evolve as edges are inserted into (removed from) a dynamic bipartite graph, it is more economical to maintain them rather than decompose the graph recursively when only a few edges change. Moreover, many applications (e.g., graph visualization) only focus on some dense (i, j)-cores. Existing solutions are simply insufficiently adaptable. They must maintain all (i, j)-cores rather than just a subset of them, which requires more effort. To solve this issue, we propose novel maintenance methods for updating expected (i, j)-cores. To estimate the influence scope of inserted (removed) edges, we first construct quasi-(i, j)-cores, which loosen the constraint of (i, j)-cores but have similar properties. Second, we present a bottom-up approach for efficiently maintaining all (i, j)-cores, from sparse to dense. Thirdly, because certain applications only focus on dense (i, j)-cores of top-n layers, we also propose a top-down approach to maintain (i, j)-cores from dense to sparse. Finally, we conduct extensive experiments to validate the efficiency of proposed approaches. Experimental results show that our maintenance solutions outperform existing approaches by one order of magnitude.

Appendix A Proofs and Algorithms

Proof for Lemma 1

Let $C = C(G_p,i,j)$ and $P = P(i,j)$, ${\hat{C}}(P,C,i,j) \subseteq P$ holds by Definition 4. Thus, we just prove $P \subseteq {\hat{C}}(P,C,i,j)$. Suppose there is a nonempty graph $A = P \setminus {\hat{C}}(P,C,i,j)$, there must be a vertex $l\ (r) \in A$ such that $d(P(C),l)<i\ (d(P(C),r)<j)$, where P(C) is the extended graph of A on C, and this vertex leads to the recursive deletion on P. Otherwise, ${\hat{C}}(P,C,i,j)$ can expand via $l\ (r)$. Since $P = C(G_c,i,j) \setminus C$, $d(P(C),l) \ge i\ (d(P(C),r) \ge j)$ always holds for $l\ (r) \in P$, which makes a contradiction. Thus, $P \subseteq {\hat{C}}(P,C,i,j)$.

In summary, $P = {\hat{C}}(P,C,i,j)$. $\square$

Proof for Theorem 1

Let $C = C(G_p,i,j)$, $F = F_b(i,j)$ and $P = P(i,j)$. Firstly, we prove ${\hat{C}}(F,C,i,j) \subseteq P$. Suppose there is a nonempty graph $A = {\hat{C}}(F,C,i,j) \setminus P$, there must be a vertex $l\ (r) \in A$ such that $d(F(C),l)<i\$ $(d(F(C),r)<j)$ and this vertex leads to recursive deletion on A, where F(C) is the extended graph of F on C. Otherwise, P can expand via $l\ (r)$, which contradicts Definition 6. According to Definition 5, $d(F(C),l) \ge i\$ $(d(F(C),r) \ge j)$ holds for an arbitrary $l\ (r) \in F$ unless there is a vertex $l'\ (r') \in C$ such that $d(C,l')< i\ (d(C,r') <j)$. Since C is the previous (i, j)-core, this makes a contradiction. As for an edge $(l,r) \in A$, since $l,r \in P$, P can expand via adding (l, r), which contradicts Definition 6. Thus, ${\hat{C}}(F,C,i,j) \subseteq P$.

Secondly, we prove $P \subseteq {\hat{C}}(F,C,i,j)$. Since $P\subseteq F$, ${\hat{C}}(P,C,i,j) \subseteq {\hat{C}}(F,C,i,$ j) holds. By Lemma 1, $P(i,j) = {\hat{C}}(P,C,i,j) \subseteq {\hat{C}}(F,C,i,j)$.

In summary, $P = {\hat{C}}(P,F,i,j)$. $\square$

Proof for Theorem 2

Since $C(G_p,i,j)=\emptyset$, we have $C(G_c,i,j) = P(i,j) \subseteq F_b(i,j)$. Since $F_b(i,j) \subseteq G_{c}$, $C(F_b(i,j),i,j) \subseteq C(G_c,i,j)$ holds. Due to $C(G_c,i,j) \subseteq F_b(i,j)$, we have $C(G_c,i,j) \subseteq C(F_b(i,j),i,j)$.

In summary, $C(F_b(i,j),i,j)=C(G_c,i,j) = P(i,j)$. $\square$

Proof for Theorem 3

Let $F = F_t(i,j)$, $P = P(i,j)$ and $C = C(G_c,i+1,j) \cup C(G_c,i,j+1) \cup C(G_p,i,j)$. Firstly, we prove $({\hat{C}}(F,C,i,j) \cup C(G_c,i+1,j) \cup C(G_c,i,j+1)) \setminus C(G_p,i,j) \subseteq P$. Since $C(G_c,i+1,j) \cup C(G_c,i,j+1) \subseteq C(G_c,i,j)$, $(C(G_c,i+1,j) \cup C(G_c,i,j+1)) \setminus C(G_p,i,j) \subseteq P$ holds. Since ${\hat{C}}(F,C,i,j) \setminus C(G_p,i,j) = \emptyset$, we just need to prove ${\hat{C}}(F,C,i,j) \subseteq P$. Suppose there is a nonempty graph $A ={\hat{C}}(F,C,i,j)\setminus P$, we have $d(F(C), l) \ge i$ or $d(F(C),r) \ge j$ for arbitrary $l,r \in A$, where F(C) is the extended graph of F on C. If $l \notin P$, there must be a vertex $l' \in C \ (r' \in C)$ such that $d(C,l')<i\ (d(C, r') < j)$, which makes a contradiction. Since $l,r \in P$, if $(l,r) \notin P$, then we have $P \subseteq P \cup (l,r)$, which contradicts Definition 6. Thus, $({\hat{C}}(F,C,i,j) \cup C(G_c,i+1,j) \cup C(G_c,i,j+1) ) \setminus C(G_p,i,j) \subseteq P$.

Secondly, we prove $P \subseteq ({\hat{C}}(F,C,i,j) \cup C(G_c,i+1,j) \cup C(G_c,i,j+1)) \setminus C(G_p,i,j)$. Suppose there is a nonempty graph $A = P \setminus (({\hat{C}}(F,C,i,j) \cup C(G_c,i+1,j) \cup C(G_c,i,j+1)) \setminus C(G_p,i,j))$. Since $(C(G_c,i+1,j) \cup C(G_c,i,j+1)) \setminus C(G_p,i,j) \subseteq P$ and ${\hat{C}}(F,C,i,j) \setminus C(G_p,i,j) = {\hat{C}}(F,C,i,j)$, $A \subseteq P \setminus {\hat{C}}(F,C,i,j)$ holds. Clearly, $A \subseteq F$. Since $A \ne \emptyset$, there must be a vertex $l\ (r) \in A$ such that $d(F(C),l)<i\ (d(F(C),r)<j)$, where F(C) is the extended graph of F on C. By Definition 7 and C is the union graph of previous (i, j)-core, $(i+1,j)$-core and $(i,j+1)$-core, $d(F(C),l) \ge i\ (F(C),r) \ge j)$ holds for $l\ (r) \in A$. Again, this is contradiction. Hence, $P \subseteq ({\hat{C}}(F,C,i,j) \cup C(G_c,i+1,j) \cup C(G_c,i,j+1)) \setminus C(G_p,i,j)$.

In summary, $P = ({\hat{C}}(F,C,i,j) \cup C(G_c,i+1,j) \cup C(G_c,i,j+1)) \setminus C(G_p,i,j)$. $\square$

Proof for Theorem 4

Let $F = F_t(i,j)$, $P = P(i,j)$ and $C = C(G_c,i+1,j) \cup C(G_c,i,j+1)$. Firstly, we prove ${\hat{C}}(F,C,i,j) \cup C \subseteq P$. Since $C(G_p,i,j) = \emptyset$, we have $P= C(G_c,i,j)$. By Definition 6, we have $C \subseteq C(G_c,i,j) = P$. Thus, we just need to prove ${\hat{C}}(F,C,i,j) \subseteq P$. Suppose there is a nonempty graph $A = {\hat{C}}(F,C,i,j)\setminus P$, we have $d(F(C),l) \ge i\ (d(F(C), r) \ge j)$ for arbitrary $l \in A\ (r \in F)$, where F(C) is the extended graph of F on C. If $l \notin P$, there must be a vertex $l' \in C\ (r' \in C)$ such that $d(C,l')< i\ (d(C,r') < j)$, which makes a contradiction because C is the union graph of $(i+1,j)$-core and $(i,j+1)$-core. Since $l,r \in P$, if $(l,r) \notin P$, then we have $P \subseteq P \cup (l,r)$, which contradicts Definition 6. Thus, ${\hat{C}}(F,C,i,j) \cup C \subseteq P$.

Secondly, we prove $P\subseteq {\hat{C}}(F,C,i,j) \cup C$. Suppose there is a nonempty graph $A = P \setminus ({\hat{C}}(F,C,i,j) \cup C)$, since $C \subseteq P$, we have $A \subseteq P \setminus {\hat{C}}(F,C,i,j)$. Clearly, $A \subseteq F$, otherwise it contradicts our assumption. Since $A \ne \emptyset$, there must be a vertex $l\ (r) \in A$ such that $d(F(C),l)<i\ (d(F(C),r)<j)$, where F(C) is the extended graph of F on C. By Definition 7 and C is the union graph of $(i+1,j)$-core and $(i,j+1)$-core, $d(F(C),l) \ge i\ (d(F(C),r) \ge j)$ holds for $l\ (r) \in A$. Again, this is a contradiction. Hence, $P \subseteq {\hat{C}}(F,C,i,j) \cup C$.

In summary, $P = {\hat{C}}(F,C,i,j) \cup C$. $\square$