Optimizing RPQs over a compact graph representation

Abstract

We propose techniques to evaluate regular path queries (RPQs) over labeled graphs (e.g., RDF). We apply a bit-parallel simulation of a Glushkov automaton representing the query over a ring: a compact wavelet-tree-based index of the graph. To the best of our knowledge, our approach is the first to evaluate RPQs over a compact representation of such graphs, where we show the key advantages of using Glushkov automata in this setting. Our scheme obtains optimal time, in terms of alternation complexity, for traversing the product graph. We further introduce various optimizations, such as the ability to process several automaton states and graph nodes/labels simultaneously, and to estimate relevant selectivities. Experiments show that our approach uses 3–5$\times$ less space, and is over 5$\times$ faster, on average, than the next best state-of-the-art system for evaluating RPQs.

A Complete running example

Fig. 14

The whole process to match the RPQ of Fig. 8 in our graph of Figs. 1 and 5

Figure 14 illustrates the whole process of matching the (2)RPQ (y,l5 $^+$ / $^{\hat{\,}}$ bus,Baq), with the NFA of Fig. 8, in the graph of Fig. 1, as represented in Fig. 5. We use a top-down tree to show branches in the process; Fig. 9 shows the states of the product graph $G_{\!\mathcal {M}}'$ we traverse (backwards in BFS). The top nodes of the tree illustrate what we have already done in previous examples (edge $1 \rightarrow 2$ of $G_{\!\mathcal {M}}'$): starting from $L_\textrm{p}[14\mathinner {.\,.}15]$ (Baq) and $D=0001$, we identified the only edge label reaching that node, l5, that is relevant in our NFA. The label l1 also appears in $L_\textrm{p}[15\mathinner {.\,.}16]$ because a transition labeled l1 reaches Baq, but $D ~ \& ~ B[\texttt {l1} ] = 0000$ because our NFA does not match it; this pruned branch is shown with a dashed arrow leading to an X. We have also seen that the only subject of those edges labeled l5 is $\texttt {BA}$, at $L_\textrm{s}[10\mathinner {.\,.}10]$, where the NFA is active at states 0110. We move to BA because $S[\texttt {BA} ]=S[4]=0000$, so $D=0110$ contains some unseen NFA states at this node. We mark $S[4]=0110$ to indicate that we have already reached $\texttt {BA}$ with those active NFA states. In part 3 of our process, we find the interval of $L_\textrm{p}$ corresponding to $L_\textrm{s}[10]=4$ (BA), $L_\textrm{p}[C_\textrm{o}[4]+1\mathinner {.\,.}C_\textrm{o}[5]] = L_\textrm{p}[11\mathinner {.\,.}14]$, completing one step.

Three symbols appear on $L_\textrm{p}[11\mathinner {.\,.}14]$ (i.e., three edge labels reach BA), but only l5 (left child) and $^{\hat{\,}}$ bus (right child) match our NFA. By l5 we reach $L_\textrm{s}[8\mathinner {.\,.}9]$ using backward search. In this interval we find two sources that, by l5, reach BA: SA (left child) and Baq (right child), both with NFA state $D=0110$ (the same as before). Conversely, by $^{\hat{\,}}$ bus, we reach $L_\textrm{s}[16\mathinner {.\,.}16]$ using backward search. There we find the only source, SA, that reaches BA, with NFA state $D=1000$. We process the three sources in BFS order, left to right:

• 1. By l5 we reach BA from SA (leftmost tree node in this level). We accept going to SA because $S[\texttt {SA} ]=S[1]=0000$ and $D=0110$ has new states, so we set $S[1]=0110$. In part 3 we obtain the interval $L_\textrm{p}[1\mathinner {.\,.}4]$ for SA. This is transition $2 \rightarrow 3$ in the product graph.

• 2. By l5 we reach BA from Baq (middle tree node in this level). Although we had already seen Baq, it was only with states $S[\texttt {Baq} ]=S[5]=0001$, so the current state $D=0110$ has some unvisited NFA states; we set $S[5]=0111$ and part 3 leads us to $L_\textrm{p}[15\mathinner {.\,.}16]$. This is transition $2 \rightarrow 4$ in the product graph.

• 3. By $^{\hat{\,}}$ bus we reach BA from SA as well (rightmost tree node in this level). Since $S[\texttt {SA} ]=S[1]=0110$ and $D=1000$, we have new states and we accept going to SA, setting $S[1]=1110$. The NFA state is still 1000 (transition $2 \rightarrow 5$ in the product graph), which contains the initial state, so we report node SA as a solution to our 2RPQ. We then continue from it, reaching $L_\textrm{p}[1\mathinner {.\,.}4]$ using part 3.

Our BFS traversal now branches from each of the tree nodes identified above:

• 1. From $L_\textrm{p}[1\mathinner {.\,.}4]$ (SA) with $D=0110$, we find edges labeled l1, l5, and $^{\hat{\,}}$ bus leading to it:

  • (a) Our NFA cannot process l2 ($D ~ \& ~ B[\texttt {l2} ]=0000$), so we abandon that edge.

  • (b) By l5 we find the source BA, but since $S[\texttt {BA} ]=S[4]=0110$ and $D=0110$, we have already visited BA with those active states, so we abandon this branch too, thus avoiding to fall into a cycle.

  • (c) By $^{\hat{\,}}$ bus we reach $L_\textrm{s}[14\mathinner {.\,.}14]$ with state $D=1000$. The only source here is $L_\textrm{s}[14]=2=$ UCh. Since $S[\texttt {UCh} ]=S[2]=0000$, we enter this state and set $S[2]=1000$ (transition $3 \rightarrow 6$ in the product graph). Furthermore, since D contains the initial state, we report UCh as the second solution to the 2RPQ.

• 2. From $L_\textrm{p}[15\mathinner {.\,.}16]$ (Baq) with $D=0110$, we find edges labeled l1 and l5 leading to it:

  • (a) Our NFA cannot process l1, so we abandon this branch.

  • (b) By l5 we reach BA again, and once again we prune the branch to avoid falling into cycles, because $S[\texttt {BA} ]=S[4]=0110$.

• 3. Finally, from $L_\textrm{p}[1\mathinner {.\,.}4]$ (SA) and $D=1000$, which we had reported, the NFA has nowhere to go, so we reject the three possible edge labels, l2, l5, and $^{\hat{\,}}$ bus. The same happens in the last tree level from $L_\textrm{p}[5\mathinner {.\,.}8]$ (UCh) after reporting it, so we finish.