Faster algorithms for quantitative verification in bounded treewidth graphs

Abstract

We consider the core algorithmic problems related to verification of systems with respect to three classical quantitative properties, namely, the mean-payoff, the ratio, and the minimum initial credit for energy property. The algorithmic problem given a graph and a quantitative property asks to compute the optimal value (the infimum value over all traces) from every node of the graph. We consider graphs with bounded treewidth—a class that contains the control flow graphs of most programs. Let n denote the number of nodes of a graph, m the number of edges (for bounded treewidth $m=O(n)$) and W the largest absolute value of the weights. Our main theoretical results are as follows. First, for the minimum initial credit problem we show that (1) for general graphs the problem can be solved in $O(n^2\cdot m)$ time and the associated decision problem in $O(n\cdot m)$ time, improving the previous known $O(n^3\cdot m\cdot \log (n\cdot W))$ and $O(n^2 \cdot m)$ bounds, respectively; and (2) for bounded treewidth graphs we present an algorithm that requires $O(n\cdot \log n)$ time. Second, for bounded treewidth graphs we present an algorithm that approximates the mean-payoff value within a factor of $1+\epsilon$ in time $O(n \cdot \log (n/\epsilon ))$ as compared to the classical exact algorithms on general graphs that require quadratic time. Third, for the ratio property we present an algorithm that for bounded treewidth graphs works in time $O(n \cdot \log (|a\cdot b|))=O(n\cdot \log (n\cdot W))$, when the output is $\frac{a}{b}$, as compared to the previously best known algorithm on general graphs with running time $O(n^2 \cdot \log (n\cdot W))$. We have implemented some of our algorithms and show that they present a significant speedup on standard benchmarks.

Appendices

A Minimum cycle

Here we present in full detail the algorithm $\mathsf {MinCycle}$ and establish its properties as stated in Theorem 2. Recall that $\mathsf {MinCycle}$ operates on a tree-decomposition $\mathrm {Tree}(G)$ of an input graph G and behaves as follows.

• 1. If G has no negative cycles, then $\mathsf {MinCycle}$ returns the weight $c^{*}$ of a minimum-weight cycle in G.

• 2. If G has negative cycles, then $\mathsf {MinCycle}$ returns a value that is at most a polynomial (in n) factor smaller than $c^{*}$.

Intuitively, $\mathsf {MinCycle}$ discovers (not necessarily simple) cycles in G and returns the minimum weight found. For reasons of efficiency, $\mathsf {MinCycle}$ does not keep track of which examined cycles are simple, hence if G has negative cycles, then $\mathsf {MinCycle}$ might return a value smaller than $c^{*}$, which corresponds to an examined non-simple cycle C that traverses some negative cycles multiple times. However, C is guaranteed to have polynomial (in n) length, and thus the weight of C is at most polynomially smaller than $c^{*}$.

The following lemma follows easily from [20, Lemma 2], and states that $\mathsf {LD}_{B}(u,v)$ is upper bounded by the smallest weight of a ${\mathsf {U}}$-shaped simple $u\rightsquigarrow v$ path in $B$.

Lemma 14

([20, Lemma 2]) For every examined bag $B$ and nodes $u,v\in B$, we have

• 1. $\mathsf {LD}_{B}(u,v)=\mathsf {wt}(P)$ for some path $P:u\rightsquigarrow v$ (and $\mathsf {LD}_{B}(u,v)=\infty$ if no such P exists),

• 2. $\mathsf {LD}_{B}(u,v)\le \min _{P:u\rightsquigarrow v}\mathsf {wt}(P)$ where P ranges over ${\mathsf {U}}$-shaped simple paths and simple cycles in $B$.

At the end of the computation, the returned value c is the weight of a (generally non-simple) cycle C, captured as a ${\mathsf {U}}$-shaped path on its smallest-level node. The cycle C can be recovered by tracing backwards the updates of line 10 performed by the algorithm, starting from the node x that performed the last update in line 13. Hence, if C traverses k distinct edges, we can write

$$\begin{aligned} c=\mathsf {wt}(C)=\sum _{i=1}^{k} k_i\cdot \mathsf {wt}(e_i) \end{aligned}$$

(2)

where each $e_i$ is a distinct edge, and $k_i$ is the number of times it appears in C.

Lemma 15

Let h be the height of $\mathrm {Tree}(G)$. For every $k_i$ in Eq. 2, we have $k_i\le 2^h$.

Proof

Note that the edge $e_i=(u_i,v_i)$ is first considered by $\mathsf {MinCycle}$ in the root bag $B_i$ of node $x_i$, where $x_i=\arg \max _{y_i\in \{u_i,v_i\}}\mathsf {Lv}(y_i)$ (line 10). As $\mathsf {MinCycle}$ backtracks from $B_i$ to the root of $\mathrm {Tree}(G)$, the edge $e_i$ can be traversed at most twice as many times in each step (because of line 10, once for each term of the sum $\mathsf {LD}_{B}(u,x)+\mathsf {LD}_{B}(x,v)$). Hence, this doubling will occur at most h times, and thus $k_i\le 2^h$. $\square$

Lemma 16

Let c be the value returned by $\mathsf {MinCycle}$, h be the height of $\mathrm {Tree}(G)$, and $c^{*}=\min _C \mathsf {wt}(C)$ over all simple cycles C in G. The following assertions hold:

• 1. If G has no negative cycles, then $c=c^{*}$.

• 2. If G has a negative cycle, then

  • (a) $c\le c^{*}$.

  • (b) $|c|=O\left( |c^{*}|\cdot n\cdot 2^h\right)$.

Proof

By Remark 1, we have that $c^{*}=\mathsf {wt}(P)$ for a ${\mathsf {U}}$-shaped path $P:x\rightsquigarrow x$. By Lemma 14, after $\mathsf {MinCycle}$ examines $B_x$, it will be $c\le \mathsf {LD}_{B_x}(x,x)\le c^{*}$, with the equalities holding if there are no negative cycles in G (by the definition of $c^{*}$, as then $\mathsf {LD}_{B_x}(x,x)$ is witnessed by a simple cycle). By line 10, c can only decrease afterwards, and again by the definition of $c^{*}$ this can only happen if there are negative cycles in G. This proves items 1 and 2a, and the remaining of the proof focuses on showing that $|c|=O\left( |c^{*}|\cdot n\cdot 2^h\right)$.

By rearranging the sum of Eq. 2, we can decompose the obtained cycle C into a set of $k'^{+}$ non-negative simple cycles $C^{+}_i$, and a set of $k'^{-}$ negative simple cycles $C^{-}_i$, and each cycle $C^{+}_i$ and $C^{-}_i$ appears with multiplicity $k^{+}_i$ and $k^{-}_i$ respectively. Then we have

$$\begin{aligned} |c|&=|\mathsf {wt}(C)| = \left| \sum _{i=1}^{k'^{+}}k^{+}_i\cdot \mathsf {wt}(C^{+}_i) + \sum _{i=1}^{k'^{-}}k^{-}_i\cdot \mathsf {wt}(C^{-}_i)\right| \le \left| \sum _{i=1}^{k'^{-}}k^{-}_i\cdot \mathsf {wt}(C^{-}_i)\right| \nonumber \\&\le \sum _{i=1}^{k^{-}}k^{-}_i\cdot |\mathsf {wt}(C^{-}_i)|\le |c^{*}|\cdot \sum _{i=1}^{k'^{-}}k^{-}_i \le |c^{*}|\cdot \sum _{i=1}^{k}k_i =O\left( |c^{*}|\cdot n\cdot 2^h\right) \end{aligned}$$

(3)

The first inequality follows from $c<0$, the third inequality holds by the definition of $c^{*}$, and the last inequality holds since the total number of (non-positive) simple cycle traversals of C cannot be more than the total number of the edge traversals. Finally, we have $\sum _{i=1}^{k} k_i=O\left( n\cdot 2^h\right)$, since $k=O(n)$, and by Lemma 15 we have $k_i\le 2^h$. $\square$

Next we discuss the time and space complexity of $\mathsf {MinCycle}$.

Lemma 17

Let h be the height of $\mathrm {Tree}(G)$. $\mathsf {MinCycle}$ accesses each bag of $\mathrm {Tree}(G)$ a constant number of times, and uses O(h) additional space.

Proof

$\mathsf {MinCycle}$ accesses each bag a constant number of times, as it performs a post-order traversal on $\mathrm {Tree}(G)$ (line 2). Because it computes the local distances in a postorder manner, the number of local distance maps $\mathsf {LD}_{B}$ it remembers is bounded by the height h of $\mathrm {Tree}(G)$. Since $\mathrm {Tree}(G)$ has bounded width, $\mathsf {LD}_{B}$ requires a constant number of words for storing a constant number of nodes and weights in each $B$. Hence the total space usage is O(h), and the result follows. $\square$

The following theorem summarizes the results of this section.

Theorem 2

Let $G=(V,E,\mathsf {wt})$ be a weighted graph of n nodes with bounded treewidth, and a balanced, binary tree-decomposition $\mathrm {Tree}(G)$ of G be given. Let $c^{*}$, be the smallest weight of a simple cycle in G. Algorithm $\mathsf {MinCycle}$ uses O(n) time and $O(\log n)$ additional space, and returns a value c such that:

• 1. If G has no negative cycles, then $c=c^{*}$.

• 2. If G has a negative cycle, then

  • (a) $c\le c^{*}$.

  • (b) $|c|=|c^{*}|\cdot n^{O(1)}$.

B The minimum ratio and mean cycle problems

Lemma 18

Let $\nu ^{*}\ne 0$ be the ratio value of G. The value $\lfloor \nu ^{*} \rfloor$ can be obtained by evaluating $O( \log |\nu ^{*}|)$ inequalities of the form $\nu ^{*}\ge \nu$.

Proof

First determine whether $\nu ^{*}> 0$, and assume w.l.o.g. that this is the case (the process is similar if $\nu ^{*}<0$). Perform an exponential search on the interval $(0,2\cdot \lfloor \nu ^{*}\rfloor )$ by a sequence of evaluations of the inequality $\nu ^{*}\ge \nu _i=2^i$. After $\log \lfloor \nu ^{*}\rfloor +1$ steps we either have $\lfloor \nu ^{*}\rfloor \in (0,1)$, or have determined a $j>0$ such that $\nu ^{*}\in [\nu _{j-1}, \nu _j]$. Then, perform a binary search in the interval $[\nu _{j-1}, \nu _j]$, until the running interval $[\ell , r]$ has length at most 1. Since $\nu _j - \nu _{j-1} = \nu _{j-1} \le \nu ^{*}$, this will happen after at most $\log \lceil \nu ^{*} \rceil$ steps. Then either $\lfloor \nu ^{*} \rfloor = \lfloor \ell \rfloor$ or $\lfloor \nu ^{*} \rfloor = \lfloor r \rfloor$, which can be determined by evaluating the inequality $\nu ^{*} \ge \lfloor r \rfloor$. A similar process can be carried out when $\nu ^{*}<0$. $\square$

Let $T_{\max }=\max _e\mathsf {wt}^{\prime }(e)$ be the largest weight of an edge with respect to $\mathsf {wt}^{\prime }$. Since $\nu ^{*}$ is a number with denominator at most $(n-1)\cdot T_{\max }$, it can be determined exactly by carrying the binary search of Lemma 18 until the length of the running interval becomes at most $\frac{1}{((n-1)\cdot T_{\max })^2}$ (thus containing a unique rational with denominator at most $(n-1)\cdot T_{\max }$). Then $\nu ^{*}$ can be obtained by using continued fractions, e.g. as in [39]. We rely in the work of Papadimitriou [44] to obtain a tighter bound.

Lemma 19

Let $\nu ^{*}\ne 0$ be the ratio value of G, such that $\nu ^{*}$ is the irreducible fraction $\frac{a}{b}\in (-1,1)$. Then $\nu ^{*}$ can be determined by evaluating $O(\log b)$ inequalities of the form $\nu ^{*}\ge \nu$.

Proof

Consider that $\nu ^{*} > 0$ (the proof is similar when $\nu ^{*} < 0$). It is shown in [44] that a rational with denominator at most b can be determined by evaluating $O(\log b)$ inequalities of the form $\nu ^{*}\ge \nu$. $\square$

Theorem 6

Let $G=(V,E,\mathsf {wt}, \mathsf {wt}^{\prime })$ be a weighted graph of n nodes with bounded treewidth, and $\lambda =\max _u |a_u\cdot b_u|$ such that $\nu ^{*}(u)$ is the irreducible fraction $\frac{a_u}{b_u}$. Let $\mathcal {T}(G)$ and $\mathcal {S}(G)$ denote the required time and space for constructing a balanced binary tree-decomposition $\mathrm {Tree}(G)$ of G with bounded width. The minimum ratio cycle problem for G can be computed in

• 1. $O(\mathcal {T}(G)+n\cdot \log (\lambda ))$ time and $O(\mathcal {S}(G)+n)$ space; and

• 2. $O(\mathcal {S}(G)+\log n)$ space.

Proof

In view of Remark 5 the graph G is strongly connected and has a minimum ratio value $\nu ^{*}$. Let $\nu ^{*}=\lfloor \nu ^{*} \rfloor + \frac{a'}{b}$ with $|\frac{a'}{b}|<1$. By Lemma 18, $\lfloor \nu ^{*} \rfloor$ can be determined by evaluating $O(\log |\nu ^{*}|)=O(\log |a|)$ inequalities of the form $\nu ^{*} \ge \nu$, and by Lemma 19, $\frac{a'}{b}$ can be determined by evaluating O(b) such inequalities. A balanced binary tree-decomposition $\mathrm {Tree}(G)$ can be constructed once in $\mathcal {T}(G)$ time and $\mathcal {S}(G)$ space, and stored in O(n) space. $\mathrm {Tree}(G)$ is also a tree-decomposition of every $G_{\nu }$ required by Lemma 11. By Theorem 2 a negative cycle in $G_{\nu }$ can be detected in O(n) time and using $O(\log n)$ space. This concludes Item 1. Item 2 is obtained by the same process, but with re-computing $\mathrm {Tree}(G)$ every time $\mathsf {MinCycle}$ traverses from a bag to a neighbor (thus not storing $\mathrm {Tree}(G)$ explicitly). $\square$