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<article xmlns:xlink="http://www.w3.org/1999/xlink">
  <front>
    <journal-meta />
    <article-meta>
      <title-group>
        <article-title>Model counting for ]2SAT problem in outerplanar graphs.</article-title>
      </title-group>
      <contrib-group>
        <contrib contrib-type="author">
          <string-name>Marco A. Lopez</string-name>
          <email>mlopezm158@alumno.uaemex.mx</email>
          <xref ref-type="aff" rid="aff1">1</xref>
        </contrib>
        <contrib contrib-type="author">
          <string-name>J. Raymundo Marcial-Romero</string-name>
          <xref ref-type="aff" rid="aff1">1</xref>
        </contrib>
        <contrib contrib-type="author">
          <string-name>Jose A. Hernandez</string-name>
          <email>xoseahernandez@uaemex.mx</email>
          <xref ref-type="aff" rid="aff1">1</xref>
        </contrib>
        <contrib contrib-type="author">
          <string-name>Guillermo De Ita</string-name>
          <email>deita@cs.buap.mx</email>
          <xref ref-type="aff" rid="aff0">0</xref>
        </contrib>
        <aff id="aff0">
          <label>0</label>
          <institution>Facultad de Ciencias de la Computacion</institution>
          ,
          <addr-line>BUAP</addr-line>
        </aff>
        <aff id="aff1">
          <label>1</label>
          <institution>Facultad de Ingenier a</institution>
          ,
          <addr-line>UAEM</addr-line>
        </aff>
      </contrib-group>
      <fpage>76</fpage>
      <lpage>87</lpage>
      <abstract>
        <p>The satis ability problem for formulas in two conjunctive Normal Form (2SAT) is solved in polynomial time, and ]2SAT which is the count version of 2SAT is ]P-complete. It has been shown that for certain types of formulas, ]2SAT can be computed in polynomial time. In this paper we de ne a new method, based on embedded cycles, to compute ]2SAT on the so-called outerplanar formulas. Our algorithm's time complexity is given by O(n + m) where n is the number of variables and m the number of clauses of the formula. Although the time complexity is similar to other methods, experimental results show that the new method is faster.</p>
      </abstract>
    </article-meta>
  </front>
  <body>
    <sec id="sec-1">
      <title>-</title>
      <p>
        Introduction
]SAT (the problem of model counting for a Boolean formula) concerns
especially to arti cial intelligence (AI), and has a direct relationship with the
automated theorem proving, as well as to approximate reasoning [1{3]. ]SAT is
a ]P complete problem, even for formulas in two conjunctive normal form, so
for complete methods, only exponential time algorithms are known. The exact
algorithm with the best bound until now was presented by Wahlstrom [
        <xref ref-type="bibr" rid="ref4">4</xref>
        ], who
provides an O(1:2377n)-time algorithm, where n is the number of variables of
the formula. Exists some formula classes where, ]2SAT can be solved in linear
time [
        <xref ref-type="bibr" rid="ref1">1</xref>
        ]. Relevant classes of this formulas are monotone formulas and cactus
formulas [
        <xref ref-type="bibr" rid="ref5">5</xref>
        ].
      </p>
      <p>In practice there are some tools, called SAT solvers to solve ]SAT in e cient
time. sharpSAT is a SAT solver reported in literature as the fastest exact and
randomized tool, it uses a component cache system that ensures a minimum
evaluation on subformulas, generated in the formula decomposition, which are
similar.</p>
      <p>
        Besides sharpSAT, in [
        <xref ref-type="bibr" rid="ref5">5</xref>
        ] they present an algorithm that uses formula
decomposition until achieve a certain type of formulas called cactus. The evidence
presented [
        <xref ref-type="bibr" rid="ref5">5</xref>
        ] shows that using formula decomposition and a cactus formula as a
base case, computing models is faster than the strategy used by sharpSAT .
      </p>
      <p>
        In [
        <xref ref-type="bibr" rid="ref6">6</xref>
        ] a method for counting models in the so calles outerplannar formulas is
presented. The method is based on a treewidth decomposition.
      </p>
      <p>
        In this paper we present a new method to count models in the so called
outerplanar formulas. This method to compute ]2SAT on outerplanar
formulas is based on a transformation in the input formula on its constraint signed
graph. Evidences show that, using outerplanar formulas as a base case is e cient
compared to [
        <xref ref-type="bibr" rid="ref5">5</xref>
        ] and sharpSAT.
      </p>
      <p>
        Another special class of graphs contained into outerplanar graphs is the class
of polygonal array graphs that has been widely used in mathematical chemistry,
since they are molecular graphs used to represent the structural formula of
chemical compounds. In particular, hexagonal arrays are the graph representations of
an important subclass of benzenoid molecules, unbranched catacondensed
benzenoid molecules, which play a distinguished role in the theoretical chemistry of
benzenoid hydrocarbons [
        <xref ref-type="bibr" rid="ref7 ref8">7, 8</xref>
        ].
      </p>
      <p>
        In our case, we are more interested in the application of counting models
on conjunctive normal form formulas as a medium to develop methods for
approximate reasoning. For example, for computing the degree of belief on
propositional formulas, or for building Bayesian models. It is relevant to know how
many models are maintained while input conjunctive normal form formulas are
being updating [
        <xref ref-type="bibr" rid="ref10 ref2 ref9">2, 9, 10</xref>
        ].
      </p>
      <p>This method works using the embedded form of outerplanar graphs, those
embedded graphs are then used to compute ]2SAT in linear time.</p>
      <p>The paper is organized as follows, in Section 2 the preliminaries are
established. In Section 3 an embedded graph representation of outerplanar formulas is
presented. In Section 4, our main procedure is presented, in section 5 the results
of the paper are shown and nally, the Conclusion.
2</p>
    </sec>
    <sec id="sec-2">
      <title>Preliminaries</title>
      <p>Let X = fx1; : : : ; xng be a set of n Boolean variables. A literal is either a variable
xi or a negated variable xi. As usual, for each xi 2 X, we write xi0 = xi and
xi1 = xi. A clause is a disjunction of di erent literals (sometimes, we also consider
a clause as a set of literals). For k 2 N , a k-clause is a clause consisting of exactly
k literals and, a ( k)-clause is a clause with at most k literals. A variable x 2 X
appears in a clause c if either the literal x1 or x0 is an element of c.</p>
      <p>A Conjunctive Normal Form (CNF) F is a conjunction of clauses (we also
call F a Conjunctive Form). A k-CNF is a CNF containing clauses with at most
k literals.</p>
      <p>We use (Y ) to express the set of variables involved in the object Y , where Y
could be a literal, a clause or a Boolean formula. Lit(F ) is the set of literals which
appear in a CNF F , i.e. if X = (F ), then Lit(F ) = X[X = fx11; x01; : : : ; x1n; x0ng.
We also denote f1; 2; : : : ; ng by [[n]].</p>
      <p>An assignment s for F is a Boolean function s : (F ) ! f0; 1g. An assignment
can be also considered as a set which does not contain complementary literals.
If x 2 s, being s an assignment, then s turns x true and x1 false, 2 f0; 1g.
Considering a clause c and assignment s as a set of literals, c is satis ed by s if
and only if c \ s 6= ;, and if for all x 2 c, x1 2 s then s falsi es c.</p>
      <p>If F1 F is a formula consisting of some clauses of F , then (F1) (F ),
and an assignment over (F1) is a partial assignment over (F ).</p>
      <p>Let F be a Boolean formula in CNF, F is satis ed by an assignment s if
each clause in F is satis ed by s. F is contradicted by s if any clause in F is
contradicted by s. A model of F is an assignment for (F ) that satis es F . We
will denote as SAT (F ) the set of models for the formula F .</p>
      <p>Given a CNF F , the SAT problem consists on determining if F has a model.
The ]SAT problem consists of counting the number of models of F de ned over
(F ). ]2-SAT denotes ]SAT for formulas in 2-CNF.
2.1</p>
      <sec id="sec-2-1">
        <title>The signed primal graph of a 2-CF</title>
        <p>
          There are some graphical representations of a CNF (see e.g. [
          <xref ref-type="bibr" rid="ref11">11</xref>
          ]), we use here
the signed primal graph of a two conjunctive normal form.
        </p>
        <p>Let F be a 2-CNF, its signed primal graph (constraint graph) is denoted by
GF = (V (F ); E(F )), with V (F ) = (F ) and E(F ) = ff (x); (y)g : fx; yg 2
F g, that is, the vertices of GF are the variables of F , and for each clause fx; yg
in F there is an edge f (x); (y)g 2 E(F ). For x 2 V (F ), (x) denotes its
degree, i.e. the number of incident edges to x. Each edge c = f (x); (y)g 2 E
is associated with an ordered pair (s1; s2) of signs, assigned as labels of the edge
connecting the literals appearing in the clause. The signs s1 and s2 are related to
the literals x and y , respectively. For example, the clause fx0; y1g determines
the labelled edge: "x + y" which is equivalent to the edge "y + x".</p>
        <p>Formally, let S = f+; g be a set of signs. A graph with labelled edges
on a set S is a pair (G; ), where G = (V; E) is a graph, and is a function
with domain E and range S. (e) is called the label of the edge e 2 E. Let
G = (V; E; ) be a signed primal graph with labelled edges on SxS. Let x and
y be vertices in V , if e = fx; yg is an edge and (e) = (s; s0), then s(resp:s0) is
called the adjacent sign to x(resp:y). We say that a 2-CNF F is a path, cycle,
a tree, or an outerplanar graph, if its signed constraint graph GF represents a
path, cycle, a tree, an outerplanar graph, respectively. We will omit the signs on
the graph if all of them are +.</p>
        <p>
          Notice that a signed primal graph of a 2-CNF can be a multigraph since two
xed variables can be involved in more than one clause of the formula forming so
parallel edges. Furthermore, a unitary clause is represented by a loop (an edge
to join a vertex to itself). A polynomial time algorithm to process parallel edges
and loops to solve ]SAT has been shown in [
          <xref ref-type="bibr" rid="ref1">1</xref>
          ].
        </p>
        <p>Let : 2-CNF ! GF be the function whose domain is the space of Boolean
formulas in 2-CNF and codomain the set of multi-graphs, is a bijection. So
any 2-CNF formula has a unique signed constraint graph associated via and
viceversa, any signed constraint graph GF has a unique formula associated.
2.2</p>
      </sec>
      <sec id="sec-2-2">
        <title>Cumulative operations</title>
        <p>
          We de ne a set of cumulative operations as macro, in this paper a macro must be
constructed using the method shown in [
          <xref ref-type="bibr" rid="ref5">5</xref>
          ], using the fact that a macro is a linear
equation of the form M = x + where and represent the models already
counted and x the models to be computed. Since the models of an outerplannar
formula always belongs to a simple cycle, M = f + ; + g, where we
omit the variable x. A macro contains four elements, so we need to perform
operations using two more elements than [
          <xref ref-type="bibr" rid="ref5">5</xref>
          ].
        </p>
        <p>A new set of equations must be constructed, associated to every pair of signs
( ; ) of an edge fx ; y g, in a graph.
(
+
;
+
8(
&gt;
) = &gt;&lt;&gt;((
&gt;(
&gt;
&gt;:((</p>
        <p>
          Counting on acyclic graphs, like tree or paths, using a di erent equation
than [
          <xref ref-type="bibr" rid="ref5">5</xref>
          ] gives a new panorama, counting in graphs like cactus graphs with always
non intersecting cycles, we can use this set of operations and de ne a macro as
the cumulative operations in a simple and non intersecting cycle.
3
        </p>
      </sec>
    </sec>
    <sec id="sec-3">
      <title>Outerplanar 2-CNF Formulas</title>
      <p>
        An outerplanar 2-CNF formula is one whose signed primal graph is ourterplanar
e.g the graph has a planar drawing for which all vertices belong to the outer
face of the drawing. Outerplanar graphs may be characterized (analogously to
Wagner's theorem for planar graphs) by the two forbidden minors K4 and K2;3,
or by their Colin de Verdiere graph invariants. They have Hamiltonian cycles if
and only if they are biconnected, in which case the outer face forms the unique
Hamiltonian cycle. Every outerplanar graph is 3-colorable, and has degeneracy
and treewidth at most 2 [
        <xref ref-type="bibr" rid="ref12">12</xref>
        ]. The outerplanar graphs are a subset of the planar
graphs, of the serial-parallel graphs, and of the circle graphs.
3.1
      </p>
      <sec id="sec-3-1">
        <title>Counting on outerplanar graphs</title>
        <p>Using the representation of an outerplanar graph as a graph with embedded
cycles, there is a method to solve the most internal cycles as simple cycles,
and replacing the set of vertices that form the cycle with a macro. In this way
it is possible to count models on any outerplanar graph its representation as
embedded graph is known.
3.2</p>
      </sec>
      <sec id="sec-3-2">
        <title>Common edges identi cation</title>
        <p>One characteristic of the outerplanar graphs is that they do not contain the
subgraph K2;3, this subgraph can be obtained with two cycles intersected by
two edges, that is, there exists two common edges between a pair of cycles. With
this we found that an outerplanar graph has at most one edge between any pair
of cycles (Fig. 1).</p>
        <p>1
4
2
3
5
6</p>
        <p>Identi cation of common edges is done by constructing an expansion tree of
the graph, which we can nd a set of back edges or cycles within the graph, this
is necessary to identify intersection edges that belongs to the expansion tree or
back edges (Fig. 2).</p>
        <p>To substitute an edge of the expansion tree for one of the back edges, for
a new construction of a embedded graph, an edge of the expansion tree can
become in one of the back edges in the embedded graph (Fig. 3), the following
considerations allow this exchange to be made between the sets of tree and back
edges:
{ If a pair of the back edges intersect on a tree edge, the back edge that closes
at a low level, on the tree, is replaced by the intersecting tree edge.
{ If two back edges intersect on a tree edge, and they form two sub-partitions,
that is, each one belongs to a di erent branch in the tree. The back edge
that closes at a low level, on the tree, is replaced by the intersecting tree
edge.
3.3</p>
      </sec>
      <sec id="sec-3-3">
        <title>Embedded graph transformation</title>
        <p>If we identify the common edges between cycles it is possible to nd an expansion
tree, which the common edges form the set of back edges, so that the back edges
form an embedded graph.</p>
        <p>By obtaining the common edges in a graph and eliminating them, it is
possible to generate a set of possible paths, obtaining Cn, a simple cycle of n vertices.
This way it is possible to generate multiple expansion trees that can be seen as
a set of embedded cycles (Fig. 4).</p>
        <p>With these graphs it can be shown that no matter which vertex the
construction of the new expansion tree begins on, adding the common edges will always
result in an embedded graph (Fig. 5).</p>
        <p>1
1
5
2
6
3
7
4
8
9
12
11</p>
        <p>10
10
11
12
9
8
4
7
3
6
2
5</p>
        <p>By constructing an embedded graph it is possible to assign an order in which
each embedded cycle must be solved, inner cycles rst. This way all cycles on
outerplanar formulas can be handled independently.</p>
      </sec>
    </sec>
    <sec id="sec-4">
      <title>Computing ]2SAT on outerplanar 2-CNF formulas</title>
      <p>
        If F consists of disconnected sub-formulas then ]2SAT (F ) = Qk
i=1 ]2SAT (Fi)
where Fi; i = 1; : : : ; k, are the disconnected sub-formulas of F [
        <xref ref-type="bibr" rid="ref3">3</xref>
        ]. The time
complexity for computing ]2SAT (F ), denoted as T (]2SAT (F )), is given by the
rule T (]2SAT (F )) = maxfT (]2SAT (Fi)) : Fi is a disconnected subformula
of F g. Thus, a rst decomposition of the formula is done via its connected
components, and from here on, we consider only outerplanar connected formulas.
      </p>
      <p>Due that we can assign an order to solve inner cycles rst, in embedded
graphs, and replace that cycle with a macro. We can compute ]2SAT in a
outerplanar graph in linear time.</p>
      <p>We built a linear equation, macro, that represents the values mi = ( +
; + ) on each cycle in the embedded graph. And a linear equation for
each node ( + ; + ).</p>
      <p>Giving the rst transformation in Fig. 5, we can handle two cycles at same
time and perform two operation sets, M1 and M2, where the vertices f5; 6; 7g 2
M1 and f8; 12; 11; 10; 9; 4g 2 M2.</p>
      <p>5
6
7
(1+0 , 0+1) - (0+0 , 0+1)
(1+1 , 1+0) - (0+1 , 0+0)
(2+1 , 1+1) - (0+0 , 0+1)</p>
      <p>The nal operation gives a new macro M1 = (2 + 1; 1 + 0), a second macro
analogously can be obtained M2 = (8+5; 5+0). A third macro can be computed,
replacing the embedded cycles, we have that the vertices f5; 7; 8; 4g 2 M3 and
fM1; M2g 2 M3 (Fig. 7).</p>
      <p>As a result we have M3 = (34 + 13; 15 + 0), to obtain a new set of values
when we replace an inner cycle by a macro we need to apply another set of
equations. If we have that Mi = ( i + i ; i + i ) and a node with values
( 1 + 1 ; 1 + 1 ), to obtain ( + ; + ):
{
{
{
{
=
=
=
=
i
i
i
i</p>
      <p>This set of new equations only needs to use when a macro is applied, then
we can compute the next macro M4 where f1; 5; 4; 3; 2g 2 M4 and M3 2 M4,
then M4 = (109 + 83; 62 + 0). Finally with the initial values of (1 + 0; 0 + 1) we
can obtain the total models in this example, ]SAT (M4) = (192; 62) = 254</p>
      <p>
        A relevant property of a macro, as de ned in this paper, is the possibility to
represent cumulative operations via symbolic variables, making macros
indistinguishable from individual operators. If subsequences of operators are repeated,
a hierarchy of macros can represent a more compactly plan than a simple
operator sequence, replacing each occurrence of a repeating subsequence with a
macro [
        <xref ref-type="bibr" rid="ref13">13</xref>
        ].
      </p>
      <p>
        The correctness of our method is based in the following Theorem from [
        <xref ref-type="bibr" rid="ref6">6</xref>
        ].
Theorem 1. Let F1 and F2 be two formulas in 2-CNF. If F1 \ F2 = fx1; x2g,
e.g. a single clause then
]2SAT (F1 [ F2) =]2SAT (F1 jfx1;x2g s)
]2SAT (F2 jfx1;x2g s)+
]2SAT (F1 jfx1;x2 1g s)
]2SAT (F1 jfx1 1;x2g s)
]2SAT (F2 jfx1;x2 1g s)+
]2SAT (F2 jfx1 1;x2g s)
Proof. In order to satisfy F1 [ F2 the clause fx1; x2g has to be satis ed, so
either fx1; x2g s or fx1; x2 1g s or fx1 1; x2g s. The computation of the
satisfying assignments of F1 [ F2 is given by
]2SAT (F1 [ F2) =]2SAT (F1 [ F2 jfx1;x2g s)+
]2SAT (F1 [ F2 jfx1;x2 1g s)+
]2SAT (F1 [ F2 jfx1 1;x2g s)
      </p>
      <p>Assigning truth values to the variables x1 and x2 to satisfy fx1; x2g in F1 [F2
gives two disconnected formula, by the hypothesis that F1 \ F2 = fx1; x2g, so
the conclusion holds. tu</p>
      <p>The previous theorem states that if we know the models of F1 where the
truth values of the variables x1 and x2 which joint F1 to another formula F2 via
a clause fx1; x2g are known, then we can substitute the models where x1 and x2
appears in F1 into those of F2 considering the truth values of x1 and x2 in F2.
5</p>
    </sec>
    <sec id="sec-5">
      <title>Results</title>
      <p>
        We implement our proposal and compare its runtime against sharpSAT which to
the best of our knowledge is the leading sequential implementation.Additionally,
in Table 1, we compare our proposal based on embedded cycles against our
previous version of markSAT based on bags [
        <xref ref-type="bibr" rid="ref6">6</xref>
        ]. Other outerplannar formulas,
Tables [
        <xref ref-type="bibr" rid="ref2 ref3 ref4">2, 3, 4</xref>
        ], represent polygonal tree graphs where each polygon has three
to eight sides. It is work to said that this implementation is sound and complete
hence the exact number of models is computed in all of them.
      </p>
      <p>Table 1 shows instances of polygonal chains, with three sides each polygon,
which provides the maximum number of edges in outerplanar graphs.</p>
      <p>Table 3 shows results on polygonal trees using a side-randomizer, which
generate graphs with polygons from three to eight sides each.</p>
      <p>Table 4 shows the running time of our proposal against sharpSAT using
outerplanar graphs as base case in a general formula decomposition.
We present a new method for model counting in outerplanar graphs, with linear
time complexity.</p>
      <p>Our procedure requires the construction of the expansion tree of the
outerplanar graphs, which in this case it is done in time O(n), the number of vertices
of the input formula. Once an expansion tree has been built a common edge
identi cation on both the tree and their back edges is done in time complexity
O(m), where m is the number of edges in the graph. A new expansion tree is
built with non common edges, in time O(n). Embedded edges can be added in
time O( n2 1 ), the maximum number of back edges in a outerplanar graph. Model
counting is done in O(n + m).</p>
    </sec>
  </body>
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