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Transcript of Art AlgoritmoRedEnergizada English
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Algorithm to Count the Number of Signed Paths in an Electrical Network via
Boolean Formulas
Guillermo De Ita, Helena G omez, Belem Merino
Benemerita Universidad Aut onoma de Puebla, M exico
Email: [email protected], [email protected], [email protected]
AbstractThis article presents a practical method to countthe different signed paths to maintain charge on each ofthe lines of energy of an electrical network. We assume acharge on each network node. We model this problem via the#2SAT problem. #2SAT consists on counting models of booleanformulas, where the input Boolean formula is in 2 conjunctiveform. To develop this method we rely on the connectivity graphof an electrical network from which we get its boolean formula
and apply recurrence equations starting from the terminalnodes up to the root node of the network. The recurrenceequations allows to carry the count to indicate the differentways to keep charge on all line of the electrical network.
Keywords-#SAT Problem; Counting models; Electrical Net-work; Graph tree;
I. INTRODUCTION
The problem of propositional satisfiability (SAT) is aclassic NP-complete problem, but there is a wide range of
research on how to identify limited instances for which the
SAT problem can be solved efficiently. The problem of
counting models of a boolean formula (#SAT) is relevantfor approximating reasoning processes tasks.
For example, for estimating the degree of reliability in
a communication network,computing degree of belief in
propositional theories, for the generation of explanations
to propositional queries, in Bayesian inference, in a
truth maintenance systems and for repairing inconsistent
databases [3], [4], [5], [9], [10]. Those previous problems
come from several AI applications such as planning, expert
systems, reasoning, etc.
#SAT is as difficult as the SAT problem, but even when
SAT can be solved in polynomial time, is not known anefficient computational method for #SAT. For example,the 2-SAT problem (SAT limited to consider ( 2)-CFs itcan be solved in linear time, however, the corresponding
counting problem #2-SAT is a #P-complete Problem.
The maximum polynomial class recognized for #2SAT
is the class ( 2, 2)-CF (conjunction of binary or unaryclauses where each variable appears twice at most) [5], [9].
Here, we extend such class for considering the topological
structure of the undirected graph induced by the restrictions
(clauses) of the formula.
We start considering signed graphs (graphs whose
edges are designated positive or negative). That class
of graphs have been very useful for modeling different
class of problems in the social sciences [6], [7], [8].
For example the signed graphs are used for modelingthe interaction among a group of persons and the type of
relationship between certain pair of individuals of the group.
We consider here an extension of the signed graphs
where all edge in the graph has associated a pair of signs.
In this way, any binary clause can be represented by such
signed edges. And we model an electric network by those
extended signed graphs.
We are interested in computing the different assignments
associated to the variables (nodes) in the network which
produces that all edge in the network will be satisfied by at
least one of its two possible signs.
The aim is to develop a method to count the different
signed paths in which an electric network held energy on
each of its electric lines, by counting the number of models
in the Boolean formula associated to the network.
I I . METHODOLOGY
Below we summarize a basic set of concepts through
which we try to improve a more complete understanding of
terms that will be used in this work.
Topology: Systematic study of the components of an
electrical network and its reciprocal links [2].
Element: Individual constituent of a network with two or
more terminals or bounds that are used for interconnection
with other elements [2].
Nodes and branches: The junction points of the
elements in a network are called nodes. Several elements
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can join together and considered as a single unit or a
whole. When several elements are joined together so
that the flow is the same in all of them are said to be
connected in series. The junctions between components
are called internal nodes. The pair of external nodes is the
nodes themselves or simply nodes. The set of elements in
series between the nodes themselves are called branches [2].
Graph: It is the geometric, topological, representation
of a network, forming a backbone of the physical layout of
the network [2]. A graph G is represented by a pair (V, E)where V is the set of nodes and E the set of edges.
A signed graph (also called sigraph) is an orderedpair = (G, ) where G = (V, E) is a graph called theunderlying graph of and : E {+, } is a functioncalled a signature or signing. E+() denotes the set ofedges from E that are mapped by to +, and E()denotes the set of edges from E that are mapped by to -.
The elements of E+() are called positive edges andthose of E() are called negative edges of . A signedgraph is all-positive (respectively, all-negative)if all of
its edges are positive (negative); further, it is said to be
homogeneous if it is either all-positive or all-negative, and
heterogeneous otherwise.
For example the signed graphs are used for modeling
the interaction among a group of persons and the type of
relationship between certain pair of individuals of the group.
Special focus is on the social inequalities. What is it about
such characteristics as sex, race, occupation, education,
and so on that leads to inequalities in social interaction? [8].
Given a set of de n Boolean variables,
X = {x1, x2,...,xn}. Its called a literal to any xvariable or the negation of it x. We use v(l) to indicate thevariable involved by the literal l.
The disjunction of different literals is called a clause.
For k N, a k clause is a clause consisting of exactlyk literals. A variable x X appears in a clause c if x or xis an element of c. Let v(c) = {x X : x appears in c}.
A conjunctive form (CF) is a conjunction of clauses.A k CF is a CF containing only k clausulas and,( k) CF denotes a CF containing clauses with atmost k literals.
Let be a 2-CF, then an assignment s for is a functions : v() {0, 1}. An assignment can also be consideredas a set of non-complementary pairs of literals. If l s,being s an assignment, then s makes l true and makes
l false. A clausula c is satisfied if and only if c s = ,
and if for all l c, l s then s falsifies c.
A CF is satisfiable by an assignment s if eachclausula in is satisfiable by s and is contradictedif it is not satisfied. s is a model of if s is a satisfiedassignment of .
Let SAT() be the set of models than has over v(). is a contradiction or unsatisfiable if SAT() = . Letv()() = |SAT()|, be the cardinality of SAT().Given a CF, the SAT problem consists in determiningif has a model. The #SAT consists of counting thenumber of models of F defined over v(). We will alsodenote v()() by #SAT() [1].
Given a signed network G, we obtain the associated
Boolean formula , this formula can be expressed as atwo Conjunctive Form (2 CF) in the following way.Let G = ((V, E), ) be the signed network, then where
v() = V and for all positive edge {x, y} in the networkthe clause (v(x), v(y)) is formed in , while for a negativeedge {x, y} the clause (v(x), v(y)) is formed in . Thismeans that the vertices of G are the variables of ,and for each clause (x, y) in there is a signed edge(v(x), v(y)) E.
III . BUILDING RECURRENCE EQUATIONS FOR
COUNTING SIGNED PATHS IN AN ELECTRICAL NETWORK
The signed graphs are very useful to represent some kind
of relationships among individuals. In this case, each person
is a node of the graph and there is an arc between {x, y}if x is in some relation to y. Many of the relationships of
interest have natural opposites, for example: likes/dislikes,
associates with/avoids, and so on [8]. For this, two different
signs {+, } are associated with each edge of the graph.
For example, if we consider that x likes y, then a positive
edge is denoted between x and y, but we do not know
what about y with x. y might dislike x or maybe like him.
In order to be more precise in the kind of relationships
between two individuals is better to consider two signs in
each edge, one sign associated with each point end of the
edge.
Furthermore, in order to consider general Boolean
formulas in 2-CF, we consider an extension of the
concept of signed graphs. Instead to consider just one
sign (+ or -) associated with each edge of a signed graph
G = ((V, E), ), we consider here that all edge in E hasassociated a pair of signs. Then the signed function has
now the type : {(+, +), (+, ), (, +), (, )} whichgives a pair of signs to each edge of G.
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In this way, all binary clause {x, y} can be representedby a signed edge in the graph in a natural way without
importance of the signs associated to the variables in the
clause. For example the clause {x, y} will be represented
as the signed edge x +
y and in this case, - is calledthe adjacent sign of x and + the adjacent sign of y in the
edge x + y.
Add more, we can consider those extended signed
graphs as electric networks. = (G, ) whereG = (V, E) is the underlying graph of and : E {(+, +), (+, ), (, +), (, )} is the signaturefunction. That means that all edge in the network is
energized in its endpoints.
We are interested here, in count the different assignments
associated to the nodes of the network in such a way that
all edge in the network will be satisfied by at least one of
its two possible signs of its endpoints.
For this, we start analyzing the way to build satisfy
assignment in the network considering first the most simple
topologies associated with the underlying graph of the
electrical networks.
A. If the Electrical Network is a Linear Path
Let us consider that the electrical network, G = (V, E)is a linear path. Let us write down its associated
formula , without a loss of generality (ordering theclauses and its literals, if it were necessary), as: =
{c1,...,cm} = {x11 , x
12 }, {x
22 , x
23 }, . . . , {x
mm1, x
mm },
where |(ci) (ci+1)| = 1, i [[m 1]], andi, i {0, 1}, i = 1,...,m.
In order to compute the number of signed paths in such
electrical network, we start with a pair of values (i, i)associated with each node i of the network. We call the
charge of node i to the pair (i, i). The value (i)indicates the number of times that node i takes the positive
value and (i) indicate the number of times that node itakes the negative value.
Let fi be a family of clauses of built as follows:
f0 = ,fi = {cj}ji, i [[m]]. Note that fi fi+1,i [[m 1]]. Let SAT(fi) = {s : s satisfies fi}, Ai ={s SAT(fi) : xi s}, Bi = {s SAT(fi) : xi s}.Let i = |Ai|; i = |Bi| and i = |SAT(fi)| = i + i.From the total number of models in i, i [[m]], there arei of which xi takes the logical value true and i models
where xi takes the logical value false.
For example, c1 = (x11 , x
12 ), f1 = {c1}, and
(1, 1) = ( 1, 1) since x1 can take one logical value
true and one logical value false and with whichever
of those values satisfies the subformula f0 while
SAT(f1) = {x11 x
12 , x
111 x
12 , x
11 x
112 }, and then
(2, 2) = (2, 1) if 1 were 1 or rather (2, 2) = (1, 2) if1 were 0.
In general, we compute the values for (i, i) associatedto each node xi, i = 2, . . ,m, according to the signs (i, i) ofthe literals in the clause ci, by the next recurrence equation:
(i, i) =
(i1 ,i1 + i1) if(i, i) = (,)(i1 + i1,i1 ) if(i, i) = (, +)(i1 ,i1 + i1) if(i, i) = (+,)(i1 + i1,i1 ) if(i, i) = (+, +)
(1)
We denote with the application of one of the fourrules of the recurrence (1), so, the expression (2, 3) (5, 2)denotes the application of one of the rules (in this case,
the rule 4), over the pair (i1, i1) = (2, 3) in order toobtain (i, i) = (i1 + i1, i1) = (5, 3).
In recurrence ( 1) the (i, i) represents the signs asso-ciated with the edge joining a child node joining with the
father node. These recurrence equations allow to carry the
count to indicate the different ways to keep the electrical
network powered. In fact, the sum i + i obtained fromthe root node of the network, indicate the total number of
Boolean formula models associated with the network and it
is as well as the number of different ways to keep the charge
on all electric line in the electrical network.
x2x1 x3 x4 x5
a)
b)
+ + + + ++- -
(1,1) (2,3)(2,1) (5,3) (8,5) =13 models
x2x1 x3 x4 x5
+ + - + +-- +
(1,1) (1,3)(2,1) (4,1) (5,1) =6 models
Figure 1. a) Counting models over a positive electrical path networkb) Counting models over a general electrical path network
Example 1: Let = {(x1, x2), (x2, x3), (x3, x4),(x4, x5)} be a path, the series (i, i), i [[5]], is computedaccording to the signs of each clause, as it is illustrated
if the figure (1a). A similar path with 5 nodes but with
different signs in the edges is shown in figure (1b).
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Example 2: Let = {(x1, x2), (x2, x3), (x3, x4),(x4, x5), (x5, x6)} be a 2-CF which conforms alinear path, the series (i, i), i [[6]], is computedas: (1, 1) = (1, 1) (2, 2) = (2, 1) since(1, 1) = ( 1, 1), and the rule 4 has to be applied. Ingeneral, applying the corresponding rule of the recurrence
(1) according to the signs expressed by (i, i), i = 3,..., 6,we have (2, 1) (3, 3) = ( 1, 3) (4, 4) =(3, 4) (5, 5) = (3, 7) (6, 6) = (10, 7), and then,#SAT() = () = 6 = 6 + 6 = 10 + 7 = 17.
If is a path , we apply (1) in order to compute(). The procedure has a linear time complexity over thenumber of variables of , since (1) is applied while we aretraversing the chain, from the initial node y0 to the final
node ym.
B. If the Electrical Network is a Tree
The charge of the nodes (i, i) for i = 1...n, arecalculated as the nodes are visited applying a post-order
traversals. The first pair of values (i, i) for the terminalnodes of the network (tree leaves) are initialized with the
value (1, 1). So, to traverse the tree in post-order, we startby the left tree first, followed by the right tree, and then
finally the root node. At each visit of a child node i 1 toa father node i, the new values (i, i) are computed forthe father node from the following formula [1]:
Algorithm Count Models ( A )
Input: A the tree defined by the depth-search overconnectivity graph of the electrical network G
Output: The number of signed paths of an electrical
network.
Procedure: Traversing A in depth-fist, and when a
node v A is left (all of its edges have been processed)assign:
1. (v, v) = (1, 1), If v is a leaf node in A
2. Ifv is a father node with a list of child nodes associated,
i.e., u1, u1,...uk, are the child nodes of v, then as we have
already visited all the child nodes, then each pair (uj , uj )j = 1,...,k has been defined based on (1), (vi , vi) isobtained by apply (2) over (i1, i1) = (uj , uj ). Thisstep is iterated until computes all the values (vj , vj ),
j = 1, , k. And finally, let v =j=1
k vj y v =j=1
k vj
3. If v is the root node of A the return (v + v)
Figure 2. The tree graph for the formula of the example 1
This procedure returns the number of signed paths for an
electrical network in time O(n + m) which is the necessarytime for traversing G in depth-first.
Example 3: Let = {(x1, x2), (x2, x3), (x2, x4),(x2, x5), (x4, x6), (x6, x7), (x6, x8)} be a 2 CF, obtainedfrom connectivity graph of an electrical network and con-
sider us that the depth-search starts in the node x1. The
tree generated by the depth-search as well as the number
of signed paths in each level of the tree is showed in
Figure 2. The procedure Count Models returns for x1 =
168, x1 = 84 and the total number of signed paths is#SAT() = 168 + 84 = 252.
IV. CONCLUSION
We show that the problem of counting the number of
models of a Boolean formula can be applied to count the
number of signed paths in an electrical network and that
the complexity of the algorithm is linear time on the length
of the network nodes.
We also have presented different linear procedures to
compute the number of energized paths in an electrical
network, based on the formula in 2-CF, obtained from thegraph of the electric network. And we will continue working
on different applications and repercussions than might have
this research in the future.
REFERENCES
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