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# In SIAM JOURNAL ON COMPUTING, 1996. Copyright SIAM AN EFFICIENT PARALLEL ALGORITHM FOR THE

In SIAM JOURNAL ON COMPUTING, 1996. Copyright SIAM

AN EFFICIENT PARALLEL ALGORITHM FOR THE GENERAL PLANAR MONOTONE CIRCUIT VALUE PROBLEM
VIJAYA RAMACHANDRANy AND HONGHUA YANGy

Abstract. A planar monotone circuit PMC is a Boolean circuit that can be embedded in the plane and that contains only AND and OR gates. Goldschlager, Cook & Dymond and others have developed NC 2 algorithms to evaluate a special layered form of a PMC. These algorithms require a large number of processors  n6 , where n is the size of the input circuit. Yang, and more recently, Delcher & Kosaraju have given NC algorithms for the general planar monotone circuit value problem. These algorithms use at least as many processors as the algorithms for the layered case. This paper gives an e cient parallel algorithm that evaluates a general PMC of size n in polylog time using only a linear number of processors on an EREW PRAM. This parallel algorithm is the best possible to within a polylog factor, and is a substantial improvement over the earlier algorithms for the problem. The algorithm uses several novel techniques to perform the evaluation, including the use of the dual of the plane embedding of the circuit to determine the propagation of values within the circuit. Key words. circuit value problem, planar monotone circuit, plane graph, dual graph, parallel algorithm, EREW PRAM AMS subject classi cations. 68Q10, 68Q15, 68Q20, 68Q22, 68Q25, 68R10, 05C10

1. Introduction. A Boolean circuit is a directed network of AND, OR and NOT gates whose wires do not form directed cycles. The problem of evaluating a Boolean circuit, given the values of its inputs, is called the circuit value problem CVP. This is a central problem in the area of algorithms and complexity. Ladner 16 has shown that CVP is P -complete under log space reductions. Some special cases of CVP have been studied, among which the monotone circuit value problem, where the Boolean circuit has only AND and OR gates, and the planar circuit value problem, where the Boolean circuit has a plane embedding, have been shown to be P -complete by Goldschlager 10 . A planar monotone circuit PMC is a Boolean circuit that is both planar and monotone. One interesting special case of CVP is the planar monotone circuit value problem PMCVP, which is the problem of evaluating a PMC. In this paper we give an e cient parallel algorithm for the PMCVP that runs in polylog time using a linear number of processors. The parallel computation model we use here is the EREW PRAM model 14 . Here is a summary of earlier results for the PMCVP. Goldschlager 7, 8 , Dymond & Cook 4 , and Mayr 17 have shown that the problem of evaluating a special layered form of PMC is in NC 2. The rst NC algorithm for the general PMCVP was given in Yang 24 ; this algorithm runs in Olog3 n time on an EREW PRAM, and uses the straight-line code parallel evaluation technique of Miller, Ramachandran & Kaltofen 18 . Recently Delcher & Kosaraju 3 have given another NC algorithm for the general PMCVP that runs in Olog4 n time using a polynomial number of processors on a CREW PRAM. All of the algorithms listed above use a large number of processors at least n6 , where n is the size of the input circuit.
 This work was supported in part by Texas Advanced Research Projects Grant 003658480 and NSF Grant CCR 90-23059. An extended abstract of this work appears in 21 . y Department of Computer Sciences, University of Texas at Austin, Austin, Texas, 78712-1188 vlr@cs.utexas.edu, yanghh@cs.utexas.edu. 1

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In earlier work Ramachandran & Yang 20  we gave an Olog2 n time EREW PRAM algorithm using a linear number of processors to evaluate a layered PMC. The algorithm we present in this paper, when restricted to evaluate a layered PMC, works with the same processor-time bounds as the one in 20 ; however, it is substantially di erent in that it works on a plane embedding of the PMC and its dual graph instead of exploiting a nice layered structure as in 20 . In one sense our algorithm can be considered to be simpler than the one in 20 since our new approach allows us to eliminate some tedious case analysis used in 20 . Our algorithm uses some ideas from 20 , as well as from 24 and 3 . In the highest level of our algorithm, we use an approach similar to that used in 3 to transform a general PMC into `face f induced subcircuits' using the terminology of 24 , these circuits are called `focused circuits' in 3 . These subcircuits are then evaluated using an algorithm to evaluate a `one-input-face PMC'. The major contribution of our paper is our e cient parallel algorithm to evaluate a one-input-face PMC, which is a PMC, not necessarily layered, all of whose input nodes are on the boundary of one face. The rest of this paper is organized as follows. In Section 3, we present our algorithm to evaluate a one-input-face PMC. The treatment in Section 3 is self-contained and does not depend on any result in 20 . In Section 4 we give an algorithm that runs in polylog time using n processors on an EREW PRAM for evaluating a face f induced subcircuit given a special type of an input assignment. This algorithm works by recursively applying the algorithm for evaluating a one-input-face PMC. Finally, in Section 5 we give an algorithm that runs in polylog time using n processors on an EREW PRAM for solving the general PMCVP by recursively applying the algorithm for evaluating a face induced subcircuit. Our results are of interest for several reasons. In designing our e cient parallel algorithm for the PMCVP we have developed a variety of e cient parallel algorithms for processing planar DAGs, especially the technique of working on the dual of a planar DAG. Other examples of algorithmic techniques based on the dual of a plane embedding can be found in 22, 13 . These tools are likely to be of use in algorithms for other problems on planar directed graphs. Our results are of interest in the context of parallel complexity since all of the earlier algorithms for the PMCVP used indirect methods such as the relationship between sequential space and parallel time 1 or the parallel evaluation of straight-line code 18 to place the problem in NC. By using direct techniques, not only are we able to place the problem in NC, but we are able to obtain a very e cient algorithm for its solution. Finally, the evaluation of circuits is a basic and important problem in computer science. Planar circuits occur very naturally in the design of integrated circuits, and the requirement that the circuit be monotone is not a restriction if the inputs are available together with their complements. Thus our e cient parallel algorithm for the evaluation of planar monotone circuits could be of practical importance.
Definition 2.1. A face of a plane graph C = V; E  is a maximal portion of the plane for which any two points may be joined by a curve such that each point of the curve neither corresponds to a vertex of C nor lies on any curve corresponding to an edge of C . The boundary of a face f in C consists of all those points x corresponding to vertices and edges of C having the property that x can be joined to a point of f by a curve, all of whose points di erent from x belong to f . By this de nition, a single edge in f belongs to the boundary of f .

2. Preliminaries.

EFFICIENT PARALLEL PLANAR MONOTONE CIRCUIT VALUE

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Definition 2.2. An embedded planar monotone circuit PMC is a plane directed acyclic graph DAG C = V; E , where i V is the set of gates or vertices in the PMC consisting of input nodes, AND gates, and OR gates, ii E is the set of directed wires or edges in the PMC, iii the fan-in or in-degree of an input node is 0, of an AND or OR gate is either 1 or 2, C may have input nodes that are in di erent faces, iv the fan-out or out-degree of an output gate is 0, of other gates is nonzero, C may have more than one output gate, but all output gates of C are in the same face. In the rest of the paper, whenever we use the term PMC, we should assume that the PMC is given with an embedding. In case an embedding is not given, we can use the algorithm in Ramachandran & Reif 19 to obtain one. We assume that the plane embedding of a PMC C is given by its combinatorial de nition: a clockwise cyclic ordering of edges incident to each vertex in C , and a counterclockwise cyclic ordered sequence of vertices and edges g0; e0 ; g1; e1; : : :; gk,1; ek,1 on the boundary of a face in C such that for any i, 0  i  k , 1, the edges ei,1 and ei are incident to vertex gi and ei,1 appears immediately before ei in the cyclic ordering of the edges incident to gi. Definition 2.3. A complete input assignment to a PMC is an assignment of values 0 or 1 to all input nodes in the PMC. A partial input assignment to a PMC is an assignment of values 0 or 1 to a subset of the input nodes in the PMC. An input node that is not assigned a value in a partial input assignment has an unknown value. A complete input assignment is a special case of a partial input assignment. Definition 2.4. The partial evaluation problem of a PMC is the problem of evaluating the value of every gate in the PMC that can be evaluated, given a partial input assignment to the PMC. The gates in a PMC that cannot be evaluated under a partial input assignment have unknown values. A PMC is completely evaluated if the value of every gate in it is either 0 or 1, it is partially evaluated otherwise. The planar monotone circuit value problem PMCVP is the problem of completely evaluating a PMC, given a complete input assignment to the PMC. A one-input-face PMC we de ne below is a PMC with the following di erences: i It is a restriction of a PMC in that all of its input nodes are on the boundary of a single face. ii It is a generalization of a PMC in that it may contain pseudo wires, which are wires that carry no value. Our algorithm may need to add pseudo wires in a PMC during the computation in subsection 3.2. Definition 2.5. A one-input-face PMC C is a variant of a PMC with the following properties. 1. C is a plane DAG consisting of input nodes, AND gates, and OR gates. 2. All input nodes of C are on the boundary of a single face fI . The in-degree of an input node is 0, of an AND or OR gate is either 1 or 2. 3. A gate in C with out-degree 0 is called an output gate. The out-degree of other gates or input nodes is nonzero. C may have more than one output gate, but all output gates of C are on the boundary of a single face fO . 4. If fI and fO are identical, then the input nodes and the output gates of C may not interlace, i.e. there exists a part of the boundary of fI which contains all input nodes but no output gates. 5. Some of the gates in C may contain a single output wire that does not carry any value and that goes into a two-input gate note that such a gate with a single

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output wire that does not carry any value is not an output gate. We call a wire that does not carry any value a pseudo wire. Further, a two-input gate g may receive at most one input from a pseudo wire, and the value of g only depends on its non-pseudo input wires. We will give a recursive algorithm in Section 3 that evaluates a one-input-face PMC of size n in Olog2 n time using n processors on an EREW PRAM, where properties 4 and 5 in De nition 2.5 are needed after the rst level of recursion. Definition 2.6. Reachi1 ; : : :; ik  for some input nodes i1 ; : : :; ik in a PMC is the part of the PMC that is reachable from i1 ; : : :; ik . Given a subcircuit P of a PMC, ReachP  is de ned to be the part of the circuit reachable from the input nodes in P . Inducedi1 ; : : :; ik  for some input nodes i1 ; : : :; ik that are on the boundary of a single face f where i1 ; : : :; ik need not be all the input nodes of f  in a PMC C is Reachi1 ; : : :; ik  augmented with a new input node set VIN and a wire set EIN which are formed as follows: if a gate x 2 C n Reachi1 ; : : :; ik  has some output wires x; y1; x; y2; : : :; x; yl , l  1 pointing to gates y1 ; y2 ; : : :; yl in Reachi1 ; : : :; ik , then we add a new input node ix to VIN and wires ix ; y1; ix ; y2; : : :; ix ; yl  to EIN . We call such Inducedi1 ; : : :; ik  a face f induced subcircuit. It is easy to see that a face f induced circuit is still a PMC. A face f induced circuit Cf is not necessarily a one-input-face PMC since the newly added input nodes can appear in faces other than f in Cf . But Cf is still simpler than a general PMC in the sense that all gates in Cf except the newly added input nodes are reachable from some input nodes on the boundary of face f , and once the values of the new input nodes are known, Cf can be transformed into a logically equivalent one-input-face PMC. We give an algorithm in Section 4 that partially evaluates a face induced circuit of size n given a special input assignment, in polylogn time using n processors on an EREW PRAM, by recursively calling the algorithm for evaluating a one-input-face PMC. In Section 5, we give an algorithm that completely evaluates a general PMC of size n in polylogn time using n processors on an EREW PRAM, by recursively calling the algorithm for partially evaluating a face induced circuit, given a special input assignment. 3. The One-Input-Face PMC. We rst consider the problem of completely evaluating a one-input-face PMC C given a complete input assignment. This treatment appears in subsections 3.1, 3.2, and 3.3. We then solve the problem of partially evaluating a one-input-face PMC in subsection 3.4. Our approach is to rst nd a set of gates in C that are guaranteed to have value 1, and then recursively evaluate the remaining smaller unevaluated subcircuits of C . In an earlier paper 20 we had considered a special case of a one-input-face PMC, namely, a layered PMC as mentioned in the introduction. In a layered PMC, a sequence of gates with value 1 at one layer guarantees that a sequence of gates at the next layer will have value 1; and the left and the right boundaries of the gates with value 1 are de ned by the starting gate and the ending gate of the sequence at each layer respectively. In a one-input-face PMC C that does not have the layered property, we do not have such a simple correspondence. In the treatment below, we work with the dual of a plane embedding of C , and de ne the left and right boundaries of the gates with value 1 in a manner that allows us to determine the propagation of the 1 values through the circuit. In subsection 3.1, we will give some de nitions and lemmas. In subsection 3.2, we will present our techniques for nding the left and right boundaries of the gates with value 1 and for simplifying the remaining circuit of C . In subsection 3.3, we will

EFFICIENT PARALLEL PLANAR MONOTONE CIRCUIT VALUE
s

5

i1

i2

i3

i4

p

t p wire auxiliary edge pseudo wire

super source or super sink input node OR gate AND gate

output gate

Fig. 1. Caug : a one-input-face PMC C augmented with a super source s and a super sink t. Here fI = fO .

give the complete algorithm for evaluating a one-input-face PMC given a complete input assignment and its complexity analysis. In subsection 3.4, we will extend the algorithm to evaluate a one-input-face PMC given a partial input assignment. Throughout Section 3, we use C to refer to a one-input-face PMC unless otherwise stated.
Definition 3.1. A gate is a source of a face f in C if it has two output wires that are on the boundary of f . A gate is a sink of a face f in C if it has exactly two input wires and both are on the boundary of f . Let fI be the face of C whose boundary contains all the input nodes and let fO be the face of C whose boundary contains all the output gates. Caug is a DAG obtained from C by adding a super source s in fI and auxiliary wires connecting s to every input node in C , and a super sink t in fO and auxiliary wires connecting every output gate to t see Figure 1. By De nition 2.5, if fI and fO are identical then the input nodes and the output gates may not interlace in C . Hence Caug is still a plane graph. In the rest of the paper, we assume that C has at least two input nodes. If C has only one input node, then by De nition 2.2 all gates in C have the same value as the only input node, which is a trivial case. The reason that we augment C to Caug is that there is a single source and a single sink in every face of Caug as shown in the following lemma. This property is crucial to many de nitions given in this subsection. Note that not every one-input-face PMC has a downward plane drawing 11 as in Figure 1, if fI and fO are di erent faces. Lemma 3.1. Every face in Caug has exactly one source and one sink. Proof. Every gate in Caug is reachable from s and reachable to t. Since Caug is a DAG, there is at least one source and one sink on the boundary of a face in Caug .

3.1. De nitions.

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Suppose a face f has two sources s1 and s2 and therefore two sinks t1 and t2 . Then we consider the following four directed paths in Caug : the path P1 from s to s1 , the path P2 from s to s2 , the path P3 from t1 to t, and the path P4 from t2 to t. The path P consisting of P1 and P2 joins the face f 0 at s1 and s2 . The path Q consisting of P3 and P4 joins the face f 0 at t1 and t2 . But s1 and s2 interlace with t1 and t2 on the boundary of f . But the two paths P and Q have to be embedded in one side of the boundary of f . A contradiction. Hence every face in Caug has exactly one source and exactly one sink. The following lemma is needed for De nition 3.3. Lemma 3.2. The output wires of a gate g in Caug are placed consecutively in the cyclic ordering of the wires around g. Proof. Let g be a gate in Caug with two input wires i1 and i2 and two output wires o1 and o2 , such that o1 and o2 interlaces with i1 and i2 in the cyclic ordering of the wires around g. Since every gate in Caug is reachable to t, there are two directed paths P1 and P2 in Caug from g to t, where P1 goes through o1 and P2 goes through o2 . Let x be the rst gate except g on P1 that is also on P2. Since Caug is acyclic, x must be the rst gate except g on P2 that is also on P1. The subpath of P1 from g to x and the subpath of P2 from g to x form an undirected cycle which divides the plane into two parts Cinside and Coutside, where Cinside is the part of the plane that is inside the cycle and Coutside is the part of the plane that is outside the cycle. Without loss of generality, assume that the super source s and the input wire i2 are in Coutside. Let i1 = g1; g. Then g1 and i1 are in Cinside and g1 is not reachable from s since otherwise there would be a directed cycle in Caug . Hence g1 is reachable from some input nodes in Cinside that cannot be reached from s in Coutside. A contradiction.
Definition 3.2. The left right input wire of a two-input gate g is the input wire of g that appears immediately after before an output wire of g in the clockwise cyclic ordering of the wires around g. Note that the source and the sink of a face f in Caug partition the boundary of f into two disjoint except at the source and the sink directed paths. Definition 3.3. The path from the source to the sink going through the left input wire of the sink is the counterclockwise boundary of the face, and the path from the source to the sink going through the right input wire of the sink is the clockwise boundary of the face.  Definition 3.4. The dual digraph Caug = V  ; E  of the plane directed primal graph Caug = V; E is de ned as follows. i For each primal face f in Caug , de ne a dual vertex f  in V  . ii For each primal edge e in Caug such that e is on the clockwise boundary of a primal face f1 and the counterclockwise boundary of another primal face f2 ,   de ne a counterclockwise dual edge e+ = f1 ; f2  in E  and a clockwise dual edge   e, = f2 ; f1  in E  .  Note that in the dual graph Caug , we introduce dual edges of both directions, clockwise and counterclockwise for each edge in the primal graph. The dual graph  Caug can also be viewed as the result of forming the undirected dual graph of Caug and replacing the dual edges by dual arcs of both directions. For convenience, for a primal face f , and a primal edge e, we will use f  ; e+ ; e, to indicate the dual vertex of f , the counterclockwise dual edge of e, and the clockwise dual edge of e respectively. In the following de nition, we de ne an auxiliary graph which can be viewed as

EFFICIENT PARALLEL PLANAR MONOTONE CIRCUIT VALUE
s

7

i1

f1*

i2

* f2

i3

f3*

i4

f6* * f5 p f10* f8* f7*

f9*

f4* * f11 t p wire auxiliary edge pseudo wire left leg right leg dual vertex

super source or super sink input node OR gate AND gate

output gate

Fig. 2. Caug and Aaug , the auxiliary dual graph. Caug consists of the solid edges. Aaug consists of the dashed edges. The two graphs overlap at s and t.

 a subgraph of Caug augmented with s and t that contains some edges called left legs and right legs. These edges aid us in de ning regions of Caug where value 1 propagates from input nodes. Thus, their de nitions are dependent on whether a gate is an AND gate or an OR gate and whether a wire is a pseudo wire, since an AND gate has value 1 if both of its inputs have value 1, an OR gate has value 1 if one of its inputs has value 1, and a pseudo wire does not pass any value.  is de ned as Definition 3.5. The auxiliary dual graph Aaug = V1 ; E1 follows see Figure 2. i V1 contains the dual vertices of all the primal faces in Caug , together with the super source s and the super sink t.  ii E1 contains all the dual edges called the left legs and the right legs de ned as follows. Let f be a primal face in Caug whose boundary does not contain t and let g be the sink of f with left input wire wl and right input wire wr . i If g is an OR gate with no pseudo input wire, then the left leg and the right leg  of f  are wl, i.e., the clockwise dual edge of wl  and wr + i.e., the counterclockwise dual edge of wr  respectively. ii If g is an AND gate with no pseudo input wire, then the left leg and the  right leg of f  are wr + and wl, respectively. iii If wl is a pseudo wire, then both the left leg and the right leg of f  are wl, .  iv If wr is a pseudo wire, then both the left leg and the right leg of f  are wr + . v Let f be a primal face whose boundary contains t or s in Caug . Then we  add an auxiliary edge f  ; t or s; f   to E1 , respectively.  that contains some dual edges of the input Aaug n fs; tg is a subgraph of Caug wires to the sinks of the faces in Caug . It is easy to see that after being augmented

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V. RAMACHANDRAN AND H. YANG

with s and t, Aaug is still a plane graph since the input nodes and the output gates of C do not interlace.   Lemma 3.3. If there is an edge e = f1 ; f2  in Aaug which is either a left leg or a right leg, then there is a directed path of length at least 1 from the sink s1 of f1 to the sink s2 of f2 in Caug . Proof. By De nition 3.5, s1 must be on either the clockwise boundary or the counterclockwise boundary of f2 in Caug . Since a gate except t has at most two input wires and therefore can be the sink of at most one face, s1 cannot be the sink of f2 . Hence there is a directed path of length at least 1 from s1 to s2 . Corollary 3.3.1. Aaug is a plane DAG whose only vertex with out-degree 0 is t. Proof. By Lemma 3.3, Aaug is acyclic and hence a DAG. It is obvious that the only vertex in Aaug with out-degree 0 is t. Definition 3.6. Two input nodes i1 and i2 in Caug are adjacent if the wire s; i1  and the wire s; i2  are adjacent in the cyclic ordering of the wires around s in Caug . Given a complete input assignment to the input nodes of Caug , a valid base B is a maximal sequence of adjacent input nodes with value 1. The left right bounding face of a valid base B is the face in Caug whose clockwise counterclockwise boundary contains an input node in B , but whose counterclockwise clockwise boundary does not see Figure 5. If all input nodes in Caug have value 1, then the left bounding face and the right bounding face are not de ned. But this is a trivial case since we know that all gates of Caug must have value 1. Definition 3.7. For a valid base B in Caug , let fl and fr be the left and right bounding faces of B respectively. The left boundary and the right boundary of B are the two directed paths Pl and Pr respectively in Aaug , such that 1 Pl and Pr start from s, 2 Pl consists of left legs and auxiliary edges and goes through s; fl , 3 Pr consists of right legs and auxiliary edges and goes through s; fr , 4 Pl and Pr end at their rst common vertex g g could be t after s see Figures 5 & 6. Definition 3.8. Given a valid base B , the left boundary Pl of B and the right boundary Pr of B divide the plane into two regions. The region whose counterclockwise boundary is Pl and whose clockwise boundary is Pr is called the internal region of B , the other region is called the external region of B see Figures 5 & 6. Lemma 3.4. Given a complete input assignment to Caug where there is only one valid base B all other input nodes have value 0, a gate g in Caug evaluates to 1 i g is in the internal region of B . Proof. Let us embed Aaug and Caug in the plane simultaneously with the same super source s and the same super sink t as in Figure 2 such that the only primal edges in Caug that cross the left and right boundaries Pl and Pr of B are the input wires to the sinks of some of the primal faces in Caug . This is provable by De nitions 3.5 and 3.7. A sink in Caug with a pseudo input wire can have only its pseudo input wire but not the other input wire crossing Pl or Pr . Further, a sink of a primal face f in Caug cannot have its two input wires crossing an edge wl in Pl  and an edge wr in Pr respectively, since otherwise f  would be the common starting  vertex of both wl and wr and therefore would be a common vertex of Pl and Pr ,  and Pr do not end at f  since w is in P  and wr is in Pr , a contradiction.  but Pl l l  and Pr from Caug and call the Therefore, if we remove all wires that cross Pl 0 0 resulting graph Caug , then every gate except the input nodes in Caug still has at least one non-pseudo input wire by the previous paragraph, and hence can still be

EFFICIENT PARALLEL PLANAR MONOTONE CIRCUIT VALUE

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reached from some input nodes without going through pseudo wires. Further, the 0 gates in the internal external region of B in Caug can be reached only by the input nodes in the internal external region. Therefore, if we remove all wires crossing Pl and Pr from Caug , the gates in the internal region of B will have value 1 and the gates in the external region of B will have value 0. We now show that this is still the case even if we do not remove the wires crossing Pl and Pr from Caug . By De nition 3.5, a wire in Caug outgoing from a gate in the external region of B and incoming to a gate in the internal region of B is either an input wire to a two-input OR gate, or a pseudo input wire to a two-input gate. Hence the gates in the internal region of B in Caug will still have value 1. A primal edge in Caug outgoing from the internal region of B to the external region of B is either an input wire to a two-input AND gate that is in the external region of B , or a pseudo wire to a two-input gate that is in the external region of B . Hence the gates in the external region of B in Caug will still have value 0. Hence the lemma holds. Corollary 3.4.1. Given a complete input assignment to Caug , if a gate g is in the internal region of a valid base in Caug , then g evaluates to 1. Proof. By Lemma 3.4 and the monotonicity of Caug . Note that the reverse of Corollary 3.4.1 need not be true if there is more than one valid base in Caug , i.e., some gates outside the internal regions of the valid bases of Caug might also be evaluated to 1. Hence our approach to evaluate a one-input-face PMC Caug is to rst nd some of the gates that evaluate to 1 based on the internal regions of the valid bases of Caug , remove them from Caug , and repeatedly evaluate the resulting Caug . 3.2. Complete Evaluation of a One-Input-Face PMC. In this subsection, we give an e cient method for computing the left and the right boundaries for all valid bases in Caug simultaneously, given a complete input assignment to Caug . The main idea is to identify for each dual vertex f  in Aaug whether it is on the left right boundary of some valid base in Caug i.e., whether BOUNl f   or BOUNr f   de ned in De nition 3.10 below is nonempty. Based on this approach, we present a technique to transform the part of Caug that has not been evaluated to 1 into several subcircuits that are one-input-face PMCs with smaller sizes, and more importantly, with geometrically decreasing number of valid bases. We rst de ne two tree structures that consist of left legs and right legs. Definition 3.9. Let Vl Vr  be the set of vertices of Aaug except s that are reachable through left right legs and auxiliary edges incoming to t from the dual vertex of the left right bounding face of some valid base in C . We de ne Tl Tr  to be the subgraph of Aaug induced by Vl Vr . For example, Figure 3 gives Tl and Tr for the circuit in Figure 1 with an input assignment 0; 1; 0; 1 to the input nodes i1 ; i2; i3 ; i4. Lemma 3.5. Both Tl and Tr are convergent trees. Proof. By Corollary 3.3.1, Aaug is a DAG. By De nition 3.5, there is exactly one left leg and one right leg, or one auxiliary edge outgoing from each vertex except s in Aaug . Hence the lemma holds. In the following de nitions and lemmas, we de ne BOUNl f   BOUNr f   and related concepts, and describe our approach to compute BOUNl f   BOUNr f  . Among the sets de ned below are two related and similar concepts, BOUNl f   BOUNr f   and BASEl f   BASEr f  ; these sets are di erent in that the latter is a superset of the former and is de ned to aid the computation of the former. Definition 3.10. PREDl f   PREDr f   is the set of the proper predecessors

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V. RAMACHANDRAN AND H. YANG
* f1 * f2 f3*

f6* * f5 f7*

f10*

f8*

f9*

f4* * f11 t auxiliary edge left leg right leg dual vertex

 Fig. 3. Tl and Tr for the circuit in Figure 1, given an input assignment 0; 1; 0; 1 to the input nodes i1 ;i2 ;i3 ;i4 . Tl consists of the light dashed edges. Tr consists of the dark dashed edges.

of f  in Tl Tr , i.e. the set of dual vertices that can reach f  through directed paths of length at least 1 in Tl Tr . We associate with each dual vertex f  in Tl Tr  the following sets of valid bases of Caug : i BASEl f   BASEr f   is the set of valid bases B such that the dual vertex of the left right bounding face of B is either f  or in PREDl f   PREDr f  . Informally, BASEl f   BASEr f   is the set of the valid bases in Caug whose left right boundaries either contain f  or a predecessor of f  in Tl Tr . ii BOUNl f   BOUNr f   is the set of valid bases B whose left right boundary contains f  . iii JOIN f   = BASEl f   BASEr f  . Informally, JOIN f   is the set of valid bases whose left and right boundaries either terminate at f  , or terminate at a predecessor of f  and the extension of the left and right boundaries rejoin at f  . iv TERM f   is the set of valid bases whose left and right boundaries terminate exactly at f  . For convenience, if a set for a dual vertex in Aaug cannot be de ned through De nition 3.10 e.g., if a dual vertex is not in Tl , we assume that it is empty. Figure 4 illustrates the above de nitions. For valid base B2 = fi4g with left bounding face f3 and right bounding face f4 in Figure 1, the left boundary of B2 is  a subpath of the path from f3 to t in Tl , the right boundary of B2 is a subpath of  to t in Tr in this case, the subpath is the single vertex f4 . Notice  the path from f4   = fB1 ; B2 g and BOUNl f4  = fB2g, between  the di erence between BASEl f4 BASEr t = fB1; B2 g and BOUNr t = , and between JOIN t = fB1 ; B2g and TERM t = . The relations among these sets are summarized in the following lemma. For convenience, we will focus on the sets with index l. The relations among the sets with index r are symmetric. Lemma 3.6. Let f  be a dual vertex in Aaug . Then   1. BASEl fp  BASEl f  , for any fp 2 PREDl f  .

EFFICIENT PARALLEL PLANAR MONOTONE CIRCUIT VALUE
(B1) l ( )j * ( ) t f1 [B1] l (B1) l ( )j ( )t [B1] l (B1) r ( )j ( ) t * f2 [B1] r (B2) l ( )j ( )t [B2]l (B2) l f7* ( )j ( )t [B2]l

11

f3*

* f5

(B2) l ( )j * ( ) t f6 [B2] l

(B1) r [B1] r

(B1) l ( )j ( )t [B1] l

(B1) l * f10 * (B1) j (B1) l f8 * (B1) t f9 (B1) r ( )j [B1] r [B1] l ( )t [B1] l (B1) r ( )j * f11 ( )t [ ]r t (B1, B2) l (B1, B2)j ( )t [ ]l

(B1, B2)l (B2) r f4* (B2) j [B2] r (B2) t [B2] l

(B1, B2) r [ ]r

auxiliary edge

left leg

right leg

dual vertex

Fig. 4. Illustrations for the sets BASEl , BASEr , BOUNl, BOUNr , JOIN , and TERM on Tl and Tr . For each node f  in Tl, the contents of l denote BASEl f , the contents of l denote BOUNlf , the contents of j denote JOIN f , and the contents of t denote TERM f . For each node f  in Tr, the contents of r denote BASEr f , and the contents of r denote BOUNrf  ,

 2. BOUNl f   BOUNl f1  = and BASEl f   BASEl f   = , for any  that does not have predecessor-successor relation with f  in T  . 1 f1 l   3. TERM f   TERM f1  = , for any f1 6= f  .  4. TERM f   = JOIN f   n fp 2PREDl f   JOIN fp .   = BASEl f   n f  2PREDl f   TERM fp ; and  5. BOUNl f p  jBOUNl f  j = jBASEl f  j , fp 2PREDl f   jTERM fp j.

P

Proof. The correctness of 1, 2, 3 follows directly from De nition 3.10. The correctness of 4 follows from De nition 3.7, De nition 3.10 and 1, 2. The correctness of 5 follows from De nition 3.10 and 1, 2, 3. Our goal is to identify the gates that are on the left and right boundaries of some   valid base. Since a left leg f1 ; f2  is on the left boundary of some valid base i  j 0 and jBOUNl f2 j 0, it su ces to compute jBOUNl f  j for  jBOUNl f1  in T  . BASEl f   and hence jBASEl f  j can be easily computed using every f l the Euler-tour technique 23 on Tl see Procedure 2 in subsection 3.3 for details. Since BASEl f   BASEr f   contains valid bases with consecutive labels modulo the total number of bases in the total order of the valid bases, it can be described succinctly by a range x, y where x, y are the numbers of the rst and the last valid bases in BASEl f   BASEr f   respectively. If we can compute jTERM f  j for every f  that is in both Tl and Tr , then we can compute jBOUNl f  j using 5 in Lemma 3.6 and the Euler-tour technique. It remains to compute jTERM f  j for every f  that is in both Tl and Tr .

12

V. RAMACHANDRAN AND H. YANG

We can try to compute jTERM f  j directly from 4 in Lemma 3.6. However, note that the sets JOIN f   = BASEl f   BASEr f   are not necessarily disjoint for di erent f  in Tl if they have predecessor-successor relation. Instead, we show in the following lemma that they satisfy some important properties, and then in Lemma 3.8 we give a formula to compute TERM f   with disjoint JOIN f   sets.
Proof.   1. By 1 in Lemma 3.6, we have both BASEl fp  BASEl f   and BASEr fp   . Therefore, JOIN fp  JOIN f  .  BASEr f  2. Without lose of generality, assume f  and fp do not have predecessor-successor  . Then by 2 in Lemma 3.6, BASEl f   BASEl fp  = . Therefore,  relation in Tl   JOIN f   = . JOIN fp Based on Lemma 3.7, we give the following de nition.  Definition 3.11. For a dual vertex f  and one of its predecessors fp in Tl   6= and JOIN fp  6= , JOIN fp  is immediately enclosed by   with JOIN f   JOIN f  , denoted by JOIN fp  I JOIN f  , i JOIN fp  JOIN f   and  on the directed path from fp to f  in T  such that  there is no dual vertex fq l JOIN fq  JOIN f  . Lemma 3.8. For a dual vertex f  that is in both Tl and Tr ,      1. TERM f  = JOIN f  n fp 2PREDl f  ^JOIN fp I JOIN f  JOIN fp , P    2. jTERM f j = jJOIN f j ,  fp 2PREDl f  ^JOIN fp I JOIN f  jJOIN fp j. Proof. By 4 in Lemma 3.6 and Lemma 3.7, 1 holds immediately. By De ni tion 3.11, none of the fp in the summation in 2 has predecessor-successor relation.  By Lemma 3.7, the sets JOIN fp  in 2 are disjoint. Hence 2 holds. The above lemmas give us the necessary tools to compute jBOUNl f  j e ciently in parallel. The algorithms that implement the computations in Lemmas 3.6 & 3.8 are given in procedures 2 & 3 and in the proof of Lemma 3.16 in subsection 3.3. Having computed the left and right boundaries of the valid bases of Caug , our next step is to identify the regions of Caug that consist of gates with value 1. In the following de nition, we de ne a separating graph Asep , which is a subgraph of Aaug that consists of the left and right boundaries of all the valid bases of Caug , and which is used to nd the regions of Caug that consists of gates with value 1. It is formally de ned as follows:

Lemma 3.7.   1. If fp is a predecessor of f  in both Tl and Tr , then JOIN fp  JOIN f  .  2. Otherwise, JOIN f   JOIN fp  = .

1. A separating graph Asep contains s and the vertices f  in Aaug for which either BOUNl f   6= or BOUNr f   6= . Asep contains t if BOUNl t 6= or BOUNr t 6= .   2. An edge f1 ; f2  of Aaug is an edge in Asep if one of the following three con ditions holds: a f1 = s and f2 is the left or right bounding face of a valid base,   6= and BOUNl f2  6= and BOUNl f1  6= TERM f1 , or c    or b BOUNl f1   6= and BOUNr f2  6= and BOUNr f1  6= TERM f1 .    BOUNr f1  Asep is a subgraph of Aaug , which is a subgraph of Caug the dual graph of Caug  augmented with s and t. Hence Asep is a plane graph. Each face in Asep is called a separating region of the primal graph Caug . Note that a separating region of Caug either is in the internal region of a valid base, or in the external region of every valid base, in which case we call it an external separating region. In the example in Figure 5, Asep consists of the left and right boundaries of

Definition 3.12.

EFFICIENT PARALLEL PLANAR MONOTONE CIRCUIT VALUE
s

13

0 i1

f1*

1 i2

* f2 internal region of B1 f5*

0 i3

* f3

1 i4

f6*

internal region of B2

p g1 * f10 g2 f8* f9* f7* f4*

g4 * f11 t p wire auxiliary edge pseudo wire

g3

left leg

right leg

dual vertex

super source or super sink input node OR gate AND gate

output gate

Fig. 5. Caug and Asep to show the left and right boundaries of B1 and B2 . Caug consists of the solid edges. The complete input assignment to the input nodes i1 ; i2 ; i3 ; i4 is 0, 1, 0, 1. B1 = fi2 g and B2 = fi4 g are two valid bases. The left bounding face and the right bounding face      of B1 are f1 and f2 respectively. The left boundary of B1 is the directed path s; f1 ;f5 ; f10 ; f8 ; f9     and the right boundary of B1 is the directed path s; f2 ; f6 ; f9 . The part of the plane inside the two boundaries is the internal region of B1 and the part of the plane outside the two boundaries is the external region of B1 . The left bounding face and the right bounding face of B2 are f3 and f4     respectively. The left boundary of B2 is the directed path s; f3 ;f6 ; f7 ; f4  and the right boundary  . A of B2 is the directed path s; f4 sep consists of all dashed edges.

B1 and B2 . The separating regions of Caug in Figure 5 are: the face in Asep        with boundary s; f1 ; f5 ; f10; f8 ; f9 ; f6 ; f2 ; s i.e., the internal region of B1 , the     face in Asep with boundary s; f3 ; f6 ; f7 ; f4 ; s i.e., the internal region of B2 , the    face in Asep with boundary s; f2 ; f6 ; f3 ; s, and the face in Asep with boundary  ; f5 ; f10; f8 ; f9 ; f6 ; f7 ; f4 ; s.        s; f1 Definition 3.13. A wire w is incoming to outgoing from a subcircuit C 0 if the head tail of w is a gate in C 0 but the tail head of w is not. Lemma 3.9. All incoming wires to an external separating region R in Caug are either wires with value 1 or pseudo wires. Any input node in R will have value 0. Proof. The wires incoming to R must come from the internal regions of some valid bases, since a wire crosses a boundary of a valid base B either from the internal region to the external region of B , or from the external region to the internal region of B . Since R is not in the internal region of any valid base of Caug , all the wires incoming to R are wires outgoing from the internal regions of some valid bases. By Corollary 3.4.1, the gates in the internal region of any valid base of Caug have value 1. Hence all incoming wires to R are either wires with value 1 or pseudo wires. Recall that Corollary 3.4.1 states that the gates in the internal region of every valid base of Caug have value 1. Hence the gates in the separating regions that are in the internal region of a valid base have value 1. In the following corollary we will extend Corollary 3.4.1 to show that in fact the gates in any separating region including external separating region that does not contain an input node with value

14

V. RAMACHANDRAN AND H. YANG
s

1 i1

* f4

0 i2

f1*

* 1 f2 i3 internal region of B1 * f7

0 i4

* f3

1 i5 internal region of B2 f9* f10*

f5*

f8*

p f6*

R

t

p wire auxiliary edge pseudo wire left leg right leg dual vertex

super source or super sink input node OR gate AND gate

output gate

Fig. 6. Caug and Asep to show the left and right boundaries of B1 and B2 . Caug consists of the solid edges. Asep consists of the dashed edges. The complete input assignment to the input nodes i1 ; i2 ; i3 ; i4 ; i5 is 1, 0, 1, 0, 1. B1 = fi3 g and B2 = fi5; i1 g are two valid bases. The left    boundary of B1 is the directed path s; f1 ;f5 ; f6  and the right boundary of B1 is the directed path        s; f2 ; f7 ;f6 . The left boundary of B2 is the directed path s;f3 ; f8 ; f9 ;f10  and the right boundary   of B2 is the directed path s; f4 ;f10 .

0 will have value 1.

lemma holds. Now we consider an external separating region R that is not in the internal region of any valid base of C . By Lemma 3.9, all incoming wires to R are either wires with value 1 or pseudo wires. Since R does not contain an input node with value 0, all the gates in R will have value 1. By Corollary 3.9.1, the problem of evaluating the one-input-face PMC C is now reduced to the problem of evaluating each subcircuit of Caug in an external separating region that contains input nodes with value 0, since the gates in other separating regions are known to have value 1. Our next step is to transform these subcircuits into one-input-face PMCs so that we can evaluate these subcircuits recursively. One nontrivial problem with a subcircuit of Caug in a separating region is that the output gates of the subcircuit may interlace with the input nodes of the subcircuit, which makes it impossible to add a super sink to the subcircuit without violating the planarity property. For example, in Figure 5, after removing the gates in the internal region of B1 and the internal region of B2 , g1 will be a new output gate and g2 will be a new input gate, and the input nodes and the output gates i1 ; g1; g2; g3; g4 are interlaced with each other in the resulting subcircuit. The following de nition and procedure give a method we will apply to solve this problem. This construction uses pseudo wires de ned in part 5 of De nition 2.5.

Corollary 3.9.1. If a separating region R of Caug does not contain an input node with value 0, then all the gates of Caug in R have value 1. Proof. If R is in the internal region of a valid base, then by Corollary 3.4.1, the

EFFICIENT PARALLEL PLANAR MONOTONE CIRCUIT VALUE

15

Procedure 1: Subcircuit transformation Input: CR , the subcircuit of Caug in an external separating region R containing at least one input node with value 0. 0 Output: CR ,0 a one-input-face PMC logically equivalent to CR . 0. Initialize CR to CR ; 1. for each wire wi in Caug incoming to CR in parallel do

let g be the gate in CR receiving wi as an input; ffnote that g must be a two-input AND gate and the sink of a facegg 0 a if the other input wire of g is in R then remove wi from CR ; b else i.e., the other input wire of g is a wire incoming to CR  0 make g a new input node with value 1 in CR ; end fifg; end fforg; 2. for each wire wo in Caug outgoing from CR in parallel do insert a one-input AND gate gwo in wo with gwo lying inside R, and remove the part of wo outgoing from gwo ;   let wo = f1 ;f   be the dual edge on the boundary of R that crosses wo ; assume wo is on the counterclockwise clockwise boundary of the face f in Caug ; let sf be the sink of f in Caug ; let w be the other edge connected to f  on the boundary of R in Asep ; a if w is an outgoing edge of f  then ffw crosses either the left input 0wire wl or the right input wire wr of sf gg attach an output wire to gwo in CR by adding a pseudo wire as follows: 0 0 ffso that gwo would not be an output gate in CR , and CR is a plane graphgg i if w crosses wl wr  then ffsf must be a two-input AND gategg 0 connect gwo and sf through a pseudo wire to replace wl wr  in CR ; end fifg; ii if w crosses wr wl  then make gwr i.e., the one-input AND gate inserted in wr at at the beginning of step 2 a two-input AND gate, 0 and connect gwo to gwr through a pseudo wire in CR ; end fifg; end fifg; b if w is an incoming edge of f  fff  is called a bottom of R in this casegg 0 then make gwo a new output gate in CR ; end fifg; end fforg;

end.

The algorithm given in Procedure 1 transforms a subcircuit of Caug in an external separating region that contains at least one input node with value 0 into a logically equivalent one-input-face PMC. Some examples of this transformation are given in 0 Figure 7. We now show that CR constructed by Procedure 1 is a one-input-face PMC that is logically equivalent to CR . 0 Lemma 3.10. CR is logically equivalent to CR . Proof. We rst show that step 1 in Procedure 1 does not change the value of the 0 gates in CR that were originally in CR . Since CR is in the external region of every valid base, by De nition 3.5, a wire outgoing from the internal region of some valid base and incoming to R must be either a pseudo wire or an input wire to a twoinput AND gate whose other input wire is not a pseudo wire; by Lemma 3.9, all the incoming wires to R are either wires with value 1 or pseudo wires. Hence removing a pseudo input wire or an input wire with value 1 to a two-input AND gate g whose other input wire is not a pseudo wire in step 1a will not change the value of g in

Definition 3.14. A circuit C 0 is logically equivalent to a circuit C , if from a partially evaluated C we construct C 0 possibly with additional gates such that for each unevaluated gate g in C , there is a gate in C 0 with the same value as g.

16

V. RAMACHANDRAN AND H. YANG
f1* wo wo* CR wl w*

R f* wr sf g wo Case 2(a)(i)

wo p wr sf C’R

f1* wo

R wo w* wr g wo Case 2(a)(ii) p gwr C’R

wo* f * CR wl sf

f1* wo

R w* w’ gwo wr Case 2(b) sf p wo* f * g w’ C’R

CR

wl

wire

auxiliary edge pseudo wire

left leg

right leg

dual vertex

super source or super sink input node OR gate AND gate

output gate

0 Fig. 7. The circuit transformation of CR to CR .

0 CR ; and the two-input AND gate g in step 1b indeed has value 1. 0 Step 2 in Procedure 1 reduces the number of new output gates in CR by adding 0 pseudo wires. Steps 2aii and 2b do not change any input to the gates of CR that were originally in CR . Step 2ai changes an input to sf by replacing wl wr  with a pseudo wire. However, since wl wr  is an incoming wire to the external separating region R, by the arguments given in the previous paragraph, wl wr  is either a pseudo input wire or an input wire with value 1, and sf is a two-input AND gate, and wr wl  is not a pseudo wire. Hence the value of sf depends only on the value of wr wl , 0 and is the same in both CR and CR . 0 is a plane DAG. Lemma 3.11. CR 0 Proof. It is easy to see that CR is still a plane graph since the pseudo wires introduced in Procedure 1 will not cross any existing wires in CR . 0 Suppose there is a directed cycle in CR . Then we map a gate g on the cycle to a gate in Caug by the following function f : f g = g if g is a gate in Caug ; f g = g2 if g is not a gate in Caug and g is a new gate inserted in wire g1 ; g2 of Caug in step 2 0 of Procedure 1. For each edge g1 ; g2 on the cycle in CR , if g1; g2 is not an edge in Caug , then we add a new edge f g1 ; f g2  to Caug , and call the augmented graph 0 0 Caug . Hence there is a cycle in Caug containing new edges. We now prove that for 0 , there is a directed path in Caug from f g1  to each new edge f g1 ; f g2  in Caug f g2 . We consider the following three cases: i Case 1. Both g1 and g2 are gates in Caug . Then f g1 ; f g2  = g1 ; g2, which is a wire in Caug . 0 ii Case 2. g1 is a newly added gate in CR , but g2 is a gate in Caug . Suppose g1 is inserted in the wire g3 ; g4 of Caug . Then g1 ; g2 is a pseudo wire added in

EFFICIENT PARALLEL PLANAR MONOTONE CIRCUIT VALUE

17

wire

input node OR gate AND gate

output gate

Fig. 8. An example of a PMC with all input nodes in a single face but output gates in di erent faces. This PMC cannot be converted into a one-input-face PMC by adding pseudo wires to the output gates, since any pseudo wire added to an output gate will create a directed cycle in this example.

step 2ai of Procedure 1 and g2 is the sink of the face whose boundary contains g4. Hence there is a directed path from f g1  = g4 to f g2  = g2 in Caug . 0 iii Case 3. g2 is a newly added gate in CR, but g1 is a gate in Caug . Then g2 is inserted in the wire g1 ; f g2 of Caug in step 2 of Procedure 1. Hence there is a directed path from f g1  = g1 to f g2  in Caug . 0 iv Case 4. Both g1 and g2 are newly added gates in CR . Then g1 ; g2 is a pseudo wire added in step 2aii of Procedure 1. Suppose g1 is inserted in the wire g3 ; g4 of Caug , and g2 is inserted in the wire g5 ; g6 of Caug . Then g5; g6 is an outgoing edge from CR , and g6 is the sink of the face whose boundary contains g4. Hence there is a directed path from f g1  = g4 to f g2  = g6 in Caug . Hence there is a directed cycle in Caug , which contradicts the fact that Caug is a 0 DAG. Hence CR is acyclic. At this point, one might wonder if it is the case that any PMC whose input nodes are on the boundary of a single face can be converted to a one-input-face PMC by adding pseudo wires to the output gates. The example in Figure 8 shows that this is not always possible when the output gates are on the boundaries of multiple faces; the 0 construction of CR exploited some special properties of a one-input-face PMC and its separating regions to guarantee that the result is a DAG, and this is not always the case when the input circuit is not a one-input-face PMC. 0 We now show that a subcircuit CR output by Procedure 1 must have all inputs in one face, and all outputs in one face, and no interlacing of inputs and outputs. This 0 will establish that CR is a one-input-face PMC.  is a bottom of a separating region R if it is the head of two edges A dual vertex f which are dual edges of Caug  on the boundary of R. See step 2b of Procedure 1 and case 2b in Figure 7. Lemma 3.12. A separating region R has at most one bottom, and if R has a bottom then R does not contain the super sink t.   Proof. Let f  be a bottom of R and let w1 and w2 be the two edges incoming  . We nd two paths P1 and P2 in Asep such that a P1 goes to f  through    to f  and P2 goes to f  through w2 and b P1 and P2 intersect with each other     w1 only at their starting vertices and their ending vertices. Let R0 be the region whose counterclockwise boundary is P1 and whose clockwise boundary is P2. Then R is inside R0 since R is a face in Asep .

18
P1*

V. RAMACHANDRAN AND H. YANG
* P2 t R * w1 f* w 2* * w1 P’2* R’’ sf (1) p wire auxiliary edge pseudo wire left leg right leg dual vertex sf (2) f* t w 2* P’ * 1 P1* R’ f’* R * P2

R’

super source or super sink input node OR gate AND gate

output gate

Fig. 9. Figures for the proof of Lemma 3.12.

We rst prove that t is not in R0, which implies that t is not in R see 1 in Figure 9. Let sf be the sink of the primal face f . If sf is t then we have proved that   t is not in R0. Otherwise, since the primal edges of w1 and w2 are outgoing from R0 , 0 note that the two input wires of sf sf and its two input wires must be outside of R   cannot be the primal edges of w1 and w2 , since only the dual edges outgoing from f  can cross the input wires of sf . Hence any outgoing edges from f  in Aaug must be outside of R0 since they cross the two input wires of sf . Hence if t were in R0 then Aaug would have contained a directed cycle, since there is a directed path from f  to t in Aaug . A contradiction. We now prove that R has at most one bottom see 2 in Figure 9. Suppose R has 0 0 another 0bottom f 0  . Let w1 and w2 be the two edges incoming to0 f 0  . We0 nd two 0 paths P1 and 0 P2 in Asep 0such that a P10  goes to f 0  through w1 and P2 goes to f 0  through w2 and b P1 and P20  intersect with each other only at their starting vertices and their ending vertices. Let R00 be the region whose counterclockwise boundary is P10  and whose clockwise boundary is P20  . Then by the proof in the previous paragraph, t is neither in R0 nor in R00. But at least one path from f  or f 0  to t will create a directed cycle in Aaug . A contradiction. 0 Corollary 3.12.1. If CR contains a bottom, then CR does not contain an 0 contains at most two newly created output gates, original output gate of Caug , and CR and the two output gates are adjacent to each other on the boundary of a face; if CR 0 does not contain a bottom, then CR does not contain any newly created output gates 0 may contain some original output gates of Caug . CR 0 0 Proof. If CR contains a bottom, then t is not in CR and hence CR does not contain an original output gate of Caug since the auxiliary wires connecting output gates to 0 t do not cross the boundary of R. Further since CR has at most one bottom, at most 0 and they are adjacent to each other on the two new output gates are created in CR boundary of a face see case 2b in Figure 7. If CR does not contain a bottom, then 0 no new output gates are created in CR by the construction in Procedure 1. 0 Lemma 3.13. All newly created input nodes in CR are on the boundary of a single
face. Proof. After removing all the gates not in R and the wires crossing the boundary

EFFICIENT PARALLEL PLANAR MONOTONE CIRCUIT VALUE

19

0 of R, all the input nodes in CR are on the boundary of a single face, which is the 0 . Further, the new faces created by the new pseudo wires added external face of CR in step 2 of Procedure 1 do not contain input nodes on their boundaries. 0 Lemma 3.14. Procedure 1 constructs a one-input face PMC CR that is logically equivalent to CR , and it runs in O1 time using a linear number of processors on an 0 input nodes in CR do not interlace. By Lemmas 3.10, 3.11, 3.13 and Corollary 3.12.1, 0 is a one-input-face PMC that is logically equivalent to CR . CR It is straightforward to see that all steps in Procedure 1 can be implemented in constant time using a linear number of processors. 0 We conclude this subsection by showing that a subcircuit CR output by Procedure 1 contains at most half the number of valid bases in Caug . Definition 3.15. We say two valid bases B and B 0 meet if the right boundary of B and the left boundary of B 0 have a common vertex. The transitive closure of the meet relation partitions the set of the valid bases in Caug into equivalence classes. 0 Lemma 3.15. The number of the valid bases in CR is at most half of the number of valid bases in Caug . 0 Proof. Let g be a newly created input node in CR . We say g is a descendant of a valid base B of Caug if an original input of g is in the internal region of B . This lemma follows from the following two claims. 0 Claim 1. Every newly created input node in CR is a descendant of at least two distinct valid bases of Caug , and the two valid bases are in the same equivalence class of Caug . 0 Claim 2. The newly created input nodes in CR that are descendants of the valid bases 0 in the same equivalence class of Caug , are in the same valid base in CR . By Claim 1, a singleton equivalence class of Caug does not generate a new input 0 node with value 1 in CR . By Claim 2, an equivalence class of Caug containing at least 0 two valid bases generates at most one valid base in CR . Hence the lemma holds. We rst prove Claim 1. Since only the sink of a face can have its input wires crossed by the dual edges in Asep , a newly created input node g must be the sink of a face f in Caug . The two original input wires w1 and w2 of g must cross a dual edge   w1 which is a left leg on the left boundary of a valid base and a dual edge w2 which  and w are is a right leg on the right boundary of a valid base respectively. Since w1 2 outgoing from the same vertex f  , they cannot be on the left and right boundaries of the same valid base since the left and right boundaries end at their rst common vertex after s. Therefore, the input wires of g are outgoing from the internal regions of at least two di erent valid bases the internal regions of several valid bases may overlap. Further, the two valid bases are in the same equivalence class since f  is a common vertex of the left boundary of one valid base and the right boundary of the other valid base. We now prove Claim 2. Since the external separating region R contains at least one input node with value 0, the boundary of R must contain the super source s. Further, s may appear on the boundary of R more than once see Figure 6, s appears twice on the boundary of the external separating region R which consists of the part of the plane between the internal region of B1 and the internal region of B2 . Since multiple appearances of s is possible, if we remove s from the boundary of R, the boundary will be divided into several connected portions, each enclosing a disjoint part of Caug in Figure 6, the two disjoint parts are the internal region of B1 and the internal region of B2 . The valid bases in one part cannot be in the same equivalence

EREW PRAM. Proof. Lemma 3.13 and Corollary 3.12.1 ensure that the output gates and the

20 V. RAMACHANDRAN AND H. YANG Algorithm 1: Complete evaluation of a one-input-face PMC Input: An embedded one-input-face PMC C and a complete input assignment to C . Output: Each gate in C is assigned a value 0 or 1. 1. if all input nodes in C have value 1 2. then assign value 1 to all gates in C ; return; else if all input nodes in C have value 0 3. then assign value 0 to all gates in C ; return; end fifg; end fifg;

4. Augment C to Caug , and construct the auxiliary dual graph Aaug ; 5. Find the edges in Aaug that are on the boundaries of valid bases of Caug see Procedure 2; 6. Construct the separating graph Asep ; 7. Remove the wires in Caug that cross the boundary edges of Asep; 8. Find the undirected connected components in the remaining Caug ; 9. for each connected component CR found in step 8 in parallel do 10. if CR does not contain input nodes with value 0 11. then assign value 1 to all gates in CR ; 0 12. else transform CR to CR using Procedure 1; 0 13. Recursively evaluate CR ; end fifg; end fforg;

end.

Procedure 2: Finding the edges in Aaug that are on the boundaries of valid bases Input: Caug , Aaug , and a complete input assignment to Caug . Output: The edges of Aaug that are on the boundaries of valid bases of Caug are marked.

2.1. Find all the valid bases in Caug , and label them in the order of the sequence in which they appear on the boundary of the input face of Caug ; 2.2. Construct Tl and Tr from Aaug ; 2.3. Compute BASEl f   and BASEr f  for each dual vertex f  in Aaug ; 2.4. Compute JOIN f  = BASEl f  BASEr f   for each dual vertex f  in Aaug ; 2.5. Find the enclosure relation I among the JOIN f  see Procedure 3; 2.6. Compute jTERMlf j and jTERMr f  j for each dual vertex f  in Aaug using Lem. 3.8; 2.7. Compute jBOUNlf j and jBOUNrf  j for each dual vertex f  in Aaug using Lem. 3.6; 2.8. Mark all dual edges f  ;g  in Aaug with jBOUNlf j 0 and jBOUNlf  j jTERM f j, or with jBOUNrf j 0 and jBOUNr f j jTERM f j;

end.

class as a valid base in a di erent part. Let P be a connected portion of the boundary of R after removing S . Let I be the set of all newly created input nodes that are descendants of the valid bases in the equivalence classes enclosed in P . Then the original input wires of the input nodes in I must cross the dual edges in P . Hence 0 the input nodes in I are adjacent on the boundary of a face in CR , and therefore are 0. in the same valid base in CR 3.3. An E cient Algorithm for the One-Input-Face PMCVP. Based on the approach we presented in the previous subsection, we give an e cient EREW PRAM algorithm, called Algorithm 1, for evaluating a one-input-face PMC. The correctness and complexity analysis of Algorithm 1 will be given in Theorem 3.1. All steps in Algorithm 1 are quite straightforward to implement except step 5, which is implemented by Procedure 2. Step 2.5 in Procedure 2 is implemented by Procedure 3, which is similar to a procedure used for the layered PMC in 20 .

Lemma 3.16. Procedure 2 i.e., step 5 in Algorithm 1 correctly nds the edges in Aaug that are on the boundaries of valid bases of Caug , and it runs in Olog n time using a linear number of processors on an EREW PRAM. Proof. The correctness of all steps except step 2.5 of Procedure 2 which imple-

ments step 5 of Algorithm 1, has been proved in Lemmas 3.6 and 3.8. We now show

EFFICIENT PARALLEL PLANAR MONOTONE CIRCUIT VALUE

21

Procedure 3: Finding  enclosure relation I for the JOIN f  the Input: Tl and JOIN f  for each dual vertex f on Tl .  Output: The enclosure forest EF  such that a dual vertex fp is the immediate predecessor of  a vertex f  in EF  i JOIN fp  I JOIN f  . 3.1. for each vertex f  with nonempty JOIN f  and with the length of the longest path from a leaf to f  in Tl being k in parallel do 3.2. Assign two triples x, ,k, f , y, k, f   for each range x;y in JOIN f ; end fforg; 3.3. Sort all triples into nondecreasing order according to the rst two elements in a triple; 3.4. for each triple x, ,k, f , where k  0 in parallel do 3.5. Find its previous triple n0 , k0 , f 0  in the sorted list; 3.6. if k0 0 and f  6= f0  then f 0 is the parent of f  in EF  ; 3.7. else if k0 0 and f 6= f 0  then f 0 is the left sibling of f in EF  ; end fifg; end fifg; end fforg; 3.8. Construct the EF  from the parent and sibling relations; end.
 the correctness of Procedure 3, which implements step 2.5 of Procedure 2. Let f1 and  be two vertices in T  such that the longest paths from a leaf to f1 and from a leaf  f2 l    to f2 are of length k1 and k2 respectively. By Lemma 3.7, f1 is a successor of f2 in  and JOIN f1  JOIN f2  i k1 k2, and for each range x2; y2 of JOIN f2     Tl  , such that x1  x2  y2  y1 in the cyclic there exists a range x1; y1 of JOIN f1    order; JOIN f1  JOIN f2  = i for each range x2; y2 of JOIN f2  and each  , x1  y1 x2  y2 in the cyclic order. Therefore, if range x1; y1 of JOIN f1 n0 ; k0; f 0  and x; ,k; f   are two consecutive triples in the sorted list, we have: 1 if k0 0 and f  6= f 0 then JOIN f   I JOIN f 0  and f 0 must be the immediate successor of f  in the EF ; 2 if k0 0 and f  6= f 0 then f  and f 0 share the common immediate successor in EF . Next, we analyze the time complexity of Procedure 2. In step 2.2, Tl Tr  can be computed using Euler-tour technique as follows. We rst remove s and all right left legs from Aaug . Then the resulting graph A0aug is a tree rooted at t, by the uniqueness of left right legs. We then mark the leaf nodes of A0aug that are the dual vertices of the left right bounding faces of valid bases of C . Finally we apply the Euler-tour technique to nd all the successors of the marked leaf nodes, and the resulting subtree of A0aug is Tl Tr . In step 2.3, since BASEl f   BASEr f   contains valid bases with consecutive labels modulo the total number of bases in the total order of the valid bases, it can be described succinctly by a range l, h where l, h are the numbers of the rst and the last valid bases in BASEl f   BASEr f   respectively. BASEl f   BASEr f   can be computed using Euler-tour technique on Tl Tr  as follows. We rst label each leaf node of Tl Tr  with the label of its corresponding valid base. Then for each vertex f  in Tl Tr , we apply the Euler-tour technique to nd the smallest label and the largest label among the leaf predecessors of f  in Tl Tr , and assign them to l and h respectively. In step 2.4, JOIN f   can be computed from BASEl f   and BASEr f   in constant time and be represented by at most two ranges. Based on the above analysis, we conclude that steps 2.2-2.4 can be implemented in Olog n time using a linear number of processors. Procedure 3 which implements step 2.5 in Procedure 2 can be implemented in Olog n time with a linear number of processors using the parallel merge sort of 2 and Euler-tour technique.

22 V. RAMACHANDRAN AND H. YANG Algorithm 2: Partial evaluation of a one-input-face PMC Input: A one-input-face PMC C and a partial input assignment to C . Output: Each gate in C that can be evaluated is assigned a value 0 or 1.

1. Assign value 1 to all input nodes with unknown value in C and apply Algorithm 1; 2. Let A be the set of the gates assigned value 0 in this solution of step 1; 3. Assign value 0 to all input nodes with unknown value in C and apply Algorithm 1; 4. Let B be the set of the gates assigned value 1 in this solution of step 3; 5. Assign value 0 to all gates in A, assign value 1 to all gates in B , and assign unknown value to the gates of C that are neither in A nor in B ;

end.

It is easy to see that all other steps of Procedure 2 can be implemented in Olog n time using a linear number of processors by computing pre x sums and applying Euler-tour and tree evaluation techniques.
Theorem 3.1. Algorithm 1 correctly solves the complete evaluation problem of a one-input-face PMC given a complete input assignment, and it runs in Olog2 n time using n processors on an EREW PRAM, where n is the size of the circuit. Proof. Steps 1-4 are quite straightforward. The correctness of step 5 is proved

by Lemma 3.16. The correctness of steps 6-13 are proved by Corollary 3.9.1 and Lemma 3.14. It is straightforward to see that all steps except steps 5, 8, and 13 in Algorithm 1 can be implemented in Olog n time using a linear number of processors. Lemma 3.16 shows that step 5 can be implemented in the same time complexity. Step 8 can be implemented in Olog n time optimally by applying the algorithm in 5 for nding connected components in a planar undirected graph. By Lemma 3.15, the number of the recursive levels needed to complete the evaluation is Olog n. Therefore, the overall time needed by Algorithm 1 is bounded by Olog2 n. Further, the total number of gates in all remaining subcircuits CR in step 12 in Algorithm 1 is less than the number of gates in the original Caug since for each newly inserted gate in CR , there is a unique gate in the internal region of a valid base being removed. Therefore, the processor bound holds. 3.4. Partial Evaluation of a One-Input-Face PMC. We extend Algorithm 1 to solve the partial evaluation problem of a one-input-face PMC in Algorithm 2.
Theorem 3.2. Algorithm 2 correctly solves the partial evaluation problem of a one-input-face PMC given a partial input assignment, and it runs in Olog2 n time using n processors on an EREW PRAM, where n is the size of the circuit. Proof. By the monotonicity of the circuit, A is a subset of the gates that should

be evaluated to 0 in the partial evaluation of C , and B is a subset of the gates that should be evaluated to 1 in the partial evaluation of C . Further, we now show that a gate g of C that is neither in A nor in B should have unknown values in the partial evaluation of C . Suppose not. Let g be a gate that should be evaluated to 0 1 in the partial evaluation of C and let g be in neither A nor B . Then g evaluates to 0 1 under every possible input assignment to the input nodes with unknown values in C . In particular, g has value 0 1 when all input nodes with unknown values are assigned value 1 0, which means g is in A B . A contradiction. Therefore, Algorithm 2 is correct. It is easy to see that the time complexity of Algorithm 2 is the same as that of Algorithm 1 since it is dominated by the two calls on Algorithm 1. 4. The Face Induced PMC. In this section, we consider a face f induced circuit Cf , which is de ned in Section 2. For convenience, we assume that Cf is

EFFICIENT PARALLEL PLANAR MONOTONE CIRCUIT VALUE

23

Algorithm 3: f -partial evaluation of a face f induced circuit Cf Input: A face f induced circuit Cf with an f -partial input assignment. Output: The solution of the f -partial evaluation problem of Cf . 1. if Cf contains only one gate then return the value of the gate end fifg;
2. 3. 4. 5.

Obtain a topological ordering of the gates in Cf ; Let m be the total number of the non-input gates in Cf ; Find g1 such that there are bm=2c non-input gates before g1 in the topological ordering; Partition the gates in Cf into two parts Pl and Ph , such that Pl contains g1 and the gates before g1 in the ordering, and Ph contains the gates after g1 in the ordering, and remove the wires of Cf pointing from gates in Pl to gates in Ph ; 6. for each gate g in Ph in parallel do if all input wires of g are removed then replace g by an input node with unknown value in Ph ; else if g is a two-input gate and only one input wire of g is removed then add an input node i with unknown value and a wire from i to g in Ph ; end fifg; end fifg; end fforg; 7. Find the undirected connected subcircuits in Pl and Ph ; 8. f 0 -partially evaluate every connected subcircuit in Pl and Ph recursively in parallel, where f 0 is the external input face of the subcircuit; ffit will be shown below that each such subcircuit is a face f 0 induced circuit with an f 0 -partial assignmentgg 9. Remove all gates that are assigned 0 or 1 in step 8 in Ph ; 10. Assign the output values of Pl to the input nodes of Ph ; 11. Partially evaluate every connected subcircuit in Ph using Algorithm 2 in parallel; ffit will be shown below that each such subcircuit is a one-input-face PMCsgg

end.

embedded with f being the external face. An f -partial input assignment to Cf is a partial input assignment where only input nodes in f can have unknown values, and the input nodes in faces other than f must have values 0 or 1. The problem of partially evaluating Cf given an f -partial input assignment is called the f -partial evaluation of Cf . Algorithm 3 gives our method to perform an f -partial evaluation of Cf . Algorithm 3 is similar to an algorithm in 3 which rst layers a face induced circuit which squares the size of the circuit and then recursively partitions the circuit at an appropriate layer. Our algorithm performs a more e cient evaluation by working on a face induced circuit directly and partitioning the circuit according to its topological ordering. It then partially evaluates each subcircuit either recursively or using Algorithm 2. Recall that a topological ordering of a digraph is a linear ordering of its vertices such that every edge in the graph points from a lower-numbered vertex to a highernumbered vertex. It is well known that a digraph has a topological ordering i it is a DAG. We now prove the correctness of Algorithm 3 and analyze its complexity. Lemma 4.1. Immediately before step 8, every connected subcircuit in Pl and Ph is a face f 0 induced circuit for some face f 0 , with an f 0 -partial input assignment. Proof. Let us add to Cf a super source s in face f and a super sink t in the output face of Cf for the purpose of the proof. We connect s to each input node in f with an edge, and connect each output gate to t with an edge. The resulting Cf is still a plane graph. Only input nodes in f can have unknown values in each connected subcircuit in Pl , since no new input nodes are created in Pl . We now show that the output gates in Pl are in the same face. By step 5, every directed path from a gate in Ph to t consists only of gates in Ph . Hence the gates in Ph can be coalesced to t and the resulting

24

V. RAMACHANDRAN AND H. YANG

no new input nodes are generated, i.e., no gate with in-degree 1 or 2 in Ph becomes a gate with in-degree 0. Let g be a gate with in-degree at least 1 in Ph just before step 9. If all gates that provide inputs to g have known values, then the value of g should be evaluated in step 8 and g should be removed in step 9. If all gates that provide inputs to g have unknown values, then the in-degree of g is not changed. If one input of a two-input gate g has unknown value and the other has known value, then the in-degree of g is 1 after step 9. Hence no new input nodes are generated in Ph in step 9. By Lemma 4.1, every connected subcircuit in Ph and Pl in step 8 is a face f 0 induced circuit for some input face f 0 with an f 0 partial input assignment. Therefore, immediately before step 11, the only input nodes left in each connected subcircuit in Ph are the input nodes in f 0 that carry unknown value. Hence every connected subcircuit in Ph is a one-input-face PMC. Theorem 4.1. Algorithm 3 correctly solves the f -partial evaluation problem of a face f induced circuit Cf , and it runs in Olog4 n time using n processors on an EREW PRAM, where n is the size of Cf . Proof. The correctness of steps 8 & 11 are shown by Lemma 4.1 and Lemma 4.2. It is straightforward to see that other steps in Algorithm 3 are correct. Step 1 takes constant time. Step 2 can be implemented in Olog3 n time using n processors on an EREW by Theorem 4.1 in Kao & Klein 12 . The connectivity of a plane undirected graph in steps 8 & 11 can be solved in Olog n time using n= log n processors on an EREW by the algorithm in Gazit 5 . Steps 3-6 & 9-10 can be implemented in Olog n time using n= log n processors. Step 11 takes Olog2 n time using n processors by Theorem 3.2. Let n0 be the number of non-input gates in the original Cf . Since the in-degree of each gate in Cf is  2, we have n0 n  3n0. Each of Ph and Pl contains at most dn0=2e non-input gates, and therefore at most 3 dn0 =2e total gates including the new input nodes. Let T n be the time needed for Algorithm 3 to partially evaluate a circuit with n gates. We have T 3n0   T 3 dn0 =2e + Olog3 n: Solving the above recurrence equation, we have T 3n0 = Olog4 n. Hence T n  T 3n0 = Olog4 n. 5. The General PMCVP. In this section, we give in Algorithm 4 our overall algorithm for evaluating a general PMC. This algorithm evaluates a general PMC recursively by decomposing it into smaller PMCs and disjoint face induced subcircuits.

one-input-face PMC. Proof. We show that after removing all gates assigned 0 or 1 in Ph in step 9,

Cf is still a plane graph. The wires outgoing from gates in Pl to gates in Ph are now incoming to t. Hence after we cut the wires outgoing from gates in Pl to t and remove t, the output gates of the connected subcircuits in Pl are in a single face, which we call f1 . Hence every connected subcircuit in Pl is still a face f induced circuit with an f -partial input assignment. Ph is C n Pl plus some new input nodes with unknown values generated in step 6. The output gates in Ph are not changed and hence are still in the same face. The new input nodes with unknown values are in the same face f1, since all gates in Pl can be coalesced to s. If there are original input nodes in f remaining in Ph which are the only input nodes in Cf that possibly carry unknown value then f1 must be identical to f . Hence every connected subcircuit in Ph is still a face f1 induced circuit with an f1 -partial input assignment. Lemma 4.2. Immediately before step 11, every connected subcircuit in Ph is a

EFFICIENT PARALLEL PLANAR MONOTONE CIRCUIT VALUE

25

1. Find the smallest k, 0  k  m, such that every connected subcircuit in C n Reachi1 ; i2 ; : :: ; ik+1  is of size  n=2 see Figure 10; Let P be a connected subcircuit of size n=2 in C n Reachi1 ;i2 ;: : :; ik  when k  1; 2. if k  1 3. then Recursively solve the complete evaluation problem for the connected subcircuits in C n ReachP  and in P n Reachik+1  whose sizes are all  n=2 in parallel see Figure 10 and Lemmas 5.1 & 5.2 for steps 3-8; ffit will be shown that each such subcircuit is a general PMC with a complete input assignmentgg 4. Completely evaluate Inducedik+1  P using Algorithm 3; ffit will be shown that each such subcircuit is a face induced circuit with a complete input assignmentgg ffnow all gates in P and C n ReachP  are completely evaluatedgg 5. Remove P from C , let o1 ;: : : ;om0 be the gates of P with wires outgoing to reachP ; ffo1 ; :: : ;om0 are on the boundary of a single face in reachP gg 6. Completely evaluate Inducedo1 ; : :: ; om0  i.e. ReachP  n P  using Algorithm 3; ffit will be shown that each such subcircuit is a face induced circuit with a complete input assignmentgg 7. else Recursively solve the complete evaluation problem for the connected subcircuits in C n Reachi1  whose sizes are all  n=2 in parallel; 8. Evaluate Inducedi1  using Algorithm 3; end fifg;

Algorithm 4: Complete evaluation of a general PMC Input: A general PMC C with input nodes i1 ; : :: ; im , and a complete input assignment. Output: Each gate in C is assigned a value 0 or 1. 0. if C contains only one gate then return the value of the gate end fifg;

end.

i 1 i 2 i 3 ik i (k+1) P reach( i 1, ... i k )

C

reach(P) i (k+1) reach( i (k+1) ) P

C

Fig. 10. A general PMC C of size n, where P is a connected subcircuit of size n=2 in C n reachi1 ; i2 ; : :: ; ik , but C n reachi1 ;i2 ;: : :; ik+1  does not contain any connected subcircuit of size n=2.

The smaller PMCs are evaluated recursively while each face induced subcircuit is evaluated by Algorithm 3. We then show the correctness and complexity of Algorithm 4 in Lemma 5.1 and Theorem 5.1. A sketch of an algorithm similar to Algorithm 4 is given in 3 .
Lemma 5.1. Each connected subcircuit in steps 3 & 7 is a general PMC with a complete input assignment.

26

V. RAMACHANDRAN AND H. YANG

Proof. Since the gates in ReachP  can be coalesced into a single gate, the output gates in C n ReachP  are in the same face. Similarly the output gates in P and P n Reachik+1  are in the same face. The input nodes in C n ReachP  are original input nodes in C . Since there is no wire in C outgoing from a gate in C n P to P , the input nodes in P n Reachik+1  are also original input nodes in C . Hence each connected subcircuit in C n ReachP  and P n Reachik+1  is a general PMC with a complete input assignment, and can be completely evaluated recursively in step 3. Similar proof holds for step 7.

a subset of the output gates in P . Inducedik+1  P are reachable from the original input ik+1. The other new input nodes in Inducedik+1  P get their value from P n Reachik+1 , which is completely evaluated in step 3. Hence Inducedik+1 P is a face f that contains ik+1  induced circuit with a complete input assignment, and can be completely evaluated using Algorithm 3 in step 4. The output gates in Inducedo1 ; : : :; om0  i.e. ReachP  n P  are the output gates in ReachP , and the output gates in ReachP  are a subset of the output gates in C and are in the same face. The input nodes o1 ; : : :; om0 in ReachP  n P are the output gates in P , and are in the same face, which we call f1 , and are completely evaluated in steps 3 & 4. All gates in ReachP  n P are reachable from the input nodes o1 ; : : :; om0 in f1 . The other input gates in ReachP  n P get values from gates in C n ReachP , which is completely evaluated in step 3. Hence Inducedo1; : : :; om0  i.e. ReachP  n P  is a face f1 induced circuit with a complete input assignment, and can be completely evaluated using Algorithm 3 in step 6. Similar proof holds for step 8. Theorem 5.1. Algorithm 4 correctly solves the PMCVP for a general PMC C , and it runs in Olog6 n time using n processors on an EREW PRAM, where n is the The reachability in steps 1, 3, and 7 can be implemented in Olog4 n time using n processors on an EREW by the multiple-source reachability algorithm for planar digraphs in Guattery & Miller 10 . The k in step 1 can be found by a binary search. Hence the total time needed in step 1 is Olog5 n. The connectivity of a plane undirected graph in steps 1, 3, and 7 can be solved in Olog n time using n processors on an EREW by the algorithm in Gazit 5 . By Theorem 4.1, steps 4-6 & 8 can be implemented in Olog4 n time using n processors on an EREW. It is easy to see that the connected subcircuits in steps 3 & 7 are of size  n=2, and the subcircuits obtained in each step are disjoint. Let T n be the time needed for Algorithm 4 to evaluate a PMC with n gates. We have
size of the circuit. Proof. The correctness of Algorithm 4 has been shown in Lemmas 5.1 & 5.2.

Lemma 5.2. Each connected subcircuit in steps 4, 6 & 8 is a face induced circuit with a complete input assignment. Proof. The output gates in Inducedik+1  P are in the same face since they are

T n = T n=2 + Olog5 n: Solving the above recurrence equation, we have T n = Olog6 n. Note that the high power in the logarithm for the running time is mainly due to the running time of the reachability algorithms in 10 and 12 . An improvement in the running time of the parallel algorithms for reachability in a plane DAG would imply an improvement in the running time of our algorithm.

EFFICIENT PARALLEL PLANAR MONOTONE CIRCUIT VALUE REFERENCES

27

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