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Information Processing Letters 85 (2003) 1–6

www.elsevier.com/locate/ipl

Monotone Boolean dualization is in co-NP[log2n]

Dimitris J. Kavvadiasa, Elias C. Stavropoulosb,∗

aUniversity of Patras, Department of Mathematics, GR-265 00 Patras, Greece

bUniversity of Patras, Computer Engineering & Informatics Department, GR-265 00 Patras, Greece

Received 6 March 2002; received in revised form 5 June 2002

Communicated by H. Ganzinger

Abstract

In 1996, Fredman and Khachiyan [J. Algorithms 21 (1996) 618–628] presented a remarkable algorithm for the problem of

checking the duality of a pair of monotone Boolean expressions in disjunctive normal form. Their algorithm runs in no(logn)

time, thus giving evidence that the problem lies in an intermediate class between P and co-NP. In this paper we show that

a modified version of their algorithm requires deterministic polynomial time plus O(log2n) nondeterministic guesses, thus

placing the problem inthe class co-NP[log2n]. Our nondeterministic version has also the advantage of having a simpler analysis

than the deterministic one.

2002 Elsevier Science B.V. All rights reserved.

Keywords: Algorithms; Computational complexity; Monotone DNF duality; Transversal hypergraph

1. Introduction

Let f(x) = f(x1,...,xN) and g(x) = g(x1,...,

xN) be a pair of monotone Boolean expressions given

by their irredundant disjunctive normal forms

f =

?

I∈F

?

i∈I

xi

and

g =

?

J∈G

?

j∈J

xj,

where F and G are the sets of prime implicants I,J ⊆

{1,...,N} of f and g, respectively. The problem of

interest here is defined as follows:

MONOTONE BOOLEAN DUALITY (or, MBD). Given

a pair of monotone Boolean expressions f and g

*Corresponding author.

E-mail addresses: djk@math.upatras.gr (D.J. Kavvadias),

estavrop@ceid.upatras.gr (E.C. Stavropoulos).

in their irredundant disjunctive normal forms, decide

whether f,g are mutually dual, i.e.,

f(x1,...,xN) = g(x1,...,xN)

for all x = (x1,...,xN) ∈ {0,1}N.

If f and g are not mutually dual, then there is a

vector x ∈ {0,1}Nsuch that, f(x1,...,xN) = g(x1,

...,xN). Obviously,suchadisqualifiercanbeguessed

and verified in time that is polynomial to the size

n = |F|+|G| of f and g. Thus, MBD is easily placed

in the class co-NP. However, its exact computational

complexity is still unknown. The most notable result

was given in 1996 by Fredman and Khachiyan [6].

In this work, the authors presented an algorithm that

checks the duality of a pair of monotone DNFs in

quasi-polynomial time no(logn). This result gives evi-

dencethat MBD lies in an intermediateclass betweenP

and co-NP.

0020-0190/02/$ – see front matter 2002 Elsevier Science B.V. All rights reserved.

PII: S0020-0190(02)00346-0

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D.J. Kavvadias, E.C. Stavropoulos / Information Processing Letters 85 (2003) 1–6

The algorithm of Fredman and Khachiyan can also

be used for enumerating the prime implicants of the

dual expression of a monotone DNF in incremental

output-subexponential time [9]. Since the size of the

dual expression may be exponentially larger than the

input one, more elaborate complexitymeasures for the

efficiency of algorithms for problems like MBD (i.e.,

with large output) must be defined, that will take into

accountnotonlythe size oftheinputbut thesize ofthe

output as well. The reader is referred to see [10,14]

for discussions of algorithms with large output and

performance criteria.

Generating the prime implicants of a monotone

DNF is equivalent to the generation of the minimal

transversals of a simple hypergraph [2,6,12] and to

the generation of the maximal models of a Boolean

expression in conjunctive normal form [11]. These

problems are central in various fields of Computer

Science (see [5,6,8] for an exposition of applications

of these problems).

The algorithm of Fredman and Khachiyan gives an

upper bound for the time complexity of MBD and im-

plies that the problem can not be co-NP-hard, unless

any co-NP-complete problem can be solved in quasi-

polynomial time. In this paper we present a nonde-

terministic version of the algorithm of Fredman and

Khachiyan. Our version uses the decomposition rules

of [6] (see next section) in a novel way and solves

the problem in deterministic polynomial time plus

O(log2n) nondeterministic steps. Having the above

time bounds, it is subsequently straightforward to ob-

tain the no(logn)deterministic time bound of [6], thus

avoiding the rather complicated analysis presented

there. Hence, we place the MONOTONE BOOLEAN

DUALITY problem in the class co-NP[log2n], the sub-

class of co-NP where only the first log2n steps are

nondeterministic. This is the complement of the class

NP[log2n], denoted β2P in [13]. Such subclasses of

NP can be defined by restricting the number of non-

deterministic steps of the computation (see [13,3]).

For a survey on limited nondeterminism, see [7]. The

same complexity result was also given independently

in [4]. Our approach differs from the one in [4] and

its analysis is much simpler. Moreover, as it is men-

tioned in [4], the same result may also be obtained by

appropriately applying Beigel and Fu’s Theorem 11

in [1].

The rest of the paper is organized as follows: In

Section 2 we present the necessary duality properties

and lemmas and shortly describe the algorithm of

Fredman and Khachiyan. In the next section we

present our nondeterministic version and give its time

complexity.Finally,inSection4someconclusionsand

directions for further research are given.

2. Overview of the Fredman and Khachiyan

algorithm

For the sake of completeness, in this section we

briefly describe the main steps of the Fredman and

Khachiyan algorithm. The reader is referred to [6]

for more details. Terminology and notation are also

borrowed from there.

Suppose that the monotone DNFs f and g are

mutually dual. Then, the following conditions hold:

I ∩J ?= ∅,

?

max?|I|: I ∈ F?? |G|,

max?|J|: J ∈ G?? |F|.

Moreover (cf. [6, Lemma 1]),

for any I ∈ F and J ∈ G,

{I: I ∈ F} =

(1)

?

{J: J ∈ G},

(2)

(3)

E =

?

I∈F

2−|I|+

?

J∈G

2−|J|? 1.

(4)

Ifanyofthese necessarydualityconditionsis violated,

then f and g are not mutually dual and a succinct

disqualifier can be found in polynomial time.

Let xi∈ {x1,...,xN}. The frequency εf

is defined as the fraction of implicants of F that xi

occurs in, i.e.,

iof xiin f

εf

i=|{I ∈ F: i ∈ I}|

Let ε ∈ (0,1]. We say that xi occurs in f with

frequency at least ε if εf

states an interesting property that holds for mutually

dual expressions (cf. [6, Lemma 2]):

|F|

.

i? ε. The following lemma

Lemma 1. Let f,g be a pair of mutually dual forms

with |F||G| ? 1. Then, there exists a variable that

occurseitherin f or g withfrequencyatleast 1/logn,

where n = |F|+|G| is the size of f and g.

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Let xi∈ {x1,...,xN}. Then, the expressions f and

g can be written as

f = xif0(y)∨f1(y)

where y = {x1,...,xi−1,xi+1,...,xN} and f0, f1,

g0, and g1are the monotone irredundant DNFs with

implicant sets

and

g = xig0(y)∨g1(y),

F0=?I \{i} | i ∈ I, I ∈ F?,

F1= {I | i / ∈ I, I ∈ F},

G0=?J \{i}| i ∈ J, J ∈ G?,

G1= {J | i / ∈ J, J ∈ G},

respectively. It was shown in [6] that f,g are mutually

dual if and only if

f1is dual to g0∨g1

Hence, the initial problem (f,g) of size n is reduced

to subproblems

and

g1is dual to f0∨f1. (5)

(f1,g0∨g1)

(g1,f0∨f1)

of smaller sizes, where xiacts like a splitting variable.

If both pairs are mutually dual, then so is the initial

one; otherwise a succinct disqualifier can be found.

Fredmanand Khachiyanpresentedan algorithm[6,

Algorithm A] that utilizes Lemma 1 and recursively

applies decomposition rule (5) to solve MBD in time

nO(log2n)for any pair of monotone disjunctive normal

forms f and g of size at most n [6, Lemma 4].

As they next show, this running time can be further

improvedif onenotices that subproblems(f1,g0∨g1)

and (g1,f0∨ f1) are not independent. Assuming,

for example, that subproblem (6) is already solved

(and, f1 is dual to g0∨ g1), then the solvability of

subproblem (7) is equivalent to the solvability of a

system of |G0| equations

g1

?y[J]?= f0

where G0is the set of prime implicants of g0, J ∈ G0,

andy[J]is thevectorobtainedbyy bythesubstitution

yj = 1 for all j ∈ J. However, each of the |G0|

Eq. (8) is equivalent to MBD for the pair of forms

(gJ

1,fJ

yj = 1,j ∈ J, and fJ

setting yj= 0,j ∈ J. Thus, the initial problem (f,g)

and(6)

(7)

?y[J]?,

(8)

0) where gJ

1is obtained from g1(y) by setting

is obtained from f0(y) by

0

has been decomposed into |G0| + 1 subproblems in

total.

A symmetric decomposition holds if one assumes

that subproblem (7) is already solved. The initial

problem (f,g) can now be decomposed into |F0| + 1

subproblems in total. We next give the decomposition

rules, as presented in [6]:

(i) Let f,g be a pair of monotone DNFs of vol-

ume v = |F||G| and let a variable xi occur in f

with frequency εf

i. Then, in polynomial time the

MBD problem for f = xif0(y) ∨ f1(y) and g =

xig0(y)∨g1(y) can be decomposedinto subprob-

lem (6) of volume |F1||G| ? (1 − εf

(1−εf

ume at most |F0||G1| = εf

i)|F||G| =

1,fJ

iv each.

i)v, plus |G0| subproblems(gJ

0) ofvol-

i|F||G| ? εf

The symmetric decompositionrule for g is as follows:

(ii) If a variable xi occurs in g with frequency εg

then in polynomial time the MBD problem for

(f,g) can be decomposed into subproblem (7) of

volume at most (1− εg

(fI

1,gI

i,

i)v, plus |F0| subproblems

iv each.

0) of volume at most εg

Finally, property (5) implies that

(iii) The MBD problem for (f,g) can be decomposed

into subproblems (6) and (7) of volumes (1 −

εf

i)v, respectively.

i)v and (1−εg

The volume v = |F||G| of f and g is an appropriate

measure for the size of the input adopted in [6] and fa-

cilitates the analysis of the algorithms. We must note

here that any variable with positive frequency may be

used in the above rules. This is a point where our ver-

sion of the algorithm differs from the the algorithm of

FredmanandKhachiyan.AlgorithmBin[6]solvesthe

MBD problem by recursively incorporating the above

decompositionrules.After a rathercomplicatedanaly-

sis, Fredman and Khachiyan show that Algorithm B

solves the MBD problem in quasi-polynomial time

no(logn):

Theorem 2 [6, Theorem 1]. The MONOTONE BOOL-

EAN DUALITY problem can be solved in n4χ(n)+O(1)

time, where χ(n)χ(n)= n.

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D.J. Kavvadias, E.C. Stavropoulos / Information Processing Letters 85 (2003) 1–6

3. The nondeterministic algorithm

In this section we present a nondeterministic al-

gorithm for checking the dualization of a pair of

monotone expressions in disjunctive normal forms.

Thedeterministicalgorithmof[6]appliesappropri-

ately the above decomposition rules, until every sub-

problem that is produced has constant size in which

case it can be solved in constant time. In contrast,

our nondeterministicversion proceedsin phases. Each

phase starts with a single subproblemon which and on

all of its descendant subproblems we apply the same

decomposition rules (in a manner explained below)

until all produced subproblems have size at most half

the size of the initial problem. At this point the duality

of the initial problem is equivalent to the duality of all

produced subproblems. We next nondeterministically

select one of the produced subproblems and check its

duality in a new phase. The above are repeated until

the selected subproblem is reduced to constant size.

We next present the way the decomposition rules

are used in each phase:

The Deterministic Phase

Input: a pair of monotone DNFs f and g satisfying

the necessary duality condition (1).

Output: a family P of subproblems (i.e., pairs of

DNFs) each of size at most half the size of the

input such that the input pair is dual if and only

if all pairs in the family are dual.

(1) Delete all redundantimplicants from F and G and

set n = |F|+|G| and v = |F||G|.

(2) Check conditions (2), (3), and (4). If any of these

conditions is violated, then f,g are not mutually

dual and a succinct disqualifier can be found in

polynomial time.

(3) If min{|F|,|G|} ? 2, the MBD problem can be

solved in polynomial time.

(4) Select a variable xisuch that

εf

or

εg

i?1/logn

Comment:Suchavariablecanbefoundinpolyno-

mialtimeanditsexistencefollowsfromLemma1.

(5) If εf

2, apply decomposition rule (i) to obtain

one subproblemof volume (1−εf

v subproblems of volume at most εf

the produced subproblems to P.

i? 1/logn.

i?1

i)v plus at most

iv each. Add

If εf

obtainonesubproblemofvolume(1−εg

most v subproblems of volume at most εg

Add the produced subproblems to P.

If εf

(iii) to obtain subproblems (6) and (7) of volumes

(1 − εf

produced subproblems to P.

(6) Apply steps 1–5 to every subproblem of volume

greater than1

2v. If such problem does not exist

(i.e., the phase has ended), return P.

i>1

2? εg

i, apply decomposition rule (ii) to

i)v plusat

iv each.

i>1

2and εg

i>1

2, apply decomposition rule

i)v and (1 − εg

i)v, respectively. Add the

We next proceed in upper bounding the number of

subproblems produced by a phase as a function of the

size of the initial problem.

Lemma 3. The number of the subproblems produced

by a Deterministic Phase is O(nlogn), where n is the

size of the initial problem.

Proof. Let (f,g) be the initial pair of expressions of

volume v. Firstly assume that variable xi occurs in

f,g with frequencies εf

both greater than1

2. Then, decomposition rule (iii) is

appliedandtheinitialproblemis decomposedintotwo

“small” subproblems, one of volume (1 − εf

and another one of volume (1 − εg

decomposition rule (iii) indicates that the current

subproblem will produce no subproblem of volume

greater that1

2v.

Suppose now that the variable xioccurs in f with

frequency

iand εg

i, respectively, that are

i)v ?1

2v. Hence,

2v

i)v ?1

1

logn? εf

i?1

2.

According to decomposition rule (i), the initial prob-

lem is decomposed into one subproblem of volume

v?= (1 − εf

ume at most εf

(1−1/logn)v. Moreover,since εf

the remaining subproblems is at most1

ter decomposition rule (i) is applied, only one “large”

(i.e., of volume more than

duced that needs further decomposition,while the rest

of the subproblems are “small” ones and are not de-

composed further until the end of the phase. It is

i)v and at most v subproblems of vol-

iv each. Since εf

i? 1/logn, v??

i?1

2v. Hence, af-

2,thevolumeof

1

2v) subproblem is pro-

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5

easy to see that the same holds for decomposition

rule (ii).

We observe therefore that in all cases at most one

subproblem of volume greater than

from a problem of size v. If such a problem is

produced, we next apply decomposition rule (i) or

(ii) to this unique subproblem which has size n?< n

and volume v?? (1 − 1/logn)v for some variable

xj of frequency, say, εj. After decomposition rule is

applied, at most 1+v subproblems are produced, one

of volume v??= (1 − εj)v?and at most v?? v of

volume at most εjv?each. See that εjv?<1

1

2v may result

2v, while

v??= (1−εj)v??

?

1−

1

logn?

1

logn

??

?2

1−

1

logn

?

v

?

?

1−

v.

The current phase lasts until the volume of every

subproblem produced is at most

of decomposition steps required for the phase to

complete is O(k) where k is the solution of the

equation:

1

2v. The number

?

1−

1

logn

−ln2

ln(1−

?k

v =1

2v ⇒

k =

1

logn)=

−ln2logn

ln(1−

1

logn)logn.

When n tends to infinity, k = O(logn). Thus, the

number of decomposition steps required for the phase

to complete is O(logn) while O(n) subproblems are

produced at each step. Consequently, at the end of the

phase there will be O(nlogn) subproblems, each of

them of volume at most1

2v.

✷

The idea is now to call a nondeterministic oracle

to guess the next subproblem for the algorithm to pro-

ceed (and the next phase to start). This nondetermin-

istic guessing is used only at the end of each phase.

Since at the end of the phase there are O(nlogn)

subproblems in total, the number of nondeterministic

guesses required is O(log(nlogn)) = O(logn).

Observe now that the number of phases is

O(logv) = O(logn) since each phase and the subse-

quent nondeterministic guessing result in a problem

at most half the size of the initial one. Thus, the total

number of nondeterministic guesses is O(log2n). We

next summarize the whole algorithm:

The Nondeterministic Algorithm

Input: a pair of monotone DNFs f and g satisfying

the necessary duality condition (1).

(1) Nondeterministically guess O(log2n) bits and

store them.

(2) Apply the Deterministic Phase for the current

problem (fc,gc) and let P be the family of the

produced subproblems.

Comment:Atthefirstrunofthealgorithm,(fc,gc)

= (f,g).

(3) Utilize the first log|P| bits stored in step 1 to

identify a subproblem in P and make it the next

current problem (fc,gc). Delete these bits from

the stored sequence of bits.

(4) Go to step 2.

We have thus proved the following theorem:

Theorem 4. The Nondeterministic Algorithm solves

MONOTONE BOOLEAN DUALITY in polynomial time

plus O(log2n) nondeterministic guesses, where n is

the size of the input.

Proof. Follows from the above discussion.

✷

Thus, O(log2n) nondeterministic guesses suffices

to find a succinct disqualifier and prove that the input

pair of monotone DNF expressions are not mutually

dual. This result places MBD to the subclass of co-NP,

the class co-NP[log2n] where only the first log2n are

nondeterministic.

Theorem 5. MONOTONE BOOLEAN DUALITY is in

co-NP[log2n].

Proof. Follows from the definition of co-NP[log2n]

and Theorem 4.

✷

The same result was independently given by Eiter

et al. in [4]. Their work is also based on the algorithms

of Fredman and Khachiyan, as our. The main differ-

ence, however, is that Eiter et al. use the algorithms

presented in [6] without any modifications while our

work uses the supporting theory and mainly the de-

composition rules (in the way presented above) in or-

der to obtain a simpler proof for the O(log2n) nonde-

terministic bound. Eiter et al. work on the recursion