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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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D.J. Kavvadias, E.C. Stavropoulos / Information Processing Letters 85 (2003) 1–6
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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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D.J. Kavvadias, E.C. Stavropoulos / Information Processing Letters 85 (2003) 1–6
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
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D.J. Kavvadias, E.C. Stavropoulos / Information Processing Letters 85 (2003) 1–6
tree T generated by Algorithm A of [6], they observe
that every node (subproblem) α in T is uniquely de-
termined by a path (in other words, by a sequence of
left and right moves) from the root (initial problem)
to α. They prove that this sequence is obtainable in
polynomial time plus O(log3n) suitably guessed bits
and thus place MBD in co-NP[log3n]. The more re-
stricted bound of O(log2n) nondeterministic guesses
was given by Eiter et al. after a more involved proof
applied on Algorithm B of [6].
4. Conclusions
In this work we presented a nondeterministic algo-
rithm for checking the duality of a pair of monotone
expressions in disjunctive normal form. The algo-
rithm utilizes the decomposition rules given by Fred-
man and Khachiyan in [6] as well as the appear-
ance of variables with appropriate frequency. Our al-
gorithm requires deterministic polynomial time plus
O(log2n) nondeterministic steps. This result places
the monotone Boolean dualization in co-NP[log2n],
the subclass of co-NP where only the first log2n steps
are nondeterministic. It also makes straightforward
the quasi-polynomial running time of the determinis-
tic version, avoiding the rather complicated analysis
presented in [6]. Future work includes further investi-
gation on the exact time complexity of the monotone
Boolean dualization, a long unresolved question.
Acknowledgements
Research by ECS was supported by the University
of Patras Research Committee (Project Caratheodory
under contract no. 1939). The authors wish to thank
the correspondingeditorforpointingoutreference[4].
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