Page 1
A Combinatorial Characterization
of Resolution Width
Albert Atserias
?
Universitat Polit` ecnica de Catalunya
Barcelona, Spain
V´ ıctor Dalmau
Universitat Pompeu Fabra
Barcelona, Spain
?
September 2, 2005
Abstract
We providea characterizationof the resolutionwidth introducedin the contextof propositionalproof
complexity in terms of the existential pebble game introduced in the context of finite model theory. The
characterizationis tight and purely combinatorial. Our first application of this result is a surprising proof
that the minimum space of refuting a 3CNF formula is always bounded from below by the minimum
width of refuting it (minus 3). This solves a wellknown open problem. The second application is the
unification of several width lower bound arguments, and a new width lower bound for the dense linear
order principle. Since we also show that this principle has resolution refutations of polynomial size,
this provides yet another example showing that the relationship between size and width cannot be made
subpolynomial.
1 Introduction
Resolution is one of the most popular proof systems for propositional logic. Since Haken [18] proved an
exponential lower bound for the smallest resolution proofs of the pigeonhole principle, its strength has been
studied in depth. The focus has been put in two related directions: (1) proving strong lower bounds for
interesting tautologies arising from combinatorial principles [29, 11, 7, 9, 4, 24, 26], and (2) the study of the
complexity of finding resolution proofs [7, 9, 3, 6]. This research is still ongoing, and it seems that further
study in both directions is necessary for a better understanding of the power of resolution.
An important step towards the understanding of the strength of resolution in a unified way was made
by BenSasson and Wigderson [9] with the introduction of the width measure. The width of a resolution
refutation is the size of the largest clause in the refutation. The main result of BenSasson and Wigderson,
building upon the workof Clegg, Edmonds and Impagliazzo [12]andBeameand Pitassi [7], isthefollowing:
if a 3CNF formula with
of width
prove size lower bounds. Indeed, if the minimal width of refuting
of requires size
previously known lower bounds for resolution in an elegant and unified way, but also showed that resolution
? variables has a resolution refutation of size
? , then it has a resolution refutation
???????????????? . This interesting result relates the size with the width in a form that is suitable to
?
is
? , then every resolution refutation
??????????????! . Equipped with this result, BenSasson and Wigderson not only rederived all
"Supported in part by CICYT TIN200404343 and by the European Commission through the RTN COMBSTRU HPRN
CT200200278.
Research partially supported by the MEC under the program ”Ramon y Cajal”, grants TIC 200204470C03 and TIC 2002
04019C03, the EU PASCAL Network of Excellence, IST2002506778, and the MODNET Marie Curie Research Training Net
work, MRTNCT2004512234.
#
1
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is automatizable in subexponential time by an extremely simple dynamic programming algorithm. We
should notice, however, that the sizewidth relationship of BenSasson and Wigderson has shown insufficient
to prove size lower bounds for some interesting cases such as the weak pigeonhole principle. In fact, Bonet
and Galesi [10] proved that the sizewidth tradeoff is essentially tight, and therefore the technique cannot
be applied to it. Indeed, the 3CNF encoding of the weak pigeonhole principle with
has
lower bound
pigeonhole principle was finally solved by Raz [24] (see also [25]) using a completely different technique.
Our goal in this paper is to establish a tight connection between the resolution width of BenSasson and
Wigderson, and the existential
of finite model theory. Research in this direction was initiated by Atserias [5] in the study of the descriptive
complexity of properties that certify unsatisfiability of random CNF formulas.
EhrenfeuchtFra¨ ıss´ e games is the generic name for the combinatorial twoplayer games that characterize
expressibility in several logics, including firstorder logic, secondorder logic, and fragments of infinitary
logic (see [14]). Among these, we encounter the existential
[19, 20] to analyze the expressive power of Datalog, a wellknown query language in database theory. The
game is played between two players, the Spoiler and the Duplicator, on two relational structures
over the same vocabulary. Each player has a set of
game, the Spoiler can make one of two different moves: either he places a free pebble over an element
of the domain of
the Duplicator must respond by placing her corresponding pebble over an element of
corresponding pebble from
respectively. If the Spoiler reaches a round in which the set of pairs of pebbled
elements is not a partial homomorphism between
and
the Duplicator wins the game.
The crucial fact that relates pebble games to resolution width is the observation, first pointed out by
Feder and Vardi [16], that the satisfiability problem of an
momorphism problem on relational structures: given two finite relational structures
vocabulary, is there a homomorphism fromto
and the clauses of represents the truthvalues
valid assignments for the clauses, and the homomorphisms from
variables to truthvalues satisfying all the clauses of
of resolution width and pebble games are intimately related. More specifically, we prove that
lution refutation of width
Thus, existential
The new characterization allows us to rederive, in a uniform way, essentially all known width lower
bounds. Moreover, an increase of insight generally reverts in the acquisition of new results. This case is
no exception. Our first application of the combinatorial characterization is a surprising result relating the
space and the width in resolution. The space measure was introduced by Esteban and Tor´ an [15] (see also
[1]). Intuitively, the minimal resolution space of refuting a CNF formula
are required to be kept in a blackboard (memory) if we insist that the refutation must be selfcontained.
This measure is referred to as the clause space by Alekhnovich, BenSasson, Razborov and Wigderson [1].
Strong space lower bounds were proved in the literature for wellknown tautologies such as the pigeonhole
principle [28, 1], Tseitin tautologies [28, 1], graph tautologies [1], and random formulas [8] to cite some.
Our new result is that the minimum space of refuting an
width of refuting minus
?
?
pigeons and
?
holes
???
????
?
? variables, but the best width lower bound that can be proved is
???
?
???
? . Note that the
?!????
?
???? from the sizewidth tradeoff is just trivial in this case. The problem about the weak
? pebble game, first introduced by Kolaitis and Vardi [19, 20] in the context
? pebble game, introduced by Kolaitis and Vardi
?
and
?
? pebbles numbered
????????????????? . In each round of the
? , or he removes a pebble from a pebbled element of
? . To each move of the Spoiler,
? , or removing her
?
?? , then he wins the game. Otherwise, we say that
? CNF formula
?
can be identified with the ho
and
represents the variables
??
over the same
?? ? Informally, the structure
?
? , the structure
?????????? and the combination of them that are
to are precisely the assignments of
??
? . Using this reformulation, we show that the concepts
?
has a reso
and
? if and only if the Spoiler wins the existential
???????
? pebble game on
?? .
? pebble games provide a purely combinatorial characterization of resolution width.
?
is the number of clauses that
? CNF formula is always bigger than the minimum
?
? ?!? . In symbols,
"
???$#
????%?&?'??? . Thus, for
? CNF formulas with small
? , space lower bounds follow at once from width lower bounds. We remark that Tor´ an [28] already obtained
this result for the restricted case of treelike resolution, but the general case remained open since then. Our
2
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result confirms the conjecture of BenSasson and Galesi [8] and answers questions posed by Esteban and
Tor´ an [15], and Alekhnovich, BenSasson, Razborov and Wigderson [1].
The second application of our result is a new width lower bound. We consider the dense linear order
principle
We show that every resolution refutation ofrequires width at least
elements of the linear order. We want to make the point that the significance of this result is not so much the
width lower bound itself, but rather, the techniques that are involved and that we discuss next.
Most of the tautologies studied in the literature, including
in order to get meaningful width lower bounds it is necessary to convert them into equivalent and short
(generally 3CNF) formulas in a preliminary step. Unfortunately, it is usually the case that the resulting
formula looses some of the intuitive appeal of the principle it expresses. Furthermore, in a width lower
bound proof, dealing with the auxiliary variables is usually simple but cumbersome and laborious. To
simplify this situation we define a variant of the pebble game, called extended pebble game, that can be
played directly over formulas with large clauses and that hides all the technical details, such as the process
of dividing large clauses, the introduction of auxiliary variables and its treatment, inside the proof of its
main property. In particular, the width lower bound for the
picture aboutby showing that it has polynomialsize resolution proofs. Thus, the
provides a new example requiring large width but having small resolution proofs (see [10, 3, 6] for further
discussion on this).
?????
?, suggested by Urquhart and used by Riis [27], and stating that no finite linear order is dense.
???
?
?
????? , where
? is the number of
???
?
?,have large initial width. Consequently,
???
?
?
is obtained this way. We complete the
???
?
?
???
?
?
principle
2 Preliminaries
Let
a set of literals. If a clause has at most
clauses. Alternatively, clauses may be viewed as disjunctions of literals, and CNF formulas may be viewed
as conjunctions of clauses. For a clause
assignment tois any function
all literals fromto
we say that undecided. Resolution is a refutation system that works with clauses. The only rule
is the socalled resolution rule:
?
be a set of propositional variables. A literal is a variable or the negation of a variable. A clause is
? literals, we call it an
? clause. An
? CNF formula is a set of
? 
? , let
?????
???
? be the set of variables that appear in it. A partial truth
? ?????????
????????? where
??????? . We say that
if it sets some literal from
? falsifies a clause
to
?
if it sets
?
? . Dually, we say that
? satisfies
??
? . In all other cases
? leaves
?
???
???
?
???
?? !?
?
?????
where
is a sequence of clauses
rule from two previous clauses in the sequence. We say that
fromis the empty clause
the number of literals in it. The width of a derivation is the maximum of the widths of its clauses.
Let
associated arity. An
and
?
and
?
are arbitrary clauses and
? is a variable. A resolution derivation from a set of initial clauses
?
?#"??????????$?&% , each of which is either a clause from
? , or follows by the resolution
is a resolution derivation of
is a refutation of
?'"??????????$?&%?(%
? . If
?
%
?? , we say that
?
"
????????$?
%
? . The width of a clause is
?
??*)+"??????????,)'%?? be a finite relational language, that is, a finite set of relation symbols with an
? structure is a tuple
??
?.
?,)?/
"
?????????,)?/
%
? where

is a set, called the universe,
)
/
0
?1'243 is a
?
0ary relation on
 , where
?
0is the arity of
)
0. Let
to
??
?.
?,)
/
"
?????????,)
/
%
? and
??
?.5
?,)?6
"
????????,)?6
%
? be
? structures. A partial homomorphism from
??
is any function
with domain
?7?89?:?;5 ,
of arity
where
structure

?
?< , such that
? defines an homomorphism from the substructure of
is a function such that for every relation symbol
then
?

? to the
? . In other words,
? )>=??
" and
@
"
????????
@?A
=B
? , if
?
@
"??????????
@CA
?D=?)?/
?E?
?
@
"
?
?????????4?
?
@CA
???F=B)
6 . If
? and
G are partial homomor
phisms, we say that
G extends
? , denoted by
?H?IG , if
J
?LK
?E?
?D?MJ
?LK
?NG
? and
?
?
@
?
?OG??
@
? for every
@
=7J
?LK
?E?
? . If
?P?QG , we also say that
? is the projection of
G to
J
?LK
?E?
? .
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The existential
Each player has a set of
one of two different moves: either he places a free pebble over an element of the domain of
a pebble from a pebbled element of
her corresponding pebble over an element of
If the Spoiler reaches a round in which the set of pairs of pebbled elements is not a partial homomorphism
between
and
of but the two corresponding pebbles are placed over different elements of
define a partial homomorphism). Otherwise, we say that the Duplicator wins the game. The next definition
formalizes this intuitive discussion:
? pebble game on
?
and
?
is played by two players: the Spoiler and the Duplicator.
? pebbles numbered
????????????????? . In each round of the game, the Spoiler can make
? , or he removes
? . To each move of the Spoiler, the Duplicator must respond by placing
? , or removing her corresponding pebble from
?
respectively.
?? , then he wins the game (note that if two different pebbles are placed on the same element
?? , then the set of pairs does not
Definition 1 ([19, 20]) Let
the Duplicator wins the
phisms from
?
be a finite relational language and let
and
?
and
?
be
? structures. We say that
of partial homomor
? pebble game on
such that
??
if there is a nonempty family
?
?
to
?
(i) If
(ii) If
?P=?? , then
?
J
?LK
?E?
?????
? .
?P=??
and
G??? , then
G?=?? .
(iii) If
?P=?? ,
?
J
?LK
?E?
?????
? , and
@
=? , then there is some
G?=??
such that
?P?QG and
@
=7J
?LK
?NG
? .
We say that
?
is a winning strategy for the Duplicator.
Intuitively, each partial homomorphism
us mention that the existential
G?=??
is a winning position for the Duplicator in the game. Let
? pebble game is known to characterize definability in the existential positive
? variable fragment of infinitary logic
formulas over the variables
quantification of a variable in
in this paper, let us state the result that links existential
???
2
??? , that is, the logic that is obtained by closing the set of atomic
?"?????????
?
2under infinitary conjunctions, infinitary disjunctions, and existential
?
"
?????????
?
2. Note that variables may be reused. Although we will not use it
? pebble games and definability in
???
2
??? .
Theorem 1 [19, 20] Let
Duplicator wins the existential
inalso holds in
?
be a finite relational language and let
?
and
?
be
? structures. Then, the
? pebble game on
?
and
?
if and only if every
???
2
??? sentence that holds
?? .
The
? variable fragments of infinitary logics have played a crucial role in the development of finite
model theory (see [17] for a good survey).
3Combinatorial Characterization as Games
It is well known that
? CNF formulas may be encoded as finite relational structures. Indeed, let
?
?
?????????
formula
domain of
set of clauses of
of the form
"
?????????????? be the finite relational language that consists of
over the propositional variables
??? relations of arity
? each. An
? CNF
?
?
"????????????
?
is encoded as an
? structure
?
??? as follows. The
?
??? is the set of variables
with
???C"????????????
?
? . For each
"
=???????????????? , the relation
?
Aencodes the
?
" negated variables. More precisely, the interpretation of
?
Aconsists of all
? tuples
???
0??
???????????
0??
???
0
? ?
?
??????????
0?!
?&=&????"????????????
?
?
?
such that
different, so our clauses have at most
?? ??
0??
?????????$ ??
0??
???
0
? ?
?
???????????
0?!
? is a clause of
? . Note that we do not require the variables to be
? literals, and not necessarily exactly
? . Next we define a particular
? CNF formula
variables
"
? whose encoding
?
?#"
?
? is of our interest. The clauses of
"
? are all the
? clauses on the
?
? and
??" that are satisfied by the truth assignment that maps
?
? to
? , and
??" to
? .
4
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We will consider the particular case of the existential
? pebble game that is played on the structures
?
??? and
?
?#"
?
? . Observe that each partial homomorphism from
partial truth assignment to the variables of
the later development, let us state it as a lemma.
?
??? to
?
?#"
?
? may be viewed as a
?
that does not falsify any clause from
? . Since this is central to
Lemma 1 Let
?
be an
? CNF formula, and let
? be a partial homomorphism from
?
??? to
?
?#"??
? . Then
? is also a partial truth assignment that does not falsify any clause from
? .
Proof: Suppose
be the
and
belong to
and
? falsifies a clause
? . Let
" be the number of negated literals in
?
and let
???
"
????????????
?'=
?
?
???
A
? tuple encoding
?
in
?
??? . Since
? falsifies
? , we must have
?
???
"
?
???????
???
???
A
?
??
?
???
A??
"
? ???????
???
???
?
? ?? . But the tuple
?
???????????????????????????
? with
" ones and
??" zeros does not
?
?
???
!
A
because it does not satisfy the clause
?????????????+ ?????
?
"
?????????
?
" with
" negated literals
???" positive literals. We conclude that
Thus, the existential
? is not a partial homomorphism; contradiction.
??
? pebble game on
?
??? and
?
?#"
?
? may be reformulated as follows.
Definition 2 Let
game on
such that
?
be an
? CNF formula. We say that the Duplicator wins the Boolean existential
if there is a nonempty familyof partial truth assignments that do not falsify any clause from
? pebble
?
?
?
(i) If
?P=?? , then
?
J
?LK
?E?
?????
? .
(ii) If
?P=??
and
G??? , then
G?=?? .
(iii) If
?=
? ,
?
J
?LK
?E?
????
? , and
? is a variable, then there is some
G?=
?
such that
??G and
??=7J
?LK
?NG
? .
We say that
?
is a winning strategy for the Duplicator.
We stress on the fact that this definition is only a particular case of the definition of winning strategy for
the existential
? pebble game defined in Section 2.
Lemma 2 If there is no resolution refutation of
tial
?
of width
? , then the Duplicator wins the Boolean existen
?????
? pebble game on
? .
Proof: Let
most
any clause in
truth assignment with empty domain (note that
is closed under projections. Now, let
be any variable not in
this case let
otherwise
the extension of
?
????#"??????????$?&%
be the set of all partial truth assignments with domain of size at most
is a winning strategy. Clearly
? be the set of all clauses having a resolution derivation from
?
of width at
? . Let
?
??? that do not falsify
? . We will see that
??
is not empty since it contains the partial
does not contain the empty clause). Clearly,
with
?
"
?????????$?
%
?
? be any partial truth assignment in
??
J
?LK
?E?
????
? , and let
?
J
?LK
?E?
? . Let us assume that there does not exist a valid extension of
? to
? in
? . In
?=?? be the clause falsified by the extension of
? that maps
? to
? . Clearly
?
?
?
?
?
???
? since
? would falsify
? . Analogously there exits some
?=?? of the form
???
?? !?
? that is falsified by
? that maps
? to
? . Note now that
?????
??? ?L??F?
?
??J
?LK
?E?
? , so
?9????? has width at most
? and belongs to
contradiction.
? . In particular,
? does not falsify it, so it cannot falsify both
?
?
?&???
? and
?
?
?
?? !?
? ;
??
Lemma 3 If the Duplicator wins the Boolean existential
lution refutation ofof width
?????
? pebble game on
? , then there is no reso
?
? .
5
Page 6
Proof: Let
will show by induction in the resolution proof of width
clause of the proof. Thus, the proof cannot be a refutation. The statement is clearly satisfied by the initial
clauses since we are dealing with partial truth assignments that do not falsify any clause of
and
any partial truth assignment in
are done since it cannot falsify it. Otherwise consider the projection
will show that
of
hypothesis
cannot falsifyeither.
?
be a winning strategy for the Duplicator for the existential
?????
? pebble game on
? . We
? that no partial truth assignment in
?
falsifies a
? . Let
??
???
?
??
?? !?
? be clauses of the proof, and let
???
be the result of applying the resolution rule. Let
? be
? . If the domain of
? does not include all the variables in
????
then we
G of
? to the variables in
is at most
?B?? . We
G (and hence
? ) does not falsify
???? . Since the width of
? ?7?
? , the domain
G has size at most
? . Therefore, there exists some extension
? of
G to
? such that
? is in
? . By induction
? does not falsify any of
?H?
???
? and
?O???? !?
? . Consequently, since
? falsifies
? or
? ,
?
????
??
Combining these two lemmas we obtain the main result of this section. We say that the Spoiler wins
the Boolean existentialif the Duplicator does not win the Boolean existential
game on
? pebble game on
?
? pebble
? .
Theorem 2 Let
Spoiler wins the Boolean existential
?
be an
? CNF formula. Then,
?
has a resolution refutation of width
? if and only if the
?????
? pebble game on
? .
We note that the Boolean existential
provides a purely combinatorial characterization of resolution width.
? pebble game does not talk about resolution at all. Thus, this
4Application: Width versus Space
In this section we show that the resolution space introduced by Esteban and Tor´ an [15] and by Alekhnovich,
BenSasson, Razborov and Wigderson [1] is tightly related to the width. Indeed, for an
the minimal space
minimal width of refuting
We start with some definitions. Let
be an
sequence of configurations
configuration
? CNF formula
? ,
"
??? of refuting
? , is always bounded from below by
????'?? , where
???? is the
? . This solves an open problem in [28, 1, 8].
?
? CNF formula. A configuration is a set of clauses. A
?
?
???
"????????????
? is a selfcontained resolution proof if
?
?
??? and for
????? , the
?
0is obtained from
?
0??
" by one of the following rules:
(i) Axiom Download:
?
0
?
?
0??
"!?
???
? for some
?=
? ,
(ii) Erasure:
?
0
?
?
0??
"
????
? for some
??=??
0??
" ,
(iii) Inference:
on two clauses from
?
0
?
?
0??
"?
???
? for some
?
that is obtained from an application of the resolution rule
?
0??
" .
The space of a selfcontained resolution proof
contained resolution refutation is a selfcontained resolution proof whose last configuration is
minimal space of refuting an unsatisfiable formula
contained resolution refutations of
?
?
???????????
? is the maximum of
?
?
0
? for
?
? ???????????? . A self
?
??
? . The
? , denoted by
"
??? , is the minimal space of all self
? . We will need the following easy lemma.
Lemma 4 (Locality Lemma [1]) Let
satisfies
? be a partial truth assignment and let
?
be a set of clauses. If
?
? , then there exists a restriction
G??? such that
?
J
?LK
?NG
??????
?
? and
G still satisfies
? .
Proof: For every
variable. Finally, let
?=
? , let
???
=P?
be a literal that is satisfied by
? . For every
??? , let
??? be the underlying
G be the projection of
? to
?????
???=??
? .
??
6
Page 7
Lemma 5 Let
existential
?
be an unsatisfiable
? CNF formula, and let
?#? . If the Duplicator wins the Boolean
is at least
????????
? pebble game on
? , then the minimal space of refuting
?
? .
Proof: Let
show that if
This will prove that
sequence of partial truth assignments
let
consider the three possible scenarios for
be an extension of
andis an
?
be a winning strategy for the Duplicator in the existential
is a selfcontained resolution proof of space less than
cannot have a resolution refutation of space less than
such that
????? ?!?
? pebble game on
? . We
?
?
???????????C%
? , then every
?
0is satisfiable.
?
? . We build, by induction on
? , a
?
0
=??
?
0satisfies
?
0and
?
J
?LK
?E?
0
??????
?
0
?. In the following,
"
0
?
?
?
0
?. Let
?
?
?
? . In order to define
?
0for
??
? , suppose that
?
0??
" has already been defined. We
?
0. Case 1:
?
0
?
?
0??
are in
"8?
???
? by an axiom download for
??=
? . Let
?P=??
?
0??
" such that all variables in
?J
?LK
?E?
? . Since
?
J
?LK
?E?
0??
"
?????
"
0??
"
?
? ,
?
0??
"?=
?
?
? clause, such an
? exists in
? . Moreover, since
? does not falsify any clause from
? , and since all variables in
Lemma, there exists some restriction
?
are defined,
? satisfies
? . Therefore,
? satisfies
?
0. Now, by the Locality
G7?? such that
?
J
?LK
?NG
????
"
0and still
G satisfies
?
0and belongs to
? . Let
the resolution rule guarantees that
?
0
??G . Case 2:
?
0
?
?
0??
"?
???
? by an inference. In this case set
?
0
?<?
0??
" . The soundness of
?
0satisfies
?
0. Of course,
?
J
?LK
?E?
0
??? ?
"
0??
"
?
"
0and
?
0
=
? . Case 3:
?
0
?
?
0??
"
????
? by a memory erasure. Obviously,
?
0??
" still satisfies
?
0since it satisfies
?
0??
" . Now, by the
Locality Lemma, there is a restriction
to
G of
?
0??
" such that
?
J
?LK
?NG
????
"
0and still
G satisfies
?
0and belongs
? . Let
?
0
?
G .
??
Theorem 3 Let
minimal space of refuting
?
be an unsatisfiable
in resolution, and
? CNF formula. Then,
"
???
#
????
???? where
in resolution.
"
??? is the
????? is the minimal width of refuting
?
Proof: Let
Duplicator wins the Boolean existential
refutingis at least
?
?
???? . Then there is no resolution refutation of width
?
?? of
? . By Theorem 2, the
? pebble game on
? . But then, by Lemma 5, the minimal space of
??
???? . Hence
"
???'#
?
???? .
??
We note that this theorem can be used to derive space lower bounds for all formulas for which width
lower bounds are known such as the pigeonhole principle, Tseitin formulas, random formulas, and so on.
5Application: Unified Width Lower Bounds
The new characterization of the width can be used to obtain width lower bounds in a simpler and unified
way. For CNF formulas whose clauses are already small, the width lower bound is obtained directly by
exhibiting a winning strategy for the Duplicator. We illustrate this point with the encoding of the pigeonhole
principle into an unsatisfiable 3CNF formula by means of auxiliary variables (the socalled standard non
deterministic extension).
We will consider theencoding the negation of the pigeonhole principle. For
every
in hole
following
? CNF formula
?
???
?
?
?
"
?
?(=???????????????
???? and
??=
??????????????
? , let
?
0???be a propositional variable meaning that pigeon
? sits
? . For every
?9=
??????????????
? ??? and
?
=
???????????????
? , let
?
0???be a new propositional variable. The
? CNF formula
?
?
0expresses that pigeon
? sits in some hole:
?
?
0??
??
0??
???
?
?
???
"
???
0???
?
"
???
0???
?
??
0???
?
?
?
0??
?
?
Finally, the
tion of all
? CNF formula
?
???
???
?
"
?
expressing the negation of the pigeonhole principle is the conjunc
?
?
0and all clauses
?
0???
2
?
??
0??
2
?
??
???
2for
?
????=&??????????????
???? ,
???
??? and
??=&???????????????
? .
Lemma 6 The Duplicator wins the Boolean existential
? pebble game on
?
???
?
?
?
"
?
.
7
Page 8
Proof: Let
?
be the set of all onetoone partial functions from
???????????????
???? into
??????????????
? . For every
@
=?? , define a partial truth assignment
(i)
??? as follows:
???
?
?
0???
?
?? if
@
????? is defined and
@
?????
??? ,
(ii)
???
?
?
0???
?
?? if
@
????? is defined and
@
?????
?
??? ,
(iii)
???
???
0???
?
?? if
@
????? is defined and
@
?????
?
? ,
(iv)
?
?
???
0???
?
?? if
@
????? is defined and
@
?????
?
? .
Let
variables. It is straightforward to check that
because, by definition, all assignments in
definition as well, and property (iii) is met because if
there is an empty hole.
???
???
?
@
=??? , and let
?
be the set of restrictions of assignments of
is a winning strategy for the Duplicator: property (i) is met
have at most
?
to all sets of at most
?
?
?
? variables in the domain, property (ii) is met by
?
?
??? has at most
?
??? variables in its domain, then
??
We claim that all width lower bounds in the literature can be easily rederived by exhibiting a winning
strategy for the Duplicator. For example, [5] provided a winning strategy for the Duplicator for random
formulas, and thus width lower bounds are also derived for them.
Our next twist is an attempt to systematize the use of the extension variables such as the
. The point is that we would like to play games on CNF formulas with arbitrarily long clauses,
and derive meaningful width lower bounds for their standard nondeterministic extensions. For an arbitrary
CNF formulawithout any restriction on the length of its clauses, let us define an equivalent
mula for
clause of length at most
be a collection of new variables. Then we define
?
0???’s in
?
???
?
?
?
"
?
?
? CNF for
?#?? . Such a formula is called the standard nondeterministic extension of
?
in [1]. For every
?
? , let
?
?????
?
?
? . For every clause
?
??
?
"
?????????
?
?
? of length
?
?
? , let
?
?
?
?
?????????
?
?
?
?
?
?
???
? as follows:
?
?
???
?
?
??
?
?
?
?
?
?
??
"
???
?
??
?
"
?
?
0
?7 ??
?
??
?
?
?
?
?
?
?
Then,
able if and only if
The aim of the following definitions is to formalize a variation on the existential
tailored for the nondeterministic extensions that we just introduced. Let
restriction on the length of its clauses. Let
truth assignment
if
?
?
??? is the conjunction of all
is.
?
?
???
? . Note that
?
?
??? is now an
? CNF formula and it is unsatisfi
?
? pebble game that is
be a CNF formula without any
?
?
be the set of propositional variables of
? . An extended partial
@is a pair
?.?
?4?
? where
??????
??
?
?????
? and
? is partial truth assignment. Moreover,
?
??
?
?
"
?$?
"
?
?????????
?
?
???$?
?
??? , then
J
?LK
?E?
? ????
"
?????????
?
??? and clause
and
?
0is satisfied by setting
?
0to
?
?
?
0
? (note that
? is always satisfied). If
@
?
?.?
?4?
? and
??
?
?
?
G
? are extended partial truth assignments,
we say that
of
it.
? is an extension of
@, denoted by
@
??? , if
??
?
?P?QG . We also say that
@is a projection
? . We say that an extended partial truth assignment
?.?
?4?
? does not falsify a clause if
? does not falsify
Definition 3 We say that the Duplicator wins the extended
of extended partial truth assignments that do not falsify any clause of
???????
? game on
?
if there is a nonempty family
such that
?
?
(i) If
?.?
?4?
?
=
?
, then
?
?
???
? .
(ii) If
?.?
?4?
?(=
?
and
?
??? , then there is some
G??? such that
?
?
?
G
?
=
?
.
(iii) If
?.?
?4?
?&=
?
,
?
?
???
? , and
??=P? , then there is some
G such that
?P?QG and
?.??
?
?
?
???
????
G
?
=
?
.
8
Page 9
(iv) If
?.?
?4?
?
=
?
,
?
?
???? , and
??=
?
has length at least
.
??? , then there is some
G and some
?P=?
such that
?P?G and
?.??
?
?
?
?$?
????
G
?
=
?
The main result about this new game is the following lemma.
Lemma 7 If the Duplicator wins the extended
existential
??? ???
? game on
? , then the Duplicator wins the Boolean
? pebble game on
?
?
??? .
Proof: Let
loss of generality that every extended partial truth assignment
for some
that are obtained from those in
assignment in, where
?
be a winning strategy for the
??????
? game on
? . We first claim that we may assume without
is such that if
?.?
?4?
? in
?
?
?
?$?
?+=Q?
and
???
?$?
?
=P??
?
??? , then
?
?
? . Indeed, let
in the following way: Let
?
? be the set of all extended partial truth assignments
?
@
?
?.?
?
?
? be an extended partial truth
?
?
??
?
?"
?
"??$?"
?
????????
?
?"
?
?
?
?$?"
?
????????
?
?
A
?
"??$?
A
?
????????
?
?
A
?
?
?
?$?
A
?
?
?
?"
???
?
?????????
?
???????
???
with
truth assignment
?"??????????$?
A
?
?? . For each choice of
?
"#=&?????????????L"???????????
?
A
=
??????????????
A
? , obtain an extended partial
@
?
?
?.?
?
?
?
? , where
?
?
??
?
?"
?
0??
?$?"
?
????????
?
?
A
?
0??
?$?
A
?
?
?
?"????
?
?????????
?
???????
????
and put all these extended partial truth assignments in
strategy for the
Now, let
assignment
?
? . It is not hard to see that
?
? is also a winning
??????
? game on
? .
@
?
?.?
?4?
? be an extended partial truth assignment. We define an ordinary partial truth
?
? as follows:
(i) The domain of
all variables
(ii) If
(iii) If
and set
the uniqueness of
?
? is the set of all
?P=P?
such that
for some
?
?
?$?
?
=7?
for some
?=
?
?
????? , together with
?
?
??such that
for some
for some
?
?
?$?
?
=7??7=P?
and
??=
? .
?
?
?$?
?
=7??=
?
?
????? , then
?
?
?
???
?
?
?
??? .
?
?
?$?
?
=? ??=
? , let
?
0be the literal of
?
??
??"??????????
?
?
? corresponding to variable
?
?
?
???
?
??
?
?? if
???
? and
?
?
???
?
??
?
?? if
?
#
? (here is where we use the assumption about
? ).
First notice that each
from
for every
and (ii) in Definition 2 are obviously satisfied. Let us consider condition (iii). Let
?
? is a partial truth assignment to the variables of
?
?
??? that does not falsify any clause
?
?
??? . Moreover, if
, every partial truth assignment
@
??? , then
?
?
?
???. Now, we construct our winning strategy
?
by including,
@
=
?
? such that
??
?
? and
?
J
?LK
?E?
??? ?
? . Conditions (i)
be such that
?=
?
?
J
?LK
?E?
???????? . Then, there exists
@
=
?
such that
??
?
? . Since
?
J
?LK
?E?
????
??? , there exists a
projection
third property of a strategy for the extended game, there is an
Thus
an extension of
with
the projection of
in its domain. In case (2), there exists some variable
This time
is an extension of
?'?
@such that
?
J
?LK
?
?
?????
?$?? and
?P?
???. Let
??
?.?
?
G
? . Let
? be an initial variable. By the
? such that
G??
? and
?.?H?
?
?
?
???
????
?
?
=
?
.
?P?
???
?
???where
?
?
?.??
?
?
?
???
????
?
? . Then, the projection of
???to the variables in
J
?LK
?E?
?
? ???
? is
? that belongs to
?
and has
? in its domain. Now let
?
?
??be an extension variable of clause
for some
?
?
?
?
?
? . We have to consider two cases: (1)
?
?
?$?
?(=7?? , and (2) otherwise. In case (1),
???to the variables in
J
?LK
?E?
?
?
?
?
?
??
? is an extension of
? that belongs to
?
and has
?
?
??
? and
? such that
G7?
? and
?.???&?
?
?
?$?
????
?
?'=
? .
?P?
?
?where
?
?
?.?B?
?
?
?
?$?
????
?
? , and the projection of
and has
?
?to the variables in
J
?LK
?E?
?C?
?
?
?
??
?
? that belongs to
?
?
?
??in its domain.
??
9
Page 10
There is a strong reason to claim that the definition of the
converse to Lemma 7 holds as we show next. Since this is stated here only for completeness and will not be
used in what follows, we only provide a sketch of the proof.
???????
? game is not arbitrary. Indeed, a sharp
Lemma 8 Ifthe Duplicator winsthe Boolean existential
wins the
??????
? pebble game on
?
?
??? , then the Duplicator
???????
? game on
? .
Proof sketch: Suppose that the Duplicator wins the Boolean existential
describe a winning strategy for the Dupicator in the
extended game on
the current extended partial truth assignment in the extended game is
assignment in the side game will be
from
if the Spoiler restricts his current extended partial truth assignment to a subset
in the definition of the extended game), then the Duplicator answers simply as she would answer in the side
game on
extended game the Spoiler requires the Duplicator to extend the current extended partial truth assignment
????
?
? pebble game on
?
?
??? . We
???????
? game on
? . In order to decide her answers in the
? , the Duplicator will play a Boolean game on
?
?
??? on the side in such a way that if
?.?
?4?
? , then the current partial truth
? . Note that
?
J
?LK
?E?
????
???
??
? and
? does not falsify any clause
? , so it does not falisfy any clause from
?
?
??? either. In the course of the play in the extended game,
?
???
(as in property (ii)
?
?!??? if the Spoiler restricted his current partial truth assignment
? to the variables in
?
. If in the
?.?
Duplicator answers simply as she would answer in the side game on
extend
to the variable
partial truth assignment
(as in property (iv) in the definition of the extended game), then the Duplicator answers as follows: using
the three remaining pebbles in side game, let us pretend that the Spoiler pebbles
the answer of the Duplicator does not satisfy
and
let the Spoiler lift the two pebbles in
process, the Duplicator will eventually reach an extension of
clause in
?4?
? with
?
?
???
? to a variable
? (as in property (iii) in the definition of the extended game), then the
?
?
??? if the Spoiler required her to
?? . Finally, if the Spoiler requires the Duplicator to extend the current extended
?.?
?4?
? with
?
?
??
? to satisfy a clause
?
?
?
"
? ??????
?
?
of length at least
???
?
?
?
? ,
?????
???E"
? , and
?
?
?
" . If
?
" , let us pretend that the Spoiler lifts the two pebbles in
?
?
?
?
?????
???E"
? and puts them in
?????
???
?
? and
?
?
?
?. Again, if the answer of the Duplicator does not satisfy
?
?,
?
?
?
" and
?????
???
?
? and put them in
?????
???
?
? and
?
?
?
?. By repeating this
? that sets some
?
0to true; otherwise some
?
?!???
? would eventually be falsified. This gives her the answer in the extended game: extend
?
to
??
?
?
?????!???
0
?
?$?
??? and
? to
?+?
?
?
?????!???
0
?
?
@
??? , where
@
?? if
?
0is positive and
@
?? if
?
0is negative. This
completes the strategy of the Duplicator.
??
The result of this lemma means that we loose essentially nothing in restricting ourselves to playing
the modified game on
principle which says that a finite linear order cannot be dense.
For every
smaller than
whose intended meaning is that
of are the following:
?
?
??? . We illustrate its use for the set of clauses expressing the dense linear order
?????
=
??????????????
? , let
?
0???be a propositional variable whose intended meaning is that
? is
? in the linear ordering. For every
?
???
???B=
??????????????
? , let
?
0????
2
be a propositional variable
? is smaller than
? , and
? is smaller than
? in the linear ordering. The clauses
???
?
?
?
?
?
!?
0???
?
?
???
0
???
?
0???
???
??
0
?E?
?
!?
0???
?
?
???
2
???
0??
2
?
?
?
!?
0???
?
?
???
2
???
0?? ???
2
???
?
??
0?????
2
???
0???
???
?
??
0?????
2
???
???
2
?????
!?
0??
2
???
0??
"
?
2
? ????????
0??
?
?
2
?.?
0??
2
?
10
Page 11
where
introduced above.
?
???
????=
???????????????
? and
?
?
?
? in (2). Since
?????
?
has large clauses, we employ the
???????
? game
Lemma 9 The Duplicator wins the extended
tation of
?E?
???????
? game on
?????
?, and therefore, every resolution refu
?
?
?.???
?
?
? requires width
????? .
on
Proof: For every linear ordering
with
?
???????????????
? , let
@??
?
?
?
?
G
? be the extended partial truth assignment
?
?
?
"??
?
?
?
?
?
?
???
?
"
??
?
?
0???
???
?
?
?
?
?????
?
?
?
?
?
??
?
?
0?????
2
???
?(?
?
?
?????
?????
?
?
?
?
??
?
?
0???
?,?
0???
?&?
???
?
?
???
??
?
?
0?????
2
?,?
0??
2
?
?
???
????????
The mapping
otherwise. During this proof, the set
is an extended partial truth assignment.
We define our winning strategy
with
extended winning strategy. We will show that condition (iii) is also satisfied. Let
element of
with
projection of
to
domain of
Let us consider now condition (iv). Let
be a clause of
since these are the only clauses of length more than three. Let
referenced incontains
for some
an arbitrary pair of indices from
such that (1)and
contains
we do the following: if
withonand such that
and that
coincides with onand such that
are done.
G is defined as
G
?
?
0???
???? if
???
? and
? otherwise, and
G
?
?
0?????
2
?
?? if
???
???
? and
?
?
in
@
?
?
?
?
?
G
? is called the domain of
@
?. By the way it is defined,
@??
?
as the set containing every extended partial truth assignment
on
?.?
?4?
?&?
@??
?
?
???
????? for every linear ordering
?
???????????????
? . Thus,
?
satisfies conditions (i) and (ii) of
?.?
?4?
?
?
@??
be any
?
?
?
??
????? . For any
?
0???, the pair
?
?
0???
???
? is in the domain of
@??, and consequently, the
@??
???&?
?
?
0???
???
??? belongs to
?
. Analogously, for every
to
be any element of
is a clause of type (7) in
be the set of indices in
for some
?
0????
2, the pair
?
?
0????
2
???
? is in the
@
?, and consequently, the projection of
@
?
???
?
?
?
0?????
2
???
??? is also in
?
.
?.?
?4?
?'?
@??
?
with
?
?
??
????? , and let
?
?
?
?.?????
?
? of length at least
?. Such a
????
?
?, say
?
?I?
0??
2,
?
???????????????
contains
? that are
? . That is,
?
?
? and
?
? if
?
?
0???????
?$????? is in
? ??? , and
?
?
? ,
?
? , and
??? if
?
?
0??
????
?
2
?
?$?9?
? is in
???? . Since
?
?
??
????? , we have
?
?
??
?
??? . Let
???
?
?????
?
?
? be
???????????????
? . We will show that there exists a linear order
?
? on
???????????????
?
??
? coincide on
? , i.e.,
?
???
and
?L? iff
?????
?
?
? for every
???
???
??=?? , and (2) the domain of
@
?
?
?
?
?,?
0??
2
? for some
? . Thus, the projection of
@
?
?to
??
?
?
?
?,?
0??
2
??? belongs to
?
. To construct
?
?
? and
? belong to
?
???
? , then we fix
?#? to be a linear ordering that coincides
??
???
?
???
?
? for some
? not in
? . It is immediate to see that such a
?
? exists
?
?
0?????
2
?,?
0??
2
? belongs to the domain of
@
?
?. Otherwise, we can find some linear ordering
?#? that
??
???
??? . In this case
?
?
0??
2
?,?
0??
2
? belongs to the domain of
@
?
?and we
??
We stress on the fact that the introduction of the new game was motivated by an attempt to generalize
the construction of winning strategies in the presence of auxiliary variables. A winning strategy for the
Duplicator in the original game on the nondeterministic extension
directly.
To complete this section, in view of the width lower bound that we just proved, it is quite interesting that
has a resolution refutation of polynomial size as we show next.
?
?
?.?????
?
? could also be easily found
???
Theorem 4 The set of clauses
?
?
???
?
?, and therefore also
?
?
?.???
?
?
? , has a resolution refutation of size
???
?
? .
11
Page 12
Proof: The idea of the proof is to derive the clauses
?
2
?????
?
!?"
?
0
???
"
?
"
?
0
???????
?
"
?
2
?
0
for every
??=
??????????????
? and
?D=
??????????????
? . Once this is done, from
?"
????? we obtain
?"
?
0for every
?
=
???????????????
? by a cut with
??
"
?
"
?
0which is derived from
!?
"
?
" and (5) (observe that
!?
"
?
" is simply (1) in
the particular case
for every
in
indices.
In order to derive
that each
?
?
?
?? ). Then we obtain
?
0??
" for every
?
=
?
?
???????????
? by a cut with (2), and
??
0??
"
?
2
?????
=
??????????????
? by a cut with (6). Then we eliminate all occurrences of all variables
?
0??
" ,
?"
?
0
???
?
?, and
?
0??
"
?
2from (7). The resulting formula would contain a copy of
???
?
?
?
" up to renaming of
?
2
???
? , do the following. We proceed by reverse induction on
? . For
??
? , note
?
?
????? is an initial clause. Suppose now we have derived every
between (6) and
and
?
2
???
? and we want to derive every
?
2
?
"
????? . First, note that
?
2
?
"
???
? is derived at once from
?
2
???
? and
?
"
?
2
?
2, which is derived by a cut
?
2
?
2(again
!?
2
?
2is (1) in the particular case
?
?
?
?? ). To derive
?
2
?
"
????? from
?
2
?????
?
2
?
"
???
? for
???
?? , we will derive the intermediate clauses
?
2
?
"
?
?
?
???$?B?
2
?
"
?????
?
?
"
???
2
???????
?
"
?
2
?
"
?
2
?
for
?&????????????? . Note that
?
2
?
"
????
??? is
?
2
?
"
????? because the range of the dots
????? is empty. Start with
?
2
????? and cut with (5) on
?
"
?
2
?
0to obtain
?
2
?
"
?????
?P?
"
?
2. Then cut this with
?
2
?
"
???
? on
?
"
?
2
to obtain
to obtain
?
2
?
"
?
???
??? . Now we derive
?
2
?
"
?
?
????
??? from
?
2
?
"
?
?
?
??? . Cut
?
2
?
"
?
?
?
??? with (6) on
?
"
? ???
2
?
2
?
"
?
?
????
???
?
?
???
2. Then cut this with (3) on
?
???
2to obtain
?
2
?
"
?
?
????
???
?
!?
2
?
0
???
???
0
?
Separately, take
?
2
????? and cut with (6) on
?
"
?
2
?
0to obtain
?
2
?
"
?????
???
2
?
0
?
Then cut the last two displayed clauses on
obtain
?
2
?
0to obtain
?
2
?
"
?
??????
?
?
?
?
???
0. Cut this with (4) on
?
??
0to
?
2
?
"
?
?
????
???
?
!?"
??
?
?
"
????
0
?
Separately, cut
displayed clause to obtain
This completes the refutation of
each.
?
2
?
"
?
?
?
??? with (5) on
?
"
????
2
to obtain
?
2
?
"
?
?
????
???
?7?"
??. Finally, cut this with the last
?
2
?
"
?
?
????
??? . Note here that
?L"
????
0is absorved by
?
2
?
"
????? in
phases of length
?
2
?
"
?
?
? ???
??? .
???
?
?. Its size
????
?
? because it consists of
????
?
?
??
Therefore, the dense linear order principle is another example of a tautology witnessing that the rela
tionship between size and width of BenSasson and Wigderson cannot be made subpolynomial. Indeed,
has resolution proofs of size
the sizewidth relationship was provided by Bonet and Galesi who showed that the minimum principle
has resolution proofs of size
for
???
?
?
???
?
? yet it requires width
?
??? to refute. A tighter lower bound to
?
"
?
???
?
? yet it requires width
?
??? to refute. We note that the width lower bound
??"
?could also be derived using Lemma 7 and a game theoretic argument.
6Conclusions
We have provided a charaterization of resolution width in terms of the existential pebble game which was
introduced in the context of finite model theory. In that context, the game was used as a tool to obtain
12
Page 13
nonexpressibility results for the existential positive fragment of infinitary logic with finitely many variables
which subsumes Datalog. Our result indicates that there is a tight connection between both areas, and
also with the area of the theory of constraint satisfaction problems since existential pebble games play an
important role there [20, 21, 13]. We think it is worth exploring further these connections. In particular, it is
an interesting project to try to establish a precise connection between definability in the existential positive
fragment of infinitary logic with finitely many variables and the proverliar games of Buss and Pudl´ ak [23],
and to interpret and use that correspondence in both fields. Also, this is a good time to mention Pudl´ ak’s
view of resolution proofs as games [22]. His view is slightly more complicated than ours (with strategies
and superstrategies), but is essentially equivalent if we restrict the size of the records to
formulation with pebble games has the benefit of being of more algebraic nature, and of establishing the link
with finite model theory.
Our results may also have some impact on the algorithmic aspects of resolution that we discuss next.
From the known results in the area of constraint satisfaction problems, it is not hard to prove that the
Duplicator wins the existential
width less than also maps homomorphically to
results, this is interpreted as follows in the context of resolution: a 3CNF formula
widthif and only if every 3CNF formulaof treewidth less than
is satisfiable. This explains why formulas of small treewidth have small resolution proofs and sheds some
more light on the recent algorithmic results of Alekhnovich and Razborov [2]. We believe these connections
should be explored further.
? in his game. The
? pebble game on structures
?
and
?
if and only if every structure of tree
[16]. In view of our
requires resolution
that maps homomorphically to
? that maps homomorphically to
??
?
?
?
??
Acknowledgments. We thank an anonymous referee for very detailed comments that improved the presen
tation.
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