A Tradeoff Between Length and Width in Resolution

: We describe a family of CNF formulas in n variables, with small initial width, which have polynomial length resolution refutations. By a result of Ben-Sasson and Wigder-son it follows that they must also have narrow resolution refutations, of width O ( √ n log n ) . We show that, for our formulas, this decrease in width comes at the expense of an increase in size


Introduction and results
Resolution is a well-known proof system for refuting propositional CNF formulas.A literal is a propositional variable or its negation.A clause is a disjunction of literals.We define a conjunctive normal form formula or CNF to be a set of clauses, which we treat semantically as though it were a conjunction of clauses.The resolution rule allows us to derive the clause C ∨ D from the two clauses C ∨ q and D ∨ ¬q, where q is any propositional variable.The weakening rule allows us to derive a clause C from any subclause D of C. A resolution refutation of a CNF F is a sequence of clauses, ending with the empty clause, where each clause either comes from F or follows from earlier clauses by resolution or weakening.
Every unsatisfiable CNF has a resolution refutation.However, interesting questions remain about the complexity of refutations.We consider two measures of complexity, length and width, and will also mention a third, space.The length (or size) of a resolution refutation Π is the number of clauses it contains.The width of Π is the maximum width of any clause in Π, where the width of a clause is just the number of literals it contains.Similarly the width of a CNF F is the maximum width of any clause in F. The space or clause space of Π is the number of clauses that need to be kept in memory while verifying Π (see Estaban and Torán [8]).
Ben-Sasson and Wigderson [6] showed an interesting and useful connection between the minimal length and minimal width of refutations.
Theorem 1.1 (Ben-Sasson and Wigderson).Let F be a CNF in n variables with width k.Suppose that F has a resolution refutation Π of length S. Then F also has a resolution refutation Π of width at most k + √ n log S.
In other words, every short refutation can be transformed into a narrow refutation.However, the transformation of Π into Π used in the proof of Theorem 1.1 may increase the length of the refutation exponentially.In this paper we address the natural question, posed for example in [4,15,14], of whether the theorem can be strengthened to guarantee that the narrow refutation Π is not substantially longer than the initial short refutation Π. 1 We show that the expected answer ("no") is correct.Our main result is the following theorem.This kind of result is known as a tradeoff between length and width.The reason for the name is that if we need a refutation of small length, we can find one; and if we need a refutation of small width, we can find one; but we must choose between small length and small width, since there is no way to minimize both in the same refutation.We briefly describe some known tradeoffs between complexity measures for resolution-see Nordström [14] for a detailed survey.
They were first studied by Ben-Sasson [4], who showed tradeoffs between space and width for resolution and between space and length for treelike resolution (in which the underlying graph of every refutation must be a tree).In particular he gave formulas of size n which have linear length treelike refutations with constant space, and which also have constant width treelike refutations, but for which for any refutation Π, the product of the width and the space of Π must be at least Ω(n/ log n), and 1 A question about the relation between length and width in the opposite direction also arises from [6].Any refutation with width w must have length at most n O(w) , since there only exist n O(w) many clauses of suitable width.Is there a family of formulas for which this bound is tight, that is, the formulas are refutable in width w, but require length n Ω(w) ?This was answered recently by Atserias, Lauria and Nordström [2]: such families do exist, for w = n c for any constant c < 1/2.for any treelike refutation Π the product of the width and the logarithm of the length of Π must be at least Ω(n/ log n).
A tradeoff between space and length for unrestricted resolution was shown by Nordström [13], and a robust system for showing such tradeoffs was developed by Ben-Sasson and Nordström in [5].For example, there are formulas of size n and constant width which have linear length refutations, and which also have refutations with space O(n/ log n), but for which any refutation with minimal space must have exponential length.
A tradeoff between length and width was also shown in [13], giving formulas of size n and constant width which have linear length refutations, and for which the minimal width of refutations is O( 3 √ n), but for which any refutation of minimal width must have exponential length.However, these parameters are not enough to answer the question about Theorem 1.1 discussed above (in particular the linear length refutations also have width O( 3 √ n)), and the formulas and method of proof in this paper are completely different.
A new tradeoff between length and width for treelike resolution was shown very recently by Razborov [17] giving, for example, formulas of width 3  √ n which have refutations of width O( 3 √ n), but for which any treelike refutation of width n 2/3−ε must have doubly exponential size.
The CNF Φ m in Theorem 1.2 is a propositional version of the coloured polynomial local search principle, or CPLS, which was introduced in [12] as a combinatorial principle as strong as reflection for resolution.It thus in some sense captures the strength of resolution, and also of first-order theories built around bounded Π 2 induction (such as Buss's theory T 2 2 [7]), as these are closely connected with resolution.We say more about this in Section 2 below.In Section 3 we formally define the CNF Φ m and prove the length upper bound, and in Section 4 we prove the width upper bound, describing two different refutations of small width.Finally in Section 5 we prove the length lower bound on refutations of small width.
The idea of the lower bound proof is, roughly, that we consider four senses in which a clause can be "narrow"-mostly these differ in which variables we are counting (see Lemma 5.4).Given a refutation Π, if Π has small width it follows immediately that every clause in Π is narrow in our first sense.If furthermore Π has subexponential length, then we can hit Π with a random restriction such that with high probability every clause in the resulting refutation is also narrow in the remaining three senses.We then use what is essentially an adversary argument to show that no such narrow refutation of the restricted CNF can exist.The restriction and the adversary argument are simpler versions of those used in the resolution length lower bound for the related formula GI 3 in [18].

Coloured polynomial local search
We write [a] for {0, . . ., a − 1}.Consider a levelled directed graph whose nodes consist of all pairs (i, x) from [a] × [b].We refer to (i, x) as node x on level i.If i < a − 1, this node has a single neighbour in the graph, node f i (x) on level i + 1.Every node in the graph is coloured with some set of colours from [0, c).Consider the following three sentences.
1. Node 0 on level 0 has no colours.THEORY OF COMPUTING, Volume 12 (5), 2016, pp.1-14 2. For every node x on every level i < a − 1, if the neighbour f i (x) of x on level i + 1 has any colour y, then x also has colour y.

Every node
x on the bottom level a − 1 has at least one colour, u(x).
Clearly, these cannot all be true at once in a finite graph.We formalize this principle as a first-order sentence, using a relation G i (x, y) to represent the presence of colour y on node (i, x).
Definition 2.1.The coloured polynomial local search principle is the universal closure of the following first-order formula with parameters a, b, c.Suppose that G i (x, y) is a three-place relation on and that f i (x) is a two-argument function, with arguments i and x, from Then the following three formulas cannot all be true: We also use CPLS as the name of the total NP search problem in which we are given the size parameters a, b and c, together with either oracles or polynomial time machines computing G, u and f , and have to find a witness that one of the three formulas above is false.If we fix c = 1 this is equivalent to the well-known polynomial local search problem PLS of Johnson, Papadimitriou and Yannakakis [10].The CPLS principle asserts that the CPLS search problem is total.
Without going into details about first-order proof systems, the CPLS principle can be proved by bounded Π 2 induction on i, starting at i = a − 1 and working towards i = 0, using the inductive hypothesis ∀x < b ∃y < c G i (x, y), that every node at level i has a colour.The short resolution refutation in the next section will have essentially this form, deriving a set of clauses expressing ∀x < b ∃y < c G i (x, y) for each i in turn, and in particular using space and width closely related to the bounds b and c on the universal and existential quantifiers.In fact, by a result of Krajíček [11] any first-order proof using a suitable form of bounded Π 2 induction can be made into a resolution refutation in a similar way (see [3] for a recent, self-contained presentation of this translation).
On the other hand, CPLS is the hardest NP search problem that is provably total using this amount of induction, in the sense that any other such search problem is reducible to CPLS.This is the main result of [12], and follows from the translation of bounded Π 2 induction into resolution mentioned above, plus the fact that 1-reflection for resolution is reducible to CPLS.Here 1-reflection for resolution is the NP search problem in which we are given (as oracles or polynomial time machines) a resolution refutation of a narrow CNF together with an assignment to its variables, and have to find a clause of the CNF that is falsified by the assignment.For more on connections of this form between proof systems, search problems and induction, see [18].
We first list the propositional variables that we will use.
1.For each i < a, x < b and y < c, there is a variable G i (x, y).
2. For each i < a, x < b and j < log b, there is a variable ( f i (x)) j , standing for the jth bit of the value of f i (x).
3. For each x < b and j < log c, there is a variable (u(x)) j , standing for the jth bit of the value of u(x).
The total number of variables is abc x to stand for the conjunction expressing that, for each j < log b, the variable ( f i (x)) j has the same value as the corresponding bit (x ) j .That is, f i (x) = x is the conjunction Similarly if y < c has binary expansion (y) 0 . . .(y) log c−1 we write u(x) = y to stand for the conjunction expressing that, for each j < log c, the variable (u(x)) j has the same value as the corresponding bit (y) j .
We will frequently write With this notation the resolution rule may take the form: from A ∧ q → C and A ∧ ¬q → D derive A → C ∨ D (where A is a conjunction).
Definition 3.1.The formula CPLS a,b,c consists of the following three sets of clauses, which we will call Axioms 1, 2 and 3.
Axiom 1.For each y < c, the clause ¬G 0 (0, y) .Axiom 2. For each i < a − 1, each pair x, x < b and each y < c, the clause Axiom 3.For each x < b and each y < c, the clause Proof.For each i, define a set of clauses expressing that every node at level i has a colour.Notice that M i has space b and width c.We construct the refutation by deriving M i for i = a − 1, . . ., 0 in turn, and then deriving the empty clause from M 0 .The details are in the following three claims.We use Claim 1 to derive M a−1 .We then keep M a−1 in memory, taking up b memory locations, while using Claim 2 to derive M a−2 .We store M a−2 , forget M a−1 , and continue.Once we have derived M 0 we use Claim 3 to reach a contradiction.The maximum space used is either b + log c + 1 while deriving M a−1 , or 2b + log b + 3 while deriving each M i from M i+1 .
Proof of Claim 1. Fix x < b.For a binary string σ of length log c or less, and a number y < c, say that y extends σ if the sequence of the first |σ | bits in the binary expansion of y equals σ , that is, if (y) j = σ j for all j < |σ |.
Axiom 3 consists of θ σ (x) for every σ of length exactly log c.So by the observation above, we can derive θ ∅ (x) from Axiom 3 using a derivation in the form of a complete binary tree of height log c.This uses length O(c), space log c + 2 and width c, the maximum width of the clauses θ σ (x).Finally, the clause θ ∅ (x) is exactly y<c G a−1 (x, y), so to show the claim we derive θ ∅ (x) for each x < b in turn.
Proof of Claim 2. Fix x < b.For x < b, let φ (x ) be the clause THEORY OF COMPUTING, Volume 12 (5), 2016, pp.1-14 The clause φ (x ) can be obtained from y<c G i+1 (x , y), which is in M i+1 , by resolving with instances of Axiom 2 for each y < c in turn.This takes length O(c), space 3 and width c + log b + 1.
We now use a similar argument to the proof of Claim 1 to derive the clause y<c G i (x, y) from all the clauses φ (x ), using a derivation in the form of a complete binary tree of height log b.To save space we do not derive all of the clauses φ (x ) together at the beginning, but only as we need them.Hence the derivation of y<c G i (x, y) takes length O(bc), space log b + 4 and width c + log b + 1.As before, to show the claim derive this for each x < b in turn.
Proof of Claim 3. Resolve y<c G 0 (0, y) with all instances of Axiom 1.
This completes the proof of Theorem 3.2.

Two narrow refutations
The main purpose of this section is to motivate the definition of the random restriction ρ in Section 5 below.We describe, in Theorems 4.1 and 4.2, two narrow strategies for the Prover in a certain Prover-Adversary game based on CPLS a,b,c (this is equivalent to, but intuitively simpler than, describing narrow resolution refutations).In Section 5 we want to show that no narrow refutation can be small, which in particular means that we should be able to show that no small strategy similar to the two outlined here can work.
Part 1 of the definition of ρ (Definition 5.1) can be seen as blocking any small strategy similar to the one outlined in Theorem 4.1, where the Prover tries to learn long paths in f , because it generates lots of cases that the Prover must be able to remember, forcing his strategy to have many nodes.Part 3 of the definition does the same for strategies similar to the one in Theorem 4.2, where the Prover tries to remember a colour on many different nodes.(See Pudlák [16] for more on this kind of approach to length lower bounds.) The Prover-Adversary game works as follows.At each turn, the Prover can ask the Adversary the value of a variable, and record the corresponding literal in his memory; alternatively, the Prover can forget a literal to allow the memory location to be re-used.The Adversary can give any answer which does not directly contradict the current contents of the Prover's memory, and the Prover wins when his memory falsifies some axiom of CPLS a,b,c .It is easy to see that a winning strategy for the Prover that requires no more than w units of memory (where a unit is enough to record one literal) can be turned into a resolution refutation of CPLS a,b,c of width w.Proof.By querying all bits of each f i (x i ) in turn, the Prover first learns a sequence x 0 , . . ., x a−1 such that x 0 = 0 and f i (x i ) = x i+1 for each i < a − 1.This requires (a − 1) log b units of memory.
The Prover then uses log c more units of memory to learn that u(x a−1 ) = y for some colour y.The Prover then queries G a−1 (x a−1 , y) and must get the answer 1, since otherwise the Adversary would violate Axiom 3. At this point the Prover can forget u(x a−1 ) = y.
For i = a − 2, . . ., 0 the Prover then queries G i (x i , y) and must get the answer 1 each time, or the Adversary would violate Axiom 2. Each time the Prover may then forget the previous value G i+1 (x i+1 , y).For i = 0 this forces the Adversary to violate Axiom 1. Proof.By a result of Atserias and Dalmau [1] bounding the minimal width of refuting a CNF in terms of the minimal space, the existence of a refutation of roughly this width follows already from the space upper bound on CPLS a,b,c shown by the refutation in Theorem 3.2.In some sense the refutation we describe here is dual to that one (see also Filmus, Lauria, Mikša, Nordström and Vinyals [9]).
For each x < b in turn, the Prover learns u(x) = y for some colour y, queries G a−1 (x, y) and must get the answer 1 (by Axiom 3), and then forgets u(x) = y.This can be done in b + log c units of memory in total.
The Prover then repeats the following process for each i = a − 2, . . ., 0. For each x < b the Prover learns f i (x) = x for some x , then queries G i (x, y), where y is the colour for which he knows G i+1 (x , y).This must get the answer 1 (by Axiom 2).The Prover then forgets f i (x) = x and goes on to the next x.Having done this for every x at level i, he forgets all the values G i+1 (x, y) from the previous level.The maximum memory used during this process is 2b + log b.
When this has reached level 0, the Prover knows that G 0 (0, y) = 1 for some colour y, contradicting Axiom 1.
The refutation in Theorem 4.1 has length at least b a , since it contains a distinct clause for every possible sequence x 0 , . . ., x a−1 .The refutation in Theorem 4.2 has length at least c b , since it contains a distinct clause corresponding to the conjunction G a−1 (0, for every possible choice of colours y 0 , . . ., y b−1 .

A length lower bound for narrow refutations
We now prove the last part of Theorem 1.2, that there is no refutation of Φ m with simultaneously small width and subexponential length.Recall that Φ m is the formula CPLS a,b,c with parameters a = b = m ε and c = m, where ε > 0 is a constant and m and m ε are both powers of two.
By subexponential we mean smaller than 2 m δ for every fixed δ > 0. By exponentially high probability we mean probability greater than 1 − 2 −m δ for some fixed δ > 0. By polynomially high probability we mean probability greater than 1 − m −δ for some fixed δ > 0. The main parameter appearing in the proof will be a rather than m, but since a = m ε this does not change these definitions.
Suppose for a contradiction that there is a refutation Π of Φ m with subexponential length and with width strictly less than m.Let p = a −3/4 and w = a 7/8 .Definition 5.1.A random restriction ρ is a partial assignment chosen in three stages, as follows.
1. Independently for each pair (i, x), with probability p put (i, x) into a set Γ. Then for each (i, x) ∈ Γ, for each y set the variable G i (x, y) independently to 0 or 1 with probability 1/2.For such (i, x) we say "G i (x, •) is set in ρ." 2. For each node x on level a − 1 with (i, a − 1) ∈ Γ, choose a random y such that G a−1 (x, y) = 1 and set all bits of u(x) to satisfy u(x) = y.For such x we say "u(x) is set in ρ." (With exponentially small probability there is no such y -in this case do nothing.) THEORY OF COMPUTING, Volume 12 ( 5), 2016, pp.1-14 3. Independently for each pair (i, x) with i < a − 1, with probability p put (i, x) into a set ∆.For each i < a − 1 let S i be the set of nodes x on level i with (i, x) ∈ ∆.Randomly choose an injection h i from S i onto a random set of nodes of size |S i | on level i + 1, and for each x ∈ S i set all bits of f i (x) according to h i .For such (i, x) we say " f i (x) is set in ρ." Definition 5.2.The CNF Φ m ρ is formed from Φ m by removing every clause containing a literal satisfied by ρ, and removing every literal falsified by ρ from the remaining clauses.Π ρ is formed from Π by the same operations.
After the restriction, Π ρ is a resolution refutation of Φ m ρ (some instances of the resolution rule in Π may have become instances of weakening in Π ρ).
We now have two goals.The first is to use the assumption about the length of Π to show that with exponentially high probability ρ simplifies Π, in that every clause in Π ρ is narrow in a certain sense.This is Lemma 5.4.The second is to show that, with polynomially high probability, not only does ρ not immediately falsify Φ m , but Φ m ρ does not even have any refutation that is narrow in the above sense.
For this we define safe configurations, which informally are certain partial assignment α such that Φ m α looks difficult to refute.In Lemma 5.5 we show that with polynomially high probability ρ does not contain certain local patterns that would make refuting Φ m ρ easy.In Lemmas 5.7 and 5.8 we show that this implies that ρ is a safe configuration, and that no safe configuration falsifies Φ m .Finally we use a sequence of safe configurations to show that Π ρ cannot be a narrow refutation of Φ m ρ, completing the proof.Lemma 5.3.With exponentially high probability, for each pair (i, x) such that G i (x, •) is set in ρ, G i (x, y) = 1 in ρ for at least one third of the colours y < c.Furthermore for each i < a, G i (x, •) is set in ρ for at most 2pa values x < b, and for each i < a − 1, f i (x) is set in ρ for at most 2pa values x < b.
Proof.This follows from the Chernoff bound and the union bound.N1.C contains a variable G i (x, y) for at most c − 1 many triples (i, x, y).N2.C contains any variable G i (x, y) for at most w many pairs (i, x).N3.C contains any variable from f i (x) for at most w many pairs (i, x).N4.C contains any variable from u(x) for at most w many values x.
Proof.Item N1 follows directly from the assumption that the width of Π is strictly less than m.This is the only place where we use this assumption.
For the remaining three items, since Π has subexponential length it is enough to show that, independently for each clause C in Π, if C is not narrow in this sense then with exponentially high probability C is satisfied by ρ, and hence does not appear in Π ρ.
For item N2, suppose that a clause C in Π contains a literal G i (x, y) or ¬G i (x, y) for more than w many pairs (i, x).For each such (i, x), the probability that such a literal is satisfied in ρ is at least p/2.Hence the probability that none of these literals in C is satisfied is at most THEORY OF COMPUTING, Volume 12 ( 5), 2016, pp.1-14 For item N3, there is a complication that, if f i (x 1 ), . . ., f i (x t ) are all the values of f set in ρ on level i, then the bits ( f i (x k )) j are not all independent, since the values assigned to f i (x) on level i are constrained to be distinct for distinct nodes x.However, we may assume that f i (x 1 ), . . ., f i (x t ) were chosen in the order shown, and that when each f i (x k ) was chosen the only constraint was that the k − 1 values already chosen on that level were excluded.By Lemma 5.3 we may assume k ≤ 2pa.Hence if a literal has the form ( f i (x)) j or ¬( f i (x)) j , if f i (x) is set in ρ then there are a/2 possible values it may take which satisfy , of which at most 2pa were excluded.Hence the probability that is satisfied is at least p(a/2 − 2pa)/a = p/2 − 2p 2 , regardless of how earlier values were set, which is at least p/3 for large a.We then argue as for item N2.
For item N4, suppose that a literal has the form (u(x)) j or ¬(u(x)) j .Then if G a−1 (x, •) is set in ρ, by the Chernoff bound we may assume that, of the c/2 possible values y of u(x) that would satisfy , for at least one third we have G a−1 (x, y) = 1.Hence for any x the probability that u(x) is set in ρ in a way that satisfies is at least p/6.We then argue as for item N2.
We define a path of length k ≥ 0 in a partial assignment α as a sequence of pairs (i, x 0 ), . . ., (i + k, x k ) such that f i+ j (x j ) = x j+1 in α for each j < k.P1.G 0 (0, •) and f 0 (0) are not set in ρ.
P2.There is no triple (i, x, x ) such that G i (x, •), G i+1 (x , •) and f i (x) are all set in ρ, with f i (x) = x .In other words, there is no path in ρ of length 1 with G set at both ends.
P3.There is no 4-tuple (i, x, x , x ) such that G i (x, •), G i+2 (x , •), f i (x) and f i+1 (x ) are all set in ρ, with f i (x) = x and f i+1 (x ) = x .That is, there is no path in ρ of length 2 with G set at both ends.
P4.There is no 4-tuple (i, x, x , x ) such that f i (x), f i+1 (x ) and f i+2 (x ) are all set in ρ, with f i (x) = x and f i+1 (x ) = x .That is, there is no path in ρ of length 3 or more.
Proof.Item P1 is true with probability (1 − p) 2 .For item P2, for any triple (i, x, x ) the probability that G i (x, •), G i+1 (x , •) and f i (x) are all set is p 3 , and the probability that f i (x) = x is 1/a.There are no more than a 3 such triples, so by the union bound the probability that there is any triple violating the condition is less than (p 3 /a)a 3 = a 2 p 3 = a −1/4 .The calculation for item P3 is similar.
For item P4, for any 4-tuple (i, x, x , x ) the probability that f i (x), f i+1 (x ) and f i+2 (x ) are set is p 3 , and the probability that f i (x) = x and f i+1 (x ) = x is 1/a 2 .There are no more than a 4 such tuples, so by the union bound the probability that there is any tuple violating the condition is less than (p 3 /a 2 )a 4 = a 2 p 3 = a −1/4 .Fix a restriction ρ which satisfies the conditions of Lemmas 5.3, 5.4 and 5.5.Definition 5.6.A safe configuration is a partial assignment α which extends ρ and satisfies the conditions listed below.We say that a colour y is present or forbidden at (i, x) if respectively G i (x, y) = 1 or G i (x, y) = 0 in α.
THEORY OF COMPUTING, Volume 12 (5), 2016, pp.1-14 S1.For each pair (i, x) either all variables belonging to f i (x) are set, or none are.Similarly for each x either all variables belonging to u(x) are set, or none are.
S2.For each level i < a, the partial assignment to the variables f i defines a partial injection.
S3.If (0, 0) and (i, x) are on the same path in α, then no colour y is present at (i, x).
S4.If (i, x) and (i , x ) are on the same path in α, then no colour y is simultaneously present at (i, x) and forbidden at (i , x ).
S5.If (i, x) and (a − 1, x ) are on the same path in α and u(x ) is set to a colour y, then the colour y is not forbidden at (i, x).
Lemma 5.7.The restriction ρ is a safe configuration.
Proof.It satisfies conditions S1 and S2 by construction.It satisfies condition S3 by item P1 of Lemma 5.5, which guarantees that (0, 0) is not on any non-trivial path.It satisfies condition S4 by items P2, P3 and P4 of Lemma 5.5.Condition S5 follows from condition S4 and the fact that, if u(x) is set in ρ, then we must have G a−1 (x, u(x)) = 1 in ρ.The next lemma will allow us to derive a contradiction from the existence of the refutation Π ρ.The empty clause at the end of the refutation is falsified by a safe configuration, namely ρ.Now suppose that a clause E in Π ρ is falsified by some safe configuration.Either E is derived from an earlier clause by weakening, or E is derived from two earlier clauses by resolution, or E is an initial clause of Φ n ρ.In both of the first two cases we can find an earlier clause in the proof which is falsified by some safe configuration-in the case of weakening this is trivial, and in the case of resolution we use Lemma 5.9.Hence we must eventually find an initial clause of Φ n ρ which is falsified by some safe configuration, contradicting Lemma 5.8.Lemma 5.9.Suppose that a clause E in Π ρ is derived from clauses C and D by a single use of the resolution rule, and that there is a safe configuration α which falsifies E. Then there is a safe configuration β which falsifies either C or D.
Proof.We write α \ ρ for the assignment γ disjoint from ρ such that α = ρ ∪ γ.By Lemma 5.4, by shrinking α as necessary we may assume without loss of generality that α \ ρ is narrow in the following sense: it sets a variable G i (x, y) for at most c − 1 many triples (i, x, y); it sets any variable G i (x, y) for at most w many pairs (i, x); it sets f i (x) for at most w many pairs (i, x); and it sets u(x) for at most w many values x.
Let q be the variable resolved on to derive E. If α already assigns a value to q, then α already falsifies either C or D, by the structure of the resolution rule.Otherwise, it is enough to show how to extend α to a safe configuration which assigns a value to q.We consider three cases.
First suppose that q has the form G i (x, y).If (i, x) is on the same path as some node at which colour y is present, or some node (a − 1, x ) such that u(x ) is set to y, we put G i (x, y) = 1.Otherwise we put G i (x, y) = 0.This does not affect conditions S1 and S2 of the definition of a safe configuration and preserves conditions S4 and S5 by construction.The only way it can falsify condition S3 is if (i, x) is on a path which contains both (0, 0) and some node (a − 1, x ) such that u(x ) is set to y.But any such path must have length a − 1, the full height of the graph.By Lemma 5.5 all paths in ρ have length 2 or less, hence by our assumption about the narrowness of α \ ρ, the longest possible path in α would consist of w + 1 many paths of length 2 from ρ linked together by w many paths of length 1 from α \ ρ, with total length 3w + 2. Now suppose that q has the form ( f i (x)) j .Say that a node (i + 1, x ) is marked if any variable G i+1 (x , y) is assigned a value, or f i+1 (x ) is set, or f i (x ) = x for some x , or i + 1 = a − 1 and u(x ) is set.In each of the four cases there are at most 2pa + w such nodes, by Lemma 5.3 and our assumption about the narrowness of α \ ρ.Hence for large a there are many unmarked nodes.Choose any unmarked node (i + 1, x ) and set f i (x) to be x .By construction, this preserves conditions S1 and S2.It preserves conditions S3 and S4 because it does not add or forbid a colour on any existing path, or join any paths together.It preserves condition S5 because we avoid nodes (a − 1, x ) for which u(x ) is set.
Finally suppose that q has the form (u(x)) j .Let π be the path containing (a − 1, x).If π contains a node (i, x) for which G i (x, •) is set in ρ, then every colour y is either forbidden or present on π, and by Lemma 5.3 at most 2/3 of colours are forbidden.If π contains no such node, then by the assumption about the narrowness of α \ ρ, at most c − 1 colours are forbidden on π.In either case, at least one colour y is not forbidden on π.Set u(x) = y.This does not affect conditions S1 to S4, and preserves condition S5 by construction.

Theorem 1 . 2 .
Fix ε > 0. Take any sufficiently large m such that both m and m ε are powers of two.There is a CNF Φ m with Θ(m 1+2ε ) variables and Θ(m 1+3ε ) clauses, of width O(log m), such that 1. Φ m has a refutation of length O(m 1+3ε ) and width m + O(log m), 2. Φ m has a refutation of width O(m ε ), 3. Φ m has no subexponential length refutation of width strictly less than m.By Theorem 1.1, it follows from item 1 of Theorem 1.2 (even without item 2) that Φ m has a refutation of width O(m 1 2 +ε √ log m).But, by item 3, as long as ε < 1/2 every such refutation requires exponential length.

Axiom 2
has width log b + 2 and Axiom 3 has width log c + 1.The total number of clauses in the formula is c + (a − 1)b 2 c + bc.Theorem 3.2.The formula CPLS a,b,c has a refutation simultaneously of length O(ab 2 c), of space 2b + log b + 3 (assuming log c ≤ b) and of width c + log b + 1.

Claim 1 .
From Axiom 3 we can derive M a−1 in length O(bc), space b + log c + 1 and width c.

Claim 2 .
For each i < a−1, from Axiom 2 and M i+1 we can derive M i in length O(b 2 c), space b+log b+3 and width c + log b + 1. Claim 3. From Axiom 1 and M 0 we can derive the empty clause in length O(c), space 3 and width c.

Theorem 4 . 1 .
CPLS a,b,c has a refutation of width a log b + log c.

Theorem 4 . 2 .
CPLS a,b,c has a refutation of width 2b + log b + log c.

Lemma 5 . 4 .
With exponentially high probability, for every clause C in Π ρ the following are true.

Lemma 5 . 5 .
With polynomially high probability, the following are all true.

Lemma 5 . 8 .
No clause in Φ m , and hence no clause in Φ m ρ, is falsified by any safe configuration.Proof.Conditions S3, S4 and S5 of the definition of safe configuration respectively guarantee that no clause from Axiom 1, 2 or 3 of CPLS a,b,c is falsified.