Chop-SAT: A New Approach to Solving SAT and Probabilistic SAT for

Agent Knowledge Bases

Thomas C. Henderson

1 a

, David Sacharny

, Amar Mitiche

, Xiuyi Fan

, Amelia Lessen

Ishaan Rajan

and Tessa Nishida

School of Computing, University of Utah, Salt Lake City, Utah, U.S.A.

Blyncsy Inc, Salt Lake City, UT, U.S.A.

Department of Telecommunications, Institut National de la Recherche Scientiﬁque, Montreal, Quebec, Canada

Nanyang Technological University, Singapore

Keywords:

Autonomous Agent Reasoning, Cutting Planes, SAT, PSAT.

Abstract:

An early approach to solve the SAT problem was to convert the disjunctions directly to equations which create

an integer programming problem with 0-1 solutions. We have independently developed a similar method

which we call Chop-SAT based on geometric considerations. Our position is that Chop-SAT provides a wide

range of geometric approaches to ﬁnd SAT and probabilistic SAT (PSAT) solutions. E.g., one potentially

powerful approach to determine that a SAT solution exists is to ﬁt the maximal volume ellipsoid and explore

it semi-major axis direction to ﬁnd an H

vertex in that direction.

1 INTRODUCTION

Given a logical sentence (or formula) from proposi-

tional calculus, the Satisﬁability Problem (SAT) is to

determine if there is a truth assignment to the logi-

cal variables that makes the sentence true. If so, the

sentence is called satisﬁable; if not, then it is unsatis-

ﬁable. SAT is the original NP-complete problem and

requires polynomial time on a nondeterministic Tur-

ing machine (Sipser, 2012). Recently, Henderson et

al. proposed a geometric approach, called Chop-SAT,

for solving SAT (see (Henderson et al., 2021)). The

method produces a convex polytope feasible region,

and the goal here is to explore properties of the feasi-

ble region which allow the solution of SAT or PSAT.

Given n logical variables (or atoms), a model (or

complete conjunction) is an assignment of 0 (false)

or 1 (true) to each atom. There are 2

models.

These models can be represented as n-tuples in n-

dimensional space, and correspond to the corners of

, the n-D hypercube. The semantics of the hy-

percube is that the i

axis corresponds to the values

which can be assigned to the i

variable, and the val-

ues can be relaxed to lie in the interval [0,1].

Given any point in H

, that point can be consid-

ered as a set of probabilities for the atoms. This al-

https://orcid.org/0000-0002-0792-3882

lows consideration of a probabilistic version of SAT

called Probabilistic SAT (PSAT) (see (Georgakopou-

los et al., 1987; Nilsson, 1986) for a detailed intro-

duction and analysis of the complexity of the prob-

lem). PSAT is deﬁned as follows; given a logical

sentence in Conjunctive Normal Form (CNF), and

a probability, p

, associated with each conjunct, C

ﬁnd a function, π : Ω → [0,1], where Ω is the set of

all complete conjunctions, and 0 ≤ π(ω

) ≤ 1, and

∑

|=C

π(ω

),ω

∈ Ω.

Chop-SAT is based on the following observation:

when an agent’s knowledge base is represented as a

CNF sentence, then every conjunct is a disjunction.

A disjunction with k

literals is made false by assign-

ing a falsifying truth value to each of the disjunction’s

literals. This assignment, along with all possible 0-1

assignments to the k = n −k

literals not in the dis-

junction, deﬁnes the models which do not satisfy the

conjunct. Geometrically, this set of models is a sub-

hypercube, H

, of the n-dimensional hypercube, H

For each conjunct there exists an (n-1)-

dimensional hyperplane, h, that separates H

from the corners of H

which satisfy the conjunct.

This leads to the notion of a chop: the hyperplane,

h, is represented so that H

is on h’s negative side

(viewing the hyperplane equation as a signed dis-

tance function) while solution corners are on h’s

Henderson, T., Sacharny, D., Mitiche, A., Fan, X., Lessen, A., Rajan, I. and Nishida, T.

Chop-SAT: A New Approach to Solving SAT and Probabilistic SAT for Agent Knowledge Bases.

DOI: 10.5220/0011614400003393

In Proceedings of the 15th International Conference on Agents and Artiﬁcial Intelligence (ICAART 2023) - Volume 3, pages 48-54

ISBN: 978-989-758-623-1; ISSN: 2184-433X

 2023 by SCITEPRESS – Science and Technology Publications, Lda. Under CC license (CC BY-NC-ND 4.0)

non-negative side. By intersecting the non-negative

half-space of h with n-dimensional space, a convex

feasible region is created which contains the solutions

for the conjunct (this holds because the intersection of

convex sets is a convex set). Applying this operation

for each conjunct in th CNF sentence results in a

convex feasible region which must contain solutions

to the CNF sentence, if any.

Our position is that Chop-SAT offers the possibil-

ity to ﬁnd efﬁcient and effective ways to solve SAT.

Chop-SAT produces a feasible region as follows:

• First, a knowledge base (KB) is deﬁned as a CNF

sentence, and each conjunct is given a probability

(for a SAT problem, this will be 1 for each con-

junct).

• Next, the Chop-SAT method is used to produce

a set of hyperplanes such that the intersection of

their non-negative half-spaces determines the fea-

sible region.

• The resulting feasible region is convex.

The feasible region represents the solution space for

the KB. If any of the original hypercube corners exist

in the feasible region, the KB is satisﬁable, otherwise

it is unsatisﬁable.

2 BACKGROUND

Gomory (Gomory, 1958) introduced the cutting plane

method as an extension to Dantzig’s linear pro-

gramming approach to solving discrete variable ex-

tremum problems (Dantzig, 1957). Gomory provided

a method so that when the objective function was

maximized (e.g., using the simplex method), if the

result was a non-integral solution, then a new con-

straint plane was found algorithmically to separate

(cut) the non-integer solution from the integer solu-

tions. Chvatal (Chvatal, 1973; Chvatal, 1985) de-

veloped this method further, proved supporting the-

orems for bounded polytopes, and applied the results

to solve combinatorial problems. Note that integer

programming is in NP (Papadimitriou, 1981).

In terms of solving SAT, Cook et al. (Cook

et al., 1987) examined the complexity of cutting plane

proofs, and in particular, those for the unsatisﬁability

of formulae in propositional calculus. This was done

in terms of resolution theorem proving using the cut-

ting planes method. Note that it is straightforward

to produce an equation from a disjunction; e.g., con-

sider (a ∨a) which results in the equation a + b ≥ 1.

Each conjunct gives rise to an equation, and the set

forms an integer programming problem where a and

b take on 0-1 values. Note that linear programming

relaxation can be used to allow the a and b to take

on values in the interval [0,1] – i.e., relaxing the 0-1

requirement. Hooker (Hooker, 1988) provided a way

to handle generalized resolution by observing that the

resolvent of two clauses corresponds to a certain cut-

ting plane in integer programming. Buss et al. (Buss

and Clote, 1996) describe an extension to Cook’s cut-

ting plane refutation approach that applies to thresh-

old logic analysis. For some recent work using cut-

ting planes, see Devriendt et al. (Devriendt et al.,

2021) who use the method to improve state-of-the-art

pseudo-Boolean optimization. All these methods are

still exponential time complexity.

3 THE Chop-SAT METHOD

We look at the SAT problem geometrically: for an n-

variable problem, the corners of the hypercube are the

only possible SAT solutions, and the interior points

are possible atom probabilities assignments. Given

a Conjunctive Normal Form (CNF) sentence, S =

∧C

∧. .. ∧C

, over a set of Boolean variables

A = {a

,...,a

}, each conjunct, C

, has at least one

truth assignment that makes it false (note that a ∨¬a

is not allowed). If there are k

literals in conjunct C

then there are 2

truth assignments that make it false,

where k = n −k

. A hyperplane can then be found for

each conjunct which separates the solutions from the

non-solutions. Moreover, this hyperplane can be po-

sitioned so as to cut the edges between non-solution

corners and their feasible region neighbors anywhere

in the range 0 < x ≤ 1 along those edges. The in-

tersection of the non-negative side of the hyperplane

with the initial feasible region (H

or ℜ

) produces

a convex feasible region. Continuing this process for

each conjunct, the result is a convex feasible region

which may or may not contain a corner of H

– i.e., a

solution for the CNF sentence.

Suppose A = {a

,...,a

} is a set of Boolean

variables, and the set of A -formulas is the inductive

set of propositional well-formed formulas over A. A

Conjunctive Normal Form (CNF) sentence over A is

deﬁned as a conjunction of a set of disjunctions of

literals.

S = C

∧C

∧. .. ∧C

= L

∨L

∨. .. ∨L

= a

or L

= ¬a

The Satisﬁability Problem is deﬁned as (Davis

et al., 1994):

Find an efﬁcient algorithm for testing an A-

formula in CNF to determine whether it is

truth-functional satisﬁable.

Chop-SAT: A New Approach to Solving SAT and Probabilistic SAT for Agent Knowledge Bases

By efﬁcient, we understand “of polynomial complex-

ity.”

3.1 Basic Approach

The SAT problem is cast as a linear programming

problem:

Minimize f ·x

Subject to: Ax ≤ c

where each constraint is given by:

−α

·x ≤ c

A solution for the SAT sentence exists iff a solution

exists for the LP problem with x ∈{0,1}

Given a set of m conjuncts, C

,i = 1 : m, each con-

junct is used to produce a hyperplane of dimension

n −1 which separates the solutions (i.e., some sub-

set of vertexes of the n-dimensional hypercube) from

non-solutions. The hyperplane for the i

conjunct is:

·x + c

= 0

Each of these hyperplanes produces an inequality:

−α

·x ≤ c

A matrix, A, is produced where each row is the 1×n-

tuple α(i). An n ×1-vector, c, is constructed where

the i

element of c is c

The way these hyperplanes are constructed, it is

now possible to run the interior-point method for lin-

ear programming to ﬁnd feasible points which min-

imize f

x for x ∈ X, where X is the feasible region

and f is an n-D vector. This allows points to be found

which lie on the feasible region boundary.

Given m conjuncts, C

,i = 1. ..m, then:

= L

∨L

∨. .. ∨L

Observe that:

• If k

= n, then this eliminates 1 solution (a vertex

of H

• If k

= n −1, then this eliminates 2 points (two

vertexes joined by an edge of H

• . . .

• If k

= 1, then this eliminates half the points in

the hypercube (all in a hyperplane of dimension

n −1).

The individual hyperplane is determined as fol-

lows. Let A = {1, 2,... ,n}, and I ⊆ A. Given

= L

∨L

∨... ∨L

, then deﬁne α

, the hyperplane

normal vector, as follows.

∀i

∈ I , α

) = 1 if L

is an atom a

, else −1

∀m /∈ I , α

(m) = 0

k α

In order to get the constant for the hyperplane

equation, a point must be found on the hyperplane.

This is selected so that the hyperplane cuts the edges

of the hypercube at a distance ξ from the center of H

This distance depends on the number k

of literals in

the conjunct:

d =k ξ

where b

is a k

-tuple of 1’s. Next:

∀i

∈ I , p(i

) = 0 if L

is an atom, else 1

∀m /∈ I , p(m) = 0

Then p is a non-solution vertex. To ﬁnd a point, q, on

the hyperplane:

q = p + dα

This allows a solution for the constant, c, in the hy-

perplane:

= −(α

·q)

This yields the hyperplane equation:

·x + c = 0

and the resulting inequality:

−α

·x ≤ c

Note that the feasible region as deﬁned above is

not necessarily bounded (e.g., when the initial fea-

sible region is ℜ

. Consider the simple case where

n = 2 and S = C

= a ∨b. Then the feasible region

resulting from this one cut is all points above the line

through the points [1;0] and [0;1]. An unbounded fea-

sible region indicates that a SAT solution exists.

4 SOME EXAMPLES

4.1 2 D One Solution

Consider the two clauses in modus ponens:

1.a

2.¬a

∨a

Then the feasible region is indicated in Figure 1 as

the black point samples (the hyperplanes are shown

as blue lines), and this region is unbounded. The hy-

perplane found for conjunct 1 (with ξ = 0.9) is:

1.0a

+ 0a

−0.9 = 0

while the hyperplane for conjunct 2 is:

−0.7071a

+ 0.7071a

+ 0.7071 = 0

The solution is: a

= 1 and a

= 1.

ICAART 2023 - 15th International Conference on Agents and Artiﬁcial Intelligence

-2

Figure 1: The Feasible Region for Unbounded Modus Po-

nens.

4.2 3 D: One Solution

As a ﬁnal example, consider the case with 3 variables

with seven clauses that chop off all corners individu-

ally except [1;1;1]. Re-writing this in CNF yields:

1. : a

∨a

∨¬a

2. : a

∨¬a

∨a

3. : a

∨¬a

4. : ¬a

∨a

5. : ¬a

∨a

∨¬a

6. : ¬a

∨¬a

∨a

7. : ¬a

∨¬a

Then the hyperplane equations are:

: −0.5774a

−0.5774a

+ 0.5774a

+ 0.5831 = 0

: −0.5774a

+ 0.5774a

−0.5774a

+ 0.5831 = 0

: −0.5774a

+ 0.5774a

+ 0.0058 = 0

: 0.5774a

−0.5774a

+ 0.5831 = 0

: 0.5774a

−0.5774a

+ 0.5774a

+ 0.0058 = 0

: 0.5774a

+ 0.5774a

−0.5774a

+ 0.0058 = 0

: 0.5774a

+ 0.5774a

+ 0.5716 = 0

The only solution is: [1;1;1]. The feasible region is

shown in Figure 2.

Figure 2: The Feasible Region for 3D with only [1;1;1] as a

Solution.

5 SOME PRELIMINARY

INVESTIGATIONS

Some useful facts have been found with respect to

feasible region properties which differentiate satisﬁ-

able from unsatisﬁable CNF sentences. For example,

given a CNF sentence, S, such that no point in the

feasible region arising from S is more than

√

n −1/2

distant from the center of H

, then S is unsatisﬁable.

Let H

be an n-D hypercube. A cut is

the removal from H

of a k-D hypercube,

⊂ H

, 0 ≤ k < n. A cut is effected as

follows. Let V = {v | v is a vertex of H

and deﬁne V

= {v | v is a vertex of H

, v /∈

V, v neighbors an element of V }. Let h

be the

(n −1)-D hyperplane deﬁned by the points in V

where:

• v ∈V is on the negative side of h

• the open edge segments in H

that lie between ele-

ments of V and V

are on the negative side of h

V n

and

• v ∈V

is on the non-negative side of h

Let h

be the n-D half-space on the non-negative side

of h

. Then a cut produces a feasible region as fol-

lows:

F = H

∩h

Given a set of cuts, C = {C

}, then the feasible region

for C is F = ∩F

, where F

is the feasible region for

Lemma 1: Given H

and a set of cuts, C , if these

cuts remove every vertex of H

, then ∀H

⊂ H

(i.e.,

edges), H

∩F =

0, where F is the feasible region of

C .

Chop-SAT: A New Approach to Solving SAT and Probabilistic SAT for Agent Knowledge Bases

Proof: Consider an arbitrary H

⊂H

, and call its end

points A and B. Some cut removes A and in so doing

also removes the open line segment (A,B) since it is

on the negative side of h

. Since some cut removes

B, then all of H

is removed.

Lemma 2: If C is a set of hyperplane cuts that re-

moves all corners of H

and results in the feasible re-

gion F , then every 2-D face of H

is either removed

or at most one point remains, and it is in the center of

the 2D square:

∀H

⊂ H

∩F =

or H

∩F is one point at the center of H

Proof: Let A,B,C,D be the set of vertexes of a 2-

D square with edges {(A,B),(B,C), (C,D),(D,A)} in

; let E be the center of the square. Consider a cut

that removes A. Any cut will at the most leave the

triangle BCD in the feasible region. Cutting B leaves

at most triangle CDE in the feasible region. Cutting

C leaves at most segment DE in the feasible region.

Finally, cutting D leaves at most the point E. This

analysis holds for every 2-D square (i.e., any H

) in

From these results we conclude that the center of

⊂H

is located at distance

√

n−k

from the center of

. Thus, corners of H

are at the maximal distance,

and centers of edges are next. But both of these types

of points are removed according to the lemmas. Thus,

the maximal possible distance is to the center of a 2-

D square (H

) and is distance

√

n−2

from the center of

5.1 Maximal Volume Inscribed

Ellipsoid

The maximal volume inscribed ellipsoid (MVE) may

play a signiﬁcant role in determining whether or not

a SAT solution exists. An ellipsoid is deﬁned by the

equation:

S = {v |v = x + Es,||s|| ≤ 1}

where x is the center of the ellipse, and E is the ellip-

soid matrix that transforms points in the unit n-sphere

to points in the ellipse. If the feasible region is un-

bounded, then that can be detected (e.g., add the hy-

perplanes for increasingly large hypercubes and lin-

ear programming will show the bounds increasing ac-

cordingly) and satisﬁability determined. Even if not,

the directions of the ellipsoids axes may point in the

direction of a solution. Importantly, the MVE can

be found in polynomial time (Khachiyan and Todd,

1990). Moreover, robust implementations exist and

can be used in practice (see (Zhang, 1998; Zhang,

2003)).

Consider the 3D example given above. Given the

center of the MVE, and the matrix deﬁning it, the ex-

treme point on the ellipsoid (and in the feasible re-

gion) can be located. Figure 3 shows some points

sampled from the feasible region (blot dots), the cen-

ter of the ellipse (red circle/star), and extreme point

on the longest ellipsoid axis (blue circle/star), and the

major ellipsoid axis (black line). We are currently

studying how well this works in higher dimensions.

Taking a look at the Matlab output for the singular

Figure 3: Demonstration of how the Maximal Volume In-

scribed Ellipsoid in the Feasible Region Can Help Find a

Solution Along its Longest Axis.

value decomposition of the ellipsoid matrix, we ﬁnd

that the major axis aligns with x = y = z line:

U =

-0.5774 -0.8165 0

-0.5774 0.4082 -0.7071

-0.5774 0.4082 0.7071

and that the most distance feasible region point along

that line is:

pe =

0.9987

5.2 PSAT Approximations

Another area of application is in determining proba-

bilities of atoms to help in the agent decision mak-

ing process. For a detailed example, consider the

Wumpus World problem popularized by Russell and

Norvig and described in (Russell and Norvig, 2009).

An agent looks for gold in a 4x4 grid where there are

pits and a Wumpus which can kill the agent. Russell

describes how to represent this using propositional

ICAART 2023 - 15th International Conference on Agents and Artiﬁcial Intelligence

logic, and in our formulation there are 80 atoms (or

logical variables) and 402 conjuncts in the CNF. The

atoms correspond to 16 for a Breeze in a cell, 16 for

gold in a cell, etc. If the probability of a pit in each

cell (except the start cell) is 20%, then Monte Carlo

can be used to determine the ground truth a priori

atom probabilities. An agent would like to estimate

the probability of danger corresponding to each of its

possible actions, and this can be done using the feasi-

ble region for the CNF sentence.

We have determined the feasible region for the

Wumpus World CNF, then found the solutions for

projecting onto all 80 axes (in both directions) ob-

taining 160 extreme points on the feasible region. We

then take the average of these to ﬁnd an approxima-

tion to the atom probabilities. Figure 4 shows both the

Monte Carlo result as well as that from Chop-SAT. It

is clear that Chop-SAT provides a very good quali-

tative approximation produced just from the rules of

the game; this is adequate when an agent must decide

based on the lower probability of danger.

Figure 4: The Monte Carlo and Chop-SAT Estimates of the

Atom Probabilities for Wumpus World.

This is simply an approximation to the atom prob-

abilities; to get the model probabilities an approxima-

tion technique can be used. E.g., assume the atom

variables are independent, then compute the top 100

model probabilities and assume the remainder are 0.

Normalize these probabilities, and it is then possible

to calculate an estimate for any logical sentence over

the variables. We are also investigating how well this

works.

Note that our approach is a polynomial time ap-

proximation (i.e., very fast) more in line with our pre-

vious work (Henderson et al., 2020), and is very dif-

ferent from standard approaches like Nilsson (Nils-

son, 1986), or more recent works like Finger and De

Bona (Finger and Bona, 2015) which rely on SAT

solvers.

6 CONCLUSIONS

The position espoused here is that Chop-SAT mer-

its further study as a way to solve both discrete and

probabilistic SAT, and allows efﬁcient and effective

solution approximation techniques. This in turn al-

lows a knowledge-based agent to improve its decision

making process. Note that the feasible region is pro-

duced very efﬁciently since converting conjuncts to

hyperplanes is linear time complexity, and obtaining

the properties of the feasible region is done using lin-

ear programming based on the interior point method

which is low polynomial time.

ACKNOWLEDGEMENTS

This work was supported in part by National Science

Foundation award 2152454.

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