SMART Seminar Series: "A polynomial algorithm to solve hard np 3 cnf-sat problems"

The P - NP conjecture
P = NP is impossible to prove
Searching invariant structures to go P
Complexity analysis of the functional descriptor approach
A polynomial algorithm for “hard” NP 3-CNF-SAT
problems ?
Prof. Marcel Rémon
Department of Mathematics, Namur University, Belgium,
Email : marcel.remon@unamur.be
and Dr. Johan Barthélemy
SMART Infrastructure Facilities, University of Wollongong,
Email : johan@uow.edu.au
29 août 2017
M.Rémon & J.Barthélemy A polynomial algorithm for “hard” NP 3-CNF-SAT problems

The Class P problems
The NP problems class
The Open Question P-NP
The 3-CNF-satisﬁability problem
The P problems class
Decision problems : A decision problem is a problem that takes as
input some string, and outputs "yes" or "no".
P problems : If there is an algorithm (say a Turing machine, or a
computer program with unbounded memory) which is able to
produce the correct answer for any input string of length n in at
most c nk steps, where k and c are constants independent of the
input string, then we say that the problem can be solved in
polynomial time and we place it in the class P.
⇔ Complexity = O(nk)

The notation NP stands for “non deterministic polynomial time”.
Definition : We define the class NP of problems by the condition
that there exists for all problem w a checking relation R ∈ P such
that for all possible input y (foreseen as a solution) the checking
relation R(y) gives 0 or 1 in a polynomial time.
We say that y is a certificate associated to the instance of problem
w.
Example : The safe lock. It takes an exponential time of find the
opening code, but the checking of the validity of the code is
straightforward.

The “P versus NP problem”, or equivalently, the question whether
P = NP or not, is an open question and is the core of this paper.
This problem is one of the seven Millennium Prize Problems in
mathematics that were stated by the Clay Mathematics Institute in
2000. A correct solution to any of the problems results in a US
$1M prize. The only solved problem is the Poincaré conjecture,
which was solved by Grigori Perelman in 2003.

A 3-CNF formula ϕ is a Boolean formula in conjunctive normal
form with exactly three literals per clause, like
ϕ := (x1 ∨ x2 ∨ ¬x3) ∧ (¬x2 ∨ x3 ∨ ¬x4) := ψ1 ∧ ψ2.
The 3-CNF-satisﬁability or 3-CNF-SAT problem is to decide
whether there exists or not logical values for the literals so that ϕ
can be true (on the previous example, ϕ = 1(True) if
x1 = ¬x2 = 1).
The 3-CNF-Satisﬁability problem is known to belong to the
NP class.

Impossibility to prove P = NP
Theorem : It is impossible to prove that P=NP in the
Deterministic framework of Mathematics. (Previous result)
The solution of the 3-CNF-SAT problem is equivalent to these two
functions Ξ and Ξ” :
(At t0) Ξ : Φn,m
O(?)
−→ {0, 1} (time to build the solutions set Sn,m)
ϕ 1 if ϕ ∈ Sn,m and 0 otherwise
(At t0 + ∆t) Ξ : Φn,m
O(nk
)
−→ {0, 1} (ϕ
?
∈ Sn,m when Sn,m is known)
ϕ 1 if ϕ ∈ Sn,m and 0 otherwise

The argument of this meta mathematical proof lies in the fact that any
operation done by Ξ in t0 can be reduced to a polynomial time operation
by Ξ in t0 + ∆t. ( 1
)
Mathematically speaking, it is impossible to make a formal or
mathematical distinction between both functions Ξ and Ξ”, as time does
not interfer with mathematics.
More precisely, if someone proves that the 3-CNF-SAT problem Ξ (or Ξ )
is non polynomial, this should stay true at any time, independantly of t,
even in t0 +∆t. The proof could not introduce time in the demonstration.
This meta mathematical argument points out the clear separation
between deterministic (e.g. mathematical) and non deterministic
problems.
1. To make it easier to understand, let us think to the Fermat’s last problem.
It takes 357 years to be solved, but NOW it only takes one operation to say that
the solution is “n = 2”.

A matrix representation of a 3-CNF formula
Lattice structure of 3-CNF-matrices
Descriptor characterization Theorem
An algorithm for the descriptor function
Searching logical invariant structures to go P
Deﬁnition : We deﬁne the 3-CNF-matrix representation of ψi as a
7 × 3 matrix giving all the solutions. For example,
[ψ1] = [(x1 ∨ x2 ∨ ¬x3)] =












x1 x2 x3
0 0 0
0 1 0
0 1 1
1 0 0
1 0 1
1 1 0
1 1 1













The 3-CNF formula ϕ will represented by a 12 × 4 matrix :
[ϕ] = [(x1 ∨ x2 ∨ ¬x3) ∧ (¬x2 ∨ x3 ∨ ¬x4)] =























x1 x2 x3 x4
0 0 0 0
0 0 0 1
0 1 0 0
0 1 1 0
0 1 1 1
1 0 0 0
1 0 0 1
1 0 1 0
1 0 1 1
1 1 0 0
1 1 1 0
1 1 1 1























This paper aims to deﬁne an algebra on this type of matrices such
that [ϕ] = [ψ1] ∧ [ψ2].

Using the neutral sign “.”, one can replace two same lines only
diﬀering by a 0 and a 1 for a variable, by a unique line with a
neutral sign for this variable :
A =












x1 x2 x3
0 0 0
0 0 1
0 1 0
0 1 1
1 0 0
1 0 1
1 1 0












≡






x1 x2 x3
0 0 .[0
1]
0 1 .
1 0 .
1 1 0






≡




x1 x2 x3
0 . .
1 0 .
1 1 0





Let A and B be two matrices and {x1, · · · , xn} the union of their
support variables. Let A and B be their extensions over
{x1, · · · , xn}. Then we deﬁne the disjunction of A and B by
A ∨ B =


x1 · · · xn
A
B


Of course, this new matrix should be reordered so that the lines are
in a ascending order, which can yield sometimes in replacing a line
with a neutral sign by two lines with a one and a zero.

Let A a matrix such that the reduction process yields to lines with
neutral sign, then A can be rewritten as the disjunction of smaller
matrices. For example,
[ψ1] =












x1 x2 x3
0 0 0
0 1 0
0 1 1
1 0 0
1 0 1
1 1 0
1 1 1












=
x1
1
∨
x1 x2
0 1
∨
x1 x2 x3
0 0 0
The block decomposition is not unique. For instance, there are 6
diﬀerent block decompositions for a 3-variables clause.

Conjunction of 3-CNF-matrices
Let A and B be two matrices, A and B their extensions to the joint set of propositional variables. Let
Ak and Bl be the one line matrices such that :
A =
Σ
A
k=1
Ak and B =
Σ
B
l=1
Bl
We deﬁne the conjunction of A and B as
A ∧ B ≡ A ∧ B =



Σ
A
k=1
Ak


 ∧



Σ
B
l=1
Bl


 =
Σ
A
k=1
Σ
B
l=1
Ak ∧ Bl =
Σ
A
k=1
Σ
B
l=1
Ck,l
where
Ck,l =
x1 xi xn
a1
k
ai
k
an
k
∧
x1 xi xn
b1
l
bi
l
bn
l
=



∅ if ∃ ci
m = “NaN"
x1 xi xn
c1
m ci
m cn
m
otherwise
and
c
i
m =



ai
k
if ai
k
= bi
l
ai
k
if ai
k
= bi
l
and bi
l
= “ · ”
bi
l
if ai
k
= bi
l
and ai
k
= “ · ”
“NaN" otherwise

Let us call ∅, the empty matrix, with no line at all and Ω, the full matrix,
as a one line matrix with only neutral signs in it.
Deﬁnitions : A semi-lattice (X, ∨) is a pair consisting of a set X and a
binary operation ∨ which is associative, commutative, and idempotent.
Let us deﬁne the two absorption laws as x = x ∨ (x ∧ y) and its dual
x = x ∧ (x ∨ y).
A lattice is an algebra (X, ∨, ∧) satisfying equations expressing
associativity, commutativity, and idempotence of ∨ and ∧, and satisfying
the two absorption equations.
Let us note A the set of all the 3-CNF-matrices. (A, ∨, ∧) is a lattice
over the set of 3-CNF-matrices with respect to the disjunction and
conjunction operators.

Descriptor characterization theorem
Every non empty 3-CNF-matrix can be characterized by a
one-line parametrised 3-CNF-matrix.
∀ [ϕ] =







x1 xi xn
s1
1 · · · sn
1
.
.
. si
j
.
.
.
s1
Σϕ
· · · sn
Σϕ







= ∅ , ∃ n functions hi : {0, 1}
i
→ {0, 1} such that
[ϕ] =
(α1,··· ,αn)∈{0,1}n
x1 · · · xi · · · xn
h1(α1) · · · hi (α1, · · · , αi ) · · · hn(α1, · · · , αn)
So, the knowledge of
h1(α1), · · · , hi (α1, · · · , αi ), · · · , hn(α1, · · · , αn) characterizes fully
[ϕ].

Example :
[ϕ] = [(x1 ∨ x2 ∨ ¬x3) ∧ (¬x2 ∨ x3 ∨ ¬x4)]
=
(α1,··· ,α4)∈{0,1}4
x1 x2 x3 x4
α1 α2 (α1 + 1)(α2 + 1)α3 + α3 α2(α3 + 1)α4 + α4 (mod 2)
=
































x1 x2 x3 x4
0 0 0 0
0 0 0 1
0 0 0 0
0 0 0 1
0 1 0 0
0 1 0 0
0 1 1 0
0 1 1 1
1 0 0 0
1 0 0 1
1 0 1 0
1 0 1 1
1 1 0 0
1 1 0 0
1 1 1 0
1 1 1 1
































α1 α2 α3 α4
0 0 0 0
0 0 0 1
← 0 0 1 0
← 0 0 1 1
0 1 0 0
← 0 1 0 1
0 1 1 0
0 1 1 1
1 0 0 0
1 0 0 1
1 0 1 0
1 0 1 1
1 1 0 0
← 1 1 0 1
1 1 1 0
1 1 1 1
Note : (α1, α2, (α1 + 1)(α2 + 1)α3 + α3, α2(α3 + 1)α4 + α4) is called the descriptor function of ϕ.

Proof :
Let the theorem be true for n − 1 and [ϕ] be a 3-CNF-matrix of dimension n. There exist two
3-CNF-matrices [ϕ1] and [ϕ2] of size n − 1 such that :
[ϕ] =
αi ∈{0,1}
x1 x2 · · · xn
0 f2(α2) · · · fn(α2, · · · , αn)
αi ∈{0,1}
x1 x2 · · · xn
1 g2(α2) · · · gn(α2, · · · , αn)
Thus
[ϕ] =
αi ∈{0,1}
x1 · · · xn
h1(α1) · · · hn(α1, · · · , αn)
where
h1(α1) = α1
hi (α1, · · · , αi ) = (α1 + 1)fi (α2, · · · , αi ) + α1gi (α2, · · · , αi ) (mod 2) for i = 1

An algorithm for the descriptor function of ϕ ∧ ψ
Let fi (α1, · · · , αi ), gi (α1, · · · , αi ) and hi (α1, · · · , αi ) be the respective descriptor functions for
ϕ, ψ and ϕ ∧ ψ. There is an impossibility of merging ϕ and ψ for some αt when
ft (·, αt ) = gt (·, αt ) (∗).
Then ht (α1, · · · , αt ) = (αt + 1) · { [ft (·, 0) + gt (·, 0)] · [ft (·, 1) · gt (·, 1)]
+ [ft (·, 0) · gt (·, 0)] } +
αt · { [ft (·, 1) + gt (·, 1)] · [ft (·, 0) + gt (·, 0)] +
[ft (·, 1) + gt (·, 1)] · [ft (·, 0) · gt (·, 0)]
+ [ft (·, 1) · gt (·, 1)] }
if [ft (·, 0) + gt (·, 0)] · [ft (·, 1) + gt (·, 1)] = 0
= αt otherwise (∗) [as ft () + gt () = 1 and ft () · gt () = 0]
but then gj (·, αj ) → g
∗
j (·, αj ) ≡ gj (·, αj ) + [ft (·, 0) + gt (·, 0)] · [ft (·, 1) + gt (·, 1)]
[with [ft (·, 0) + gt (·, 0)] · [ft (·, 1) + gt (·, 1)] = function(·, αj ) = 1]
The impossibility (*) of merging ft () and gt () yields to a new descriptor function for some αj , j < t.
And gj (·, αj ) → gj (·, αj ) + 1 bypasses the value of αj giving the incompatibility between ϕ and ψ. The
algorithm stops if both values for some αi should be bypassed : ⇒ hi (αi ) = ∅ ⇔ ϕ ∧ ψ = ∅

General complexity for functional descriptors
Complexity theorem over solutions listing
Theorem :
The complexity of the functional descriptor approach for a 3-CNF-SAT
problem with m clauses and n propositional variables is
O(m n2
max1≤t≤n max1≤j≤m lenj (ht) )
where lenj (ht) is the number of terms in ht(·) when the j ﬁrst clauses
are considered.
Note : lenj (ht ) highly depends on the order for considering the clauses.
0

5000

10000

15000

20000

25000

30000

0
20
40
60
80
100
120

3-‐CNF-‐Clauses

len(h_t)
for
20
variables
and
120
clauses

h_1

h_2

h_18

h_19

h_20

0

20

40

60

80

100

120

140

160

0
10
20
30
40
50
60
70
80
90
100

3-‐CNF
Clauses

len(h_t)
for
20
variables
and
120
clauses

h_9

h_10

h_11

h_12

h_13

h_14

h_15

h_16

h_17

h_18

h_19

h_20

Complexity for the same dataset before and after a sorting algorithm.

Sub-problems with limited maxj lenj (ht)
lenj (ht) can be approximated via a easy algorithm. Going through
the original problem ϕ, one can ﬁlter the clauses so that a
sub-problem with a limited maxj lenj (ht) is found.
0"
500"
1000"
1500"
2000"
2500"
3000"
3500"
4000"
0" 500" 1000" 1500" 2000" 2500" 3000" 3500" 4000" 4500" 5000"
20#000#clauses#with#5#000#proposi2onal#variables#
##W(xt)#
Filter of the clauses via the approximation of W (xt ) = log2(lenj (ht )).

Comparison between predicted (green line) and exact lenht () for 50 variables.
So we have a P sub-problem, and we can have many such
sub-problems. These problems normally get many solutions,
including all the solutions of the original problem.

Complexity theorem for listing the solutions of a 3-CNF ϕ
We consider a 3-CNF problem ϕ with n propositional variables. We
suppose Hϕ is computed and available. Let Σϕ = # Sϕ be the
number of solutions for ϕ. Then the complexity needed to list all
Σϕ solutions from Hϕ is O(2 n Σϕ).
Let Sϕ = {¯s1, · · · ,¯sΣϕ } be the set of solutions with
¯sj = (s1
j , · · · , sn
j ) and si
j ∈ {0, 1}. We can describe the solutions as
leafs of a tree.

No solution (1,· · · )h1(α1) ≡ 0
h2(0, α2) = 1
h3(0, 1, α3) ≡ 1
¯s3 = (0, 1, 1)
No solution (0,1,0,· · · )
h2(0, α2) = 0
h3(0, 0, α3) = 1
¯s2 = (0, 0, 1)
h3(0, 0, α3) = 0
¯s1 = (0, 0, 0)
Tree representation of the solutions for ϕ

Proof : For each node, one needs to ﬁnd whether
Im ht(· · · , αt) = {0, 1}, {0} or {1}, where “ · · · ” represents the branch
to the node. This takes O(2 × # nodes) operations. Each solution
corresponds to a leaf of the tree, and the branch to it contains n nodes.
So, the maximal number of nodes for Σϕ solutions is n × Σϕ. Therefore,
the complexity for listing the solutions of a 3-CNF problem ϕ is
O(2 n Σϕ).
So, for each P sub-problem, one can list the solutions. This will
take polynomial time if the sub-problem is “hard”, in the sense
that :
Σϕ = O(nK
) for some constant K
Note : the problem is said to be “hard” in the sense that the probability
to get a solution at random [=
Σϕ
2n ] tends to zero as n tends to inﬁnity.
The hardiest 3-CNF-SAT problems are the one without solution.

• If the sub-problem ϕ∗ found is a “hard” problem, the 3-CNF-SAT
problem ϕ is solved in polynomial time as we have just to validate
the Σϕ∗ ∈ O(nK ) solutions [which takes a polynomial time as ϕ is
NP ] to get all the solutions of ϕ (if there exists any).
• If no “hard” sub-problem can be found, merge the descriptor
functions of non “hard” sub-problems, retaining from each only the
lines hi (·) with the highest number of coeﬃcients. This represents
the maximal complexity in terms of constraint for the variable xi .
This new descriptor mixed function Hϕ∗ still corresponds to a
sub-problem of ϕ. We still have to prove that ϕ∗ is “hard”.
[Proof : As Hϕ∗ is composed of the joining of horizontal parts of the previous trees, any solution of ϕ
will be a branch of it, as the branch corresponding to this solution is common to all trees and then to all
horizontal parts of them]

Thank you for your attention and remarks.

SMART Seminar Series: "A polynomial algorithm to solve hard np 3 cnf-sat problems"

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