1. Gábor Kusper University of Linz RISC Austria
Proving by Assignment Trees that SAT Solvers are Non-Polynomial in Unit Propagation Framework with 1 Selection and Cache
Gábor Kusper
University of Linz
RISC
Austria
2004

2. Outline Representations of SAT Unit Propagation and Resolution
DPLL and GUS
Unit Propagation Framework

3. Motivation Theorema, http://www.theorema.org/ A1, ..., An G.
A1, ..., An, G is unsatisfiable.

4. Boolean Satisfiability (SAT)
Identify truth assignment that satisfies boolean formula or prove it does not exist.
SAT: Boolean Satisfiability if the boolean formula is in CNF form.
Well-known NP-complete problem.

5. Conjunctive Normal Form (CNF)
Clause
F = ( a  c )  ( b  c )  (¬a  ¬b  ¬c )
Negative Literal
Pozitive Literal

6. Representations F = ( a  c )  ( b  c )  (¬a  ¬b  ¬c )
S = { { a, c }, { b, c }, {¬a, ¬b, ¬c } }
+ x +
M = x + +
True
True
True a b c
False
False
False

7. Clause Classification
x - +
x x x
x - x
+ - +
2-Clause
Empty Clause
Unit Clause
Clear Clause

8. Clause Set Classification
x - +
x x x
x - x
+ - +
Unsatisfiable Clause Set
Empty Clause Set,
Satisfiable

9. Resolution Unit Propagation
Simplifying SAT
Resolution
Unit Propagation

10. Resolution if A, B are clauses and a is a literal, then
res(A, B, a) := A  B \ {a, ¬a}.
- + + x
+ + x +
x + + +
res(A, B, a) A B a

11. Unit Propagation (UP) If S is a clause set and a is a literal, then
UP(S, { a }) := { C \ { a } | C  S  a  C }.
x - +
- - x
+ x x
+ + +
Unit Clause
x - x
UP by +xx

12. SAT Solvers
DPLL
GUS

13. BCP
Boolean Constraint Propagation (BCP): Iterated application of unit propagation.
x - +
- - x
+ x x
+ + +
x - x
UP by +xx
UP by x-x

14. DPLL Procedure Davis & Putnam & Logemann & Loveland Procedure
Function DPLL(S : Clause Set) : Boolean
Begin
S := BCP(S);
if S =  then return True;
if   S then return False;
Let a be a variable in S;
return DPLL( S  { { a } } )  DPLL( S  { { ¬a } } );
End

15. Example - - x x + + - x - + x x - x x - - x - - x x + + x + + - x -

16. Assignment We say for short
clause disjunctive set of literals.
assignment conjunctive set of literals.
The intended meaning of assignment { a, b } is that we assign true to a and false to b.
If C is a clause, then C is an assignment.
If A is an assignment, then A is a clause.
If S is a clause set, then S is an assignment set.
If T is an assignment set, then T is a clause set.

17. Hyper-Unit Propagation (HUP)
If S is a clause set and A is an assignment, then
HUP(S, A) := { C \ A | C  S  C  A =  }.
x - +
- - x
+ + +
x x +
x x x
HUP by ++x
x - +
- - x
+ + +
x x +
x x x
x - +
x - x
UP by +xx
UP by x+x

18. Sub-Model (sm) If C is a clause and c is a literal in C, then
sm(C, c) := C \ { c }  { c }.
sm(C, c) is an assignment.
(a  b  c)  (a  b)  c.
+ + + x
- - + x
sm(C, c) C c

19. Expansion Rule If HUP(S, A) is unsat, then {A}  S  S.
If C  S, c  C and HUP(S, sm(C, c)) is unsat, then {sm(C, c)}  S  S, i.e.,
{C \ {c}}  S  S,
because res(C, sm(C, c)) = C \ {c}.

20. GUS General Unicorn-SAT(GUS) Function GUS(S : Clause Set) : Boolean
Begin
if S =  then return True;
Let C be a minimal clause in S;
For each c  C do
If ( GUS( HUP(S, sm(C, c) ) ) ) then return True; od
return False;
End

21. Example - - x x + + - x - + - x x - x x x + x x + x x - - x - x x +

22. Unit Propagation Framework
Unit Propagation Framework with 1 Selection and Cache
Limitation Lemma

23. Strategy Set & Function
If C is a clause and T is an assignment set, then
T is a strategy set generated by C,
if { C }  T is unsatisfiable.
If C is a clause and F is a function, that has the type
“function F( C : clause) : assignment set”, and
F(C) is a strategy set generated by C, then
F is a strategy function.

24. Example x x x x + x x x C C T T + + + + + + + + C - - - + + x x x C

25. Strategy Function of DPLL
Function DPLL_STF( C : Clause ) : Assignment Set
Begin
if ( C is empty ) then return ;
if ( C is unit ) then return { C };
Let c be a literal in C;
return { { c }, { c } };
End

26. Example
x x x x
+ x x x
+x x x C T C T
+ x x x
- x x x C T

27. Strategy Function of GUS
Function GUS_STF( C : Clause ) : Assignment Set
Begin
T := ;
While (C is not empty) do
Let c be a literal in C;
T := T  { sm(C, c) };
C := C \ { c }; od
return T;
End

28. Example x x x x + x x x C C T T + + + + C - - - + - - + x - + x x

29. UPFw1S
Function UPFw1S(S : Clause Set, STF : Strategy Function) : Boolean
Begin
if ( S =  ) then return True;
if (   S ) then return False;
Let C be a clause in S; // 1 Selection
T := STF ( C ); // T is a strategy set
For each A  T do
if ( UPFw1S( HUP(S, A), STF ) ) then return True; od
return False;
End

30. UPFw1S is a SAT solver S S1 C S2 A1 A2 Ak Sk HUP by A1 HUP by A2
We know that if T is a strategy set generated by C,
then { C }  T is unsatisfiable.
Assume T = { A1, A2, ..., Ak } is strategy set generated by C  S.
HUP by A1 S
HUP by A2 S1
HUP by Ak C S2
A1
A2
Ak Sk

31. Simulated DPLL & GUS
Function Simulated_DPLL( S : Clause Set) : Assignment
Begin
return UPFw1S(S, DPLL_STF);
End
Function Simulated_GUS( S : Clause Set) : Assignment
return UPFw1S(S, GUS_STF);

32. k-Clause Set
[2^k] := the set of all possible k-clauses on the same variables.
For all k : [2^k] is unsatisfiable.
Assume we have 3 variables.
+ + x
[4]= + - x
- + x
- - x
x + +
[4]= x + -
x - +
x - -
+ x +
[4]= + x -
- x +
- x - or
[1]= x x x

33. HUP on k-Clause Set HUP by ++x HUP by ++x
HUP([2^k], A) = [2^(k-m)], where m = |A|.
HUP by ++x
[8]
[2]
+ + +
+ + -
+ - +
[8]= + - -
- + +
- + -
- - +
- - -
HUP by ++x
x x + =[2]
x x -

34. DPLL and GUS on [2] [2] [1] + - [2] [1] + - HUP by + DPLL: UP steps: 1

35. DPLL and GUS on [4] [2] [4] ++ -x [2] +x [1] [4] ++ +- [2] -x
HUP by +x
[2] 1
[4]
DPLL:
HUP by -x ++
UP steps: 4 -x
[2] 1 +x
HUP by -+
[1]
[4]
GUS:
HUP by +x
UP steps: 4 ++ +-
[2] 1 -x

36. DPLL and GUS on [8] [4] [8] +++ -xx [4] +xx [1] [8] +++ [2] ++- +-x
HUP by +xx
[4] 4
[8]
DPLL:
HUP by -xx
+++
UPs: 10
-xx
[4] 4
+xx
HUP by --+
[1]
[8]
GUS:
HUP by -+x
HUP by +xx
+++
UPs: 11 -1
[2] 1
++-
+-x
[4] 4
-xx

37. Limit for Strategy Sets
If T is strategy set generated by C and C is a k-clause, then T subsumes at least 2^k – 1 k-clauses on the variables of C.
This holds, because { C }  T is unsatisfiable.
T = - + x = - + +
- x x
+ + +
+ + -
+ - +
+ - -
+ + +
- - +
- + x
+ x x C T

38. Limit for Children [1] [8] +++ [2] ++- +-x [4] -xx HUP by --+
If we break down [2^k] by a strategy set to
[2^k1], [2^k2], ..., [2^km] then
2^k1 + 2^k ^km  2^k – 1.
This holds, because of Limit for Strategy Sets.
HUP by --+
[1]
[8]
HUP by -+x
HUP by +xx
+++
[2]
++-
+-x
[4]
-xx

39. Assignment Tree (A-tree)
A, Cs is an A-tree if A is an assignment and Cs is a set of A-trees.
We can represent a run of UPFw1S by an A-tree.
xxx
HUP by --+
[1]
[8]
--+
-+x
+xx
HUP by -+x
HUP by +xx
+++
[2]
++-
xx+
x-+
x+x
+-x
[4]
-xx
xx+

40. Unit Assignment Tree xxx xxx --+ -+x +xx -xx -xx +xx xx+ x-+ x+x x-x
(a)  , Cs is a unit A-tree if Cs is a set of unit A-trees.
(b)  A, Cs is a unit A-tree if A is a unit and
Cs is a set of unit A-trees.
convert to
unit A-tree
xxx
xxx
--+
-+x
+xx
-xx
-xx
+xx
xx+
x-+
x+x
x-x
x+x
x-x
x+x
xx+
xx+
xx+
xx+
xx+

41. Cached Unit Assignment Tree
xxx
convert to
cached unit A-tree
-xx
-xx
+xx
xxx
x-x
x+x
x-x
x+x
-xx
+xx
xx+
xx+
xx+
xx+
x-x
x+x
x-x
x+x
xx+
xx+
xx+
xx+

42. Auxiliary Lemmas
The number of edges of a cached unit A-tree equals to the number of used UP steps in the run of UPFw1SC, if this cached unit A-tree corresponds to this run of the framework.
A well-know graph theory states that if a tree has N nodes, then it has N-1 edges.

43. Number of Nodes of a CUA-Tree
Let AT be a A-tree, which represent a run of UPFw1SC on [2^n], n > 0.
We convert AT to a cached unit A-tree. Let CUAT this cached unit A-tree.
Then CUAT has 2^n + 2^(n-1) – 1 nodes.
Proof by Induction:
For k = 1 is true, because [2] has 2 nodes.
Let us assume that it holds for k = n-1.
We show that it holds for k = n. We assume that n > 1.
The child nodes of root correspond to a run of UPFw1SC on [2^(n-1)].
Since n > 1, the root has at least two child nodes.
Hence, it has at least 2 * (2^(n-1)+2^(n-2) – 1) + 1 nodes.

44. Limitation Lemma
There is no SAT solver algorithm in the Unit Propagation Framework with 1 Selection and Cache, which can show that [2^n] is unsatisfiable using fewer UP steps than 2^n + 2^(n-1) - 2.
It holds, because each cache unit A-tree, which correspond to a run of this framework on [2^n] has at least 2^n + 2^(n-1) – 1 nodes.

45. Hence
Therefore, each SAT solver in this framework uses at least 2^n + 2^(n-1) – 2 UP steps to show that [2^n] is unsatisfiable.
Hence, SAT solvers are non-polynomial in Unit Propagation Framework with 1 Selection and Cache.

46. Thank you for your attention!
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