Satisfiability Checking
The Omega Test
Prof. Dr. Erika Ábrahám
RWTH Aachen University
Informatik 2
LuFG Theory of Hybrid Systems
WS 14/15
The Omega test
Goal: Decide satisfiability for conjunctions of linear constraints of the
form
X
ai xi ≥ b
0≤i≤n
over integers.
Original application:
Program optimizations done by a compiler.
Extension of Fourier-Motzkin variable elimination:
Pick one variable and eliminate it.
Continue until all variables but one are eliminated.
Satisfiability Checking — Prof. Dr. Erika Ábrahám (RWTH Aachen University)
WS 14/15
2 / 16
Overview of the Omega test
?
Check
REAL shadow
No
solution? UNSAT
Possible solution
?
Check
DARK shadow
Integer - SAT
solution?
No integer solution
?
Check
GREY shadow
Integer - SAT
solution?
No integer solution
?
UNSAT
Satisfiability Checking — Prof. Dr. Erika Ábrahám (RWTH Aachen University)
WS 14/15
3 / 16
The real shadow
Check
REAL shadow
Assume we eliminate variable z
For each pair of upper/lower bound:
β ≤ bz
cβ ≤ cbz
cz
cbz
≤ γ (b, c > 0)
≤ bγ
Constraint for real shadow:
cβ ≤ bγ
Satisfiability Checking — Prof. Dr. Erika Ábrahám (RWTH Aachen University)
WS 14/15
4 / 16
The real shadow: Example I
Check
REAL shadow
Eliminate x:
4y ≤ 2x
1.5
1.25
1
0.75
0.5
0.25
!0.25
y
4y ≤ −2x + 6
@
R
@
4y " 2x
4y " 6!2x
4y # 1
1.5 2
4y ≤ 6 − 4y
8y ≤ 6
x
3.5
4y ≤ 2x
Real Shadow:
8y ≤ 6
4y ≥ 1
4y ≤ −2x + 6
No integer solution
0.5
1
4y ≥ 1
2.5
3
y ≤ 0.75
y ≥ 0.25
=⇒ Original problem
has no solution
Satisfiability Checking — Prof. Dr. Erika Ábrahám (RWTH Aachen University)
WS 14/15
5 / 16
The real shadow: Example II
Let’s eliminate y instead:
Check
REAL shadow
1 ≤ 4y
H
1.5
1.25
1
0.75
0.5
0.25
!0.25
4y ≤ 2x
j
H
1 ≤ 2x
y
1 ≤ 4y
4y " 2x
0.5
1
4y " 6!2x
4y # 1
1.5 2
2.5
3
H
x
3.5
4y ≤ −2x + 6
j
H
1 ≤ −2x + 6
4y ≤ −2x + 6
Real Shadow:
1 ≤ 2x
1 ≤ −2x + 6
4y ≥ 1
Integer solution!
4y ≤ 2x
x ≥ 0.5
x ≤ 2.5
But original problem
has no integer solution!
Satisfiability Checking — Prof. Dr. Erika Ábrahám (RWTH Aachen University)
WS 14/15
6 / 16
From real to dark shadow
?
Check
REAL shadow
No
solution? UNSAT
Possible solution
?
Check
DARK shadow
An integer solution for the REAL shadow does not guarantee that
there is an integer solution for the original problem.
Thus, we check the DARK shadow next.
Satisfiability Checking — Prof. Dr. Erika Ábrahám (RWTH Aachen University)
WS 14/15
7 / 16
Idea of the dark shadow
Check
DARK shadow
Idea of the DARK shadow:
β
β
b
≤ bz
≤ z
|:b
cz
z
≤
≤
γ |:c
γ
c
z ∈N
How to compute the dark shadow?
Try to prove that there is an integer z between
Satisfiability Checking — Prof. Dr. Erika Ábrahám (RWTH Aachen University)
β
b
and γc .
WS 14/15
8 / 16
Dark shadow: Proof by contradiction
Check
DARK shadow
Assume there is no integer z between
Let i := b βb c
i ∈Z
β
i
b
| {z }
≥ b1
β
b
and γc . Then:
i +1
| {z }
γ
c
-
≥ c1
Satisfiability Checking — Prof. Dr. Erika Ábrahám (RWTH Aachen University)
WS 14/15
9 / 16
Dark shadow: Proof by contradiction
β
i
b
| {z }
≥ b1
β
b
−i
i + 1 − γc
β
b
+1−
γ
c
≥
≥
≥
1
b
1
c
1
b
+
1
c
cβ + cb − bγ ≥ c + b
cβ − bγ ≥ −cb + c + b
i +1
| {z }
γ
c
-
≥ c1
|·c ·b
| − cb
| · (−1)
bγ − cβ ≤ cb − c − b
Satisfiability Checking — Prof. Dr. Erika Ábrahám (RWTH Aachen University)
WS 14/15
10 / 16
Dark shadow: Proof by contradiction
From previous slide:
↔
↔
↔
bγ − cβ
¬(bγ − cβ
¬(bγ − cβ
¬(bγ − cβ
|
≤
>
≥
≥
cb − c − b
cb − c − b)
cb − c − b + 1)
(c − 1)(b − 1))
{z
}
*
Thus, if * holds, we know that there must be an integer solution.
If c = 1 or b = 1, then this is the same as the real shadow.
This case is called an exact projection.
Satisfiability Checking — Prof. Dr. Erika Ábrahám (RWTH Aachen University)
WS 14/15
11 / 16
Example for the dark shadow
Check
DARK shadow
Eliminate y with the dark shadow:
2y ≤ 2x + 1
y
1.5
2y " 1#2x
4y ≥ 3
Hj
H
2y " 5!2x
4(2x + 1) − 2 · 3 ≥ (2 − 1)(4 − 1)
1.25
1
0.75
4y ≥ 3
4y $ 3
0.5
HH
j
2y ≤ −2x + 5
4(−2x + 5) − 2 · 3 ≥ (2 − 1)(4 − 1)
0.25
0.5
1
1.5
2
x
2y ≤ −2x + 5
Dark Shadow:
x
x
4y ≥ 3
=⇒ Integer solution!
!0.25
2y ≤ 2x + 1
Satisfiability Checking — Prof. Dr. Erika Ábrahám (RWTH Aachen University)
≥ 5/8
≤ 11/8
WS 14/15
12 / 16
From dark to grey shadow
?
Check
REAL shadow
No
solution? UNSAT
Possible solution
?
Check
DARK shadow
Integer - SAT
solution?
No integer solution
?
Check
GREY shadow
No integer solution in the DARK shadow does not guarantee that
there is no integer solution for the original problem.
Thus, we check the GREY shadow next.
Satisfiability Checking — Prof. Dr. Erika Ábrahám (RWTH Aachen University)
WS 14/15
13 / 16
The grey shadow
Check
GREY shadow
Idea of the Grey shadow
If the real shadow R has integer solutions,
but the dark shadow D does not, search R \ D.
In R:
Not in D:
↔
⇒
bγ ≥ cbz ≥ cβ
cb − c − b ≥ bγ − cβ
cb − c − b + cβ ≥ bγ
cb − c − b + cβ ≥ cbz ≥ cβ
(cb − c − b)/c + β ≥ bz ≥ β
Satisfiability Checking — Prof. Dr. Erika Ábrahám (RWTH Aachen University)
|:c
WS 14/15
14 / 16
The grey shadow
Check
GREY shadow
Try all values of z such that
(cb − c − b)/c + β ≥ bz ≥ β
Optimization: find the largest coefficient c in any upper bound and try
the following for each lower bound bz ≥ β:
bz = β + i
for (cb − c − b)/c ≥ i ≥ 0
As before, combine this with the original problem, and solve
recursively.
Satisfiability Checking — Prof. Dr. Erika Ábrahám (RWTH Aachen University)
WS 14/15
15 / 16
Example of the grey shadow
Eliminate y :
c = 4, b = 4, β = 3
Check
GREY shadow
y
1.75
1.5 4 y " 2 # 3 x
4 y " 11 ! 3 x
1.25
1
0.75
4y $ 3
0.5
0.25
!0.25
0.5
1
1.5
2
2.5
4y ≤ 3x + 2
4y ≤ −3x + 11
3
x
New constraint:
4y = 3 + i for
2 ≥ i ≥ 0:
4y = 3
4y = 4
4y = 5
=⇒
Integer solution
with 4y = 4
4y ≥ 3
Satisfiability Checking — Prof. Dr. Erika Ábrahám (RWTH Aachen University)
WS 14/15
16 / 16