Conflict-driven cutting-plane learning
The devil is in the details
Jo Devriendt
jodevriendt.com
Acknowledgment:
MIAO group at Lund/Copenhagen
slides: slides.com/jod/cdcpl-devil
How to solve?
"Generate-and-test" search tree...
...maybe we get lucky with good heuristics?
- variables
- values
- find a variable-value assignment that satisfies a number of constraints
Combinatorial search
Combinatorial search
Suppose the problem is UNSATisfiable:
- heuristics don't help
- strong pruning (propagation) helps, but often kicks in only after deciding on many variables
How to derive that no candidate solution satisfies all constraints?
Proof construction
x + y - z \geq 1 \\
w - y + z \geq 1 \\
-x -w \geq -1 \\
To show UNSAT of above constraints,
just add all of them together.
0 \geq 1
No need to search over candidate solutions!
Dual reasoning
besides searching over variable-value assignments,
search over constraint combinations
Preliminaries
- Boolean variables: take value \(0\) or \(1\)
- Negated variables: $$\overline{x}=1-x \text{ or } \overline{x}+x=1$$
- variable or its negation is a literal
- Linear inequalities (positive coefficients): $$x+2\overline{y}+3z \geq 4$$
- Propositional clause: $$x \vee \overline{y} \vee z$$ is equivalent to $$x+\overline{y}+z \geq 1$$
x + y + z \geq 1\\
\overline{x} + y + z \geq 1\\
y + \overline{z} \geq 1 \\
\overline{y} + z\geq 1 \\
\overline{y} + \overline{z}\geq 1 \\
\(z\)
\(\overline{z}\)
Naive SAT solver
- Depth-first search
- Unit propagation:
propagate single non-falsified literal in clause
\(x\)
\(y\)
\(\overline{y}\)
\(\overline{z}\)
\(z\)
x + y + z \geq 1\\
\overline{x} + y + z \geq 1\\
y + \overline{z} \geq 1 \\
\overline{y} + z\geq 1 \\
\overline{y} + \overline{z}\geq 1 \\
Naive SAT solver
- Depth-first search
- Unit propagation:
propagate single non-falsified literal in clause
\(x\)
\(y\)
\(z\)
\(\overline{z}\)
\(\overline{x}\)
\(y\)
\(\overline{y}\)
x + y + z \geq 1\\
\overline{x} + y + z \geq 1\\
y + \overline{z} \geq 1 \\
\overline{y} + z\geq 1 \\
\overline{y} + \overline{z}\geq 1 \\
Naive SAT solver
- Depth-first search
- Unit propagation:
propagate single non-falsified literal in clause
\(\overline{y}\)
\(\overline{z}\)
\(z\)
\(x\)
\(y\)
\(z\)
\(\overline{z}\)
x + y + z \geq 1\\
\overline{x} + y + z \geq 1\\
y + \overline{z} \geq 1 \\
\overline{y} + z\geq 1 \\
\overline{y} + \overline{z}\geq 1 \\
Preliminaries
Decision literals
Search trail: the current assignment, as a list of literals in chronological order, paired with their decision status
Reason constraint
| d | d | p |
\overline{y}
\overline{z}
x
Conflict constraint
Propagated literal
\(\overline{x}\)
\(y\)
\(\overline{y}\)
\(\overline{y}\)
\(\overline{z}\)
\(z\)
\(x\)
\(y\)
\(z\)
\(\overline{z}\)
Conflict-driven clause learning
- Depth-first search
- Unit propagation:
propagate last non-falsified variable in clause - Clause learning:
- from each conflict, get the conflict clause
- iterate backwards over the trail
- add the reason to the conflict
- stop when the conflict clause propagates at a previous decision point, add the current conflict as a learned clause
- backtrack to the propagating decision point, continue search
Conflict-driven clause learning
\(x\)
\(y\)
\(z\)
\(\overline{z}\)
\textcolor{red}{\overline{y}\geq 1}
x + y + z \geq 1\\
\overline{x} + y + z \geq 1\\
y + \overline{z} \geq 1 \\
\overline{y} + z\geq 1 \\
\overline{y} + \overline{z}\geq 1 \\
z
p
y
d
x
d
- from each conflict, get the conflict clause
- iterate backwards over the trail
- add the reason to the conflict
- stop when the conflict clause propagates at a previous decision point, add the current conflict as a learned clause
- backtrack to the propagating decision point, continue search
\(\overline{y}\)
\(\overline{z}\)
\(x\)
\(\overline{x}\)
\overline{y} \geq 1
y \geq 1
y + z \geq 1
0 \geq 1
Conflict-driven clause learning
x + y + z \geq 1\\
\overline{x} + y + z \geq 1\\
y + \overline{z} \geq 1 \\
\overline{y} + z\geq 1 \\
\overline{y} + \overline{z}\geq 1 \\
x
p
\overline{z}
p
\overline{y}
p
- from each conflict, get the conflict clause
- iterate backwards over the trail
- add the reason to the conflict
- stop when the conflict clause propagates at a previous decision point, add the current conflict as a learned clause
- backtrack to the propagating decision point, continue search
Crucial invariants
- addition eliminates propagated literal from conflict
- at every step, conflict is falsified by trail
(contains only false literals)
\frac{z+X+W\geq 1 ~~~~~ \overline{z}+ Y +W \geq 1}{X+Y+W \geq 1}
\frac{z+\overline{x} \geq 1 ~~~~~ \overline{z}+ y \geq 1}{\overline{x}+y\geq 1}
"add" the clauses
\frac{z+\overline{x}+\textcolor{red}{w} \geq 1 ~~~~~ \overline{z}+ y+\textcolor{red}{w} \geq 1}{\overline{x}+y+\textcolor{red}{2}w\geq 1}
\frac{}{\lceil\frac{\overline{x}+y+2w}{2}\rceil\geq \lceil\frac{1}{2}\rceil}
\frac{~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~}{\overline{x}+y+\textcolor{red}{w}\geq 1}
Resolution rule
In case of double literals:
Divide lhs and rhs by 2, ceil:
\(\overline{y}\)
\(\overline{z}\)
\(x\)
\(\overline{x}\)
\overline{y} \geq 1
y \geq 1
y + z \geq 1
0 \geq 1
Conflict-driven clause learning
x + y + z \geq 1\\
\overline{x} + y + z \geq 1\\
y + \overline{z} \geq 1 \\
\overline{y} + z\geq 1 \\
\overline{y} + \overline{z}\geq 1 \\
x
p
\overline{z}
p
\overline{y}
p
- every "addition" is an application of resolution rule
- all learned clauses form a resolution proof
Automatic proof construction (dual reasoning) during search!
Conflict-driven clause learning
Crucial invariants
- addition eliminates propagated literal from conflict
- at every step, conflict is falsified by trail
Mid-presentation summary
- from search conflicts, derive constraints
- apply resolution rule to conflict clause and reason clauses
- build resolution proofs
Can we do better?
- Resolution is not only proof system
- Field of proof complexity
- given an unsatisfiable formula, what is the shortest proof possible?
- Resolution is the weakest of all proof systems!
- there exist simple formulas where smallest resolution proof is exponential [H85]
- other proof systems are polynomial
- lower bound to solve time!
- E.g., cutting plane (CP) proof system
Cutting plane proofs
Rules:
- Boolean axiom
- Addition
- Multiplication
- Division
- Trivial inconsistency:
\frac{}{x \geq 0} \;\;\;\; \frac{}{\overline{x} \geq 0}
\frac{
\sum_i a_i x_i \geq b \;\;\;\; c \in \mathbb{N}^+
}
{
\sum_i ca_i x_i \geq cb
}
\frac{
\sum_i a_i x_i \geq b \;\;\;\; \sum_i c_i x_i \geq d
}
{
\sum_i (a_i+c_i) x_i \geq b+d
}
\frac{
\sum_i a_i x_i \geq b \;\;\;\; c \in \mathbb{N}^+
}
{
\sum_i \lceil\frac{a_i}{c}\rceil x_i \geq \lceil \frac{b}{c} \rceil
}
0 \geq 1
Can simulate resolution rule
Conflict-driven cutting-plane learning
Can we construct cutting plane proof during search, similar to resolution via CDCL?
- add reason constraints to conflict constraint
- eliminating propagated variables
- keeping constraint falsified by trail
- add learned constraint to continue search
Let's see...
- Input: 0-1 linear constraints
- Depth-first search
- with decisions and propagated literals as a trail
Conflict-driven cutting-plane learning
| d | d | d |
\overline{y}
v
\overline{x}
2\overline{y}+2z+\overline{v} \geq 3\\
Slack
Given a trail, the slack of the constraint is the difference between the lhs and the rhs (assuming unknown literals as true)
\textcolor{red}{2}+2z+\textcolor{red}{0} \geq 3\\
slack =
2+2+0-3 =
1
- slack<0 means constraint is conflicting
- all unknown literals with coefficient greater than slack propagate to true
Conflict-driven cutting-plane learning
x+y+2z \geq 2\\
w+y+2\overline{z} \geq 2\\
Problem
\textcolor{red}{2}\\
\textcolor{red}{2}\\
constraints
slack
trail
x
d
w
d
\overline{y}
d
Conflict-driven cutting-plane learning
x+y+2z \geq 2\\
w+y+2\overline{z} \geq 2\\
Problem
\textcolor{red}{1}\\
\textcolor{red}{1}\\
constraints
slack
trail
x
d
w
d
\overline{y}
d
z
p
Conflict-driven cutting-plane learning
x+y+2z \geq 2\\
w+y+2\overline{z} \geq 2\\
Problem
\textcolor{red}{1}\\
\textcolor{red}{-1}\\
constraints
slack
trail
x
d
w
d
\overline{y}
d
z
p
- add reason constraint to conflict
- use division/multiplication to eliminate propagated literal
- keep constraint conflicting (slack<0)
w+x+2y \geq 2\\
\textcolor{red}{0}\\
Slack is sub-additive: slack of summed constraints is at most sum of slacks
Can we get the reason constraint to slack 0 with the coefficient of the propagated variable identical to the conflict coefficient?
Conflict-driven cutting-plane learning
x+y+2z \geq 2\\
w+y+2\overline{z} \geq 2\\
round-to-one [NE18]
\textcolor{red}{1}\\
\textcolor{red}{-1}\\
constraints
slack
trail
x
d
w
d
\overline{y}
d
z
p
- adding variable axioms \(x \geq 0\) for falsified literal \(x\) does not change slack
- adding variable axioms \(\overline{x} \geq 0\) for non-falsified literal \(x\) does not change slack
- dividing by a common divisor \(d\) of all coefficients yields \(\lfloor\frac{slack}{d}\rfloor\)
- slack is always lower than propagating coefficient
Divide by propagating coefficient, after ensuring all coefficients are divisible
Conflict-driven cutting-plane learning
x+y+2z \geq 2\\
w+y+2\overline{z} \geq 2\\
round-to-one [NE18]
\textcolor{red}{1}\\
\textcolor{red}{-1}\\
constraints
slack
trail
x
d
w
d
\overline{y}
d
z
p
Divide by propagating coefficient, after ensuring all coefficients are divisible
1) non-falsified: add \(\overline{x}\geq 0\)
2) falsified: add \(y \geq 0\)
divisor
2y+2z \geq 1\\
y+z \geq 1\\
3) divide by 2
4) multiply with conflict coefficient
2y+2z \geq 2\\
5) add to conflict
w+3y \geq 2\\
\textcolor{red}{-1!}\\
\textcolor{red}{1}\\
\textcolor{red}{0}\\
\textcolor{red}{0}\\
Conflict-driven cutting-plane learning
x+y+2z \geq 2\\
w+y+2\overline{z} \geq 2\\
\textcolor{red}{1}\\
\textcolor{red}{-1}\\
constraints
slack
trail
x
d
w
d
\overline{y}
d
z
p
1) non-falsified: add \(\overline{x}\geq 0\)
2) falsified: add \(y \geq 0\)
divisor
2y+2z \geq 1\\
y+z \geq 1\\
3) divide by 2
4) multiply with conflict coefficient
2y+2z \geq 2\\
5) add to conflict
w+3y \geq 2\\
\frac{}{x \geq 0} \;\;\;\; \frac{}{\overline{x} \geq 0}
\frac{
\sum_i a_i x_i \geq b \;\;\;\; c \in \mathbb{N}^+
}
{
\sum_i ca_i x_i \geq cb
}
\frac{
\sum_i a_i x_i \geq b \;\;\;\; \sum_i c_i x_i \geq d
}
{
\sum_i (a_i+c_i) x_i \geq b+d
}
\frac{
\sum_i a_i x_i \geq b \;\;\;\; c \in \mathbb{N}^+
}
{
\sum_i \lceil\frac{a_i}{c}\rceil x_i \geq \lceil \frac{b}{c} \rceil
}
Building cutting plane proofs!
\textcolor{red}{-1!}\\
\textcolor{red}{1}\\
\textcolor{red}{0}\\
\textcolor{red}{0}\\
Conflict-driven cutting-plane learning
round-to-one [NE18]
Divide by propagating coefficient, after ensuring all coefficients are divisible, then multiply by conflict coefficient.
Rest of search routine: roughly same as CDCL
Efficient implementation: gitlab.com/JoD/exact
Applications
- Find unsatisfiability
- prove optimality of solution
- find intersection of all solutions
- prove logical consequence
- Currently used to generate explanations for constraint programs with CPMpy (faster than Google OR-tools)
- Used as backend solver for IDP-like system
- Verification of bit-level multiplier circuits [LBDEN20]
- Formal verification of clausal decomposition of constraints with VeriPB
- Can perform well when linear relaxation of problem is bad - improved a MIPLIB instance
What now?
Round-to-one is not the final answer
- Divisors different from propagating coefficient?
- yes, all divisors greater than slack should work
- How many axioms to add?
- [EN18] adds maximal amount, but rather as little as needed?
- Manipulate conflict instead of propagating constraint?
- definitely possible!
- Other sensible ways of transforming constraint?
- yes, e.g., saturation rule and mixed-integer-rounding cut
strictly stronger than regular division!
Way more degrees of freedom than in CDCL ...
... thesis topic :)
What does Exact do?
Efficient implementation: gitlab.com/JoD/exact
- Multiply reason constraint with conflict coefficient
- If propagating coefficient suffices as divisor, use that
- Else, try to find multiple of propagating coefficient that yields a divisor of conflict coefficient
- Else, round to one
- Use axiom addition sparingly
- Except when reducing small coefficients as long as \(\lceil\frac{rhs}{divisor}\rceil\) remains unchanged.
Thanks for listening!
Questions?
[H85] The intractibility of resolution - Haken
[CCT87] On the complexity of cutting-plane proofs - Cook, Coullard, Turán
[MS96] GRASP - a new search algorithm for satisfiability - Marques-Silva, Sakallah
[BS97] Using CSP lookback techniques to solve real-world SAT instances - Bayardo, Schrag
[MMZZM01] Chaff: Engineering an efficient SAT solver - Moskewicz, Madigan, Zhao, Zhang, Malik
[CK05] A fast pseudo-Boolean constraint solver - Chai, Kuehlmann
[SS06] Pueblo: A hybrid pseudo-Boolean SAT solver - Sheini, Sakallah
[LP10] The Sat4j library, release 2.2 - Le Berre, Parrain
[EN18] Divide and conquer: Towards faster pseudo-boolean solving - Elffers, Nordström
[D20] Watched Propagation for 0-1 Integer Linear Constraints - Devriendt
[DGN20] Learn to Relax: Integrating 0-1 Integer Linear Programming with Pseudo-Boolean Conflict-Driven Search - Devriendt, Gleixner, Nordström
[SDNS20] Theoretical and Experimental Results for Planning with Learned Binarized NeuralNetwork Transition Models - Say, Devriendt, Nordström, Stuckey
[LBDEN20] Verifying Properties of Bit-vector Multiplication Using Cutting Planes Reasoning - Liew, Beame, Devriendt, Elffers, Nordström
RoundingSat
[LBDEN20]
- Verification of various bit-level multiplier circuits
- All instances are unsatisfiable
Conflict-driven cutting-plane learning
By Jo Devriendt
