Download CS 497: Section EA

UIUC CS 497: Section EA
Lecture #2
Reasoning in Artificial Intelligence
Professor: Eyal Amir
Spring Semester 2004
Last Time
• Propositional logic as a language for
representing knowledge
• Did not touch on reasoning procedures
• Defined language, signature, models
• From homeworks: you should know
– Soundness; Completeness theorem
– Deduction theorem
– De Morgan Laws
Today
• Reasoning procedures for propositional
logic
– Checking Satisfiability (SAT) using DPLL
– Proving entailment using Resolution
• Application du jour: Formal Verification
• Applications we will not touch
– AI planning, graph algorithms, cryptography,
…
SAT via Generate and Test
• If we have a truth table of KB, then we can
check that KB satisfiable by looking at it.
• Problem: n propositional symbols  2n
rows in truth table
– Checking interpretation I takes time O(|KB|)
– Generating table is expensive: O(2n |KB|) time
• Observation: SAT requires us to look only
for one model
Clausal Form
• Every formula can be reformulated into an
equivalent CNF formula (conjunction of
clauses).
• Examples (using De Morgan Laws):
( a  b )  ( a  b )
 (a  a)  (a  b)  (b  a)  (b  b)
Clausal Form
• Every formula can be reformulated into an
equivalent CNF formula (conjunction of
clauses).
• Examples:
face _ door _ t1  turn _ cl _ 90 _ t1  face _ door _ t 2
face _ door _ t1  turn _ cl _ 90 _ t1  face _ door _ t 2
Clausal Form
• Every formula can be reformulated into an
equivalent CNF formula (conjunction of
clauses).
• Examples:
face _ door _ t1  move _ fwd _ t1 
at _ corridor _ t 2  face _ door _ t 2 
face _ door _ t1  move _ fwd _ t1  at _ corridor _ t 2
face _ door _ t1  move _ fwd _ t1  face _ door _ t 2
Propagating a Truth-Value
• KB in CNF, and we observe p=TRUE
• Then, removing clauses with p positive
from KB gives an equivalent theory.
• Example:
KB
Observe
abc
a  b  d
a
Propagating a Truth-Value
• KB in CNF, and we observe p=TRUE
• Then, removing negative instances of p
from KB gives an equivalent theory.
• Example:
KB
Observe
abc
a  b  d
a
DPLL Search Procedure for CNF
1. If no clauses in KB, return T
2. If a clause in KB is empty (FALSE), return F
3. If KB has a unit clause C with prop. p, then
return DPLL(KB,p←polarity(p,C))
4. Choose an uninstantiated variable p
5. If DPLL(KB, p←TRUE) returns T, return T
6. If DPLL(KB, p←FALSE) returns T, return T
7. Return F
DPLL in Action
abc
a  c  d
a  e  f
c  e  f
c  d  f
On board
DPLL in Action
abc
a  c  d
a  e  f
 c  e  f
 c  a   f
On board
Note: we could know
without thorough checking
that this KB is satisfiable
DPLL in Action
ac
a  c  f
a  e  f
c  e  f
a   f
c f
On board
Related: SAT Solving
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•
•
•
•
•
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Order of selection of variables (lecture #5)
Stochastic local search (paper #1)
Binary Decision Diagrams (paper #2)
Strategies other than unit (paper #23)
2-SAT is solvable in linear time
Smart backtracking (paper #21)
Clauses/Vars in Random SAT (paper #22)
SAT via probabilistic models (paper #15)
Take a Breath
• Until now: SAT solving
– Search in the space of models
• From now: Resolution theorem proving
– Search in the space of proofs
• Later: Formal Verification
Resolution Theorem Proving
•
Given:
– KB – a set of propositional sentences
– Query Q – a logical sentence
•
Calling procedure:
1. Add Q to KB
2. Convert KB into clausal form
3. Run theorem prover. If we prove
contradiction, return T. Otherwise, return F.
Resolution Theorem Proving
1. Add Q to KB
2. Convert KB into clausal form
3. Run theorem prover. If we prove
contradiction, return T. Otherwise, return F.
KB  Q
╨
╨
Deduction theorem:
KB
Q
iff
FALSE
Resolution Theorem Proving
1. Add Q to KB
2. Convert KB into clausal form
3. Run theorem prover. If we prove
contradiction, return T. Otherwise, return F.
KB  Q
╨
╨
Deduction theorem:
KB
Q
iff
FALSE
Propositional Resolution
• Resolution inference rule:
C1: p1  C1’
C1  p1  C1’
C2: p1  C2’
C2  p1  C2’
-------------------C3: C1’  C2’
Propositional Resolution
• Resolution algorithm (saturation):
1. While there are unresolved C1,C2:
(1) Select C1, C2 in KB
(2) If C1, C2 are resolvable, resolve them
into a new clause C3
(3) Add C3 to KB
C1: p1  C1’
(4) If C3={ } (empty clause), C2: p1  C2’
we got a contradiction.
--------------------
2. STOP
C3: C1’  C2’
Resolution in Action
ac
a  c  f
a  e  f
c  e  f
a   f
c f
On board
KB
C1: p1  C1’
C2: p1  C2’
-------------------C3: C1’  C2’
Negated Query
Resolution in Action
abc
a  c  d
a  e  f
c  e  f
c  d  f
c
On board
KB
C1: p1  C1’
C2: p1  C2’
-------------------C3: C1’  C2’
Negated Query
Properties of Resolution
• Running time for n variables, m clauses:
– Resolving two clauses:
O(n)
– Finding two resolvable clauses:
O(1)
– Overall algorithm:
O(3nn)
Properties of Resolution
• Theorem: Resolution is sound
– Resolving clauses in KB generates valid
consequences of KB
• Theorem: Resolution is refutation
complete
╨
– Resolution of KB with Q yields the empty
clause iff KB
Q
Properties of Resolution
• Resolution does not always generate Q
KB = { {a,b}, {a,b}, {b,c} }
Q = b  c = {b,c}
a
╨
• Theorem: Resolution always generates a
clause that subsumes Q iff KB
Q
Example: Resolving KB generates b
Simple Enhancements
• Remove subsumed clauses
– { p } subsumes { p , q }
– { p , q } subsumes { p , q, r }
– { p } does not subsume { p , q }
• Contract same literals
– { p , p , q } becomes { p , q }
• Unit resolution: resolve unit clauses first
Related to Prop. Resolution
• Clause selection for resolution (lecture #5,
paper #19)
• Consequence finding (paper #3)
• Prime implicates/implicants (paper #4)
Resolution vs SAT
• SAT solvers can find models
• Resolution sometimes better at finding
contradictions
• With resolution it is easier to explain and
provide a proof
Summary So Far
• Finding models using DPLL
• Resolution theorem proving allows us to
find contradictions and explanation.
– The deduction theorem tells us how to ask
queries from either SAT solvers or Resolution
Application: Hardware Verification
x1
x2
AND
f1
f3
not
f5
not
x3
AND
f2
OR
f4
Question: Can we set this boolean cirtuit to TRUE?
f5(x1,x2,x3) = a function of the input signal
Application: Hardware Verification
x1
x2
AND
f1
f3
not
f5
not
x3
AND
f2
OR
f4
SAT(f5) ?
Question: Can we set this boolean cirtuit to TRUE?
f5(x1,x2,x3) = f3  f4 = f1  (f2  x3) =
(x1  x2)  (x2  x3)
M[x1]=FALSE
M[x2]=FALSE
M[x3]=FALSE
Hardware Verification
• Questions in logical circuit verification
– Equivalence of circuits
– Arrival of the circuit to a state (required a
temporal model, which gets propositionalized)
– Achieving an output from the circuit
Summary
• SAT checking using DPLL (instantiate,
propagate, backtrack)
• Entailment/SAT checking using Resolution
(create more and more clauses until KB is
saturated)
• Formal verification uses mainly SAT
checking such as DPLL, but also
sometimes resolution
Next Time
•
•
FOL Resolution
Homework:
1. Read readings (incl. application reading)
2. Make sure you know:
– Deduction theorem for FOL
– Language of FOL
– Soundness, completeness, and incompleteness
theorems
– Models of first-order logic