Logical Agents
CSC 411: AI
Fall 2013
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Outline
Knowledge-based agents
Wumpus world
Logic in general – models and entailment
Propositional (Boolean) logic
Equivalence, validity, satisfiability
Inference rules and theorem proving
• Forward chaining
• Backward chaining
• Resolution
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Knowledge bases
Inference engine ← Domain-independent algorithms
Knowledge base ← Domain-specific content
A knowledge base is a set of sentences in a formal
language.
The declarative approach to building an agent (or other
system):
• T ELL it what it needs to know.
• Then it can A SK itself what to do.
Knowledge level versus implementation level.
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A simple knowledge-based agent
KB-AGENT(percept)
static t = 0
T ELL(KB, M AKE -P ERCEPT-S ENTENCE(percept, t))
action = A SK(KB, M AKE -ACTION -Q UERY(t))
T ELL(KB, M AKE -ACTION -S ENTENCE(action, t))
t = t+1
return action
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A simple knowledge-based agent
The agent must be able to
• Represent states, actions, etc.
• Incorporate new percepts
• Update internal representations of the world
• Deduce hidden properties of the world
• Deduce appropriate actions
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Wumpus world PEAS description
4
Performance measure:
Stench
Bree z e
Gold:
Death:
Taking a step:
Using an arrow:
3
+1000
−1000
−1
−10
Stench
Bree z e
PIT
PIT
Bree z e
Gold
2
Bree z e
Stench
Bree z e
1
PIT
Bree z e
3
4
START
1
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Wumpus world PEAS description
Environment:
Smelly when adjacent to Wumpus.
Breezy when adjacent to pit.
Glitter iff gold is in same square.
Shooting kills Wumpus if you are
facing it.
Shooting uses up only arrow.
Grabbing picks up gold if in same
square.
Releasing drops the gold in same
square.
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4
Stench
Bree z e
3
Stench
Bree z e
PIT
PIT
Bree z e
Gold
2
Bree z e
Stench
Bree z e
1
PIT
Bree z e
3
4
START
1
2
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Wumpus world PEAS description
4
Sensors:
Stench, Breeze, Glitter,
Bump, Scream
Actuators:
Left turn, Right turn, Forward, Grab, Release, Shoot
Stench
Bree z e
3
Stench
Bree z e
PIT
PIT
Bree z e
Gold
2
Bree z e
Stench
Bree z e
1
PIT
Bree z e
3
4
START
1
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Wumpus world characterization
Fully Observable? No, only local perception.
Deterministic? Yes, outcomes are exactly specified.
Episodic? No, sequential at the level of actions.
Static? Yes, the Wumpus and pits do not move.
Discrete? Yes.
Single-agent? Yes, the Wumpus is essentially a natural
feature.
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Exploring a Wumpus world
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Exploring a Wumpus world
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Exploring a Wumpus world
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Exploring a Wumpus world
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Exploring a Wumpus world
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Exploring a Wumpus world
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Exploring a Wumpus world
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Exploring a Wumpus world
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Logic in general
Logics are formal languages for representing information
such that conclusions can be drawn.
Syntax defines the sentences in the language; semantics
define the “meaning” (i.e., truth values) of sentences.
The language of arithmetic:
x + 2 ≥ y is a sentence
x2 + y > () is not a sentence
x + 2 ≥ y is true in a world where x = 7, y = 1
x + 2 ≥ y is false in a world where x = 0, y = 6
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Entailment
The knowledge base KB entails sentence α if and only if
α is true in all worlds where KB is true: KB |= α.
Said differently, a set of premises KB logically entails a
conclusion α if and only if every interpretation that
satisfies the premises also satisfies the conclusion.
{p} |= (p ∨ q)
{p, q} |= (p ∧ q)
Entailment is a relationship between sentences (i.e.
syntax), based on semantics.
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Models
Logicians typically think in terms of models, formally
structured worlds with respect to which truth can be
evaluated.
We can think of a model as a possible world.
In the semantics of arithmetic, x + y = 4 is true in a world
where x is 2 and y is 2, but false in a world where x is 1
and y is 10.
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Models
We say m is a model of a sentence α if α is true in m.
M(α) is the set of all models of α.
KB |= α iff M(KB) ⊆ M(α).
{p, q} |= p because the worlds in which p is true is a
superset of the worlds in which p is true and q is true.
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Entailment in the Wumpus world
You detect nothing in [1,1].
You move right.
You detect a breeze in [2,1].
Consider possible models for
KB assuming only pits:
3 (Boolean) possibilities →
8 possible models.
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Wumpus models
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Wumpus models
KB = Wumpus-world rules + observations
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Wumpus models
KB = Wumpus-world rules + observations
α1 = “[1,2] is safe.” Does KB |= α1 ?
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Wumpus models
KB = Wumpus-world rules + observations
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Wumpus models
KB = Wumpus-world rules + observations
α2 = “[2,2] is safe.” Does KB |= α2 ?
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Inference
KB `i α means that sentence α can be derived from KB
by procedure i.
Soundness: i is sound if whenever KB `i α, it is also true
that KB |= α.
Completeness: i is complete if whenever KB |= α, it is
also true that KB `i α.
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Propositional logic: Syntax
The proposition symbols P1 , P2 , etc. are sentences.
If S is a sentence, ¬S is a sentence (negation).
If S1 and S2 are sentences. . .
S1 ∧ S2 is a sentence (conjunction).
S1 ∨ S2 is a sentence (disjunction).
S1 ⇒ S2 is a sentence (implication).
S1 ⇔ S2 is a sentence (biconditional).
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Propositional logic: Semantics
Each model specifies true or false for each proposition
symbol.
P1,2 P2,2 P3,1
False True False
With these symbols, 8 possible models can be enumerated
automatically.
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Propositional logic: Semantics
Rules for evaluating truth with respect to a model m:
¬S is true iff S is false.
S1 ∧ S2 is true iff S1 is true and S2 is true.
S1 ∨ S2 is true iff S1 is true or S2 is true.
S1 ⇒ S2 is true iff S1 is false or S2 is true.
S1 ⇔ S2 is true iff S1 ⇒ S2 is true and S2 ⇒ S1 is true.
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Propositional logic: Semantics
A simple recursive process can evaluate an arbitrary
sentence:
¬P1,2 ∧ (P2,2 ∨ P3,1 ) = not false ∧ ( true ∨ false )
= true ∧ ( true ∨ false )
= true ∧ true
= true
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Truth tables for connectives
P
False
False
True
True
Q
False
True
False
True
¬P P ∧ Q P ∨ Q P ⇒ Q P ⇔ Q
True False False
True
True
True False True
True
False
False False True
False
False
False True True
True
True
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Wumpus world sentences
Let Pi,j be true if there is a pit in [i, j].
Let Bi,j be true if there is a breeze in [i, j].
Suppose we know
P1,1 and B1,2 and B2,1 .
We also know that pits cause breezes in adjacent squares.
Then
B1,2 ⇔ (P1,1 ∨ P2,2 ∨ P1,3 ).
B2,1 ⇔ (P1,1 ∨ P2,2 ∨ P3,1 ).
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Truth tables for inference
B1,1
F
F
B2,1
F
F
P1,1
F
F
P1,2
F
F
F
T
F
F
F
F
T
T
F
F
F
F
F
T
F
F
T
T
T
T
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P2,1
F
F
...
F
...
F
F
...
T
...
T
P2,2
F
F
P3,1
F
T
KB
F
F
α1
T
T
F
F
F
T
F
T
F
F
T
T
T
T
F
F
F
T
T
T
F
F
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Inference by enumeration
TT-E NTAILS ?(KB, α)
symbols = a list of symbols in KB and α
return TT-C HECK -A LL(KB, α, symbols, {})
TT-C HECK -A LL(KB, α, symbols, model)
if E MPTY ?(symbols)
if PL-T RUE ?(KB, model)
return PL-T RUE ?(α, model)
else return True
else p = First(symbols)
rest = Rest(symbols)
return TT-C HECK -A LL(KB, α, rest, E XTEND(p, True, model))
and TT-C HECK -A LL(KB, α, rest, E XTEND(p, False, model))
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Inference by enumeration
Depth-first enumeration of all models is sound and
complete.
For n symbols, time complexity is O(2n ); space
complexity is O(n).
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Logical equivalence
Two sentences are logically equivalent iff they are true in
same models:
α ≡ β iff α |= β and β |= α.
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Logical equivalence
(α ∧ β)
(α ∨ β)
((α ∧ β) ∧ γ)
((α ∨ β) ∨ γ)
¬(¬α)
(α ⇒ β)
¬(α ∧ β)
¬(α ∨ β)
(α ∧ (β ∨ γ))
(α ∧ (β ∨ γ))
≡
≡
≡
≡
≡
≡
≡
≡
≡
≡
(β ∧ α)
(β ∨ α)
(α ∧ (β ∧ γ))
(α ∨ (β ∨ γ))
α
(¬β ⇒ ¬α)
(¬α ∨ ¬β)
(¬α ∧ ¬β)
((α ∧ β) ∨ (α ∧ γ))
((α ∨ β) ∧ (α ∨ γ))
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Commutativity of ∧
Commutativity of ∨
Associativity of ∧
Associativity of ∨
Double negation
Contraposition
de Morgan
de Morgan
Distributivity, ∧/∨
Distributivity, ∨/∧
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Validity
A sentence is valid if it is true in all models, e.g.,
True
A ∨ ¬A
A⇒A
(A ∧ (A ⇒ B)) ⇒ B
Validity is connected to inference via the Deduction
Theorem:
KB |= α iff (KB ⇒ α) is valid.
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Satisfiability
A sentence is satisfiable if it is true in some model.
A sentence is unsatisfiable if it is true in no model.
Satisfiability is also connected to inference:
KB |= α iff (KB ∧ ¬α) is unsatisfiable.
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Proof methods: Inference rules
One kind of proof method involves the application of
inference rules.
• Rules give sound generation of new sentences from
old.
• A proof is a sequence of inference rule applications.
(We can use rules as operators in search.)
• This typically requires transformation of sentences
into some normal form.
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Proof methods: Model checking
Another approach is model checking, e.g.,
• Truth table enumeration (as we’ve seen).
• Smarter backtracking (potentially more efficient).
• Heuristic search or optimization in model space
(sound but incomplete).
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Exercise: Bike riding
“You often ride your bike, but only on days when the
weather is nice. Whenever you ride, you wear a helmet.
Wearing your helmet always messes up your hair beyond
easy repair, but you like to be presentable: you never
arrive at work with messy hair, and you never go out on
dinner dates with messy hair.”
What propositions should we define for this story?
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Exercise: Bike riding
“You often ride your bike, but only on days when the
weather is nice. Whenever you ride, you wear a helmet.
Wearing your helmet always messes up your hair beyond
easy repair, but you like to be presentable: you never
arrive at work with messy hair, and you never go out on
dinner dates with messy hair.”
One possible set: NiceOut, BikeRide, Helmet, MessyHair,
Work, DinnerDate.
Develop sentences to represent this story.
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Exercise: Bike riding
“You call me on your cell phone and tell me you’re out on
a dinner date. If I wonder whether you rode your bike,
should I ask you or figure it out on my own?”
How can I express this as a sentence in logic?
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Resolution
We start with Conjunctive Normal Form (CNF).
Call a logical variable or its complement a literal.
A, B, ¬A, ¬B
Call a disjunction of literals (including a literal standing
alone) a clause.
A, (B ∨ ¬A)
A sentence consisting of a conjunction of clauses is in
CNF.
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Resolution
The resolution rule for CNF:
a1 ∨ . . . ∨ ak
b1 ∨ . . . ∨ bn
a1 ∨ . . . ∨ aj−1 ∨ aj+1 ∨ . . . ∨ ak ∨ b1 ∨ . . . ∨ bm−1 ∨ bm+1 ∨ . . . ∨ bn
where aj and bm are complementary literals.
Resolution is sound and complete for propositional logic.
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Resolution in Wumpus world
P1,3 ∨ P2,2
P1,3
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¬P2,2
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Resolution in Wumpus world
P1,3 ∨ P2,2
P1,3
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¬P2,2
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Conversion to CNF
B1,1 ⇔ (P1,2 ∨ P2,1 )
Eliminate ⇔, replacing α ⇔ β with (α ⇒ β) ∧ (β ⇒ α).
(B1,1 ⇒ (P1,2 ∨ P2,1 )) ∧ ((P1,2 ∨ P2,1 ) ⇒ B1,1 )
Eliminate ⇒, replacing α ⇒ β with ¬α ∨ β.
(¬B1,1 ∨ P1,2 ∨ P2,1 ) ∧ (¬(P1,2 ∨ P2,1 ) ∨ B1,1 )
...
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Conversion to CNF
...
(¬B1,1 ∨ P1,2 ∨ P2,1 ) ∧ (¬(P1,2 ∨ P2,1 ) ∨ B1,1 )
Move ¬ inwards, using de Morgan’s rule:
(¬B1,1 ∨ P1,2 ∨ P2,1 ) ∧ ((¬P1,2 ∧ ¬P2,1 ) ∨ B1,1 )
Apply distributivity law (∧ over ∨) and flatten:
(¬B1,1 ∨ P1,2 ∨ P2,1 ) ∧ (¬P1,2 ∨ B1,1 ) ∧ (¬P2,1 ∨ B1,1 )
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Exercise: Bike riding
In CNF:
¬NiceOut ⇒ ¬BikeRide
BikeRide ⇒ Helmet
Helmet ⇒ MessyHair
¬(MessyHair ∧ DinnerDate)
¬(MessyHair ∧ Work)
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Exercise: Bike riding
In CNF:
¬NiceOut ⇒ ¬BikeRide
BikeRide ⇒ Helmet
Helmet ⇒ MessyHair
¬(MessyHair ∧ DinnerDate)
¬(MessyHair ∧ Work)
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NiceOut ∨ ¬BikeRide
Helmet ∨ ¬BikeRide
¬Helmet ∨ MessyHair
¬DinnerDate ∨ ¬MessyHair
¬Work ∨ ¬MessyHair
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Exercise: Bike riding
In CNF:
¬NiceOut ⇒ ¬BikeRide
BikeRide ⇒ Helmet
Helmet ⇒ MessyHair
¬(MessyHair ∧ DinnerDate)
¬(MessyHair ∧ Work)
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NiceOut ∨ ¬BikeRide
Helmet ∨ ¬BikeRide
¬Helmet ∨ MessyHair
¬DinnerDate ∨ ¬MessyHair
¬Work ∨ ¬MessyHair
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Exercise: Bike riding
In CNF:
¬NiceOut ⇒ ¬BikeRide
BikeRide ⇒ Helmet
Helmet ⇒ MessyHair
¬(MessyHair ∧ DinnerDate)
¬(MessyHair ∧ Work)
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NiceOut ∨ ¬BikeRide
Helmet ∨ ¬BikeRide
¬Helmet ∨ MessyHair
¬DinnerDate ∨ ¬MessyHair
¬Work ∨ ¬MessyHair
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Exercise: Bike riding
In CNF:
¬NiceOut ⇒ ¬BikeRide
BikeRide ⇒ Helmet
Helmet ⇒ MessyHair
¬(MessyHair ∧ DinnerDate)
¬(MessyHair ∧ Work)
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NiceOut ∨ ¬BikeRide
Helmet ∨ ¬BikeRide
¬Helmet ∨ MessyHair
¬DinnerDate ∨ ¬MessyHair
¬Work ∨ ¬MessyHair
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Exercise: Bike riding
In CNF:
¬NiceOut ⇒ ¬BikeRide
BikeRide ⇒ Helmet
Helmet ⇒ MessyHair
¬(MessyHair ∧ DinnerDate)
¬(MessyHair ∧ Work)
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NiceOut ∨ ¬BikeRide
Helmet ∨ ¬BikeRide
¬Helmet ∨ MessyHair
¬DinnerDate ∨ ¬MessyHair
¬Work ∨ ¬MessyHair
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Exercise: Bike riding
A truth table, for enumeration
NiceOut
T
T
F
F
BikeRide ¬NiceOut ⇒ ¬BikeRide
T
T
T
T
F
F
T
T
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NiceOut ∨ ¬BikeRide
T
T
F
T
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Resolution algorithm
The strategy is proof by contradiction.
To prove α, we show that there exist no models in which
KB ∧ ¬α is true.
That is, we prove KB ∧ ¬α is unsatisfiable.
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Resolution algorithm
PL-R ESOLUTION(KB, α)
clauses = clauses in CNF representation of KB ∧ ¬α
empty = the empty clause
new = the empty set
while True
for Ci , Cj in clauses
resolvent = PL-R ESOLVE(Ci , Cj )
if resolvent is the empty set
return True
new = new ∪ resolvent
if new ⊆ clauses
return False
clauses = clauses ∪ new
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Exercise: Bike riding
“You call me on your cell phone and tell me you’re out on
a dinner date. . . ”
NiceOut ∨ ¬BikeRide
Helmet ∨ ¬BikeRide
¬Helmet ∨ MessyHair
¬DinnerDate ∨ ¬MessyHair
¬Work ∨ ¬MessyHair
DinnerDate
I’ll guess you didn’t ride your bike (¬BikeRide). Can I
prove it?
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Exercise: Bike riding
Recall. . .
a1 ∨ . . . ∨ ak
b1 ∨ . . . ∨ bn
a1 ∨ . . . ∨ aj−1 ∨ aj+1 ∨ . . . ∨ ak ∨ b1 ∨ . . . ∨ bm−1 ∨ bm+1 ∨ . . . ∨ bn
Prove ¬BikeRide by resolution.
To start: Negate ¬BikeRide. . .
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Exercise: Bike riding
Helmet ∨ ¬BikeRide
BikeRide
Helmet
¬Helmet ∨ MessyHair
Helmet
MessyHair
MessyHair
¬DinnerDate ∨ ¬MessyHair
¬DinnerDate
¬DinnerDate
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DinnerDate
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Forward and backward chaining
Horn Form: KB is a conjunction of Horn clauses.
A Horn clause is an implication of a conjunction to a
positive literal:
C, (B ⇒ A), (C ∧ D ⇒ B)
Modus Ponens (for Horn Form) is complete for Horn
KBs.
α1 . . . αn
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α1 ∧ . . . ∧ αn ⇒ β
β
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Forward chaining
Q
A
B
P
L∧M
B∧L
A∧P
A∧B
P
⇒
⇒
⇒
⇒
⇒
Q
P
M
L
L
M
L
A
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B
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Forward chaining
Fire any rule whose premises are satisfied in the KB.
Add its conclusion to the KB.
Continue until the query is found.
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Forward chaining algorithm
PL-FC-E NTAILS ?(KB, q)
count = a table, where count[c] is # of symbols in c’s premise
inferred = a table, where inferred[s] is initially false
agenda = a list of symbols known to be true in KB
while not Empty?(agenda)
p = Pop(agenda)
if p is q return True
if inferred[p] is False
inferred[p] = True
for c such that p ∈ c.premise
count[c] = count[c] − 1
if count[c] is 0
Push(c.conclusion, agenda)
return False
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Forward chaining example
Agenda: A, B.
P⇒Q:1
L∧M ⇒P: 2
B∧L⇒M : 2
A∧P⇒L: 2
A∧B⇒L: 2
Pop A.
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Forward chaining example
After processing A.
Agenda: B.
P⇒Q:1
L∧M ⇒P: 2
B∧L⇒M : 2
A∧P⇒L: 1
A∧B⇒L: 1
Pop B.
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Forward chaining example
After processing B.
P⇒Q:1
L∧M ⇒P: 2
B∧L⇒M : 1
A∧P⇒L: 1
A∧B⇒L: 0
Agenda: L.
Pop L.
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Forward chaining example
After processing L.
P⇒Q:1
L∧M ⇒P: 1
B∧L⇒M : 0
A∧P⇒L: 1
A∧B⇒L: 0
Agenda: M.
Pop M.
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Forward chaining example
After processing M.
P⇒Q:1
L∧M ⇒P: 0
B∧L⇒M : 0
A∧P⇒L: 1
A∧B⇒L: 0
Agenda: P.
Pop P.
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Forward chaining example
After processing P.
P⇒Q:0
L∧M ⇒P: 0
B∧L⇒M : 0
A∧P⇒L: 0
A∧B⇒L: 0
Agenda: Q, L.
Pop Q.
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Forward chaining example
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Forward chaining example
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Forward chaining example
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Forward chaining example
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Forward chaining example
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Forward chaining example
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Forward chaining example
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Forward chaining example
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Proof of completeness
FC derives every atomic sentence that is entailed by KB.
FC reaches a fixed point where no new atomic sentences
are derived.
Consider the final state to be a model m, assigning
true/false to symbols.
Every clause in the original KB is true in m
a1 ∧ . . . ∧ ak ⇒ b
Then m is a model of KB.
If KB |= q, q is true in every model of KB, including m.
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Backward chaining
Work backward from query q.
To prove q by BC,
• Check if q is known already, or
• Prove by BC all premises of some rule concluding q.
Avoid loops by checking if a new subgoal is already on
the goal stack.
Avoid repeated work by checking if a new subgoal has
already been proved true or has already failed.
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Backward chaining example
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Backward chaining example
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Backward chaining example
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Backward chaining example
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Backward chaining example
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Backward chaining example
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Backward chaining example
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Backward chaining example
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Backward chaining example
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Backward chaining example
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Forward vs backward chaining
FC is data-driven, automatic processing, e.g., object
recognition, routine decisions.
It may do lots of work that is irrelevant to the goal.
BC is goal-driven, appropriate for problem solving, e.g.,
Where are my keys? How do I get CS degree?
The complexity of BC can be much less than linear in size
of KB.
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Efficient propositional inference
Two families of efficient algorithms for propositional
inference are
• Complete backtracking search algorithms, such as
DPLL, and
• Incomplete local search algorithms, such as
WalkSAT.
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Logical Agents
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The DPLL algorithm: Concepts
A clause is true if any literal is true.
A sentence is false if any clause is false.
A unit clause contains only one literal.
A pure symbol always appears in the same positive or
negative form in all clauses.
(A ∨ ¬B), (¬B ∨ ¬C), (C ∨ A)
A and B are pure; C is not.
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The DPLL algorithm: Heuristics
Notice when a clause can be true if one of its literals is
true.
Notice that a sentence is false if one of its clauses is false.
Make the single literal in unit clauses true.
Make pure symbol literals true.
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Logical Agents
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The DPLL algorithm
DPLL-S ATISFIABLE ?(s)
clauses = the set of clauses in the CNF representation of s
symbols = propositional symbols in s
return DPLL(clauses, symbols, {})
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Logical Agents
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The DPLL algorithm
DPLL(clauses, symbols, model)
if A LL -T RUE(clauses, model) return True
if S OME -FALSE(clauses, model) return False
p, value = F IND -P URE -S YMBOL(clauses, symbols, model)
if p found
return DPLL(clauses, symbols \ p, [p = value | model])
p, value = F IND -U NIT-C LAUSE(clauses, model)
if p found
return DPLL(clauses, symbols \ p, [p = value | model])
p = First(symbols)
return DPLL(clauses, symbols \ p, [p = True | model]) or
DPLL(clauses, symbols \ p, [p = False | model])
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The WalkSAT algorithm
WalkSat is an incomplete local search algorithm.
Its evaluation function is the min-conflict heuristic of
minimizing the number of unsatisfied clauses.
It balances between greediness and randomness.
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The WalkSAT algorithm
WALK SAT(clauses, p, max-flips)
model = a random assignment of True/False to symbols in clauses
for j = 1 to max-flips
if S ATISFIES ?(model, clauses)
return model
clause = C HOOSE -R ANDOM -FALSE -C LAUSE(model)
if Sample(0, 1) < p
model = F LIP(Random-Element(clause), clause, model)
else symbol = A RG -M AX(clause, M AXIMIZE -S ATISFIED)
model = F LIP(symbol, clause, model)
return Failure
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Hard satisfiability problems
Consider random 3-CNF sentences.
(¬D ∨ ¬B ∨ C) ∧ (B ∨ ¬A ∨ ¬C) ∧ (¬C ∨ ¬B ∨ E)∨
(E ∨ ¬D ∨ B) ∨ (B ∨ E ∨ ¬C)
Let
m = number of clauses and
n = number of symbols.
Hard problems seem to cluster near a critical point,
m/n = 4.3.
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Hard satisfiability problems
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Hard satisfiability problems
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Inference in the Wumpus world
¬P1,1
¬W1,1
Bx,y ⇔ (Px,y+1 ∨ Px,y−1 ∨ Px+1,y ∨ Px−1,y )
Sx,y ⇔ (Wx,y+1 ∨ Wx,y−1 ∨ Wx+1,y ∨ Wx−1,y )
W1,1 ∨ W1,2 ∨ . . . ∨ W4,4
¬W1,1 ∨ W1,2
¬W1,1 ∨ W1,3
...
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Logical Agents
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A propositional Wumpus world agent
PL-W UMPUS -AGENT(percept)
static: KB, x, y, orientation, visited, action, plan
U PDATE(x, y, orientation, visited, action)
if percept.stench
T ELL(KB, stenchx,y )
else T ELL(KB, ¬stenchx,y )
if percept.breeze
T ELL(KB, breezex,y )
else T ELL(KB, ¬breezex,y )
if percept.glitter
return action = Grab
...
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Logical Agents
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A propositional Wumpus world agent
PL-W UMPUS -AGENT(percept)
...
else if not Empty?(plan)
return action = Pop(plan)
else for j, k on F RINGE(visited)
if A SK(KB, Pj,k ∧ ¬Wj,k ) is True or
A SK(KB, Pj,k ∨ Wj,k ) is False
plan = P LAN -PATH(x, y, orientation, j, k,visited)
return action = Pop(plan)
else return action = C HOOSE -R ANDOM -M OVE()
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Logical Agents
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Expressiveness limitation of
propositional logic
KB contains “physics” sentences for every single square.
For every time t and every x, y location Lx,y ,
Lx,y (t) ∧ FacingRight(t) ∧ Forward(t) ⇒ Lx+1,y (t + 1),
etc.
Rapid proliferation of clauses.
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Logical Agents
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Summary
Logical agents apply inference to a knowledge base to
derive new information and make decisions.
• Syntax: Formal structure of sentences.
• Semantics: Truth of sentences with respect models.
• Entailment: Necessary truth of one sentence given
another.
• Inference: Deriving sentences from other sentences.
• Soundness: Derivations produce only entailed
sentences.
• Completeness: Derivations can produce all entailed
sentences.
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Logical Agents
Last observations
An agent, as in Wumpus world, needs to represent partial
and negated information, reason by cases, etc.
Resolution is complete for propositional logic.
Forward and backward chaining run in linear time; they
are complete for Horn clauses.
Propositional logic lacks expressive power.
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Logical Agents
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