Summary of previous lecture
Knowledge
and reasoning 2
Knowledge-based agents
Knowledge
base
deduce
Action
Inside our
agent
TDDC65 Artificial intelligence and Lisp
The real
world
Peter Dalenius
petda@ida.liu.se
Department of Computer and Information Science
Linköping University
Percept
execute
perceive
1. How can we represent knowledge?
2. How can we draw conclusions?
Summary of previous lecture
Logic
Truth table
What is the value?
It depends on the interpretation!
P
Q
P∧Q⇒P
F
F
T
F
T
T
T
F
T
T
T
T
Summary of previous lecture
What is a good argument?
In
order to prove that it is correct to
draw the conclusion S when we know
KB, we can do either of the following:
Is it satisfiable?
Can it ever be true?
P∧Q⇒P
– Prove KB ╞ S (entailment) using either a
truth table or reasoning with interpretations
– Prove KB ├ S using a sound and complete
proof system, e.g. resolution
Is it a tautology?
Is it always true?
Does it entail P ∧ Q?
Is it correct to draw the conclusion P ∧ Q?
Is it equivalent to P ⇒ Q?
Do these two formulas always have the same value?
Summary of previous lecture
The Wumpus world
Representing the Wumpus world
There is at least one Wumpus in the cave:
W1,1 ∨ W1,2 ∨ W1,3 ∨ … ∨ W4,4
If there is a breeze in square (x, y) then there
is a pit in one of the four adjacent squares:
PIT
PIT
Bx,y ⇔ Px,y+1 ∨ Px,y-1 ∨ Px+1,y ∨ Px-1,y
This is a template for a number of formulas.
PIT
There is only one Wumpus:
That’s a tricky one to represent…
1
We need a more expressive logic!
What’s new in first order logic?
Propositional
logic (used so far)
– Each symbol represents a proposition, a
single statement that says something
about the world.
– Several things are hard to formulate.
First
order predicate logic
– An extension to propositional logic.
– Helps us to formulate more complex
statements.
We can talk about objects (e.g. the different
rooms in the cave) denoted by object
constants. Example: A, Room1, John
The objects can have properties or relations
to each other. These are called predicates.
Example: Empty(Room1), Father(John, Mary)
We have functions that can return objects.
Example: nextRoom(Room1), fatherOf(Mary)
We have quantifiers to say general things.
∀ means ”for all” and ∃ means ”exists”.
p.247
Sentences in first order logic
Sentences in first order logic
A term is an expression that refers to an
object, either a constant, a variable, or a
function application. Example: Room1, x,
motherOf(John). Note that a term is not a
sentence in itself.
An atomic sentence states a fact. It is formed
from a predicate symbol followed by a list of
terms. Example: Mother(Anna, John),
Long(tailOf(Cat56))
A complex sentence is formed by connectives,
just like in propositional logic. Example:
Shining(Sun) ∧ Angry(brother(John))
A quantified sentence says something about
an entire collection of objects.
∀x P(x) means that P(x) is true for all objects
∃x P(x) means that P(x) is true for some objects
p.248
More examples of formulas
The sky is blue and the grass is green
Blue(Sky) ∧ Green(Grass)
Color(Sky, Blue) ∧ Color(Grass, Green)
John is Mary’s father
fatherOf(Mary) = John
Father(John, Mary)
There are nice people
Everyone has a father
∃x(People(x) ∧ Nice(x))
It is ok to use equality.
p.249
Connections between ∀ and ∃
Nobody
Move ¬ inwards, and
likes carrots:
the quantifier changes.
Compare to DeMorgan.
¬∃x Likes(x, Carrots)
∀x ¬Likes(x, Carrots)
Everybody likes ice cream:
∀x Likes(x, IceCream)
¬¬∀x Likes(x, IceCream)
¬∃x ¬Likes(x, IceCream)
∀x∃y(Father(y,x))
p.252
2
Interpretations in first order logic
What do first order sentences/formulas
actually mean?
An interpretation in first order logic consists of
four parts:
– The domain, i.e. the set of objects we are talking
about.
– A definition of all the constant symbols.
– A definition of all the predicate symbols (properties
and relations).
– A definition of all the function symbols.
Working with N5
Is the following formula true in N5?
∃x∀y(Greater(x,y) ∨ x = y)
The formula actually says that there exists a
greatest number. If we let x = 5, is the
following formula true?
∀y(Greater(x,y) ∨ x = y)
We can try all possible values of y and see if
the inner formula is true (which it is).
Sample interpretation N5
The
domain is {0, 1, 2, 3, 4}.
is one constant symbol:
There
– Smallest has the value 0
There
are two function symbols:
– plus(x, y) is defined as addition mod 5
– minus(x, y) is defined as subtraction mod 5
There
is one predicate symbol:
– Greater(x, y) is defined as true whenever x
is greater than y
Working with N5
Which
of the following formulas are true
in the interpretation N5?
∀x∀y Greater(plus(x, y), x)
∀x∀y∀z (plus(x, y) = z) ⇒ (minus(z, y) = x)
∃x∃y (minus(x, y) = Smallest)
Experiences from first order logic
1. Convert to propositional logic
Much
easier to express knowledge
compared to propositional logic. ☺
Much more complex interpretations,
which makes it harder to draw
conclusions (no truth tables). We
need a sound and complete proof
system! We already have resolution for
propositional logic. Can we use that?
Yes, we can use resolution if we transform
our first order formulas into propositional
formulas. This is called propositionalization.
Example:
Instead of ∀x Car(x) ⇒ Blue(x)
we write Blue(C1) ∧ Blue(C2) ∧ Blue(C3)
if the domain consists of cars C1, C2 and C3.
We will soon see a method for doing this…
p.274
3
2. Matching atomic sentences
In propositional logic it’s easy to
match positive and negative literals.
P
R
Unification
¬P, ¬R
¬R
In first order logic, it’s not clear how we can match
atomic sentences with different inner structure.
P(f(a), b)
¬P(x, b), R(x, c)
Atomic sentences with different terms can be
matched by unification. This is a form of pattern
matching that binds variables to values.
The two sentences in the example match if the
variable x is bound to the term f(a). This must also be
done in the resulting sentence.
Q(b, c)
P(f(a), b)
¬P(x, b), R(x, c)
???
{x = f(a)} R(f(a), c)
p.275
Resolution in first order logic
If we know KB, is S
a valid conclusion?
KB ├ S
the same as
Transformation
1.
Is the set of formulas
KB ∪ {¬S} unsatisfiable?
transformation
New set of quantifierfree formulas in CNF
2.
3.
4.
5.
resoluiion (using unification)
6.
If we can prove contradiction, then the
answer to our original question is yes!
Our goal is to transform a first
order formula to a propositional
formula in CNF so we can use
resolution.
Rewrite ⇔ and ⇒ according to rules.
Move ¬ inwards.
Standardize variables.
Skolemize (drop existential qualifiers).
Drop universal quantifiers.
Distribute ∨ over ∧.
Note that the resulting formula is not equivalent to the original
one, but it will be satisfiable if and only if the original one is.
p.296
Skolemize with ∃ at the beginning
Skolemize with ∃ inside
Assume
that the following formula is
∃x∀y(Greater(x,y) ∨ x = y)
there exists and x, we can choose that
x and call it a, so the formula can be
written
If
∀y(Greater(a,y) ∨ a = y)
a
is called a Skolem constant.
Assume that the following formula is true
∀x∃y(Greater(x,y)) ∨ x = Smallest
true
When ∃ is inside an ∀, we cannot introduce a
Skolem constant. The y that x is greater than,
might not be the same y for all x.
Instead we introduce a Skolem function f, so
the formula becomes
∀x(Greater(x,f(x))) ∨ x = Smallest
4
Transformation example
A complete example
∀x∃y(P(x,y) ⇒ ¬∀x(Q(x,y) ∨ P(x,y)))
rewrite ⇒
∀x∃y(¬P(x,y) ∨ ¬∀x(Q(x,y) ∨ P(x,y)))
move ¬ inwards
∀x∃y(¬P(x,y) ∨ ∃x ¬(Q(x,y) ∨ P(x,y)))
∀x∃y(¬P(x,y) ∨ ∃x (¬Q(x,y) ∧ ¬P(x,y)))
standardize variables
∀x∃y(¬P(x,y) ∨ ∃z (¬Q(z,y) ∧ ¬P(z,y)))
Skolemize
∀x(¬P(x,f(x)) ∨ (¬Q(g(x),f(x)) ∧ ¬P(g(x),f(x))))
drop universal quantifiers
¬P(x,f(x)) ∨ (¬Q(g(x),f(x)) ∧ ¬P(g(x),f(x)))
distribute ∨ over ∧
(¬P(x,f(x)) ∨ ¬Q(g(x),f(x))) ∧ (¬P(x,f(x)) ∨ ¬P(g(x),f(x)))
set of clauses
{¬P(x,f(x)) ∨ ¬Q(g(x),f(x)), ¬P(x,f(x)) ∨ ¬P(g(x),f(x))}
From ”Horses are animals”, it follows that
”The head of a horse is the head of an
animal”. Demonstrate that this inference is
valid by carrying out the following steps:
– Translate the premise and the conclusion to the
language of first order logic using the predicates
HeadOf(h, x), Horse(x) and Animal(x).
– Negate the conclusion. Convert the premise and
the negated conclusion into CNF.
– Use resolution to show that the conclusion follows
from the premise (i.e. deduce a contradiction from
the negated conclusion).
1. Formulas in first order logic
2. Transform to CNF
The
premise
The
The
conclusion
∀x(Horse(x) ⇒ Animal(x))
∀x∃y(HeadOf(y,x) ∧ Horse(x) ⇒ Animal(x))
2. Transform to CNF
The
premise
∀x(Horse(x) ⇒ Animal(x))
∀x(¬Horse(x) ∨ Animal(x))
¬Horse(x) ∨ Animal(x)
{¬Horse(x) ∨ Animal(x)}
convert ⇒
drop ∀
set of clauses
3. Resolution (using unification)
negated conclusion
¬∀x∃y(HeadOf(y,x) ∧ Horse(x) ⇒ Animal(x))
¬Horse(x) ∨ Animal(x)
convert ⇒
¬∀x∃y(¬(HeadOf(y,x) ∧ Horse(x)) ∨ Animal(x))
move ¬ inwards
∃x∀y ¬(¬(HeadOf(y,x) ∧ Horse(x)) ∨ Animal(x))
∃x∀y(HeadOf(y,x) ∧ Horse(x) ∧ ¬Animal(x))
Skolemize
HeadOf(y,a)
Horse(a)
¬Animal(a)
{x = a}
Animal(a)
∀y(HeadOf(y,a) ∧ Horse(a) ∧ ¬Animal(a))
HeadOf(y,a) ∧ Horse(a) ∧ ¬Animal(a)
drop ∀
set of clauses
{HeadOf(y,a), Horse(a), ¬Animal(a)}
5
Formal languages
Language
What exists in the world?
(Ontological commitment)
Knowledge engineering
What an agent
believes about facts
(Epistemological
commitment)
1.
2.
3.
Propositional logic
facts
true/false/unknown
First order logic
facts, objects, relations
true/false/unknown
Temporal logic
facts, objects, relations, times
true/false/unknown
Probability theory
facts
degree of belief 0..1
Fuzzy logic
facts with degree of truth 0..1
known interval value
4.
5.
6.
7.
Identify the task.
Assemble the relevant knowledge.
Decide on a vocabulary of predicates,
functions, and constants.
Encode general knowledge about the
domain.
Encode a description of the specific problem
instance.
Pose queries to the inference procedure and
get answers.
Debug the knowledge base.
p.244
Situation calculus
p.261
Situation calculus
Both propositional and first order logic by
default assumes a static world that does not
change. But an intelligent agent has to be
able to reason about the result of actions and
how they change the world.
We will briefly discuss situation calculus,
implemented in first order logic, that gives us
that possibility. We will use the Wumpus
world as example.
A situation encode everything there is to
know about the world at a given time. The
function Result(a, s) names the new situation
after doing action a in situation s.
Fluents are functions and predicates that vary
from one situation to the next. Example:
¬Holding(G1, S0)
Eternal functions and predicates indicate
what is true in all situations. Example:
Gold(G1)
p.328
Situations
Describing actions
The
simplest way to describe actions is
by possibility axioms and effect axioms.
Examples of possibility axioms:
At(A,x1,s) ∧ Adjacent(x1,x2) ⇒ Poss(Go(x1,x2),s)
Gold(g) ∧ At(A,x,s) ∧ At(g,x,s) ⇒ Poss(Grab(g),s)
Examples
of effect axioms:
Poss(Go(x1,x2),s) ⇒ At(A,x2,Result(Go(x1,x2),s))
Poss(Grab(g),s) ⇒ Holding(g,Result(Grab(g),s))
6
Example of initial knowledge base
Searching for gold
At(A,[1,1],S0)
At(G1,[1,2],S0)
Gold(G1)
Adjacent([1,1],[1,2])
Adjacent([1,2],[1,1])
agent is at [1,1]
object G1 is at [1,2]
object G1 is the gold
[1,1] is next to [1,2]
[1,2] is next to [1,1]
Plus possibility and effect axioms…
The frame problem
Our first step is Go([1,1],[1,2]) which is
possible according to the knowledge base,
including possibility and effect axioms.
Our next step is Grab(G1) since the gold is in
[1,2], but this is not possible! From the
knowledge base and the axioms we cannot
deduce that the gold is in [1,2] in the situation
S1 = Result(Go([1,1],[1,2],S0). Intuitively this
should be obvious, but it is not stated in our
axioms!
Solving the frame problem
The effect axiom says what changes, but they
don’t say what stays the same (e.g. that the
gold is still in [1,2]). This is an example of the
frame problem.
In the real world, most things stay the same
from one situation to another. How can we
encode this knowledge?
If we have A actions and F fluent predicates
in our language, we need AF frame axioms
(where most of them say that a fluent is not
affected at all). This seems difficult!
Instead of writing out the effect of each
action, we consider how each fluent predicate
evolves over time. We use successor-state
axioms instead:
action is possible ⇒
(fluent is true in result state ⇔ action’s effect made it true
∨ it was true before and action left it alone)
Example:
Poss(a, s) ⇒
(At(A,y,Result(a,s)) ⇔ a = Go(x,y)
∨ (At(A,y,s) ∧ a ≠ Go(y,z)))
p.331
Problems in situation calculus
Representational frame problem
You should be able to…
– How can we represent what stays the same?
Inferential frame problem
– How can we find the result of a sequence of n actions in a
reasonable time?
Ramification problem
– How can we represent implicit effect, e.g. that the gold
moves with the agent when grabbed?
Qualification problem
– What happens if the agent crashes inside the cave?
Explain basic concepts (e.g. tautology,
satisfiability)
Formulate sentences in both propositional
and first order logic
Decide if a formula is true or false in a given
interpretation
Use resolution in both propositional and first
order logic
Briefly discuss advanced concepts (e.g.
different kinds of logic)
7