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Artificial Intelligence. Knowledge Representation Problem. Knowledge bases. Knowledge base = set of sentences in a formal language. Stages of Knowledge Use. Acquisition structure of facts integration of old & new knowledge Retrieval (recall) roles of linking and chunking

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Artificial intelligence

Artificial Intelligence

Knowledge Representation Problem


Knowledge bases
Knowledge bases

  • Knowledge base = set of sentences in a formal language


Stages of knowledge use
Stages of Knowledge Use

  • Acquisition

    • structure of facts

    • integration of old & new knowledge

  • Retrieval (recall)

    • roles of linking and chunking

    • means of improving recall efficiency


Representation
Representation

  • Set of syntactic and semantic conventions which make it possible to describe things

  • Syntax

    • specific symbols allowed and rules allowed

  • Semantics

    • how meaning is associated with symbol arrangements allowed by syntax


Knowledge representation schemas
Knowledge Representation Schemas

  • Logic based representation – first order predicate logic, Prolog

  • Procedural representation – rules, production system

  • Network representation – semantic networks, conceptual graphs

  • Structural representation – scripts, frames, objects


Conceptual graphs
Conceptual Graphs

  • each concept has got its type and an instance

    general concept – a concept with a wildcard instance

    specific concept – a concept with a concrete instance

  • there exsists a hierarchy of types subtype:

  • concept w is specialisation of concept v iftype(v)>type(w) or instance(w)::type(v)

colour

dog:*X

brown

colour

dog:Emma

brown

animal

dog

cat


Types of knowledge
Types of Knowledge

  • Objects

    • both physical & concepts

  • Events

    • usually involve time

    • maybe cause & effect relationships

  • Performance

    • how to do things

  • META Knowledge

    • knowledge about how to use knowledge



Basic connectives and truth tables

statements (propositions): declarative sentences that are

either true or false--but not both.

Eg. Ahmed Hassan wrote Gone with the Wind.

2+3=5.

not statements:

What a beautiful morning!

Get up and do your exercises.


Fundamentals of Logic

"The number x is an integer."

is not a statement because its truth value cannot be determined until a numerical value is assigned for x.


Propositional logic
Propositional logic

  • Logical constants: true, false

  • Propositional symbols: P, Q, S, ... (atomic sentences)

  • Sentences are combined by connectives:

    ...and [conjunction]

    ...or [disjunction]

    ...implies [implication / conditional]

    ..is equivalent [biconditional]

    ...not [negation]

  • Literal: atomic sentence or negated atomic sentence


Truth Tables

p q

0 0 0 0 0 1 1

0 1 0 1 1 1 0

1 0 0 1 1 0 0

1 1 1 1 0 1 1


Examples of pl sentences
Examples of PL sentences

  • P means “It is hot.”

  • Q means “It is humid.”

  • R means “It is raining.”

  • (P  Q)  R

    “If it is hot and humid, then it is raining”

  • Q  P

    “If it is humid, then it is hot”

  • A better way:

    Hot = “It is hot”

    Humid = “It is humid”

    Raining = “It is raining”


Example

s: Aya goes out for a walk.

t: The moon is out.

u: It is snowing.

: If the moon is out and it is not snowing, then

Aya goes out for a walk.

If it is snowing and the moon is not out, then Aya

will not go out for a walk.


Logical Equivalence

p

q

0

0

1

1

0

1

0

1

1

1

0

0

1

1

0

1

1

1

0

1


Logical equivalence
Logical equivalence

  • Two sentences are logically equivalent} iff true in same models: α ≡ ß iff α╞ βand β╞ α


Tables of logical equivalences

Identity laws

Like adding 0

Domination laws

Like multiplying by 0

Idempotent laws

Delete redundancies

Double negation

“I don’t like you, not”

Commutativity

Like “x+y = y+x”

Associativity

Like “(x+y)+z = y+(x+z)”

Distributivity

Like “(x+y)z = xz+yz”

De Morgan

Tables of Logical Equivalences


Tables of logical equivalences1

Excluded middle

Negating creates opposite

Definition of implication in terms of Not and Or

Tables of Logical Equivalences


Fundamentals of Logic

A compound statement is called a tautology(T0) if it is

true for all truth value assignments for its component statements.

If a compound statement is false for all such assignments, then it is called a contradiction(F0).

: tautology

: contradiction


Propositional logic 2 more defn
Propositional Logic - 2 more defn…

A tautology is a proposition that’s always TRUE.

A contradiction is a proposition that’s always FALSE.

T

T

F

F


Tautology example
Tautology example

Demonstrate that

[¬p(p q )]q

is a tautology in two ways:

  • Using a truth table – show that [¬p(p q )]q is always true

  • Using a proof (will get to this later).







Derivational proof techniques
Derivational Proof Techniques

EG: consider the compound proposition

(p p )  ((sr)t) )  (qr )

Q: Why is this a tautology?


Derivational proof techniques1
Derivational Proof Techniques

A: Part of it is a tautology (p p ) and the disjunction of True with any other compound proposition is still True:

(p p )  ((sr)t ))  (qr )

  • T  ((sr)t ))  (qr )

  • T

    Derivational techniques formalize the intuition of this example.


Tautology by proof
Tautology by proof

[¬p(p q )]q


Tautology by proof1
Tautology by proof

[¬p(p q )]q

[(¬pp)(¬pq)]q Distributive


Tautology by proof2
Tautology by proof

[¬p(p q )]q

[(¬pp)(¬pq)]q Distributive

[ F (¬pq)]q ULE


Tautology by proof3
Tautology by proof

[¬p(p q )]q

[(¬pp)(¬pq)]q Distributive

[ F (¬pq)]q ULE

[¬pq ]q Identity


Tautology by proof4
Tautology by proof

[¬p(p q )]q

[(¬pp)(¬pq)]q Distributive

[ F (¬pq)]q ULE

[¬pq ]q Identity

¬[¬pq ] q ULE


Tautology by proof5
Tautology by proof

[¬p(p q )]q

[(¬pp)(¬pq)]q Distributive

[ F (¬pq)]q ULE

[¬pq ]q Identity

¬[¬pq ] q ULE

[¬(¬p)¬q ] q DeMorgan


Tautology by proof6
Tautology by proof

[¬p(p q )]q

[(¬pp)(¬pq)]q Distributive

[ F (¬pq)]q ULE

[¬pq ]q Identity

¬[¬pq ] q ULE

[¬(¬p)¬q ] q DeMorgan

[p ¬q ] q Double Negation


Tautology by proof7
Tautology by proof

[¬p(p q )]q

[(¬pp)(¬pq)]q Distributive

[ F (¬pq)]q ULE

[¬pq ]q Identity

¬[¬pq ] q ULE

[¬(¬p)¬q ] q DeMorgan

[p ¬q ] q Double Negation

p [¬q q ]Associative


Tautology by proof8
Tautology by proof

[¬p(p q )]q

[(¬pp)(¬pq)]q Distributive

[ F (¬pq)]q ULE

[¬pq ]q Identity

¬[¬pq ] q ULE

[¬(¬p)¬q ] q DeMorgan

[p ¬q ] q Double Negation

p [¬q q ]Associative

p [q ¬q ]Commutative


Tautology by proof9
Tautology by proof

[¬p(p q )]q

[(¬pp)(¬pq)]q Distributive

[ F (¬pq)]q ULE

[¬pq ]q Identity

¬[¬pq ] q ULE

[¬(¬p)¬q ] q DeMorgan

[p ¬q ] q Double Negation

p [¬q q ]Associative

p [q ¬q ]Commutative

p T ULE


Tautology by proof10
Tautology by proof

[¬p(p q )]q

[(¬pp)(¬pq)]q Distributive

[ F (¬pq)]q ULE

[¬pq ]q Identity

¬[¬pq ] q ULE

[¬(¬p)¬q ] q DeMorgan

[p ¬q ] q Double Negation

p [¬q q ]Associative

p [q ¬q ]Commutative

p T ULE

T Domination


Examples
Examples

  • “I don’t study well and fail” is logically equivalent to “If I study well, then I don’t fail”

  • Write a C program that represents the compound proposition (pq)r


Use truth table to find
Use truth table to find

- P  Q  P  R

P  -Q  R  P  R  - Q

A  B  -C  D  E  F


Limitations of propositional logic
Limitations of propositional logic

  • So far we studied propositional logic

  • Some English statements are hard to model in propositional logic:

  • “If your roommate is wet because of rain, your roommate must not be carrying any umbrella”

  • Pathetic attempt at modeling this:

  • RoommateWetBecauseOfRain => (NOT(RoommateCarryingUmbrella0) AND NOT(RoommateCarryingUmbrella1) AND NOT(RoommateCarryingUmbrella2) AND …)


Problems with propositional logic
Problems with propositional logic

  • No notion of objects

  • No notion of relations among objects

  • RoommateCarryingUmbrella0 is instructive to us, suggesting

    • there is an object we call Roommate,

    • there is an object we call Umbrella0,

    • there is a relationship Carrying between these two objects

  • Formally, none of this meaning is there

    • Might as well have replaced RoommateCarryingUmbrella0 by P



Constants
Constants

  • Constants refer to objects, functions and relationships.

    Ahmed, Mona, loves, happy,

  • Simple sentences express relationships among objects.

    loves(Ahmed, Mona)

    They are called atoms.

  • Compound sentences capture relationships among relations.

    loves(x,y) Þ loves(y,x)

    loves(x,y) Ù loves(y,x) Þ happy(x)

  • Relations can be unary as well.

    tall(Tomy)


Elements of first order logic
Elements of first-order logic

  • Objects:can give these names such as Umbrella0, Person0, John, Earth, …

  • Relations:Carrying(., .), IsAnUmbrella(.)

    • Carrying(Person0, Umbrella0), IsUmbrella(Umbrella0)

    • Relations with one object = unary relations = properties

  • Functions: Roommate(.)

    • Roommate(Person0)

  • Equality: Roommate(Person0) = Person1


Example with functions
Example with Functions

E.g. Mona loves her dog.

loves(Mona, dog_of (Mona))

Note: We are allowed to relate sentences only.

So, we can say:

loves(Mona, dog_of (Mona)) Ù

loves(Mona, cat_of (Mona))

But not,

loves(Mona,

dog_of (Mona) Ù cat_of (Mona))

E.g. How about saying that Ahmed has a big nose?

Ahmed is an object and

nose_of (Ahmed)

is a function that constructs an object from the argument object.

Then, we can write:

big(nose_of (Ahmed))


First order logic1
First-Order Logic: ,

  • The language that we have described so far, consisting of atoms and the connectives (,,,,,) is typically called predicate logic.

  • To extend it to first-order logic, we need to add quantifiers.

  • The purpose of quantifiers is to allow us to say things about sets of objects.

  • To say that Heba loves everything we write:

    x. loves (Heba, x)

    We can think of  as a big conjunction. For example, if there are only three objects Heba, dog, and cat, what the above asserts is:

    loves (Heba, dog)  loves (Heba, cat)  loves (Heba, Heba)

  • To say that Hassan loves something we write:

    x. loves (Hassan, x)

    We can think of  as a big disjunction. For example, if there are only three objects as above, then what we are asserting is:

    loves (Hassan, dog)  loves (Hassan, cat)  loves (Hassan, Hassan)


  • First Order Predicate Logic

    • enriched by variables, predicates, functions

    • quantifiers , 

      friends(father(david),father(andrew))

       Y friends(Y, petr)

       X likes(X,ice_cream)

       X  Y  Z parent(X,Y)  parent(X,Z)  siblings(Y,Z)


Reasoning about many objects at once
Reasoning about many objects at once

  • Variables: x, y, z, … can refer to multiple objects

  • New operators “for all” and “there exists”

    • Universal quantifier and existential quantifier

  • for all x: CompletelyWhite(x) => NOT(PartiallyBlack(x))

    • Completely white objects are never partially black

  • there exists x: PartiallyWhite(x) AND PartiallyBlack(x)

    • There exists some object in the world that is partially white and partially black


Practice converting english to first order logic
Practice converting English to first-order logic

  • “John has an umbrella”

  • there exists y: (Has(John, y) AND IsUmbrella(y))

  • “Anything that has an umbrella is not wet”

  • for all x: ((there exists y: (Has(x, y) AND IsUmbrella(y))) => NOT(IsWet(x)))

  • “Any person who has an umbrella is not wet”

  • for all x: (IsPerson(x) => ((there exists y: (Has(x, y) AND IsUmbrella(y))) => NOT(IsWet(x))))


More practice converting english to first order logic
More practice converting English to first-order logic

  • “John has at least two umbrellas”

  • there exists x: (there exists y: (Has(John, x) AND IsUmbrella(x) AND Has(John, y) AND IsUmbrella(y) AND NOT(x=y))

  • “John has at most two umbrellas”

  • for all x, y, z: ((Has(John, x) AND IsUmbrella(x) AND Has(John, y) AND IsUmbrella(y) AND Has(John, z) AND IsUmbrella(z)) => (x=y OR x=z OR y=z))


Even more practice converting english to first order logic
Even more practice converting English to first-order logic…

  • “Duke’s basketball team defeats any other basketball team”

  • for all x: ((IsBasketballTeam(x) AND NOT(x=BasketballTeamOf(Duke))) => Defeats(BasketballTeamOf(Duke), x))

  • “Every team defeats some other team”

  • for all x: (IsTeam(x) => (there exists y: (IsTeam(y) AND NOT(x=y) AND Defeats(x,y))))


Reverse translation logic…

  • Translate the following into English.

  • x hesitates(x)  lost(x)

  • He who hesitates is lost.

  • x business(x)  like(x,Showbusiness)

  • There is no business like show business.

  • x glitters(x)  gold(x)

  • Not everything that glitters is gold.

  • x t person(x)  time(t)  canfool(x,t)

  • You can fool some of the people all the time.


Translating english to fol
Translating English to FOL logic…

Every gardener likes the sun.

x gardener(x)  likes(x,Sun)

You can fool some of the people all of the time.

x t person(x) time(t)  can-fool(x,t)

You can fool all of the people some of the time.

x t (person(x)  time(t) can-fool(x,t))

x (person(x)  t (time(t) can-fool(x,t)))

All purple mushrooms are poisonous.

x (mushroom(x)  purple(x))  poisonous(x)

No purple mushroom is poisonous.

x purple(x)  mushroom(x)  poisonous(x)

x (mushroom(x)  purple(x)) poisonous(x)

There are exactly two purple mushrooms.

x y mushroom(x)  purple(x)  mushroom(y)  purple(y) ^ (x=y)  z (mushroom(z)  purple(z))  ((x=z)  (y=z))

Clinton is not tall.

tall(Clinton)

X is above Y iff X is on directly on top of Y or there is a pile of one or more other objects directly on top of one another starting with X and ending with Y.

x y above(x,y) ↔ (on(x,y) z (on(x,z)  above(z,y)))

Equivalent

Equivalent


Resolution for first order logic
Resolution for first-order logic logic…

  • for all x: (NOT(Knows(John, x)) OR IsMean(x) OR Loves(John, x))

    • John loves everything he knows, with the possible exception of mean things

  • for all y: (Loves(Jane, y) OR Knows(y, Jane))

    • Jane loves everything that does not know her

  • What can we unify? What can we conclude?

  • Use the substitution: {x/Jane, y/John}

  • Get: IsMean(Jane) OR Loves(John, Jane) OR Loves(Jane, John)

  • Complete (i.e., if not satisfiable, will find a proof of this), if we can remove literals that are duplicates after unification

    • Also need to put everything in canonical form first


Properties of quantifiers
Properties of quantifiers logic…

  • x y is the same as yx

  • x y is the same as yx

  • x y is not the same as yxx y Loves(x,y)

    • “There is a person who loves everyone in the world” yx Loves(x,y)

    • “Everyone in the world is loved by at least one person”

  • Quantifier duality: each can be expressed using the other x Likes(x,IceCream) x Likes(x,IceCream) x Likes(x,Broccoli) xLikes(x,Broccoli)


Using fol
Using FOL logic…

  • Brothers are siblings

  •  x,y Brother(x,y)  Sibling(x,y)

  • One's mother is one's female parent

  •  m,c Mother(c) = m  (Female(m)  Parent(m,c))

  • “Sibling” is symmetric

  •  x,y Sibling(x,y)  Sibling(y,x)

  • A first cousin is a child of a parent’s sibling

  •  x,y FirstCousin(x,y)  p,ps Parent(p,x) Sibling(ps,p)  Parent(ps,y)


An example
An example logic…

  • Sameh is a lawyer.

  • Lawyers are rich.

  • Rich people have big houses.

  • Big houses are a lot of work.

  • We would like to conclude that Sameh’s house is a lot of work.

  • Natural languages are ambiguous so we can have different axiomatizations.


Axiomatization 1
Axiomatization 1 logic…

  • lawyer(Sameh)

  • x lawyer(x)  rich(x)

  • x rich(x)  y house(x,y)

  • x,y rich(x)  house(x,y)  big(y)

  • x,y ( house(x,y)  big(y)  work(y) )

  • 3 and 4, say that rich people do have at least one house and all their houses are big.

  • Conclusion we want to show:

    house(Sameh, S_house)  work(Sameh, S_house)

  • Or, do we want to conclude that John has at least one house that needs a lot of work? I.e.

  • y house(Sameh,y)  work(y)


Amir and the cat
Amir and the cat logic…

  • Everyone who loves all animals is loved by someone.

  • Anyone who kills an animal is loved by no one.

  • Mohamed loves all animals.

  • Either Mohamed or Amir killed the cat, who is named SoSo.

  • Did Amir kill the cat?


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