Artificial Intelligence

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 means of improving recall efficiency

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
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 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 if type(v)>type(w) or instance(w)::type(v) dog:*X colour brown dog:Emma colour brown animal dog cat

Types of Knowledge Objects Events Performance META Knowledge
both physical & concepts Events usually involve time maybe cause & effect relationships Performance how to do things META Knowledge knowledge about how to use knowledge

Proposition logic

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 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

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 1 1 1 1 1

Logical equivalence Two sentences are logically equivalent} iff true in same models: α ≡ ß iff α╞ β and 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 L3

Tables of Logical Equivalences
Excluded middle Negating creates opposite Definition of implication in terms of Not and Or

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…
A tautology is a proposition that’s always TRUE. A contradiction is a proposition that’s always FALSE. p p p  p p  p T F T F

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).

Tautology by truth table
p q ¬p p q ¬p (p q ) [¬p (p q )]q T F

Derivational Proof Techniques
EG: consider the compound proposition (p p )  ((sr)t) )  (qr ) Q: Why is this a tautology?

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. L3

Tautology by proof [¬p (p q )]q L3

Tautology by proof [¬p (p q )]q  [(¬p p)(¬p q)]q Distributive

 [ F  (¬p q)]q ULE L3

 [ F  (¬p q)]q ULE  [¬p q ]q Identity L3

 [ F  (¬p q)]q ULE  [¬p q ]q Identity  ¬ [¬p q ]  q ULE L3

 [ F  (¬p q)]q ULE  [¬p q ]q Identity  ¬ [¬p q ]  q ULE  [¬(¬p) ¬q ]  q DeMorgan L3

 [ F  (¬p q)]q ULE  [¬p q ]q Identity  ¬ [¬p q ]  q ULE  [¬(¬p) ¬q ]  q DeMorgan  [p  ¬q ]  q Double Negation L3

 [ 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 L3

 [ 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 L3

 [ 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 L3

 [ 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 L3

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 - P  Q  P  R P  -Q  R  P  R  - Q
A  B  -C  D  E  F

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
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

First-Order Logic Syntax

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
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
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 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
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
“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
“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…
“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 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
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
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
x y is the same as y x x y is the same as y x x y Loves(x,y) y x Loves(x,y) “Everyone in the world is loved by at least one person” x Likes(x,Broccoli) x Likes(x,Broccoli)

Using FOL Brothers are siblings One's mother is one's female parent
 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 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 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 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?

first-order 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

Converting sentences to CNF
1. Eliminate all ↔ connectives (P ↔ Q)  ((P  Q) ^ (Q  P)) 2. Eliminate all  connectives (P  Q)  (P  Q) 3. Reduce the scope of each negation symbol to a single predicate P  P (P  Q)  P  Q (P  Q)  P  Q (x)P  (x)P (x)P  (x)P 4. Standardize variables: rename all variables so that each quantifier has its own unique variable name

Converting sentences 5. Eliminate existential quantification by introducing Skolem constants/functions (x)P(x)  P(c) c is a Skolem constant (a brand-new constant symbol that is not used in any other sentence) (x)(y)P(x,y)  (x)P(x, f(x)) since  is within the scope of a universally quantified variable, use a Skolem function f to construct a new value that depends on the universally quantified variable f must be a brand-new function name not occurring in any other sentence in the KB. E.g., (x)(y)loves(x,y)  (x)loves(x,f(x)) In this case, f(x) specifies the person that x loves

Types of Inference Model Checking Forward chaining with modus ponens
Backward chaining with modus ponens

Model Checking Enumerate all possible worlds
Restrict to possible worlds in which the KB is true Check whether the goal is true in those worlds or not

Inference as Search State: current set of sentences
Operator: sound inference rules to derive new entailed sentences from a set of sentences Can be goal directed if there is a particular goal sentence we have in mind Can also try to enumerate every entailed sentence

Generalized Modus Ponens

Modus Ponens - special case of Resolution
p Þ q p q Sunday Þ Dr Yasser is teaching AI Sunday Dr Yasser teaching AI Using the tricks: p Þ q Þ p p Þ p Ù q q, i.e. q

Sound rules of inference
Each can be shown to be sound using a truth table RULE PREMISE CONCLUSION Modus Ponens A, A  B B And Introduction A, B A  B And Elimination A  B A Double Negation A A Unit Resolution A  B, B A Resolution A  B, B  C A  C

An example 2. Eliminate  3. Reduce scope of negation
(x)(P(x)  ((y)(P(y)  P(f(x,y)))  (y)(Q(x,y)  P(y)))) 2. Eliminate  (x)(P(x)  ((y)(P(y)  P(f(x,y)))  (y)(Q(x,y)  P(y)))) 3. Reduce scope of negation (x)(P(x)  ((y)(P(y)  P(f(x,y))) (y)(Q(x,y)  P(y)))) 4. Standardize variables (x)(P(x)  ((y)(P(y)  P(f(x,y))) (z)(Q(x,z)  P(z)))) 5. Eliminate existential quantification (x)(P(x) ((y)(P(y)  P(f(x,y))) (Q(x,g(x))  P(g(x))))) 6. Drop universal quantification symbols (P(x)  ((P(y)  P(f(x,y))) (Q(x,g(x))  P(g(x)))))

Two broad kinds of rule system
forward chaining systems, and backward chaining systems. In a forward chaining system you start with the initial facts, and keep using the rules to draw new conclusions (or take certain actions) given those facts In a backward chaining system you start with some hypothesis (or goal) you are trying to prove, and keep looking for rules that would allow you to conclude that hypothesis, perhaps setting new subgoals to prove as you go.

Forward chaining Proofs start with the given axioms/premises in KB, deriving new sentences until the goal/query sentence is derived This defines a forward-chaining inference procedure because it moves “forward” from the KB to the goal [eventually]

Forward chaining Idea: fire any rule whose premises are satisfied in the KB, add its conclusion to the KB, until query is found

Forward chaining example

Backward chaining Proofs start with the goal query, find rules with that conclusion, and then prove each of the antecedents in the implication Keep going until you reach premises Avoid loops: check if new sub-goal is already on the goal stack Avoid repeated work: check if new sub-goal Has already been proved true Has already failed

Backward Chaining Is Tom faster than someone?

Forward chaining example
KB: allergies(X)  sneeze(X) cat(Y)  allergic-to-cats(X)  allergies(X) cat(Felix) allergic-to-cats(Lise) Goal: sneeze(Lise)

Exercise You go to the doctor and for insurance reasons they perform a test for a horrible disease You test positive The doctor says the test is 99% accurate Do you worry?

Reduction to propositional inference
Suppose the KB contains just the following: x King(x)  Greedy(x)  Evil(x) King(Ali) Greedy(Ali) Brother(Saad, Ali) Instantiating the universal sentence in all possible ways, we have: King(John)  Greedy(John)  Evil(John) King(Richard)  Greedy(Richard)  Evil(Richard) King(John) Greedy(John) Brother(Richard,John) The new KB is propositionalized: proposition symbols are King(John), Greedy(John), Evil(John), King(Richard), etc.

An example 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.

Axiomatization 1 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 Sameh has at least one house that needs a lot of work? I.e. y house(Sameh,y)  work(y)

Hassan and the cat Every animal owner is an animal lover
Everyone who loves all animals is loved by someone. Anyone who kills an animal is loved by no one. Mustafa owns a dog. Either Mustafa or Hassan killed the cat, who is named SoSo. Did Hassan kill the cat?

Practice example Did Hassan kill the cat
Mustafa owns a dog. Every dog owner is an animal lover. No animal lover kills an animal. Either Hassan or Mustafa killed the cat, who is named SoSo . Did Hassan kill the cat? These can be represented as follows: A. (x) Dog(x)  Owns(Mustafa ,x) B. (x) ((y) Dog(y)  Owns(x, y))  AnimalLover(x) C. (x) AnimalLover(x)  ((y) Animal(y)  Kills(x,y)) D. Kills(Mustafa ,SoSo)  Kills(Hassan,SoSo) E. Cat(SoSo) F. (x) Cat(x)  Animal(x) G. Kills(Hassan, SoSo) GOAL

Add the negation of query:
Convert to clause form A1. (Dog(D)) A2. (Owns(Mustafa,D)) B. (Dog(y), Owns(x, y), AnimalLover(x)) C. (AnimalLover(a), Animal(b), Kills(a,b)) D. (Kills(Mustafa,SoSo), Kills(Hassan,SoSo)) E. Cat(SoSo) F. (Cat(z), Animal(z)) Add the negation of query: G: (Kills(Hassan, SoSo))

The resolution refutation proof
R1: G, D, {} (Kills(Mustafa,SoSo)) R2: R1, C, {a/Mustafa, b/SoSo} (~AnimalLover(Mustafa), ~Animal(SoSo)) R3: R2, B, {x/Mustafa} (~Dog(y), ~Owns(Mustafa, y), ~Animal(SoSo)) R4: R3, A1, {y/D} (~Owns(Mustafa, D), ~Animal(SoSo)) R5: R4, A2, {} (~Animal(SoSo)) R6: R5, F, {z/SoSo} (~Cat(SoSo)) R7: R6, E, {} FALSE

The proof tree G D {} R1: K(J,T) C {a/J,b/T} R2: AL(J)  A(T) B
{x/J} R3: D(y)  O(J,y)  A(T) A1 {y/D} R4: O(J,D), A(T) A2 {} R5: A(T) F {z/T} R6: C(T) A {} R7: FALSE

Umbrellas in first-order logic
You know the following things: You have exactly one other person living in your house, who is wet If a person is wet, it is because of the rain, the sprinklers, or both If a person is wet because of the sprinklers, the sprinklers must be on If a person is wet because of rain, that person must not be carrying any umbrella There is an umbrella that “lives in” your house, which is not in its house An umbrella that is not in its house must be carried by some person who lives in that house You are not carrying any umbrella Can you conclude that the sprinklers are on?

Example knowledge base
The law says that it is a crime for an American to sell weapons to hostile nations. The country Nono, an enemy of America, has some missiles, and all of its missiles were sold to it by Colonel West, who is American. Prove that Col. West is a criminal

Example knowledge base
... it is a crime for an American to sell weapons to hostile nations: American(x)  Weapon(y)  Sells(x,y,z)  Hostile(z)  Criminal(x) Owns(Nono,M1) and Missile(M1) … all of its missiles were sold to it by Colonel West Missile(x)  Owns(Nono,x)  Sells(West,x,Nono) Missile(x)  Weapon(x) An enemy of America counts as "hostile“: Enemy(x,America)  Hostile(x) American(West) Enemy(Nono,America)

Resolution proof: definite clauses
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Rule-Based Systems Also known as “production systems” or “expert systems” Rule-based systems are one of the most successful AI paradigms Used for synthesis (construction) type systems Also used for analysis (diagnostic or classification) type systems

Rule-Based Systems Instead of representing knowledge in a relatively declarative, static way (as a bunch of things that are true), rule-based system represent knowledge in terms of a bunch of rules that tell you what you should do or what you could conclude in different situations. A rule-based system consists of a bunch of IF-THEN rules, a bunch of facts, and some interpreter controlling the application of the rules, given the facts.

IF (lecturing X) AND (marking-practicals X) THEN ADD (overworked X)
IF (month february) THEN ADD (lecturing ali) IF (month february) THEN ADD (marking-practicals ali) IF (overworked X) OR (slept-badly X) THEN ADD (bad-mood X) IF (bad-mood X) THEN DELETE (happy X) IF (lecturing X) THEN DELETE (researching X)

Rule Based Reasoning The advantages of rule-based approach:
The ability to use Good performance Good explanation The disadvantage are Cannot handle missing information Knowledge tends to be very task dependent

Other Reasoning There exist some other approaches as:
Case-Based Reasoning Model-Based Reasoning Hybrid Reasoning Rule-based + case-based Rule-based + model-based Model-based + case-based

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