# CS344: Artificial Intelligence

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CS344: Artificial Intelligence
Pushpak Bhattacharyya CSE Dept., IIT Bombay Lecture 12, 13– Predicate Calculus and Knowledge Representation

Logic and inferencing Vision NLP Search Reasoning Learning Knowledge Robotics Expert Systems Planning Obtaining implication of given facts and rules -- Hallmark of intelligence

Inferencing through Deduction Induction
Deduction (General to specific) Induction (Specific to General) Abduction (Conclusion to hypothesis in absence of any other evidence to contrary) Deduction Given: All men are mortal (rule) Shakespeare is a man (fact) To prove: Shakespeare is mortal (inference) Induction Given: Shakespeare is mortal Newton is mortal (Observation) Dijkstra is mortal To prove: All men are mortal (Generalization)

If there is rain, then there will be no picnic
Fact1: There was rain Conclude: There was no picnic Deduction Fact2: There was no picnic Conclude: There was no rain (?) Induction and abduction are fallible forms of reasoning. Their conclusions are susceptible to retraction Two systems of logic 1) Propositional calculus 2) Predicate calculus

Propositions Stand for facts/assertions Declarative statements As opposed to interrogative statements (questions) or imperative statements (request, order) Operators => and ¬ form a minimal set (can express other operations) - Prove it. Tautologies are formulae whose truth value is always T, whatever the assignment is

Model In propositional calculus any formula with n propositions has 2n models (assignments) - Tautologies evaluate to T in all models. Examples: 1) 2) e Morgan with AND

Semantic Tree/Tableau method of proving tautology
Start with the negation of the formula - α - formula α-formula β-formula - β - formula α-formula - α - formula

Example 2: Contradictions in all paths X (α - formula) (α - formulae)
B C Contradictions in all paths B C

Exercise: Prove the backward implication in the previous example

Inferencing in PC Backward chaining Resolution Forward chaining

Knowledge Procedural Declarative Declarative knowledge deals with factoid questions (what is the capital of India? Who won the Wimbledon in 2005? Etc.) Procedural knowledge deals with “How” Procedural knowledge can be embedded in declarative knowledge

Example: Employee knowledge base
Employee record Emp id : 1124 Age : 27 Salary : 10L / annum Tax : Procedure to calculate tax from basic salary, Loans, medical factors, and # of children

Text Knowledge Representation

A Semantic Graph bought student June computer new past tense agent
in: modifier a: indefinite the: definite student past tense agent bought object time computer new June modifier The student bought a new computer in June.

Representation of Knowledge
UNL representation Representation of Knowledge Ram is reading the newspaper

UNL: a United Nations project
Dave, Parikh and Bhattacharyya, Journal of Machine Translation, 2002 Started in 1996 10 year program 15 research groups across continents First goal: generators Next goal: analysers (needs solving various ambiguity problems) Current active language groups UNL_French (GETA-CLIPS, IMAG) UNL_Hindi (IIT Bombay with additional work on UNL_English) UNL_Italian (Univ. of Pisa) UNL_Portugese (Univ of Sao Paolo, Brazil) UNL_Russian (Institute of Linguistics, Moscow) UNL_Spanish (UPM, Madrid)

Knowledge Representation
UNL Graph - relations read agt obj Ram newspaper

Knowledge Representation
UNL Graph - UWs read(icl>interpret) obj agt Ram(iof>person) newspaper(icl>print_media)

Knowledge Representation
UNL graph - attributes @entry @present @progress read(icl>interpret) obj agt @def newspaper(icl>print_media) Ram(iof>person) Ram is reading the newspaper

The boy who works here went to school
Another Example The boy who works here went to school plt agt @ past school(icl>institution) go(icl>move) boy(icl>person) work(icl>do) here @ entry plc :01

Predicate Calculus

Predicate Calculus: well known examples
Man is mortal : rule ∀x[man(x) → mortal(x)] shakespeare is a man man(shakespeare) To infer shakespeare is mortal mortal(shakespeare)

Forward Chaining/ Inferencing
man(x) → mortal(x) Dropping the quantifier, implicitly Universal quantification assumed man(shakespeare) Goal mortal(shakespeare) Found in one step x = shakespeare, unification

Backward Chaining/ Inferencing
man(x) → mortal(x) Goal mortal(shakespeare) x = shakespeare Travel back over and hit the fact asserted man(shakespeare)

Resolution - Refutation
man(x) → mortal(x) Convert to clausal form ~man(shakespeare) mortal(x) Clauses in the knowledge base man(shakespeare) mortal(shakespeare)

Resolution – Refutation contd
Negate the goal ~man(shakespeare) Get a pair of resolvents

Resolution Tree

Search in resolution Heuristics for Resolution Search
Goal Supported Strategy Always start with the negated goal Set of support strategy Always one of the resolvents is the most recently produced resolute

Inferencing in Predicate Calculus
Forward chaining Given P, , to infer Q P, match L.H.S of Assert Q from R.H.S Backward chaining Q, Match R.H.S of assert P Check if P exists Resolution – Refutation Negate goal Convert all pieces of knowledge into clausal form (disjunction of literals) See if contradiction indicated by null clause can be derived

P converted to Draw the resolution tree (actually an inverted tree). Every node is a clausal form and branches are intermediate inference steps.

Terminology Pair of clauses being resolved is called the Resolvents. The resulting clause is called the Resolute. Choosing the correct pair of resolvents is a matter of search.

Predicate Calculus Rules
Introduction through an example (Zohar Manna, 1974): Problem: A, B and C belong to the Himalayan club. Every member in the club is either a mountain climber or a skier or both. A likes whatever B dislikes and dislikes whatever B likes. A likes rain and snow. No mountain climber likes rain. Every skier likes snow. Is there a member who is a mountain climber and not a skier? Given knowledge has: Facts Rules

Predicate Calculus: Example contd.
Let mc denote mountain climber and sk denotes skier. Knowledge representation in the given problem is as follows: member(A) member(B) member(C) ∀x[member(x) → (mc(x) ∨ sk(x))] ∀x[mc(x) → ~like(x,rain)] ∀x[sk(x) → like(x, snow)] ∀x[like(B, x) → ~like(A, x)] ∀x[~like(B, x) → like(A, x)] like(A, rain) like(A, snow) Question: ∃x[member(x) ∧ mc(x) ∧ ~sk(x)] We have to infer the 11th expression from the given 10. Done through Resolution Refutation.

Club example: Inferencing
member(A) member(B) member(C) Can be written as

Negate–

Now standardize the variables apart which results in the following
member(A) member(B) member(C)

10 7 12 5 4 13 14 2 11 15 16 13 2 17

Assignment Prove the inferencing in the Himalayan club example with different starting points, producing different resolution trees. Think of a Prolog implementation of the problem Prolog Reference (Prolog by Chockshin & Melish)

From predicate calculus
Problem-2 From predicate calculus

A “department” environment
Dr. X is the HoD of CSE Y and Z work in CSE Dr. P is the HoD of ME Q and R work in ME Y is married to Q By Institute policy staffs of the same department cannot marry All married staff of CSE are insured by LIC HoD is the boss of all staff in the department

Diagrammatic representation
CSE ME Dr. P Dr. X Z Y R Q married

Questions on “department”
Who works in CSE? Is there a married person in ME? Is there somebody insured by LIC?

Problem-3 (Zohar Manna, Mathematical Theory of Computation, 1974)
From Propositional Calculus

Tourist in a country of truth-sayers and liers
Facts and Rules: In a certain country, people either always speak the truth or always lie. A tourist T comes to a junction in the country and finds an inhabitant S of the country standing there. One of the roads at the junction leads to the capital of the country and the other does not. S can be asked only yes/no questions. Question: What single yes/no question can T ask of S, so that the direction of the capital is revealed?

Diagrammatic representation
Capital S (either always says the truth Or always lies) T (tourist)

Deciding the Propositions: a very difficult step- needs human intelligence
P: Left road leads to capital Q: S always speaks the truth

Meta Question: What question should the tourist ask
The form of the question Very difficult: needs human intelligence The tourist should ask Is R true? The answer is “yes” if and only if the left road leads to the capital The structure of R to be found as a function of P and Q

A more mechanical part: use of truth table
Q S’s Answer R T Yes F No

Get form of R: quite mechanical
From the truth table R is of the form (P x-nor Q) or (P ≡ Q)

Get R in English/Hindi/Hebrew…
Natural Language Generation: non-trivial The question the tourist will ask is Is it true that the left road leads to the capital if and only if you speak the truth? Exercise: A more well known form of this question asked by the tourist uses the X-OR operator instead of the X-Nor. What changes do you have to incorporate to the solution, to get that answer?

From Propositional Calculus
Problem-4 From Propositional Calculus

Another tourist example: this time in a restaurant setting in a different country (Manna, 1974)
Facts: A tourist is in a restaurant in a country when the waiter tells him: “do you see the three men in the table yonder? One of them is X who always speaks the truth, another is Y who always lies and the third is Z who sometimes speaks the truth and sometimes lies, i.e., answers yes/no randomly without regard to the question. Question: Can you (the tourist) ask three yes/no questions to these men, always indicating who should answer the question, and determine who of them is X, who y and who Z?

Solution: Most of the steps are doable by humans only
Number the persons: 1, 2, 3 1 can be X/Y/Z 2 can be X/Y/Z 3 can be X/Y/Z Let the first question be to 1 One of 2 and 3 has to be NOT Z. Critical step in the solution: only humans can do?

Now cast the problem in the same setting as the tourist and the capital example
Solving by analogy Use of previously solved problems Hallmark of intelligence

Analogy with the tourist and the capital problem
Find the direction to the capital  Find Z; who amongst 1, 2 and 3 is Z? Ask a single yes/no question to S (the person standing at the junction)  Ask a single yes/no question to 1 Answer forced to reveal the direction of the capital  Answer forced to reveal who from 1,2,3 is Z

Question to 1 Ask “Is R true” and the answer is yes if and only if 2 is not Z Propositions P: 2 is not Z Q: 1 always speaks the truth, i.e., 1 is X

Use of truth table as before
P Q 1’s Answer R T Yes F No

Question to 1: the first question
Is it true that 2 is not Z if and only if you are X?

Analysis of 1’s answer Ans= yes
Case 1: 1 is X/Y (always speaks the truth or always lies) 2 is indeed not Z (we can trust 1’s answer) Case 2: 1 is Z 2 is indeed not Z (we cannot trust 1’s answer; but that does not affect us)

Ans= no Case 1: 1 is X/Y (always speaks the truth or always lies) 2 is Z; hence 3 is not Z Case 2: 1 is Z 3 is not Z Note carefully: how cleverly Z is identified. Can a machine do it?

Next steps: ask the 2nd question to determine X/Y
Once “Not Z” is identified- say 2, ask him a tautology Is P≡P If yes, 2 is X If no, 2 is Y

Ask the 3rd Question Ask 2 “is 1 Z” If 2 is X If 2 is Y (always lies)
Ans=yes, 1 is Z Ans=no, 1 is Y If 2 is Y (always lies) Ans=yes, 1 is X Ans=no, 1 is Z 3 is the remaining person

What do these examples show?
Logic systematizes the reasoning process Helps identify what is mechanical/routine/automatable Brings to light the steps that only human intelligence can perform These are especially of foundational and structural nature (e.g., deciding what propositions to start with) Algorithmizing reasoning is not trivial