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

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

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Inferencing through − 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)

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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

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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

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

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

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Example 2: BC BC Contradictions in all paths X α-formula (α - formulae) (β - formulae) (α - formula)

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Exercise: Prove the backward implication in the previous example

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Inferencing in PC Resolution Forward chaining Backward chaining

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Knowledge Declarative Procedural 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

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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

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Text Knowledge Representation

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A Semantic Graph 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.

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UNL representation Ram is reading the newspaper Representation of Knowledge

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UNL: a United Nations project 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) Dave, Parikh and Bhattacharyya, Journal of Machine Translation, 2002

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Knowledge Representation Ram read newspaper agt obj UNL Graph - relations

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

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

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

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Predicate Calculus

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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)

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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

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

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

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

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Resolution Tree

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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

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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

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1. P 2. converted to 3. Draw the resolution tree (actually an inverted tree). Every node is a clausal form and branches are intermediate inference steps.

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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.

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Predicate Calculus 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

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Predicate Calculus: Example contd. Let mc denote mountain climber and sk denotes skier. Knowledge representation in the given problem is as follows: 1. member(A) 2. member(B) 3. member(C) 4. ∀ x[member(x) → (mc(x) ∨ sk(x))] 5. ∀ x[mc(x) → ~like(x,rain)] 6. ∀ x[sk(x) → like(x, snow)] 7. ∀ x[like(B, x) → ~like(A, x)] 8. ∀ x[~like(B, x) → like(A, x)] 9. like(A, rain) 10. like(A, snow) 11. Question: ∃ x[member(x) ∧ mc(x) ∧ ~sk(x)] We have to infer the 11 th expression from the given 10. Done through Resolution Refutation.

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Club example: Inferencing 1. member(A) 2. member(B) 3. member(C) 4. – Can be written as – 5. – 6. – 7. –

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8. – 9. 10. 11. – Negate–

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Now standardize the variables apart which results in the following 1. member(A) 2. member(B) 3. member(C) 4. 5. 6. 7. 8. 9. 10. 11.

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7 10 125 13 4 142 11 15 1613 17 2

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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)

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Problem-2 From predicate calculus

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A “department” environment 1. Dr. X is the HoD of CSE 2. Y and Z work in CSE 3. Dr. P is the HoD of ME 4. Q and R work in ME 5. Y is married to Q 6. By Institute policy staffs of the same department cannot marry 7. All married staff of CSE are insured by LIC 8. HoD is the boss of all staff in the department

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Diagrammatic representation Dr. X Y Z Dr. P Q R married CSEME

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

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

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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?

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

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

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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

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A more mechanical part: use of truth table PQS’s Answer R TTYesT TF F FTNoF FF T

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

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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?

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Problem-4 From Propositional Calculus

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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?

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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?

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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

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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

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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

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Use of truth table as before PQ1’s Answer R TTYesT TF F FTNoF FF T

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

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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)

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Analysis of 1’s answer (contd) 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?

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Next steps: ask the 2 nd 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

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Ask the 3 rd Question Ask 2 “is 1 Z” If 2 is X 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

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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

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