CS344: Introduction to Artificial Intelligence. Pushpak Bhattacharyya CSE Dept., IIT Bombay Lecture 10 Club and Circuit

Transcription

1 CS344: Introduction to Artificial Intelligence Pushpak Bhattacharyya CSE Dept., IIT Bombay Lecture 10 Club and Circuit Examples

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

3 Resolution Refutation contd Negate the goal ~mortal(shakespeare Get a pair of resolvents ~ mortal ( shakespear e ~ man ( x mortal ( x ~ man ( shakespear e man ( shakespear e

4 Resolution Tree Resolvent 1 Resolvent 2 Resolute

5 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

6 Inferencing in Predicate Calculus Forward chaining Given P, P Q, to infer Q P, match LHSof L.H.S Assert Q from R.H.S Backward chaining Q, Match hrhs R.H.S of P Q 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

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

8 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.

9 Himalayan Club example 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

10 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.

11 Club example: Inferencing 1. member(a 2. member(b 3. member(c x[ member( x ( mc( x sk( x] Can be written as [ member ( x ( mc( x sk( x] ~ member( x mc( x sk( x x [ sk ( x lk ( x, snow ] ~ sk( x lk( x, snow x [ mc ( x ~ lk ( x, rain ] ~ mc( x ~ lk( x, rain x [ like ( A, x ~ lk ( B, x ] ~ like( A, x ~ lk( B, x

12 8. x[~ lk( A, x lk( B, x] 9. lk( A, rain lk( A, x lk( B, x lk( A, snow x [ member ( x mc ( x ~ sk ( x ] Negate x[~ member( x ~ mc( x sk( x]

13 Now standardize the variables apart which results in the following 1. member(a 2. member(b 3. member(c 4. ~ member ( x mc ( x1 sk ( x ~ 2 sk( x2 lk( x, snow mc ( x3 ~ lk ( x, rain ~ 3 ~ like( A, x ~ lk( B, x 4 4 lk ( A, x lk ( B, x 5 5 lk( A, rain lk(a (, snow ~ member( x ~ mc( x6 sk( x 6 6

14 ~ 4 7 like( A, x ~ lk( B, x 4 lk( A, snow ~ lk ( B, snow ~ sk( x2 lk( x2, snow 5 ~ ( ( 1 ( 1 13 ~ sk( B member x mc x sk x 4 1 ~ member( B mc( B member(b ~ member ( x ~ mc ( x6 sk ( x 6 6 mc(b 15 ~ member ( B sk ( B ~ sk ( B ~ member ( B member(b 17 2

15 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

16 Application of Predicate Calculus Systematic Inferencing Knowledge Representation ti Puzzles Circuit Verification Robotics Intelligent DB

17 Circuit Verification Does the circuit meet the specs? Are there faults? are they locatable?

18 Example : 2-bit full adder C1 X2 X1 Y C X 1, X 2 : inputs; C 1 : prev. carry; C 2 : next carry; Y: output t

19 K-Map c1 x2x1 0 1 Y Y = C1( X1 X 2 + C1( X1 X 2 = ( C1 ( X 1 X 2

20 K-Map (contd.. c1 x2x1 0 1 C C 2 = X 2 X 1 + C1( X 1 + X 2

21 Circuit

22 Verification First task (most difficult Building blocks : predicates Circuit observation : Assertion on terminals

23 Predicates & Functions Function 1 signal(t t is a terminal ; signal takes the value 0 or 1 Function 2 type(x x is a circuit element; type(x takes the value AND, OR, NOT, XOR Predicate 3 connected(t1,t2 t1 is an output terminal and t2 is an input terminal Function 3 In(n,x n th input of ckt element x Function 4 Out(x ( Output of ckt element x

24 Alternate Full Adder Circuit

25 Functions type(x : takes values AND, OR NOT and XOR, where X is a gate. in(n, X : the value of signal at the n th input of gate X. out(x : output of gate X. signal(t : state at terminal t = 1/0 Predicates connected(t1,t2: true, if terminal t1 and t2 are connected

26 General Properties Commutativity: t 1 1,,t 2 [connected(t 1,t 2 connected(t 2,t 1 ] By definition of connection: t 1,t 2 [connected(t 1,t 2 { signal(t 1 = signal(t 1 }]

27 Gate properties 1. OR definition: X [{type(x = OR} {(out(x = 1 y (in(y, X = 1}] 2. AND definition: iti X [{type(x = AND} {(out(x = 1 y (in(y, X = 1}]

28 Gate properties contd 1. XOR definition: X [{type(x = XOR} {(out(x = 1 (in(1, X in(2, X}] 2. NOT definition: iti X [{type(x = NOT} {out(x in(1, X} ( no_of_inpu t(x = 1]

29 Some necessary functions a. no_of_input(x, takes values from N. b. Count_ls(x, returns #ls in the input of X [{type(x = XOR} X {(out(x = 1 odd((count_ls(x}]

30 Circuit specific properties Connectivity: connected(x 1, in(1,a 1, connected(x 1,in(2,A 1, connected(out(a 1, in(1, A 2, connected(c1, in(2, A 2, connected(y, out(a 2 Circuit elements: type(a 1 = XOR, type(a 2 = XOR, type(a 3 = AND