1. Module 4 KNOWLEDGE AND REASONING

2. Knowledge based agent We design agents that can form representations of the world Agents that use a process of inference to derive new representations about the world and agents that use these new representations to deduce what to do A knowledge-based agent needs to know many things: –the current state of the world –how to infer unseen properties of the world from percepts –how the world evolves over time –what it wants to achieve –what its own actions do in various circumstances

3. Knowledge Base (set of representations of facts about the world) The knowledge level or epistemological level is the most abstract; we can describe the agent by saying what it knows. For example, an automated taxi might be said to know that the Golden Gate Bridge links San Francisco and Marin County. The logical level is the level at which the knowledge is encoded into sentences. For example, the taxi might be described as having the logical sentence Links(GGBridge, SF, Marin) in its knowledge base. The implementation level is the level that runs on the agent architecture. It is the level at which there are physical representations of the sentences at the logical level. A sentence such as Links(GGBridge, SF, Marin) could be represented in the KB by the string "Links (GGBridge, SF, Marin)" contained in a list of strings; or by a "1" entry in a three-dimensional table indexed by road links and location pairs; or by a complex set of pointers connecting machine addresses corresponding to the individual symbols. The choice of implementation is very important to the efficient performance of the agent, but it is irrelevant to the logical level and the knowledge level.

4. 4 The Inference engine derives new sentences from the input and KB The inference mechanism depends on representation in KB The agent operates as follows: 1. It receives percepts from environment 2. It computes what action it should perform (by IE and KB) 3. It performs the chosen action (some actions are simply inserting inferred new facts into KB). Knowledge Base Inference Engine Input from environment Output (actions) Learning (KB update)

5. Example – Knowledge Based Agent A classical AI problem An early computer game An agent had to explore a cave made of series of interconnected rooms In one room, there was a Wumpus who would kill agent if it entered the room Some rooms have pits – agent would die if it entered it Agent had 1 arrow to kill Wumpus Goal – Locate gold and return to start without getting killed.

7. The Wumpus World environment The Wumpus computer game The agent explores a cave consisting of rooms connected by passageways. Lurking somewhere in the cave is the Wumpus, a beast that eats any agent that enters its room. Some rooms contain bottomless pits that trap any agent that wanders into the room. Occasionally, there is a heap of gold in a room. The goal is: – to collect the gold and –exit the world –without being eaten

8. Wumpus World PEAS description Performance measure: –gold +1000, death -10000 –-1 per step, -10 for using the arrow Environment: –Squares adjacent to wumpus are smelly –Squares adjacent to pit are breezy –Glitter iff gold is in the same square –Shooting kills wumpus if you are facing it, it screams –Shooting uses up the only arrow –Grabbing picks up gold if in same square –Releasing drops the gold in same square –You bump if you walk into a wall Actuators: Left turn, Right turn, Forward, Grab, Climb, Shoot, Die Sensors: Stench, Breeze, Glitter, Bump, Scream

9. Wumpus world characterization Fully Observable No – only local perception Deterministic Yes – outcomes exactly specified Episodic No – since no need to think ahead Static Yes – Wumpus and Pits do not move Discrete Yes Single-agent Yes – Wumpus is essentially a natural feature

10. The first step taken by the agent in the wumpus world, (a)The initial situation, after percept [None, None, None, None, None]. (b)After one move, with percept [None, Breeze, None, None, None]. [  stench,  breeze,  glitter,  bump,  scream][1,1] [stench,  breeze,  glitter,  bump,  scream] [1,2] [  stench, breeze,  glitter,  bump,  scream] [2,1] → Safe cell

11. Two later stages in the progress of the agent, (a) After the third move, with percept [Stench, None, None, None, None]. (b) After the fifth move, with percept [Stench, Breeze, Glitter, None, None].

12. Propositional logic: Syntax (limited expressiveness) Propositional logic is the simplest logic – illustrates basic ideas The proposition symbols P 1, P 2 etc are sentences –If S is a sentence,  S is a sentence (negation) –If S 1 and S 2 are sentences, S 1  S 2 is a sentence (conjunction) –If S 1 and S 2 are sentences, S 1  S 2 is a sentence (disjunction) –If S 1 and S 2 are sentences, S 1  S 2 is a sentence (implication) –If S 1 and S 2 are sentences, S 1  S 2 is a sentence (biconditional)

13. Symbols of propositional logic Logical constants TRUE and FALSE are sentences Proposition symbols P and Q are sentences A, V,,, are logical connectives parenthesis ( )

14. Ambiguity P A Q V R (P A Q) V R P A (Q V R) Order of precedence (highest to lowest)

15. Propositional logic: Semantics Each model specifies true/false for each proposition symbol E.g. P 1,2 P 2,2 P 3,1 falsetruefalse With these symbols, 8 possible models, can be enumerated automatically. Rules for evaluating truth with respect to a model m:  S is true iff S is false S 1  S 2 is true iff S 1 is true and S 2 is true S 1  S 2 is true iff S 1 is true or S 2 is true S 1  S 2 is true iffS 1 is false orS 2 is true i.e., is false iffS 1 is true andS 2 is false S 1  S 2 is true iffS 1  S 2 is true and S 2  S 1 is true Simple recursive process evaluates an arbitrary sentence, e.g.,  P 1,2  (P 2,2  P 3,1 ) = true  (true  false) = true  true = true

16. Truth tables for connectives If P then Q If same then true else false If P then Q and If Q then P

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

19. Inference Rules

20. DNF (SOP) to CNF (POS) for resolution refutation Disjunctive Normal Form (P  Q  R) V (X   Y  Z) Conjunctive Normal Form ( P V X)  (P V  Y)  (P V Z)  (…..) Every sentence of propositional logic is logically equivalent to a conjunction of disjuncts of literals

21. Conversion to CNF Eliminate α ß to (α  ß )  (ß  α ) Eliminate α  ß to  α V ß CNF requires  to appear only with literals For this apply  (  α) ≡ α (double-negative elimination)  (α  ß) ≡  α V  ß (de Morgan)  (α V ß) ≡  α   ß (de Morgan) Now we have a sentence with nested  and V operators applied to literals. Apply distributivity law distributing V over 

22. Resolution Inference Rule for CNF “If A or B or C is true, but not A, then B or C must be true.” “If A is false then B or C must be true, or if A is true then D or E must be true, hence since A is either true or false, B or C or D or E must be true.” Simplification

23. Q~Q v P~P v ~Q v R premises P ~Q v R R theorem Proof by resolution

24. Horn Clauses Resolution in general can be exponential in space and time. If we can reduce all clauses to “Horn clauses” resolution is linear in space and time A clause with at most 1 positive literal.e.g. Every Horn clause can be rewritten as an implication with a conjunction of positive literals in the premises and a single positive literal as a conclusion. e.g.

25. Wumpus world in propositional logic Facts are propositions: –e.g., W 44 “Wumpus is at square (4,4) –96 propositions (16 each for wumpus, stench, pit, breeze, gold, glitter) to represent a particular cave General knowledge in sentences: –e.g., W 44  (S 44  S 43  S 34 ) “if the wumpus is at (4,4), there is stench at (4,4), (4,3) and (3,4)” –“many” sentences

26. Some Atomic Propositions S12 = There is a stench in cell (1,2) B34 = There is a breeze in cell (3,4) W22 = The Wumpus is in cell (2,2) V11 = We have visited cell (1,1) OK11 = Cell (1,1) is safe. etc Some rules (R1) ~S11 => ~W11 ^ ~W12 ^ ~W21 (R2) ~S21 => ~W11 ^ ~W21 ^ ~W22 ^ ~W31 (R3) ~S12 => ~W11 ^ ~W12 ^ ~W22 ^ ~W13 (R4) S12 => W13 v W12 v W22 v W11 etc Wumpus world in propositional logic

28. Problems with propositional agent There are too many propositions to handle eg. The simple rule “don’t go forward if wumpus is in front of you” can be stated by set of 64 rules in propositional logic What if the world changes over time

29. Question bank 1. Represent each of these sentences in propositional logic. a. If I take History, I cannot take Economics. b. I must take either Japanese or Italian but not both. c. I must take at least two of cs520, cs552, and cs564. 2. Use truth tables to show the following sentences are valid, and thus that the equivalences hold: P ^ (Q ^ R) (P ^ Q) ^ R Associativity of conjunction P \/ (Q \/ R) (P \/ Q) \/ RAssociativity of disjunction P ^ Q Q ^ P Commutativity of conjunction P \/ Q Q \/ P Commutativity of disjunction P ^ (Q \/ R) (P ^ Q) \/ ( P ^ R) Distributivity of ^ over \/ P \/ (Q ^ R) (P \/ Q) ^ ( P \/ R) Distributivity of \/ over ^ ~(P ^ Q) ~Q \/ ~P de Morgan’s Law ~(P \/ Q) ~Q ^ ~P de Morgan’s Law P => Q ~Q => ~P Contraposition ~ ~P P Double negation P => Q ~P \/ Q P Q (P => Q) ^ (Q => P) P Q (P ^ Q) \/ (~Q ^ ~P) P ^ ~P False P \/ ~P True

30. Question bank Explain the steps involved in converting the propositional logic statement into CNF with a suitable example

31. Pros and cons of propositional logic Propositional logic is declarative: pieces of syntax correspond to facts Propositional logic allows partial/disjunctive/negated information (unlike most data structures and databases) Propositional logic is compositional: meaning of B 1,1  P 1,2 is derived from meaning of B 1,1 and of P 1,2 Propositional logic is context-independent (unlike natural language, where meaning depends on context)  Propositional calculus requires explicit enumeration All elephants are gray: elephant1-is-gray^elephant1-is-gray^….  logic has very limited expressive power (unlike natural language)

32. First-order predicate logic Propositional logic assumes the world contains facts First-order logic (like natural language) assumes the world contains –Objects: people, houses, numbers, colors, baseball games, wars, … –Relations: brother of, bigger than, inside, part of, has color, occurred after, owns… –Properties: red, round, bogus, prime, multistoried... –Functions: father of, best friend, one more than, plus, …

33. Examples "One plus two equals three" Objects: one, two, three, one plus two Relation: equals Function: plus. (One plus two is a name for the object that is obtained by applying the function plus to the objects one and two. Three is another name for this object.) "Squares neighboring the wumpus are smelly." Objects: wumpus, square Property: smelly Relation: neighboring. "Evil King John ruled England in 1200." Objects: John, England, 1200 Relation: ruled Properties: evil, king.

34. Syntax of FOL: Basic elements ConstantsKingJohn, 2, NUS,... PredicatesBrother, GreaterThan, Round... FunctionsSqrt, LeftLegOf, FatherOf, Cosin... Variablesx, y, a, b,... Connectives , , , ,  Equality= Quantifiers , 

35. Syntax of predicate logic Atomic sentence =predicate (term 1,...,term n ) or term 1 = term 2 Term = function (term 1,...,term n ) or constant or variable Term with no variables is called ground term E.g. Brother(KingJohn,RichardTheLionheart) E.g. > (Length(LeftLegOf(Richard)), Length(LeftLegOf(KingJohn))) E.g. GreaterThan ( Length ( LeftLegOf ( John ) ), ( Length ( LeftLegOf ( Eric ) ) ) )

36. Complex sentences Complex sentences are made from atomic sentences using connectives  S, S 1  S 2, S 1  S 2, S 1  S 2, S 1  S 2, E.g. Sibling(KingJohn,Richard)  Sibling(Richard,KingJohn)

37. Equality term 1 = term 2 is true under a given interpretation if and only if term 1 and term 2 refer to the same object e.g., definition of Sibling in terms of Parent:  x,y Sibling(x,y)  [  (x = y)   m,f  (m = f)  Parent(m,x)  Parent(f,x)  Parent(m,y)  Parent(f,y)]

38. Universal quantification  Everyone at NUS is smart:  x At(x,NUS)  Smart(x)  x P is true in a model m iff P is true with x being each possible object in the model Roughly speaking, equivalent to the conjunction of instantiations of P Example: At(KingJohn,NUS)  Smart(KingJohn)  At(Richard,NUS)  Smart(Richard)  At(Jesus,NUS)  Smart(Jesus)  … Example: “All elephants are gray” (  x )(elephant(x)  color(x, GRAY))

39. Existential quantification  Someone at NUS is smart:  x At(x,NUS)  Smart(x)  x P is true in a model m iff P is true with x being some possible object in the model Roughly speaking, equivalent to the disjunction of instantiations of P Example: At(KingJohn,NUS)  Smart(KingJohn)  At(Richard,NUS)  Smart(Richard)  At(Jesus,NUS)  Smart(Jesus) ... Example: “Someone wrote Computer Chess” (  x write(x, COMPUTER-CHESS))

40. Properties of quantifiers  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 “There is a person who loves everyone in the world”  x  y Loves(x,y) “Everyone in the world is loved by at least one person”  y  x Loves(x,y) Quantifier duality(De Morgan’s law for quantification): each can be expressed using the other  x Likes(x,IceCream)OR  x  Likes(x,IceCream)  x Likes(x,Broccoli) OR  x  Likes(x,Broccoli)

43. x loves y Loves(x,y) Everybody loves Jerry  x Loves (x, Jerry) Everybody loves somebody  x  y Loves (x, y) There is somebody whom somebody loves  y  x Loves (x, y) Nobody loves everybody   x  y Loves (x, y) There is somebody whom Rita doesn’t love  y  Loves (Rita, y) There is exactly one person whom everybody loves  y(  x Loves(x,y)   z((  w Loves (w,z)  z=y)) There are exactly two people whom Lynn Loves  x  y ((x  y)  Loves(Lynn,x)  Loves(Lynn,y)  z( Loves (Lynn,z)  (z=x  z=y))) Everybody loves himself or herself  x Loves(x,x) There is someone who loves no one besides herself or himself  x  y Loves(x,y)  (x=y) Examples on FOPL

44. Example Tony, Mike, John belong to mountaineering club. Every member of the mountaineering club is either a skier or mountain climber or both. No mountain climber likes rain, and all skiers like snow. Mike dislikes whatever Tony likes and likes whatever Tony dislikes. Represent above statements in FOPL

45. Wumpus world in FOL the objects in the environment –terms: constants, variables, functions constants: –times: 0, 1, 2,... –headings: R, L, D, U –coordinates: 1, 2, 3, 4 –locations: 16 squares –percepts: Stench, Breeze, Glitter, Bump, Scream, None –actions: Turn(Left), Turn(Right), Forward, Grab, Shoot –Agent, Wumpus functions: –Square(x,y) // [x,y] –Home(Wumpus) –Perception(s, b, g, h, y) –Heading(t), Location(t)

46. Wumpus world in FOL predicates: (true or false) –properties (of one term/object) Breezy(t) // agent feeling breeze at time t Breeze(s) // breeze blowing on square s Pit(s), Gold(s), etc. Time(x) // object x is a time Coordinate(x), Action(x), Heading(x), etc. –relations (of multiple terms/objects) At(s,t) // agent on square s at time t Adjacent (r,s) // squares r and s are adjacent Alive(x,t) // x is alive at time t Percept(p,t) // perception at time t BestAction(a,t) // action a to take at time t term = term: (true or false) –Home(Wumpus) = [3,3] –Heading(5) = U

47. Wumpus world in FOL the knowledge base –General knowledge –Facts that are fixed –Facts that may change over time Atomic sentences Compound sentences

48. Exploring Wumpus world in FOL Deciding the best action: Need to reason about the cave conditions Diagnostic rules (generate hidden causes from observed effects) ∀ s Breezy(s) ⇒ ∃ r Adjacent(r,s)  Pit(r) Causal rules (some fact causes certain percepts to be generated) ∀ r Pit(r) ⇒ [ ∀ s Adjacent(r,s) ⇒ Breezy(s)]

49. Inference Rules Modus ponens Modus Tollens And elimination And introduction

50. Inference Rules involving Quantifiers

51. An Example Proof of Resolution 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 “West is a criminal”Criminal(West) Eq. 1 Eq. 2 Eq. 3

52. Assumed FOL Missiles are weapons An enemy of America counts as hostile West is an American Nono is a country Nono an enemy of America Eq. 4 Eq. 5 Eq. 6 Eq. 7 Eq. 9 Eq. 8 Equations 1 to 9 form knowledge base

53. Apply Existential elimination to Eq. 2 From Eq. 10 and AND elimination From Eq. 4 and Universal Elimination From Eq. 12, 13 and Modus Ponens From Eq. 3 and universal elimination From Eq. 15, 10 and Modus Ponens From eq. 1 and universal elimination (three times) Eq. 10 Eq. 11 Eq. 12 Eq. 13 Eq. 14 Eq. 15 Eq. 16 Eq. 17

54. From Eq. 5 and universal elimination From Eq. 8, 18 and Modus Ponens From Eq. 6, 7, 14, 16, 19 and And-introduction From Eq. 17, 20 and Modus Ponens Eq. 18 Eq. 19 Eq. 20 Eq. 21

55. Crime problem revisited using canonical form Put original knowledge base to horn form

56. The Proof (just in four steps)

57. Answering Questions by Matching Data Base of Ground Instances Person(John) Person(Mary) Dog(Fido) Likes(John,Mary) Not(Likes (Mary, John)) Likes(Fido, Mary) Likes(Fido, John) Likes(Fido, Fido) Owners(Fido, John, Mary) Possible Queries Person(x) Likes(Mary,x) Likes(x,x) Likes(x,y) Not(Likes(z, John)) Owners(Fido,x,y) Simple Matching Token match on predicates Variables only for query arguments  const-const or const- variable match Zero, one, or multiple answers

58. Unification Q(x) P(y)  FAIL P(x) P(y)  x/y P(Marcus) P(y)  Marcus/y P(Marcus) P(Julius)  FAIL P(x,x) P(y,y)  (y/x) P(y,z)  (z/y, y/x)

59. Algorithm: Unify(L1,L2) 1. If L1 or L2 is a variable or constant, then: a) If L1 and L2 are identical, then return NIL. b) Else if L1 is a variable, then if L1 occurs in L2 then return FAIL, else return {(L2/L1)}. c) Else if L2 is a variable, then if L2 occurs in L1 then return FAIL, else return {(L1/L2)}. d) Else return FAIL. 2. If the initial predicate symbols in L1 and L2 are not identical, then return FAIL. 3. If L1 and L2 have a different number of arguments, then return FAIL 4. Set SUBST to NIL. 5. For i  1 to number of arguments in L1: a) Call Unify with the ith argument of L1 and the ith argument of L2, putting result in S. b) If S = FAIL then return FAIL. c) If S is not equal to NIL then: i. Apply S to the remainder of both L1 and L2. ii. SUBST := APPEND(S, SUBST). 6. Return SUBST.

60. Resolution Proof Each "vee" in the proof tree represents a resolution step: the two sentences at the top are the premises, and the one at the bottom is the conclusion or resolvent. The substitution is shown for each resolution To Prove S(A)

61. Resolution with refutation (proof by contradiction) or reductio ad absurdum To Prove S(A) Assume S(A) => false

62. Conversion to Normal Form Eliminate Implications Moveinwards Standardize variables Move quantifiers left Skolemize (remove existential quantifiers by elimination) Everyone has a heartEveryone has the same heart Mapping person to heart… F is a skolem function

63. Conversion to Normal Form (contd.) Drop prefix Distribute A over V Flatten nested conjunctions and disjunctions Convert disjunctions to implications (optional)

64. Resolution (refutation)

66. 1.Marcus was a man. man(Marcus) 2.Marcus was a Pompeian. Pompeian(Marcus) 3.All Pompeians were Romans.  x: Pompeian(x)  Roman(x) 4.Caesar was a ruler. ruler(Caesar) 5.All Romans were either loyal to Caesar or hated him.  x: Roman(x)  loyalto(x, Caesar) V hate(x,Caesar) 6.Everyone is loyal to someone.  x:  y: loyalto(x,y) 7.People only try to assassinate rulers they aren't loyal to.  x:  y:person(x)  ruler(y)  tryassassinate(x,y)  loyalto(x,y) 8.Marcus tried to assassinate Caesar. tryassassinate(Marcus, Caesar) 9.All men are people.  x: man(x)  person(x)

67. A Resolution Proof Axioms in clause form: 1. man(Marcus) 2. Pompeian(Marcus) 3.  Pompeian(x 1 ) v Roman(x 1 ) 4. Ruler(Caesar) 5.  Roman(x 2 ) v loyalto(x 2, Caesar) v hate(x 2, Caesar) 6. loyalto(x 3, f1(x 3 )) 7.  man(x 4 ) v  ruler(y 1 ) v  tryassassinate(x 4, y 1 ) v  loyalto (x 4, y 1 ) 8. tryassassinate(Marcus, Caesar)

68. Resolution Proof cont. Prove: hate(Marcus, Caesar)  hate(Marcus, Caesar)  Roman(Marcus) V loyalto(Marcus,Caesar) Marcus/x 2 5 3 2 7 1 4 8 Marcus/x 1  Pompeian(Marcus) V loyalto(Marcus,Caesar) loyalto(Marcus,Caesar) Marcus/x 4, Caesar/y 1  man(Marcus) V  ruler(Caesar) V  tryassassinate(Marcus, Caesar)  ruler(Caesar) V  tryassassinate(Marcus, Caesar)  tryassassinate(Marcus, Caesar)

69. Assignment 1Marcus was a man 2Marcus was a Pompeian 3Marcus was born in 40 A.D. 4All men are mortal 5All Pompeians died when volcano erupted in 79 A.D. 6No mortal lives longer than 150 years 7It is now 1991 8Alive means not dead 9If someone dies, then he is dead at all later times Prove by resolution that Marcus is not alive now

70. Using Resolution with Equality and Reduce Axioms in clause form: 1.man(Marcus) 2.Pompeian(Marcus) 3.Born(Marcus, 40) 4.  man(x 1 ) V mortal(x 1 ) 5.  Pompeian(x 2 ) V died(x 2,79) 6.erupted(volcano, 79) 7.  mortal(x 3 ) V  born(x 3, t 1 ) V  gt(t 2 —t 1, 150) V dead(x 3, t 2 ) 8.Now=2002 9.  alive(x 4, t 3 ) V  dead (x 4, t 3 ) 10.  dead(x 5, t 4 ) V alive (x 5, t 4 ) 11.  died (x 6, t 5 ) V  gt(x 6, t 5 ) V dead(x 6, t 6 ) Prove:  alive(Marcus, now)

71. Assignment From “Horse are animals”, it follows that “The head of the Horse is the head of an animal”. Demonstrate that this inference is valid by carrying out the following steps: (i)Translate the premise and the conclusion into the language of the first order logic. Use three predicates HeadOf(h,x), Horse(x) and Animal(x) (ii) Negate the conclusion and convert the premise and negated conclusion into conjunctive normal form. (iii) Use resolution to show that the conclusion follows from the premise.

72. Question Bank Represent each of these sentences in first-order predicate calculus (FOPL). –All basketballs are made of rubber. –Every flower located in Sue’s yard is red. –At least one question on HW4 was hard. –All birds can fly except for penguins and ostriches or unless they have a broken wing. –John’s boss is Mary. –Selling a book changes who owns it but not who wrote it. –James father is married to king Johan’s mother –For all x and all y, if x is the parent of y then y is the child of x –There is someone who is loved by everyone –There is noone who does not like icecream –Spot has atleast two sisters –Not all students take History and Biology –Only one student failed in both History and Biology –There is a barber who shaves all men in town who do not shave themselves –Well, I like Sandy and donot like Sandy –One more outburst like that and you will be in contempt of court –Only one student failed History. –The best score in History was better than the best score in Biology. –Every person who dislikes all vegetarians is smart. –No person likes a smart vegetarian. –There is a woman who likes all men who are not vegetarians. –No person likes a professor unless the professor is smart. –Politicians can fool some of the people all of the time, and they can fool all of the people some of the time, but they can’t fool all of the people all of the time. –All Germans speak the same languages

73. University Questions Consider the following axioms: All people who are graduating are happy All happy people smile Someone is graduating i.Represent these statements in FOL ii.Convert each formula to clause form iii.Prove that “Is someone smiling?” using resolution technique. Draw the resolution tree.

74. Forward Chaining Data driven reasoning Agent derives conclusion from incoming percepts It starts from known facts (positive literals) If all the premises of an implication are known then its conclusion is added to set of known facts This process continues till query added or no further inferences made

75. Forward chaining proof

78. Forward chaining algorithm

79. Backward Chaining Goal directed reasoning Works backward from query Find those implications in KB that conclude Q

80. Backward chaining example

87. Backward chaining algorithm

88. Knowledge engineering Process in FOL 1.Identify the task (what to talk about) 2.Assemble the relevant knowledge 3.Decide on a vocabulary of predicates, functions, and constants 4.Encode general knowledge about the domain 5.Encode a description of the specific problem instance 6.Pose queries to the inference procedure and get answers 7.Debug the knowledge base

89. The digital circuit domain One-bit full adder Input bits to be added Carry bit XOR gates AND gatesOR gate Sum Carry bit for next adder

90. The electronic circuit domain One-bit full adder 0 1 1 1 0 0 1 1

91. The electronic circuits domain 1.Identify the task –Does the circuit actually add properly? (circuit verification) 2.Assemble the relevant knowledge –Composed of wires and gates; Types of gates (AND, OR, XOR, NOT) –Irrelevant: size, shape, color, cost of gates 3.Decide on a vocabulary –Alternatives: Type(X 1 ) = XOR Type(X 1, XOR) XOR(X 1 )

92. Gates or circuits can have one or more input terminals and one or more output terminals. E.g. the function In(1,X1) denotes the first input terminal for gate X1. Connectivity between gates can be represented by a predicate: Connected(Out(1,X1),In(1,X2)) Signal has to be on or off – Signal(t) this function takes terminal as an argument and returns the signal value for that terminal. Two constants – 1 and 0. Note: We could use a unary predicate On(t); However, this would make it difficult to ask the question – what are the possible values of the signals at the output terminals of circuit ?

93. The electronic circuits domain 4- Encode general knowledge of the domain If two terminals are connected they have the same signal –  t 1,t 2 Connected(t 1, t 2 )  Signal(t 1 ) = Signal(t 2 ) Connected is a commutative predicate –  t 1,t 2 Connected(t 1, t 2 )  Connected(t 2, t 1 ) The signal at every terminal is either 1 or 0 –  t (Signal(t) = 1)  (Signal(t) = 0) –1 ≠ 0

94.  An OR gate’s output is 1 iff any of its inputs is 1 –  g Type(g) = OR  Signal(Out(1,g)) = 1   n Signal(In(n,g)) = 1  An AND gate’s output is 0 iff any of its inputs is 0 –  g Type(g) = AND  Signal(Out(1,g)) = 0   n Signal(In(n,g)) = 0  An XOR gate’s output is 1 iff its inputs are different –  g Type(g) = XOR  Signal(Out(1,g)) = 1  Signal(In(1,g)) ≠ Signal(In(2,g))  A NOT gate’s output is different from its input –  g Type(g) = NOT  Signal(Out(1,g)) ≠ Signal(In(1,g))

95. The electronic circuits domain 5.Encode the specific problem instance Type(X 1 ) = XOR Type(X 2 ) = XOR Type(A 1 ) = AND Type(A 2 ) = AND Type(O 1 ) = OR Connected(Out(1,X 1 ),In(1,X 2 ))Connected(In(1,C 1 ),In(1,X 1 )) Connected(Out(1,X 1 ),In(2,A 2 ))Connected(In(1,C 1 ),In(1,A 1 )) Connected(Out(1,A 2 ),In(1,O 1 )) Connected(In(2,C 1 ),In(2,X 1 )) Connected(Out(1,A 1 ),In(2,O 1 )) Connected(In(2,C 1 ),In(2,A 1 )) Connected(Out(1,X 2 ),Out(1,C 1 )) Connected(In(3,C 1 ),In(2,X 2 )) Connected(Out(1,O 1 ),Out(2,C 1 )) Connected(In(3,C 1 ),In(1,A 2 ))

96. The electronic circuits domain 6 - Pose queries to the inference procedure What are the possible sets of values of the input of the adder circuit that lead to a signal of 0 for the sum output bit and a value of 1 for the carry output terminal?  i 1,i 2,i 3 (Signal(In(1,C 1 ))= i 1 )  (Signal(In(2,C 1 )) = i 2 )  (Signal(In(3,C 1 )) = i 3 )  (Signal(Out(1,C 1 )) = 0  Signal(Out(2,C 1 )) = 1) Answer: (i1 = 1; i2 = 1; i3 = 0) or ((i1 = 1; i2 = 0; i3 = 1) or (i1 = 0; i2 = 1; i3 = 0) What are the possible sets of values of all the terminals of the adder circuit?  i 1,i 2,i 3,o 1,o 2 (Signal(In(1,C 1 )) = i 1 )  (Signal(In(2,C 1 )) = i 2 )  (Signal(In(3,C 1 )) = i 3 )  (Signal(Out(1,C 1 )) = o 1  Signal(Out(2,C 1 )) = o 2 ) 7 - Debug the knowledge base May have omitted assertions like 1 ≠ 0

97. Question Bank Analyze in detail the shopping domain, exercise general ontology and show how domain knowledge can be used by a shopping agent.