1. 1 CS 2710, ISSP 2160 The Situation Calculus KR and Planning Some final topics in KR

2. Situation Calculus Planning in propositional logic: Section 7.7 through 7.7.2 Section 10.4.2 Handouts 2

3. Other topics in KR Semantic Networks: 12.5.1 [Description Logic: 12.5.2: we didn’t cover this] [Satisfiability and WalkSat: Intro to Section 7.6; 7.6.2: we didn’t cover this] 3

4. 4 Actions, Situations, and Events The Situation Calculus The robot is in the kitchen. –in(robot,kitchen) He walks into the living room. –in(robot,livingRoom) in(robot,kitchen,2:02pm) in(robot,livingRoom,2:17pm) But what if you are not sure when it was? We can do something simpler than rely on time stamps… The Situation Calculus is a logic formalism for representing and reasoning about dynamic domains.

5. 5 Situation Calculus Ontology Actions: terms, such as “forward” and “turn(right))” Situations: terms; initial situation, say s0, and all situations that are generated by applying an action to a situation. result(a,s) names the situation resulting when action a is done in situation s.

6. 6 Situation Calculus Ontology continued Fluents: functions and predicates that vary from one situation to the next. By convention, the situation is the last argument of the fluent. ~holding(robot,gold,s0) Atemporal or eternal predicates and functions do not change from situation to situation. gold(g1). lastName(wumpus,smith). adjacent(livingRoom,kitchen).

7. 7 Sequences of Actions Also useful to reason about action sequences All S resultSeq([],S) = S All A,Se,S resultSeq([A|Se],S) = resultSeq(Se,result(A,S)) resultSeq([a,b,c],so) is result(c,result(b,result(a,s0)

8. 8 Modified Wumpus World Fluent predicates: at(O,X,S) and holding(O,S) –In our simple world, only the agent can hold a piece of gold, so for simplicity, only the gold and situation are arguments Initial situation: at(agent,[1,1],s0) ^ at(g1,[1,2],s0) But we want to exclude possibilities from the initial situation too…

9. 9 Initial KB All O,X (at(O,X,s0)  [(O=agent ^ X = [1,1]) v (O=g1 ^ X = [1,2])]) All O ~holding(O,s0) Eternals: –gold(g1) ^ adjacent([1,1],[1,2]) ^ adjacent([1,2],[1,1]) etc.

10. 10 Goal: g1 is in [1,1] Planning by answering the query: Exists S at(g1,[1,1],resultSeq(S,s0)) Solution: At(g1,[1,1],resultSeq([go([1,1],[1,2]),grab(g1),go([1,2],[1,1])],s0)) Let’s look at what has to go in the KB for such queries to be answered...

11. 11 Possibility and Effect Axioms Possibility axioms: –Preconditions  poss(A,S) Effect axioms: –poss(A,S)  changes that result from that action

12. 12 Axioms for our Wumpus World For brevity: we will omit universal quantifies that range over entire sentence. S ranges over situations, A ranges over actions, O over objects (including agents), G over gold, and X,Y,Z over locations.

13. 13 Possibility Axioms The possibility axioms that an agent can –go between adjacent locations, –grab a piece of gold in the current location, and –release gold it is holding

14. 14 Effect Axioms If an action is possible, then certain fluents will hold in the situation that results from executing the action –Going from X to Y results in being at Y –Grabbing the gold results in holding the gold –Releasing the gold results in not holding it

15. 15 Frame Problem We run into the frame problem Effect axioms say what changes, but don’t say what stays the same A real problem, because (in a non-toy domain), each action affects only a tiny fraction of all fluents

16. 16 Frame Problem (continued) One solution approach is writing explicit frame axioms, such as: (at(O,X,S) ^ ~(O=agent) ^ ~holding(O,S))  at(O,X,result(Go(Y,Z),S)) If something is at X in S, and it is not the agent, and also it is not something the agent holds, then O is still at X if the agent moves somewhere. F fluents and A actions: O(FA) axioms needed We can do something more efficient than this

17. 17 Frame Problem What stays the same? A actions, F fluents, and E effects/action (worst case). Typically, E << F That is, the effects of an action are typically only a small set of all the things that could change Want O(AE) versus O(AF) solution

18. 18 “Solving” the Frame Problem For each fluent, have successor-state axioms: Action is possible  (fluent is true in result state  action’s effect made it true v it was true before and action left it alone) Each of the E effects of each of the A actions is mentioned exactly once, so O(AE) axioms needed Note: we will return to this point later, after going through the wumpus world example

19. 19 Initial KB (reminder) All O,X at(O,S,s0)  [O=agent ^ X = [1,1]) v (O=g1 ^ X = [1,2])] All O ~holding(O,s0) Eternals: –gold(g1) ^ adjacent([1,1],[1,2]) ^ adjacent([1,2],[1,1]). Trace through reasoning so far on board; state space handed out At this point, we are switching to variables being small case, constants upper case, following the text

20. 4-5 are Successor-State Axioms 1.At(Agent, x, s)  Adjacent(x,y)  Poss(Go(x,y),s) 2.Gold(g)  At(Agent,x,s)  At(g, x, s)  Poss(Grab(g),s) 3.Holding(g,s)  Poss(Release(g),s) 4.Poss(a,s)  Holding(g,Result(a,s))  a = Grab(g) v (Holding(g,s)  a  Release(g))) 5.Poss(a,s)  (At(o,y,Result(a,s))  (a = Go(x,y)  (o = Agent v Holding(o,s))) v (At(o,y,s)  ¬(  z y  z  a = Go(y,z)  (o = Agent v Holding(o,s)))) 20

21. More explicit version; Replaced existential with universal in 5 1.All x,y,s ((At(Agent, x, s)  Adjacent(x,y))  Poss(Go(x,y),s) 2.All g,x,s ((Gold(g)  At(Agent,x,s)  At(g, x, s))  Poss (Grab(g),s)) 3.All g,s (Holding(g,s)  Poss(Release(g),s)) 4.All a,s,g (Poss(a,s)  (Holding(g,Result(a,s))  (a = Grab(g) v (Holding(g,s)  a  Release(g)))) 5.All a,s,o,y,z (Poss(a,s)  (At(o,y,Result(a,s))  ((a = Go(x,y)  (o = Agent v Holding(o,s))) v (At(o,y,s)  ¬(a = Go(y,z)  y  z  (o = Agent v Holding(o,s))))))) –Justification: previous 5 has ¬(  z … the change is justified because this is equivalent to all z ¬… 21

22. Same as previous, but without comments 1.All x,y,s ((At(Agent, x, s)  Adjacent(x,y))  Poss(Go(x,y),s) 2.All g,x,s ((Gold(g)  At(Agent,x,s)  At(g, x, s))  Poss (Grab(g),s)) 3.All g,s (Holding(g,s)  Poss(Release(g),s)) 4.All a,s,g (Poss(a,s)  (Holding(g,Result(a,s))  (a = Grab(g) v (Holding(g,s)  a  Release(g)))) 5.All a,s,o,y,z (Poss(a,s)  (At(o,y,Result(a,s))  ((a = Go(x,y)  (o = Agent v Holding(o,s))) v (At(o,y,s)  ¬(a = Go(y,z)  y  z  (o = Agent v Holding(o,s))))))) 22

23. A return to complexity Each of the E effects of each of the A actions is mentioned exactly once, so O(AE) axioms needed Notes: –an effect may be to make a fluent true (add it) or to make it false (delete it) –Counting axioms is a bit arbitrary, since a single axiom may mention a disjunction of add effects and/or a disjunction of delete effects (see the holding axiom for the blocks world) –It is true that each of the add or delete effects of each action is mentioned once 23

24. A return to complexity Each of the E effects of each of the A actions is mentioned exactly once, so O(AE) axioms needed total action mentions Another schema for successor state axioms: Action is possible  (fluent is true in result state  (action is one of the actions that makes it true (add effect) v it was true before and action is not one of the actions that makes it false (delete effect)) Thus, for each fluent (possible effect), we mention exactly once each action that has it as an effect (either making it true or making it false) Or, for each action, it is mentioned exactly once in a successor state axiom for each of its effects 24

25. Fall 2014 In class exercise – the blocks world [handout] 25

26. 26 Qualification Problem Ensuring that all necessary conditions for an action’s success have been specified. No complete solution in logic. KR/planning designers have to decide how much detail to go into.

27. What did we see? A sophisticated KR scheme Important Problems in planning: –Addressed by successor-state axioms (Reiter 1991) Frame Problem (what stays the same?) Ramification Problem (implicit effects, such as that gold moves too if the agent moves and it is holding the gold) –Not addressed completely in logic Qualification Problem Concepts for planning, such as fluents and situations Planning as search 27

28. Stopped here, Fall 2014 28

29. 29 Semantic Networks Graphical aids for visualizing the knowledge base Efficient algorithms for inferring properties based on category membership Often, correspond to a subset of first-order logic Many variants All distinguish among individual objects, categories of objects and relations among objects

30. 30 Example See figure 12.5 (next slide) Specify what edges and nodes mean In Figure 12.5, indivs and categories look the same memberOf(indiv,category) sisterOf(indiv,indiv) subsetOf(category,category) hasMother(indiv,indiv)

31. 31

32. 32 Semantic Networks Is hasMother(persons,femalePersons) consistent with the representation? Nope: hasMother is a relation between individuals cat1-- label  cat2 means: all X (X in cat1  (all Y label(X,Y)  Y in cat2)) (Note: this does not say that each person has a mother)

33. 33 Semantic Networks cat – label  value All X (X in cat  label(X,value))

34. 34 Inheritance Inheritance is efficient and convenient Trace paths from individuals to categories, inheriting properties as you go In Figure 12.5, how many legs does John have? Most specific (nearest) information wins

35. 35 Semantic Networks In this type of semantic network, only binary relations are possible A richer representation is possible by reifying propositions and events (example: SNePS) This forces creation of a rich ontology of reified concepts; many current ideas originated in semantic network systems

36. 36 Description Logics http://en.wikipedia.org/wiki/Description_logic This won’t be tested on the exam, but I want you to know what description logics are Subset of full first order logic; a family of logics of increasing expressiveness; well studied; most are decidable; good link between theory and practice.

37. We stopped here Fall 2012. 37

38. Proof methods Proof methods divide into (roughly) two kinds: Application of inference rules: Legitimate (sound) generation of new sentences from old. –Resolution –Forward & Backward chaining Model checking Searching through truth assignments. Improved backtracking: Davis--Putnam-Logemann-Loveland (DPLL) Heuristic search in model space: Walksat.

39. Model Checking Two families of efficient algorithms: Complete backtracking search algorithms: DPLL algorithm. You read this on your own for the midterm. Incomplete local search algorithms –WalkSAT algorithm

40. The DPLL algorithm Determine if an input propositional logic sentence (in CNF) is satisfiable. This is just backtracking search for a CSP. Improvements: 1.Early termination A clause is true if any literal is true. A sentence is false if any clause is false. 2.Pure symbol heuristic Pure symbol: always appears with the same "sign" in all clauses. e.g., In the three clauses (A   B), (  B   C), (C  A), A and B are pure, C is impure. Make a pure symbol literal true 3 Unit clause heuristic Unit clause: only one literal in the clause The only literal in a unit clause must be true. Note: literals can become a pure symbol or a unit clause when other literals obtain truth values. e.g.

41. The WalkSAT algorithm Incomplete, local search algorithm Evaluation function: The min-conflict heuristic of minimizing the number of unsatisfied clauses Balance between greediness and randomness See figure 7.18 (on your own)

42. WrapUp: Situation Calculus Planning in propositional logic: Section 7.7 through 7.7.2 Section 10.4.2 Handouts 42

43. WrapUp: Other topics in KR Semantic Networks: 12.5.1 [not covered Fall 2014] 43