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Slide 1 Logic: Domain Modeling /Proofs + Top-Down Proofs Jim Little UBC CS 322 – CSP October 22, 2014
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Slide 2 Lecture Overview Recap Using Logic to Model a Domain (Electrical System) Reasoning/Proofs (in the Electrical Domain) Top-Down Proof Procedure
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Slide 3 Soundness & completeness of proof procedures A proof procedure X is sound … A proof procedure X is complete…. BottomUp for PDCL is sound and complete. We proved this in general even for domains represented by thousands of propositions and corresponding KB with millions of definite clauses ! (KB ⊦ P g) → (KB ⊧ g) (KB ⊧ g) ← (KB ⊦ P g)
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Slide 4 Lecture Overview Recap Using PDCL Logic to Model a Domain (Electrical System) Reasoning/Proofs (in the Electrical Domain) Top-Down Proof Procedure
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Slide 5 Electrical Environment / up /down
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Slide 6 Let’s define relevant propositions / up /down For each wire w For each circuit breaker cb For each switch s For each light l For each outlet p 7 2 3 x 2 2 How many interpretations?
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Slide 7 Let’s now tell system knowledge about how the domain works / up /down live_l 1 ← live_w 0 ← live_w 1 ← live_w_0 live_w_1 ∧ up_s_2. live_w_2 ∧ down_s_2. live_w_3 ∧ up_s_1.
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Slide 8 More on how the domain works…. / up /down live_w 2 ← live_w 3 ∧ down_s 1. live_l 2 ← live_w 4. live_w 4 ← live_w 3 ∧ up_s 3. live_p 1 ← live_w 3..
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Slide 9 More on how the domain works…. / up /down live_w 3 ← live_w 5 ∧ ok_cb 1. live_p 2 ← live_w 6. live_w 6 ← live_w 5 ∧ ok_cb 2. live_w 5 ← live_outside.
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Slide 10 What else we may know about this domain? That some simple propositions are true / up /down
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Slide 11 What else we may know about this domain? That some additional simple propositions are true down_s 1. up_s 2. up_s 3. ok_cb 1. ok_cb 2. live_outside. / up /down
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Slide 12 All our knowledge….. down_s 1. up_s 2. up_s 3. ok_cb 1. ok_cb 2. live_outside / up /down live_l 1 ← live_w 0 live_w 0 ← live_w 1 ∧ up_s 2. live_w 0 ← live_w 2 ∧ down_s 2. live_w 1 ← live_w 3 ∧ up_s 1. live_w 2 ← live_w 3 ∧ down_s 1. live_l 2 ← live_w 4. live_w 4 ← live_w 3 ∧ up_s 3. live_p 1 ← live_w 3. live_w 3 ← live_w 5 ∧ ok_cb 1. live_p 2 ← live_w 6. live_w 6 ← live_w 5 ∧ ok_cb 2. live_w 5 ← live_outside.
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Slide 13 Lecture Overview Recap Using Logic to Model a Domain (Electrical System) Reasoning/Proofs (in the Electrical Domain) Top-Down Proof Procedure
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Slide 14 What Semantics is telling us Our KB (all we know about this domain) is going to be true only in a subset of all possible 2^19 interpretations What is logically entailed by our KB are all the propositions that are true in all those models This is what we should be able to derive given a sound and complete proof procedure
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Slide 15 If we apply the bottom-up (BU) proof procedure down_s 1. up_s 2. up_s 3. ok_cb 1. ok_cb 2. live_outside live_l 1 ← live_w 0 live_w 0 ← live_w 1 ∧ up_s 2. live_w 0 ← live_w 2 ∧ down_s 2. live_w 1 ← live_w 3 ∧ up_s 1. live_w 2 ← live_w 3 ∧ down_s 1. live_l 2 ← live_w 4. live_w 4 ← live_w 3 ∧ up_s 3. live_p 1 ← live_w 3.. live_w 3 ← live_w 5 ∧ ok_cb 1. live_p 2 ← live_w 6. live_w 6 ← live_w 5 ∧ ok_cb 2. live_w 5 ← live_outside. live_l 1 ? No live_w 2 ? Yes Green are in C
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Slide 16 Lecture Overview Recap Using Logic to Model a Domain (Electrical System) Reasoning/Proofs (in the Electrical Domain) Top-Down Proof Procedure
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Bottom-up vs. Top-down KB G is proved if G C When does BU look at the query G ? Bottom-up D. Only at the beginning A. In every loop iterationB. Never C. Only at the end C Slide 17
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Bottom-up vs. Top-down Key Idea of top-down: search backward from a query G to determine if it can be derived from KB. KB C G is proved if G C When does BU look at the query G? At the end Bottom-up Top-down TD performs a backward search starting at G KB answer Query G Slide 18
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Slide 19 Top-down Ground Proof Procedure Key Idea: search backward from a query G to determine if it can be derived from KB.
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Slide 20 Top-down Proof Procedure: Basic elements Notation : An answer clause is of the form: yes ← a 1 ∧ a 2 ∧ … ∧ a m Rule of inference (called SLD Resolution) Given an answer clause of the form: yes ← a 1 ∧ a 2 ∧ … ∧ a m and the clause: a i ← b 1 ∧ b 2 ∧ … ∧ b p You can generate the answer clause yes ← a 1 ∧ … ∧ a i-1 ∧ b 1 ∧ b 2 ∧ … ∧ b p ∧ a i+1 ∧ … ∧ a m i.e., resolving a i with a i ← b 1 ∧ b 2 ∧ … ∧ b p Express query as an answer clause (e.g., query a 1 ∧ a 2 ∧ … ∧ a m ) yes ← a 1 ∧ a 2 ∧ … ∧ a m
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Slide 21 Rule of inference: Examples Rule of inference (called SLD Resolution) Given an answer clause of the form: yes ← a 1 ∧ a 2 ∧ … ∧ a m and the KB clause: a i ← b 1 ∧ b 2 ∧ … ∧ b p You can generate the answer clause yes ← a 1 ∧ … ∧ a i-1 ∧ b 1 ∧ b 2 ∧ … ∧ b p ∧ a i+1 ∧ … ∧ a m yes ← b ∧ c.b ← k ∧ f. yes ← e ∧ f.e. KB clause -> yes ← k ∧ f ∧ c. -> yes ← f.
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Slide 22 (successful) Derivations An answer is an answer clause with m = 0. That is, it is the answer clause yes ←. A (successful) derivation of query “?q 1 ∧ … ∧ q k “ from KB is a sequence of answer clauses γ 0, γ 1,…,γ n such that γ 0 is the answer clause yes ← q 1 ∧ … ∧ q k γ i is obtained by resolving γ i-1 with a clause in KB, and γ n is an answer. An unsuccessful derivation….. yes ←. yes ← a ∧ b.
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Slide 23 Example: derivations a ← e ∧ f. a ← b ∧ c.b ← k ∧ f. c ← e.d ← k. e. f ← j ∧ e. f ← c. j ← c. Query: a (two ways) a ← b ∧ c. a ← e ∧ f. a ← k ∧ f ∧ c. a ← f. FAIL since a ← c. k cannot be eliminated a ← e. yes ←. yes ← a. yes ← a. KB
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Slide 24 Example: derivations k ← e. a ← b ∧ c.b ← k ∧ f. c ← e.d ← k. e. f ← j ∧ e. f ← c. j ← c. Query: b ∧ e A. Provable by TDB. It depends C. Not Provable by TD KB
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Slide 25 Course Big Picture Environment Problem Inference Planning Deterministic Stochastic Search Arc Consistency Search Value Iteration Var. Elimination Constraint Satisfaction Logics STRIPS Belief Nets Vars + Constraints Decision Nets Markov Processes Var. Elimination Static Sequential Representation Reasoning Technique SLS
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Slide 26 Learning Goals for today’s class You can: Model a relatively simple domain with propositional definite clause logic (PDCL) Trace query derivation using SLD resolution rule of inference
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