1. 10/28 Homework 3 returned Homework 4 socket opened (No office hours today) Where hard problems are Phase Transition

2. Davis-Putnam-Logeman-Loveland Procedure  detect failure

3. DPLL Example Clauses (p,s,u) (~p, q) (~q, r) (q,~s,t) (r,s) (~s,t) (~s,u) Pick p; set p=true unit propagation (p,s,u) satisfied (remove) p;(~p,q)  q derived; set q=T (~p,q) satisfied (remove) (q,~s,t) satisfied (remove) q;(~q,r)  r derived; set r=T (~q,r) satisfied (remove) (r,s) satisfied (remove) pure literal elimination in all the remaining clauses, s occurs negative set ~s=True (i.e. s=False) At this point all clauses satisfied. Return p=T,q=T;r=T;s=False s was not Pure in all clauses (only The remaining ones)

4. Model-checking by Stochastic Hill-climbing Start with a model (a random t/f assignment to propositions) For I = 1 to max_flips do –If model satisfies clauses then return model –Else clause := a randomly selected clause from clauses that is false in model With probability p whichever symbol in clause maximizes the number of satisfied clauses /*greedy step*/ With probability (1-p) flip the value in model of a randomly selected symbol from clause /*random step*/ Return Failure Remarkably good in practice!! Clauses 1. (p,s,u) 2. (~p, q) 3. (~q, r) 4. (q,~s,t) 5. (r,s) 6. (~s,t) 7. (~s,u) Consider the assignment “all false” -- clauses 1 (p,s,u) & 5 (r,s) are violated --Pick one—say 5 (r,s) [if we flip r, 1 (remains) violated if we flip s, 4,6,7 are violated] So, greedy thing is to flip r we get all false, except r otherwise, pick either randomly

5. Phase Transition in SAT Theoretically we only know that phase transition ratio occurs between 3.26 and 4.596. Experimentally, it seems to be close to 4.3 (We also have a proof that 3-SAT has sharp threshold)

6. http://www.ipam.ucla.edu/publications/ptac2002/ptac2002_dachlioptas_formulas.pdf Progress in nailing the bound.. (just FYI) Not discussed in class

7. A longer proof for 5.19 bound (FYI) Not discussed in class Probability that a 3-sat clause is satisfied is 7/8 3 vars; 2 3 possible assignments only one is all false P(X >= t  t

8. An easy upper bound for 3-sat transition (optional) Suppose there are n variables and c clauses Probability that a random assignment satisfied a clause if 7/8 –Each clause contains 3 variables; so 2 3 possible assignments for the variables. Of these, just one, the all false one, makes the clause false Probability that all c clauses are satisfied is (7/8) c (assuming clause independence; holds when n>>3) There are 2 n possible random assignments. So, the number of assignments which will satisfy the entire 3SAT instance is 2 n (7/8) c This is the expected number of satisfying assignments We want to know when the expected num of satisfying assignments becomes less than 1 (i.e., unsatisfiable) –2 n (7/8) c < 1 –n + c log 2 7/8 < 0 Taking log to the base 2 on both sides –n < - c log 2 7/8 –n < c log 2 8/7 –c/n > 1/log 2 8/7 = 1/0.1926 = 5.1921 Not discussed in class

9. CSP: SAT with multi-valued variables It is easy to generalize the SAT problem to handle non-boolean variables –CSP problem Given a set of discrete variables and their domains And a set of constraints (expressed as legal value combinations that can be taken by various subsets of the variables) –Clausal constraints (such as p=>q ) can be seen in this way too Find a model (an assignment of domain values to the variables) that satisfies all constraints –SAT is a boolean CSP

10. We can model any CSP problem also as a SAT problem Consider the variables “WA-is-red” “WA-is-green”…. |V|*|D| boolean variables --but this leads to some loss of structure

11. Planning as CSP Variables: propositions at each level Domains: the actions supporting them at that level + NULL value Goal constraints: Certain Constraints: Activity constraints:

12. We will mostly talk about discrete domain variables All CSPs can be compiled to binary CSPs (by introducing additional variables)  Most early work on CSP concentrated on binary CSPs

14. 10/30 Counseling today during office hours

15. Most of the SAT improvements work for CSP too. Main differences are: --Variable selection heuristics CSP solvers typically consider most constrained variable first heuristic (i.e., heuristic with the smallest domain) --Value selection heuristics CSP solvers consider “least constraining value first” heuristic -- Lookahead Instead of unit propagation, CSP solvers use variety of constraint propagation algorithms (forward checking, arc-consistency, path consistency) Backtracking Search As is the case for SAT, basic search For CSP is in the space of partial Assignments --extend the assignment by selecting a variable and considering all its values (in different branches) --prune any (even partial) assignment if it violates a constraint Red Violates

16. Variable Ordering Strategies Notice that for “boolean CSPs” (SAT), most constrained variable heuristics are less effective (since all variables have domains of size 2 (normal), 1 (have a specific value) or 0 (backtrack)

17. Value Ordering Heuristics

18. Lookahead Variable ordering can be improved in the presence of FC DVO (Dynamic variable ordering) --Consider the variable with the smallest remaining domain next

20. K-consistency Forward checking is a “weak” form of look-ahead --stronger look-ahead are based on the notion of enforcing k-consistency A CSP is k-consistent, if any legal assignment of values for any k-1 sized subset of variables can be extended to cover any kth variable. If a CSP is n-consistent, then it can be solved without backtracking! [Make sure you get this…] Enforcing higher than 2 or 3 consistency is rarely worth it

21. 2-consistency Forward checking will stop thinking everyone has non-empty domains. However, NT is blue; if you propagate that we know that SA cannot be blue. This means SA is empty domain Graphplan mutex propagation can be Seen as a form of 3-consistency Enforcement..

22. Consistency enforcement as inference A:{1,2} B:{1,2} A<B A=1 V A=2 B=1 V B=2 ~(A=1) V ~(B=1) ~(A=2) V ~(B=1) ~(A=2) V ~(B=2) A=2 V ~(B=1) ~(B=1) V ~(B=1) = ~(B=1) A:{1,2} B:{1,2} A<B Currently, B=2 A=1 V A=2 B=1 V B=2 ~(A=1) V ~(B=1) ~(A=2) V ~(B=1) ~(A=2) V ~(B=2) B=2 ~(A=2) 1-level “unit resolution” One of the resolvers is Derived from A’s domain Constraint. The other is a Inter-variable constraint of Size 2

23. Inference/ Theorem Proving Satisfaction “Conditioning” Inference satisfaction Inference/Satisfaction (Conditioning) Duality “Try to explicate hidden structure” “Try to split cases (disjunction) into search tree (by committing)”

24. Summary of Propositional Logic Syntax Semantics (entailment) Entailment computation –Model-theoretic Using CSP techniques –Proof-theoretic Resolution refutation –Heuristics to limit type of resolutions »Set of support Connection to CSP –K-consistency can be seen as a form of limited inference

25. Why FOPC If your thesis is utter vacuous Use first-order predicate calculus. With sufficient formality The sheerest banality Will be hailed by the critics: "Miraculous!"

34. Left-leg-of Tarskian Interpretations

37. Inference in first order logic For “ground” sentences (i.e., sentences without any quantification), all the old rules work directly –P(a,b)=> Q(a); P(a,b) |= Q(a) –~P(a,b) V Q(a) resolved with P(a,b) gives Q(a) What about quantified sentences? –Universal Instantiation (a universally quantified statement entails every instantiation of it) Can we combine these (so we can avoid unnecessary instantiations?) Yes. Generalized modus ponens Needs UNIFICATION