10/28 homework 3 returned homework 4 socket opened (no office hours today) where hard problems are...
Post on 22-Dec-2015
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10/28
Homework 3 returned
Homework 4 socket opened
(No office hours today)
Where hard problems are
Phase Transition
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Davis-Putnam-Logeman-Loveland Procedure
detect failure
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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 (onlyThe remaining ones)
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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
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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)
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http://www.ipam.ucla.edu/publications/ptac2002/ptac2002_dachlioptas_formulas.pdf
Progress in nailing the bound.. (just FYI)
Not discussed in class
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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 23 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 2n possible random assignments. So, the number of assignments which will satisfy the entire 3SAT instance is 2n (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)
– 2n (7/8)c < 1– n + c log2 7/8 < 0
• Taking log to the base 2 on both sides– n < - c log2 7/8 – n < c log2 8/7– c/n > 1/log2 8/7 = 1/0.1926 = 5.1921
Not discussed in class
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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
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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
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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
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10/30
Counseling today during office hours
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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 SearchAs is the case for SAT, basic search For CSP is in the space of partialAssignments --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
RedViolates
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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)
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Value Ordering Heuristics
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Lookahead
Variable ordering can be improved in the presence of FC
DVO (Dynamic variable ordering) --Consider the variable with the smallest remaining domain next
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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 beSeen as a form of 3-consistency Enforcement..
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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
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Why FOPC
If your thesis is utter vacuousUse first-order predicate calculus.With sufficient formalityThe sheerest banality
Will be hailed by the critics: "Miraculous!"
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Left-leg-of
Tarskian Interpretations
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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
)(),()(),( bQbaentailsPxQyxyPx
)(|),();(),( bqbaPxQyxyPx