Answer Set Programming: A new Paradigm for Knowledge Representation and Constraint Programming

1
Answer Set ProgrammingA new Paradigm for
Knowledge Representation and Constraint
Programming

• Russell and Norvig 10.7
• Lecture Notes for Cmput 366
• (Some slides ,especially those with pictures,
  are taken from C. Barals talk at AAAI05)

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Intelligent Agent

• Can acquire knowledge through various means such
  as learning from experience, observations,
  reading, etc., and
• Can reason with this knowledge to make plans,
  explain observations, achieve goals, etc.

3
To learn knowledge and to reason with it

• we need to know how to represent knowledge in a
  computer readable format.
• McCarthy 1959 in Programs with commonsense
• In order for a program to be capable of
  learning something it must first be capable of
  being told it.

4
Importance of KR

• KR is the starting point of building intelligent
  entities (or AI systems), and leads to the next
  steps of acquiring knowledge and reasoning with
  knowledge.

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What does KR entail?

• We need languages and corresponding methodologies
  to represent various kinds of knowledge.

6
Importance of inventing suitable KR languages

• Development of a suitable knowledge
  representation language and methodology is as
  important to AI systems
• as
• Calculus is to Physics and Engineering.

7
Historical perspective

• AI pioneers (especially McCarthy and Minsky)
  realized the importance of KR to AI.
• McCarthy 1959 Programs with commonsense
• (perhaps the first paper on logical AI).
• Minsky 1974 A framework for representing
  knowledge.

John McCarthy
Marvin Minksy
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What are the properties of a good KR language.

• To start with should be non-monotonic
• i.e., allow revision of conclusion in presence of
  new knowledge.
• Hayes 1973 (Computation and Deduction) mentions
  monotonicity (calls it extension property) and
  notes that rules of default do not satisfy it.
• Minsky 1974 (A framework for representing
  knowledge)
• criticizes monotonicity of logistic
  systems.

Pat Hayes
Marvin Minsky
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Inadequacy of first order logic

• They are monotonic More information one has,
  more consequences one gets.
• Human communication is typically based on closed
  world assumption.

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An Example of Closed World Assumption

• ground-wet ? watering.
• ground-wet ? raining.
• In an open world, there could be others that
  cause ground-wet (we simply dont know, or have
  not said).
• But in a closed world, what we said is all that
  we know, for Horn clauses, this is called Clark
  Completion,
• Ground-wet ? watering ? raining

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Problem with Clark Completion

• a ? a.
• When completed, it becomes
• a ? a
• Two models a and . The first model doesnt
  seem to make sense (how can we have a?)
• The desired model should be since there is
  no way to establish a, hence a is (believed to
  be) false.

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Transitive Closure graph reachability

• reach(a).
• reach(X) ? reach(Y), edge(Y,X).
• a and b are reachable but c and d are not.
• But c is not reachable, neither is d. Can we
  infer these?

a
b
edge(a,b).
c
d
edge(c,d). edge(d,c)
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Reachability Clark completion (cf. page 355 of
RN)

• ?XY edge(X,Y) ? ((Xa ?Y b) ? (X c ? Yd) ?
  (X d ? Y c))
• ?X reach(X) ? (X a or ?Y (reach(Y) ?
  edge(Y,X))
• Equality axioms.
• edge(a,b), edge(c,d), edge(d,c), reach(a),
  reach(b) is a model.
• But so is edge(a,b), edge(c,d), edge(d,c),
  reach(a), reach(b), reach(c), reach(d) .
• Hence one can not conclude reach(c), reach(d).
• Need to go beyond first order logic.

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Pre-1980 history of non-monotonic logics from
Minkers 93 survey

• THNOT in PLANNER Hewitt in 1969
• Prolog Colmerauer et al. 1973
• Circumscription McCarthy 1977
• Default Reasoning Reiter
  1978
• Closed World Assumption (CWA) Reiter 1978
• Negation as failure
  Clark 1978
• Truth maintenance systems Doyle
  1979
• AIJ Volume 13, 1980, a special issue

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Circumscription

• Only consider minimal models for the
  circumscribed predicates
• E.g.
• bird(X) ? ab(X) ? flies(X)
• To circumscribe predicate ab, we can assume
  ab(X)
• unless ab(X) is known to be true. Thus, in lack
  of
• information about a bird being abnormal, we
  conclude it
• flies.

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Circumscription

• bird(X) ? ab(X) ? flies(X)
• bird(tweety)
• Models (after propositionalizing)
• M1bird(tweety), ab(tweety),flies(tweet
  y)
• M2bird(tweety), ab(tweety)
• M3bird(tweety), flies(tweety)
• M3 is smaller than others wrt predicate ab.
  Thus,
• flies(tweety) follows from the given formulas
  under circumscription.

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Default Logic

• We write default rules.
• E.g.
• bird(X) ab(X)
• -----------------------
• flies(X)
• Reads if X is a bird, and it can be consistently
  assumed
• that it is not abnormal, then it
  flies.

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Have we invented calculus of KR yet?

• What basic properties should it have?
• have a simple and intuitive syntax and semantics
• be non-monotonic
• allow us to represent and reason with incomplete
  information and
• allow us to express and answer problem solving
  queries such as planning queries, explanation
  queries and diagnostic queries.

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Have we invented calculus of KR yet? -
continued.

• What properties will make it useful?
• should have building block results
• should have interpreters for reasoning with the
  language
• should have existing applications and
• should have systems that can learn knowledge in
  this language.

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Is ASP a good candidate?

• An ASP program (late 1980s) is a collection of
  rules of the form
• A0 or or Al ? B1, , Bm, not C1, , not Cn.
• where Ais, Bjs and Cks are literals.

Michael Gelfond
Jack Minker
Vladimir Lifschitz
Ray Reiter
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Is ASP a good candidate?

• Its syntax uses the intuitive If-then form.
• It is non-monotonic.
• Can express defaults and their exceptions.
• Can represent and reason with incomplete
  information.
• Can express and answer problem solving queries.
• Large body of building block results.
• Various implementations Smodels, DLV, Prolog.
• Many applications built using it.
• Learning systems Progol.
• Its initial paper among the top 5 AI source
  documents in terms of citeseer citation.

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How ASP differs from

• Prolog ordering matters in Prolog can not
  handle cycles with not has extra-logical
  features does not have disjunction and classical
  negation and is not declarative.
• Logic Programming is a class of languages and
  many different semantics are proposed for not.
• Classical Logic
• Classical logic is monotonic.
• ? in AnsProlog, which helps in expressing
  causality, is not reverse implication.
• Disjunction symbol or in AnsProlog is
  non-classical.
• The negation as failure symbol not in AnsProlog
  is non-classical.

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Normal program

• A normal program in ASP is a collection of rules
  of the form
• A ? B1, , Bm, not C1, , not Cn.
• where A, Bjs and Cks are function-free atoms.
• If the body is empty, we write
• A ?.
• Or simply
• A.

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Semantics

• A function-free program can be grounded (called
  propositionalization in textbook)
• p(X) ? q(X), not s(X) . Function-free
• p(X) ? q(f(X)), not s(X). Not function-free

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Semantics

• Suppose we have constants a,b,c in our program,
  the rule
• p(X) ? q(X), not s(X).
• is a compact representation of three ground rules
• p(a) ? q(a), not s(a).
• p(b) ? q(b), not s(b).
• p(c) ? q(c), not s(c).

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Semantics

• Informally, a stable model M of a ground program
  P is a set of ground atoms such that
• Every rule is satisfied, i.e., for any rule in P
• A ? B1, , Bm, not C1, , not Cn.
• if Bjs are satisfied (Bjs are in M) and Cjs
  are also satisfied (not Cj is satisfied if Cj is
  not in M), then A is in M.
• Every A ? M can be derived from a rule by a
  non-circular reasoning.

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Examples

• P1 a ? a.
• M a is not a stable model but M is.
• P2 a ? not b.
• a is the only stable model
• P3 a ? not a.
• It has no stable model

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Examples

• P4 a ? not b. b ? not a.
• Two stable models a and b.

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Examples

• P4 a ? not b. b ? not a.
• Two stable models a and b.
• P5 a ? not b. b ? not a. a ? not a.
• a is the only stable model.

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Does tweety fly?

• fly(X) ? bird(X), not ab(X).
• ab(X) ? penguin(X).
• bird(X) ? penguin(X).
• bird(tweety).
• We conclude fly(tweety).
• But if we add
• penguin(tweety).
• We can no longer conclude fly(tweety)
• and conclude fly(tweety), by virtue of CWA.

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Constraints for disallowing

• The head of a rule may be empty
• ? B1, , Bm, not C1, , not Cn.
• It says no stable model may contain all Bjs and
  none of Cjs.

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Generate-and-constrain first generate

• To specify both possibilities a is in a solution
  or not, we can use a dummy a
• a ? not a.
• a ? not a.
• Two stable models a, a the latter
  represents that a is not in solution

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Generate-and-constrain first generate

• To specify all subsets of a,b,c, we can write
• a ? not a. b ? not b. c ? not
  c.
• a ? not a. b ? not b. c ? not
  c.
• Eight stable models each corresponding to a
  subset, e.g. a, b,c represents that a is in
  it, but not b, nor c.

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Generate-and-constrain then constrain

• Any subset of a,b,c such that a and b cannot be
  together.
• a ? not a. b ? not b. c ? not
  c.
• a ? not a. b ? not b. c ? not
  c.
• ? a ,b.
• What if we want to say whenever a is in a stable
  model, so is b?

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Hamiltonian Cycle

• Given a set of facts defining the vertices and
  edges of a directed graph and a starting vertex
  v0, find a path that visits every vertex exactly
  once.

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Hamiltonian Cycle

• Any edge could be on such a path. We use in(U,V)
  to represent that edge(U,V) is on such a path.
• in(U,V) ? edge(U,V), not out(U,V).
• out(U,V) ? edge(U,V), not in(U,V).
• out(U,V) is a dummy representing edge(U,V) is not
  on such a path.

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Hamiltonian Cycle

• A path must be chained to form a sequence over
  the edges on it
• reachable(V) ? in(v0,V).
• reachable(V) ? reachable(U), in(U,V).

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Hamiltonian Cycle

• A vertex cannot be visited more than once.
• This can be defined as no more than one edge on
  such a path that goes into any vertex (similarly
  out of such an edge)
• ? edge(U,V),in(U,V), edge(W,V)in(W,V), U ? W.
• ? edge(U,V),in(U,V), edge(U,W),in(U,W), V ?
  W.

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Hamiltonian Cycle

• Dont forget to say that every vertex must be
  reached.
• ? vertex(U), not reachable(U).

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3-colorability

• Whether 3 colors, say red, blue, and yellow, are
  sufficient to color a map
• A map is represented by a graph, with facts about
  nodes and arc as given, e.g,
• vertex(a).
• vertex(b).
• arc(a,b).

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3-colorability

• Every vertex must be colored with exactly one
  color
• color(V,r) ? vertex(V), not color(V,b), not
  color(V,y).
• color(V,b) ? vertex(V), not color(V,r), not
  color(V,y).
• color(V,y) ? vertex(V), not color(V,b), not
  color(V,r).
• No adjacent vertices may be colored with the same
  color
• ? vertex(V), vertex(U), arc(V,U),col(C ),
  color(V,C),
• color(U,C).
• Of course, we need to say what colors are
• col(r). col(b). col(y).

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3-colorability

• A different encoding
• color(V,C) ? node(V), col(C), not
  otherColor(V,C).
• otherColor(V,C) ? node(V), col(C), not
  color(V,C).
• ? node(V), col(C1), col(C2), color(V,C1),
  color(V,C2), C1? C2.
• ? node(V), col(C), not color(V,C).
• ? node(V), node(U), V ? U, arc(V,U), col(C ),
  color(V,C), color(U,C).

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So, what exactly is a stable model of a normal
program P

• Idea you guess a set of atoms and verify it is
  indeed exactly the set of atoms that can be
  derived (page 357 of textbook)
• Reduct of P w.r.t. M h ? b1, , bm
• h? b1, , bm, not c1, , not cn is in P
• and no ci is in M
• M is a stable model of P iff the set of (atomic)
  consequences of the reduct of P is precisely M

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Stable model

• P
• a ? not b.
• b ? not a.
• M a is a stable model, since the reduct of
  P wrt. M is
• a ?.
• its set of (atomic) consequences is precisely M
  itself.

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Stable model

• Why
• a ? not a.
• has no stable model?
• The empty set is not a stable model. (Why?)
• If Ma were a stable model, the reduct of
  program wrt a is the empty set, whose (atomic)
  consequences is also empty, not the same as M.

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Extensions Cardinality constraint

• A cardinality constraint is of form
• L a1, , am, not b1, , not bk U
• The constraint is satisfied in a model if the
  cardinality of the subset of the literals
  satisfied by the model is between integers L and
  U, inclusive.
• A cardinality constraint can be used anywhere in
  a rule.
• E.g. P 0a, b, not d2 ?.
• a is a stable model, but is a,b a stable
  model?

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Cardinality constraint

• Generate all subsets of a,b,c,d such that
  whenever a is in it so is b
• 0a, b, c, d4 ?.
• b ? a.
• As 4 is the max number of literals that may be
  satisfied, you may omit it for simplicity
• 0a, b, c, d ?.

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Cardinality constraint

• Generate all subsets of a,b,c,d such that if a
  is not in it, then b is in it.
• 0a, b, c, d ?.
• b ? not a.
• Are they stable models?
• M1 a,b,c M2 b,c,d,e

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ASP Systems

• Smodels (Helsinki Univ. of Tech.)
• DLV (Vienna Univ. of Tech.)
• ASSAT (HK Univ. of Sci. and Tech.)
• Cmodel (U. of Texas at Austin)

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The Smodels System

• An efficient system for computing answer sets of
  normal programs (later exteneded for disjunctive
  programs).
• Consists of two parts
• Lparse ground a program
• Smodels compute the stable models of the
  grounded program, based on DPLL.

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Smodels

• Syntax largely borrowed from Prolog.
• a - not b.
• b - not a.
• - a.
• A number of language constructs for convenience

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Conditional Literals in Smodels

• A short hand to express a set take the form
• l d
• where l is an atom and d a domain predicate.
• E.g. Set a vertex v to exactly one color among
• red, blue and yellow
• 1 setColor(v,C) color(C)1.
• color(red). color(blue). color(yellow).
• is equivalent to
• 1setColor(v,red), setColor(v,blue),
  setColor(v,yellow) 1.

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N-colorability

• every vertex is colored with exactly one color.
• 1 setColor(V,C) col(C) 1 - vertex(V).
• facts representing colors
• col(1..colors).
• no adjacent vertices are colored with the same
  color
• - vertex(V), vertex(U), arc(V,U), col(C ), U?V,
  setColor(V,C), setColor(U,C).
• Typical command line
• lparse -c colors3 coloring.lp smodels

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Conditional literals in Smodels

• Example
• 1 p(I,J) d(I,J) 1.
• d(I,J) - d(I),d(J).
• d(1..2).
• The first rule above is equivalent to
• 1 p(1,1),p(1,2),p(2,1),p(2,2) 1.

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Conditional literals in Smodels

• Note the difference with the following program
• 1 p(I,J) d(I) 1 - d(J).
• d(1..2).
• The first rule above is equivalent to
• 1 p(I,1) d(I) 1 - d(1).
• 1 p(I,2) d(I) 1 - d(2).
• which are equivalent to
• 1 p(1,1),p(2,1) 1 - d(1).
• 1 p(1,2),p(2,2) 1 - d(2).

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Hamiltonian Cycle Revisited

• Any subset of edges can be on such a path
• in(U,V) ? edge(U,V), not out(U,V).
• out(U,V) ? edge(U,V), not in(U,V).
• Now can be programmed as
• 0 in(U,V) edge(U,V) .

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Wumpus World

• There is exactly one wumpus
• 1 at(I,J,wumpus) room(I,J) 1.
• room(I,J) - col(I), row(J).
• For a 4 by 4 grid, this is equivalent to exactly
  one atom being true in the set of 16
• 1 at(1,1,wumpus), at(1,2,wumpus),. 1.

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Wumpus World

• One or more adjacent rooms has a pit if breeze at
  current room (cf. Assignment 3)
• 1 at(Ni,Nj,pit) adjacent(I,J,Ni,Nj) -
• room(I,J),
  
• sensor(I,J,none,breeze).

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N-queens problem

• hide.
• show q(X,Y).
• d(1..queens).
• 1 q(X,Y)d(Y) 1 - d(X).
• - d(X), d(Y), d(X1), q(X,Y), q(X1,Y), X1 ! X.
• - d(X), d(Y), d(Y1), q(X,Y), q(X,Y1), Y1 ! Y.
• - d(X), d(Y), d(X1), d(Y1), q(X,Y), q(X1,Y1),
• X ! X1, Y ! Y1, abs(X - X1) abs(Y - Y1).
• - d(X), not hasq(X).
• hasq(X) - d(X), d(Y), q(X,Y).
• Typical command line
• lparse -c queens8 queens.lp smodels

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Weight constraints

• We can replace cardinality by weights
• L l1, w1 , lm wm U
• where each li is an atom or a not_atom. Its
  satisfied when the sum of the satisfied lis is
  between L and U.
• When all wi 1, it becomes a cardinality
  constraint.
• (We dont need to use weight constraints in this
  course)

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Classic Negation

• safe ? ? train. vs. safe ? not
  train.
• Use a new name e.g., no_train, to represent it
• safe ? no_train.
• of course, they cannot be both in a stable model.
• ? train, no_train.