PROBLEM SOLVING AND SEARCH IN ARTIFICIAL INTELLIGENCE€¦ · Artiﬁcial Intelligence,...

Artificial Intelligence, Computational Logic

PROBLEM SOLVING AND SEARCHIN ARTIFICIAL INTELLIGENCE

Lecture 6 ASP Modelling, Tree Decomposition

Sarah Gaggl

Dresden, 13th May 2014

Agenda1 Introduction2 Uninformed Search versus Informed Search (Best First Search, A*

Search, Heuristics)3 Constraint Satisfaction4 Answer Set Programming (ASP)5 Structural Decomposition Techniques (Tree/Hypertree Decompositions)6 Local Search, Stochastic Hill Climbing, Simulated Annealing7 Tabu Search8 Evolutionary Algorithms/ Genetic Algorithms9 Adversial Search and Game Playing

TU Dresden, 13th May 2014 PSSAI slide 2 of 73

Modeling and Interpreting

Problem

Logic Program

Solution

Stable Models

?-

6

Modeling Interpreting

Solving


Modeling• For solving a problem class C for a problem instance I,

encode1 the problem instance I as a set PI of facts and2 the problem class C as a set PC of rules

such that the solutions to C for I can be (polynomially) extractedfrom the stable models of PI ∪ PC

• PI is (still) called problem instance• PC is often called the problem encoding

• An encoding PC is uniform, if it can be used to solve all itsproblem instancesThat is, PC encodes the solutions to C for any set PI of facts


ASP solving process

Problem

LogicProgram Grounder Solver Stable

Models

Solution

- - -

?

6


Solving


ASP solving process

Problem


Models

Solution

- - -

?

6


Solving


ASP solving process

Problem


Models

Solution

- - -

?

6


Solving6

ElaboratingTU Dresden, 13th May 2014 PSSAI slide 13 of 73

Basic methodology

MethodologyGenerate and Test (or: Guess and Check)

Generator Generate potential stable model candidates(typically through non-deterministic constructs)

Tester Eliminate invalid candidates(typically through integrity constraints)

NutshellLogic program = Data + Generator + Tester ( + Optimizer)


Satisfiability testing

• Problem Instance: A propositional formula φ in CNF• Problem Class: Is there an assignment of propositional variables to true

and false such that a given formula φ is true

• Example: Consider formula

(a ∨ ¬b) ∧ (¬a ∨ b)

• Logic Program:

Generator Tester Stable models{ a, b } ← ← not a, b

← a, not bX1 = {a, b}X2 = {}






(a ∨ ¬b) ∧ (¬a ∨ b)

• Logic Program:


← a, not bX1 = {a, b}X2 = {}


The n-Queens Problem

5 Z0Z0Z4 0Z0Z03 Z0Z0Z2 0Z0Z01 Z0Z0Z

1 2 3 4 5

• Place n queens on an n× nchess board

• Queens must not attack oneanother

Q Q Q

Q Q


Defining the Field

queens.lp

row(1..n).col(1..n).

• Create file queens.lp

• Define the field– n rows– n columns


Defining the Field

Running . . .

$ gringo queens.lp --const n=5 | claspAnswer: 1row(1) row(2) row(3) row(4) row(5) \col(1) col(2) col(3) col(4) col(5)SATISFIABLE

Models : 1Time : 0.000

Prepare : 0.000Prepro. : 0.000Solving : 0.000


Placing some Queens

queens.lp

row(1..n).col(1..n).{ queen(I,J) : row(I) : col(J) }.

• Guess a solution candidate

by placing some queens on the board


Placing some Queens

Running . . .

$ gringo queens.lp --const n=5 | clasp 3Answer: 1row(1) row(2) row(3) row(4) row(5) \col(1) col(2) col(3) col(4) col(5)Answer: 2row(1) row(2) row(3) row(4) row(5) \col(1) col(2) col(3) col(4) col(5) queen(1,1)Answer: 3row(1) row(2) row(3) row(4) row(5) \col(1) col(2) col(3) col(4) col(5) queen(2,1)SATISFIABLE

Models : 3+...


Placing some Queens: Answer 1

Answer 1

5 Z0Z0Z4 0Z0Z03 Z0Z0Z2 0Z0Z01 Z0Z0Z

1 2 3 4 5



Answer 2

5 Z0Z0Z4 0Z0Z03 Z0Z0Z2 0Z0Z01 L0Z0Z

1 2 3 4 5



Answer 3

5 Z0Z0Z4 0Z0Z03 Z0Z0Z2 QZ0Z01 Z0Z0Z

1 2 3 4 5


Placing n Queens

queens.lp

row(1..n).col(1..n).{ queen(I,J) : row(I) : col(J) }.:- not n { queen(I,J) } n.

• Place exactly n queens on the board


Placing n Queens

Running . . .

$ gringo queens.lp --const n=5 | clasp 2Answer: 1row(1) row(2) row(3) row(4) row(5) \col(1) col(2) col(3) col(4) col(5) \queen(5,1) queen(4,1) queen(3,1) \queen(2,1) queen(1,1)Answer: 2row(1) row(2) row(3) row(4) row(5) \col(1) col(2) col(3) col(4) col(5) \queen(1,2) queen(4,1) queen(3,1) \queen(2,1) queen(1,1)...


Placing n Queens: Answer 1

Answer 1

5 L0Z0Z4 QZ0Z03 L0Z0Z2 QZ0Z01 L0Z0Z

1 2 3 4 5


Placing n Queens: Answer 2

Answer 2

5 Z0Z0Z4 QZ0Z03 L0Z0Z2 QZ0Z01 LQZ0Z

1 2 3 4 5


Horizontal and vertical Attack

queens.lp

row(1..n).col(1..n).{ queen(I,J) : row(I) : col(J) }.:- not n { queen(I,J) } n.:- queen(I,J), queen(I,JJ), J != JJ.

:- queen(I,J), queen(II,J), I != II.

• Forbid horizontal attacks

• Forbid vertical attacks



queens.lp

row(1..n).col(1..n).{ queen(I,J) : row(I) : col(J) }.:- not n { queen(I,J) } n.:- queen(I,J), queen(I,JJ), J != JJ.:- queen(I,J), queen(II,J), I != II.

• Forbid horizontal attacks• Forbid vertical attacks



Running . . .

$ gringo queens.lp --const n=5 | claspAnswer: 1row(1) row(2) row(3) row(4) row(5) \col(1) col(2) col(3) col(4) col(5) \queen(5,5) queen(4,4) queen(3,3) \queen(2,2) queen(1,1)...


Horizontal and vertical Attack: Answer 1

Answer 1

5 Z0Z0L4 0Z0L03 Z0L0Z2 0L0Z01 L0Z0Z

1 2 3 4 5


Diagonal Attack

queens.lp

row(1..n).col(1..n).{ queen(I,J) : row(I) : col(J) }.:- not n { queen(I,J) } n.:- queen(I,J), queen(I,JJ), J != JJ.:- queen(I,J), queen(II,J), I != II.:- queen(I,J), queen(II,JJ), (I,J) != (II,JJ), I-J == II-JJ.:- queen(I,J), queen(II,JJ), (I,J) != (II,JJ), I+J == II+JJ.

• Forbid diagonal attacks


Diagonal Attack

Running . . .

$ gringo queens.lp --const n=5 | claspAnswer: 1row(1) row(2) row(3) row(4) row(5) \col(1) col(2) col(3) col(4) col(5) \queen(4,5) queen(1,4) queen(3,3) \queen(5,2) queen(2,1)SATISFIABLE

Models : 1+Time : 0.000

Prepare : 0.000Prepro. : 0.000Solving : 0.000


Diagonal Attack: Answer 1

Answer 1

5 ZQZ0Z4 0Z0ZQ3 Z0L0Z2 QZ0Z01 Z0ZQZ

1 2 3 4 5


Optimizing

queens-opt.lp

1 { queen(I,1..n) } 1 :- I = 1..n.1 { queen(1..n,J) } 1 :- J = 1..n.:- 2 { queen(D-J,J) }, D = 2..2*n.:- 2 { queen(D+J,J) }, D = 1-n..n-1.

• Encoding can be optimized• Much faster to solve


Traveling Salesperson

node(1..6).

edge(1,2;3;4). edge(2,4;5;6). edge(3,1;4;5).edge(4,1;2). edge(5,3;4;6). edge(6,2;3;5).

cost(1,2,2). cost(1,3,3). cost(1,4,1).cost(2,4,2). cost(2,5,2). cost(2,6,4).cost(3,1,3). cost(3,4,2). cost(3,5,2).cost(4,1,1). cost(4,2,2).cost(5,3,2). cost(5,4,2). cost(5,6,1).cost(6,2,4). cost(6,3,3). cost(6,5,1).



node(1..6).

edge(1,2;3;4). edge(2,4;5;6). edge(3,1;4;5).edge(4,1;2). edge(5,3;4;6). edge(6,2;3;5).

cost(1,2,2). cost(1,3,3). cost(1,4,1).cost(2,4,2). cost(2,5,2). cost(2,6,4).cost(3,1,3). cost(3,4,2). cost(3,5,2).cost(4,1,1). cost(4,2,2).cost(5,3,2). cost(5,4,2). cost(5,6,1).cost(6,2,4). cost(6,3,3). cost(6,5,1).



1 { cycle(X,Y) : edge(X,Y) } 1 :- node(X).1 { cycle(X,Y) : edge(X,Y) } 1 :- node(Y).

reached(Y) :- cycle(1,Y).reached(Y) :- cycle(X,Y), reached(X).

:- node(Y), not reached(Y).

#minimize [ cycle(X,Y) = C : cost(X,Y,C) ].


Agenda1 Introduction2 Uninformed Search versus Informed Search (Best First Search, A*

Search, Heuristics)3 Constraint Satisfaction4 Answer Set Programming (ASP)5 Structural Decomposition Techniques (Tree/Hypertree Decompositions)6 Local Search, Stochastic Hill Climbing, Simulated Annealing7 Tabu Search8 Evolutionary Algorithms/ Genetic Algorithms9 Adversial Search and Game Playing


Fixed-Parameter Tractability (FPT) –Motivation

Some Observations• For intractable problems, computational costs often depend primarily on

some problem parameters rather than on the mere size of the instances.• Many hard problems become tractable if some problem parameter is fixed

or bounded by a fixed constant.• Typical parameters for graphs: treewidth and cliquewidth.

– Meta-theorems allow for rather easy proofs of FPT results w.r.t.these parameters

– Dedicated dynamic algorithms required for practical realization!

FPT is one branch in the area of Parameterized Complexity• Downey & Fellows: Parameterized Complexity. Springer, 1999• Flum & Grohe: Parameterized Complexity Theory. Springer, 2006• Niedermeier: Invitation to Fixed-Parameter Algorithms. OUP, 2006


Introduction• Many instances of constraint satisfaction problems can be solved in

polynomial time if their treewidth (or hypertree width) is small.• Solving of problems with bounded width includes two phases:

– Generate a (hyper)tree decomposition with small width;– Solve a problem (based on generated decomposition) with a

particular algorithm such as for example dynamic programming.• The efficiency of solving of problem based on its (hyper)tree

decomposition depends from the width of (hyper)tree decomposition.• It is of high importance to generate (hyper)tree decompositions with small

width.


CSP: Map-Coloring

WesternAustralia

NorthernTerritory

SouthAustralia

Queensland

New South Wales

Victoria

Tasmania

Variables WA, NT, Q, NSW, V, SA, TDomains Di = {red, green, blue}

Constraints: adjacent regions must have different colors e.g., WA 6= NT, or(WA, NT) ∈ {(red, green), (red, blue), (green, red), (green, blue), . . .}


Constraint Graph

Binary CSP: each constraint relates at most two variablesConstraint graph: nodes are variables, arcs show constraints

Victoria

WA

NT

SA

Q

NSW

V

T

General-purpose CSP algorithms use the graph structureto speed up search. E.g., Tasmania is an independent sub-problem!TU Dresden, 13th May 2014 PSSAI slide 51 of 73

Tree-structured CSPs

AB

CD

E

F

TheoremIf the constraint graph has no loops, the CSP can be solved in O(n d2) time.

• Compare to general CSPs, where worst-case time is O(dn)

• This property also applies to logical and probabilistic reasoning: animportant example of the relation between syntactic restrictions and thecomplexity of reasoning.


CSP: SAT Problem(x1 ∨ x2 ∨ ¬x3) ∧ (x1 ∨ ¬x4 ∨ x5 ∨ x6) ∧ · · · ∧ (x3 ∨ x4 ∨ x7 ∨ x8) . . .Possible CSP fomulation:

Variables x1, x2, x3, . . .Domains 0, 1

Constraints – C1: (x1 ∨ x2 ∨ ¬x3)→ true– C2: (x1 ∨ ¬x4 ∨ x5 ∨ x6)→ true– . . .


CSP and Hypergraph

(x1 ∨ x2 ∨ ¬x3) ∧ (x1 ∨ ¬x4 ∨ x5 ∨ x6) ∧ (x3 ∨ x4 ∨ x7 ∨ x8) . . .

In general worst case complexity: 2NumberOfVariables = 219


Hypergraph and its Primal Graph


CSP and (Hyper)treewidth• In general exponential worst case complexity.• Can we solve this instance more efficiently (or in polynomial time)?• Yes, if it has a small (hyper) treewidth!!!


Tree Decomposition

DefinitionTree Decomposition Let G = (V, E) be a graph. A tree decomposition of G is apair (T,χ), where T = (I, F) is a tree with node set I and edge set F, andχ = {χi : i ∈ I} is a family of subsets of V, one for each node of T, such that

1⋃

i∈I χi = V,

2 for every edge (v, w) ∈ E, there is an i ∈ I with v ∈ χi and w ∈ χi, and3 for all i, j, k ∈ I, if j is on the path from i to k in T, then χi ∩ χk ⊆ χj.

The width of a tree decompostion is maxi∈I |χi| − 1.The treewidth of a graph G, denoted by tw(G), is the minimum width over allpossible tree decompositions of G.


Tree Decomposition - Example

All pairs of vertices that are connected appear in some node of the tree.Connectedness condition for vertices


Elimination Ordering• For the given problem find the tree decomposition with minimal width ->

NP hard.• There exists a perfect elimination ordering which produces tree

decomposition with treewidth (smallest width).• Tree decomposition problem→ search for the best elimination ordering of

vertices!• Permutation Problem→ similar to TSP.

Possible elimination ordering for graph in previous slide:10, 9, 8, 7, 2, 3, 6, 1, 5, 4


Perfect Elimination Ordering

Vertex 10 is eliminated from the graph. All neighbors of 10 are connected and atree node is created that contains vertex 10 and its neighbors.

Elimination ordering: 10, 9, 8, 7, 2, 3, 6, 1, 5, 4


Perfect Elimination Ordering ctd.

The tree decomposition node withvertices [7,9,10] is connected with thetree decomposition node which iscreated when the next vertex whichappears in [7,9,10] is eliminated (in thiscase vertex 9)

Vertex 9 is eliminated from the graph. All neighbors of vertex 9 are connectedand a new tree node is created.



Tree Decomposition of a Graph

Width: max(vertices in tree node)−1 = 3.

Treewidth: minimal width over all possible tree decompostions.


Bounded Treewidth for CSP

If a graph has treewidth k, and we are given the corresponding treedecomposition, then the problem can be solved in O(ndk+1) time.

n - number of variables,

d - maximum domain size of any variable in the CSP.

But, finding the decomposition with minimal treewidth is NP-hard.

→ Heuristic methods work well in practice!


References

Martin Gebser, Benjamin Kaufmann Roland Kaminski, and TorstenSchaub.Answer Set Solving in Practice.Synthesis Lectures on Artificial Intelligence and Machine Learning.Morgan and Claypool Publishers, 2012.doi=10.2200/S00457ED1V01Y201211AIM019.

Michael Gelfond and Vladimir Lifschitz.Classical negation in logic programs and disjunctive databases.New Generation Comput., 9(3–4):365–386, 1991.

• See also: http://potassco.sourceforge.net


http://potassco.sourceforge.net

PROBLEM SOLVING AND SEARCH IN ARTIFICIAL INTELLIGENCE€¦ · Artiﬁcial Intelligence,...

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