april 13, 2006 map coloring with logic two formulations of map coloring tim hinrichs stanford logic...

April 13, 2006

Map Coloring with LogicTwo Formulations of Map Coloring

Tim Hinrichs

Stanford Logic Group

April 13, 2006 Map Coloring with Logic 2

Map Coloring


Map Coloring

• Color the nodes of the graph so that– no two adjacent nodes are the same color


Student Formulation

Premises:color(X,C) => region(X)color(X,C) => hue(C)adjacent(X1,X2) ^ color(X1,C) => -color(X2,C)region(X) => Y.color(X,Y)

<ground atoms for adjacent, region, hue>

Query:TUVXYZ | color(r1,T) ^ color(r2,U) ^ color(r3,V) ^ …

- This is not the formulation used in the CSP/LP literature.- LP engines today do not understand this formulation.

r3 r6

r4 r2 r5

r1


Logic Programming Formulation

• Axiomsnext(red,blue)next(red,green)next(red,yellow)next(blue,red)next(blue,green)…

• QueryR1 R2 R3 R4 R5 R6 | next(R1,R2) ^ next(R1,R3) ^ next(R1,R5) ^

next(R1,R6) ^ next(R2,R3) ^ next(R2,R4) ^ next(R2,R5) ^ next(R2,R6) ^ next(R3,R4) ^ next(R3,R6) ^ next(R5,R6)

This is the typical formulation used in the literature.This form (deductive) is required by LP.

r3 r6

r4 r2 r5

r1


Map Coloring with Logic

• Two formulations1. The one students produce in CS157 every year

2. The one used in the literature [McCarthy82]

• (1) is more natural (and non-deductive)

(2) is more efficient (and deductive)

• Question for the day:– How do we translate (1) into (2)?


Agenda

1. Consistency to Deduction: Reformulate non-deductive problem into deductive problem (a form of closing a theory)

2. Database Reformulation: turn output of (1) into the Logic Programming formulation.

3. Bilevel Reasoning: speeding up (1)


Running Example

• Example: a very simple graph and two colors– 3 nodes, 2 edges– red and blue

Solutions:Problem:


Constraint Satisfaction

• Definition (Constraint Satisfaction Problem):Input:

<V, DV, CV>

V: set of variables v1,v2,…

DV: domain for each variable -- dom(vi)

CV: constraints on values of variables

Output: Assignment of values to variables so that

1. val(vi) in dom(vi)2. all the constraints are satisfied


Map Coloring as a CSP

• Variables (V): nodes

• Domains (DV): same for each variablefor every i, dom(vi) = the set of colors

• Constraints (CV): mathematically, a set of permissible tables, i.e. for every pair of variables x and y, a table for all the allowed values of x and y.


Example

V: {R1, R2, R3}

DV: dom(R1) = {r, b} dom(R2) = {r, b} dom(R3) = {r, b}

R1 R2

r b

b r

CV:

r1 r2 r3

R2 R3

r b

b r


Example Solution

V: {R1, R2, R3}


R1 R2

r b

b r

CV:

r1 r2 r3

R2 R3

r b

b r


All Solutions

V: {R1, R2, R3}


R1 R2

r b

b r

CV:

r1 r2 r3

R2 R3

r b

b r


All Solutions as a DB Query

• The set of all solutions can also be computed using the natural join operator.

r1 r2 r3

R1 R2

r b

b r

R2 R3

r b

b r=

R1 R2 R3

r b r

b r b


Constraint Representations

• Representing constraints as permissible tables can be expensive and cumbersome.

• Often, constraints are more naturally and economically represented as logical sentences.


Student Formulation

Premises:color(X,C) => region(X)color(X,C) => hue(C)adjacent(X1,X2) ^ color(X1,C) => -color(X2,C)region(X) => Y.color(X,Y)adjacent(r1,r2) ^ adjacent(r2,r3) ^ hue(r) ^ hue(b)region(r1) ^ region(r2) ^ region(r3)

Entailment Query:XYZ | color(r1,X) ^ color(r2,Y) ^ color(r3,Z)

Models:

r1 r2 r3


Student Formulation

Premises:color(X,C) => region(X)color(X,C) => hue(C)adjacent(X1,X2) ^ color(X1,C) => -color(X2,C)region(X) => Y.color(X,Y)adjacent(r1,r2) ^ adjacent(r2,r3) ^ hue(r) ^ hue(b)region(r1) ^ region(r2) ^ region(r3)

Build a model:XYZ | color(r1,X) ^ color(r2,Y) ^ color(r3,Z)

Models:

r1 r2 r3


Logic Programming Formulation

• Is different than the student formulation

• [McCarthy82] borrowed from Pereira and Porto [1980]– McCarthy started with their formulation and showed various

reformulations that improved efficiency.

– Served as an exploration of Kowalski’s doctrine:Algorithm = Logic + Control

• Originally used to illustrate issues in Constraint Satisfaction/Logic Programming


LP Formulation

• Axioms

next(r,b)next(b,r)

• Entailment queryR1 R2 R3 | next(R1,R2) ^ next(R2,R3)

r1 r2 r3

Deductive


Problem Statement• Produce an algorithm that converts

Axioms:color(X,C) => region(X)color(X,C) => hue(C)adjacent(X1,X2) ^ color(X1,C) => -color(X2,C)region(X) => Y.color(X,Y)…

Build Model: XYZ | color(r1,X) ^ color(r2,Y) ^ color(r3,Z)

intoAxioms:

next(red,blue)next(blue,red)

Entailment Query: XYZ | next(X,Y) ^ next(Y,Z)


Herbrand Logic

• Assume no function constants

• The axioms include a – DCA over all ground terms – UNA over all ground terms

• Assume these hold implicitly– Finite Herbrand Logic


An Observation: Studentcolor(X,C) => region(X)color(X,C) => hue(C)adjacent(X,Y) ^ color(X,C) => -color(Y,C)

• Write the last with V instead of =>

-adjacent(X,Y) V -color(X,C) V -color(Y,C)

• Except for typing, these constraints are entirely negative– Entail what colorings are invalid.

– Every other coloring is valid.


Student Coloringsadjacent(X,Y) ^ color(X,C) => -color(Y,C)

• Invalid answers (entailed) XYZ | -(color(r1,X) ^ color(r2,Y) ^ color(r3,Z))

• Remaining answers

…


Idea

• To reformulate from Student to LP,1. Compute all invalid colorings (which are entailed).

2. Compute the complement of that set.

=

…


Implementation

: the set of Student constraintsQuery: color(r1,X) ^ color(r2,Y) ^ color(r3,Z)

• Reify the notion of an invalid coloring.(*) invalid(X,Y,Z) <=>

-(color(r1,X) ^ color(r2,Y) ^ color(r3,Z))

• Ask for all the invalid colorings that are logically entailed.

XYZ | ^ (*) |= invalid(X,Y,Z)


Implementation

XYZ | invalid(X,Y,Z)

• The following expression is logically entailed and represents the set of all invalid colorings:

X=r1 v X=r2 v X=r3 v Y=r1 v Y=r2 v Y=r3 v

Z=r1 v Z=r2 v Z=r3 v X=Y v Y=Z

X Y Z


Implementation

• To compute an expression that represents the complement of that set, it turns out that all we need to do is negate that expression.

X≠r1 ^ X≠r2 ^ X≠r3 ^Y≠r1 ^ Y≠r2 ^ Y≠r3 ^Z≠r1 ^ Z≠r2 ^ Z≠r3 ^ X≠Y ^ Y≠Z

• The above (with unique names axioms) entails all the valid colorings.

X Y Z


Progress: Both are deductive• Student after reformulation: all deductive answers to

XYZ | X≠r1 ^ X≠r2 ^ X≠r3 ^ Y≠r1 ^ Y≠r2 ^ Y≠r3 ^ Z≠r1 ^ Z≠r2 ^ Z≠r3 ^ X≠Z ^ Y≠Z

UNA[r,b,r1,r2,r3]

• LP formulation: all deductive answers toXYZ | next(X,Y) ^ next(Y,Z)next(r,b)next(b,r)

X Y Z


Student* as Database Query

• Because Student* is now a query over a complete theory (≠), it can be viewed as a DB query.

XYZ | X≠r1 ^ X≠r2 ^ X≠r3 ^

Y≠r1 ^ Y≠r2 ^ Y≠r3 ^Z≠r1 ^ Z≠r2 ^ Z≠r3 ^X≠Z ^ Y≠Z

r1 r2

r1 r3

r1 r

r1 b

r2 r1

r2 r3

r2 r

r2 b

r3 r1

…

≠


Database Reformulation

• Given one database query, find an equivalent query that is more efficient.– Includes performing transformations on the data itself, with a

space limit.

• Theorem [Chirkova]: There are infinitely many distinct transformations.

• Theorem [Chirkova]: There is an algorithm that given the query constructs a finite space of queries that is guaranteed to include the optimal one.


Back to Map Coloring

• Given the queryX≠r1 ^ X≠r2 ^ X≠r3 ^ Y≠r1 ^ … Z≠r3 ^ X≠Y ^ Y≠Z

one of the transformations Chirkova’s work considers isview(X,Y) <= X≠r1 ^ X≠r2 ^ X≠r3 ^ Y≠r1 ^ Y≠r2 ^ Y≠r3 ^

X≠Y

which can be used to rewrite the above query asview(X,Y) ^ view(Y,Z)

X Y Z


Comparison

• Student**XYZ | view(X,Y) ^ view(Y,Z)view(r,b)view(b,r)

• Logic ProgrammingXYZ | next(X,Y) ^ next(Y,Z)next(r,b)next(b,r)

X Y Z


Overview• Starting with the Student formulation,

XYZ | color(r1,X) ^ color(r2,Y) ^ color(r3,Z) color(X,C) => region(X) color(X,C) => hue(C) adjacent(X1,X2) ^ color(X1,C) => -color(X2,C)

use logical entailment to find an expression for all the invalid colorings and negate it. XYZ | X≠r1 ^ X≠r2 ^ X≠r3 ^ … ^ X≠Y ^ Y≠Z

• Then apply Chirkova’s work to produce the LP formulation, or one that is at least as good as LP. XYZ | view(X,Y) ^ view(Y,Z)


Another Example

• N-Queens: place N queens on an NxN chessboard s.t.– No queen can be attacked by another queen.

That is– No two queens share the same row– No two queens share the same column– No two queens share the same diagonal


8 Queens Solution


N-Queens Constraints

• All queens in different rowsX≠Y ^ row(X,R) => -row(Y,R)

• All queens in different columnsX≠Y ^ col(X,C) => -col(Y,C)

• All queens in different diagonalsX≠Y ^ diag(X,D1,D2)

=> -diag(Y,D1,D2)


N-Queens Diagonals• Row, Col is enough to

compute the two diagonals

diag(X,pos,D) <= row(X,R) ^ col(X,C) ^ sub(C, R, D)

diag(X,neg,D) <= row(X,R) ^ col(X,C) ^ add(C, R, D)

• Diagonals represented by their slope and y-intercepts

0

1

2

3

4


4 Queens Query

• Satisfiability Query:

(X1 Y1 X2 Y2 X3 Y3 X4 Y4) |

row(q1,X1) ^ col(q1,Y1) ^row(q2,X2) ^ col(q2,Y2) ^row(q3,X3) ^ col(q3,Y3) ^row(q4,X4) ^ col(q4,Y4)


Efficiency in N-Queens

• The expression that represents all invalid positions can become quite large and might take a long time to compute because of the add and sub facts.– About 150 conjuncts

• Idea: produce an expression in terms of add and sub, as well as =.


Bilevel Reasoning • Separate constraints from data.

X≠Y ^ row(X,R) => -row(Y,R)X≠Y ^ col(X,C) => -col(Y,C)X≠Y ^ diag(X,D1,D2) => -diag(Y,D1,D2)diag(X,pos,D) <= row(X,R) ^ col(X,C) ^ sub(C, R, D)diag(X,neg,D) <= row(X,R) ^ col(X,C) ^ add(C, R, D)add(1,1,2)add(1,1,3)...add(4,4,8)sub(1,1,0)sub(1,2,-1)...sub(4,4,0)



X≠Y ^ row(X,R) => -row(Y,R)X≠Y ^ col(X,C) => -col(Y,C)X≠Y ^ diag(X,D1,D2) => -diag(Y,D1,D2)diag(X,pos,D) <= row(X,R) ^ col(X,C) ^ sub(C, R, D)diag(X,neg,D) <= row(X,R) ^ col(X,C) ^ add(C, R, D)add(1,1,2)add(1,1,3)...add(4,4,8)sub(1,1,0)sub(1,2,-1)...sub(4,4,0)


X≠Y ^ row(X,R) => -row(Y,R)X≠Y ^ col(X,C) => -col(Y,C)X≠Y ^ diag(X,D1,D2) => -diag(Y,D1,D2)diag(X,pos,D) <= row(X,R) ^ col(X,C) ^ sub(C, R, D)diag(X,neg,D) <= row(X,R) ^ col(X,C) ^ add(C, R, D)


add(1,1,2)add(1,1,3)...add(4,4,8)sub(1,1,0)sub(1,2,-1)...sub(4,4,0)


Implementation

’: the set of n-queens constraints w/o add/sub

Query: row(q1,X1) ^ col(q2,Y1) ^ row(q3,X2) ^ ...

• Reify the notion of an invalid coloring.(*) invalid(X1,Y1,X2,Y2,X3,Y3,X4,Y4) <=>

-(row(q1,X1) ^ col(q2,Y1) ^ row(q3,X2) ^ ...)

• Ask for all the invalid colorings where add/sub can be assumed, i.e. use residues (a form of abduction).XYZ | ’ ^ (*) |= invalid(X,Y,Z)


4-Queens Partial Reformulation

goal(X1,Y1,X2,Y2,X3,Y3,X4,Y4) <=

X1 ≠ X2

Y1 ≠ Y2(-sub(Y2,X2,Z) v -sub(Y1,X1,Z))(-add(Y2,X2,Z)) v -add(Y1,X1,Z))

X1 ≠ X3

Y1 ≠ Y3

X2 ≠ X3

Y2 ≠ Y3(-sub(Y3,X3,Z) v -sub(Y1,X1,Z))(-sub(Y3,X3,Z) v -sub(Y2,X2,Z))(-add(Y3,X3,Z) v -add(Y1,X1,Z))(-add(Y3,X3,Z) v -add(Y2,X2,Z))...


Consistency to Deduction• Computing the sentences that are consistent with the

premises and making them deductive consequences. • Definition (query consistency to deduction):

qC2D[, (xbar)] = {tbar | (tbar) is satisfiable}

• In map coloring, = the set of map coloring axioms(xbar) = color(r1,X) ^ color(r2,Y) ^ color(r3,Z) qC2D = {<r,b,r>, <b,r,b>}


Descriptions• Definition (Description): e(xbar) is a description for the set of

n-tuples S, where |xbar| = n if and only if|= e(tbar) if tbar S|= -e(tbar) if tbar S

• Example: X=Y V Y=Z describes {rrr,rrb,brr,bbr,rbb,bbb} in FHL

|=H r=r V r=r |=H r=r V r=b |=H b=r V r=r

|=H b=b V b=r |=H r=b V b=b |=H b=b V b=b

and|=H -(r=b V b=r)

|=H -(b=r V r=b)


Algorithm

• Input : a finite set of FHL sentences

(xbar): a query in the vocabulary of Delta

1. Using algorithm Alg, compute an expression e(xbar) in terms of equality such that for any model M in R,

|=M e(tbar) if and only if |= -(tbar)

2. Output -e(xbar)


Theorems

• Theorem (Soundness and Completeness): Let Alg(, (xbar)) = -e(xbar).

|=M -e(tbar) if and only if (tbar) qC2D[, (xbar)]Proof: Nontrivial

• Theorem (Decidability): qC2D[, (xbar)] for FHL is decidable.Proof: Walk over Herbrand models and if |=M , collect all tbar

s.t. |=M (tbar)


Relationship to CWA

• CWA[] includes -p(tbar) whenever |# p(tbar)

• CWA[] is inconsistent whenever – Delta entails some disjunction d1 v … v dn and

– Delta entails none of the di.

• Adding the map coloring answers to the Student formulation is inconsistent.


Relationship to CWA

• Student axioms entail a disjunctive answer:color(r1,b) ^ color(r2,r) ^ color(r3,b)

v

color(r1,r) ^ color(r2,b) ^ color(r3,r)

– but they entail neither disjunct alone.

• Adding both answers to the axioms produces an inconsistencycolor(r1,b) (by the first answer) and

color(r2,b) (by the second answer)


Relationship to CWA

• CWA: for every ground p, add -p if p is not entailed.

• C2D: for every ground p, add p if -p is not entailed.


Map Coloring with Logic[McCarthy82] McCarthy.

Coloring Maps and the Kowalski Doctrine. Stanford Tech Report 1982.

[Pereira80] Pereira and Porto. Selective Backtracking for Logic Programs. CADE 1980.

[Chirkova2002] Chirkova. Automated Database Restructuring. Stanford thesis 2002.


Scrap starts here



color(X,C) => region(X)color(X,C) => hue(C)adjacent(X1,X2) ^ color(X1,C) => -color(X2,C)adjacent(r1,r2)adjacent(r2,r3)hue(r)hue(b)region(r1)region(r2)region(r3)



color(X,C) => region(X)color(X,C) => hue(C)adjacent(X1,X2) ^ color(X1,C) => -color(X2,C)

adjacent(r1,r2)adjacent(r2,r3)hue(r)hue(b)region(r1)region(r2)region(r3)


Bilevel Map coloring

• It turns out that separating hue off from the rest of the axioms is an interesting case.

color(X,C) => region(X)color(X,C) => hue(C)adjacent(X1,X2) ^ color(X1,C) => -color(X2,C)adjacent(r1,r2)adjacent(r2,r3)region(r1)region(r2)region(r3)

hue(r)hue(b)


Partial Reformulation• The algorithm presented earlier started with axioms

talking about hues, regions, adjacency, and coloradjacent(X,Y) ^ color(X,C) => -color(Y,C)

to produceXYZ | X≠r1 ^ X≠r2 ^ X≠r3 ^


but if we separate hue we should get the following resultXYZ | hue(X) ^ hue(Y) ^ hue(Z) ^ X≠Y ^ Y≠Z

X Y Z


Implementation

• Epilog (the Logic Group’s theorem prover) produces an expression that represents the set of all invalid colorings, in terms of hue:

XYZ | invalid(X,Y,Z), wrt hue=> -hue(X) v -hue(Y) v -hue(Z) v X=Y v Y=Z

• Negating (just as before) gives ushue(X) ^ hue(Y) ^ hue(Z) ^ X≠Y ^ Y≠Z

X Y Z



• Given: a schema R and a set of queries Q={Q1,…,Qn}– Compute a set of views on R that minimize the cost of

evaluating all the Qi.

(Each relation in R is assigned a size.)

• Rada Chirkova [Chirkova2002]: conjunctive queries.– Decidable.

– Triply exponential upper bound (in Q).

– Exponential lower bound (in Q).



R T

Eval(Q,T) < Eval(Q,R)

Find views V on R producing schema T such that

Viewset V


Example

• Schema: parent(x,y)

• Query: XY | samefamily(X,Y)

samefamily(X,Y) <= anc(X,Z) ^ anc(Y,Z)

anc(X,Y) <= parent(X,Y)anc(X,Y) <= anc(X,Z) ^ anc(Z,Y)

• Can be evaluated more efficiently as: samefamily(X,Y) <=

foundingfather(X,F) ^ foundingfather(Y,F)


Map Coloring in Math• 1852: Guthrie makes Four Colour Conjecture:

Any planar graph can be colored using only four colors.

• Definition(Planar graph): Graph that can be drawn on the plane without edges crossing.

• Incorrect proofs given by Kempe (1879) and Tait (1880). Kempe’s was accepted for over a decade.

• 1977: Appel and Haken wrote a program to exhaustively check a large number of cases. They (debatably) proved the conjecture.


Map Coloring in AI

• Used to illustrate Constraint Satisfaction techniques, e.g. Russell and Norvig

• Trying to color a particular graph– Not trying to show every graph with particular

characteristics can be colored.


Student Formulation (2)

Premises:color(X,C) => region(X)color(X,C) => hue(C)adjacent(X1,X2) ^ color(X1,C) => -color(X2,C)

adjacent(r1,r2) ^ adjacent(r2,r3) ^ hue(r) ^ hue(b)region(r1) ^ region(r2) ^ region(r3)

Build a model:

r1 r2 r3

…


DB Reformulation Again• Given the query

hue(X) ^ hue(Y) ^ hue(Z) ^ X≠Y ^ Y≠Z

one of the views Rada’s work considers is

view(X,Y) <= hue(X) ^ hue(Y) ^ X≠Y

which can be used to rewrite the above query as

view(X,Y) ^ view(Y,Z)

X Y Z


Back to Map coloring• In Student*, the schema consists of a single relation: ≠.

• The query XYZ | X≠r1 ^ X≠r2 ^ X≠r3 ^


• Unfortunately, this query includes constants, which Rada’s work does not cover.– Notice that this query is defined in terms of -hue.

X Y Z


Efficiency

• It is unsettling and inefficient to have the answer defined in terms of -hue when it is more natural to define it in terms of hue.

hue(X) ^hue(Y) ^hue(Z) ^X ≠ Y ^Y ≠ Z

– Independent of which colors are available

– No constants

X Y Z


A Technical Note about Descriptions

• e(X,Y,Z) <=> X=Y V Y=Z describes S = {rrr,rrb,brr,bbr,rbb,bbb}

For every tuple tbar in S, |=H e(tbar)

For every tuple tbar not in S, |=H -e(tbar)

• In Herbrand logic, suppose we know that 1. |=H e(tbar) implies tbar in S2. e(xbar) is built on a single relation constant: =. Then e(xbar) is a description of S.


Issues

• Need efficient algorithm for computing all solutions in FHL that terminates.– Would settle for termination on clausal form.

• OSHL[Zhu,Plaisted]

• Model Evolution[Baumgartner]

• Computing the complement via negation worked well for map coloring, but in general it may be expensive when converting to clausal form (CNF to DNF).


An Observation about McCarthy

XYZ | next(X,Y) ^ next(Y,Z)

• The valid colorings are entailed


Complexity Analysis

• CS157hue(X) ^ hue(Y) ^ hue(Z) ^ X≠Y ^ Y≠Z hue(red) red ≠ bluehue(blue) blue ≠ red

• McCarthynext(X,Y) ^ next(Y,Z)next(red,blue)next(blue,red)

• Compare on trees. C=# of colors, N=# of nodes


Asymptotic Complexity Results for Trees

1. hue(X) ^ hue(Y) ^ hue(Z) ^ X≠Y ^ Y≠Z• O(CN)

2. hue(X) ^ hue(Y) ^ X≠Y ^ hue(Z) ^ Y≠Z• O(CN-1)

3. next(X,Y) ^ next(Y,Z)• O(CN-2)

4. Same as 3 but indexed on first argument of next.• O(CN-3)


Back to FOL

• If we are using first-order deduction procedures to answer finite Herbrand entailment queries, need a DCA and UNA.

• [Reiter80] showed that for * queries, DCA is unnecessary.


Looks can be deceiving

• Suppose there were object constants besides red and blue, e.g. paperclip

• New axioms:-r1(X) V -r2(X)-r3(X) V -r2(X)

r1(X) V r2(X) V r3(X) => hue(X)hue(red)hue(blue)

-hue(paperclip)


Running the algorithm:

• After conversion:

goal(X,Y,Z) <=> X ≠ paperclip ^X ≠ Y ^Y ≠ paperclip ^Y ≠ Z ^Z ≠ paperclip

1. Linear in the size of -hue(x)2. Not the same as the McCarthy formulation


Solving CSPs as DB queries

• Can every CSP be formulated as a DB query?– Map coloring can be formulated as a DB query.

• When is it beneficial to solve a CSP by translating it into a DB query?– Under some conditions, it is beneficial for map coloring.

• Can the transformation be automated?


Language of Study

• Option 1 (Mathematics): Explore algorithms of the following sort.– Input: <V, DV, CV>, i.e. a set, a set of functions, and a set of

tables.

– Output: <D, (x1,…,xn)>, i.e. a database and a query over that database.

• Option 2 (Logic): Explore algorithms of the following sort.– Input: Description of a CSP in logic.

– Output: Description of the database and the query in logic.


CS157 Orderings

1. hue(X) ^ hue(Y) ^ hue(Z) ^ X≠Y ^ Y≠Z• Walk hue space and check each point.

2. X≠Y ^ Y≠Z ^ hue(X) ^ hue(Y) ^ hue(Z)• Walk ≠ space and check each point.

3. hue(X) ^ hue(Y) ^ X≠Y ^ hue(Z) ^ Y≠Z• Walk hue space, pruning as we go.

4. X≠Y ^ hue(X) ^ hue(Y) ^ Y≠Z ^ hue(Z)• Walk ≠ space, pruning as we go.


On the Complexity of Query Evaluation

• Complexity depends heavily on indexing.

• Four options1. Relational: index only on the relation of facts

2. Relational + attach : eval ≠ in code

3. Arguments: index on reln and each argument position

4. Arguments + attach


Some Connections

• Some connections to other work

– CWA– Applicability to Datalog queries


Closed World Assumption

• qC2D and CWA are closely related

• In fact, a generalization of qC2D and CWA are nearly equivalent. – If we could compute CWA efficiently, that might help

us compute qC2D efficiently.


A Property of qC2D

• Consider adding our solutions to the map coloring axioms.-r1(X) <= r2(X)-r2(X) <= r3(X)r1(r) ^ r2(b) ^ r3(r)r1(b) ^ r2(r) ^ r3(b)

• These axioms are inconsistent. – Very similar to the inconsistency that can be produced

by Closed World Assumption.


Closed World

• Definition (Set-based CWA): Let P={p1,…,pn} be a set of closed sentences.

sCWA[, P] = {-pi | |# pi}

• Example: = {q(a), q(b)}

P = {q(a), q(c)}

sCWA[, P] = {-q(c)}

• sCWA generalizes the standard definition for CWA


sC2D

• For comparison to sCWA, we define sC2D.

• Definition (sC2D): Let P={p1,…,pn} be a set of sentences.

sC2D[, P] = {pi | pi is satisfiable}

• Theorem: Let {p1,…,pn} be a set of ground atoms.

sCWA[, {p1,…,pn}] = sC2D[, {-p1,…,-pn}]


An Application• Suppose we have some set of Datalog sentences and a

negative query.:- not r(X), not s(Y), not p(Z)

• One approach:goal(X,Y,Z) :- univ(X), univ(Y), univ(Z),

not r(X), not s(Y), not p(Z)

• Cost: |univ|3

• If we had an explicit definition for the negative theory, the cost might be substantially less.


An Application (2)

Data:p(a,b) p(b,a) p(c,a) p(d,a)p(a,c) p(b,c) p(c,b) p(d,b)p(a,d) p(b,d) p(c,d) p(d,c)

Query: univ(X), univ(Y), not p(X,Y)Cost: |univ|2

Query: univ(X), not p(X,X)Cost: |univ|


Computing the Complement

In Map Coloring• Expression for invalid

colorsX=Y V Y=Z

• Negation of that expressionX#Y ^ Y#Z

• Final reformulationgoal(X,Y,Z) <= X#Y ^ Y#Z

More generally• Expression for entailed

negations of query(xbar)

• Negation of that expression-(xbar)

• Final reformulationgoal(xbar) <= -(xbar)


Questions

• Every relation in a theory with a DCA can be defined entirely in terms of equality.

• Question: When can we use fullfinds to compute that definition? (logically justified)

• Question: When should we use fullfinds to compute that definition? (computability/complexity justified)


Consistency

• No coloring of the map is entailed by the axioms.– But both colorings are consistent with the axioms.

• How do we compute consistency?– With a DCA, consistency is decidable.

• How do we compute consistency efficiently?– Sort of sounds like abduction.


Abduction

• Definition (Abduction): explains |= in the language L if and only if

1. |= 2. is consistent

3. is in L

• Let be the map coloring axioms

– r1(r) ^ r2(b) ^ r3(r) explains |= XYZ. r1(X) ^ r2(Y) ^ r3(Z)


Attempt #3

• Residue attempts to compute abductive consequences

-r1(X) <= r2(X)-r2(X) <= r3(X)goal(X,Y,Z) <= r1(X) ^ r2(Y) ^ r3(Z)

? (fullresidue '(goal ?x ?y ?z) 'mapc)Call: (GOAL ?X ?Y ?Z) | Call: (R1 ?X) | Fail: (R1 ?X)Fail: (GOAL ?X ?Y ?Z)

• Equality does not help.


An observation

• The constraints only talk about the negative part of the theory.

-r1(X) <= r2(X) or -r1(X) | -r2(X)-r2(X) <= r3(X) or -r2(X) | -r3(X)

• The query only talks about the positive part of the theory. XYZ. r1(X) ^ r2(Y) ^ r3(Z)

• If we could define the negative part of the theory and then negate it, we would have an explicit definition for the positive theory, on which we could use entailment tools.


Reformulation and Completion

1. Temporarily add the goal to the original axioms.-r1(X) <= r2(X)-r2(X) <= r3(X)goal(X,Y,Z) <=> r1(X) ^ r2(Y) ^ r3(Z)

2. Capture the negative query in terms of equality.-goal(X,Y,Z) <= X=Y | Y=Z

3. Apply predicate completion-goal(X,Y,Z) <=> X=Y | Y=Z goal(X,Y,Z) <=> X#Y ^ Y#Z

4. Assuming UNAxioms, ask for goal(X,Y,Z).


Step 4

goal(X,Y,Z) <= X#Y ^ Y#Z r # b b # r

? (fullfindx '(?x ?y ?z) '(goal ?x ?y ?z) 'mapcc)Call: (GOAL ?X ?Y ?Z) | Call: (NOT (= ?X ?Y)) | Exit: (NOT (= R B)) | Call: (NOT (= B ?Z)) | Exit: (NOT (= B R))Exit: (GOAL R B R)


Step 2 -r1(X) <= r2(X)

-r2(X) <= r3(X)

goal(X,Y,Z) <= r1(X) ^ r2(Y) ^ r3(Z)-goal(X,Y,Z) <= -r1(X)-goal(X,Y,Z) <= -r2(X)-goal(X,Y,Z) <= -r3(X)

r1(X) <= goal(X,Y,Z)r2(Y) <= goal(X,Y,Z)r3(Z) <= goal(X,Y,Z)

? (fullfinds '(?x ?y ?z) '(not (goal ?x ?y ?z)) 'mapc)((?u ?u ?w) (?u ?w ?w))


Back to the CSP Version

V: {v1,v2,v3}

DV: {{r, b},{r, b},{r, b}}

CV:

v1 v2

r b

b r

v2 v3

r b

b r

goal(X,Y,Z) <= X#Y ^ Y#Zr # bb # r


CS157 Formulation (3)

Premises: Models:X=red | X=blue-r1(X) <= r2(X)-r2(X) <= r3(X)

X.r1(X)

X.r2(X)

X.r3(X)

Satisfaction Query: XYZ | r1(X) ^ r2(Y) ^ r3(Z)

– All resolution-style t.p. return something like

In general, adding these sentencesmay cause inconsistency.

k1, k2, k3


Example

V: {v1,…,v7}

DV:

dom(v1) = {r,y,b}

…

dom(v7) = {r,y,b}

v1 v2

r y

r b

y b

y r

b y

b r

1

24

5

3

6

7

v6 v7

r y

r b

y b

y r

b y

b r

CV: ,…,


Example

V: {v1,…,v7}

DV:

dom(v1) = {r,y,b}

…

dom(v7) = {r,y,b}

v1 v2

r y

r b

y b

y r

b y

b r

v6 v7

r y

r b

y b

y r

b y

b r

CV: ,…,

1

24

5

3

6

7


Pereira, Porto, and McCarthy

• Peireira and Porto [1980] treated map coloring in the context of Logic Programming

goal <= next(V1,V2), next(V1,V4), next(V1,V5), next(V2,V3), next(V3,V4), next(V3,V5), next(V3, V7), next(V4,V5), next(V5,V6), next(V6,V7)

next(r,y) next(y,b) next(b,y)

next(r,b) next(y,r) next(b,r)


Why does this work?

• Reformulation (Step 2) rewrites the negation of the query in terms of just equality. – For larger map colorings, there will be O(n2) answers

to -goal(X,Y,Z), where n = # of regions.

• At an abstract level, it enumerates all the cases where -goal(X,Y,Z) is directly entailed

• Everything else is thrown into goal.

april 13, 2006 map coloring with logic two formulations of map coloring tim hinrichs stanford logic...

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