Prolog lecture 6 - University of Cambridge · Prolog programs can be viewed as constraint...

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Prolog lecture 6 ● Solving Sudoku puzzles ● Constraint Logic Programming ● Natural Language Processing

Transcript of Prolog lecture 6 - University of Cambridge · Prolog programs can be viewed as constraint...

Page 1: Prolog lecture 6 - University of Cambridge · Prolog programs can be viewed as constraint satisfaction problems Prolog is limited to the single equality constraint: – two terms

Prolog lecture 6

● Solving Sudoku puzzles● Constraint Logic Programming● Natural Language Processing

Page 2: Prolog lecture 6 - University of Cambridge · Prolog programs can be viewed as constraint satisfaction problems Prolog is limited to the single equality constraint: – two terms

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Playing Sudoku

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Make the problem easier

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We can model this problem in Prolog using list permutations

Each row must be a permutation of [1,2,3,4]

Each column must be a permutation of [1,2,3,4]

Each 2x2 box must be a permutation of [1,2,3,4]

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Represent the board as a list of lists

[[A,B,C,D], [E,F,G,H], [I,J,K,L], [M,N,O,P]]

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The sudoku predicate is built from simultaneous perm constraintssudoku([[X11,X12,X13,X14],[X21,X22,X23,X24], [X31,X32,X33,X34],[X41,X42,X43,X44]]) :- %rows perm([X11,X12,X13,X14],[1,2,3,4]), perm([X21,X22,X23,X24],[1,2,3,4]), perm([X31,X32,X33,X34],[1,2,3,4]), perm([X41,X42,X43,X44],[1,2,3,4]), %cols perm([X11,X21,X31,X41],[1,2,3,4]), perm([X12,X22,X32,X42],[1,2,3,4]), perm([X13,X23,X33,X43],[1,2,3,4]), perm([X14,X24,X34,X44],[1,2,3,4]), %boxes perm([X11,X12,X21,X22],[1,2,3,4]), perm([X13,X14,X23,X24],[1,2,3,4]), perm([X31,X32,X41,X42],[1,2,3,4]), perm([X33,X34,X43,X44],[1,2,3,4]).

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Scale up in the obvious way to 3x3

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Brute-force is impractically slow

There are very many valid grids:6670903752021072936960 ≈ 6.671 × 1021

Our current approach does not encode the interrelationships between the constraints

For more information on Sudoku enumeration:http://www.afjarvis.staff.shef.ac.uk/sudoku/

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Prolog programs can be viewed as constraint satisfaction problems

Prolog is limited to the single equality constraint:– two terms must unify

We can generalise this to include other types of constraint

Doing so leads to Constraint Logic Programming– and a means to solve Sudoku problems (p319)

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Consider variables taking values from domains with constraints

Given:– the set of variables– the domains of each variable– constraints on these variables

We want to find:– an assignment of values to variables satisfying the

constraints

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Sudoku can be expressed as constraints

A {1,2,3,4} B {1,2,3,4}C {1,2,3,4} D {1,2,3,4}E {1,2,3,4} F {1,2,3,4}G {1,2,3,4} H {1,2,3,4}I {1,2,3,4} J {1,2,3,4}K {1,2,3,4} L {1,2,3,4}M {1,2,3,4} N {1,2,3,4}O {1,2,3,4} P {1,2,3,4}

First, we express the variables and domains

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Express Sudoku as a Constraint Graph

{1,2,3,4}{1,2,3,4}

{1,2,3,4}

{1,2,3,4}

{1,2,3,4}

{1,2,3,4}{1,2,3,4}

{1,2,3,4}{1,2,3,4}{1,2,3,4}

{1,2,3,4}

{1,2,3,4}

{1,2,3,4}

{1,2,3,4}{1,2,3,4}

{1,2,3,4}

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Constraints: All variables in rows are different

{1,2,3,4}{1,2,3,4}

{1,2,3,4}

{1,2,3,4}

{1,2,3,4}

{1,2,3,4}{1,2,3,4}

{1,2,3,4}{1,2,3,4}{1,2,3,4}

{1,2,3,4}

{1,2,3,4}

{1,2,3,4}

{1,2,3,4}{1,2,3,4}

{1,2,3,4}

Edges represent inequalities between variables

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Constraints: All variables in columns are different

{1,2,3,4}{1,2,3,4}

{1,2,3,4}

{1,2,3,4}

{1,2,3,4}

{1,2,3,4}{1,2,3,4}

{1,2,3,4}{1,2,3,4}{1,2,3,4}

{1,2,3,4}

{1,2,3,4}

{1,2,3,4}

{1,2,3,4}{1,2,3,4}

{1,2,3,4}

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Constraints: All variables in boxes are different

{1,2,3,4}{1,2,3,4}

{1,2,3,4}

{1,2,3,4}

{1,2,3,4}

{1,2,3,4}{1,2,3,4}

{1,2,3,4}{1,2,3,4}{1,2,3,4}

{1,2,3,4}

{1,2,3,4}

{1,2,3,4}

{1,2,3,4}{1,2,3,4}

{1,2,3,4}

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All constraints shown together

{1,2,3,4}{1,2,3,4}

{1,2,3,4}

{1,2,3,4}

{1,2,3,4}

{1,2,3,4}{1,2,3,4}

{1,2,3,4}{1,2,3,4}{1,2,3,4}

{1,2,3,4}

{1,2,3,4}

{1,2,3,4}

{1,2,3,4}{1,2,3,4}

{1,2,3,4}

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Reduce domains according to initial values

{1,2,3,4}{1,2,3,4}

{4}*

{1,2,3,4}

{1,2,3,4}

{2}*{1,2,3,4}

{1,2,3,4}{1,2,3,4}{1,2,3,4}{1}*

{1,2,3,4}

{1,2,3,4}

{3}*{1,2,3,4}

{1,2,3,4}

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When a domain changes we update its constraints

{1,2,3,4}{1,2,3,4}

{4}*

{1,2,3,4}

{1,2,3,4}

{2}*{1,2,3,4}

{1,2,3,4}{1,2,3,4}{1,2,3,4}{1}*

{1,2,3,4}

{1,2,3,4}

{3}*{1,2,3,4}

{1,2,3,4}

The asterisk (*) notation reminds us that all constraints which connect

to this variable need updating

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{1,2,3,4}{1,2,3,4}

{1,2,3,4}

{1,2,3,4}

{1,2,3,4}{1,2,3,4}{1,2,3,4}

{1,2,3,4}

{1,2,3,4}

{1,2,3,4}

{1,2,3,4}{1,2,3,4}

{4}*

{2}*{1}*

{3}*

Update constraints connected to CWe will remove 4 from the

domain of A, B and D

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{1,2,3}*{1,2,3}*

{1,2,3}*

{1,2,3,4}

{1,2,3,4}{1,2,3,4}{1,2,3,4}

{1,2,3,4}

{1,2,3,4}

{1,2,3,4}

{1,2,3,4}{1,2,3,4}

{4}*

{2}*{1}*

{3}*

Update constraints connected to CWe add asterisks to A, B and D

...but will defer looking at them

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{1,2,3}*{1,2,3}*

{1,2,3}*

{1,2,3,4}

{1,2,3,4}{1,2,3,4}{1,2,3,4}

{1,2,3,4}

{1,2,3,4}

{1,2,3,4}

{1,2,3,4}{1,2,3,4}

{4}*

{2}*{1}*

{3}*

Update constraints connected to CNow examine column

constraints

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{1,2,3}*{1,2,3}*

{1,2,3}*

{1,2,3,4}

{1,2,3}*{1,2,3,4}{1,2,3,4}

{1,2,3,4}

{1,2,3,4}

{1,2,3,4}

{1,2,3}*{1,2,3,4}

{4}*

{2}*{1}*

{3}*

Update constraints connected to C

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{1,2,3}*{1,2,3}*

{1,2,3}*

{1,2,3,4}

{1,2,3}*{1,2,3,4}{1,2,3,4}

{1,2,3,4}

{1,2,3,4}

{1,2,3,4}

{1,2,3}*{1,2,3,4}

{4}*

{2}*{1}*

{3}*

Update constraints connected to CNote that D and G have

already had their domains updated

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{1,2,3}*{1,2,3}*

{1,2,3}*

{1,2,3,4}

{1,2,3}*{1,2,3}*{1,2,3,4}

{1,2,3,4}

{1,2,3,4}

{1,2,3,4}

{1,2,3}*{1,2,3,4}

{4}*

{2}*{1}*

{3}*

Update constraints connected to CWe have exhausted C's

constraints for now

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{1,2,3}*{1,2,3}*

{1,2,3}*

{1,2,3,4}

{1,2,3}*{1,2,3}*{1,2,3,4}

{1,2,3,4}

{1,2,3,4}

{1,2,3,4}

{1,2,3}*{1,2,3,4}

{4}

{2}*{1}*

{3}*

Update constraints connected to F

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{1,2,3}*{1,2,3}*

{1,2,3}*

{1,3,4}*

{1,3}*{1,3}*{1,2,3,4}

{1,2,3,4}

{1,2,3,4}

{1,2,3,4}

{1,2,3}*{1,2,3,4}

{4}

{2}*{1}*

{3}*

Update constraints connected to F

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{1,2,3}*{1,2,3}*

{1,2,3}*

{1,3,4}*

{1,3}*{1,3}*{1,2,3,4}

{1,2,3,4}

{1,2,3,4}

{1,2,3,4}

{1,2,3}*{1,2,3,4}

{4}

{2}*{1}*

{3}*

Update constraints connected to F

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{1,2,3}*{1,3}*

{1,2,3}*

{1,3,4}*

{1,3}*{1,3}*{1,2,3,4}

{1,3,4}

{1,2,3,4}

{1,2,3,4}

{1,2,3}*{1,2,3,4}

{4}

{2}*{1}*

{3}*

Update constraints connected to F

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{1,2,3}*{1,3}*

{1,2,3}*

{1,3,4}*

{1,3}*{1,3}*{1,2,3,4}

{1,3,4}*

{1,2,3,4}

{1,2,3,4}

{1,2,3}*{1,2,3,4}

{4}

{2}*{1}*

{3}*

Update constraints connected to F

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{1,3}*{1,3}*

{1,2,3}*

{1,3,4}*

{1,3}*{1,3}*{1,2,3,4}

{1,3,4}*

{1,2,3,4}

{1,2,3,4}

{1,2,3}*{1,2,3,4}

{4}

{2}*{1}*

{3}*

Update constraints connected to FWe have exhausted F's

constraints for now

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{1,3}*{1,3}*

{1,2,3}*

{1,3,4}*

{1,3}*{1,3}*{1,2,3,4}

{1,3,4}*

{1,2,3,4}

{1,2,3,4}

{1,2,3}*{1,2,3,4}

{4}

{2}{1}*

{3}*

Update constraints connected to K

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{1,3}*{1,3}*

{1,2,3}*

{1,3,4}*

{1,3}*{1,3}*{2,3,4}*

{3,4}*

{2,3,4}*

{1,2,3,4}

{1,2,3}*{1,2,3,4}

{4}

{2}{1}*

{3}*

Update constraints connected to K

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{1,3}*{1,3}*

{1,2,3}*

{1,3,4}*

{1,3}*{1,3}*{2,3,4}*

{3,4}*

{2,3,4}*

{1,2,3,4}

{1,2,3}*{1,2,3,4}

{4}

{2}{1}*

{3}*

Update constraints connected to K

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{1,3}*{1,3}*

{1,2,3}*

{1,3,4}*

{3}*{1,3}*{2,3,4}*

{3,4}*

{2,3,4}*

{1,2,3,4}

{2,3}*{1,2,3,4}

{4}

{2}{1}*

{3}*

Update constraints connected to K

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{1,3}*{1,3}*

{1,2,3}*

{1,3,4}*

{3}*{1,3}*{2,3,4}*

{3,4}*

{2,3,4}*

{1,2,3,4}

{2,3}*{1,2,3,4}

{4}

{2}{1}*

{3}*

Update constraints connected to K

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{1,3}*{1,3}*

{1,2,3}*

{1,3,4}*

{3}*{1,3}*{2,3,4}*

{3,4}*

{2,3,4}*

{1,2,3,4}

{2,3}*{2,3,4}*

{4}

{2}{1}*

{3}*

Update constraints connected to K

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{1,3}*{1,3}*

{1,2,3}*

{1,3,4}*

{3}*{1,3}*{2,3,4}*

{3,4}*

{2,3,4}*

{1,2,3,4}

{2,3}*{2,3,4}*

{4}

{2}{1}

{3}*

Update constraints connected to KWe have exhausted K's

constraints for now

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{1,3}*{1,3}*

{1,2,3}*

{1,3,4}*

{3}*{1,3}*{2,3,4}*

{3,4}*

{2,3,4}*

{1,2,3,4}

{2,3}*{2,3,4}*

{4}

{2}{1}

{3}*

Update constraints connected to D

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{1,3}*{1,3}*

{1,2,3}*

{1,3,4}*

{3}*{1,3}*{2,3,4}*

{3,4}*

{2,3,4}*

{1,2,3,4}

{2,3}*{2,3,4}*

{4}

{2}{1}

{3}*

Update constraints connected to DNo values can be eliminated directly

(we should see the answer though!)

Page 40: Prolog lecture 6 - University of Cambridge · Prolog programs can be viewed as constraint satisfaction problems Prolog is limited to the single equality constraint: – two terms

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{1,3}*{1,3}*

{1,2,3}*

{1,3,4}*

{3}*{1,3}*{2,3,4}*

{3,4}*

{2,3,4}*

{1,2,3,4}

{2,3}*{2,3,4}*

{4}

{2}{1}

{3}*

Update constraints connected to D

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{1,3}*{1,3}*

{1,2}*

{1,3,4}*

{3}*{1,3}*{2,3,4}*

{3,4}*

{2,3,4}*

{1,2,3,4}

{2,3}*{2,3,4}*

{4}

{2}{1}

{3}*

Change can occur in source domainSingle 3 at G eliminates D's 3

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{1,3}*{1,3}*

{1,2}*

{1,3,4}*

{3}*{1,3}*{2,3,4}*

{3,4}*

{2,3,4}*

{1,2,3,4}

{2,3}*{2,3,4}*

{4}

{2}{1}

{3}*

Change can occur in source domainIf the source domain changes we mark all its

constraints for update again

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{3}{1}

{2}

{4}

{3}{1}{2}

{4}

{3}

{1}

{2}{4}

{4}

{2}{1}

{3}

Iterate the algorithm to convergence (no further changes occur)

Why will the algorithm eventually converge?

Page 44: Prolog lecture 6 - University of Cambridge · Prolog programs can be viewed as constraint satisfaction problems Prolog is limited to the single equality constraint: – two terms

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Outcome 1: Single valued domains

{3}{1}

{2}{4}

{3}{1}{2}

{4}

{3}{1}

{2}{4}

{4}

{2}{1}

{3}

We have found a unique solution to the problem

Page 45: Prolog lecture 6 - University of Cambridge · Prolog programs can be viewed as constraint satisfaction problems Prolog is limited to the single equality constraint: – two terms

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Outcome 2: Some empty domains

VariablesA {1}B {1,2}

C {1,2}

ConstraintsA ≠B, A ≠C, B ≠C

A

BC

{1}

{1,2}{1,2}

Our constraints are shown to be inconsistent– therefore there is no solution to this problem

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Outcome 2: Some empty domains

VariablesA {1}B {1,2}

C {1,2}

ConstraintsA ≠B, A ≠C, B ≠C {2}{2}

Our constraints are shown to be inconsistent– therefore there is no solution to this problem

A

BC

{1}

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Outcome 2: Some empty domains

VariablesA {1}B {1,2}

C {1,2}

ConstraintsA ≠B, A ≠C, B ≠C {2}{}

Our constraints are shown to be inconsistent– therefore there is no solution to this problem

A

BC

{1}

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Outcome 3: Some multivalued domains

{3}{1}

{2}

{4}

{1,2,3}{1,2,3}{1,2}

{4}

{1,2,3}

{1,2}

{1,2}{1,2,3,4}

{4}

{2}{1,2,3}

{3}

Not all combinations of these variable assignment possibilities are global solutions though...

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Outcome 3: Hypothesise labellings

To find global solutions from the narrowed domains we hypothesise a solution in a domain and propagate the changes

Backtrack if something goes wrong

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:- use_module(library(bounds)).

valid4(L) :- L in 1..4, all_different(L).

sudoku2([[X11,X12,X13,X14],[X21,X22,X23,X24], [X31,X32,X33,X34],[X41,X42,X43,X44]]) :- valid4([X11,X12,X13,X14]),valid4([X21,X22,X23,X24]), valid4([X31,X32,X33,X34]),valid4([X41,X42,X43,X44]), valid4([X11,X21,X31,X41]),valid4([X12,X22,X32,X42]), valid4([X13,X23,X33,X43]),valid4([X14,X24,X34,X44]), valid4([X11,X12,X21,X22]),valid4([X13,X14,X23,X24]), valid4([X31,X32,X41,X42]),valid4([X33,X34,X43,X44]), labeling([],[X11,X12,X13,X14,X21,X22,X23,X24, X31,X32,X33,X34,X41,X42,X43,X44]).

Using CLP in Prolog

Rows

Cols

Boxes

Chapter 14 of the textbook has more information

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51

Prolog can be used for parsing context-free grammars (p555)

Here is a simple grammar:

Terminal symbols:a, b, c

Non-terminal symbols: s

s → 'a' 'b's → 'a' 'c's → s s

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52

Parsing by consumption

Write a predicate for each non-terminal that:– consumes as much of the first list as is necessary to

match the non-terminal, and– returns the remaining elements in the second list

These predicate evaluations will thus be true:– s([a,b],[])– s([a,b,c,d],[c,d])

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53

A Prolog program that accepts sentences from our grammar

s → 'a' 'b's → 'a' 'c's → s s% match a single character

c([X|T],X,T).

% grammar predicatess(In,Out) :- c(In,a,In2), c(In2,b,Out).s(In,Out) :- c(In,a,In2), c(In2,c,Out).s(In,Out) :- s(In,In2), s(In2,Out).

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54

Prolog DCG syntax

Prolog provides us with a shortcut for encoding Definite Clause Grammar (DCG) syntax.

This will both test and generate:– s([a,c,a,b],[]).– s(A,[]).

s → 'a' 'b's → 'a' 'c's → s s

s --> [a],[b].s --> [a],[c].s --> s,s.

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55

Building a parse tree

% match a single characterc([X|T],X,T).

% grammar predicatess(ab,In,Out) :- c(In,a,In2), c(In2,b,Out).s(ac,In,Out) :- c(In,a,In2), c(In2,c,Out).s(t(A,B),In,Out) :- s(A,In,In2), s(B,In2,Out).

:- s(Result,[a,c,a,b,a,b],[]).

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56

Prolog's DCG syntax helps us again

s(ab) --> [a],[b].s(ac) --> [a],[c].s(t(A,B)) --> s(A),s(B).

Unfortunately the DCG syntax is not part of the ISO Prolog standard– Almost all modern compilers will include it though

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57

Parsing Natural Language(back to Prolog's roots)

This is a very limited English grammar subset.

Things get complicated very quickly!– see the Natural Language

Processing course next year (Prolog is not a pre-requisite)

s --> np,vp.np --> det,n.vp --> v.vp --> v,np.

n --> [cat].n --> [dog].v --> [eats].det --> [the].

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58

the dog eats the cat

det noun

verb

det noun

np

np

vp

s

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59

We can also handle agreement

We consider only third-person constructions here!

s(N) --> np(N),vp(N).np(N) --> det,n(N).vp(N) --> v(N).vp(N) --> v(N),np(_).

n(s) --> [cat].n(s) --> [dog].n(p) --> [cats].v(s) --> [eats].v(p) --> [eat].det --> [the].

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60

the cats eat the dog

det n(p)

verb(p)

det n(s)

np(p)

np(s)

vp(p)

s

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61

Real Natural Language Processing

Things get much more complicated very quickly

Ambiguities, special cases and noise all make the approach we have demonstrated hard to scale– Although people have definitely tried!

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62

Prolog has lasting influence

Languages that have been influenced by Prolog:– Mercury

● Compiled language that takes advantage of knowledge about predicate determinism

– Erlang● Massively concurrent programming

Projects– Overlog

● Declarative networking– XSB

● Declarative database: tabled resolution, HiLog

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63

Closing Remarks

Declarative programming is different to Functional or Procedural programming– Foundations of Computer Science & Programming in Java

Prolog is built on logical deduction– formal explanation in Logic & Proof

It can provide concise implementations of algorithms such as sorting or graph search– Algorithms I & Algorithms II

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64

Closing Remarks

Foundations of Functional Programming (Part IB)– Building computation from first principles

Databases (Part 1B)– Find out more about representing data and SQL

Artificial Intelligence (Part 1B)– Search, constraint programming and more

C & C++ (Part 1B)– Doing useful stuff in the real world

Natural Language Processing (Part II)– Parsing natural language

Page 65: Prolog lecture 6 - University of Cambridge · Prolog programs can be viewed as constraint satisfaction problems Prolog is limited to the single equality constraint: – two terms

EndC

You shoot yourself in the foot.

C++ You accidentally create a dozen instances of yourself and shoot

them all in the foot. Providing emergency medical care is impossible since you can't tell which are bitwise copies and which

are just pointing at others and saying, "That's me over there."

Prolog You explain in your program that you want to be shot in the foot. The interpreter figures out all the possible ways to do it, but then

backtracks completely, instead destroying the gun.


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