PLVol1

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Prolog Programming

slides written by

Dr W.F. Clocksin

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The Plan

An example program Syntax of terms

Some simple programs

Terms as data structures, unification The Cut

Writing real programs

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What is Prolog?

Prolog is the most widely used languageto have been inspired by logicprogramming research. Some features:

Prolog uses logical variables. These are

not the same as variables in otherlanguages. Programmers can use them asholes in data structures that aregradually filled in as computationproceeds.

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More

Unification is a built-in term-manipulation method that passesparameters, returns results, selects andconstructs data structures.

Basic control flow model is backtracking.

Program clauses and data have the sameform.

The relational form of procedures makesit possible to define reversibleprocedures.

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More

Clauses provide a convenient way toexpress case analysis andnondeterminism.

Sometimes it is necessary to use control

features that are not part of logic.

A Prolog program can also be seen as arelational database containing rules as

well as facts.

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What a program looks like

/* At the Zoo */

elephant(george).

elephant(mary).

panda(chi_chi).

panda(ming_ming).

dangerous(X) :- big_teeth(X).

dangerous(X) :- venomous(X).

guess(X, tiger) :- stripey(X), big_teeth(X), isaCat(X).

guess(X, koala) :- arboreal(X), sleepy(X).

guess(X, zebra) :- stripey(X), isaHorse(X).

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Prolog is a declarative language

Clauses are statements about what istrue about a problem, instead ofinstructions how to accomplish thesolution.

The Prolog system uses the clauses towork out how to accomplish the solutionby searching through the space ofpossible solutions.

Not all problems have pure declarativespecifications. Sometimes extralogicalstatements are needed.

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Example: Concatenate lists a and b

list procedure cat(list a, list b){

list t = list u = copylist(a);while (t.tail != nil) t = t.tail;t.tail = b;return u;

}

In an imperative language

In a declarative language

In a functional languagecat(a,b) if b = nil then a

else cons(head(a),cat(tail(a),b))

cat([], Z, Z).cat([H|T], L, [H|Z]) :- cat(T, L, Z).

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Complete Syntax of Terms

Term

Constant VariableCompound Term

Atom Number

alpha17

gross_payjohn_smithdyspepsia+=/=12Q&A

0

1571.6182.04e-27-13.6

likes(john, mary)book(dickens, Z, cricket)

f(x)[1, 3, g(a), 7, 9]-(+(15, 17), t)15 + 17 - t

XGross_pay

Diagnosis_257_

Names an individual Stands for an individualunable to be named when

program is written

Names an individualthat has parts

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Compound Terms

parents(spot, fido, rover)

The parents of Spot are Fido and Rover.

Functor (an atom) of arity 3. components (any terms)

It is possible to depict the term as a tree:

parents

roverfidospot

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Compound Terms

=/=(15+X, (0*a)+(2

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More about operators

Any atom may be designated an operator. The

only purpose is for convenience; the only effectis how the term containing the atom is parsed.Operators are syntactic sugar.

We wont be designating operators in this

course, but it is as well to understand them,because a number of atoms have built-indesignations as operators.

Operators have three properties: position,

precedence and associativity.

more

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Examples of operator properties

Position Operator Syntax Normal SyntaxPrefix: -2 -(2)

Infix: 5+17 +(17,5)

Postfix: N! !(N)

Associativity: left, right, none.

X+Y+Z is parsed as (X+Y)+Z

because addition is left-associative.

Precedence: an integer.

X+Y*Z is parsed as X+(Y*Z)

because multiplication has higher precedence.

These are all the

same as thenormal rules of

arithmetic.

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The last point about CompoundTerms

Constants are simply compound terms of arity 0.

badger

means the same asbadger()

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Structure of Programs

Programs consist of procedures.

Procedures consist of clauses.

Each clause is a fact or a rule.

Programs are executed by posing queries.

An example

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Example

elephant(george).

elephant(mary).elephant(X) :- grey(X), mammal(X), hasTrunk(X).

Procedure forelephant

Predicate

Clauses

Rule

Facts

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Example

?- elephant(george).

yes

?- elephant(jane).

no

Queries

Replies

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Clauses: Facts and Rules

Head Body This is a rule.

Head This is a fact.

ifprovided that

turnstile

Full stop at the end.

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Body of a (rule) clause contains goals.

likes(mary, X) :- human(X), honest(X).

Head Body

Goals

Exercise: Identify all the parts ofProlog text you have seen so far.

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Interpretation of Clauses

Clauses can be given a declarative reading or aprocedural reading.

H :- G1, G2, , Gn.

That H is provable follows fromgoals G1, G2, , Gnbeing provable.

To execute procedure H, theprocedures called by goals G1, G2,, Gnare executed first.

Declarative reading:

Procedural reading:

Form of clause:

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male(bertram).male(percival).

female(lucinda).

female(camilla).

pair(X, Y) :- male(X), female(Y).

?- pair(percival, X).?- pair(apollo, daphne).

?- pair(camilla, X).

?- pair(X, lucinda).

?- pair(X, X).

?- pair(bertram, lucinda).

?- pair(X, daphne).

?- pair(X, Y).

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Worksheet 2

drinks(john, martini).

drinks(mary, gin).

drinks(susan, vodka).

drinks(john, gin).

drinks(fred, gin).

pair(X, Y, Z) :-

drinks(X, Z),

drinks(Y, Z).

?- pair(X, john, martini).?- pair(mary, susan, gin).

?- pair(john, mary, gin).

?- pair(john, john, gin).

?- pair(X, Y, gin).

?- pair(bertram, lucinda).

?- pair(bertram, lucinda, vodka).

?- pair(X, Y, Z).

This definition forces X and Y to be distinct:

pair(X, Y, Z) :- drinks(X, Z), drinks(Y, Z), X \== Y.

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Worksheet 3

berkshire

wiltshire

surrey

hampshire sussex

kent

How to represent this relation?Note that borders are symmetric.

(a) Representing a symmetric relation.

(b) Implementing a strange ticket condition.

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WS3

border(sussex, kent).

border(sussex, surrey).

border(surrey, kent).border(hampshire, sussex).

border(hampshire, surrey).

border(hampshire, berkshire).

border(berkshire, surrey).

border(wiltshire, hampshire).

border(wiltshire, berkshire).

This relation represents

one direction of border: What about the other?

(a) Say border(kent, sussex).


(b) Say

adjacent(X, Y) :- border(X, Y).

adjacent(X, Y) :- border(Y, X).

(c) Say

border(X, Y) :- border(Y, X).

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WS3

valid(X, Y) :- adjacent(X, Z), adjacent(Z, Y)

Now a somewhat strange type of discount ticket. For theticket to be valid, one must pass through an intermediate

county.

A valid ticket between a start and end county obeys the

following rule:

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WS3


border(sussex, surrey).

border(surrey, kent).

border(hampshire, sussex).

border(hampshire, surrey).

border(hampshire, berkshire).border(berkshire, surrey).

border(wiltshire, hampshire).

border(wiltshire, berkshire).

adjacent(X, Y) :- border(X, Y).

adjacent(X, Y) :- border(Y, X).

?- valid(wiltshire, sussex).?- valid(wiltshire, kent).

?- valid(hampshire, hampshire).?- valid(X, kent).?- valid(sussex, X).?- valid(X, Y).

valid(X, Y) :-adjacent(X, Z),

adjacent(Z, Y)

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Worksheet 4

a(g, h).

a(g, d).

a(e, d).

a(h, f).a(e, f).

a(a, e).

a(a, b).

a(b, f).

a(b, c).

a(f, c).

arc

a

ed

gh

f

cb

path(X, X).path(X, Y) :- a(X, Z), path(Z, Y).

Note that Prolog can

distinguish between

the 0-ary constanta

(the name of a node)and the 2-ary

functora (the nameof a relation).

?- path(f, f).?- path(a, c).?- path(g, e).?- path(g, X).?- path(X, h).

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But what happens if

a(g, h).a(g, d).

a(e, d).

a(h, f).

a(e, f).

a(a, e).

a(a, b).

a(b, f).

a(b, c).

a(f, c).

a(d, a).

a

ed

gh

f

cb

path(X, X).path(X, Y) :- a(X, Z), path(Z, Y).

This program worksonly for acyclic graphs.

The program mayinfinitely loop given acyclic graph. We needto leave a trail of

visited nodes. This isaccomplished with adata structure (to beseen later).

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Unification

Two terms unify if substitutions can be made forany variables in the terms so that the terms aremade identical. If no such substitution exists,the terms do not unify.

The Unification Algorithm proceeds by recursivedescent of the two terms.

Constants unify if they are identical

Variables unify with any term, including othervariables

Compound terms unify if their functors andcomponents unify.

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Examples

The terms f(X, a(b,c)) and f(d, a(Z, c)) unify.

Z c

ad

f

b c

aX

f

The terms are made equal if d is substituted for X,and b is substituted for Z. We also say X isinstantiated to d and Z is instantiated to b, or X/d,Z/b.

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Examples

The terms f(X, a(b,c)) and f(Z, a(Z, c)) unify.

Z c

aZ

f

b c

aX

f

Note that Z co-refers within the term.Here, X/b, Z/b.

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Examples

The terms f(c, a(b,c)) and f(Z, a(Z, c)) do notunify.

Z c

aZ

f

b c

ac

f

No matter how hard you try, these two termscannot be made identical by substituting termsfor variables.

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Exercise

Do terms g(Z, f(A, 17, B), A+B, 17) andg(C, f(D, D, E), C, E) unify?

A B

+f

g

Z 17

A B17

Cf

g

C E

D ED

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Exercise

First write in the co-referring variables.

A B

+f

g

Z 17

A B17

Cf

g

C E

D ED

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Exercise

A B

+f

g

Z 17

A B17

Cf

g

C E

D ED

Z/C, C/Z

Now proceed by recursive descent

We go top-down, left-to-right, butthe order does not matter as long as

it is systematic and complete.

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Exercise

A B

+f

g

Z 17

A B17

Cf

g

C E

D ED

Z/C, C/Z, A/D, D/A

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Exercise

A B

+f

g

Z 17

A B17

Cf

g

C E

D ED

Z/C, C/Z, A/17, D/17

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Exercise

A B

+f

g

Z 17

A B17

Cf

g

C E

D ED

Z/C, C/Z, A/17, D/17, B/E, E/B

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Exercise

A B

+f

g

Z 17

A B17

Cf

g

C E

D ED

Z/17+B, C/17+B, A/17, D/17, B/E, E/B

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Exercise

A B

+f

g

Z 17

A B17

Cf

g

C E

D ED

Z/17+17, C/17+17, A/17, D/17, B/17, E/17

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Exercise Alternative Method

Z/C

A B

+f

g

Z 17

A B17

Cf

g

C E

D ED

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Z/C

A B

+f

g

C 17

A B17

Cf

g

C E

D ED

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A/D, Z/C

A B

+f

g

C 17

A B17

Cf

g

C E

D ED

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D/17, A/D, Z/C

D B

+f

g

C 17

D B17

Cf

g

C E

D ED

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D/17, A/17, Z/C

17 B

+f

g

C 17

17 B17

Cf

g

C E

17 E17

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B/E, D/17, A/17, Z/C

17 B

+f

g

C 17

17 B17

Cf

g

C E

17 E17

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B/E, D/17, A/17, Z/C

17 E

+f

g

C 17

17 E17

Cf

g

C E

17 E17

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C/17+E, B/E, D/17, A/17, Z/C

17 E

+f

g

C 17

17 E17

Cf

g

C E

17 E17

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C/17+E, B/E, D/17, A/17, Z/17+E

17 E

+f

g

+17

17 E17

+f

g

+ E

17 E17

17 E

17 EE17

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E/17, C/17+E, B/E, D/17, A/17, Z/C

17 E

+f

g

+17

17 E17

+f

g

+ E

17 E17

17 E

17 EE17

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E/17, C/17+17, B/17, D/17, A/17, Z/C

17 17

+f

g

+17

17 1717

+f

g

+ 17

17 1717

17 17

17 171717

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