prolog syntax + unification

cs774 (Prasad) L2PrologUnify 1

Prolog syntax + Unification

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Prolog Syntax– Terms (Data objects)

• Atoms– (symbolic) constants

» E.g., anna, x_25, ‘Tom_’, ::=, <--->, etc

– Numbers

» E.g., -25, 100.25e+5, etc

– Variables

» E.g., X, Result, _23, _, AnonymousVariable, etc

• Structures» E.g., f(X,23,g(‘Tom’,2)), etc

– Formulae (Predicate-logic clauses)

• Facts

• Rules


Using variableseq(X,X).

?-eq(a,b).

No

?-eq(b,b).

Yes

?-eq(A,B).

A = B.

p(X,Y).

?-p(a,b).

Yes

q(_,_).

?-q(a,b).

• Lexical scope of a variable is one clause (fact/rule).

• Multiple named variable occurrences can be used to capture “equality” constraints.– Bindings to distinct variables

are independent (but can be same too).

• Multiple anonymous variable occurrences are considered distinct.


Anonymous Variable

• p(X,Y). is equivalent to p(_,_).– Informally, it is equivalent to:

For all x, for all y: p(x, y) holds.

• ?- p(a,_). – Informally, it is equivalent to:

There exists x, such that p(a, x) holds.


(cont’d)

• eq(X,X). • ?- eq(1,Ans).

Ans = 1– answer extraction

• q(_,_). • ?- q(1,Ans).

Ans unbound– interpreted as: for all x, q(1,x) holds


Structures (~ records)


add(5, mul(a,b))

functor[name/arity]

arguments

Structures encode labelled trees (or DAGs if they contain variables).

Matching

• Basic operation (cf. assignment)

• t1 matches t2 if– t1 is (syntactically) identical to t2, or– if there exists bindings to variables such that

the two terms become identical after substituition.


Inductive Definition of Matching Terms

• If S and T are constants, then S and T match only if they are the same constants.

• If S is a variable and T is anything, then they match, and S is instantiated to T.– (Similarly, if T is a variable.)

• … (cont’d)…


(cont’d)

• If S and T are structures, then S and T match only if:– S and T have the same principal functor, and– all their corresponding components match,

recursively.

• The resulting instantiation is determined by matching of the components.


ExamplesTerm Term Matching Substituition

a b No match

a X { X <- a}

f(X) g(Y) No match

f(X,g(a)) f(b,g(Y)) {X <- b, Y <- a}

f(X,g(a)) f(a,g(X)) {X <- a}

f(X,g(a)) f(b,g(Y)) {X <- b, Y <- a}

f(X) f(Y) {X <- Y} or {Y <- X}Identical upto variable renaming

f(g(X,a), g(b,X))

f(Z,Z)No match

Conflicting substituition


Extended Example

f(X, g(a))

f(Y, g(Y))

{X<-Y} {X<-a, Y<-a}

Common instance: f(a,g(a))


(cont’d)

f(X, h(b), g(X))

f(a, h(Y), g(Y))

{X<-a} {Y<-b} Not unifiable

Common instance: ??


Occur’s Check Problem

f(X, g(X))

f(Y, Y) • These terms cannot be unified because there

is no finite term substituition {X<- t} such that X = g(X).

• In practice, this can cause non-terminating computation, requiring special handling.


(Cont’d)

• ?-member(a,[f(a),a]).• Yes• ?-member(X,[f(X),X]).• Yes

Infinite loop:

X = f(f(f(…)))


CS784 (Prasad) L167AG 15

Executable Specification in Prolog

type(i,int).

type(x,real).

type(+(E,F),T) :-

type(E,T), type(F,T).

type(+(E,F),real) :- type(E,T1),type(F,T2), T1 \= T2.

• Type Checking ?- type(+(i,x),real).• Type Inference ?- type(+(x,x),T).

CS784 (Prasad) L167AG 16

Alternative : with conditional

type(i,int).

type(x,real).

type(+(E,F),T) :- type(E,TT),

( (TT == real) ->

T = real ;

type(F,T)

).

• Type Checking ?- type(+(i,x),real).• Type Inference ?- type(+(x,x),T).

prolog syntax + unification

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