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Counting in PrologRules that involve counting often usethe is predicate to evaluate anumeric value.Consider the relation len(L,N) thatis true if the length of list L is N.len([],0).

len([_|T],N) :- len(T,M), N is M+1.

| ?- len([1,2,3],X).

X = 3

| ?- len(Y,2).

Y = [_10903,_10905]

The symbols _10903 and _10905 are“internal variables” created as neededwhen a particular value is not forcedin a solution.

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Debugging PrologCare is required in developing andtesting Prolog programs because thelanguage is untyped; undeclaredpredicates or relations are simplytreated as false.Thus in a definition like adj([A,B|_]) :- A=B.

adj([_,B|T]) :- adk([B|T]).

| ?- adj([1,2,2]).no

(Some Prolog systems warn when anundefined relation is referenced, butmany others don’t).

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Similarly, givenmember(A,[A|_]).

member(A,[_|T]) :- member(A,[T]).

| ?- member(2,[1,2]).

Infinite recursion! (Why?)

If you’re not sure what is going on,Prolog’s trace feature is very handy.The commandtrace.

turns on tracing. (notrace turnstracing off).Hence| ?- trace.yes

[trace]

| ?- member(2,[1,2]).

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(1) 0 Call: member(2,[1,2]) ?

(1) 1 Head [1->2]:member(2,[1,2]) ?

(1) 1 Head [2]:member(2,[1,2]) ?

(2) 1 Call: member(2,[[2]]) ?

(2) 2 Head [1->2]:member(2,[[2]]) ?

(2) 2 Head [2]:member(2,[[2]]) ?

(3) 2 Call: member(2,[[]]) ?

(3) 3 Head [1->2]:member(2,[[]]) ?

(3) 3 Head [2]: member(2,[[]])?

(4) 3 Call: member(2,[[]]) ?

(4) 4 Head [1->2]:member(2,[[]]) ?

(4) 4 Head [2]: member(2,[[]])?

(5) 4 Call: member(2,[[]]) ?

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Termination Issues in PrologSearching infinite domains (likeintegers) can lead to non-termination, with Prolog trying everyvalue.Considerodd(1).

odd(N) :- odd(M), N is M+2.

| ?- odd(X).

X = 1 ;

X = 3 ;

X = 5 ;X = 7

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A query | ?- odd(X), X=2.

going into an infinite search,generating each and every oddinteger and finding none is equal to2!The obvious alternative,odd(2) (which is equivalent toX=2, odd(X) ) also does an infinite,but fruitless search.We’ll soon learn that Prolog doeshave a mechanism to “cut off”fruitless searches.

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Definition Order can MatterIdeally, the order of definition offacts and rules should not matter.But,in practice definition order canmatter. A good general guideline is todefine facts before rules. To see why,consider a very complete database ofmotherOf relations that goes back asfar asmotherOf(cain,eve).

Now we defineisMortal(X) :- isMortal(Y), motherOf(X,Y).

isMortal(eve).

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These definitions state that the firstwoman was mortal, and allindividuals descended from her arealso mortal.But when we try as trivial a query as| ?- isMortal(eve).

we go into an infinite search!Why?Let’s trace what Prolog does when itsees| ?- isMortal(eve).

It matches with the first definitioninvolving isMortal , which isisMortal(X) :- isMortal(Y), motherOf(X,Y).

It sets X=eve and tries to solveisMortal(Y), motherOf(eve,Y).

It will then expand isMortal(Y) into

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isMortal(Z), motherOf(Y,Z).

An infinite expansion ensues.The solution is simple—place the“base case” fact that terminatesrecursion first.If we useisMortal(eve).

isMortal(X) :- isMortal(Y), motherOf(X,Y).yes

| ?- isMortal(eve).

yes

But now another problem appears!If we ask| ?- isMortal(clarkKent).

we go into another infinite search!Why?

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The problem is that Clark Kent is fromthe planet Krypton, and hence won’tappear in our motherOf database.Let’s trace the query.It doesn’t match isMortal(eve) .We next tryisMortal(clarkKent) :- isMortal(Y), motherOf(clarkKent,Y).

We try Y=eve , but eve isn’t Clark’smother. So we recurse, getting:isMortal(Z), motherOf(Y,Z),motherOf(clarkKent,Y).

But eve isn’t Clark’s grandmothereither! So we keep going further back,trying to find a chain of descendentsthat leads from eve to clarkKent .No such chain exists, and there is no

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limit to how long a chain Prolog willtry.There is a solution though!We simply rewrite our recursivedefinition to be isMortal(X) :- motherOf(X,Y),isMortal(Y).

This is logically the same, but now wework from the individual X backtoward eve , rather than from evetoward X. Since we have nomotherOf relation involvingclarkKent , we immediately stop oursearch and answer no!

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Extra-logical Aspects ofProlog

To make a Prolog program moreefficient, or to represent negativeinformation, Prolog needs featuresthat have a procedural flavor. Theseconstructs are called “extra-logical”because they go beyond Prolog’s coreof logic-based inference.

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The CutThe most commonly used extra-logical feature of Prolog is the “cutsymbol,” “!”A ! in a goal, fact or rule “cuts off”backtracking.In particular, once a ! is reached (andautomatically matched), we may notbacktrack across it. The rule we’veselected and the bindings we’vealready selected are “locked in” or“frozen.”For example, givenx(A) :- y(A,B), z(B) , ! , v(B,C).

once the ! is hit we can’t backtrackto resatisfy y(A,B) or z(B) in someother way. We are locked into this

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rule, with the bindings of A and Balready in place.We can backtrack to try varioussolutions to v(B,C) .It is sometimes useful to have several! ’s in a rule. This allows us to find apartial solution, lock it in, find afurther solution, then lock it in, etc.For example, in a rulea(X) - b(X), !, c(X,Y), ! , d(Y).

we first try to satisfy b(X) , perhapstrying several facts or rules thatdefine the b relation. Once we have asolution to b(X) , we lock it in, alongwith the binding for X.Then we try to satisfy c(X,Y) , usingthe fixed binding for X, but perhapstrying several bindings for Y untilc(X,Y) is satisfied.

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We then lock in this match usinganother ! .Finally we check if d(Y) can besatisfied with the binding of Y alreadyselected and locked in.

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When are Cuts Needed?A cut can be useful in improvingefficiency, by forcing Prolog to avoiduseless or redundant searches.Consider a query likemember(X,list1),

member(X,list2), isPrime(X).

This asks Prolog to find an X that is inlist1 and also in list2 and also isprime.X will be bound, in sequence, to eachvalue in list1 . We then check if X isalso in list2 , and then check if X isprime.Assume we find X=8 is in list1 andlist2 . isPrime(8) fails (of course).We backtrack to member(X,list2)and try to resatisfy it with the samevalue of X.

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But clearly there is never any point intrying to resatisfy member(X,list2) .Once we know a value of X is inlist2 , we test it using isPrime(X) .If it fails, we want to go right back tomember(X,list1) and get adifferent X.To create a version of member thatnever backtracks once it has beensatisfied we can use ! .We definemember1(X,[X|_]) :- !.member1(X,[_|Y]) :- member1(X,Y).

Our query is nowmember(X,list1), member1(X,list2), isPrime(X).

(Why isn’t member1 used in bothterms?)

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Expressing NegativeInformation

Sometimes it is useful to state rulesabout what can’t be true. This allowsus to avoid long and fruitlesssearches.fail is a goal that always fails. It canbe used to represent goals or resultsthat can never be true.Assume we want to optimize ourgrandMotherOf rules by stating thata male can never be anyone’sgrandmother (and hence a completesearch of all motherOf andfatherOf relations is useless).A rule to do this isgrandMotherOf(X,GM) :- male(GM), fail.

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This rule doesn’t do quite what wehope it will!Why?The standard approach in Prolog is totry other rules if the current rule fails.Hence we need some way to “cut off”any further backtracking once thisnegative rule is found to beapplicable.This can be done using grandMotherOf(X,GM) :- male(GM),!, fail.

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Other Extra-LogicalOperators

• assert and retract

These operators allow a Prologprogram to add new rules duringexecution and (perhaps) later removethem. This allows programs to learnas they execute.• findall

Called as findall(X,goal,List)where X is a variable in goal . Allpossible solutions for X that satisfygoal are found and placed in List .For example,findall(X,(append(_,[X|_],[-1,2,-3,4]),(X<0)), L).

L = [-1,-3]

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• var and nonvar

var(X) tests whether X is unbound(free).nonvar(Y) tests whether Y is bound(no longer free).These two operators are useful intailoring rules to particularcombinations of bound and unboundvariables.For example, the rulegrandMotherOf(X,GM) :- male(GM),!, fail.

might backfire if GMis not yet bound.We could set GM to a person forwhom male(GM) is true, then failbecause we don’t want grandmotherswho are male!

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To remedy this problem. we use therule only when GM is bound. Our rulebecomesgrandMotherOf(X,GM) :- nonvar(GM), male(GM),!, fail.