'if' in prolog?

Is there a way to do an if in prolog, e.g. if a variable is 0, then to do some actions (write text to the terminal). An else isn't even needed, but I can't find any documentation of if.

10 Answers

Yes, there is such a control construct in ISO Prolog, called ->. You use it like this:

( condition -> then_clause ; else_clause ) 

Here is an example that uses a chain of else-if-clauses:

( X < 0 -> writeln('X is negative. That's weird! Failing now.'), fail ; X =:= 0 -> writeln('X is zero.') ; writeln('X is positive.') ) 

Note that if you omit the else-clause, the condition failing will mean that the whole if-statement will fail. Therefore, I recommend always including the else-clause (even if it is just true).

A standard prolog predicate will do this.

 isfive(5). 

will evaluate to true if you call it with 5 and fail(return false) if you run it with anything else. For not equal you use \=

isNotEqual(A,B):- A\=B. 

Technically it is does not unify, but it is similar to not equal.

Learn Prolog Now is a good website for learning prolog.

Edit: To add another example.

isEqual(A,A). 

Prolog predicates 'unify' -

So, in an imperative langauge I'd write

function bazoo(integer foo) { if(foo == 5) doSomething(); else doSomeOtherThing(); } 

In Prolog I'd write

bazoo(5) :- doSomething. bazoo(Foo) :- Foo =/= 5, doSomeOtherThing. 

which, when you understand both styles, is actually a lot clearer.
"I'm bazoo for the special case when foo is 5"
"I'm bazoo for the normal case when foo isn't 5"

First, let's recall some classical first order logic:

"If P then Q else R" is equivalent to "(P and Q) or (non_P and R)".

How can we express "if-then-else" like that in Prolog?

Let's take the following concrete example:

If X is a member of list [1,2] then X equals 2 else X equals 4.

We can match above pattern ("If P then Q else R") if ...

• condition P is list_member([1,2],X),
• negated condition non_P is non_member([1,2],X),
• consequence Q is X=2, and
• alternative R is X=4.

To express list (non-)membership in a pure way, we define:

 list_memberd([E|Es],X) :- ( E = X ; dif(E,X), list_memberd(Es,X) ). non_member(Es,X) :- maplist(dif(X),Es). 

Let's check out different ways of expressing "if-then-else" in Prolog!

1. (P,Q ; non_P,R)

    ?- (list_memberd([1,2],X), X=2 ; non_member([1,2],X), X=4). X = 2 ; X = 4. ?- X=2, (list_memberd([1,2],X), X=2 ; non_member([1,2],X), X=4), X=2. X = 2 ; false. ?- (list_memberd([1,2],X), X=2 ; non_member([1,2],X), X=4), X=2. X = 2 ; false. ?- X=4, (list_memberd([1,2],X), X=2 ; non_member([1,2],X), X=4), X=4. X = 4. ?- (list_memberd([1,2],X), X=2 ; non_member([1,2],X), X=4), X=4. X = 4. 

   Correctness score 5/5. Efficiency score 3/5.

2. (P -> Q ; R)

    ?- (list_memberd([1,2],X) -> X=2 ; X=4). false. % WRONG ?- X=2, (list_memberd([1,2],X) -> X=2 ; X=4), X=2. X = 2. ?- (list_memberd([1,2],X) -> X=2 ; X=4), X=2. false. % WRONG ?- X=4, (list_memberd([1,2],X) -> X=2 ; X=4), X=4. X = 4. ?- (list_memberd([1,2],X) -> X=2 ; X=4), X=4. false. % WRONG 

   Correctness score 2/5. Efficiency score 2/5.

3. (P *-> Q ; R)

    ?- (list_memberd([1,2],X) *-> X=2 ; X=4). X = 2 ; false. % WRONG ?- X=2, (list_memberd([1,2],X) *-> X=2 ; X=4), X=2. X = 2 ; false. ?- (list_memberd([1,2],X) *-> X=2 ; X=4), X=2. X = 2 ; false. ?- X=4, (list_memberd([1,2],X) *-> X=2 ; X=4), X=4. X = 4. ?- (list_memberd([1,2],X) *-> X=2 ; X=4), X=4. false. % WRONG 

   Correctness score 3/5. Efficiency score 1/5.

(Preliminary) summary:

1. (P,Q ; non_P,R) is correct, but needs a discrete implementation of non_P.

2. (P -> Q ; R) loses declarative semantics when instantiation is insufficient.

3. (P *-> Q ; R) is "less" incomplete than (P -> Q ; R), but still has similar woes.

Luckily for us, there are alternatives: Enter the logically monotone control construct if_/3!

We can use if_/3 together with the reified list-membership predicate memberd_t/3 like so:

 ?- if_(memberd_t(X,[1,2]), X=2, X=4). X = 2 ; X = 4. ?- X=2, if_(memberd_t(X,[1,2]), X=2, X=4), X=2. X = 2. ?- if_(memberd_t(X,[1,2]), X=2, X=4), X=2. X = 2 ; false. ?- X=4, if_(memberd_t(X,[1,2]), X=2, X=4), X=4. X = 4. ?- if_(memberd_t(X,[1,2]), X=2, X=4), X=4. X = 4. 

Correctness score 5/5. Efficiency score 4/5.

I found this helpful for using an if statement in a rule.

max(X,Y,Z) :- ( X =< Y -> Z = Y ; Z = X ). 

Thanks to

There are essentially three different ways how to express something like if-then-else in Prolog. To compare them consider char_class/2. For a and b the class should be ab and other for all other terms. One could write this clumsily like so:

char_class(a, ab). char_class(b, ab). char_class(X, other) :- dif(X, a), dif(X, b). ?- char_class(Ch, Class). Ch = a, Class = ab ; Ch = b, Class = ab ; Class = other, dif(Ch, a), dif(Ch, b). 

To write things more compactly, an if-then-else construct is needed. Prolog has a built-in one:

?- ( ( Ch = a ; Ch = b ) -> Class = ab ; Class = other ). Ch = a, Class = ab. 

While this answer is sound, it is incomplete. Just the first answer from ( Ch = a ; Ch = b ) is given. The other answers are chopped away. Not very relational, indeed.

A better construct, often called a "soft cut" (don't believe the name, a cut is a cut is a cut), gives slightly better results (this is in YAP):

?- ( ( Ch = a ; Ch = b ) *-> Class = ab ; Class = other ). Ch = a, Class = ab ; Ch = b, Class = ab. 

Alternatively, SICStus has if/3 with very similar semantics:

?- if( ( Ch = a ; Ch = b ), Class = ab , Class = other ). Ch = a, Class = ab ; Ch = b, Class = ab. 

So the last answer is still suppressed. Now enter library(reif) for SICStus, YAP, and SWI. Install it and say:

?- use_module(library(reif)). ?- if_( ( Ch = a ; Ch = b ), Class = ab , Class = other ). Ch = a, Class = ab ; Ch = b, Class = ab ; Class = other, dif(Ch, a), dif(Ch, b). 

Note that all the if_/3 is compiled away to a wildly nested if-then-else for

char_class(Ch, Class) :- if_( ( Ch = a ; Ch = b ), Class = ab , Class = other ). 

which expands in YAP 6.3.4 to:

char_class(A,B) :- ( A\=a -> ( A\=b -> B=other ; ( A==b -> B=ab ) ; A=b, B=ab ; dif(A,b), B=other ) ; ( A==a -> B=ab ) ; A=a, B=ab ; dif(A,a), ( A\=b -> B=other ; ( A==b -> B=ab ) ; A=b, B=ab ; dif(A,b), B=other ) ). 

The best thing to do is to use the so-called cuts, which has the symbol !.

if_then_else(Condition, Action1, Action2) :- Condition, !, Action1. if_then_else(Condition, Action1, Action2) :- Action2. 

The above is the basic structure of a condition function.

To exemplify, here's the max function:

max(X,Y,X):-X>Y,!. max(X,Y,Y):-Y=<X. 

I suggest reading more documentation on cuts, but in general they are like breakpoints. Ex.: In case the first max function returns a true value, the second function is not verified.

PS: I'm fairly new to Prolog, but this is what I've found out.

Prolog program actually is big condition for "if" with "then" which prints "Goal is reached" and "else" which prints "No sloutions was found". A, Bmeans "A is true and B is true", most of prolog systems will not try to satisfy "B" if "A" is not reachable (i.e. X=3, write('X is 3'),nl will print 'X is 3' when X=3, and will do nothing if X=2).

( A == B -> writeln("ok") ; writeln("nok") ), 

The else part is required

You should read Learn Prolog Now! Chapter 10.2 Using Cut. This provides an example:

max(X,Y,Z) :- X =< Y,!, Y = Z.

to be said,

Z is equal to Y IF ! is true (which it always is) AND X is <= Y.