succ not getting removed on pred

Question

/* Define a Prolog predicate replicate/3 which corresponds to
 * the Haskell function of the same name, except that the numeric
 * argument is expressed symbolically.
 *
 * For example, replicate(s(s(s(0))),a,[a,a,a]) should be satisfied.
 */

So far I've come to this solution:

replicate(0,_,[]).
replicate(X,Y,[Y|Z]) :- replicate(p(X),Y,Z).

but the problem is that the s(s(s(0))) is not getting reduced by the pred function. it results into p(p(p(s(s(s(0))))))

could you guys help me out?

Prolog doesn't evaluate like that. Try `replicate(s(X), Y, [Y|Z]) :- replicate(X, Y, Z).` instead. — Daniel Lyons, Aug 22 '18 at 16:21
No problem. :) Just pay attention to Prolog, it's evaluation scheme is *very* different than other languages like Haskell. :) — Daniel Lyons, Aug 22 '18 at 16:34
@false the OP referred to the Haskell library function, as a reference. — Will Ness, Aug 23 '18 at 16:52
@Will: Indeed, but why would anyone who searches tag [tag:haskell], want to find this question. — false, Aug 23 '18 at 17:28
@false that's how I understand tagging: to reflect what's in the question. — Will Ness, Aug 23 '18 at 17:30
@false we have five slots to fill. I pick the five most pertinent things about the question and use them as tags. Haskell here seems to be the least pertinent, but still pertinent. I don't have anything to replace it with. — Will Ness, Aug 23 '18 at 17:34
Let us [continue this discussion in chat](https://chat.stackoverflow.com/rooms/178687/discussion-between-will-ness-and-false). — Will Ness, Aug 24 '18 at 13:28

Will Ness · Accepted Answer · 2018-08-23T06:41:25.107

This is Haskell's replicate coded with the (deprecated) n+k patterns:

replicate 0 _ = []
replicate (n+1) a = a : x where x = replicate n a

This directly corresponds to the Prolog definition:

replicate(0, _, []).
replicate(s(N), A, [A | X]) :- replicate(N, A, X).

We just move the result into the arguments list, and make it the last argument to the predicate:

    x = replicate n a    ----->     replicate(N, A, X).

The pattern matching is the same. What's not the same, is that Prolog is not an expression-oriented language. There are no expressions which get evaluated before being used as arguments in the next function call; instead, there are terms which are auto-quoted, always, used as is as arguments to predicates.

... and what's further not the same, is that `replicate(N, s(0), [A,B])` determines `N` *and* `A` and `B`. — false, Aug 23 '18 at 12:51

succ not getting removed on pred

1 Answers1