Prolog - How to find the letters?

Author

How to find the letters?

Konrad Den Ende

2004-10-01, 8:59 am

I have just started with Prolog and tried following code
letter(C) :- (C >= 0'A , C =< 0'Z).

What i expected the software to do when i call
?- letter(X).
is to find 65, then 66, then 67, then... and then 91 and then
simply no. To my surprise it tells me that there's an error of
instantiating the variable at the first position of ">=".
What did i miss?

--

Kindly
Konrad
---------------------------------------------------
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---------------------------------------------------

Mikael Brockman

2004-10-01, 3:57 pm

"Konrad Den Ende" <tmp1@viltersten.com> writes:

> I have just started with Prolog and tried following code
> letter(C) :- (C >= 0'A , C =< 0'Z).
>
> What i expected the software to do when i call
> ?- letter(X).
> is to find 65, then 66, then 67, then... and then 91 and then
> simply no. To my surprise it tells me that there's an error of
> instantiating the variable at the first position of ">=".
> What did i miss?

You can't have free variables in math expressions. If you've
got R = 5 and you tell Prolog to give you R + 10, it'll just
change R to 5 and send 5 + 10 to the math guy. The math guy
is into maths *only*. All the searching & backtracking stuff,
he doesn't want any of it. Sorry.

Matthias Kretschmer

2004-10-01, 3:57 pm

Konrad Den Ende schrieb:
> I have just started with Prolog and tried following code
> letter(C) :- (C >= 0'A , C =< 0'Z).
>
> What i expected the software to do when i call
> ?- letter(X).
> is to find 65, then 66, then 67, then... and then 91 and then
> simply no. To my surprise it tells me that there's an error of
> instantiating the variable at the first position of ">=".
> What did i miss?
>

you have to define your predicate in another way (>=, =<, etc. are
comparing functions that want the two arguments instantiated, so no free
variables are allowed here).

Try to make a predicate between(A, X, B) which is successful for values
X between A and B, so you can define letter/1:
letter(X) :- between(0'A, X, 0'Z).

With an appropriate design of between/3, X can be a free variable.
Something like that counts all values between A and B and tests if X is
one of them (e.g. make the intervall smaller and check if X is a bound
of the intervall).

--
Matthias

Stefan Ljungstrand

2004-10-01, 3:57 pm

On Fri, 1 Oct 2004, Konrad Den Ende wrote:

> I have just started with Prolog and tried following code
> letter(C) :- (C >= 0'A , C =< 0'Z).
>
> What i expected the software to do when i call
> ?- letter(X).
> is to find 65, then 66, then 67, then... and then 91 and then
> simply no. To my surprise it tells me that there's an error of
> instantiating the variable at the first position of ">=".
> What did i miss?

How should it know that C is only expected to be an integer ?

Also, the comparisions require their arguments to be ground expresssions.
(If not, one can get a run-time instantation error, as you noticed.)
(Also, is/2 have similar problems, in this sense.)

However, one can do things like this with Constraint Logic Programming
(CLP), e.g. CLP(FD) :

SICStus 3.11.1 (x86-linux-glibc2.3): Fri Feb 20 18:38:25 CET 2004
| ?- use_module(library(clpfd)).
.......
yes
| ?- C #>= 0'A,C #=< 0'Z.
C in 65..90 ? ;
no
| ?- C #>= 0'A,C #=< 0'Z,indomain(C).
C = 65 ? ;
C = 66 ? ;
C = 67 ? ;
C = 68 ? ;
C = 69 ? ;
C = 70 ?
yes
| ?- C #>= 0'A,C #=< 0'Z,indomain(C),put(C),fail.
ABCDEFGHIJKLMNOPQRSTUVWXYZ
no
| ?-

> --
>
> Kindly
> Konrad

--
Stefan Lj
md9slj

The infinity that can be finitely expressed is not the true infinity

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