CODEWITHSUNDEEP
prologclpfd
colymore
colymore

Reputation: 12316

Prolog 'how to return' a value

I known prolog doesn't return values but i need to update the value of some variable and get it in console with this procedure:

max(A,B,C) :- (A>B -> C is A ; C is B).                                                   
maxAltura([],RES).
maxAltura([c(X,Y)|[]],RES) :- max(RES,Y, SUM).
maxAltura([c(X,Y)|R1],RES) :- RES>Y, maxAltura(R1,RES).
maxAltura([c(X,Y)|R1],RES) :- RES<Y, maxAltura(R1,Y).
maxAltura([c(X,Y)|R1],RES) :- RES=:=Y, maxAltura(R1,Y).

It just takes a list of tuples and gives the max value of second element of that tupes.

This is my output

    maxAltura([c(1,8),c(5,0),c(6,4),c(10,0),c(11,10),c(12,0)],0).
   Call: (7) maxAltura([c(1, 8), c(5, 0), c(6, 4), c(10, 0), c(11, 10), c(12, 0)], 0) ? creep
   Call: (8) 0>8 ? creep
   Fail: (8) 0>8 ? creep
   Redo: (7) maxAltura([c(1, 8), c(5, 0), c(6, 4), c(10, 0), c(11, 10), c(12, 0)], 0) ? creep
   Call: (8) 0<8 ? creep
   Exit: (8) 0<8 ? creep
   Call: (8) maxAltura([c(5, 0), c(6, 4), c(10, 0), c(11, 10), c(12, 0)], 8) ? creep
   Call: (9) 8>0 ? creep
   Exit: (9) 8>0 ? creep
   Call: (9) maxAltura([c(6, 4), c(10, 0), c(11, 10), c(12, 0)], 8) ? creep
   Call: (10) 8>4 ? creep
   Exit: (10) 8>4 ? creep
   Call: (10) maxAltura([c(10, 0), c(11, 10), c(12, 0)], 8) ? creep
   Call: (11) 8>0 ? creep
   Exit: (11) 8>0 ? creep
   Call: (11) maxAltura([c(11, 10), c(12, 0)], 8) ? creep
   Call: (12) 8>10 ? creep
   Fail: (12) 8>10 ? creep
   Redo: (11) maxAltura([c(11, 10), c(12, 0)], 8) ? creep
   Call: (12) 8<10 ? creep
   Exit: (12) 8<10 ? creep
   Call: (12) maxAltura([c(12, 0)], 10) ? creep
   Call: (13) max(10, 0, _G4361) ? creep
   Call: (14) 10>0 ? creep
   Exit: (14) 10>0 ? creep
   Call: (14) _G4359 is 10 ? creep
   Exit: (14) 10 is 10 ? creep
   Exit: (13) max(10, 0, 10) ? creep
   Exit: (12) maxAltura([c(12, 0)], 10) ? creep
   Exit: (11) maxAltura([c(11, 10), c(12, 0)], 8) ? creep
   Exit: (10) maxAltura([c(10, 0), c(11, 10), c(12, 0)], 8) ? creep
   Exit: (9) maxAltura([c(6, 4), c(10, 0), c(11, 10), c(12, 0)], 8) ? creep
   Exit: (8) maxAltura([c(5, 0), c(6, 4), c(10, 0), c(11, 10), c(12, 0)], 8) ? creep
   Exit: (7) maxAltura([c(1, 8), c(5, 0), c(6, 4), c(10, 0), c(11, 10), c(12, 0)], 0) ? creep
true .

As you can see it keeps 10 as max value, witch it's correct by i need something like MAX=10. Why just gives me true?

Upvotes: 0

Views: 267

Answers (2)

tas
tas

Reputation: 8140

You can alter the given solution to work in both directions by using library(clpfd). As long as the list is sufficiently instantiated is/2 and >/2 work fine but as soon as you try to use them with unbound variables you're in trouble. Consider the following example: 

   ?- maxAltura(L,M).
L = [c(_A,M)] ? ;
     ERROR at  clause 1 of user:max/3 !!
     INSTANTIATION ERROR- =:=/2: expected bound value

Fortunately, this can be remedied using clpfd. In my original answer I suggested to simply replace is/2 and >/2 in max/3 by #=/2 and #> like so:

:- use_module(library(clpfd)).

max(A,B,C) :- A#>B -> C #= A ; C #= B.

% use @CapelliC's maxAltura/2 here

However, as pointed out by @mat and @false in the comments this yields too specific/incomplete answers. So I recommend you rather define max/3 as suggested in the comments, e.g.:

:- use_module(library(clpfd)).

max(A,B,C) :- max(A,B) #= C.

% use @CapelliC's maxAltura/2 here

The example query you used still works as expected:

   ?- maxAltura([c(1,8),c(5,0),c(6,4),c(10,0),c(11,10),c(12,0)],M).
M = 10 ? ;
no

The above query with both arguments being variables now works as well:

   ?- maxAltura(L,M).
L = [c(_A,M)] ? ;
L = [c(_A,_B),c(_C,_D)],
M#>=_B,
M#=max(_D,_B),
M#>=_D ? ;
L = [c(_A,_B),c(_C,_D),c(_E,_F)],
M#>=_B,
M#=max(_G,_B),
M#>=_G,
_G#>=_D,
_G#>=_F,
_G#=max(_F,_D) ? 
...

Upvotes: 2

CapelliC
CapelliC

Reputation: 60014

this simplified version binds the maximum as last argument.

max(A,B,C) :- A>B -> C is A ; C is B.

maxAltura([c(_,Y)],Y).
maxAltura([c(_,Y)|R1],RES) :- maxAltura(R1,T), max(T,Y,RES).

Note that max/3 is useless: you can write

maxAltura([c(_,Y)|R1],RES) :- maxAltura(R1,T), RES is max(T,Y).

Also, with library(aggregate), you can simplify further:

maxAltura(L,MaxY) :- aggregate(max(Y), X^member(c(X,Y),L), MaxY).

Upvotes: 1

Related Questions