Prolog - Einstein Riddle

Question

I'm trying to solve the Einstein Riddle using Prolog. I'm sorry for my English. The task is:

In the block of flats is 5 flats. (ground floor a 1., 2., 3., 4. floor)
Family Mullerovi have 4 children.
An administrator lives on the middle floor.
Mrs. Meierova pays for her flat 60 crowns less than a doctor on 3. floor.
Family living on the hightest floor have 5 children.
Family Kernovi have 1 child than teacher.
Mr Kaufmann works as a sales representative.
The teacher pays for her flat 1740 crowns.
On the hightest floor janitor lives.
The doctor pays for his flat 1800 crowns.
In the flat, which costs 1760 crowns, arent any children.
On the 2.floor the family have twice more children than family on the 3.floor.
Mr Hanz lives directly above family Kernovi.
In the flat under the janitor live 2 children.
Mullerovi pay for their flat 1770 crowns.
The roof flat is cheaper 10 crowns than then ground floor.
Sales representative pays 30 crowns more than janitor does.

My code looks like, I don't know how to solve problems with hire:

person('Name', 'Job','Children','Hire','Floor').

houses(Hs):-
   length(Hs),
   member(person(mullerovi,_,4,1770,_),Hs),
   member(person(_,administrator,_,_,2),Hs),
   member(person(_,doctor,_,1800,_),Hs),
   pays_less(person(_,doctor,_,Hire,_),person('Meinova',_,_,Hire-60,_),Hs),
   member(person(_,_,5,_,4),Hs),
   have_one_children_less(person('Kernovi',_,Children-1,_,_),person(_,teacher,Children,_,_),Hs),
   member(person('Kaufmann','Sales representative',_,_,_),Hs),
   member(person(_,teacher,_,1740,_),Hs),
   member(person(_,janitor,_,_,4),Hs),
   member(person(_,_,0,1760,_),Hs),
   have_2x_children(
         person(_,_,2*Children,_,2),
         person(_,_,Children,_,3),Hs),
   lives_above(person('Hanz',_,_,_,_),person('Kernovi',_,_,_,_),Hs),
   lives_above(person(_,janitor,_,_,_),person(_,_,2,_,_),Hs),
   pays_more(
         person(_,_,_,Hire+10,4),
         person(_,_,_,Hire,'Ground floor'),Hs),
   pays_more(
         person(_,'Sale representative',_,Hire+30,_),
         person(_,janitor,_,Hire,_),Hs).

Thanks for any advices.

You're making the assumption that prolog will arithmetically evaluate `Hire+10`. It doesn't. It's trying to unify against a term in the form of `X+Y`. Prolog does structural evaluation, not arithmetic. — Enigmativity, Apr 24 '16 at 07:44
`length(Hs),` makes no sense. You really need to put more effort into it first. — false, Apr 24 '16 at 10:30
Possible duplicate of [Einstein Riddle using Prolog](http://stackoverflow.com/questions/36816529/einstein-riddle-using-prolog) — lurker, Apr 24 '16 at 10:59
@lurker - It's close to being a duplicate, but there is a little more complexity here that isn't covered in the other question. — Enigmativity, Apr 24 '16 at 11:39
@WillNess so you are starting your own CLP-implementation :-) — false, Jul 10 '16 at 14:01
Possible duplicate of [Einstein Riddle with List of terms](http://stackoverflow.com/questions/36743498/einstein-riddle-with-list-of-terms) — , Dec 20 '16 at 23:19

Will Ness · Answer 1 · 2016-07-11T17:00:53.723

"In the block of flats is 5 flats" translates as length(Hs,5).

Then, pays_less( person(_,doctor,_,Hire,_), person('Meinova',_,_,Hire-60,_),Hs) is undefined; it ought to be

.....
member( person(_,doctor,_,Hire,_), Hs), 
member( person('Meinova',_,_,Hire2,_), Hs), 
.....
Hire2 is Hire-60,  
.....

or something like that.

Acquaint yourself with is/2 operator and its preconditions. B=80, A = B-60 succeeds, unifying A with -(80,60). You want B=80, A is B-60 which succeeds and unifies A with 20 (but would fail if B were not yet a numerical expression).

The line Hire2 is Hire-60, should be placed as high as possible in the predicate, for maximal efficiency.

Rather `when( ( ground(Hire), ground(Hire2) ), Hire2 =:= Hire-60 )` — false, Jul 10 '16 at 10:25

Prolog - Einstein Riddle

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