Find mutual element in different facts in swi-prolog

Question

i am trying to have an input of a list of movies and find the actors who play at the same movies. Question : Given a list of movies, display their link by a set of stars using recursion. these are an example of facts :

fact(movie,actor).

starsin(a,bob).
starsin(b,bob).
starsin(c,bob).

starsin(a,maria).
starsin(b,maria).
starsin(c,maria).

starsin(a,george).
starsin(b,george).
starsin(c,george).

Example of input and out put :

?- sameActors([a,b,c],Y).
 Y = bob,maria,george.

Rule written so far :

sameActors(Actors,Movies) :-
      findall(Stars,(member(movie,Actors),starsin(movie,Stars)),Name),
      sum_list(Name,Movies).

I am quite new and can't find any similar solution online for my problem, i don't understand what i am doing wrong , or what i need to add.

Answer 1

Here is another one (I finally found a way)

No recursion, just a relative of findall, setof/3:

Given a database of "actors starring in movies":

starsin(a,bob).
starsin(c,bob).

starsin(a,maria).
starsin(b,maria).
starsin(c,maria).

starsin(a,george).
starsin(b,george).
starsin(c,george).
starsin(d,george).

We do some reflection (described in setof/3 inside setof/3 not working, but why?), and then:

subselect(Ax,MovIn) :- 
   setof(Mx,starsin(Mx,Ax),MovAx), subset(MovIn, MovAx).
actors_appearing_in_movies(MovIn,ActOut) :- 
   setof(Ax, subselect(Ax,MovIn) , ActOut).

This has the right feel of being an RDBMS operation!

Testing!

Note that for the empty set of movies, we get all the actors. This is marginally right: all the actors star in all the movies of the empty set.

Testing consists in running these goals and observing that they succeed:

actors_appearing_in_movies([],ActOut),
permutation([bob, george, maria],ActOut),!. 

actors_appearing_in_movies([a],ActOut),
permutation([bob, george, maria],ActOut),!.

actors_appearing_in_movies([a,b],ActOut),
permutation([george, maria],ActOut),!.

actors_appearing_in_movies([a,b,c],ActOut),
permutation([george, maria],ActOut),!.

actors_appearing_in_movies([a,b,c,d],ActOut),
permutation([george],ActOut),!.

Bonus Round: In R

Completely unrelated, but I thought about how to do that in R.

After some fumbling:

# Load tidyverse dplyr

library(dplyr)

# Create a data frame ("tibble") with our raw data using `tribble`

t <- tribble(
        ~movie, ~actor
        ,"a"   , "bob"
        ,"c"   , "bob"
        ,"a"   , "maria"
        ,"b"   , "maria"
        ,"c"   , "maria"
        ,"a"   , "george"
        ,"b"   , "george"
        ,"c"   , "george"
        ,"d"   , "george")

# The function!

actors_appearing_in_movies <- function(data, movies_must) {
   # (movie,actor) pairs of actors active in movies we are interested in 
   t1 <- data %>% filter(is.element(movie, movies_must))

   # (actor, (movies)) pairs of actors and the movies they appear in
   # for movies we are interested in 
   t2 <- t1 %>% group_by(actor) %>% summarize(movies = list(unique(movie)))   

   # Retain only those which appear in all movies
   t3 <- t2 %>% rowwise() %>% filter(setequal(movies_must,movies))

   # Select only the actor column
   # ("Select" works columnwise, not rowwise as in SQL)   
   t4 <- t3 %>% select(actor)

   return(t4)
}

Results?

The above approach has a different opinion on who is in the empty movie set:

> actors_appearing_in_movies(t, c())
# A tibble: 0 x 1
# … with 1 variable: actor <chr>

But:

> actors_appearing_in_movies(t, c("a"))

# A tibble: 3 x 1
  actor 
  <chr> 
1 bob   
2 george
3 maria 

> actors_appearing_in_movies(t, c("a","b"))

# A tibble: 2 x 1
  actor 
  <chr> 
1 george
2 maria 

> actors_appearing_in_movies(t, c("a","b","c"))

# A tibble: 2 x 1
  actor 
  <chr> 
1 george
2 maria 

> actors_appearing_in_movies(t, c("a","b","c","d"))

# A tibble: 1 x 1
  actor 
  <chr> 
1 george