Gathering results of methods in one List with Prolog

up vote
1
down vote

favorite

So I am trying to

I am defining the sets with is_a(b, a), is_a(c, a), which for simplicity would look visually something like this:

      a

  b       c

d   e    f  g

I want to give in the list [b, c] and as a result get the list [d, e, f, g]

At the moment when I give in a node or a variable, then it can find everything that is underneath it with this method:

find_nodes(Root, Root) :-

    + is_a(_, Root).

find_nodes(Root, X) :-

    is_a(Node, Root),

    find_nodes(Node, X).

Which when run gives me the result I need :

?- find_nodes(b, X).

X = d.

X = e.

But it is not in a list, so I have tried :

?- all_nodes([b, c], X).



all_nodes(, _).

all_nodes([H|T], [R|Res]):-

    findall(L, find_nodes(H, L), R),

    all_nodes(T, Res).

Which gives me - X = [[d, e], [f, g]|_4040], which consists of lists within lists, but I need just 1 list, that would be X = [d, e, f, g].

What am I doing wrong here?

EDIT

Like @lurker said findall returns a list and adding list to a list will give the result I get right now.

The one thing I also tried was using:

all_nodes(, _).

all_nodes([H|T], [R|Res]):-

    find_nodes(H, R),

    all_nodes(T, Res).

But well that one does not work either because It only gives me 1 element, which in this case is d and then f.

edited Nov 8 at 1:22

asked Nov 8 at 1:04

user9441470

findall(...) has a list as its third argument. So when you call findall(L, find_nodes(H, L), R). R is a list. So [R|Res] is a list whose first element is itself a list, R. Your other issue is your base case. all_nodes(, _) says that all nodes of an empty list is anything you want since _ is an anonymous variable. It can be anything.
– lurker
Nov 8 at 1:06

@lurker all_nodes(, _) didnt really know what else to put there, probably could also work. But how can I do it without using findall because when i do it without and call the function recursively while replacing findall to just find_nodes then it takes only the d and goes to the next head.
– user9441470
Nov 8 at 1:14

1

Well, _ does not work because it really does mean that your rule says, finding the nodes for the empty list is anything is a successful result, which obviously isn't logical. would be more logical, would it not? Isn't the result of finding all nodes in an empty tree an empty list specifically?
– lurker
Nov 8 at 1:59

1

You can use findall. You just can't directly use the result the way you are trying to use it. You would need to concatenate the result to your current list of known nodes.
– lurker
Nov 8 at 2:00

1

SWI Prolog also has append(ListOfLists, List) you can check out.
– lurker
Nov 8 at 2:07

add a comment |

up vote
1
down vote

favorite

So I am trying to

I am defining the sets with is_a(b, a), is_a(c, a), which for simplicity would look visually something like this:

      a

  b       c

d   e    f  g

I want to give in the list [b, c] and as a result get the list [d, e, f, g]

At the moment when I give in a node or a variable, then it can find everything that is underneath it with this method:

find_nodes(Root, Root) :-

    + is_a(_, Root).

find_nodes(Root, X) :-

    is_a(Node, Root),

    find_nodes(Node, X).

Which when run gives me the result I need :

?- find_nodes(b, X).

X = d.

X = e.

But it is not in a list, so I have tried :

?- all_nodes([b, c], X).



all_nodes(, _).

all_nodes([H|T], [R|Res]):-

    findall(L, find_nodes(H, L), R),

    all_nodes(T, Res).

Which gives me - X = [[d, e], [f, g]|_4040], which consists of lists within lists, but I need just 1 list, that would be X = [d, e, f, g].

What am I doing wrong here?

EDIT

Like @lurker said findall returns a list and adding list to a list will give the result I get right now.

The one thing I also tried was using:

all_nodes(, _).

all_nodes([H|T], [R|Res]):-

    find_nodes(H, R),

    all_nodes(T, Res).

But well that one does not work either because It only gives me 1 element, which in this case is d and then f.

edited Nov 8 at 1:22

asked Nov 8 at 1:04

user9441470

findall(...) has a list as its third argument. So when you call findall(L, find_nodes(H, L), R). R is a list. So [R|Res] is a list whose first element is itself a list, R. Your other issue is your base case. all_nodes(, _) says that all nodes of an empty list is anything you want since _ is an anonymous variable. It can be anything.
– lurker
Nov 8 at 1:06

@lurker all_nodes(, _) didnt really know what else to put there, probably could also work. But how can I do it without using findall because when i do it without and call the function recursively while replacing findall to just find_nodes then it takes only the d and goes to the next head.
– user9441470
Nov 8 at 1:14

1

Well, _ does not work because it really does mean that your rule says, finding the nodes for the empty list is anything is a successful result, which obviously isn't logical. would be more logical, would it not? Isn't the result of finding all nodes in an empty tree an empty list specifically?
– lurker
Nov 8 at 1:59

1

You can use findall. You just can't directly use the result the way you are trying to use it. You would need to concatenate the result to your current list of known nodes.
– lurker
Nov 8 at 2:00

1

SWI Prolog also has append(ListOfLists, List) you can check out.
– lurker
Nov 8 at 2:07

add a comment |

up vote
1
down vote

favorite

So I am trying to

I am defining the sets with is_a(b, a), is_a(c, a), which for simplicity would look visually something like this:

      a

  b       c

d   e    f  g

I want to give in the list [b, c] and as a result get the list [d, e, f, g]

At the moment when I give in a node or a variable, then it can find everything that is underneath it with this method:

find_nodes(Root, Root) :-

    + is_a(_, Root).

find_nodes(Root, X) :-

    is_a(Node, Root),

    find_nodes(Node, X).

Which when run gives me the result I need :

?- find_nodes(b, X).

X = d.

X = e.

But it is not in a list, so I have tried :

?- all_nodes([b, c], X).



all_nodes(, _).

all_nodes([H|T], [R|Res]):-

    findall(L, find_nodes(H, L), R),

    all_nodes(T, Res).

Which gives me - X = [[d, e], [f, g]|_4040], which consists of lists within lists, but I need just 1 list, that would be X = [d, e, f, g].

What am I doing wrong here?

EDIT

Like @lurker said findall returns a list and adding list to a list will give the result I get right now.

The one thing I also tried was using:

all_nodes(, _).

all_nodes([H|T], [R|Res]):-

    find_nodes(H, R),

    all_nodes(T, Res).

But well that one does not work either because It only gives me 1 element, which in this case is d and then f.

edited Nov 8 at 1:22

asked Nov 8 at 1:04

user9441470

So I am trying to

I am defining the sets with is_a(b, a), is_a(c, a), which for simplicity would look visually something like this:

      a

  b       c

d   e    f  g

I want to give in the list [b, c] and as a result get the list [d, e, f, g]

At the moment when I give in a node or a variable, then it can find everything that is underneath it with this method:

find_nodes(Root, Root) :-

    + is_a(_, Root).

find_nodes(Root, X) :-

    is_a(Node, Root),

    find_nodes(Node, X).

Which when run gives me the result I need :

?- find_nodes(b, X).

X = d.

X = e.

But it is not in a list, so I have tried :

?- all_nodes([b, c], X).



all_nodes(, _).

all_nodes([H|T], [R|Res]):-

    findall(L, find_nodes(H, L), R),

    all_nodes(T, Res).

Which gives me - X = [[d, e], [f, g]|_4040], which consists of lists within lists, but I need just 1 list, that would be X = [d, e, f, g].

What am I doing wrong here?

EDIT

Like @lurker said findall returns a list and adding list to a list will give the result I get right now.

The one thing I also tried was using:

all_nodes(, _).

all_nodes([H|T], [R|Res]):-

    find_nodes(H, R),

    all_nodes(T, Res).

But well that one does not work either because It only gives me 1 element, which in this case is d and then f.

list prolog

edited Nov 8 at 1:22

asked Nov 8 at 1:04

user9441470

edited Nov 8 at 1:22

asked Nov 8 at 1:04

user9441470

edited Nov 8 at 1:22

asked Nov 8 at 1:04

user9441470

asked Nov 8 at 1:04

user9441470

asked Nov 8 at 1:04

user9441470

findall(...) has a list as its third argument. So when you call findall(L, find_nodes(H, L), R). R is a list. So [R|Res] is a list whose first element is itself a list, R. Your other issue is your base case. all_nodes(, _) says that all nodes of an empty list is anything you want since _ is an anonymous variable. It can be anything.
– lurker
Nov 8 at 1:06

@lurker all_nodes(, _) didnt really know what else to put there, probably could also work. But how can I do it without using findall because when i do it without and call the function recursively while replacing findall to just find_nodes then it takes only the d and goes to the next head.
– user9441470
Nov 8 at 1:14

1

Well, _ does not work because it really does mean that your rule says, finding the nodes for the empty list is anything is a successful result, which obviously isn't logical. would be more logical, would it not? Isn't the result of finding all nodes in an empty tree an empty list specifically?
– lurker
Nov 8 at 1:59

1

You can use findall. You just can't directly use the result the way you are trying to use it. You would need to concatenate the result to your current list of known nodes.
– lurker
Nov 8 at 2:00

1

SWI Prolog also has append(ListOfLists, List) you can check out.
– lurker
Nov 8 at 2:07

add a comment |

findall(...) has a list as its third argument. So when you call findall(L, find_nodes(H, L), R). R is a list. So [R|Res] is a list whose first element is itself a list, R. Your other issue is your base case. all_nodes(, _) says that all nodes of an empty list is anything you want since _ is an anonymous variable. It can be anything.
– lurker
Nov 8 at 1:06

@lurker all_nodes(, _) didnt really know what else to put there, probably could also work. But how can I do it without using findall because when i do it without and call the function recursively while replacing findall to just find_nodes then it takes only the d and goes to the next head.
– user9441470
Nov 8 at 1:14

1

Well, _ does not work because it really does mean that your rule says, finding the nodes for the empty list is anything is a successful result, which obviously isn't logical. would be more logical, would it not? Isn't the result of finding all nodes in an empty tree an empty list specifically?
– lurker
Nov 8 at 1:59

1

You can use findall. You just can't directly use the result the way you are trying to use it. You would need to concatenate the result to your current list of known nodes.
– lurker
Nov 8 at 2:00

1

SWI Prolog also has append(ListOfLists, List) you can check out.
– lurker
Nov 8 at 2:07

findall(...) has a list as its third argument. So when you call findall(L, find_nodes(H, L), R). R is a list. So [R|Res] is a list whose first element is itself a list, R. Your other issue is your base case. all_nodes(, _) says that all nodes of an empty list is anything you want since _ is an anonymous variable. It can be anything.
– lurker
Nov 8 at 1:06

@lurker all_nodes(, _) didnt really know what else to put there, probably could also work. But how can I do it without using findall because when i do it without and call the function recursively while replacing findall to just find_nodes then it takes only the d and goes to the next head.
– user9441470
Nov 8 at 1:14

Well, _ does not work because it really does mean that your rule says, finding the nodes for the empty list is anything is a successful result, which obviously isn't logical. would be more logical, would it not? Isn't the result of finding all nodes in an empty tree an empty list specifically?
– lurker
Nov 8 at 1:59

You can use findall. You just can't directly use the result the way you are trying to use it. You would need to concatenate the result to your current list of known nodes.
– lurker
Nov 8 at 2:00

SWI Prolog also has append(ListOfLists, List) you can check out.
– lurker
Nov 8 at 2:07

add a comment |

1 Answer
1

active

oldest

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up vote
1
down vote

accepted

You can take advantage of the de facto standard findall/4 (*) predicate to solve the problem. This predicate is a variant of the standard findall/3 predicate that allows passing a tail for the list of solutions collected by the predicate. For example:

?- findall(N, (N=1; N=2; N=3), L, [4,5]).

L = [1, 2, 3, 4, 5].

In the following solution, I have renamed predicates and variables for clarity and modified your node leaf predicate:

is_a(a, b).

is_a(a, c).



is_a(b, d).

is_a(b, e).



is_a(c, f).

is_a(c, g).



leaf(Leaf, Leaf) :-

    + is_a(Leaf, _).

leaf(Node, Leaf) :-

    is_a(Node, Child),

    leaf(Child, Leaf).



all_nodes(, ).

all_nodes([Node| Nodes], Leaves):-

    findall(Leaf, leaf(Node, Leaf), Leaves, Tail),

    all_nodes(Nodes, Tail).

Sample calls:

?- all_nodes([b, c], X).

X = [d, e, f, g].



?- all_nodes([a], X).

X = [d, e, f, g].



?- all_nodes([b], X).

X = [d, e].

(*) It's a built-in predicate in GNU Prolog, JIProlog, Lean Prolog, O-Prolog, SICStus Prolog, SWI-Prolog, XSB, and YAP (possibly others).

edited Nov 8 at 10:28

answered Nov 8 at 9:31

Paulo Moura

10.6k21325

With this method if I am calling the leaf node all_nodes([d, e], X) then it returns X = [d, e], is there a way to not return the nodes that were called, even if they are leaves?.
– user9441470
Nov 8 at 10:54

1

Try modifying the findall/4 goal argument to (leaf(Node, Leaf), Node == Leaf).
– Paulo Moura
Nov 8 at 10:57

It does not work at all with this
– user9441470
Nov 8 at 10:59

1

Yes, it does. Likely you made a typo modifying the findall/4 goal argument. You want findall(Leaf, (leaf(Node, Leaf), Node == Leaf), Leaves, Tail). With this change your get L = for the query all_nodes([d, e], L).
– Paulo Moura
Nov 8 at 11:03

add a comment |

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

up vote
1
down vote

accepted

?- findall(N, (N=1; N=2; N=3), L, [4,5]).

L = [1, 2, 3, 4, 5].

In the following solution, I have renamed predicates and variables for clarity and modified your node leaf predicate:

is_a(a, b).

is_a(a, c).



is_a(b, d).

is_a(b, e).



is_a(c, f).

is_a(c, g).



leaf(Leaf, Leaf) :-

    + is_a(Leaf, _).

leaf(Node, Leaf) :-

    is_a(Node, Child),

    leaf(Child, Leaf).



all_nodes(, ).

all_nodes([Node| Nodes], Leaves):-

    findall(Leaf, leaf(Node, Leaf), Leaves, Tail),

    all_nodes(Nodes, Tail).

Sample calls:

?- all_nodes([b, c], X).

X = [d, e, f, g].



?- all_nodes([a], X).

X = [d, e, f, g].



?- all_nodes([b], X).

X = [d, e].

(*) It's a built-in predicate in GNU Prolog, JIProlog, Lean Prolog, O-Prolog, SICStus Prolog, SWI-Prolog, XSB, and YAP (possibly others).

edited Nov 8 at 10:28

answered Nov 8 at 9:31

Paulo Moura

10.6k21325

With this method if I am calling the leaf node all_nodes([d, e], X) then it returns X = [d, e], is there a way to not return the nodes that were called, even if they are leaves?.
– user9441470
Nov 8 at 10:54

1

Try modifying the findall/4 goal argument to (leaf(Node, Leaf), Node == Leaf).
– Paulo Moura
Nov 8 at 10:57

It does not work at all with this
– user9441470
Nov 8 at 10:59

1

Yes, it does. Likely you made a typo modifying the findall/4 goal argument. You want findall(Leaf, (leaf(Node, Leaf), Node == Leaf), Leaves, Tail). With this change your get L = for the query all_nodes([d, e], L).
– Paulo Moura
Nov 8 at 11:03

add a comment |

up vote
1
down vote

accepted

?- findall(N, (N=1; N=2; N=3), L, [4,5]).

L = [1, 2, 3, 4, 5].

In the following solution, I have renamed predicates and variables for clarity and modified your node leaf predicate:

is_a(a, b).

is_a(a, c).



is_a(b, d).

is_a(b, e).



is_a(c, f).

is_a(c, g).



leaf(Leaf, Leaf) :-

    + is_a(Leaf, _).

leaf(Node, Leaf) :-

    is_a(Node, Child),

    leaf(Child, Leaf).



all_nodes(, ).

all_nodes([Node| Nodes], Leaves):-

    findall(Leaf, leaf(Node, Leaf), Leaves, Tail),

    all_nodes(Nodes, Tail).

Sample calls:

?- all_nodes([b, c], X).

X = [d, e, f, g].



?- all_nodes([a], X).

X = [d, e, f, g].



?- all_nodes([b], X).

X = [d, e].

(*) It's a built-in predicate in GNU Prolog, JIProlog, Lean Prolog, O-Prolog, SICStus Prolog, SWI-Prolog, XSB, and YAP (possibly others).

edited Nov 8 at 10:28

answered Nov 8 at 9:31

Paulo Moura

10.6k21325

With this method if I am calling the leaf node all_nodes([d, e], X) then it returns X = [d, e], is there a way to not return the nodes that were called, even if they are leaves?.
– user9441470
Nov 8 at 10:54

1

Try modifying the findall/4 goal argument to (leaf(Node, Leaf), Node == Leaf).
– Paulo Moura
Nov 8 at 10:57

It does not work at all with this
– user9441470
Nov 8 at 10:59

1

Yes, it does. Likely you made a typo modifying the findall/4 goal argument. You want findall(Leaf, (leaf(Node, Leaf), Node == Leaf), Leaves, Tail). With this change your get L = for the query all_nodes([d, e], L).
– Paulo Moura
Nov 8 at 11:03

add a comment |

up vote
1
down vote

accepted

?- findall(N, (N=1; N=2; N=3), L, [4,5]).

L = [1, 2, 3, 4, 5].

In the following solution, I have renamed predicates and variables for clarity and modified your node leaf predicate:

is_a(a, b).

is_a(a, c).



is_a(b, d).

is_a(b, e).



is_a(c, f).

is_a(c, g).



leaf(Leaf, Leaf) :-

    + is_a(Leaf, _).

leaf(Node, Leaf) :-

    is_a(Node, Child),

    leaf(Child, Leaf).



all_nodes(, ).

all_nodes([Node| Nodes], Leaves):-

    findall(Leaf, leaf(Node, Leaf), Leaves, Tail),

    all_nodes(Nodes, Tail).

Sample calls:

?- all_nodes([b, c], X).

X = [d, e, f, g].



?- all_nodes([a], X).

X = [d, e, f, g].



?- all_nodes([b], X).

X = [d, e].

(*) It's a built-in predicate in GNU Prolog, JIProlog, Lean Prolog, O-Prolog, SICStus Prolog, SWI-Prolog, XSB, and YAP (possibly others).

edited Nov 8 at 10:28

answered Nov 8 at 9:31

Paulo Moura

10.6k21325

?- findall(N, (N=1; N=2; N=3), L, [4,5]).

L = [1, 2, 3, 4, 5].

In the following solution, I have renamed predicates and variables for clarity and modified your node leaf predicate:

is_a(a, b).

is_a(a, c).



is_a(b, d).

is_a(b, e).



is_a(c, f).

is_a(c, g).



leaf(Leaf, Leaf) :-

    + is_a(Leaf, _).

leaf(Node, Leaf) :-

    is_a(Node, Child),

    leaf(Child, Leaf).



all_nodes(, ).

all_nodes([Node| Nodes], Leaves):-

    findall(Leaf, leaf(Node, Leaf), Leaves, Tail),

    all_nodes(Nodes, Tail).

Sample calls:

?- all_nodes([b, c], X).

X = [d, e, f, g].



?- all_nodes([a], X).

X = [d, e, f, g].



?- all_nodes([b], X).

X = [d, e].

(*) It's a built-in predicate in GNU Prolog, JIProlog, Lean Prolog, O-Prolog, SICStus Prolog, SWI-Prolog, XSB, and YAP (possibly others).

edited Nov 8 at 10:28

answered Nov 8 at 9:31

Paulo Moura

10.6k21325

edited Nov 8 at 10:28

answered Nov 8 at 9:31

Paulo Moura

10.6k21325

answered Nov 8 at 9:31

Paulo Moura

10.6k21325

answered Nov 8 at 9:31

Paulo Moura

10.6k21325

With this method if I am calling the leaf node all_nodes([d, e], X) then it returns X = [d, e], is there a way to not return the nodes that were called, even if they are leaves?.
– user9441470
Nov 8 at 10:54

1

Try modifying the findall/4 goal argument to (leaf(Node, Leaf), Node == Leaf).
– Paulo Moura
Nov 8 at 10:57

It does not work at all with this
– user9441470
Nov 8 at 10:59

1

Yes, it does. Likely you made a typo modifying the findall/4 goal argument. You want findall(Leaf, (leaf(Node, Leaf), Node == Leaf), Leaves, Tail). With this change your get L = for the query all_nodes([d, e], L).
– Paulo Moura
Nov 8 at 11:03

add a comment |

With this method if I am calling the leaf node all_nodes([d, e], X) then it returns X = [d, e], is there a way to not return the nodes that were called, even if they are leaves?.
– user9441470
Nov 8 at 10:54

1

Try modifying the findall/4 goal argument to (leaf(Node, Leaf), Node == Leaf).
– Paulo Moura
Nov 8 at 10:57

It does not work at all with this
– user9441470
Nov 8 at 10:59

1

Yes, it does. Likely you made a typo modifying the findall/4 goal argument. You want findall(Leaf, (leaf(Node, Leaf), Node == Leaf), Leaves, Tail). With this change your get L = for the query all_nodes([d, e], L).
– Paulo Moura
Nov 8 at 11:03

With this method if I am calling the leaf node all_nodes([d, e], X) then it returns X = [d, e], is there a way to not return the nodes that were called, even if they are leaves?.
– user9441470
Nov 8 at 10:54

Try modifying the findall/4 goal argument to (leaf(Node, Leaf), Node == Leaf).
– Paulo Moura
Nov 8 at 10:57

It does not work at all with this
– user9441470
Nov 8 at 10:59

Yes, it does. Likely you made a typo modifying the findall/4 goal argument. You want findall(Leaf, (leaf(Node, Leaf), Node == Leaf), Leaves, Tail). With this change your get L = for the query all_nodes([d, e], L).
– Paulo Moura
Nov 8 at 11:03

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