Graph representations in Prolog - list

Question

Consider the following graph
and that it is described by the below Prolog term :
graph([connected(a,[b,c]), connected(b,[a,c]), connected(c,[a,b,d]), connected(d,[c]) ]).
I would like to define a predicate which transforms the above connections into a list of the corresponding pairs. In other words, a predicate which yields
[[a,b],[a,c],[b,c],[c,d]] for the above term-graph.
Could you please advise how to do it ?
My attempt so far is the following :
map 2-neighbor vertex to pairs :
map2ne(adjacent(H,[K|T]),Pair) :-
append([H],[K],L),
append([H],T,M),
append([L],[M],Pair).
This runs ok.
map 3-neighbor vertex to pairs :
map3n(adjacent(H,[K,L|T]),Pair) :-
append([H],[K],A1),
append([H],[L],A2),
append([A1],[A2],Z),
append([H],T,M),
append(Z,[M],Pair).
This also runs ok.
But when I try to extend it to n-neighbor vertex, then it fails :
mapmany(adjacent(H, [K|_]),Pair) :-
append([H],[K],L),
append(L,[],Pair),
mapmany(adjacent(H,[K|_]),M),
append(M,Pair,Pair).
And also the below fails, which was intented to map many n-neighbor vertices to pairs :
mapping(Map,Pairs) :-
select(X,Map,Y),
mapmany(X,PairX),
append([PairX],Pairs),
mapping(Y,Pairs).

Answer 1

If you're going to use a solution based on setof/3, I strongly recommend defining an auxiliary predicate. This predicate should define exactly what we want a set of. When we want to define "the set of all edges in the graph", mathematically we might say something like "Edges is the set of all Edge terms where Edge is an edge in Graph".
We can write this very directly as follows:
graph_edges(Graph, Edges) :-
setof( Edge,
graph_edge(Graph, Edge),
Edges ).
It remains to define graph_edge/2 itself. The core of this can be lifted from slago's solution:
graph_edge(Graph, Edge) :-
member(connected(V, Ns), Graph),
member(W, Ns),
edge(V, W, Edge).
The advantages of having this as a separate predicate are:
the setof call is easier to read
the predicate itself has a nice descriptive name
the predicate can be tested in isolation
the predicate can be reused
no ^ signs anywhere, which have no meaning in Prolog except for complicating setof calls that don't use an auxiliary predicate
no worrying about "existential quantification", which has no meaning in Prolog except for complicating setof calls that don't use an auxiliary predicate

Answer 2

There are too many flaws in your code:
The adjacency list defined by graph/1 is composed of terms of the form connected(Vertex, Neighbors); however, your code deals with an adjacency list of terms of the form adjacent(Vertex, Neighbors).
Predicate append/3 should not be used to create all lists; for example, instead of append([H], [K], L), you should use L = [H, K].
In Prolog, it is more idiomatic to represent a pair of items V and W as V-W, instead of [V,W].
By the answer you expect for the example given (i.e., [a-b,a-c,b-c,c-d]), a single term V-W (i.e., {V,W}) represents both the edges (V,W) and (W,V). So, to avoid redundancy, you must exclusively choose V-W or W-V to put in your answer (without loss of generality, you can choose the term where V precedes W).
To to create an edge, you can do the following:
edge(V, W, Edge) :-
( V #< W
-> Edge = V-W
; Edge = W-V ).
Examples:
?- edge(a, b, Edge).
Edge = a-b.
?- edge(b, a, Edge).
Edge = a-b.
To create all edges connecting a vertex V to its neighbors Ns, without duplicates, just ask:
?- V=a, Ns=[b,c,d], setof(E, W^Ns^(member(W,Ns), edge(V,W,E)), Edges).
V = a,
Ns = [b, c, d],
Edges = [a-b, a-c, a-d].
Notice that the construct Var^Goal tells setof/3 not to bind variable Var in Goal (in other words, indicates that Var is existentially quantified).
Generalizing this idea, we have:
graph_edges(Graph, Edges) :-
setof( Edge,
V^Ns^W^( member(connected(V, Ns), Graph),
member(W, Ns),
edge(V, W, Edge)),
Edges ).
graph([connected(a, [b, c]),
connected(b, [a, c]),
connected(c, [a, b, d]),
connected(d, [c])]).
Example:
?- graph(G), graph_edges(G, E).
G = [connected(a, [b, c]), connected(b, [a, c]), connected(c, [a, b, d]), connected(d, [c])],
E = [a-b, a-c, b-c, c-d].
LIBRARY UGRAPHS
In SWI-Prolog, a trivial solution would be to use the predicate edges/2 from library(ugraphs). Be careful though, because the representation of undirected graphs on which the predicate edge/2 is based is different from the one you are considering (an undirected graph in the library(ugraphs) is represented by a list of vertex pairs where the order of the vertices in these pairs matters). For example:
?- edges([a-[b,c], b-[a,c], c-[a,b,d], d-[c]], E).
E = [a-b, a-c, b-a, b-c, c-a, c-b, c-d, d-c].