minimum in list of lists in prolog - list

Question

hello i have a list like this:
[[3,[a,b,c,d]],[2,[a,b,d]],[5,[d,e,f]]]
list of lists...
i want to find the minimum number on inner list
in this case i want to return D=2 and L=[a,b,d]
i tried this code:
minway([[N|L]],N,L).
minway([[M|L1]|L2],D,_):- M<D, minway(L2,M,L1).
minway([[M|_]|L2],D,L):- M>=D, minway(L2,D,L).
but i got error:
</2: Arguments are not sufficiently instantiated
Exception: (8) minway([[3,[a,b,c,d]],[2,[a,b,d]],[5,[d,e,f]]], _G7777, _G7778) ?
creep
for this run sentence:
minway([[3,[a,b,c,d]],[2,[a,b,d]],[5,[d,e,f]]],D,L).
the result need to be:
D=2.
L=[a,b,d].
where my problem?
and how to fix it?
tnx a lot

Answer 1

First, switch to a better data representation: Instead of [Key,Value], use Key-Value!
Then, define minway_/3 based on
iwhen/2,
ground/1,
keysort/2, and
member/2, like so:
minway_(Lss, N, Ls) :-
iwhen(ground(Lss), (keysort(Lss,Ess), Ess = [N-_|_], member(N-Ls, Ess))).
Sample query using SICStus Prolog 4.5.0:
| ?- minway_([3-[a,b,c,d],2-[a,b,d],5-[d,e,f],2-[x,t,y]], N, Ls).
N = 2, Ls = [a,b,d] ? ;
N = 2, Ls = [x,t,y] ? ;
no

Answer 2

There are a couple of fundamental issues.
One is in your problem lies in your representation of a list. Your predicates seem to assume that, for example, [3, [a,b,c]] is represented as [3 | [a,b,c]] but it is not. The list [3 | [a,b,c]] is the list with 3 as the head, and [a,b,c] as the rest of the list or the tail. In other words, [3 | [a,b,c]] is [3, a, b, c].
And, so, your base case would be:
minway([[N,L]], N, L).
The second issue is in your other predicate clauses. There's no starting point for D. In other words, it's never given a value to start with, so you get an instantiation error. You cannot compare N > D if one of the variables doesn't have a value.
When doing a minimum or maximum from scratch, a common approach is to start by assuming the first element is the candidate result, and then replace it if you find a better one on each step of the recursion. It also means you need to carry with you the last candidate at each recursive call, so that adds extra arguments:
minway([[N,L]|T], D, R) :-
minway(T, N, L, D, R).
minway([], D, R, D, R). % At the end, so D, R is the answer
minway([[N,L]|T], Dm, Rm, D, R) :-
( N < Dm
-> minway(T, N, L, D, R) % N, L are new candidates if N < Dm
; minway(T, N, Dm, Rm, D, R) % Dm, Rm are still best candidate
).
In Prolog, you can simplify this a little since Prolog has a more general term comparison operator, #<, #>, etc, which is smart about comparing more complex terms. For example, [2, [d,e,f]] #< [3, [a,b,c]] is true since 2 < 3 is true. We can then write:
minway([H|T], D, R) :-
minway(T, H, D, R).
minway([], [D, R], D, R).
minway([H|T], M, D, R) :-
( H #< M
-> minway(T, H, D, R)
; minway(T, M, D, R)
).

Answer 3

You can do this by using the minimum predicate. Findall can be very helpful.
min([X],X).
min([H|T],Min):-
min(T,TMin),
H>TMin,
Min is TMin.
min([H|T],Min):-
min(T,TMin),
H=<TMin,
Min is H.
minway(List,D,L):-
findall(Value,member([Value,_],List),VList),
min(VList,Min),
D=Min,
findall(EList,member([Min,EList],List),L).
?-minway([[3,[a,b,c,d]],[2,[a,b,d]],[5,[d,e,f]]],D,L).
D = 2,
L = [[a, b, d]]

Answer 4

Try library(aggregate):
?- aggregate_all(min(X,Y),
member([X,Y], [[3,[a,b,c,d]],
[2,[a,b,d]],
[5,[d,e,f]]]),
min(D,L)).
D = 2,
L = [a, b, d].
See also here:
Aggregation operators on backtrackable predicates
https://www.swi-prolog.org/pldoc/man?section=aggregate