0

I have this code :

link(a,b,4). 
link(a,c,2). 
link(b,g,5). 
link(c,g,6). 
link(c,d,5). 
link(d,g,3). 

path(S,D,TDist):- 
    link(S,D,TDist). 
path(S,D,TDist):- 
    link(S,X,TD1), path(X,D,TD2), TDist=TD1+TD2.

This will follow a depth first search strategy, but the result is that it will give me all the paths, and it won't show which is the shortest. Is it possible to still use that strategy and find the shortest path? if not, what search strategy to use? and how can I implement it.

false
  • 10,264
  • 13
  • 101
  • 209
John Sall
  • 1,027
  • 1
  • 12
  • 25

1 Answers1

1

I think there are problems with your code:

  • TDist=TD1+TD2 doesn't compute the sum, use is/2 instead, at least when a path is returned.

  • It will loop if the graph contains cycles, but assuming the data actually is a DAG, we can ignore by now.

  • We can't say what the actual path will be, just its value.

Anyway, library(aggregate) can be used to find the shortest path. For instance

?- aggregate(min(D), path(a,g,D), D).
D = 8.

Or, since gnu-prolog doesn't have library(aggregate), take the first element computed by setof/3:

?- setof(D, path(a,g,D), [Min|Rest]).
Min = 8,
Rest = [9, 10].
CapelliC
  • 59,646
  • 5
  • 47
  • 90
  • Do you think there's any other way to do this? by combining the goal code into the code itself? So only what I will put in the goal is path(a, g, D). – John Sall Nov 26 '18 at 18:48
  • sorry, could you elaborate ? I don't understand the question... of course, the changes required to get a more functional path/3 are described as problems. a) use is/2 b) check if a node already has been visited. Here you have some choices, the most basic is to add a Visited accumulator to path/3, that become path/4 and use \+member(Node,Visited) as a filter before the recursive call. c) Visited can also be returned 'tagging' it by the distance, to know the actual path found. There are plenty of answer on SO, but Iet me know if you are still lost. – CapelliC Nov 26 '18 at 19:14