Related
Previously I wrote the predicate, reach(Arrivals,Departure) to find a list of names of departure points from which I can get to a given point arrivals. Now, how can I make sure that the total cost from the calculated route does not exceed a given limit, for example now it`s work like this:
?-reach(krum, Departure).
Answer: Departure = [uzhorod].
in this example, the total cost of the route = 6000, how can it be implemented if 7000 is entered, it is false and if 6000 is true.
This is my facts:
trip(01, kuiv, odessa, 1500).
trip(02, kuiv, lviv, 700).
trip(08, lviv, zaporizhya, 700).
trip(03, uzhorod, krum, 6000).
trip(04, vunohradiv, odessa, 2540).
trip(05, ternopil, kuiv, 3800).
trip(06, zaporizhya, donetsk, 900).
trip(07, lytsk, mariupol, 7500).
trip(Id, Departure, Arrivals, Price)
This is my code:
reachable(D, D, _).
reachable(Departure, Arrival, Visited) :-
trip(_, Departure, Point, _),
\+ member(Point, Visited),
reachable(Point, Arrival, [Point|Visited]).
reachable(Departure, Arrival) :-
reachable(Departure, Arrival, [Departure]).
reach(Arrival, Departures) :-
setof(
Departure,
reachable(Departure, Arrival),
Departures
).
I'm trying to write a predicate that calculates which destination a group of friends will visit.
The friends list their countries of preferences like this
choice(marie, [peru,greece,vietnam]).
choice(jean, [greece,peru,vietnam]).
choice(sasha, [vietnam,peru,greece]).
choice(helena,[peru,vietnam,greece]).
choice(emma, [greece,peru,vietnam]).
I want to write a predicate called where that takes 2 arguments to perform the calculation.
The formula I have in mind is that the first country is worth 3 points, the second one is worth 2 points, and the last one is worth 1 point.
Here's an example of what I'm trying to achieve.
?- where([marie,jean,sasha,helena,emma],Country).
peru .
So far I have this
where([], X).
where([H|T], N) :- choice(H, [A|B]), where(T,N).
It lets me iterate through all the different friends and shows their choices but I can't iterate through the list of choices and assign points to the destinations.
How should I go about iterating through the list of choices for each friend and assigning points to calculate the best destination?
While this will solve your problem, I know it uses many predicates that you have not seen. So think of this an opportunity to excel and learn a lot.
Even if you don't understand it all, there is enough detail and intermediate results in the test that you should be able to navigate your way to a proper solution you create.
Also this is by no means efficient, it was just a quick proof of concept I did to see how this could be done.
choice(marie, [peru,greece,vietnam]).
choice(jean, [greece,peru,vietnam]).
choice(sasha, [vietnam,peru,greece]).
choice(helena,[peru,vietnam,greece]).
choice(emma, [greece,peru,vietnam]).
destinations(Destinations) :-
findall(D1,choice(_,D1),D2),
flatten(D2,D3),
list_to_set(D3,Destinations).
init_weights(Destinations,Weights) :-
empty_assoc(Assoc),
init_weights(Destinations,Assoc,Weights).
init_weights([],Weights,Weights).
init_weights([H|T],Assoc0,Weights) :-
put_assoc(H,Assoc0,0,Assoc1),
init_weights(T,Assoc1,Weights).
update_weights([C1,C2,C3],Weights0,Weights) :-
del_assoc(C1,Weights0,Value0,Weights1),
Value1 is Value0 + 3,
put_assoc(C1,Weights1,Value1,Weights2),
del_assoc(C2,Weights2,Value2,Weights3),
Value3 is Value2 + 2,
put_assoc(C2,Weights3,Value3,Weights4),
del_assoc(C3,Weights4,Value4,Weights5),
Value5 is Value4 + 1,
put_assoc(C3,Weights5,Value5,Weights).
person_weight(Person,Weights0,Weights) :-
choice(Person,[C1,C2,C3]),
update_weights([C1,C2,C3],Weights0,Weights).
people(People) :-
findall(Person,choice(Person,_),People).
choice(Destination) :-
destinations(Destinations),
init_weights(Destinations,Weights0),
people(People),
update_choices(People,Weights0,Weights1),
cross_ref_assoc(Weights1,Weights),
max_assoc(Weights, _, Destination),
true.
cross_ref_assoc(Assoc0,Assoc) :-
assoc_to_list(Assoc0,List0),
maplist(key_reverse,List0,List),
list_to_assoc(List,Assoc).
key_reverse(Key-Value,Value-Key).
update_choices([],Weights,Weights).
update_choices([Person|People],Weights0,Weights) :-
person_weight(Person,Weights0,Weights1),
update_choices(People,Weights1,Weights).
Tests
:- begin_tests(destination).
test(destinations) :-
destinations([peru, greece, vietnam]).
test(init_weights) :-
destinations(Destinations),
init_weights(Destinations,Weights),
assoc_to_list(Weights,[greece-0, peru-0, vietnam-0]).
test(update_weights) :-
destinations(Destinations),
init_weights(Destinations,Weights0),
update_weights([peru,greece,vietnam],Weights0,Weights),
assoc_to_list(Weights,[greece-2,peru-3,vietnam-1]).
test(person_weight) :-
destinations(Destinations),
init_weights(Destinations,Weights0),
person_weight(jean,Weights0,Weights),
assoc_to_list(Weights,[greece-3,peru-2,vietnam-1]).
test(people) :-
people([marie,jean,sasha,helena,emma]).
test(update_choices) :-
destinations(Destinations),
init_weights(Destinations,Weights0),
people(People),
update_choices(People,Weights0,Weights),
assoc_to_list(Weights,[greece-10,peru-12,vietnam-8]).
test(cross_ref_assoc) :-
List0 = [1-a,2-b,3-c],
list_to_assoc(List0,Assoc0),
cross_ref_assoc(Assoc0,Assoc),
assoc_to_list(Assoc,[a-1,b-2,c-3]).
test(choice) :-
choice(peru).
:- end_tests(destination).
As suggested by GuyCoder, you need an accumulator to sum each person preferences, and foldl/N allows to does exactly this.
choice(marie, [peru,greece,vietnam]).
choice(jean, [greece,peru,vietnam]).
choice(sasha, [vietnam,peru,greece]).
choice(helena,[peru,vietnam,greece]).
choice(emma, [greece,peru,vietnam]).
where(People,Where) :-
foldl([Person,State,Updated]>>(choice(Person,C),update(State,C,Updated)),
People,
[0=greece,0=peru,0=vietnam],
Pref),
aggregate(max(S,S=W),member(S=W,Pref),max(_,_=Where)).
% sort(Pref,Sorted),
% last(Sorted,_=Where).
update(S0,[A,B,C],S3) :-
update(S0,3,A,S1),
update(S1,2,B,S2),
update(S2,1,C,S3).
update(L,V,C,U) :-
append(X,[Y=C|Z],L),
P is Y+V,
append(X,[P=C|Z],U).
I have left commented the last two goals replaced by the single goal aggregate/3, so you can try to understand the syntax...
I'm having some issues with the findall/3 in Prolog.
Facts:
%
country(dublin,ireland).
country(cork,ireland).
country(london,uk).
country(rome,italy).
country(moscow,russia).
country(hongkong,china).
country(amsterdam,holland).
country(berlin,germany).
country(paris,france).
country(newyork,usa).
country(chicago,usa).
country(sao_paulo,brazil).
country(rio,brazil).
I need to write predicate to show the connections from city X to city Y (one by one). X and Y are the two inputs (city) and T is the output. Each solution T is a list of all the cities connecting X to Y (X and Y included).
Example:
| ?- trip(rome,dublin,T).
T=[rome,london,dublin] ;
//first solution T=[rome,paris,dublin];
//second solution
my try is
path(X,Y,[X|Y]):- edge(X,Y).
path(X,Y,[]):- edge(X,Z),not(member(Z,V)),path(Z,Y,[Z|V]).
Any ideas would be greatly appreciated.
Cheers,
I think your question is: "how can I display all possible trip between two country ?". If so I advice you to do something like that:
trip(X,Y,T) :- X \= Y,
path(X, Y, [X], T).
path(Departure, Arrival, Visited, go(Departure, Arrival)) :- direct(Departure, Arrival) , !.
path(Departure, Arrival, Visited, go(Departure, Intermediate, GO)) :-
direct(Departure, Intermediate),
Intermediate \= Departure,
\+ member(Intermediate,Visited),
path(Intermediate, Arrival, [Intermediate|Visited], GO).
So I also advice you to modify the facts in your knowledge base like:
direct(rome, moskow).
direct(moskow, paris).
direct(paris, chicago).
direct(sau_paolo, rio).
direct(rio, honkong).
and so on.
What you did is logically uncorrect. How can you know if you can travel from a city to another only knowing where they are? There should be some facts that state that is possible travel from a city to another, e.g. direct(rome, milan) (or travel(rome, milan), you choose the functor).
I hope to have help you.
If you have some dubt, just ask ;)
by use prolog
Timetable of air routes
Each air route can be described by a structure:
country of departure, country of landing, date of departure, time of departure, duration, tariffs.
Tariff can described by a structure:
price for business class, price for economy class.
Implement the following rules:
• Find all routes from the given country
• Find all routes to the given country with duration less then than the given number
• Find all routes with the cheapest price for economy class
• Find all prices for flying from one given country to another given country
First, sorry for posting the whole program, but as I don't know were the problem is I don't know which parts are irrelevant. These are two slightly different implementations of the same logic puzzle in SWI-Prolog, the first one succeeds the second one fails and I can't find the reason for the failure.
The puzzle:
4 persons are having a diner:
Donna, Doreen, David, Danny
the woman (Donna,Doreen) are sitting vis-a-vis.
the men (David,Danny) are sitting vis-a-vis.
Each of them picked a unique meal and beverage.
1) Doreen sits next to the person that ordered risotto.
2) the salad came with a coke.
3) the person with the lasagna sits vis-a-vis the person with the milk.
4) david never drinks coffee.
5) donna only drinks water.
6) danny had no appetite for risotto.
who ordered the pizza?
I choose the following approach
table with positions:
1
4 O 2
3
domain: positions{1,2,3,4}
variables: persons, meals, beverages
First the inefficient succeeding implementation:
solution(Pizza, Doreen, Donna, David, Danny) :-
% assignment of unique positions to the variables
unique(Doreen,Donna,David,Danny),
unique(Lasagna,Pizza,Risotto,Salad),
unique(Water,Coke,Coffee,Milk),
% general setting
vis_a_vis(Donna,Doreen),
vis_a_vis(David,Danny),
% the six constraints
next_to(Doreen,Risotto),
Salad = Coke,
vis_a_vis(Lasagna,Milk),
\+ David = Coffee,
Donna = Water,
\+ Danny = Risotto.
unique(X1,X2,X3,X4) :-
pos(X1),
pos(X2),
\+ X1 = X2,
pos(X3),
\+ X1 = X3, \+ X2 = X3,
pos(X4),
\+ X1 = X4, \+ X2 = X4, \+ X3 = X4.
right(1,2).
right(2,3).
right(3,4).
right(4,1).
vis_a_vis(1,3).
vis_a_vis(3,1).
vis_a_vis(2,4).
vis_a_vis(4,2).
next_to(X,Y) :- right(X,Y).
next_to(X,Y) :- right(Y,X).
pos(1).
pos(2).
pos(3).
pos(4).
This works and gives the right result. But when I try to reorder the clauses of the solution procedure to be more efficient (this is the second implementation)
solution(Pizza, Doreen, Donna, David, Danny) :-
% general setting
vis_a_vis(Donna,Doreen),
vis_a_vis(David,Danny),
% the six constraints
Salad = Coke,
vis_a_vis(Lasagna,Milk),
\+ David = Coffee,
Donna = Water,
\+ Danny = Risotto,
% assignment of unique positions to the variables
unique(Doreen,Donna,David,Danny),
unique(Lasagna,Pizza,Risotto,Salad),
unique(Water,Coke,Coffee,Milk).
%% all other predicates are like the ones in the first implementation
I get a unassigned variable warning when trying to load the file:
Warning: /home/pizza.pl:28:
Singleton variable in \+: Coffee
and the computation returns false. But shouldn't it return the same result?
I see no reason for the difference...
the warning is due to the fact that Coffe and Risotto are unbound when the negation is executed. If you replace \+ David = Coffee, by David \= Coffee, you will avoid the warning, but the solution cannot will not be computed. Should be clear indeed that since Coffee is unbound, David \= Coffee will always fail. You can use dif/2, the solution will work and will be more efficient. I've named solution1/2 your first snippet, and solution2/5 this one (using dif/2):
solution2(Pizza, Doreen, Donna, David, Danny) :-
% general setting
vis_a_vis(Donna,Doreen),
vis_a_vis(David,Danny),
% the six constraints
next_to(Doreen,Risotto), % note: you forgot this one
Salad = Coke,
vis_a_vis(Lasagna,Milk),
dif(David, Coffee),
Donna = Water,
dif(Danny, Risotto),
% assignment of unique positions to the variables
unique(Doreen,Donna,David,Danny),
unique(Lasagna,Pizza,Risotto,Salad),
unique(Water,Coke,Coffee,Milk).
a small test:
?- time(aggregate_all(count,solution1(P,A,B,C,D),N)).
% 380,475 inferences, 0.058 CPU in 0.058 seconds (100% CPU, 6564298 Lips)
N = 8.
?- time(aggregate_all(count,solution2(P,A,B,C,D),N)).
% 10,626 inferences, 0.002 CPU in 0.002 seconds (100% CPU, 4738996 Lips)
N = 8.
I am trying to create a prolog rule which will generate all the people in a social network using S number degrees of separation.
This is the rule that i have made but it is only printing empty lists. Can somebody please help me into helping me understand why this is happening and me where i am going wrong?:
socialN(_,N):- N<1,!.
socialN(_,N,_,_):- N<1,!.
socialN(P1,Separation,S1,S):-
(message(P1,P2,_); message(P2,P1,_)),
D is Separation-1,
\+(member(P2,S1)),
append(P2,S1,S2),socialN(P1,D,S2,S),!.
socialN(P2,Separation,S,S).
These are the facts:
message(allan, steve, 2013-09-03).
message(nayna, jane, 2013-09-03).
message(steve, jane, 2013-09-04).
message(steve, allan, 2013-09-04).
message(mark, martin, 2013-09-04).
message(martin, steve, 2013-09-04).
message(allan, martin, 2013-09-05).
E.g. Mark’s network includes just Martin for 1 degree of separation; it includes Martin, Steve and Allan for 2 degrees of separation; and Martin, Steve, Allan and Jane for 3.
I see you are using append and member, so I suppose you are trying to build up a list of people. I was a bit surprised that you were not using findall. Like this:
allDirectLinks(P1, L) :- findall(P2, directlyLinked(P1, P2), L).
directlyLinked(P1, P1).
directlyLinked(P1, P2) :- message(P1, P2, _).
directlyLinked(P1, P2) :- message(P2, P1, _).
From there, you can write a recursive function to find the indirect links:
socialN(0, P, [P]) :- !.
socialN(N, P1, L3) :-
N>0, !,
N1 is N-1,
socialN(N1, P1, L1)
maplist(allDirectLinks, L1, L2),
append(L2, L3).
For example, this yields in Y a list of people separated 2 steps or less from Mark:
socialN(2, mark, X), list_to_set(X, Y).
Please note, Mark himself is included in the resulting list (being a 'level 0' link); I suppose it cannot be too hard to filter that out afterwards.
I hope this makes sense; I am a bit rusty, haven't done any Prolog in 25 years.
EDIT: explanation of the rules I defined:
directlyLinked: true if there is a message between two persons (regardless of the direction of the message)
allDirectLinks: accumulates into list L all persons directly linked to a given person P1; just read the manual about findall
socialN: builds up a list of people connected to a given person (P) at a distance less than or equal to a given distance (N)
socialN(0, ...): at distance 0, every person is linked to himself
socialN(N, ...): makes a recursive call to get a list of connections at distance N-1, then uses maplist to apply allDirectLinks to every connection found, and finally uses append to concatenate the results together.