I recently started learning Prolog for fun. I found the following murder mystery puzzle. Since I don't know much about Prolog except the basics, I cannot really evaluate the solution provided in the link, however, it didn't seem particularly nice to me. My solution is not enough to generate the correct answers so I'm looking for some pointers as to how to get there or if it's at all possible to get there with my approach. Here's the puzzle just in case the link goes down:
To discover who killed Mr. Boddy, you need to learn where each person
was, and what weapon was in the room. Clues are scattered throughout
the quiz (you cannot solve question 1 until all 10 are read).
To begin, you need to know the suspects. There are three men (George,
John, Robert) and three women (Barbara, Christine, Yolanda). Each
person was in a different room (Bathroom, Dining Room, Kitchen, Living
Room, Pantry, Study). A suspected weapon was found in each room (Bag,
Firearm, Gas, Knife, Poison, Rope). Who was found in the kitchen?
Clue 1: The man in the kitchen was not found with the rope, knife, or
bag. Which weapon, then, which was not the firearm, was found in the
kitchen?
Clue 2: Barbara was either in the study or the bathroom; Yolanda was
in the other. Which room was Barbara found in?
Clue 3: The person with the bag, who was not Barbara nor George, was
not in the bathroom nor the dining room. Who had the bag in the room
with them?
Clue 4: The woman with the rope was found in the study. Who had the
rope?
Clue 5: The weapon in the living room was found with either John or
George. What weapon was in the living room?
Clue 6: The knife was not in the dining room. So where was the knife?
Clue 7: Yolanda was not with the weapon found in the study nor the
pantry. What weapon was found with Yolanda?
Clue 8: The firearm was in the room with George. In which room was the
firearm found?
It was discovered that Mr. Boddy was gassed in the pantry. The suspect
found in that room was the murderer. Who, then, do you point the
finger towards?
Here's the link to the author's solution.
Here's my attempted solution:
male(george).
male(john).
male(robert).
female(barbara).
female(christine).
female(yolanda).
person(X) :- male(X).
person(X) :- female(X).
room(kitchen).
room(bathroom).
room(diningroom).
room(livingroom).
room(pantry).
room(study).
weapon(bag).
weapon(firearm).
weapon(gas).
weapon(knife).
weapon(poison).
weapon(rope).
/*
Clue 1: The man in the kitchen was not found with
the rope, knife, or bag.
Which weapon, then, which was not the firearm,
was found in the kitchen?
*/
/* X is Weapon, Y is Room, Z is Person */
killer(X, Y, Z) :-
room(Y) = room(kitchen),
male(Z),
dif(weapon(X), weapon(rope)),
dif(weapon(X), weapon(knife)),
dif(weapon(X), weapon(bag)),
dif(weapon(X), weapon(firearm)).
/*
Clue 2: Barbara was either in the study or the bathroom;
Yolanda was in the other.
Which room was Barbara found in?
*/
/* It was easy to deduce the following from other data */
killer(X, Y, Z) :-
female(Z) = female(barbara),
room(study) = room(Y).
killer(X, Y, Z) :-
female(Z) = female(yolanda),
room(bathroom) = room(Y).
/*
Clue 3: The person with the bag, who was not Barbara nor
George, was not in the bathroom nor the dining room.
Who had the bag in the room with them?
*/
killer(X, Y, Z) :-
weapon(bag) = weapon(X),
dif(room(Y), room(bathroom)),
dif(room(Y), room(diningroom)),
dif(person(Z), male(george)),
dif(person(Z), female(barbara)).
/*
Clue 4: The woman with the rope was found in the study.
Who had the rope?
*/
killer(X, Y, Z) :-
weapon(rope) = weapon(X),
room(study) = room(Y),
female(Z).
/*
Clue 5: The weapon in the living room was found with either
John or George. What weapon was in the living room?
*/
killer(X, Y, Z) :-
room(Y) = room(livingroom),
dif(male(Z), male(robert)).
/*
Clue 6: The knife was not in the dining room.
So where was the knife?
*/
killer(X, Y, Z) :-
weapon(knife) = weapon(X),
room(Y) \= room(diningroom).
/*
Clue 7: Yolanda was not with the weapon found
in the study nor the pantry.
What weapon was found with Yolanda?
*/
killer(X, Y, Z) :-
female(yolanda) = female(Z),
dif(room(study), room(Y)),
dif(room(pantry), room(Y)).
/*
Clue 8: The firearm was in the room with George.
In which room was the firearm found?
*/
killer(X, Y, Z) :-
weapon(firearm) = weapon(X),
male(george) = male(Z).
/*
It was discovered that Mr. Boddy was gassed in the pantry.
The suspect found in that room was the murderer.
Who, then, do you point the finger towards?
*/
killer(X, Y, Z) :-
room(Y) = room(pantry),
weapon(X) = weapon(gas).
I took a more positive approach to this problem. Rather than trying any form of negation I went with just plain unification.
Key is this predicate pair:
members([],_).
members([M|Ms],Xs) :- select(M,Xs,Ys),members(Ms,Ys).
This is a basic permutation predicate. It will take a list of the first argument and try to unify against all permutations of second list.
Now a lot of the rules became quite easy to express:
For example, clue 1:
clue1(House) :- members([[P,kitchen,_],[_,_,rope],[_,_,knife],[_,_,bag],[_,_,firearm]],House),man(P).
So this meant that the rope, knife, bag and firearm were all members of the house, but in different rooms than the kitchen. Prolog would keep backtracking util it found a fit for these items.
Here's my full solution:
man(george).
man(john).
man(robert).
woman(barbara).
woman(christine).
woman(yolanda).
members([],_).
members([M|Ms],Xs) :- select(M,Xs,Ys),members(Ms,Ys).
clue1(House) :- members([[P,kitchen,_],[_,_,rope],[_,_,knife],[_,_,bag],[_,_,firearm]],House),man(P).
clue2(House) :- member([barbara,study,_],House), member([yolanda,bathroom,_],House).
clue2(House) :- member([barbara,bathroom,_],House), member([yolanda,study,_],House).
clue3(House) :- members([[_,_,bag],[barbara,_,_],[george,_,_]],House),members([[_,_,bag],[_,bathroom,_],[_,dining_room,_]],House).
clue4(House) :- members([[P,study,rope]],House),woman(P).
clue5(House) :- members([[john,living_room,_]],House).
clue5(House) :- members([[george,living_room,_]],House).
clue6(House) :- members([[_,_,knife],[_,dining_room,_]],House).
clue7(House) :- members([[yolanda,_,_],[_,study,_],[_,pantry,_]],House).
clue8(House) :- member([george,_,firearm],House).
clue9(House,P) :- members([[P,pantry,gas]],House).
solve(X) :-
House = [[_,bathroom,_],[_,dining_room,_],[_,kitchen,_],[_,living_room,_],[_,pantry,_],[_,study,_]],
clue1(House),
clue2(House),
clue3(House),
clue4(House),
clue5(House),
clue6(House),
clue7(House),
clue8(House),
clue9(House,X),
members([[george,_,_],[john,_,_],[robert,_,_],[barbara,_,_],[christine,_,_],[yolanda,_,_]],House),
members([[_,_,bag],[_,_,firearm],[_,_,gas],[_,_,knife],[_,_,poison],[_,_,rope]],House),
write(House),
true.
That gave me:
?- solve(X).
[[yolanda,bathroom,knife],[george,dining_room,firearm],[robert,kitchen,poison],[john,living_room,bag],[christine,pantry,gas],[barbara,study,rope]]
X = christine .
Edit: See an improved version of the reference solution at https://swish.swi-prolog.org/p/crime_constraints.pl.
I agree that the solution you linked to is ugly, but it does use the right approach. Yours isn't quite going in the right direction. Some remarks:
/* X is Weapon, Y is Room, Z is Person */
Why not use the variable names Weapon, Room, and Person then? It makes your program much easier to read.
weapon(rope) = weapon(X)
This is exactly equivalent to just writing X = rope or rope = X.
But apart from these there are other two big problems with the way you are approaching this puzzle:
First, you are not modeling relationships between your objects as data. For example, for "The woman with the rope was found in the study." you have this clause:
killer(X, Y, Z) :-
weapon(rope) = weapon(X),
room(study) = room(Y),
female(Z).
This does indeed have three solutions that you can interpret as "a relation killer(rope, study, barbara), killer(rope, study, christine), or killer(rope, study, yolanda)", but your program doesn't know how to interpret it that way. You don't actually construct data that expresses this relationship. This is what the solution you linked to does correctly: It models rooms and weapons as variables which can be bound to atoms representing persons. Thus it can express this clue as woman(Rope) ("the person with the Rope is a woman") and Rope = Study ("the rope and the study are associated with the same person").
The second big problem is that you are modeling all clues as different clauses of the same predicate. This is wrong because in Prolog the different clauses of a predicate express a choice: Something holds if the first clause holds or the second clause holds or the third clause holds, etc. But you want to express that the first clue holds and the second clue holds and the third clue holds, etc. And "and" is expressed by combining the different conditions with , in the body of one clause. This is why the linked solution has different predicates clue1, clue2, etc., all of which are called from the body of one big predicate.
Derive Rules from the clues in sequence
Each person was in a different room (Bathroom, Dining Room, Kitchen,
Living Room, Pantry, Study). A suspected weapon was found in each room
(Bag, Firearm, Gas, Knife, Poison, Rope).
unique(A,B,C,D,E,F) :-
A \= B, A \= C, A \= D, A \= E, A \= F,
B \= C, B \= D, B \= E, B \= F,
C \= D, C \= E, C \= F,
D \= E, D \= F,
E \= F.
suspicious(pwr(george,WA,RA), pwr(john,WB,RB), pwr(robert,WC,RC), pwr(barbara,WD,RD), pwr(christine,WE,RE), pwr(yolanda,WF,RF)) :-
weapon(WA), weapon(WB), weapon(WC), weapon(WD), weapon(WE), weapon(WF),
unique(WA,WB,WC,WD,WE,WF),
room(RA), room(RB), room(RC), room(RD), room(RE), room(RF),
unique(RA,RB,RC,RD,RE,RF).
Now let us examine
Clue 1: The man in the kitchen was not found with the rope, knife, or
bag. Which weapon, then, which was not the firearm, was found in the
kitchen?
clue1(L) :-
oneof(pwr(P,W,kitchen),L),
male(P),
weapon(W),
W \= rope, W \= knife, W \= bag, W \= firearm.
We do this for each of the 8 clues and finally
It was discovered that Mr. Boddy was gassed in the pantry. The suspect
found in that room was the murderer. Who, then, do you point the
finger towards?
killer(X, L) :- member(pwr(X,gas,pantry),L).
resolved(X) :-
suspicious(A,B,C,D,E,F),
L = [A,B,C,D,E,F],
clue1(L),
clue2(L),
clue3(L),
clue4(L),
clue5(L),
clue6(L),
clue7(L),
clue8(L),
killer(X, L).
The full program could be found and run. The inference is rather slow (but faster than the authors solution).
Why consider it a better design to use relations instead of Variable bindings?
I understand a prolog program as a ruleset to derive knowledge. That means:
Each relation in prolog should describe a relation in the domain
Adding entities (Weapons, Persons, Rooms) to the world should not make the ruleset obsolete. The problem has not changed (we only extended the world) so the rules and queries need not to be touched.
Extending the problem (e.g. by adding a seventh location) should have minimal impact
Not every aspect is optimal in the referenced solution, some may be better expressed if one is more familiar with prolog.
Why do I think that a ruleset should be robust to world changes?
I used datalog in program analysis. That means that each relation in source code (or byte code) was modeled as facts and the rules inferred types, security vulnerabilities, design patterns etc. There were multiple millions of facts and multiple thousands of ruleset code. Adding an entity (e.g. a source code line, a type annotation) should not drive me to reimplement the ruleset code (which was quite hard to write it correctly).
Why do I think that using implicit relations is bad code?
Consider this code from the reference solution, it is totally misleading:
clue1(Bathroom, Dining, Kitchen, Livingroom, Pantry, Study, Bag, Firearm, Gas, Knife, Poison, Rope) :-
man(Kitchen), // a man is a kitchen?
\+Kitchen=Rope, // a kitchen is not a rope?
\+Kitchen=Knife, // a kitchen is not a knife?
\+Kitchen=Bag, // a kitchen is not a bag
\+Kitchen=Firearm. // a kitchen is not a firearm
Ok the variable names are ugly, better readable would be
clue1(InBathroom, InDiningroom, InKitchen, InLivingroom, InPantry, InStudy, WithBag, WithFirearm, WithGas, WithKnife, WithPoison, WithRope) :-
man(InKitchen), // (person) in the kitchen is a man - ok
\+Kitchen=Rope, // (person) in the kitchen is not
(person) with a rope - better than above
\+Kitchen=Knife, // ...
\+Kitchen=Bag, // ...
\+Kitchen=Firearm. // ...
But we misuse the equal relation for an explicit one. There is a clear indicator: Variables containing predicates in their names are probably implicit relations. "personInKitchen" is a (logical) predicate "in" connecting two substantives "person" and "kitchen".
As comparison a model with lists and function symbols (suspect/3 is the relational function that connects persons to weapons and rooms, Suspects is the list of suspects):
clue1(Suspects) :-
member(suspect(Person,Weapon,Room),Suspects),
male(Person), // The man (Person)
Room = kitchen, // in the Kitchen (Room)
Weapon \= rope, // was not found with the (Weapon) rope
Weapon \= knife, // (Weapon) knife
Weapon \= bag, // (Weapon) bag
Weapon \= firearm.// (Weapon) firearm
Summary
So if you use prolog for private purpose, I do not mind "misusing" Variables to come to a quick solution. But if your ruleset and your data grows it seems to me quite essential to model all relations explicitly.
I am working through sample questions while studying, using SWI-Prolog. I have reached the last section of this question, where I have to recursively (I hope) compare elements of a list containing 'researcher' structures to determine whether or not the researchers have the same surname, and, if they do, return the Forename and Surname of the group leader for that list.
There is only one list that meets this criteria and it has four members, all with the same surname. However, the correct answer is returned FOUR times. I feel my solution is inelegant and is lacking. Here is the question:
The following Prolog database represents subject teaching teams.
% A research group structure takes the form
% group(Crew, Leader, Assistant_leader).
%
% Crew is a list of researcher structures,
% but excludes the researcher structures for Leader
% and Assistant_leader.
%
% researcher structures take the form
% researcher(Surname, First_name, expertise(Level, Area)).
group([researcher(giles,will,expertise(3,engineering)),
researcher(ford,bertha,expertise(2,computing))],
researcher(mcelvey,bob,expertise(5,biology)),
researcher(pike,michelle,expertise(4,physics))).
group([researcher(davis,owen,expertise(4,mathematics)),
researcher(raleigh,sophie,expertise(4,physics))],
researcher(beattie,katy,expertise(5,engineering)),
researcher(deane,fergus,expertise(4,chemistry))).
group([researcher(hardy,dan,expertise(4,biology))],
researcher(mellon,paul,expertise(4,computing)),
researcher(halls,antonia,expertise(3,physics))).
group([researcher(doone,pat,expertise(2,computing)),
researcher(doone,burt,expertise(5,computing)),
researcher(doone,celia,expertise(4,computing)),
researcher(doone,norma,expertise(2,computing))],
researcher(maine,jack,expertise(3,biology)),
researcher(havilland,olive,expertise(5,chemistry))).
Given this information, write Prolog rules (and any additional predicates required) that can be used to return the following:
the first name and surname of any leader whose crew members number more than one and who all have the same surname. [4 marks]
This is the solution I presently have using recursion, though it's unnecessarily inefficient as for every member of the list, it compares that member to every other member. So, as the correct list is four members long, it returns 'jack maine' four times.
surname(researcher(S,_,_),S).
checkSurname([],Surname):-
Surname==Surname. % base case
checkSurname([Researcher|List],Surname):-
surname(Researcher,SameSurname),
Surname == SameSurname,
checkSurname(List,SameSurname).
q4(Forename,Surname):-
group(Crew,researcher(Surname,Forename,_),_),
length(Crew,Length),
Length > 1,
member(researcher(SameSurname,_,_),Crew),
checkSurname(Crew,SameSurname).
How could I do this without the duplicate results and without redundantly comparing each member to every other member each time? For every approach I've taken I am snagged each time with 'SameSurname' being left as a singleton, hence having to force use of it twice in the q4 predicate.
Current output
13 ?- q4(X,Y).
X = jack,
Y = maine ; x4
A compact and efficient solution:
q4(F, S) :-
group([researcher(First,_,_), researcher(Second,_,_)| Crew], researcher(S, F, _), _),
\+ (member(researcher(Surname, _, _), [researcher(Second,_,_)| Crew]), First \== Surname).
Example call (resulting in a single solution):
?- q4(X,Y).
X = jack,
Y = maine.
You are doing it more complicated than it has to be. Your q4/2 could be even simpler:
q4(First_name, Surname) :-
group(Crew, researcher(Surname, First_name, _E), _A),
length(Crew, Len), Len > 1,
all_same_surname(Crew).
Now you only need to define all_same_surname/1. The idea is simple: take the surname of the first crew member and compare it to the surnames of the rest:
all_same_surname([researcher(Surname, _FN, _E)|Rest]) :-
rest_same_surname(Rest, Surname).
rest_same_surname([], _Surname).
rest_same_surname([researcher(Surname, _FN, _E)|Rest), Surname) :-
rest_same_surname(Rest, Surname).
(Obviously, all_same_surname/1 fails immediately if there are no members of the crew)
This should be it, unless I misunderstood the problem statement.
?- q4(F, S).
F = jack,
S = maine.
How about that?
Note: The solution just takes the most straight-forward approach to answering the question and being easy to write and read. There is a lot of stuff that could be done otherwise. Since there is no reason not to, I used pattern matching and unification in the heads of the predicates, and not comparison in the body or extra predicates for extracting arguments from the compound terms.
P.S. Think about what member/2 does (look up its definition in the library, even), and you will see where all the extra choice points in your solution are coming from.
Boris did answer this question already, but I want to show the most concise solution I could come with. It's just for the educational purposes (promoting findall/3 and maplist/2):
q4(F, S) :-
group(Crew, researcher(S, F, _), _),
findall(Surname, member(researcher(Surname, _, _), Crew), Surnames),
Surnames = [ First, Second | Rest ],
maplist(=(First), [ Second | Rest ]).
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.
I've looked through the similar questions but can't find anything that's relevant to my problem. I'm struggling to find an algorithm or set of 'loops' that will find a path from CityA to CityB, using a database of
distance(City1,City2,Distance)
facts. What I've managed to do so far is below, but it always backtracks at write(X), and then completes with the final iteration, which is what I want it to do but only to a certain extent.
For example, I don't want it to print out any city names that are dead ends, or to use the final iteration. I want it to basically make a path from CityA to CityB, writing the name of the cities it goes to on the path.
I hope somebody can help me!
all_possible_paths(CityA, CityB) :-
write(CityA),
nl,
loop_process(CityA, CityB).
loop_process(CityA, CityB) :-
CityA == CityB.
loop_process(CityA, CityB) :-
CityA \== CityB,
distance(CityA, X, _),
write(X),
nl,
loop_process(X, CityB).
I tried to demonstrate how you can achieve what you're working on so that you can understand better how it works. So since your OP wasn't very complete, I took some liberties ! Here are the facts I'm working with :
road(birmingham,bristol, 9).
road(london,birmingham, 3).
road(london,bristol, 6).
road(london,plymouth, 5).
road(plymouth,london, 5).
road(portsmouth,london, 4).
road(portsmouth,plymouth, 8).
Here is the predicate we will call to find our paths, get_road/4. It basically calls the working predicate, that has two accumulators (one for the points already visited and one for the distance we went through).
get_road(Start, End, Visited, Result) :-
get_road(Start, End, [Start], 0, Visited, Result).
Here is the working predicate,
get_road/6 : get_road(+Start, +End, +Waypoints, +DistanceAcc, -Visited, -TotalDistance) :
The first clause tells that if there is a road between our first point and our last point, we can end here.
get_road(Start, End, Waypoints, DistanceAcc, Visited, TotalDistance) :-
road(Start, End, Distance),
reverse([End|Waypoints], Visited),
TotalDistance is DistanceAcc + Distance.
The second clause tells that if there is a road between our first point and an intermediate point, we can take it and then solve get_road(intermediate, end).
get_road(Start, End, Waypoints, DistanceAcc, Visited, TotalDistance) :-
road(Start, Waypoint, Distance),
\+ member(Waypoint, Waypoints),
NewDistanceAcc is DistanceAcc + Distance,
get_road(Waypoint, End, [Waypoint|Waypoints], NewDistanceAcc, Visited, TotalDistance).
Usage is as follows :
?- get_road(portsmouth, plymouth, Visited, Distance).
And yields :
Visited = [portsmouth, plymouth],
Distance = 8 ;
Visited = [portsmouth, london, plymouth],
Distance = 9 ;
Visited = [portsmouth, plymouth, london, plymouth],
Distance = 18 ;
false.
I hope it will be helpful to you.
Please separate the pure part from the impure (I/O, like write/1, nl/0 but also (==)/2 and (\==)/2). As long as they are entirely interlaced with your pure code you cannot expect much.
Probably you want a relation between a starting point, an end point and a path in between.
Should that path be acyclic or do you permit cycles?
To ensure that an element X does not occur in a list Xs use the goal maplist(dif(X),Xs).
You do not need any further auxiliary predicates to make this a nice relation!
You should return a successful list as an Out variable in all_possible_paths. Then write out that list. Don't do both in the same procedure.