I have a large numbers of facts that are already in my file (position(M,P)), M is the name and P is the position of the player , I am asked to do a player_list(L,N), L is the list of players and N is the size of this list. I did it and it works the problem is that it gives the list without the names it gives me numbers and not names
player_list([H|T],N):- L = [H|T],
position(H,P),
\+ member(H,L),
append(L,H),
player_list(T,N).
what I get is:
?- player_list(X,4).
X = [_9176, _9182, _9188, _9194] .
so what should I do ?
You could use an additional list as an argument to keep track of the players you already have. This list is empty at the beginning, so the calling predicate calls the predicate describing the actual relation with [] as an additional argument:
player_list(PLs,L) :-
pl_l_(PLs,L,[]). % <- actual relation
The definition you posted is missing a base case, that is, if you already have the desired amount of players, you can stop adding others. In this case the number of players to add is zero otherwise it is greater than zero. You also have to describe that the head of the list (PL) is a player (whose position you don't care about, so the variable is preceded by an underscore (_P), otherwise the goal is just like in your code) and is not in the accumulator yet (as opposed to your code, where you check if PL is not in L) but in the recursive call it is in the accumulator. You can achieve the latter by having [PL|Acc0] in the recursive goal, so you don't need append/2. Putting all this together, your code might look something like this:
pl_l_([],0,_). % base case
pl_l_([PL|PLs],L1,Acc0) :-
L1 > 0, % number of players yet to add
L0 is L1-1, % new number of players to add
position(PL,_P), % PL is a player and
\+ member(PL,Acc0), % not in the accumulator yet
pl_l_(PLs,L0,[PL|Acc0]). % the relation holds for PLs, L0 and [PL|Acc0] as well
With respect to your comment, I assume that your code contains the following four facts:
position(zlatan,center).
position(rooney,forward).
position(ronaldo,forward).
position(messi,forward).
Then your example query yields the desired results:
?- player_list(X,4).
X = [zlatan,rooney,ronaldo,messi] ? ;
X = [zlatan,rooney,messi,ronaldo] ? ;
...
If you intend to use the predicate the other way around as well, I suggest the use of CLP(FD). To see why, consider the most general query:
?- player_list(X,Y).
X = [],
Y = 0 ? ;
ERROR at clause 2 of user:pl_l_/3 !!
INSTANTIATION ERROR- =:=/2: expected bound value
You get this error because >/2 expects both arguments to be ground. You can modify the predicate pl_l_/3 to use CLP(FD) like so:
:- use_module(library(clpfd)).
pl_l_([],0,_).
pl_l_([PL|PLs],L1,Acc0) :-
L1 #> 0, % <- new
L0 #= L1-1, % <- new
position(PL,_P),
\+ member(PL,Acc0),
pl_l_(PLs,L0,[PL|Acc0]).
With these modifications the predicate is more versatile:
?- player_list([zlatan,messi,ronaldo],Y).
Y = 3
?- player_list(X,Y).
X = [],
Y = 0 ? ;
X = [zlatan],
Y = 1 ? ;
X = [zlatan,rooney],
Y = 2 ?
...
Related
I have understood the theory part of Recursion. I have seen exercises but I get confused. I've tried to solve some, some I understand and some I don't. This exercise is confusing me. I can't understand why, so I use comments to show you my weak points. I should have power (X,N,P) so P=X^N.
Some examples:
?- power(3,5,X).
X = 243
?- power(4,3,X).
X = 64
?- power(2,4,X).
X = 16
The solution of this exercise is: (See comments too)
power(X,0,1). % I know how works recursion,but those numbers 0 or 1 why?
power(X,1,X). % X,1,X i can't get it.
power(X,N,P) :- % X,N,P if only
N1 is N-1, % N1=N-1 ..ok i understand
power(X,N1,P1), % P1 is used to reach the the P
P is P1*X. % P = P1*X
What I know recursion, I use a different my example
related(X, Y) :-
parent(X, Z),
related(Z, Y).
Compare my example with the exercise. I could say that my first line, what I think. Please help me out with it is a lot of confusing.
related(X, Y) :- is similar to power(X,N,P) :- . Second sentence of my example parent(X, Z), is similar to N1 is N-1, and the third sentence is related(Z, Y). similar to power(X,N1,P1), and P is P1*X..
Let's go over the definition of the predicate step by step. First you have the fact...
power(X,0,1).
... that states: The 0th power of any X is 1. Then there is the fact...
power(X,1,X).
... that states: The 1st power of any X is X itself. Finally, you have a recursive rule that reads:
power(X,N,P) :- % P is the Nth power of X if
N1 is N-1, % N1 = N-1 and
power(X,N1,P1), % P1 is the N1th power of X and
P is P1*X. % P = P1*X
Possibly your confusion is due to the two base cases that are expressed by the two facts (one of those is actually superfluous). Let's consider the following queries:
?- power(5,0,X).
X = 1 ;
ERROR: Out of local stack
The answer 1 is certainly what we expect, but then the predicate loops until it runs out of stack. That's certainly not desirable. And this query...
?- power(5,1,X).
X = 5 ;
X = 5 ;
ERROR: Out of local stack
... yields the correct answer twice before running out of stack. The reason for the redundant answer is that the recursive rule can reduce any given N to zero and to one thus yielding the same answer twice. If you look at the structure of your recursive rule, it is obvious that the first base case is sufficient, so let's remove the second. The reason for looping out of stack is that, after N becomes zero, the recursive rule will search for other solutions (for N=-1, N=-2, N=-3,...) that do not exist. To avoid that, you can add a goal that prevents the recursive rule from further search, if N is equal to or smaller than zero. That leaves you with following definition:
power(X,0,1). % the 0th power of any X is 1
power(X,N,P) :- % P is the Nth power of X if
N > 0, % N > 0 and
N1 is N-1, % N1 = N-1 and
power(X,N1,P1), % P1 is the N1th power of X and
P is P1*X. % P = P1*X
Now the predicate works as expected:
?- power(5,0,X).
X = 1 ;
false.
?- power(5,1,X).
X = 5 ;
false.
?- power(5,3,X).
X = 125 ;
false.
I hope this alleviates some of your confusions.
I would like to create a predicate that returns the element that most often appears, if there are more than one with the same number of occurrences the first:
occ([a,b,c,a,a,a,b],M).
yes M = a
occ([a,b,c,a,b],M).
yes M = a
Note that in Prolog you would generally create rules, not functions to solve this.
There are a number of ways to approach this, I'll provide two.
Recursion
One way is to recurse over the list, keeping a running count of occurrences, and with each call recording what the current max is, an example of use of an accumulator:
% find the X with the most occurrences N in a list L
occ(X,N,L) :-
occ(L,max(null,0),[],max(X,N)).
%% occ(+L, +CurrentMax, +Counts, +FinalMax) is det.
%
% recurse through L, using CurrentMax accumulator to
% store current candidate as a term `max(X,N)`
%
% Counts is a list in which we accumulate counts of
% occurrences to far, as list of element-count pairs X-N
%
% The final argument is unified with the CurrentMax
% accumulator as the base case
occ([], max(Xm, Nm), _, max(Xm, Nm)).
occ([X|L], max(Xm, Nm), Counts, FinalMax) :-
% get the current count of X
( select(X-N, Counts, CountsT)
->
N1 is N+1
;
N1 = 1,
CountsT = Counts),
% make a new list of counts with the
% original entry for X replaced by a new
% one with count of N+1
Counts2 = [X-N1 | CountsT],
% recurse, using either new current best candidate
% or same one depending on whether count is exceeded.
% in case of tie, use same one, thus prioritizing first result
( N1 > Nm
->
occ(L, max(X,N1), Counts2, FinalMax)
;
occ(L, max(Xm,Nm), Counts2, FinalMax)).
Example:
?- occ(X,N,[a,b,c,a,b]).
X = a,
N = 2.
Higher-order aggregate operations
An alternative approach is to use higher-order aggregate predicates. Arguably this leads to more declarative code, although tastes will vary. If you are using SWI-Prolog you can use the aggregate library. We can start with a rule to count occurrences in a list (note I'm going to switch from your occ/2 to more explicit predicates here):
% count number N of instances of X in a list L
element_count(X,N,L) :-
aggregate(count,member(X,L),N).
If you don't want to or can't use aggregate/3 then have a look at the answers to this question previously asked on stack overflow.
Next we can use aggregate/3 to find the maximum number for N, plus a "witness" (i.e. the value of X with the highest value):
% count number N of instances of X in a list L, for highest N
max_element_count(X,N,L) :-
aggregate(max(N1,X1),element_count(X1,N1,L),max(N,X)).
(I'll leave it to you to make an equivalent implementation of this rule if you're not using the aggregate library)
Let's try it:
?- max_element_count(X,N,[a,b,c,a,a,a,b]).
X = a,
N = 4.
With a tie it seems to satisfy your criterion of using the first occurrence in the case of tie-breakers:
?- max_element_count(X,N,[a,b,c,a,b]).
X = a,
N = 2.
But this is not in fact guaranteed - we just happen to choose a here as it is alphabetically before b. Let's try:
?- max_element_count(X,N,[b,a,c,a,b]).
X = a,
N = 2.
Oops!
This time we will find the first member of the list whose number of occurrences is equal to the max:
max_element_count2(X,N,L) :-
aggregate(max(N1),X1,element_count(X1,N1,L),N),
member(X,L),
element_count(X,N,L),
!.
This assumes that member/2 will unify with elements in order, which is the behavior I have always seen with Prologs but don't know off the top of my head if it is mandated by the standard.
To demonstrate:
?- max_element_count2(X,N,[b,a,c,a,b]).
X = b,
N = 2.
I'd like to assert facts about all members of a List in prolog, and have any resulting unification retained. As an example, I'd like to assert that each list member is equal to five, but none of the below constructs does this:
?- L=[X,Y,Z], forall(member(E,L), E=5).
L = [_h27057686,_h27057704,_h27057722]
X = _h27057686
Y = _h27057704
Z = _h27057722
yes
?- L=[X,Y,Z], foreach(member(E,L), E=5).
L = [_h27057686,_h27057704,_h27057722]
X = _h27057686
Y = _h27057704
Z = _h27057722
yes
I would like a way to pose the query such that X=5,Y=5, and Z=5.
There is a lot of terminology that you might be getting wrong, or I am misunderstanding you.
"Equal to" is not the same as "could unify", or "unify", but it depends how you mean it.
With SWI-Prolog, from the top level:
?- X == 5.
false. % the free variable X is not the integer 5
?- unifiable(X, 5, U).
U = [X=5]. % you could unify X with 5, then X will be 5
?- X = 5.
X = 5. % X unifies with 5 (and is now bound to the integer 5)
The comment by CapelliC already has the answer that you are most likely after: given a list of variables (either free or not), make so that each variable in the list is bound to the integer 5. This is best done by unification (the third query above). The maplist simply applies the unification to each element of the list.
PS. In case you are wondering how to read the maplist(=(5), L):
These three are equivalent:
maplist(=(5), [X,Y,Z])
maplist(=, [5,5,5], [X,Y,Z])
X=5, Y=5, Z=5
And of course X=5 is the same as =(X,5).
I'm trying to solve this problem in SWI Prolog, and my code currently looks like this:
insert(L1,X,L2):-
COUNTER = 1,
NEXT = 1,
insert_plus(L1,COUNTER,NEXT,X,L2).
insert_plus([],_,_,_,[]).
insert_plus([H|T],COUNTER,NEXT,X,[H|T1]) :- % don't insert
COUNTER \= NEXT,
insert_plus(T,COUNTER+1,NEXT,X,T1).
insert_plus([H|T],COUNTER,NEXT,X,[H|[X|T]]) :- % DO insert
COUNTER = NEXT,
insert_plus(T,COUNTER+1,NEXT*2,X,T).
Can someone explain why this does not always work as expected?
?- insert([1,2,3,4,5,6,7],9,X).
X = [1,9,2,3,4,5,6,7]. % BAD! expected: `X = [1,9,2,9,3,4,9,5,6,7]`
Prolog doesn't evaluate expressions, it proves relations. So arithmetic must be carried away explicitly. Here
...
insert_plus(T, COUNTER+1, NEXT, X, T1).
you need
...
SUCC is COUNTER+1,
insert_plus(T, SUCC, NEXT, X, T1).
the same problem - with both COUNTER and NEXT - occurs in the last rule.
The absolute bare minimum that you need to change is:
insert_plus([],_,_,_,[]).
insert_plus([H|T],COUNTER,NEXT,X,[H|T1]) :-
COUNTER =\= NEXT, % `(=\=)/2` arithmetic not-equal
insert_plus(T,COUNTER+1,NEXT,X,T1).
insert_plus([H|T],COUNTER,NEXT,X,[H|[X|T1]]) :- % use `T1`, not `T`
COUNTER =:= NEXT, % `(=:=)/2` arithmetic equal
insert_plus(T,COUNTER+1,NEXT*2,X,T1). % use `T1` (as above)
Sample query:
?- insert([1,2,3,4,5,6,7],9,X).
X = [1,9,2,9,3,4,9,5,6,7]. % expected result
In addition to the above changes I recommend you take advise that #CapelliC gave
in his answer concerning arithmetic expression evaluation using the builtin Prolog predicate (is)/2...
... or, even better, use clpfd!
just started programming with prolog and I'm having a few issues. The function I have is supposed to take a value X and copy it N number of times into M. My function returns a list of N number of memory locations. Here's the code, any ideas?
duple(N,_,M):- length(M,Q), N is Q.
duple(N,X,M):- append(X,M,Q), duple(N,X,Q).
Those are not memory adresses. Those are free variables. What you see is their internal names in your prolog system of choice. Then, as #chac pointed out (+1 btw), the third clause is not really making sense! Maybe you can try to tell us what you meant so that we can bring light about how to do it correctly.
I'm going to give you two implementations of your predicate to try to show you correct Prolog syntax:
duple1(N, X, L) :-
length(L, N),
maplist(=(X), L).
Here, in your duple1/3 predicate, we tell prolog that the length of the resulting list L is N, and then we tell it that each element of L should be unified with X for the predicate to hold.
Another to do that would be to build the resulting list "manually" through recursion:
duple2(0, _X, []).
duple2(N, X, [X|L]) :-
N > 0,
NewN is N - 1,
duple1(NewN, X, L).
Though, note that because we use >/2, is and -/2, ie arithmetic, we prevent prolog from using this predicate in several ways, such as:
?- duple1(X, Y, [xyz, xyz]).
X = 2,
Y = xyz.
This worked before, in our first predicate!
Hope this was of some help.
I suppose you call your predicate, for instance, in this way:
?- duple(3,xyz,L).
and you get
L = [_G289, _G292, _G295] ;
ERROR: Out of global stack
If you try
?- length(X,Y).
X = [],
Y = 0 ;
X = [_G299],
Y = 1 ;
X = [_G299, _G302],
Y = 2 ;
X = [_G299, _G302, _G305],
Y = 3 ;
X = [_G299, _G302, _G305, _G308],
Y = 4 .
...
you can see what's happening:
your query will match the specified *M*, displaying a list of M uninstantiated variables (memory locations), then continue backtracking and generating evee longer lists 'til there is stack space. Your second rule will never fire (and I don't really understand its purpose).
A generator is easier to write in this way:
duple(N,X,M) :- findall(X,between(1,N,_),M).
test:
?- duple(3,xyz,L).
L = [xyz, xyz, xyz].