I am trying to count a whole Europe population using this fucntion:
printEuropePopulation(A) :-
country(_, X, _, _),
printEuropePopulation(A+X),
true.
printEuropePopulation(A) :- write(A).
And program gives me
Out of local stack
when I call my func
printEuropePopulation(0)
Facts looks like:
country("Slovenia",20,2009,27300).
country("Finland",338,5238,35500).
country("France",549,60876,33800).
How can I fix it? Thanks in advance.
You cannot do this with a recursive predicate I think? It is easier to collect all the facts and then find the sum.
total_population(Total) :-
findall(P, country_population(_, P), Ps),
list_sum(Ps, Total).
In your definition you are taking the area it seems, but I take the population in thousands:
country_population(C, P) :- country(C, _, P, _).
If you want to find the sum you can do a recursive predicate because now you have all the population counts (the second argument?) in a list.
list_sum([], 0).
list_sum([X|Xs], Sum) :-
list_sum(Xs, S0),
Sum is S0 + X.
Now you can query it to find the sum of the second arguments of country
?- total_population(P).
P = 68123.
But if you are using SWI-Prolog you can also use one library which is called "aggregate" because you can use it to aggregate for example the sum:
?- aggregate_all(sum(P), country(_, _, P, _), Total_population).
Total_population = 68123.
Related
Basically, I want to be able to check to see if at least one value in a list satisfies some predicate.
What I have so far:
need(x,y).
check_list(X,[H|T]) :-
need(H,X).
And so this works fine so long as I only have one value in the list. I'm not sure how to make it check the other values. When I try and use recursion I eventually find an element that satisfies the second predicate but it then goes back up the stack which will eventually cause it to be false.How can I make it 'break' essentially?
The backtracking you are seeing during recursion is Prolog attempting to find more ways for the predicate to succeed. This is a fundamental Prolog behavior and is what makes it useful. It seeks to find all of the solutions.
In your case, you only want to confirm one solution to the problem of, An element in the list that meets a specific criterion. For this, you could use a cut:
check_list(X, [H|_]) :-
need(X, H), !. % Don't backtrack after success
check_list(X, [_|T]) :-
check_list(X, T).
Or you could use once/1 which is specifically designed to handle cases where you only want a single solution:
check_list(X, [H|_]) :-
need(X, H).
check_list(X, [_|T]) :-
check_list(X, T).
check_list_once(X, L) :- once(check_list(X, L)).
Here is an example of what you can do.
I want to check is numbers are odd.
is_even(X) :-
X mod 2 =:= 0.
check_list(L, CL) :-
include(is_even, L, CL).
with result
?- check_list([1,2,3,4,5], L).
L = [2, 4].
?- check_list([1,3,5], L).
L = [].
You can use simple recursion:
need(x,y).
check_list(X,[H|T]) :-
( need(H,X) -> true;
check_list(X,T) ).
You can see in the examples below that this definition is deterministic:
?- check_list(y,[1,2,3]).
false.
?- check_list(y,[x,2,3]).
true.
?- check_list(y,[1,2,x]).
true.
?- check_list(Y,[1,2,x]).
Y = y.
?- check_list(Y,[1,2,3]).
false.
?- check_list(Y,[1,x,3]).
Y = y.
?- check_list(Y,[1,X,3]).
Y = y,
X = x.
?- check_list(Y,[1,2,3]), Y = x.
false.
?- check_list(Y,[1,2,3]), Y = y.
false.
?- check_list(Y,[1,2,3]).
false.
?- check_list(Y,[1,2,x]), Y = y.
Y = y.
Though if you want your queries to have uninstantiated variables e.g check_list(Y,[1,2,x]). and you add another fact need(x,z). Then:
?- check_list(Y,[1,2,x]).
Y = y.
Returns only one result and not Y = z. You could use if_/3 from library reif if you want a better definition of check_list/3.
The idea is as follows: Suppose I have a list P = [(1,0),(4,3)] or similar. I want to evaluate the polynomial that's defined by this list in the manner: 1X^0 + 4X^3.
To do this, I've written the following:
evaluate(P,X,Y) :- evaluate(P,X,Y,0).
evaluate([],_,S,S).
evaluate([P1,P2|Ps],X,Y,S) :-
S1 is S+P1*X^P2,
evaluate(Ps,X,Y,S1).
Which is supposed to succeed when Y is the sum of the polynomial P, given x=X.
The problem is that when I try and run this code, I get the error:
is/2: Arithmetic: `(',')/2' is not a function
But I have no idea where this is coming from or how to fix it.
I did try splitting the S1 is up in to its segments, but doing that didn't help.
EDIT: Ok, I found out that it's about the way the list is written down. How do I work with tuples in this way within the bounds of Prolog?
Your problem is that your data structure for each item in the list is a tuple as you noted and where you access the values of tuple in the list is not correct.
This
evaluate([P1,P2|Ps],X,Y,S) :-
should be
evaluate([(P1,P2)|Ps],X,Y,S) :-
Notice the parenthesis around P1,P2.
When I run with the change I get
?- evaluate([(1,0),(4,3)],5,Y).
Y = 501.
Also it is common to put the output arguments at the end,
evaluate_01(P,X,Y,0).
as
evaluate_01(P,X,0,Y).
and then change the other predicates as necessary.
evaluate_02(P,X,Y) :- evaluate_02(P,X,0,Y).
evaluate_02([],_,S,S).
evaluate_02([(P1,P2)|Ps],X,S,Y) :-
S1 is S+P1*X^P2,
evaluate_02(Ps,X,S1,Y).
As an interesting option, this can be done with maplist/3 and sumlist/2:
evaluate_poly(Poly, X, R) :-
maplist(evaluate_term(X), Poly, EvaluatedTerms),
sumlist(EvaluatedTerms, R).
evaluate_term(X, (Coeff, Power), TermValue) :-
TermValue is Coeff * (X ^ Power).
I have written the following in Prolog (I am using version 7.4.0-rc1), trying to define a predicate insertPermutation/2 which is true if and only if both arguments are lists, one a permutation of the other.
delete(X,[X|T],T). % Base case, element equals head.
delete(X,[A|B],[A|C]) :- delete(X,B,C). % And/or repeat for the tail.
insert(X,Y,Z) :- delete(X,Z,Y). % Inserting is deletion in reverse.
insertPermutation([],[]). % Base case.
insertPermutation([H|T],P) :- insertPermutation(Q,T), insert(H,Q,P). % P permutation of T, H inserted.
I have already been made aware that delete is not a good name for the above helper predicate. We are required to write these predicates, and we cannot use the built-in predicates. This is why I wrote the above code in this way, and I chose the name I did (because I first wrote it to delete an element). It is true if and only if the third argument is a list, equal to the list in the second argument with the first instance of the first argument removed.
The insertPermutation predicate recursively tests if P equals a permutation of the tail of the first list, with the head added in any position in the permutation. This way it works to the base case of both being empty lists.
However, the permutation predicate does not behave the way I want it to. For instance, to the query
?- insertPermutation([1,2,2],[1,2,3]).
Prolog does not return false, but freezes. To the query
?- insertPermutation(X,[a,b,c]).
Prolog responds with
X = [a, b, c] ;
X = [b, a, c] ;
X = [c, a, b] ;
X = [a, c, b] ;
X = [b, c, a] ;
X = [c, b, a] ;
after which it freezes again. I see these problems are related, but not how. Can someone point out what case I am missing?
Edit: Two things, this is homework, and I need to solve this problem using an insert predicate. I wrote this one.
The answer is to change the last line
% P permutation of T, H inserted.
insertPermutation([H|T],P) :-
insertPermutation(Q,T),
insert(H,Q,P).
% P permutation of T, H inserted.
insertPermutation(P,[H|T]) :-
insertPermutation(Q,T),
insert(H,Q,P).
The use cases only needed to check if the first element is a permutation of the latter, not the other way around (or vice versa). Anti-climatic, but the answer to my problem.
I was required to write a set of functions for problems in class. I think the way I wrote them was a bit more complicated than they needed to be. I had to implement all the functions myself, without using and pre-defined ones. I'd like to know if there are any quick any easy "one line" versions of these answers?
Sets can be represented as lists. The members of a set may appear in any order on the list, but there shouldn't be more than one
occurrence of an element on the list.
(a) Define dif(A, B) to
compute the set difference of A and B, A-B.
(b) Define cartesian(A,
B) to compute the Cartesian product of set A and set B, { (a, b) |
a∈A, b∈B }.
(c) Consider the mathematical-induction proof of the
following: If a set A has n elements, then the powerset of A has 2n
elements. Following the proof, define powerset(A) to compute the
powerset of set A, { B | B ⊆ A }.
(d) Define a function which, given
a set A and a natural number k, returns the set of all the subsets of
A of size k.
(* Takes in an element and a list and compares to see if element is in list*)
fun helperMem(x,[]) = false
| helperMem(x,n::y) =
if x=n then true
else helperMem(x,y);
(* Takes in two lists and gives back a single list containing unique elements of each*)
fun helperUnion([],y) = y
| helperUnion(a::x,y) =
if helperMem(a,y) then helperUnion(x,y)
else a::helperUnion(x,y);
(* Takes in an element and a list. Attaches new element to list or list of lists*)
fun helperAttach(a,[]) = []
helperAttach(a,b::y) = helperUnion([a],b)::helperAttach(a,y);
(* Problem 1-a *)
fun myDifference([],y) = []
| myDifference(a::x,y) =
if helper(a,y) then myDifference(x,y)
else a::myDifference(x,y);
(* Problem 1-b *)
fun myCartesian(xs, ys) =
let fun first(x,[]) = []
| first(x, y::ys) = (x,y)::first(x,ys)
fun second([], ys) = []
| second(x::xs, ys) = first(x, ys) # second(xs,ys)
in second(xs,ys)
end;
(* Problem 1-c *)
fun power([]) = [[]]
| power(a::y) = union(power(y),insert(a,power(y)));
I never got to problem 1-d, as these took me a while to get. Any suggestions on cutting these shorter? There was another problem that I didn't get, but I'd like to know how to solve it for future tests.
(staircase problem) You want to go up a staircase of n (>0) steps. At one time, you can go by one step, two steps, or three steps. But,
for example, if there is one step left to go, you can go only by one
step, not by two or three steps. How many different ways are there to
go up the staircase? Solve this problem with sml. (a) Solve it
recursively. (b) Solve it iteratively.
Any help on how to solve this?
Your set functions seem nice. I would not change anything principal about them except perhaps their formatting and naming:
fun member (x, []) = false
| member (x, y::ys) = x = y orelse member (x, ys)
fun dif ([], B) = []
| dif (a::A, B) = if member (a, B) then dif (A, B) else a::dif(A, B)
fun union ([], B) = B
| union (a::A, B) = if member (a, B) then union (A, B) else a::union(A, B)
(* Your cartesian looks nice as it is. Here is how you could do it using map: *)
local val concat = List.concat
val map = List.map
in fun cartesian (A, B) = concat (map (fn a => map (fn b => (a,b)) B) A) end
Your power is also very neat. If you call your function insert, it deserves a comment about inserting something into many lists. Perhaps insertEach or similar is a better name.
On your last task, since this is a counting problem, you don't need to generate the actual combinations of steps (e.g. as lists of steps), only count them. Using the recursive approach, try and write the base cases down as they are in the problem description.
I.e., make a function steps : int -> int where the number of ways to take 0, 1 and 2 steps are pre-calculated, but for n steps, n > 2, you know that there is a set of combinations of steps that begin with either 1, 2 or 3 steps plus the number combinations of taking n-1, n-2 and n-3 steps respectively.
Using the iterative approach, start from the bottom and use parameterised counting variables. (Sorry for the vague hint here.)
how can I accomplish this:
Give a tail-recursive definition for each of the following predicates.
power(X,Y,Z): XY=Z.
gcd(X,Y,Z): The greatest common divisor of X and Y is Z.
sum(L,Sum): Sum is the sum of the elements in L.
so far I have done this but not sure if that's correct
power(_,0,1) :- !.
power(X,Y,Z) :- Y1 is Y - 1,power(X,Y1,Z1),Z is X * Z1.
sum(void,0).
sum(t(V,L,R),S) :- sum(L,S1),sum(R,S2), S is V + S1 + S2.
These are not tail recursive. You can write tail recursive variants by using an accumulator, see this answer.
Your sum is over a tree, which is unusual, normally one would use a list. In Prolog [] is the empty list and [X|R] is the pattern for a nonempty list with the head X and the tail R.