I have a variable X that may contain multiple values: X = 1; X = 4; X = 7...
These values map to a list that contain x,y,z, or w. Each one of these value/list pairs are split into multiple facts, so I could have:
map(2,[x,y]).
map(3,[x]).
map(9,[y,w]).
I'm trying to write a program that, given X, I can look up these lists and count how many occurences of x,y,z, or w there are.
This is my attempt:
count(A,B,C,D,X) :- A = 0, B = 0, C = 0, D = 0,
check_list(X,x,A),
check_list(X,y,B),
check_list(X,z.C),
check_list(X,w,D).
check_list(X,Element,Counter) :-
map(X, List),
member(List, Element),
S is Counter + 1,
Counter = S.
The idea behind my program is I call check_list to check if there is a member that contains x,y,z,w for every possible value of X. If there is that member, I will increment the counter. I then want the values of A,B,C,D to have A = number of occurrences of x, B = number of occurrences of y, etc etc.
You are using Prolog variables wrong. Variables cannot change their values once they are instantiated unless Prolog backtracks to a choice-point previous to the instantiation. For example, in the rule for count/5 you unify A with zero and then you expect that satisfying check_list(X,x,A) will bind A to the number of occurrences of x, but A is not a free variable at that point.
So, you have to remove A = 0, ..., D = 0 from the first rule.
Next, you need a predicate that can be used to find the number of occurrences of an element in a list. You can use findall/3 for that:
occurrences(X, List, N):- findall(_, member(X, List), O), length(O, N).
Or you can write it yourself:
occurrences(_, [], 0).
occurrences(X, [X|Tail], N):-!, occurrences(X, Tail, N1), N is N1 + 1.
occurrences(X, [_|Tail], N):-occurrences(X, Tail, N).
Related
After doing some Prolog in uni and doing some exercises I decided to go along somewhat further although I got to admit I don't understand recursion that well, I get the concept and idea but how to code it, is still a question for me. So that's why I was curious if anyone knows how to help tackle this problem.
The idea is given a number e.g. 45, check whether it is possible to make a list starting with 1 going n+1 into the list and if the sum of the list is the same as the given number.
So for 45, [1,2,3,4,5,6,7,8,9] would be correct.
So far I tried looking at the [sum_list/2][1] implemented in Prolog itself but that only checks whether a list is the same as the number it follows.
So given a predicate lijstSom(L,S) (dutch for listSum), given
?- lijstSom(L, 45)
L = [1,2,3,4,5,6,7,8,9];
False
My Idea was something along the line of for example if S = 45, doing steps of the numbers (increasing by 1) and subtracting it of S, if 0 is the remainder, return the list, else return false.
But for that you need counters and I find it rather hard to grasp that in recursion.
EDIT:
Steps in recursion.
Base case empty list, 0 (counter nr, that is minus S), 45 (S, the remainder)
[1], 1, 44
[1,2], 2, 42
[1,2,3], 3, 39
I'm not sure how to read the example
?- lijstSom(L, 45)
L = [1,2,3,4,5,6,7,8,9],
False
...but think of the predicate lijstSom(List, Sum) as relating certain lists of integers to their sum, as opposed to computing the sum of lists of integers. Why "certain lists"? Because we have the constraint that the integers in the list of integers must be monotonically increasing in increments of 1, starting from 1.
You can thus ask the Prolog Processor the following:
"Say something about the relationship between the first argument of lijstSom/2 and the second argument lijstSom/2 (assuming the first is a list of monotonically increasing integers, and the second an integer):
lijstSom([1,2,3], Sum)
... should return true (because yes, there is at least one solution) and give Sum = 6 (because it constructs the solution, too ... we are some corner of Construtivism here.
lijstSom(L, 6)
... should return true (because yes, there is at least one solution) and give the solution [1,2,3].
lijstSom([1,2,3], 6)
... should return true (because yes, [1,2,3] has a sum 6); no further information is needed.
lijstSom(L, S)
... should an infinite series of true and pairs of solution ("generate the solutions").
L = [1], S = 1;
L = [1,2], S = 3;
L = [1,2,3], S = 6;
...
lijstSom([1,2,3], 7)
...should return false ("fail") because 7 is not in a relation lijstSom with [1,2,3] as 7 =/= 1+2+3.
One might even want things to have Prolog Processor say something interesting about:
lijstSom([1,2,X], 6)
X = 3
or even
lijstSom([1,2,X], S)
X = 3
S = 6
In fact, lijstSom/2 as near to mathematically magical as physically possible, which is to say:
Have unrestricted access to the full table of list<->sum relationships floating somewhere in Platonic Math Space.
Be able to find the correct entry in seriously less than infinite number of steps.
And output it.
Of course we are restricted to polynomial algorithms of low exponent and finite number of dstinguishable symbols for eminently practical reasons. Sucks!
So, first define lijstSom(L,S) using an inductive definition:
lijstSom([a list with final value N],S) ... is true if ... lijstSom([a list],S-N and
lijstSom([],0) because the empty list has sum 0.
This is nice because it gives the recipe to reduce a list of arbitrary length down to a list of size 0 eventually while keeping full knowledge its sum!
Prolog is not good at working with the tail of lists, but good with working with the head, so we cheat & change our definition of lijstSom/2 to state that the list is given in reverse order:
lijstSom([3,2,1], 6)
Now some code.
#= is the "constain to be equal" operator from library(clpfd). To employ it, we need to issue use_module(library(clpfd)). command first.
lijstSom([],0).
lijstSom([K|Rest],N) :- lijstSom([Rest],T), T+K #= N.
The above follows the mathematical desiderate of lijstSom and allows the Prolog Processor to perform its computation: in the second clause, it can compute the values for a list of size A from the values of a list of size A-1, "falling down" the staircase of always decreasing list length until it reaches the terminating case of lijstSom([],0)..
But we haven't said anything about the monotonically decreasing-by-1 list.
Let's be more precise:
lijstSom([],0) :- !.
lijstSom([1],1) :- ! .
lijstSom([K,V|Rest],N) :- K #= V+1, T+K #= N, lijstSom([V|Rest],T).
Better!
(We have also added '!' to tell the Prolog Processor to not look for alternate solutions past this point, because we know more about the algorithm than it will ever do. Additionally, the 3rd line works, but only because I got it right after running the tests below and having them pass.)
If the checks fail, the Prolog Processor will says "false" - no solution for your input. This is exactly what we want.
But does it work? How far can we go in the "mathematic-ness" of this eminently physical machine?
Load library(clpfd) for constraints and use library(plunit) for unit tests:
Put this into a file x.pl that you can load with [x] alias consult('x') or reload with make on the Prolog REPL:
:- use_module(library(clpfd)).
lijstSom([],0) :-
format("Hit case ([],0)\n"),!.
lijstSom([1],1) :-
format("Hit case ([1],1)\n"),!.
lijstSom([K,V|Rest],N) :-
format("Called with K=~w, V=~w, Rest=~w, N=~w\n", [K,V,Rest,N]),
K #= V+1,
T+K #= N,
T #> 0, V #> 0, % needed to avoid infinite descent
lijstSom([V|Rest],T).
:- begin_tests(listsom).
test("0 verify") :- lijstSom([],0).
test("1 verify") :- lijstSom([1],1).
test("3 verify") :- lijstSom([2,1],3).
test("6 verify") :- lijstSom([3,2,1],6).
test("0 construct") :- lijstSom(L,0) , L = [].
test("1 construct") :- lijstSom(L,1) , L = [1].
test("3 construct") :- lijstSom(L,3) , L = [2,1].
test("6 construct") :- lijstSom(L,6) , L = [3,2,1].
test("0 sum") :- lijstSom([],S) , S = 0.
test("1 sum") :- lijstSom([1],S) , S = 1.
test("3 sum") :- lijstSom([2,1],S) , S = 3.
test("6 sum") :- lijstSom([3,2,1],S) , S = 6.
test("1 partial") :- lijstSom([X],1) , X = 1.
test("3 partial") :- lijstSom([X,1],3) , X = 2.
test("6 partial") :- lijstSom([X,2,1],6) , X = 3.
test("1 extreme partial") :- lijstSom([X],S) , X = 1, S = 1.
test("3 extreme partial") :- lijstSom([X,1],S) , X = 2, S = 3.
test("6 extreme partial") :- lijstSom([X,2,1],S) , X = 3, S = 6.
test("6 partial list") :- lijstSom([X|L],6) , X = 3, L = [2,1].
% Important to test the NOPES
test("bad list", fail) :- lijstSom([3,1],_).
test("bad sum", fail) :- lijstSom([3,2,1],5).
test("reversed list", fail) :- lijstSom([1,2,3],6).
test("infinite descent from 2", fail) :- lijstSom(_,2).
test("infinite descent from 9", fail) :- lijstSom(_,9).
:- end_tests(listsom).
Then
?- run_tests(listsom).
% PL-Unit: listsom ...................... done
% All 22 tests passed
What would Dijkstra say? Yeah, he would probably bitch about something.
I am fairly new to functional programming and I do not understand my error here. I am trying to make a function that takes an integer list and returns both the sum of the even elements and the sum of the odd elements. The error I am getting is in line 1, and it states: "Error: right-hand-side of clause doesn't agree with function result type [overload conflict] ...". I don't understand the error, and I would appreciate any help in understanding my error.
fun add(nil) = 0
| add([x]) = x
| add(x :: xs) =
let
val evenList = xs;
val oddList = x :: xs
in
(hd evenList + add(tl(tl(evenList))), hd oddList + add(tl(tl(oddList))))
end;
The reason for the type error is that the function should return a pair, but your base cases don't.
I suspect you got to that code by thinking about skipping every other element, dividing the list by skipping.
There's a different way to approach this.
Consider the list [a,b,c,d].
Counting from 1, the elements are numbered
1 2 3 4
a b c d
Now consider the positions in the tail of the list.
They are
1 2 3
b c d
That is, odd positions in the tail are even positions in the entire list, and even positions in the tail are odd in the entire list.
This means that if we recursively compute "odds and evens" in the tail, we will get the sums from the tail, where its "odds" is our "evens", and if we add our head to the tail's "evens", we will get the "odds" we want.
All we need now is a good base case – and the sums of an empty list must be (0, 0).
Something like this:
fun add [] = (0,0)
| add (x::xs) = case add xs of
(odds, evens) => (x + evens, odds)
or, you can deconstruct the recursive result with a let-binding instead of case:
fun add [] = (0,0)
| add (x::xs) = let val (odds, evens) = add xs
in
(x + evens, odds)
end
I have the following file with the predicate attends which symbolizes that every student attends a certain course
(the first argument : Student_ID, the second argument : Course_ID).
attends(476, c216).
attends(478, c216).
attends(484, c216).
attends(487, c216).
attends(491, c216).
What I want to do is create a predicate-function which creates an examination schedule consisting of 3 lists (A,B,C) with courses.
Each list symbolizes a week. Then for instance to find the optimal schedule in order it suits most students, it prints out all the different permutations of courses in pairs of 3-3-2:
The list A is week one with courses [c204,c209,c210] in the first case bellow.
List B is week 2 etc...
?- schedule(A,B,C).
A = [c204,c209,c210],
B = [c212,c214,c216],
C = [c217,c218];
A = [c204,c209,c210],
B = [c212,c214,c216],
C = [c218,c217];
Problem 1:
So how can I take the attends/2 predicates and convert only the second argument to a List, in a manner that the list will contain all the courses that have been declared.
For example: L = [c212,c213...].
Problem 2:
The permutations will be done with a custom function called k_permutation/3:
delete(E,L,NL):-
append(L1,[E|L2],L),
append(L1,L2,NL).
k_permutation(0,_,[]).
k_permutation(K,L1,[X|T2]) :-
K > 0,
K1 is K - 1,
delete(X,L1,L2),
k_permutation(K1,L2,T2).
But for some reason this custom function (k_permutation/3) runs for infinite time. Is there something wrong with the functions recursion?
How should the function be used?
Well as for problem 1 the easy way would be:
collect_courses(L1):- findall(Course, attends(_,Course), L), sort(L,L1).
L will have all courses that appear in attends/2, so it will have duplicates, that's the reason we're using sort/2 which removes duplicates.
As for problem 2, first of all Swi-Prolog already has a definition of delete/3 predicate so I suggest that you rename it. Apart from that the k_permutations/2 works fine:
?- k_permutation(2,[1,2,3],L).
L = [1, 2] ;
L = [1, 3] ;
L = [2, 1] ;
L = [2, 3] ;
L = [3, 1] ;
L = [3, 2] ;
false.
I've been implementing higher order functions recursively with .foldRight() like any, all, and takeWhile as practice, but dropWhile has been elusive. _Collections.kt has the imperative way but I couldn't convert it to a recursive structure.
For reference, this is takeWhile
fun takeWhile(list:List<Int>, func:(Int) -> Boolean):List<Int> = list.foldRight(emptyList(),
{ next:Int, acc:List<Int> -> if (func(next)) acc.plus(next) else emptyList() })
First, let's outline the idea of the solution.
With foldRight, you can only process the items one by one from right to left, maintaining an accumulator.
The problem is, for an item at position i, the dropWhile logic makes a decision whether to include the item into the result or not based on whether there is an item at position j <= i that does not satisfy the predicate (include if yes). This means you cannot simply maintain the result items: for some items you already processed, you don't know if they should actually be included.
Example:
(we're processing the items right-to-left, so the prefix is unknown to us)
... (some unknown items) ... ... ... ... a b c d <--- (right-to-left)
predicate satisfied: T T F T
As we discover more items on the left, there are two possibilities:
We found the beginning of the sequence, and there were no items that gave F on the predicate:
(the sequence start) y z a b c d <--- (right-to-left)
predicate satisfied: T T T T F T
-------
drop
In this case, the prefix y z a b should be dropped.
We found an item that does not satisfy the predicate:
... (some unknown items) ... w z a b c d <--- (right-to-left)
predicate satisfied: F T T T F T
-------
include
Only at this point we know for sure that we need to include the items w z a b, we could not do that earlier because there could be the beginning of the sequence instead of item w, and then we should have dropped z a b.
But note that in both cases we are certain that the items c d are to be included into the result: that's because they have c with F predicate in front of them.
Given this, it becomes clear that, when processing the items right-to-left, you can maintain a separate list of items that are not certain to be included into the result and are either to be dropped or to be included when a false predicate result is encountered, together with the item that gave such false result.
My implementation:
I used a pair of two lists for the accumulator, where the first list is for the items that are certain to be included, and the second one for those which are not.
fun <T> List<T>.myDropWhile(predicate: (T) -> Boolean) =
foldRight(Pair(emptyList<T>(), emptyList<T>())) { item, (certain, uncertain) ->
if (predicate(item))
Pair(certain, uncertain + item) else
Pair(certain + uncertain + item, emptyList())
}.first.reversed()
Example:
val ints = listOf(0, 0, 0, 1, 0, 2, 3, 0, 0, 4)
println(ints.myDropWhile { it == 0 }) // [1, 0, 2, 3, 0, 0, 4]
See: runnable demo of this code with more tests.
Note: copying a read-only list by doing uncertain + item or certain + uncertain + item in each iteration gives O(n^2) worst-case time complexity, which is impractical. Using mutable data structures gives O(n) time.
I'm new in Prolog and I have some problem understanding how the recursion works.
The think I want to do is to create a list of numbers (to later draw a graphic).
So I have this code :
nbClassTest(0, _).
nbClassTest(X, L) :-
numberTestClass(A,X),
append([A], L, L),
X is X - 1,
nbClassTest(X, L).
But it keeps giving me 'false' as an answer and I don't understand why it doesn't fill the list. It should end if X reaches 0 right?
The numberTestClass(A,X), gives me a number (in the variable A) for some X as if it was a function.
You should build the list without appending, because it's rather inefficient.
This code could do:
nbClassTest(0, []).
nbClassTest(X, [A|R]) :-
numberTestClass(A, X),
X is X - 1,
nbClassTest(X, R).
or, if your system has between/3, you can use an 'all solutions' idiom:
nbClassTest(X, L) :-
findall(A, (between(1, X, N), numberTestClass(A, X)), R),
reverse(R, L).
the problem is that you use the same variable for the old and the new list. right now your first to append/3 creates a list of infinite length consisting of elements equal to the value of A.
?-append([42],L,L).
L = [42|L].
?- append([42],L,L), [A,B,C,D|E]=L.
L = [42|L],
A = B, B = C, C = D, D = 42,
E = [42|L].
then, if the next A is not the same with the previous A it will fail.
?- append([42],L,L), append([41],L,L).
false.
there is still on more issue with the code; your base case has an non-instantiated variable. you might want that but i believe that you actually want an empty list:
nbClassTest(0, []).
nbClassTest(X, L) :-
numberTestClass(A,X),
append([A], L, NL),
X is X - 1,
nbClassTest(X, NL).
last, append/3 is kinda inefficient so you might want to avoid it and build the list the other way around (or use difference lists)
It fails because you use append in wrong way
try
nbClassTest(0, _).
nbClassTest(X, L) :-
numberTestClass(A,X),
append([A], L, Nl),
X is X - 1,
nbClassTest(X, Nl).
append concatenate 2 lists so there is no such list which after adding to it element still will be same list.