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i have ERROR in this code in prolog can you help me?
div(X):- X mod 3 =:= 0.
div(X):- X mod 3 =:= 0 ...
div(X):- X mod 3 =:=1 ...
2 mod 3 is not 0 or 1, so there is no way to do div(2), so it fails.
Maybe you wanted
X mod 3 =:= 0 % remainder 0
X mod 3 =\= 0 % remainder not 0
?
Whenever I run my code, I get an error that the arguments are not instantiated.
ads(X,Z):- mod(X,2) =:= 0, Z is Z+X.
ads(X,Z) :- mod(N,2) =\= 0,Z is Z.
sum_of_nums(0,0,0).
sum_of_nums(X,Y,Z) :- X=<Y, ad(X,Z), sum_of_nums(X+1,Y,Z).
I want to be able to add numbers from X to Y but only the even ones. Don't know why it doesn't work.
First, there are some tiny errors, your compiler should indicate, like the unnecessary singleton variables in the second clause. After repairing this, and replacing ads by ad we have:
ad(X,Z):- mod(X,2) =:= 0, Z is Z+X.
ad(X,Z) :- mod(X,2) =\= 0,Z is Z.
sum_of_nums(0,0,0).
sum_of_nums(X,Y,Z) :- X=<Y, ad(X,Z), sum_of_nums(X+1,Y,Z).
?- sum_of_nums(1,2,S).
error(instantiation_error,(is)/2).
To locate this error, I will insert some goals false such that the
resulting program still produces this instantiation error.
ad(X,Z):- mod(X,2) =:= 0, Z is Z+X, false.
ad(X,Z) :- false, mod(X,2) =\= 0,Z is Z.
sum_of_nums(0,0,0) :- false.
sum_of_nums(X,Y,Z) :- X=<Y, ad(X,Z), false, sum_of_nums(X+1,Y,Z).
?- sum_of_nums(1,2,S).
error(instantiation_error,(is)/2).
Therefore, you have an error in this part already.
It's the Z is Z+X. On the right hand side of (is)/2 you always
need to have variables that are instantiated (known). And Z is not
known.
Variables are a bit different in Prolog. You cannot reassign them.
And also, writing this in Prolog directly doesn't really show what the
language is good at.
sum_of(X, Y, 0) :-
X > Y.
sum_of(X1, Y, Z1) :-
X1 =< Y,
( X1 mod 2 =:= 0 -> S = X1 ; S = 0 ),
X2 is X1+1,
sum_of(X2, Y, Z2),
Z1 is Z2+S.
A more idiomatic way would be:
?- between(1,6,N).
N = 1
; N = 2
; N = 3
; N = 4
; N = 5
; N = 6.
?- between(1,6,N), N mod 2 =:= 0.
N = 2
; N = 4
; N = 6.
?- findall(N, ( between(1,6,N), N mod 2 =:= 0 ), Ns).
Ns = [2,4,6].
?- findall(N, ( between(1,6,N), N mod 2 =:= 0 ), Ns), sum_list(Ns,Sum).
Ns = [2,4,6], Sum = 12.
sum_of(X,Y,Sum) :-
findall(N, ( between(X,Y,N), N mod 2 =:= 0 ), Ns),
sum_list(Ns,Sum).
Getting error in Prolog.
I want the 1st list to have positive numbers divisible by 2 and 2nd list to have negative number divisible by 3.
This is the code I've written so far but I'm getting the output as 'false'
split([],[],[]).
split([X|L],[X|L1],L2):- X >=0, X mod 2 =:= 0, split(L,L1,L2).
split([X|L],L1,[X|L2]):- X<0, X mod 3 =:= 0, split(L,L1,L2).
test case - split([1,-2,-3,4,-17,3],L1,L2).
output - false
Can someone please tell me where am I going wrong?
You did not specify a clause for values that are not positive and dividable by two, or negative and dividable by three. In that case you probably do not want to include it in any list, so:
split([],[],[]).
split([X|L],[X|L1],L2):-
X >= 0,
X mod 2 =:= 0,
split(L,L1,L2).
split([X|L],L1,[X|L2]):-
X < 0,
X mod 3 =:= 0,
split(L,L1,L2).
split([X|L], L1, L2) :-
\+ ((X >= 0, X mod 2 =:= 0); (X < 0, X mod 3 =:= 0)),
split(L, L1, L2).
I am trying to print all the even numbers from 1 to 10 using Prolog, and here is what I have tried:
printn(10,0):- write(10),!.
printn(X,Sum):-
( X mod 2 =:= 0 -> Sum is X+Sum, Next is X+1, nl, printn(Next);
Next is X+1, printn(Next) ).
but it returns false.
You don't need to create the list with the numbers from the beginning, it is better to examine numbers once:
print(X,Y):-print_even(X,Y,0).
print_even(X, X, Sum):-
( X mod 2 =:= 0 -> Sum1 is X+Sum;
Sum1 = Sum
), print(Sum1).
print_even(X, Y, Sum):-
X<Y, Next is X+1,
( X mod 2 =:= 0 -> Sum1 is X+Sum, print_even(Next, Y, Sum1);
print_even(Next, Y, Sum)
).
Keep in mind that in Prolog Sum is Sum+1 always fails you need to use a new variable e.g Sum1.
Example:
?- print(1,10).
30
true ;
false.
The most useful way of obtaining Prolog output is to capture the solution in a variable, either individually through backtracking, or in a list. The idea of "printing", which carries over from using other languages allows for formatting, etc, but is not considered the best way to express a solution.
In Prolog, you want to express your problem as a relation. For example, we might say, even_with_max(X, Max) is true (or succeeds) if X is an even number less than or equal to Max. In Prolog, when reasoning with integers, the CLP(FD) library is what you want to use.
:- use_module(library(clpfd)).
even_up_to(X, Max) :-
X in 1..Max,
X mod 2 #= 0, % EDIT: as suggested by Taku
label([X]).
This will yield:
3 ?- even_up_to(X, 10).
X = 2 ;
X = 4 ;
X = 6 ;
X = 8 ;
X = 10.
If you then want to collect into a list, you can use: findall(X, even_up_to(X), Evens).
What error do you have? Here is my solution:
Create list [1...10]
Filter it, excluding odd numbers
Sum elements of the list
Code:
sumList([], 0).
sumList([Head|Tail], Sum) :-
sumList(Tail, Rest),
Sum is Head + Rest.
isOdd(X) :-
not((X mod 2) =:= 0).
sumOfEvenNums(A, B, Out) :-
numlist(A, B, Numbers),
exclude(isOdd, Numbers, Even_numbers),
sumList(Even_numbers, Out).
Now you can call sumOfEvenNums(1, 10, N)
In ECLiPSe, you can write with iterator:
sum_even(Sum):-
( for(I,1,10),fromto(0,In,Out,Sum)
do
(I mod 2 =:= 0 -> Out is In + I;Out is In)
)
With library(aggregate):
evens_upto(Sum) :-
aggregate(sum(E), (between(1, 10, E), E mod 2 =:= 0), Sum).
Thanks to #CapelliC for the inspiration.
How to write mod/3 function for successor arithmetic (Peano's Numbers) in prolog?
Think of it this way:
If you want to find the mod of 10 and 4, you divide 10 by 4 and return the reminder. But since divide is multiple subtractions we will use the multiple subtraction logic here.
For Example: 10 mod 4 is same as 10-4 mod 4 which is 6 mod 4 which is again same as 6-4 mod 4 = 2 mod 4. Since the first element (2) is less than the second (4), we terminate the program here and return first element (2).
mod(_, 0, 0). Says that anything mod 0 is 0.
mod(0, _ , 0). Says that 0 mod anything is 0.
mod(X, s(0), 0). Says that anything mod 1 is 0.
This is the tricky part:
mod(A, B, N) :- minus(A, B, R), (R #< B -> N = R ; mod(R, B, N)). This uses the multiple minus logic. If first removes second from first and then checks if first is smaller that the second. If yes, recursively call the mod function. If not return the first element.
s(0).
s(X):- X.
plus(0, Y, Y).
plus(s(X), Y, s(Z)):- plus(X , Y , Z).
minus(A, B, C) :- plus(C, B, A).
mod(_, 0, 0).
mod(0, _ , 0).
mod(X, s(0), 0).
mod(A, B, N) :- minus(A, B, R), (R #< B -> N = R ; mod(R, B, N)).
Thank you #Toby for the edit request.