I'm trying to define the division in prolog using the remainder theorem and the well-ordering principle.
I've got thus far:
less(0, s(0)).
less(0, s(B)) :- less(0, B).
less(s(A), s(s(B))) :- less(A, s(B)).
add(A,0,A) :- nat(A).
add(A,s(B),s(C)) :- add(A,B,C). % add(A,B+1,C+1) = add(A,B,C)
add2(A,0,A).
add2(A,s(B),s(C)) :- add2(A,B,C). % add(A,B+1,C+1) = add(A,B,C)
times(A,0,0).
times(A,s(B),X) :- times(A,B,X1),
add(A,X1,X).
eq(0,0).
eq(s(A), s(B)) :- eq(A, B).
% A / B = Q (R) => A = B * Q + R
div(A, B, Q, R) :- less(R, B), eq(A, add(times(Q, R), R)).
But the definition of div is somehow wrong. Could someone please give me a hint?
PS: I shouldn't be using eq, but I couldn't get is or = to work.
In SWI-Prolog, you can try ?- gtrace, your_goal. to use the graphical tracer and see what goes wrong. Instead of eq(A, add(times(Q, R), R)), you should write for example: times(Q, R, T), add(T, R, A), since you want to use the "times/3" and "add/3" predicates, instead of just calling the "eq/2" predicate with a compound term consisting of "add/2" and "times/2" as its second argument. There are other problems with the code as well, for example, the definition of nat/1 is missing, but I hope this helps somewhat.
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 brand new to prolog and I feel like there is a concept that I am failing to understand, which is preventing me from grasping the concept of recursion in prolog. I am trying to return S, which is the sum of the square of each digit, taken as a list from an integer that is entered by the user in a query. E.g The user enters 12345, I must return S = (1^2)+(2^2)+(3^2)+(4^2)+(5^2) = 55.
In my program below, I understand why the each segment of the calculation of S is printed multiple time as it is part of the recursive rule. However, I do not understand how I would be able to print S as the final result. I figured that I could set a variable = to the result from sos in the second rule and add it as a parameter for intToList but can't seem to figure this one out. The compiler warns that S is a singleton variable in the intToList rule.
sos([],0).
sos([H|T],S) :-
sos(T, S1),
S is (S1 + (H * H)),
write('S is: '),write(S),nl.
intToList(0,[]).
intToList(N,[H|T]) :-
N1 is floor(N/10),
H is N mod 10,
intToList(N1,T),
sos([H|T],S).
The issue with your original code is that you're trying to handle your call to sos/2 within your recursive clause for intToList/2. Break it out (and rename intToList/2 to something more meaningful):
sosDigits(Number, SoS) :-
number_digits(Number, Digits),
sos(Digits, SoS).
Here's your original sos/2 without the write, which seems to work fine:
sos([], 0).
sos([H|T], S) :-
sos(T, S1),
S is (S1 + (H * H)).
Or better, use an accumulator for tail recursion:
sos(Numbers, SoS) :-
sos(Numbers, 0, SoS).
sos([], SoS, SoS).
sos([X|Xs], A, SoS) :-
A1 is A + X*X,
sos(Xs, A1, SoS).
You can also implement sos/2 using maplist/3 and sumlist/2:
square(X, S) :- S is X * X.
sos(Numbers, SoS) :- maplist(square, Numbers, Squares), sumlist(Squares, SoS).
Your intToList/2 needs to be refactored using an accumulator to maintain correct digit order and to get rid of the call to sos/2. Renamed as explained above:
number_digits(Number, Digits) :-
number_digits(Number, [], Digits).
number_digits(Number, DigitsSoFar, [Number | DigitsSoFar]) :-
Number < 10.
number_digits(Number, DigitsSoFar, Digits) :-
Number >= 10,
NumberPrefix is Number div 10,
ThisDigit is Number mod 10,
number_digits(NumberPrefix, [ThisDigit | DigitsSoFar], Digits).
The above number_digits/2 also handles 0 correctly, so that number_digits(0, Digits) yields Digit = [0] rather than Digits = [].
You can rewrite the above implementation of number_digits/3 using the -> ; construct:
number_digits(Number, DigitsSoFar, Digits) :-
( Number < 10
-> Digits = [Number | DigitsSoFar]
; NumberPrefix is Number div 10,
ThisDigit is Number mod 10,
number_digits(NumberPrefix, [ThisDigit | DigitsSoFar], Digits)
).
Then it won't leave a choice point.
Try this:
sos([],Accumulator,Accumulator).
sos([H|T],Accumulator,Result_out) :-
Square is H * H,
Accumulator1 is Accumulator + Square,
sos(T,Accumulator1,Result_out).
int_to_list(N,R) :-
atom_chars(N,Digit_Chars),
int_to_list1(Digit_Chars,Digits),
sos(Digits,0,R).
int_to_list1([],[]).
int_to_list1([Digit_Char|Digit_Chars],[Digit|Digits]) :-
atom_number(Digit_Char,Digit),
int_to_list1(Digit_Chars,Digits).
For int_to_list I used atom_chars which is built-in e.g.
?- atom_chars(12345,R).
R = ['1', '2', '3', '4', '5'].
And then used a typical loop to convert each character to a number using atom_number e.g.
?- atom_number('2',R).
R = 2.
For sos I used an accumulator to accumulate the answer, and then once the list was empty moved the value in the accumulator to the result with
sos([],Accumulator,Accumulator).
Notice that there are to different variables for the accumulator e.g.
Accumulator1 is Accumulator + Square,
sos(T,Accumulator1,Result_out).
this is because in Prolog variables are immutable, so one can not keep assigning new values to the same variable.
Here are some example runs
?- int_to_list(1234,R).
R = 30.
?- int_to_list(12345,R).
R = 55.
?- int_to_list(123456,R).
R = 91.
If you have any questions just ask in the comments under this answer.
I'm writing a prolog predicate which replace an element with another in a given atom. The predicate I wrote is like this:
replace_var(Expr0, Var, Val, Expr) :-
Expr0 =.. Chars,
chars_replaced(Chars, Rs),
Expr =.. Rs.
chars_replaced(Chars, Rs) :-
maplist(rep, Chars, Rs).
rep(Var,Val).
rep(C, C) :- dif(C,var).
The result I want it to return is something like:
-?replace_var(hello, l, w, X).
X = hewwo.
The problem is about the rep() predicate. I don't know how to write it or how to pass the Val and Var to the predicate.
Please give me some suggestions. Thanks!
this is wrong
Expr0 =.. Chars
you need instead
atom_chars(Expr0, Chars)
and this one really puzzle me
rep(Var,Val).
rep(C, C) :- dif(C,var).
what do you mean, specially the second one ?
anyway, the whole could be
replace_var(Expr0, Var, Val, Expr) :-
atom_chars(Expr0, Cs),
maplist(rep(Var, Val), Cs, Ts),
atom_chars(Expr, Ts).
rep(C, T, C, T).
rep(_, _, C, C).
disclaimer: untested code
I want to ask about prolog programming. I have:
byCar(auckland,hamilton).
byCar(valmont,metz).
byTrain(metz,frankfurt).
byPlane(frankfurt,bangkok).
byPlane(bangkok,auckland).
travell(From,To,go(From,To,car)) :- byCar(From,To).
travell(From,To,go(From,To,train)) :- byTrain(From,To).
travell(From,To,go(From,To,plane)) :- byPlane(From,To).
travell(From,To,go(From,Step,Via,Go)) :- travell(From,Step,go(From,Step,Via)),travell(Step,To,Go).
Then I ask in winprolog:
?- travell(valmont,hamilton,Go).
It answered
Go = go(valmont,metz,car,go(metz,frankfurt,train,go(frankfurt,bangkok,plane,go(bangkok,auckland,plane,go(auckland,hamilton,car)))))
My question is, is it possible to count how many 'go' it have? like my example before it has 5 'go'
would be easier if you could uniform go/3 and go/4 arity. Anyway, try
count_last_arg(Struct,Func,Count) :-
count_last_arg(Struct,Func,0,Count).
count_last_arg(Struct,Func,N,Count) :-
Struct =.. L, last(L, LL), LL =.. [Func|_]
-> N1 is N+1,
count_last_arg(LL,Func,N1,Count)
; Struct =.. [Func|_]
-> Count is N+1
; Count is N.
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.