I try to check the correctness of student mathematical expression using Prolog (SWI-Prolog). So, for example if the student were asked to add three variable x, y, and z, and there's a rule that the first two variable that must be added are: x and y (in any order), and the last variable that must be added is z then I expect that prolog can give me true value if the student's answer is any of these:
x+y+z
(x+y)+ z
z+(x+y)
z+x+y
y+x+z
and many other possibilities.
I use the following rule for this checking:
addData :-
assert(variable(v1)),
assert(variable(v2)),
assert(variable(v3)),
assert(varName(v1,x)),
assert(varName(v2,y)),
assert(varName(v3,z)),
assert(varExpr(v1,x)),
assert(varExpr(v2,y)),
assert(varExpr(v3,z)).
add(A,B,R) :- R = A + B.
removeAll :- retractall(variable(X)),
retractall(varName(X,_)),
retractall(varExpr(X,_)).
checkExpr :-
% The first two variable must be x and y, in any combination
( (varExpr(v1,AExpr), varExpr(v2,BExpr));
(varExpr(v2,AExpr), varExpr(v1,BExpr))
),
add(AExpr, BExpr, R1),
% store the expression result as another variable, say v4
retractall(variable(v4)),
retractall(varName(v4, _)),
retractall(varExpr(v4, _)),
assert(variable(v4)),
assert(varName(v4, result)),
assert(varExpr(v4, R1)),
% add the result from prev addition with Z (in any combination)
( (varExpr(v3,CExpr), varExpr(v4,DExpr));
(varExpr(v4,CExpr), varExpr(v3,DExpr))
),
add(CExpr, DExpr, R2),
R2 = z + x + y. % will give me false
% R2 = z + (x + y). % will give me true
% Expected: both should give me true
checkCorrect :- removeAll,
addData,
checkExpr.
You should try to specify a grammar and write a parser for your expressions.
Avoid assert/retract, that make the program much more difficult to understand, and attempt instead to master the declarative model of Prolog.
Expressions are recursive data structures, using operators with known precedence and associativity to compose, and parenthesis to change specified precedence where required.
See this answer for a parser and evaluator, that accepts input from text. In your question you show expressions from code. Then you are using Prolog' parser to do the dirty work, and can simply express your requirements on the resulting syntax tree:
expression(A + B) :-
expression(A),
expression(B).
expression(A * B) :-
expression(A),
expression(B).
expression(V) :-
memberchk(V, [x,y,z]).
?- expression(x+y+(x+z*y)).
true .
edit: we can provide a template of what we want and let Prolog work out the details by means of unification:
% enumerate acceptable expressions
checkExpr(E) :-
member(E, [F = A + D, F = D + A]),
F = f,
A = c * N,
N = 1.8,
D = d.
And so on...
Test:
?- checkExpr(f=(c*1.8)+d).
true.
?- checkExpr(f=(c*1.8)+e).
false.
?- checkExpr(f=d+c*1.8).
true.
Related
I'm currently learning SWI-Prolog. I want to implement a function factorable(X) which is true if X can be written as X = n*b.
This is what I've gotten so far:
isTeiler(X,Y) :- Y mod X =:= 0.
hatTeiler(X,X) :- fail,!.
hatTeiler(X,Y) :- isTeiler(Y,X), !; Z is Y+1, hatTeiler(X,Z),!.
factorable(X) :- hatTeiler(X,2).
My problem is now that I don't understand how to end the recursion with a fail without backtracking. I thought the cut would do the job but after hatTeilerfails when both arguments are equal it jumps right to isTeiler which is of course true if both arguments are equal. I also tried using \+ but without success.
It looks like you add cuts to end a recursion but this is usually done by making rule heads more specific or adding guards to a clause.
E.g. a rule:
x_y_sum(X,succ(Y,1),succ(Z,1)) :-
x_y_sum(X,Y,Z).
will never be matched by x_y_sum(X,0,Y). A recursion just ends in this case.
Alternatively, a guard will prevent the application of a rule for invalid cases.
hatTeiler(X,X) :- fail,!.
I assume this rule should prevent matching of the rule below with equal arguments. It is much easier just to add the inequality of X and Y as a conditon:
hatTeiler(X,Y) :-
Y>X,
isTeiler(Y,X),
!;
Z is Y+1,
hatTeiler(X,Z),
!.
Then hatTeiler(5,5) fails automatically. (*)
You also have a disjunction operator ; that is much better written as two clauses (i drop the cuts or not all possibilities will be explored):
hatTeiler(X,Y) :- % (1)
Y > X,
isTeiler(Y,X).
hatTeiler(X,Y) :- % (2)
Y > X,
Z is Y+1,
hatTeiler(X,Z).
Now we can read the rules declaratively:
(1) if Y is larger than X and X divides Y without remainder, hatTeiler(X,Y) is true.
(2) if Y is larger than X and (roughly speaking) hatTeiler(X,Y+1) is true, then hatTeiler(X, Y) is also true.
Rule (1) sounds good, but (2) sounds fishy: for specific X and Y we get e.g.: hatTeiler(4,15) is true when hatTeiler(4,16) is true. If I understand correctly, this problem is about divisors so I would not expect this property to hold. Moreover, the backwards reasoning of prolog will then try to deduce hatTeiler(4,17), hatTeiler(4,18), etc. which leads to non-termination. I guess you want the cut to stop the recursion but it looks like you need a different property.
Coming from the original property, you want to check if X = N * B for some N and B. We know that 2 <= N <= X and X mod N = 0. For the first one there is even a built-in called between/2 that makes the whole thing a two-liner:
hT(X,B) :-
between(2, X, B),
0 is (X mod B).
?- hT(12,X).
X = 2 ;
X = 3 ;
X = 4 ;
X = 6 ;
X = 12.
Now you only need to write your own between and you're done - all without cuts.
(*) The more general hasTeiler(X,X) fails because is (and <) only works when the right hand side (both sides) is variable-free and contains only arithmetic terms (i.e. numbers, +, -, etc).
If you put cut before the fail, it will be freeze the backtracking.
The cut operation freeze the backtracking , if prolog cross it.
Actually when prolog have failed, it backtracks to last cut.
for example :
a:- b,
c,!,
d,
e,!,
f.
Here, if b or c have failed, backtrack do not freeze.
if d or f have failed, backtrack Immediately freeze, because before it is a cut
if e have failed , it can backtrack just on d
I hope it be useful
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 ?
...
Like the title states, I am trying to return the sum of the returned values from sub predicates but it's not working. Here is my code:
addlistnum([],[],X).
addlistnum(digits(Y,[A|T]),digits(F,[B|T]),X) :-
X is Y + F.
digits(Num, List) :-
digits(0, List, Num).
digits(Num, [], Num).
digits(N, [A|As], Num) :-
N1 is N * 10 + A,
digits(N1, As, Num).
The sub predicate works fine. It converts list to an integer. Now I want to sum the converted values.
Example:
?- digits(X,[3,3,3]).
X = 333. % works as expected
Building on that, addlistnum([3,3,3,3],[2,2,2],X) is supposed to produce X = 3555 (as 3555 is 3333 + 222), but I get false instead.
I also tried:
addlistnum([],[],X).
addlistnum([A|T],[B|T],X) :-
X is Y + F,
digits(Y,[A|T]),
digits(F,[B|T]).
It simply returns false, which gives no information about is wrong.
Problem is at these rules:
addlistnum([],[],X).
addlistnum(digits(Y,[A|T]),digits(F,[B|T]),X) :-
X is Y + F.
Second one is, "addition of two list is the addition of the integer conversion of these list":
addlistnum(A,B,X) :-
digits(NA,A),
digits(NB,B),
X is NA + NB.
first one is not necessary, "digits" for an empty list is zero, thus, this rule also covers "addition of two empty list is zero"
so I just got started with Prolog this semester, and got the homework to implement a pretty basic d(function, variable, derivative) which I did like this:
d(X,X,1) :- !.
d(C,X,0) :- atomic(C). %, (C \= X).
d(X**E,X,E*X**(E-1)).
d(U+V,X,A+B) :- d(U,X,A), d(V,X,B).
d(U-V,X,A-B) :- d(U,X,A), d(V,X,B).
d(U*V,X,DU*V+U*DV) :- d(U,X,DU), d(V,X,DV).
d(U/V,X,(DU*V-U*DV)/(V*V)) :- d(U,X,DU), d(V,X,DV).
I know this is not complete, but it covers all the tasks required in the exercise.
However,
?- d((x*x+2*x+3)/(3*x),x,R).
leads to
R = ((1*x+x*1+ (0*x+2*1)+0)* (3*x)- (x*x+2*x+3)* (0*x+3*1))/ (3*x* (3*x)).
which doesn't look pretty at all. is/2 unfortunately doesn't like my x as it is not a number...
Is there a simple solution to achieve a cleaner result?
I would rather see this as two separate problems:
First, get derivation right (you're probably getting close, depending on your concrete requirements).
Then, work on simplifying expressions on an algebraic level. Exploit algebraic identities, see if applying the laws of commutativity / associativity / distributivity on some subexpressions enable their rewriting into something equivalent (but simpler / more compact).
As a starting point, you may want to look at the somewhat related question "Replacing parts of expression in prolog".
Here's a simplistic sketch how you could do the simplification—using iwhen/2 to safeguard against insufficient instantiation:
expr_simplified(A, B) :-
iwhen(ground(A), xpr_simplr(A,B)).
xpr_simplr(A, B) :-
( atomic(A)
-> A = B
; ( A = X+0 ; A = 0+X ; A = 1*X ; A = X*1 )
-> xpr_simplr(X, B)
; ( A = 0*_ ; A = _*0 )
-> B = 0
; A = X+X
-> B = X*2
; A = X*X
-> B = X**2
; A = X**1
-> B = X
; A =.. [F|Xs0], % defaulty catch-all
maplist(xpr_simplr, Xs0, Xs),
B =.. [F|Xs]
).
Let's see what it does with the expression you gave. We apply expr_simplified/2 until we reach a fixed point:
?- A = ((1*x+x*1+(0*x+2*1)+0)*(3*x)-(x*x+2*x+3)*(0*x+3*1))/(3*x*(3*x)),
expr_simplified(A,B),
expr_simplified(B,C),
expr_simplified(C,D).
A = ((1*x+x*1+(0*x+2*1)+0)*(3*x)-(x*x+2*x+3)*(0*x+3*1))/(3*x*(3*x)),
B = ((x+x+(0+2))*(3*x)-(x**2+2*x+3)*(0+3))/(3*x)**2,
C = ((x*2+2)*(3*x)-(x**2+2*x+3)*3)/(3*x)**2,
D = C. % fixed point reached
As imperfect as the simplifier is, the expression got a lot more readable.
a possibility to get a number is to replace each instance of variable x with a value, visiting the derived tree. You should do writing a clause to match each binary operator, or use a generic visit, like
set_vars(E, Vs, Ev) :-
E =.. [F,L,R],
set_vars(L, Vs, Lv),
set_vars(R, Vs, Rv),
Ev =.. [F,Lv,Rv].
set_vars(V, Vs, N) :- memberchk(V=N, Vs).
set_vars(V, _, V).
that yields
?- d((x*x+2*x+3)/(3*x),x,R), set_vars(R,[x=5],E), T is E.
R = ((1*x+x*1+ (0*x+2*1)+0)* (3*x)- (x*x+2*x+3)* (0*x+3*1))/ (3*x* (3*x)),
E = ((1*5+5*1+ (0*5+2*1)+0)* (3*5)- (5*5+2*5+3)* (0*5+3*1))/ (3*5* (3*5)),
T = 0.29333333333333333
but, there is a bug in your first clause, that once corrected, will allow to evaluate directly the derived expression:
d(X,V,1) :- X == V, !.
...
now, we can throw away the utility set_vars/3, so
?- d((T*T+2*T+3)/(3*T),T,R), T=8, V is R.
T = 8,
R = ((1*8+8*1+ (0*8+2*1)+0)* (3*8)- (8*8+2*8+3)* (0*8+3*1))/ (3*8* (3*8)),
V = 0.3177083333333333.
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!