SWI-Prolog. Check correctness of mathematical expression

Question

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.

Answer 1

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.