Prolog solving prefix arithmetic expression with unknown variable

问题

I want to make an arithmetic solver in Prolog that can have +,-,*,^ operations on numbers >= 2. It should also be possible to have a variable x in there. The input should be a prefix expression in a list.

I have made a program that parses an arithmetic expression in prefix format into a syntax tree. So that:

?- parse([+,+,2,9,*,3,x],Tree).
Tree = plus(plus(num(2), num(9)), mul(num(3), var(x))) .

(1) At this stage, I want to extend this program to be able to solve it for a given x value. This should be done by adding another predicate evaluate(Tree, Value, Solution) which given a value for the unknown x, calculates the solution.

Example:

?- parse([*, 2, ^, x, 3],Tree), evaluate(Ast, 2, Solution).
Tree = mul(num(2), pow(var(x), num(3))) ,
Solution = 16.

I'm not sure how to solve this problem due to my lack of Prolog skills, but I need a way of setting the var(x) to num(2) like in this example (because x = 2). Maybe member in Prolog can be used to do this. Then I have to solve it using perhaps is/2

Edit: My attempt to solving it. Getting error: 'Undefined procedure: evaluate/3 However, there are definitions for: evaluate/5'

evaluate(plus(A,B),Value,Sol) --> evaluate(A,AV,Sol), evaluate(B,BV,Sol), Value is AV+BV.
evaluate(mul(A,B),Value,Sol) --> evaluate(A,AV,Sol), evaluate(B,BV,Sol), Value is AV*BV.
evaluate(pow(A,B),Value,Sol) --> evaluate(A,AV,Sol), evaluate(B,BV,Sol), Value is AV^BV.
evaluate(num(Num),Value,Sol) --> number(Num).
evaluate(var(x),Value,Sol) --> number(Value).

(2) I'd also want to be able to express it in postfix form. Having a predicate postfixform(Tree, Postfixlist)

Example:

?- parse([+, *, 2, x, ^, x, 5 ],Tree), postfix(Tree,Postfix).
Tree = plus(mul(num(2), var(x)), pow(var(x), num(5))) ,
Postfix = [2, x, *, x, 5, ^, +].

Any help with (1) and (2) would be highly appreciated!

回答1:

You don't need to use a grammar for this, as you are doing. You should use normal rules.

This is the pattern you need to follow.

evaluate(plus(A,B),Value,Sol) :- 
   evaluate(A, Value, A2),
   evaluate(B, Value, B2),
   Sol is A2+B2.

And

evaluate(num(X),_Value,Sol) :- Sol = X.
evaluate(var(x),Value,Sol) :- Sol = Value.

来源：https://stackoverflow.com/questions/19916466/prolog-solving-prefix-arithmetic-expression-with-unknown-variable