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Symbolic simplification in Haskell (using recursion?)
我心中所想的简化是：
0*e = e*0 = 0
1*e = e*1 = 0+e = e+0 = e-0 = e
并简化常量子表达式，例如 Plus(Const 1) (Const 2) 将变为 Const 3。我不希望变量（或变量和常量）被合并： Var "st" 是不同于 Var "s" 的变量。
例如，simplify(Plus (Var "x") (Const 0))= Var "x"的中文翻译为：简化（加（变量“x”）（常量0））= 变量“x”。
嗯，你不能将模式匹配应用于各个情况吗？
simplify (Plus (Const 0) (Expr x)) = simplify (Expr x)
simplify (Plus (Expr x) (Const 0)) = simplify (Expr x)
simplify (Mult (Const 0) _) = Const 0
simplify (Mult _ (Const 0)) = Const 0
– … and so on
编辑：当然可以…添加了递归。
我不太了解 Haskell，但基本上你需要进行表达式树遍历。
the tree is EXP: (operator) (EXP) (EXP) EXP: (const) EXP: (var)
then your simplify becomes heres the psuedo code
simplify(Exp e)
if (e is const) return e
else if (e is var) return e
else
{//encode simplification rules
Exp left = simplify(e.left)
Exp right = simplify(e.right)
if(operator is PLUS)
{
if(left == 0) return right;
if(right == 0) return left;
}
else if(operator is MULT)
{
if(left == 1) return right;
if(right == 1) return left;
if(left == 0) return 0;
if(right == 0) return 0;
}
//and so on for other operators
}
这有点像 Java 但我认为这个想法在这里了，本质上你将不得不进行树遍历。