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java - 简化二叉表达式树的简洁方法

转载 作者:行者123 更新时间:2023-11-30 10:30:53 26 4
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我的程序的目标是显示数学表达式的符号导数。在创建表示导数的新树后,很可能会留下多余的术语

例如,下面的树没有被简化。

Example of binary expression tree

0 + 5 * (x * 5) 可以重写为 25 * x

我的程序使用很多很多 ifelse block 通过检查常量乘以常量等来减少树。然后,它相应地重新排列子树。

这是简化树的递归函数的一小部分:

if(root.getVal().equals("*")) {

if(root.getLeftChild().getVal().equals("1")) {
return root.getRightChild();
}
else if(root.getRightChild().getVal().equals("1")) {
return root.getLeftChild();
}
else if(root.getLeftChild().getVal().equals("0")) {
return root.getLeftChild();
}
else if(root.getRightChild().getVal().equals("0")) {
return root.getRightChild();
}
else if(root.getLeftChild().getVal().equals("*")) {
if(root.getRightChild().getType().equals("constant")) {
if(root.getLeftChild().getLeftChild().getType().equals("constant")) { // Ex: (5*x)*6 ==> 30*x
int num1 = Integer.parseInt(root.getRightChild().getVal());
int num2 = Integer.parseInt(root.getLeftChild().getLeftChild().getVal());
OpNode mult = new OpNode("*");
mult.setLeftChild(new ConstNode(String.valueOf(num1 * num2)));
mult.setRightChild(root.getLeftChild().getRightChild());

return mult;
}
...
...
...
...

该函数效果很好,除了我需要调用它几次以确保树完全缩减(以防缩减打开另一种缩减可能性)。然而,它有 200 行,而且还在不断增长,这让我相信一定有更好的方法来做到这一点。

最佳答案

解决此问题的一种典型方法是 visitor pattern .任何时候您需要遍历递归结构,在每个节点上应用逻辑,这取决于节点的“类型”,这种模式是一个很好的工具。

对于这个特定问题,特别是在 Java 中,我首先将您的表达式“抽象语法树”更直接地表示为类型层次结构。

我已经整理了一个简单的示例,假设您的 AST 处理 +、-、*、/以及文字数字和命名变量。我将我的 Visitor 称为 Folder——我们有时将此名称用于替换(“折叠”)子树的类似访问者。 (想想:编译器中的优化或脱糖过程。)

处理“我​​有时需要重复简化”的技巧是进行深度优先遍历:在我们简化他们的父级之前,所有子级都得到完全简化。

示例如下(免责声明:我讨厌 Java,所以我不保证这是该语言中最“惯用”的实现):

interface Folder {
// we could use the name "fold" for all of these, overloading on the
// argument type, and the dispatch code in each concrete Expression
// class would still do the right thing (selecting an overload using
// the type of "this") --- but this is a little easier to follow
Expression foldBinaryOperation(BinaryOperation expr);
Expression foldUnaryOperation(UnaryOperation expr);
Expression foldNumber(Number expr);
Expression foldVariable(Variable expr);
}

abstract class Expression {
abstract Expression fold(Folder f);

// logic to build a readable representation for testing
abstract String repr();
}

enum BinaryOperator {
PLUS,
MINUS,
MUL,
DIV,
}

enum UnaryOperator {
NEGATE,
}

class BinaryOperation extends Expression {
public BinaryOperation(BinaryOperator operator,
Expression left, Expression right)
{
this.operator = operator;
this.left = left;
this.right = right;
}

public BinaryOperator operator;
public Expression left;
public Expression right;

public Expression fold(Folder f) {
return f.foldBinaryOperation(this);
}

public String repr() {
// parens for clarity
String result = "(" + left.repr();
switch (operator) {
case PLUS:
result += " + ";
break;
case MINUS:
result += " - ";
break;
case MUL:
result += " * ";
break;
case DIV:
result += " / ";
break;
}
result += right.repr() + ")";
return result;
}
}

class UnaryOperation extends Expression {
public UnaryOperation(UnaryOperator operator, Expression operand)
{
this.operator = operator;
this.operand = operand;
}

public UnaryOperator operator;
public Expression operand;

public Expression fold(Folder f) {
return f.foldUnaryOperation(this);
}

public String repr() {
String result = "";
switch (operator) {
case NEGATE:
result = "-";
break;
}
result += operand.repr();
return result;
}
}

class Number extends Expression {
public Number(double value)
{
this.value = value;
}

public double value;

public Expression fold(Folder f) {
return f.foldNumber(this);
}

public String repr() {
return Double.toString(value);
}
}

class Variable extends Expression {
public Variable(String name)
{
this.name = name;
}

public String name;

public Expression fold(Folder f) {
return f.foldVariable(this);
}

public String repr() {
return name;
}
}

// a base class providing "standard" traversal logic (we could have
// made Folder abstract and put these there
class DefaultFolder implements Folder {
public Expression foldBinaryOperation(BinaryOperation expr) {
// recurse into both sides of the binary operation
return new BinaryOperation(
expr.operator, expr.left.fold(this), expr.right.fold(this));
}

public Expression foldUnaryOperation(UnaryOperation expr) {
// recurse into operand
return new UnaryOperation(expr.operator, expr.operand.fold(this));
}

public Expression foldNumber(Number expr) {
// numbers are "terminal": no more recursive structure to walk
return expr;
}

public Expression foldVariable(Variable expr) {
// another non-recursive expression
return expr;
}
}

class Simplifier extends DefaultFolder {
public Expression foldBinaryOperation(BinaryOperation expr) {
// we want to do a depth-first traversal, ensuring that all
// sub-expressions are simplified before their parents...
// ... so begin by invoking the superclass "default"
// traversal logic.
BinaryOperation folded_expr =
// this cast is safe because we know the default fold
// logic never changes the type of the top-level expression
(BinaryOperation)super.foldBinaryOperation(expr);

// now apply our "shallow" simplification logic on the result
switch (folded_expr.operator) {
case PLUS:
// x + 0 => x
if (folded_expr.right instanceof Number
&& ((Number)(folded_expr.right)).value == 0)
return folded_expr.left;

// 0 + x => x
if (folded_expr.left instanceof Number
&& ((Number)(folded_expr.left)).value == 0)
return folded_expr.right;
break;

case MINUS:
// x - 0 => x
if (folded_expr.right instanceof Number
&& ((Number)(folded_expr.right)).value == 0)
return folded_expr.left;

// 0 - x => -x
if (folded_expr.left instanceof Number
&& ((Number)(folded_expr.left)).value == 0) {
// a weird case: we need to construct a UnaryOperator
// representing -right, then simplify it
UnaryOperation minus_right = new UnaryOperation(
UnaryOperator.NEGATE, folded_expr.right);
return foldUnaryOperation(minus_right);
}
break;

case MUL:
// 1 * x => x
if (folded_expr.left instanceof Number
&& ((Number)(folded_expr.left)).value == 1)
return folded_expr.right;

case DIV:
// x * 1 => x
// x / 1 => x
if (folded_expr.right instanceof Number
&& ((Number)(folded_expr.right)).value == 1)
return folded_expr.left;
break;
}

// no rules applied
return folded_expr;
}

public Expression foldUnaryOperation(UnaryOperation expr) {
// as before, go depth-first:
UnaryOperation folded_expr =
// see note in foldBinaryOperation about safety here
(UnaryOperation)super.foldUnaryOperation(expr);

switch (folded_expr.operator) {
case NEGATE:
// --x => x
if (folded_expr.operand instanceof UnaryOperation
&& ((UnaryOperation)folded_expr).operator ==
UnaryOperator.NEGATE)
return ((UnaryOperation)folded_expr.operand).operand;

// -(number) => -number
if (folded_expr.operand instanceof Number)
return new Number(-((Number)(folded_expr.operand)).value);
break;
}

// no rules applied
return folded_expr;
}

// we don't need to implement the other two; the inherited defaults are fine
}

public class Simplify {
public static void main(String[] args) {
Simplifier simplifier = new Simplifier();

Expression[] exprs = new Expression[] {
new BinaryOperation(
BinaryOperator.PLUS,
new Number(0.0),
new Variable("x")
),

new BinaryOperation(
BinaryOperator.PLUS,
new Number(17.3),
new UnaryOperation(
UnaryOperator.NEGATE,
new UnaryOperation(
UnaryOperator.NEGATE,
new BinaryOperation(
BinaryOperator.DIV,
new Number(0.0),
new Number(1.0)
)
)
)
),
};

for (Expression expr: exprs) {
System.out.println("Unsimplified: " + expr.repr());

Expression simplified = expr.fold(simplifier);
System.out.println("Simplified: " + simplified.repr());
}
}
}

输出:

> java Simplify

Unsimplified: (0.0 + x)
Simplified: x
Unsimplified: (17.3 + --(0.0 / 1.0))
Simplified: 17.3

关于java - 简化二叉表达式树的简洁方法,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/43532261/

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