564 lines
18 KiB
Dart
564 lines
18 KiB
Dart
import 'dart:math';
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import 'package:flutter/foundation.dart'; // For kDebugMode
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import 'package:math_expressions/math_expressions.dart';
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import 'models/calculation_step.dart';
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/// 帮助解析一元一次方程 ax+b=cx+d 的辅助类
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class LinearEquationParts {
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final double a, b, c, d;
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LinearEquationParts(this.a, this.b, this.c, this.d);
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}
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class SolverService {
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/// 主入口方法，识别并分发任务
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CalculationResult solve(String input) {
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// 预处理输入字符串
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final cleanInput = input.replaceAll(' ', '').toLowerCase();
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// 对包含x的方程进行预处理，展开表达式
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String processedInput = cleanInput;
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if (processedInput.contains('x') && processedInput.contains('(')) {
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processedInput = _expandExpressions(processedInput);
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}
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// 1. 检查是否为二元一次方程组 (格式: ...;...)
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if (processedInput.contains(';') &&
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processedInput.contains('x') &&
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processedInput.contains('y')) {
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return _solveSystemOfLinearEquations(processedInput);
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}
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// 2. 检查是否为一元二次方程 (包含 x^2 或 x²)
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if (processedInput.contains('x^2') || processedInput.contains('x²')) {
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return _solveQuadraticEquation(processedInput.replaceAll('x²', 'x^2'));
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}
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// 3. 检查是否为一元一次方程 (包含 x 但不包含 y 或 x^2)
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if (processedInput.contains('x') && !processedInput.contains('y')) {
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return _solveLinearEquation(processedInput);
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}
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// 4. 如果都不是，则作为简单表达式计算
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try {
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return _solveSimpleExpression(input); // 使用原始输入以保留运算符
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} catch (e) {
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if (kDebugMode) {
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print(e);
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}
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throw Exception('无法识别的格式。请检查您的方程或表达式。');
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}
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}
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/// ---- 求解器实现 ----
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/// 1. 求解简单表达式
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CalculationResult _solveSimpleExpression(String input) {
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final steps = <CalculationStep>[];
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steps.add(
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CalculationStep(
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title: "第一步：表达式求值",
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explanation: "这是一个标准的数学表达式，我们将直接计算其结果。",
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formula: input,
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),
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);
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// **FIXED**: Correct usage of the math_expressions library
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Parser p = Parser();
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Expression exp = p.parse(input);
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ContextModel cm = ContextModel();
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final result = exp.evaluate(EvaluationType.REAL, cm);
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return CalculationResult(steps: steps, finalAnswer: result.toString());
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}
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/// 2. 求解一元一次方程
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CalculationResult _solveLinearEquation(String input) {
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final steps = <CalculationStep>[];
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steps.add(
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CalculationStep(title: "原方程", explanation: "这是一元一次方程。", formula: input),
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);
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// 解析方程: ax+b=cx+d
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final parts = _parseLinearEquation(input);
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final a = parts.a, b = parts.b, c = parts.c, d = parts.d;
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// 移项合并
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final newA = a - c;
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final newD = d - b;
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steps.add(
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CalculationStep(
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title: "第一步：移项",
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explanation: "将所有含 x 的项移到等式左边，常数项移到右边。",
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formula:
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"${a}x ${c >= 0 ? '-' : '+'} ${c.abs()}x = $d ${b >= 0 ? '-' : '+'} ${b.abs()}",
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),
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);
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steps.add(
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CalculationStep(
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title: "第二步：合并同类项",
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explanation: "合并等式两边的项。",
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formula: "${newA}x = $newD",
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),
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);
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// 求解x
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if (newA == 0) {
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if (newD == 0) {
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return CalculationResult(steps: steps, finalAnswer: "有无穷多解");
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} else {
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return CalculationResult(steps: steps, finalAnswer: "无解");
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}
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}
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final x = newD / newA;
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steps.add(
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CalculationStep(
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title: "第三步：求解 x",
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explanation: "两边同时除以 x 的系数 ($newA)。",
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formula: "x = $newD / $newA",
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),
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);
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return CalculationResult(steps: steps, finalAnswer: "x = $x");
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}
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/// 3. 求解一元二次方程 (升级版)
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CalculationResult _solveQuadraticEquation(String input) {
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final steps = <CalculationStep>[];
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// 整理成 ax^2+bx+c=0 的形式并提取系数
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final eqParts = input.split('=');
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if (eqParts.length != 2) throw Exception("方程格式错误，应包含一个 '='。");
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final leftCoeffs = _parsePolynomial(eqParts[0]);
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final rightCoeffs = _parsePolynomial(eqParts[1]);
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// 移项合并
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final a = (leftCoeffs[2] ?? 0) - (rightCoeffs[2] ?? 0);
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final b = (leftCoeffs[1] ?? 0) - (rightCoeffs[1] ?? 0);
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final c = (leftCoeffs[0] ?? 0) - (rightCoeffs[0] ?? 0);
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if (a == 0) {
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// 如果 a=0, 这实际上是一个一元一次方程
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return _solveLinearEquation("${b}x+${c}=0");
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}
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// **FIXED**: Corrected typo 'ax' to 'a'
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steps.add(
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CalculationStep(
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title: "第一步：整理方程",
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explanation: "将方程整理成标准形式 ax^2+bx+c=0。",
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formula:
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"${a}x^2 ${b >= 0 ? '+' : ''} ${b}x ${c >= 0 ? '+' : ''} $c = 0",
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),
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);
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// 尝试因式分解 (十字相乘法)
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if (a == a.round() && b == b.round() && c == c.round()) {
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final factors = _tryFactorization(a.toInt(), b.toInt(), c.toInt());
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if (factors != null) {
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steps.add(
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CalculationStep(
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title: "第二步：因式分解法 (十字相乘)",
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explanation: "我们发现可以将方程分解为两个一次因式的乘积。",
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formula: factors.formula,
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),
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);
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steps.add(
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CalculationStep(
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title: "第三步：求解",
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explanation: "分别令每个因式等于 0，解出 x。",
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formula: "解得 ${factors.solution}",
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),
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);
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return CalculationResult(steps: steps, finalAnswer: factors.solution);
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} else {
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steps.add(
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CalculationStep(
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title: "第二步：选择解法",
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explanation: "尝试因式分解失败，我们选择使用求根公式法。",
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formula: "\$\\Delta = b^2 - 4ac\$",
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),
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);
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}
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} else {
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steps.add(
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CalculationStep(
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title: "第二步：选择解法",
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explanation: "系数包含小数，我们使用求根公式法。",
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formula: "\$\\Delta = b^2 - 4ac\$",
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),
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);
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}
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// 使用求根公式法
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final delta = b * b - 4 * a * c;
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steps.add(
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CalculationStep(
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title: "第三步：计算判别式 (Delta)",
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explanation: "\$\\Delta = ($b)^2 - 4 \\cdot ($a) \\cdot ($c) = $delta",
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// **FIXED**: Removed unnecessary braces for linter warning
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formula: "\$\\Delta = $delta",
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),
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);
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if (delta > 0) {
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final x1 = (-b + sqrt(delta)) / (2 * a);
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final x2 = (-b - sqrt(delta)) / (2 * a);
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steps.add(
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CalculationStep(
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title: "第四步：应用求根公式",
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explanation:
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"因为 \$\\Delta > 0\$，方程有两个不相等的实数根。公式: \$x = \\frac{-b \\pm \\sqrt{\\Delta}}{2a}\$",
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formula:
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"x₁ = ${x1.toStringAsFixed(4)}, x₂ = ${x2.toStringAsFixed(4)}",
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),
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);
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return CalculationResult(
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steps: steps,
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finalAnswer:
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"x₁ = ${x1.toStringAsFixed(4)}, x₂ = ${x2.toStringAsFixed(4)}",
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);
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} else if (delta == 0) {
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final x = -b / (2 * a);
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steps.add(
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CalculationStep(
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title: "第四步：应用求根公式",
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explanation: "因为 \$\Delta = 0\$，方程有两个相等的实数根。",
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formula: "x₁ = x₂ = ${x.toStringAsFixed(4)}",
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),
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);
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return CalculationResult(
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steps: steps,
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finalAnswer: "x₁ = x₂ = ${x.toStringAsFixed(4)}",
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);
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} else {
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steps.add(
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CalculationStep(
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title: "第四步：判断解",
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explanation: "因为 \$\Delta < 0\$，该方程在实数范围内无解。",
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formula: "无实数解",
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),
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);
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return CalculationResult(steps: steps, finalAnswer: "无实数解");
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}
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}
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/// 4. 求解二元一次方程组
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CalculationResult _solveSystemOfLinearEquations(String input) {
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final steps = <CalculationStep>[];
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final equations = input.split(';');
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if (equations.length != 2) throw Exception("格式错误, 请用 ';' 分隔两个方程。");
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final p1 = _parseTwoVariableLinear(equations[0]);
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final p2 = _parseTwoVariableLinear(equations[1]);
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double a1 = p1[0], b1 = p1[1], c1 = p1[2];
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double a2 = p2[0], b2 = p2[1], c2 = p2[2];
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steps.add(
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CalculationStep(
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title: "原始方程组",
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explanation: "这是一个二元一次方程组，我们将使用加减消元法求解。",
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formula:
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"${a1}x ${b1 >= 0 ? '+' : ''} ${b1}y = $c1 ---(1)\n${a2}x ${b2 >= 0 ? '+' : ''} ${b2}y = $c2 ---(2)",
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),
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);
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final det = a1 * b2 - a2 * b1;
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if (det == 0) {
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if (a1 * c2 - a2 * c1 == 0) {
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return CalculationResult(steps: steps, finalAnswer: "有无穷多解");
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} else {
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return CalculationResult(steps: steps, finalAnswer: "无解");
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}
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}
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final newA1 = a1 * b2, newC1 = c1 * b2;
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final newA2 = a2 * b1, newC2 = c2 * b1;
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// **FIXED**: Wrapped calculations in braces {} for string interpolation
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steps.add(
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CalculationStep(
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title: "第一步：消元",
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explanation: "为了消去变量 y，将方程(1)两边乘以 $b2\$，方程(2)两边乘以 $b1\$。",
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formula:
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"${newA1}x ${b1 * b2 >= 0 ? '+' : ''} ${b1 * b2}y = $newC1 ---(3)\n${newA2}x ${b1 * b2 >= 0 ? '+' : ''} ${b1 * b2}y = $newC2 ---(4)",
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),
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);
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final xCoeff = newA1 - newA2;
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final constCoeff = newC1 - newC2;
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steps.add(
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CalculationStep(
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title: "第二步：相减",
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explanation: "将方程(3)减去方程(4)，得到一个只含 x 的方程。",
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formula:
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"($newA1 - $newA2)x = $newC1 - $newC2 \n=> ${xCoeff}x = $constCoeff",
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),
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);
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final x = constCoeff / xCoeff;
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steps.add(
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CalculationStep(
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title: "第三步：解出 x",
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explanation: "求解上述方程得到 x 的值。",
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formula: "x = $x",
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),
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);
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// **FIXED**: Added a check for b1 to avoid division by zero if we substitute into an equation without y.
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if (b1.abs() < 1e-9) {
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// If b1 is very close to 0, use the second equation
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final yCoeff = b2;
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final yConst = c2 - a2 * x;
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final y = yConst / yCoeff;
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steps.add(
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CalculationStep(
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title: "第四步：回代求解 y",
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explanation: "将 x = $x 代入原方程(2)中。",
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// **FIXED**: Corrected string interpolation for calculations and substitutions
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formula:
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"${a2}(${x}) + ${b2}y = $c2 \n=> ${a2 * x} + ${b2}y = $c2 \n=> ${b2}y = $c2 - ${a2 * x} \n=> ${b2}y = ${c2 - a2 * x}",
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),
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);
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steps.add(
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CalculationStep(
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title: "第五步：解出 y",
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explanation: "求解得到 y 的值。",
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formula: "y = $y",
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),
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);
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return CalculationResult(steps: steps, finalAnswer: "x = $x, y = $y");
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} else {
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// Default case, substitute into the first equation
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final yCoeff = b1;
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final yConst = c1 - a1 * x;
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final y = yConst / yCoeff;
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steps.add(
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CalculationStep(
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title: "第四步：回代求解 y",
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explanation: "将 x = $x 代入原方程(1)中。",
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formula:
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"${a1}(${x}) + ${b1}y = $c1 \n=> ${a1 * x} + ${b1}y = $c1 \n=> ${b1}y = $c1 - ${a1 * x} \n=> ${b1}y = ${c1 - a1 * x}",
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),
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);
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steps.add(
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CalculationStep(
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title: "第五步：解出 y",
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explanation: "求解得到 y 的值。",
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formula: "y = $y",
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),
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);
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return CalculationResult(steps: steps, finalAnswer: "x = $x, y = $y");
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}
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}
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/// ---- 辅助函数 ----
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/// 展开表达式，例如 (x-1)(x+2) -> x^2+x-2
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String _expandExpressions(String input) {
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String result = input;
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// 循环直到没有更多的表达式可以展开
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while (true) {
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String oldResult = result;
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// 模式1: k*(ax+b)^2, 例如 4(x-1)^2 or (x-1)^2
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final powerMatch = RegExp(
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r'(-?\d*\.?\d*)?\(([^)]+)\)\^2',
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).firstMatch(result);
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if (powerMatch != null) {
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final kStr = powerMatch.group(1);
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double k = 1.0;
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if (kStr != null && kStr.isNotEmpty) {
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if (kStr == '-') {
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k = -1.0;
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} else {
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k = double.parse(kStr);
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}
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}
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final factor = powerMatch.group(2)!;
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final coeffs = _parsePolynomial(factor);
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final a = coeffs[1] ?? 0;
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final b = coeffs[0] ?? 0;
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// (ax+b)^2 = a^2*x^2 + 2ab*x + b^2
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final newA = k * a * a;
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final newB = k * 2 * a * b;
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final newC = k * b * b;
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final expanded =
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"${newA}x^2${newB >= 0 ? '+' : ''}${newB}x${newC >= 0 ? '+' : ''}${newC}";
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result = result.replaceFirst(powerMatch.group(0)!, "($expanded)");
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continue; // 重新开始循环以处理嵌套表达式
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}
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// 模式2: (ax+b)(cx+d), 例如 (x-3)(x+2)
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final factorMulMatch = RegExp(
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r'\(([^)]+)\)\(([^)]+)\)',
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).firstMatch(result);
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if (factorMulMatch != null) {
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final factor1 = factorMulMatch.group(1)!;
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final factor2 = factorMulMatch.group(2)!;
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final coeffs1 = _parsePolynomial(factor1);
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final coeffs2 = _parsePolynomial(factor2);
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final a = coeffs1[1] ?? 0;
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final b = coeffs1[0] ?? 0;
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final c = coeffs2[1] ?? 0;
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final d = coeffs2[0] ?? 0;
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// (ax+b)(cx+d) = ac*x^2 + (ad+bc)*x + bd
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final newA = a * c;
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final newB = a * d + b * c;
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final newC = b * d;
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final expanded =
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"${newA}x^2${newB >= 0 ? '+' : ''}${newB}x${newC >= 0 ? '+' : ''}${newC}";
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result = result.replaceFirst(factorMulMatch.group(0)!, "($expanded)");
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continue; // 重新开始循环
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}
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// 如果在这次迭代中没有改变，则跳出循环
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if (result == oldResult) {
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break;
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}
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}
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return result;
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}
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LinearEquationParts _parseLinearEquation(String input) {
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final parts = input.split('=');
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if (parts.length != 2) throw Exception("方程格式错误，应包含一个'='。");
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final leftCoeffs = _parsePolynomial(parts[0]);
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final rightCoeffs = _parsePolynomial(parts[1]);
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final a = leftCoeffs[1] ?? 0.0;
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final b = leftCoeffs[0] ?? 0.0;
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final c = rightCoeffs[1] ?? 0.0;
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final d = rightCoeffs[0] ?? 0.0;
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return LinearEquationParts(a, b, c, d);
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}
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Map<int, double> _parsePolynomial(String side) {
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final coeffs = <int, double>{};
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final pattern = RegExp(
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r'([+-]?(?:\d*\.?\d*))?x(?:\^(\d+))?|([+-]?\d*\.?\d+)',
|
||
);
|
||
var s = side.startsWith('+') || side.startsWith('-') ? side : '+$side';
|
||
|
||
for (final match in pattern.allMatches(s)) {
|
||
if (match.group(3) != null) {
|
||
coeffs[0] = (coeffs[0] ?? 0) + double.parse(match.group(3)!);
|
||
} else {
|
||
int power = match.group(2) != null ? int.parse(match.group(2)!) : 1;
|
||
String coeffStr = match.group(1) ?? '+';
|
||
double coeff = 1.0;
|
||
if (coeffStr.isNotEmpty && coeffStr != '+') {
|
||
if (coeffStr == '-') {
|
||
coeff = -1.0;
|
||
} else {
|
||
coeff = double.parse(coeffStr);
|
||
}
|
||
} else if (coeffStr == '-') {
|
||
coeff = -1.0;
|
||
}
|
||
coeffs[power] = (coeffs[power] ?? 0) + coeff;
|
||
}
|
||
}
|
||
return coeffs;
|
||
}
|
||
|
||
List<double> _parseTwoVariableLinear(String equation) {
|
||
final parts = equation.split('=');
|
||
if (parts.length != 2) throw Exception("方程 $equation 格式错误");
|
||
final c = double.tryParse(parts[1]) ?? 0.0;
|
||
|
||
double a = 0, b = 0;
|
||
final xMatch = RegExp(r'([+-]?\d*\.?\d*)x').firstMatch(parts[0]);
|
||
if (xMatch != null) {
|
||
final coeff = xMatch.group(1);
|
||
if (coeff == null || coeff == '+') {
|
||
a = 1.0;
|
||
} else if (coeff == '-') {
|
||
a = -1.0;
|
||
} else {
|
||
a = double.tryParse(coeff) ?? 0.0;
|
||
}
|
||
}
|
||
final yMatch = RegExp(r'([+-]?\d*\.?\d*)y').firstMatch(parts[0]);
|
||
if (yMatch != null) {
|
||
final coeff = yMatch.group(1);
|
||
if (coeff == null || coeff == '+') {
|
||
b = 1.0;
|
||
} else if (coeff == '-') {
|
||
b = -1.0;
|
||
} else {
|
||
b = double.tryParse(coeff) ?? 0.0;
|
||
}
|
||
}
|
||
return [a, b, c];
|
||
}
|
||
|
||
({String formula, String solution})? _tryFactorization(int a, int b, int c) {
|
||
if (a == 0) return null;
|
||
int ac = a * c;
|
||
for (int i = 1; i <= sqrt(ac.abs()); i++) {
|
||
if (ac % i == 0) {
|
||
int j = ac ~/ i;
|
||
if (check(i, j, b)) return formatFactor(i, j, a);
|
||
if (check(-i, -j, b)) return formatFactor(-i, -j, a);
|
||
if (check(i, -j, b)) return formatFactor(i, -j, a);
|
||
if (check(-i, j, b)) return formatFactor(-i, j, a);
|
||
}
|
||
}
|
||
return null;
|
||
}
|
||
|
||
// **FIXED**: Simplified check method
|
||
bool check(int m, int n, int b) {
|
||
return m + n == b;
|
||
}
|
||
|
||
({String formula, String solution}) formatFactor(int m, int n, int a) {
|
||
int common = gcd(n.abs(), a.abs());
|
||
int num = n ~/ common;
|
||
int den = a ~/ common;
|
||
|
||
final a1 = den;
|
||
final c1 = num;
|
||
final a2 = a ~/ den;
|
||
final c2 = m ~/ a2;
|
||
|
||
// **FIXED**: Correctly handle coefficients of 1
|
||
final f1Part1 = a1 == 1 ? 'x' : '${a1}x';
|
||
final f1 = c1 == 0 ? f1Part1 : "$f1Part1 ${c1 >= 0 ? '+' : ''} $c1";
|
||
|
||
final f2Part1 = a2 == 1 ? 'x' : '${a2}x';
|
||
final f2 = c2 == 0 ? f2Part1 : "$f2Part1 ${c2 >= 0 ? '+' : ''} $c2";
|
||
|
||
// **FIXED**: Renamed variables to lowerCamelCase
|
||
final int x1Num = -c1, x1Den = a1;
|
||
final int x2Num = -c2, x2Den = a2;
|
||
|
||
final sol1 = x1Den == 1 ? "$x1Num" : "$x1Num/$x1Den";
|
||
final sol2 = x2Den == 1 ? "$x2Num" : "$x2Num/$x2Den";
|
||
|
||
final solution = x1Num * x2Den == x2Num * x1Den
|
||
? "x₁ = x₂ = $sol1"
|
||
: "x₁ = $sol1, x₂ = $sol2";
|
||
|
||
return (formula: "(${f1})(${f2}) = 0", solution: solution);
|
||
}
|
||
|
||
int gcd(int a, int b) => b == 0 ? a : gcd(b, a % b);
|
||
}
|