677 lines
20 KiB
Dart
677 lines
20 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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stepNumber: 1,
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title: '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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GrammarParser p = GrammarParser();
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Expression exp = p.parse(input);
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final result = RealEvaluator().evaluate(exp);
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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(
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stepNumber: 0,
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title: '原方程',
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explanation: '这是一元一次方程。',
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formula: '\$\$$input\$\$',
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),
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);
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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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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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stepNumber: 1,
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title: '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 ? '-' : '+'} ${b.abs()}\$\$',
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),
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);
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steps.add(
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CalculationStep(
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stepNumber: 2,
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title: '合并同类项',
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explanation: '合并等式两边的项。',
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formula: '\$\$${newA}x = $newD\$\$',
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),
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);
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if (newA == 0) {
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return CalculationResult(
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steps: steps,
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finalAnswer: newD == 0 ? '有无穷多解' : '无解',
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);
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}
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final x = newD / newA;
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steps.add(
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CalculationStep(
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stepNumber: 3,
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title: '求解 x',
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explanation: '两边同时除以 x 的系数 ($newA)。',
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formula: '\$\$x = \\frac{$newD}{$newA}\$\$',
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),
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);
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return CalculationResult(steps: steps, finalAnswer: '\$\$x = $x\$\$');
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}
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/// 3. 求解一元二次方程 (升级版)
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CalculationResult _solveQuadraticEquation(String input) {
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final steps = <CalculationStep>[];
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final eqParts = input.split('=');
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if (eqParts.length != 2) throw Exception("方程格式错误，应包含一个 '='。");
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final leftCoeffs = _parsePolynomial(eqParts[0]);
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final rightCoeffs = _parsePolynomial(eqParts[1]);
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final a = (leftCoeffs[2] ?? 0) - (rightCoeffs[2] ?? 0);
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final b = (leftCoeffs[1] ?? 0) - (rightCoeffs[1] ?? 0);
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final c = (leftCoeffs[0] ?? 0) - (rightCoeffs[0] ?? 0);
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if (a == 0) {
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return _solveLinearEquation('${b}x+$c=0');
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}
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steps.add(
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CalculationStep(
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stepNumber: 1,
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title: '整理方程',
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explanation: r'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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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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stepNumber: 2,
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title: '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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stepNumber: 3,
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title: '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: factors.solution,
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),
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);
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steps.add(
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CalculationStep(
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stepNumber: 4,
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title: '化简结果',
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explanation: '将分数化简到最简形式，并将负号写在分数外面。',
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formula: 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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}
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}
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steps.add(
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CalculationStep(
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stepNumber: 2,
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title: '选择解法',
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explanation: '无法进行因式分解，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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final delta = b * b - 4 * a * c;
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steps.add(
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CalculationStep(
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stepNumber: 3,
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title: '计算判别式 (Delta)',
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explanation:
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'\$\$\\Delta = b^2 - 4ac = ($b)^2 - 4 \\cdot ($a) \\cdot ($c) = $delta\$\$',
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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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stepNumber: 4,
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title: '应用求根公式',
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explanation:
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r'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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'\$\$x_1 = ${x1.toStringAsFixed(4)}, \\quad x_2 = ${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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'\$\$x_1 = ${x1.toStringAsFixed(4)}, \\quad x_2 = ${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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stepNumber: 4,
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title: '应用求根公式',
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explanation: r'因为 $\Delta = 0$，方程有两个相等的实数根。',
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formula: '\$\$x_1 = x_2 = ${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: '\$\$x_1 = x_2 = ${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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stepNumber: 4,
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title: '判断解',
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explanation: r'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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stepNumber: 0,
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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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'''
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\\begin{cases}
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${a1}x ${b1 >= 0 ? '+' : ''} ${b1}y = $c1 & (1) \\
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${a2}x ${b2 >= 0 ? '+' : ''} ${b2}y = $c2 & (2)
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\\end{cases}
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''',
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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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return CalculationResult(
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steps: steps,
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finalAnswer: a1 * c2 - a2 * c1 == 0 ? '有无穷多解' : '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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steps.add(
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CalculationStep(
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stepNumber: 1,
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title: '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 $b1。',
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formula:
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'''
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\\begin{cases}
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${newA1}x ${b1 * b2 >= 0 ? '+' : ''} ${b1 * b2}y = $newC1 & (3) \\
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${newA2}x ${b1 * b2 >= 0 ? '+' : ''} ${b1 * b2}y = $newC2 & (4)
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\\end{cases}
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''',
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),
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);
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final xCoeff = newA1 - newA2;
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final constCoeff = newC1 - newC2;
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steps.add(
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CalculationStep(
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stepNumber: 2,
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title: '相减',
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explanation: '将方程(3)减去方程(4)，得到一个只含 x 的方程。',
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formula:
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'\$\$($newA1 - $newA2)x = $newC1 - $newC2 \\Rightarrow ${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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stepNumber: 3,
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title: '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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if (b1.abs() < 1e-9) {
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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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stepNumber: 4,
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title: '回代求解 y',
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explanation: '将 x = $x 代入原方程(2)中。',
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formula:
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'''
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\\begin{aligned}
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$a2($x) + ${b2}y &= $c2 \\
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${a2 * x} + ${b2}y &= $c2 \\
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${b2}y &= $c2 - ${a2 * x} \\
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${b2}y &= ${c2 - a2 * x}
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\\end{aligned}
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''',
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),
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);
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steps.add(
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CalculationStep(
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stepNumber: 5,
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title: '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(
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steps: steps,
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finalAnswer: '\$\$x = $x, \\quad y = $y\$\$',
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);
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} else {
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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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stepNumber: 4,
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title: '回代求解 y',
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explanation: '将 x = $x 代入原方程(1)中。',
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formula:
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'''
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\\begin{aligned}
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$a1($x) + ${b1}y &= $c1 \\
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${a1 * x} + ${b1}y &= $c1 \\
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${b1}y &= $c1 - ${a1 * x} \\
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${b1}y &= ${c1 - a1 * x}
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\\end{aligned}
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''',
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),
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);
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steps.add(
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CalculationStep(
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stepNumber: 5,
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title: '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(
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steps: steps,
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finalAnswer: '\$\$x = $x, \\quad y = $y\$\$',
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);
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}
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}
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/// ---- 辅助函数 ----
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String _expandExpressions(String input) {
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String result = input;
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int maxIterations = 10; // Prevent infinite loops
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int iterationCount = 0;
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while (iterationCount < maxIterations) {
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String oldResult = result;
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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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k = kStr == '-' ? -1.0 : double.parse(kStr);
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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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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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iterationCount++;
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continue;
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}
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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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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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iterationCount++;
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continue;
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}
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// Handle expressions like x(expr) or (expr)x or coeff(expr)
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final termFactorMatch = RegExp(
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r'([+-]?(?:\d*\.?\d*)?x?)\(([^)]+)\)',
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).firstMatch(result);
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if (termFactorMatch != null) {
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final termStr = termFactorMatch.group(1)!;
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final factorStr = termFactorMatch.group(2)!;
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// Skip if the term is just a sign or empty
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if (termStr == '+' || termStr == '-' || termStr.isEmpty) {
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break;
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}
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// Parse the term (coefficient and x power)
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final termCoeffs = _parsePolynomial(termStr);
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final factorCoeffs = _parsePolynomial(factorStr);
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final termA = termCoeffs[1] ?? 0; // x coefficient
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final termB = termCoeffs[0] ?? 0; // constant term
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final factorA = factorCoeffs[1] ?? 0; // x coefficient
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final factorB = factorCoeffs[0] ?? 0; // constant term
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// Multiply: (termA*x + termB) * (factorA*x + factorB)
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final newA = termA * factorA;
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final newB = termA * factorB + termB * factorA;
|
||
final newC = termB * factorB;
|
||
|
||
final expanded =
|
||
'${newA}x^2${newB >= 0 ? '+' : ''}${newB}x${newC >= 0 ? '+' : ''}$newC';
|
||
result = result.replaceFirst(termFactorMatch.group(0)!, '($expanded)');
|
||
iterationCount++;
|
||
continue;
|
||
}
|
||
|
||
if (result == oldResult) break;
|
||
iterationCount++;
|
||
}
|
||
|
||
if (iterationCount >= maxIterations) {
|
||
throw Exception('表达式展开过于复杂，请简化输入。');
|
||
}
|
||
|
||
return result;
|
||
}
|
||
|
||
LinearEquationParts _parseLinearEquation(String input) {
|
||
final parts = input.split('=');
|
||
if (parts.length != 2) throw Exception("方程格式错误，应包含一个'='。");
|
||
|
||
final leftCoeffs = _parsePolynomial(parts[0]);
|
||
final rightCoeffs = _parsePolynomial(parts[1]);
|
||
|
||
return LinearEquationParts(
|
||
(leftCoeffs[1] ?? 0.0),
|
||
(leftCoeffs[0] ?? 0.0),
|
||
(rightCoeffs[1] ?? 0.0),
|
||
(rightCoeffs[0] ?? 0.0),
|
||
);
|
||
}
|
||
|
||
Map<int, double> _parsePolynomial(String side) {
|
||
final coeffs = <int, double>{};
|
||
final pattern = RegExp(
|
||
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 != '+') {
|
||
coeff = coeffStr == '-' ? -1.0 : 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.isEmpty || 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.isEmpty || 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;
|
||
int absAc = ac.abs();
|
||
for (int d = 1; d <= sqrt(absAc).toInt(); d++) {
|
||
if (absAc % d == 0) {
|
||
int d1 = d;
|
||
int d2 = absAc ~/ d;
|
||
List<int> signs1 = ac >= 0 ? [1, -1] : [1, -1];
|
||
List<int> signs2 = ac >= 0 ? [1, -1] : [1, -1];
|
||
for (int s1 in signs1) {
|
||
for (int s2 in signs2) {
|
||
int m = s1 * d1;
|
||
int n = s2 * d2;
|
||
if (check(m, n, b)) return formatFactor(m, n, a);
|
||
m = s1 * d1;
|
||
n = s2 * (-d2);
|
||
if (check(m, n, b)) return formatFactor(m, n, a);
|
||
m = s1 * (-d1);
|
||
n = s2 * d2;
|
||
if (check(m, n, b)) return formatFactor(m, n, a);
|
||
m = s1 * (-d1);
|
||
n = s2 * (-d2);
|
||
if (check(m, n, b)) return formatFactor(m, n, a);
|
||
}
|
||
}
|
||
}
|
||
}
|
||
return null;
|
||
}
|
||
|
||
bool check(int m, int n, int b) => m + n == b;
|
||
|
||
({String formula, String solution}) formatFactor(int m, int n, int a) {
|
||
// Roots are -m/a and -n/a
|
||
int g1 = gcd(m.abs(), a.abs());
|
||
int root1Num = -m ~/ g1;
|
||
int root1Den = a ~/ g1;
|
||
|
||
int g2 = gcd(n.abs(), a.abs());
|
||
int root2Num = -n ~/ g2;
|
||
int root2Den = a ~/ g2;
|
||
|
||
String sol1 = _formatFraction(root1Num, root1Den);
|
||
String sol2 = _formatFraction(root2Num, root2Den);
|
||
|
||
// For formula, show (a x + m)(x + n/a) or simplified
|
||
String f1 = a == 1 ? 'x' : '${a}x';
|
||
f1 = m == 0 ? f1 : '$f1 ${m >= 0 ? '+' : ''} $m';
|
||
|
||
String f2;
|
||
if (n % a == 0) {
|
||
int coeff = n ~/ a;
|
||
f2 = 'x ${coeff >= 0 ? '+' : ''} $coeff';
|
||
if (coeff == 0) f2 = 'x';
|
||
} else {
|
||
f2 = 'x ${n >= 0 ? '+' : ''} \\frac{$n}{$a}';
|
||
}
|
||
|
||
String formula = '\$\$($f1)($f2) = 0\$\$';
|
||
|
||
String solution;
|
||
if (root1Num * root2Den == root2Num * root1Den) {
|
||
solution = '\$\$x_1 = x_2 = $sol1\$\$';
|
||
} else {
|
||
solution = '\$\$x_1 = $sol1, \\quad x_2 = $sol2\$\$';
|
||
}
|
||
|
||
return (formula: formula, solution: solution);
|
||
}
|
||
|
||
String _formatFraction(int num, int den) {
|
||
if (den == 0) return 'undefined';
|
||
|
||
// Handle sign: make numerator positive, put sign outside
|
||
bool isNegative = (num < 0) != (den < 0);
|
||
int absNum = num.abs();
|
||
int absDen = den.abs();
|
||
|
||
// Simplify fraction
|
||
int g = gcd(absNum, absDen);
|
||
absNum ~/= g;
|
||
absDen ~/= g;
|
||
|
||
if (absDen == 1) {
|
||
return isNegative ? '-$absNum' : '$absNum';
|
||
} else {
|
||
String fraction = '\\frac{$absNum}{$absDen}';
|
||
return isNegative ? '-$fraction' : fraction;
|
||
}
|
||
}
|
||
|
||
int gcd(int a, int b) => b == 0 ? a : gcd(b, a % b);
|
||
}
|