1407 lines
44 KiB
Dart
1407 lines
44 KiB
Dart
import 'dart:developer' show log;
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import 'dart:math' hide log;
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import 'package:rational/rational.dart';
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import 'package:simple_math_calc/calculator.dart';
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import 'package:simple_math_calc/parser.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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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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// Parse the input to get LaTeX-formatted version
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final parser = Parser(input);
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final parsedExpr = parser.parse();
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final latexInput = parsedExpr.toString().replaceAll('*', '\\cdot');
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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: '\$\$$latexInput\$\$',
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),
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);
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// 检查是否为特殊三角函数值，可以返回精确结果
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final exactTrigResult = getExactTrigResult(input);
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if (exactTrigResult != null) {
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return CalculationResult(
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steps: steps,
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finalAnswer: '\$\$$exactTrigResult\$\$',
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);
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}
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// 预处理输入，将三角函数的参数从度转换为弧度
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String processedInput = convertTrigToRadians(input);
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try {
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// 使用自定义解析器解析表达式
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final parser = Parser(processedInput);
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final expr = parser.parse();
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// 对表达式进行求值
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final evaluatedExpr = expr.evaluate();
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// 获取数值结果 - 需要正确进行类型转换
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double result;
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if (evaluatedExpr is IntExpr) {
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result = evaluatedExpr.value.toDouble();
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} else if (evaluatedExpr is DoubleExpr) {
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result = evaluatedExpr.value;
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} else if (evaluatedExpr is FractionExpr) {
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result = evaluatedExpr.numerator / evaluatedExpr.denominator;
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} else {
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// 如果无法完全求值为数值，尝试简化并转换为字符串
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final simplified = evaluatedExpr.simplify();
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return CalculationResult(
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steps: steps,
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finalAnswer: '\$\$${simplified.toString()}\$\$',
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);
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}
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// 尝试将结果格式化为几倍根号的形式
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final formattedResult = formatSqrtResult(result);
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return CalculationResult(
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steps: steps,
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finalAnswer: '\$\$$formattedResult\$\$',
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);
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} catch (e) {
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throw Exception('无法解析表达式: $input');
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}
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}
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/// 2. 求解一元一次方程
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CalculationResult _solveLinearEquation(String input) {
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final steps = <CalculationStep>[];
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// Parse the input to get LaTeX-formatted version
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final parser = Parser(input);
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final parsedExpr = parser.parse();
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final latexInput = parsedExpr.toString().replaceAll('*', '\\cdot');
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steps.add(
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CalculationStep(
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stepNumber: 0,
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title: '原方程',
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explanation: '这是一元一次方程。',
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formula: '\$\$$latexInput\$\$',
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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 = _rationalFromDouble(a) - _rationalFromDouble(c);
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final newD = _rationalFromDouble(d) - _rationalFromDouble(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 ? '-' : '+'} ${b.abs()}\$\$',
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),
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);
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steps.add(
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CalculationStep(
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stepNumber: 2,
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title: '合并同类项',
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explanation: '合并等式两边的项。',
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formula:
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'\$\$${newA.toDouble().toStringAsFixed(4)}x = ${newD.toDouble().toStringAsFixed(4)}\$\$',
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),
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);
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if (newA == Rational.zero) {
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return CalculationResult(
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steps: steps,
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finalAnswer: newD == Rational.zero ? '有无穷多解' : '无解',
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);
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}
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final x = newD / newA;
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steps.add(
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CalculationStep(
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stepNumber: 3,
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title: '求解 x',
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explanation: '两边同时除以 x 的系数 ($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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// Keep original equation for display
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final originalEquation = _formatOriginalEquation(input);
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// Parse coefficients symbolically (kept for potential future use)
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// final leftCoeffsSymbolic = _parsePolynomialSymbolic(eqParts[0]);
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// final rightCoeffsSymbolic = _parsePolynomialSymbolic(eqParts[1]);
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// final aSymbolic = _subtractCoefficients(
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// leftCoeffsSymbolic[2] ?? '0',
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// rightCoeffsSymbolic[2] ?? '0',
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// );
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// final bSymbolic = _subtractCoefficients(
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// leftCoeffsSymbolic[1] ?? '0',
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// rightCoeffsSymbolic[1] ?? '0',
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// );
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// final cSymbolic = _subtractCoefficients(
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// leftCoeffsSymbolic[0] ?? '0',
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// rightCoeffsSymbolic[0] ?? '0',
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// );
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// Also get numeric values for calculations
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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: originalEquation,
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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，我们选择使用配方法。',
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formula: r'配方法：$x^2 + \frac{b}{a}x + \frac{c}{a} = 0$',
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),
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);
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// Step 1: Divide by a if a ≠ 1
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String currentEquation;
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if (a == 1) {
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currentEquation =
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'x^2 ${b >= 0 ? "+" : ""}${b}x ${c >= 0 ? "+" : ""}$c = 0';
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} else {
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final aStr = a == -1 ? '-' : a.toString();
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currentEquation =
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'\\frac{1}{$aStr}(x^2 ${b >= 0 ? "+" : ""}${b}x ${c >= 0 ? "+" : ""}$c) = 0';
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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: '方程变形',
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explanation: a == 1 ? '方程已经是标准形式。' : '将方程两边同时除以 $a。',
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formula: '\$\$${currentEquation}\$\$',
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),
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);
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// Step 2: Move constant term to the other side
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final constantTerm = c / a;
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steps.add(
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CalculationStep(
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stepNumber: 4,
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title: '移项',
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explanation: '将常数项移到方程右边。',
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formula: '\$\$x^2 ${b >= 0 ? "+" : ""}${b}x = ${-constantTerm}\$\$',
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),
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);
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// Step 3: Complete the square
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final halfCoeff = b / (2 * a);
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final completeSquareTerm = halfCoeff * halfCoeff;
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final completeStr = completeSquareTerm >= 0
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? '+${completeSquareTerm}'
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: completeSquareTerm.toString();
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final xTerm = halfCoeff >= 0 ? "+${halfCoeff}" : halfCoeff.toString();
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final rightSide = "${-constantTerm} ${completeStr}";
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steps.add(
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CalculationStep(
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stepNumber: 5,
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title: '配方',
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explanation:
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'在方程两边同时加上 \$(\\frac{b}{2a})^2 = ${completeSquareTerm}\$ 以配成完全平方。',
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formula: '\$\$(x ${xTerm})^2 = $rightSide\$\$',
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),
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);
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// Step 4: Simplify right side
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final rightSideValue = -constantTerm + completeSquareTerm;
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final rightSideStrValue = rightSideValue >= 0
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? rightSideValue.toString()
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: '(${rightSideValue})';
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steps.add(
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CalculationStep(
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stepNumber: 6,
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title: '化简',
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explanation: '合并右边的常数项。',
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formula:
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'\$\$(x ${halfCoeff >= 0 ? "+" : ""}${halfCoeff})^2 = $rightSideStrValue\$\$',
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),
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);
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// Step 5: Take square root - check for symbolic representation
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final symbolicSqrt = _getSymbolicSquareRoot(rightSideValue);
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final sqrtStr = rightSideValue >= 0
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? (symbolicSqrt ?? sqrt(rightSideValue.abs()).toString())
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: '${sqrt(rightSideValue.abs()).toString()}i';
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steps.add(
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CalculationStep(
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stepNumber: 7,
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title: '开方',
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explanation: '对方程两边同时开平方。',
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formula:
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'\$\$x ${halfCoeff >= 0 ? "+" : ""}${halfCoeff} = \\pm $sqrtStr\$\$',
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),
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);
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// Step 6: Solve for x - use symbolic forms when possible
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if (rightSideValue >= 0) {
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final roots = _calculateSymbolicRoots(a, b, rightSideValue, symbolicSqrt);
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steps.add(
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CalculationStep(
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stepNumber: 8,
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title: '解出 x',
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explanation: '分别取正负号，解出 x 的值。',
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formula: roots.formula,
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),
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);
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return CalculationResult(steps: steps, finalAnswer: roots.finalAnswer);
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} else {
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// Complex roots
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final imagPart = sqrt(rightSideValue.abs());
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steps.add(
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CalculationStep(
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stepNumber: 8,
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title: '解出 x',
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explanation: '方程在实数范围内无解，但有虚数解。',
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formula:
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'\$\$x_1 = ${-halfCoeff} + ${imagPart}i, \\quad x_2 = ${-halfCoeff} - ${imagPart}i\$\$',
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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 = ${-halfCoeff} + ${imagPart}i, \\quad x_2 = ${-halfCoeff} - ${imagPart}i\$\$',
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);
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}
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}
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/// 4. 求解二元一次方程组
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CalculationResult _solveSystemOfLinearEquations(String input) {
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final steps = <CalculationStep>[];
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final equations = input.split(';');
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if (equations.length != 2) throw Exception("格式错误, 请用 ';' 分隔两个方程。");
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final p1 = _parseTwoVariableLinear(equations[0]);
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final p2 = _parseTwoVariableLinear(equations[1]);
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double a1 = p1[0], b1 = p1[1], c1 = p1[2];
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double a2 = p2[0], b2 = p2[1], c2 = p2[2];
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steps.add(
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CalculationStep(
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stepNumber: 0,
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title: '原始方程组',
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explanation: '这是一个二元一次方程组，我们将使用加减消元法求解。',
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formula:
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'''
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\$\$
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\\begin{cases}
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${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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);
|
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final det =
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_rationalFromDouble(a1) * _rationalFromDouble(b2) -
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_rationalFromDouble(a2) * _rationalFromDouble(b1);
|
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if (det == Rational.zero) {
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final infiniteCheck =
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_rationalFromDouble(a1) * _rationalFromDouble(c2) -
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_rationalFromDouble(a2) * _rationalFromDouble(c1);
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return CalculationResult(
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steps: steps,
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finalAnswer: infiniteCheck == Rational.zero ? '有无穷多解' : '无解',
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);
|
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}
|
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final newA1 = _rationalFromDouble(a1) * _rationalFromDouble(b2);
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final newC1 = _rationalFromDouble(c1) * _rationalFromDouble(b2);
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final newA2 = _rationalFromDouble(a2) * _rationalFromDouble(b1);
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final newC2 = _rationalFromDouble(c2) * _rationalFromDouble(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 $b1。',
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formula:
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'''
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\$\$
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\\begin{cases}
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${newA1.toDouble().toStringAsFixed(2)}x ${b1 * b2 >= 0 ? '+' : ''} ${(b1 * b2).toStringAsFixed(2)}y = ${newC1.toDouble().toStringAsFixed(2)} & (3) \\\\
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${newA2.toDouble().toStringAsFixed(2)}x ${b1 * b2 >= 0 ? '+' : ''} ${(b1 * b2).toStringAsFixed(2)}y = ${newC2.toDouble().toStringAsFixed(2)} & (4)
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\\end{cases}
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\$\$
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''',
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),
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);
|
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||
final xCoeff = newA1 - newA2;
|
||
final constCoeff = newC1 - newC2;
|
||
|
||
steps.add(
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||
CalculationStep(
|
||
stepNumber: 2,
|
||
title: '相减',
|
||
explanation: '将方程(3)减去方程(4)，得到一个只含 x 的方程。',
|
||
formula:
|
||
'\$\$(${newA1.toDouble().toStringAsFixed(2)} - ${newA2.toDouble().toStringAsFixed(2)})x = ${newC1.toDouble().toStringAsFixed(2)} - ${newC2.toDouble().toStringAsFixed(2)} \\Rightarrow ${xCoeff.toDouble().toStringAsFixed(2)}x = ${constCoeff.toDouble().toStringAsFixed(2)}\$\$',
|
||
),
|
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);
|
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|
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final x = constCoeff / xCoeff;
|
||
steps.add(
|
||
CalculationStep(
|
||
stepNumber: 3,
|
||
title: '解出 x',
|
||
explanation: '求解上述方程得到 x 的值。',
|
||
formula: '\$\$x = $x\$\$',
|
||
),
|
||
);
|
||
|
||
if (b1.abs() < 1e-9) {
|
||
final yCoeff = b2;
|
||
final yConst = c2 - a2 * x.toDouble();
|
||
final y = yConst / yCoeff;
|
||
steps.add(
|
||
CalculationStep(
|
||
stepNumber: 4,
|
||
title: '回代求解 y',
|
||
explanation: '将 x = ${x.toDouble().toStringAsFixed(4)} 代入原方程(2)中。',
|
||
formula:
|
||
'''
|
||
\$\$
|
||
\\begin{aligned}
|
||
$a2(${x.toDouble().toStringAsFixed(4)}) + ${b2}y &= $c2 \\\\
|
||
${a2 * x.toDouble()} + ${b2}y &= $c2 \\\\
|
||
${b2}y &= $c2 - ${a2 * x.toDouble()} \\\\
|
||
${b2}y &= ${c2 - a2 * x.toDouble()}
|
||
\\end{aligned}
|
||
\$\$
|
||
''',
|
||
),
|
||
);
|
||
steps.add(
|
||
CalculationStep(
|
||
stepNumber: 5,
|
||
title: '解出 y',
|
||
explanation: '求解得到 y 的值。',
|
||
formula: '\$\$y = ${y.toStringAsFixed(4)}\$\$',
|
||
),
|
||
);
|
||
return CalculationResult(
|
||
steps: steps,
|
||
finalAnswer:
|
||
'\$\$x = ${x.toDouble().toStringAsFixed(4)}, \\quad y = ${y.toStringAsFixed(4)}\$\$',
|
||
);
|
||
} else {
|
||
final yCoeff = b1;
|
||
final yConst = c1 - a1 * x.toDouble();
|
||
final y = yConst / yCoeff;
|
||
steps.add(
|
||
CalculationStep(
|
||
stepNumber: 4,
|
||
title: '回代求解 y',
|
||
explanation: '将 x = ${x.toDouble().toStringAsFixed(4)} 代入原方程(1)中。',
|
||
formula:
|
||
'''
|
||
\$\$
|
||
\\begin{aligned}
|
||
$a1(${x.toDouble().toStringAsFixed(4)}) + ${b1}y &= $c1 \\\\
|
||
${a1 * x.toDouble()} + ${b1}y &= $c1 \\\\
|
||
${b1}y &= $c1 - ${a1 * x.toDouble()} \\\\
|
||
${b1}y &= ${c1 - a1 * x.toDouble()}
|
||
\\end{aligned}
|
||
\$\$
|
||
''',
|
||
),
|
||
);
|
||
steps.add(
|
||
CalculationStep(
|
||
stepNumber: 5,
|
||
title: '解出 y',
|
||
explanation: '求解得到 y 的值。',
|
||
formula: '\$\$y = ${y.toStringAsFixed(4)}\$\$',
|
||
),
|
||
);
|
||
return CalculationResult(
|
||
steps: steps,
|
||
finalAnswer:
|
||
'\$\$x = ${x.toDouble().toStringAsFixed(4)}, \\quad y = ${y.toStringAsFixed(4)}\$\$',
|
||
);
|
||
}
|
||
}
|
||
|
||
/// 检查表达式是否可绘制（包含变量x且可以被求值）
|
||
bool isGraphableExpression(String expression) {
|
||
try {
|
||
// 移除空格并转换为小写
|
||
String cleanExpr = expression.replaceAll(' ', '').toLowerCase();
|
||
|
||
// 如果以 y= 开头，去掉前缀
|
||
if (cleanExpr.startsWith('y=')) {
|
||
cleanExpr = cleanExpr.substring(2);
|
||
}
|
||
|
||
// 不能包含等号（方程而不是函数表达式）
|
||
if (cleanExpr.contains('=')) {
|
||
return false;
|
||
}
|
||
|
||
// 必须包含变量x
|
||
if (!cleanExpr.contains('x')) {
|
||
return false;
|
||
}
|
||
|
||
// 尝试展开表达式（如果包含括号）
|
||
String processedExpr = cleanExpr;
|
||
if (processedExpr.contains('(')) {
|
||
processedExpr = _expandExpressions(processedExpr);
|
||
}
|
||
|
||
// 尝试解析表达式
|
||
final parser = Parser(processedExpr);
|
||
final expr = parser.parse();
|
||
|
||
// 测试在几个点上是否可以求值
|
||
final testPoints = [-1.0, 0.0, 1.0];
|
||
for (final x in testPoints) {
|
||
try {
|
||
final substituted = expr.substitute('x', DoubleExpr(x));
|
||
final evaluated = substituted.evaluate();
|
||
if (evaluated is DoubleExpr &&
|
||
evaluated.value.isFinite &&
|
||
!evaluated.value.isNaN) {
|
||
// 至少有一个点可以求值就算成功
|
||
return true;
|
||
}
|
||
} catch (e) {
|
||
// 继续测试其他点
|
||
continue;
|
||
}
|
||
}
|
||
|
||
return false;
|
||
} catch (e) {
|
||
return false;
|
||
}
|
||
}
|
||
|
||
/// 准备函数表达式用于绘图（展开因式形式）
|
||
String prepareFunctionForGraphing(String expression) {
|
||
// 移除空格并转换为小写
|
||
String cleanExpr = expression.replaceAll(' ', '').toLowerCase();
|
||
|
||
// 如果以 y= 开头，去掉前缀
|
||
if (cleanExpr.startsWith('y=')) {
|
||
cleanExpr = cleanExpr.substring(2);
|
||
}
|
||
|
||
// 如果表达式包含括号，进行展开
|
||
if (cleanExpr.contains('(')) {
|
||
cleanExpr = _expandExpressions(cleanExpr);
|
||
}
|
||
|
||
// 清理格式：移除不必要的.0后缀和简化格式
|
||
cleanExpr = cleanExpr
|
||
.replaceAll('.0', '') // 移除所有.0
|
||
.replaceAll('+0', '') // 移除+0
|
||
.replaceAll('-0', '') // 移除-0
|
||
.replaceAll('1x^2', 'x^2') // 1x^2 -> x^2
|
||
.replaceAll('1x', 'x'); // 1x -> x
|
||
|
||
// 移除开头的+号
|
||
if (cleanExpr.startsWith('+')) {
|
||
cleanExpr = cleanExpr.substring(1);
|
||
}
|
||
|
||
return cleanExpr;
|
||
}
|
||
|
||
/// ---- 辅助函数 ----
|
||
|
||
String _expandExpressions(String input) {
|
||
String result = input;
|
||
int maxIterations = 10; // Prevent infinite loops
|
||
int iterationCount = 0;
|
||
|
||
while (iterationCount < maxIterations) {
|
||
String oldResult = result;
|
||
|
||
final powerMatch = RegExp(
|
||
r'(-?\d*\.?\d*)?\(([^)]+)\)\^2',
|
||
).firstMatch(result);
|
||
if (powerMatch != null) {
|
||
final kStr = powerMatch.group(1);
|
||
double k = 1.0;
|
||
if (kStr != null && kStr.isNotEmpty) {
|
||
k = kStr == '-' ? -1.0 : double.parse(kStr);
|
||
}
|
||
|
||
final factor = powerMatch.group(2)!;
|
||
final coeffs = _parsePolynomial(factor);
|
||
final a = coeffs[1] ?? 0;
|
||
final b = coeffs[0] ?? 0;
|
||
|
||
final newA = k * a * a;
|
||
final newB = k * 2 * a * b;
|
||
final newC = k * b * b;
|
||
|
||
final expanded =
|
||
'${newA}x^2${newB >= 0 ? '+' : ''}${newB}x${newC >= 0 ? '+' : ''}$newC';
|
||
result = result.replaceFirst(powerMatch.group(0)!, expanded);
|
||
iterationCount++;
|
||
continue;
|
||
}
|
||
|
||
final factorMulMatch = RegExp(
|
||
r'\(([^)]+)\)\(([^)]+)\)',
|
||
).firstMatch(result);
|
||
if (factorMulMatch != null) {
|
||
final factor1 = factorMulMatch.group(1)!;
|
||
final factor2 = factorMulMatch.group(2)!;
|
||
log('Expanding: ($factor1) * ($factor2)');
|
||
|
||
final coeffs1 = _parsePolynomial(factor1);
|
||
final coeffs2 = _parsePolynomial(factor2);
|
||
log('Coeffs1: $coeffs1, Coeffs2: $coeffs2');
|
||
|
||
final a = coeffs1[1] ?? 0;
|
||
final b = coeffs1[0] ?? 0;
|
||
final c = coeffs2[1] ?? 0;
|
||
final d = coeffs2[0] ?? 0;
|
||
log('a=$a, b=$b, c=$c, d=$d');
|
||
|
||
final newA = a * c;
|
||
final newB = a * d + b * c;
|
||
final newC = b * d;
|
||
log('newA=$newA, newB=$newB, newC=$newC');
|
||
|
||
final expanded =
|
||
'${newA}x^2${newB >= 0 ? '+' : ''}${newB}x${newC >= 0 ? '+' : ''}$newC';
|
||
log('Expanded result: $expanded');
|
||
|
||
result = result.replaceFirst(factorMulMatch.group(0)!, expanded);
|
||
iterationCount++;
|
||
continue;
|
||
}
|
||
|
||
// Handle expressions like x(expr) or (expr)x or coeff(expr)
|
||
final termFactorMatch = RegExp(
|
||
r'([+-]?(?:\d*\.?\d*)?x?)\(([^)]+)\)',
|
||
).firstMatch(result);
|
||
if (termFactorMatch != null) {
|
||
final termStr = termFactorMatch.group(1)!;
|
||
final factorStr = termFactorMatch.group(2)!;
|
||
|
||
// Skip if the term is just a sign or empty
|
||
if (termStr == '+' || termStr == '-' || termStr.isEmpty) {
|
||
break;
|
||
}
|
||
|
||
// Parse the term (coefficient and x power)
|
||
final termCoeffs = _parsePolynomial(termStr);
|
||
final factorCoeffs = _parsePolynomial(factorStr);
|
||
|
||
final termA = termCoeffs[1] ?? 0; // x coefficient
|
||
final termB = termCoeffs[0] ?? 0; // constant term
|
||
|
||
final factorA = factorCoeffs[1] ?? 0; // x coefficient
|
||
final factorB = factorCoeffs[0] ?? 0; // constant term
|
||
|
||
// Multiply: (termA*x + termB) * (factorA*x + factorB)
|
||
final newA = termA * factorA;
|
||
final newB = termA * factorB + termB * factorA;
|
||
final newC = termB * factorB;
|
||
|
||
final expanded =
|
||
'${newA == 1
|
||
? ''
|
||
: newA == -1
|
||
? '-'
|
||
: newA}x^2${newB >= 0 ? '+' : ''}${newB}x${newC >= 0 ? '+' : ''}$newC';
|
||
result = result.replaceFirst(termFactorMatch.group(0)!, expanded);
|
||
iterationCount++;
|
||
continue;
|
||
}
|
||
|
||
if (result == oldResult) break;
|
||
iterationCount++;
|
||
}
|
||
|
||
if (iterationCount >= maxIterations) {
|
||
throw Exception('表达式展开过于复杂，请简化输入。');
|
||
}
|
||
|
||
// 清理展开后的表达式格式
|
||
result = _cleanExpandedExpression(result);
|
||
|
||
// 检查是否为方程（包含等号），如果是的话，将右边的常数项移到左边
|
||
if (result.contains('=')) {
|
||
final parts = result.split('=');
|
||
if (parts.length == 2) {
|
||
final leftSide = parts[0];
|
||
final rightSide = parts[1];
|
||
|
||
// 解析左边的多项式
|
||
final leftCoeffs = _parsePolynomial(leftSide);
|
||
final rightCoeffs = _parsePolynomial(rightSide);
|
||
|
||
// 计算标准形式 ax^2 + bx + c = 0 的系数
|
||
// A = B 转换为 A - B = 0，所以右边的系数要取相反数
|
||
final a = (leftCoeffs[2] ?? 0) - (rightCoeffs[2] ?? 0);
|
||
final b = (leftCoeffs[1] ?? 0) - (rightCoeffs[1] ?? 0);
|
||
final c = (leftCoeffs[0] ?? 0) - (rightCoeffs[0] ?? 0);
|
||
|
||
// 构建标准形式的方程
|
||
String standardForm = '';
|
||
if (a != 0) {
|
||
standardForm +=
|
||
'${a == 1
|
||
? ''
|
||
: a == -1
|
||
? '-'
|
||
: a}x^2';
|
||
}
|
||
if (b != 0) {
|
||
standardForm += b > 0 ? '+${b}x' : '${b}x';
|
||
}
|
||
if (c != 0) {
|
||
standardForm += c > 0 ? '+$c' : '$c';
|
||
}
|
||
|
||
// 移除开头的加号
|
||
if (standardForm.startsWith('+')) {
|
||
standardForm = standardForm.substring(1);
|
||
}
|
||
|
||
// 如果所有系数都为0，则方程恒成立
|
||
if (standardForm.isEmpty) {
|
||
standardForm = '0';
|
||
}
|
||
|
||
result = '$standardForm=0';
|
||
}
|
||
}
|
||
|
||
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>{};
|
||
|
||
// 如果输入包含括号，去掉括号
|
||
var cleanSide = side;
|
||
if (cleanSide.startsWith('(') && cleanSide.endsWith(')')) {
|
||
cleanSide = cleanSide.substring(1, cleanSide.length - 1);
|
||
}
|
||
|
||
// 扩展模式以支持 sqrt 函数
|
||
final pattern = RegExp(
|
||
r'([+-]?(?:\d*\.?\d*|sqrt\(\d+\)))x(?:\^(\d+))?|([+-]?(?:\d*\.?\d*|sqrt\(\d+\)))',
|
||
);
|
||
var s = cleanSide.startsWith('+') || cleanSide.startsWith('-')
|
||
? cleanSide
|
||
: '+$cleanSide';
|
||
|
||
for (final match in pattern.allMatches(s)) {
|
||
if (match.group(0)!.isEmpty) continue; // Skip empty matches
|
||
|
||
if (match.group(3) != null) {
|
||
// 常数项
|
||
final constStr = match.group(3)!;
|
||
final constValue = _parseCoefficientWithSqrt(constStr);
|
||
coeffs[0] = (coeffs[0] ?? 0) + constValue;
|
||
} else {
|
||
// x 的幂次项
|
||
int power = match.group(2) != null ? int.parse(match.group(2)!) : 1;
|
||
String coeffStr = match.group(1) ?? '+';
|
||
final coeff = _parseCoefficientWithSqrt(coeffStr);
|
||
coeffs[power] = (coeffs[power] ?? 0) + coeff;
|
||
}
|
||
}
|
||
return coeffs;
|
||
}
|
||
|
||
/// 解析包含 sqrt 函数的系数
|
||
double _parseCoefficientWithSqrt(String coeffStr) {
|
||
if (coeffStr.isEmpty || coeffStr == '+') return 1.0;
|
||
if (coeffStr == '-') return -1.0;
|
||
|
||
// 检查是否包含 sqrt 函数
|
||
final sqrtMatch = RegExp(r'sqrt\((\d+)\)').firstMatch(coeffStr);
|
||
if (sqrtMatch != null) {
|
||
final innerValue = int.parse(sqrtMatch.group(1)!);
|
||
|
||
// 对于完全平方数，直接返回整数结果
|
||
final sqrtValue = sqrt(innerValue.toDouble());
|
||
final rounded = sqrtValue.round();
|
||
if ((sqrtValue - rounded).abs() < 1e-10) {
|
||
// 检查是否有系数
|
||
final coeffPart = coeffStr.replaceFirst(sqrtMatch.group(0)!, '');
|
||
if (coeffPart.isEmpty) return rounded.toDouble();
|
||
if (coeffPart == '-') return -rounded.toDouble();
|
||
|
||
final coeff = double.parse(coeffPart);
|
||
return coeff * rounded;
|
||
}
|
||
|
||
// 对于非完全平方数，计算数值但保持高精度
|
||
final nonPerfectSqrtValue = sqrt(innerValue.toDouble());
|
||
|
||
// 检查是否有系数
|
||
final coeffPart = coeffStr.replaceFirst(sqrtMatch.group(0)!, '');
|
||
if (coeffPart.isEmpty) return nonPerfectSqrtValue;
|
||
if (coeffPart == '-') return -nonPerfectSqrtValue;
|
||
|
||
final coeff = double.parse(coeffPart);
|
||
return coeff * nonPerfectSqrtValue;
|
||
}
|
||
|
||
// 普通数值
|
||
return double.parse(coeffStr);
|
||
}
|
||
|
||
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();
|
||
|
||
// Try all divisors of abs(ac) and consider both positive and negative factors
|
||
for (int d = 1; d <= sqrt(absAc).toInt(); d++) {
|
||
if (absAc % d == 0) {
|
||
int d1 = d;
|
||
int d2 = absAc ~/ d;
|
||
|
||
// Try all sign combinations for the factors
|
||
// We need m * n = ac and m + n = b
|
||
List<int> signCombinations = [1, -1];
|
||
|
||
for (int sign1 in signCombinations) {
|
||
for (int sign2 in signCombinations) {
|
||
int m = sign1 * d1;
|
||
int n = sign2 * d2;
|
||
if (m + n == b && m * n == ac) {
|
||
return formatFactor(m, n, a);
|
||
}
|
||
|
||
// Also try the swapped version
|
||
m = sign1 * d2;
|
||
n = sign2 * d1;
|
||
if (m + n == b && m * n == ac) {
|
||
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);
|
||
|
||
/// 格式化原始方程，保持符号形式
|
||
String _formatOriginalEquation(String input) {
|
||
// Parse the equation and convert to LaTeX
|
||
String result = input.replaceAll(' ', '');
|
||
|
||
// 确保方程格式正确
|
||
if (!result.contains('=')) {
|
||
result = '$result=0';
|
||
}
|
||
|
||
final parts = result.split('=');
|
||
if (parts.length == 2) {
|
||
// Check if the equation is already in standard polynomial form
|
||
// If it doesn't contain parentheses and looks like a standard polynomial,
|
||
// return it as-is to avoid unnecessary parsing
|
||
final leftSide = parts[0];
|
||
final rightSide = parts[1];
|
||
|
||
// If left side is a standard polynomial (no parentheses, only x^2, x, and constants)
|
||
// and right side is 0, return the original
|
||
if (_isStandardPolynomial(leftSide) &&
|
||
(rightSide == '0' || rightSide.isEmpty)) {
|
||
result = '$leftSide=0';
|
||
return '\$\$$result\$\$';
|
||
}
|
||
|
||
try {
|
||
final leftParser = Parser(parts[0]);
|
||
final leftExpr = leftParser.parse();
|
||
final rightParser = Parser(parts[1]);
|
||
final rightExpr = rightParser.parse();
|
||
|
||
// Get the string representation and clean it up
|
||
String leftStr = leftExpr.toString().replaceAll('*', '\\cdot');
|
||
String rightStr = rightExpr.toString().replaceAll('*', '\\cdot');
|
||
|
||
// Clean up unnecessary parentheses
|
||
leftStr = _cleanParentheses(leftStr);
|
||
rightStr = _cleanParentheses(rightStr);
|
||
|
||
result = '$leftStr=$rightStr';
|
||
} catch (e) {
|
||
// Fallback to original if parsing fails
|
||
result = result.replaceAll('sqrt(', '\\sqrt{');
|
||
result = result.replaceAll(')', '}');
|
||
}
|
||
} else {
|
||
try {
|
||
final parser = Parser(result.split('=')[0]);
|
||
final expr = parser.parse();
|
||
|
||
// Get the string representation and clean it up
|
||
String exprStr = expr.toString().replaceAll('*', '\\cdot');
|
||
exprStr = _cleanParentheses(exprStr);
|
||
|
||
result = '$exprStr=0';
|
||
} catch (e) {
|
||
// Fallback
|
||
result = result.replaceAll('sqrt(', '\\sqrt{');
|
||
result = result.replaceAll(')', '}');
|
||
}
|
||
}
|
||
|
||
return '\$\$$result\$\$';
|
||
}
|
||
|
||
/// 检查字符串是否为标准多项式形式（不含括号，只有x^2、x和常数项）
|
||
bool _isStandardPolynomial(String expr) {
|
||
// Remove spaces
|
||
final cleanExpr = expr.replaceAll(' ', '');
|
||
|
||
// If it contains parentheses, it's not standard
|
||
if (cleanExpr.contains('(') || cleanExpr.contains(')')) {
|
||
return false;
|
||
}
|
||
|
||
// Check if it matches the pattern of a standard polynomial
|
||
// Should only contain: digits, x, ^, +, -, and spaces (already removed)
|
||
final validChars = RegExp(r'^[0-9x\^\+\-\.]*$');
|
||
if (!validChars.hasMatch(cleanExpr)) {
|
||
return false;
|
||
}
|
||
|
||
// Should not have complex expressions like x*x or 2x*3
|
||
if (cleanExpr.contains('*') || cleanExpr.contains('/')) {
|
||
return false;
|
||
}
|
||
|
||
// Should have proper x^2 format (not xx or x2)
|
||
if (cleanExpr.contains('x^2') ||
|
||
cleanExpr.contains('x^3') ||
|
||
cleanExpr.contains('x^4')) {
|
||
// This is likely a polynomial
|
||
return true;
|
||
}
|
||
|
||
// Check for simple terms like x, 2x, x+1, etc.
|
||
final termPattern = RegExp(
|
||
r'^[+-]?(?:\d*\.?\d*)?x?(?:\^\d+)?(?:[+-][+-]?(?:\d*\.?\d*)?x?(?:\^\d+)?)*$',
|
||
);
|
||
return termPattern.hasMatch(cleanExpr);
|
||
}
|
||
|
||
/// 清理不必要的括号
|
||
String _cleanParentheses(String expr) {
|
||
// 移除最外层的括号，如果它们不影响运算顺序
|
||
if (expr.startsWith('(') && expr.endsWith(')')) {
|
||
String inner = expr.substring(1, expr.length - 1);
|
||
|
||
// 检查移除括号是否会改变含义
|
||
// 简单检查：如果内部没有运算符，或者只有加减号，可以移除
|
||
if (!inner.contains('+') &&
|
||
!inner.contains('-') &&
|
||
!inner.contains('*') &&
|
||
!inner.contains('/')) {
|
||
return inner;
|
||
}
|
||
|
||
// 如果内部表达式是简单的，可以移除括号
|
||
// 例如：(x+1) 可以变成 x+1, 但 (x+1)*(x-1) 不能移除
|
||
final operators = RegExp(r'[+\-*/]');
|
||
final matches = operators.allMatches(inner).toList();
|
||
|
||
// 如果只有一个运算符且是加减号，可以移除
|
||
if (matches.length == 1 && (inner.contains('+') || inner.contains('-'))) {
|
||
return inner;
|
||
}
|
||
}
|
||
|
||
return expr;
|
||
}
|
||
|
||
/// 清理展开后的表达式格式
|
||
String _cleanExpandedExpression(String expr) {
|
||
String result = expr;
|
||
|
||
// 移除不必要的.0后缀
|
||
result = result.replaceAll('.0', '');
|
||
|
||
// 移除+0和-0
|
||
result = result.replaceAll('+0', '');
|
||
result = result.replaceAll('-0', '');
|
||
|
||
// 简化系数为1的情况
|
||
result = result.replaceAll('1x^2', 'x^2');
|
||
result = result.replaceAll('1x', 'x');
|
||
|
||
// 移除开头的+号
|
||
if (result.startsWith('+')) {
|
||
result = result.substring(1);
|
||
}
|
||
|
||
// 处理连续的运算符
|
||
result = result.replaceAll('++', '+');
|
||
result = result.replaceAll('+-', '-');
|
||
result = result.replaceAll('-+', '-');
|
||
result = result.replaceAll('--', '+');
|
||
|
||
return result;
|
||
}
|
||
|
||
Rational _rationalFromDouble(double value, {int maxPrecision = 12}) {
|
||
// 限制小数精度，避免无限循环小数
|
||
final str = value.toStringAsFixed(maxPrecision);
|
||
|
||
if (!str.contains('.')) {
|
||
return Rational.parse(str);
|
||
}
|
||
|
||
final parts = str.split('.');
|
||
final integerPart = parts[0];
|
||
final fractionalPart = parts[1];
|
||
|
||
final numerator = BigInt.parse(integerPart + fractionalPart);
|
||
final denominator = BigInt.from(10).pow(fractionalPart.length);
|
||
|
||
return Rational(numerator, denominator);
|
||
}
|
||
|
||
/// 检查数值是否可以表示为符号平方根形式
|
||
String? _getSymbolicSquareRoot(double value) {
|
||
if (value <= 0) return null;
|
||
|
||
// 对于完全平方数，直接返回整数平方根
|
||
final sqrtValue = sqrt(value);
|
||
final intSqrt = sqrtValue.toInt();
|
||
if ((sqrtValue - intSqrt).abs() < 1e-10) {
|
||
return intSqrt.toString();
|
||
}
|
||
|
||
// 检查是否可以表示为 k√m 的形式，其中 m 不是完全平方数
|
||
// 遍历可能的 k 值，从大到小
|
||
for (int k = sqrt(value).toInt(); k >= 2; k--) {
|
||
final kSquared = k * k;
|
||
if (kSquared > value) continue;
|
||
|
||
final remaining = value / kSquared;
|
||
final remainingSqrt = sqrt(remaining);
|
||
final intRemainingSqrt = remainingSqrt.toInt();
|
||
|
||
// 检查剩余部分是否为完全平方数
|
||
if ((remainingSqrt - intRemainingSqrt).abs() < 1e-10) {
|
||
// 找到匹配：value = k² * m，其中 m 是完全平方数
|
||
if (intRemainingSqrt == 1) {
|
||
return k.toString(); // k√1 = k
|
||
} else {
|
||
return '$k\\sqrt{$intRemainingSqrt}';
|
||
}
|
||
}
|
||
}
|
||
|
||
// 特殊情况：检查是否为简单的分数形式，如 48 = 16*3 = 4²*3
|
||
// 对于 value = 48, k = 4, remaining = 48/16 = 3, sqrt(3) ≈ 1.732, intRemainingSqrt = 1
|
||
// 但 1.732 != 1, 所以上面的循环不会匹配
|
||
// 我们需要检查 remaining 是否是整数且不是完全平方数
|
||
final intValue = value.toInt();
|
||
if (value == intValue.toDouble()) {
|
||
// 尝试找到最大的完全平方因子
|
||
int maxK = 1;
|
||
for (int k = 2; k * k <= intValue; k++) {
|
||
if (intValue % (k * k) == 0) {
|
||
maxK = k;
|
||
}
|
||
}
|
||
|
||
if (maxK > 1) {
|
||
final remaining = intValue ~/ (maxK * maxK);
|
||
if (remaining > 1) {
|
||
return '$maxK\\sqrt{$remaining}';
|
||
}
|
||
}
|
||
}
|
||
|
||
return null; // 无法用简单符号形式表示
|
||
}
|
||
|
||
/// 计算符号形式的二次方程根
|
||
({String formula, String finalAnswer}) _calculateSymbolicRoots(
|
||
double a,
|
||
double b,
|
||
double discriminant,
|
||
String? symbolicSqrt,
|
||
) {
|
||
final halfCoeff = b / (2 * a);
|
||
final denominator = 2 * a;
|
||
|
||
String formula;
|
||
String finalAnswer;
|
||
|
||
if (symbolicSqrt != null) {
|
||
// 使用符号形式
|
||
final sqrtExpr = symbolicSqrt;
|
||
|
||
// 计算根：(-b ± sqrt(discriminant)) / (2a)
|
||
final root1Expr = _formatSymbolicRoot(-b, sqrtExpr, denominator, true);
|
||
final root2Expr = _formatSymbolicRoot(-b, sqrtExpr, denominator, false);
|
||
|
||
formula = '\$\$x_1 = $root1Expr, \\quad x_2 = $root2Expr\$\$';
|
||
finalAnswer = '\$\$x_1 = $root1Expr, \\quad x_2 = $root2Expr\$\$';
|
||
} else {
|
||
// 回退到数值计算
|
||
final sqrtValue = sqrt(discriminant);
|
||
final x1 = -halfCoeff + sqrtValue;
|
||
final x2 = -halfCoeff - sqrtValue;
|
||
|
||
formula =
|
||
'\$\$x_1 = ${-halfCoeff} + $sqrtValue = $x1, \\quad x_2 = ${-halfCoeff} - $sqrtValue = $x2\$\$';
|
||
finalAnswer = '\$\$x_1 = $x1, \\quad x_2 = $x2\$\$';
|
||
}
|
||
|
||
return (formula: formula, finalAnswer: finalAnswer);
|
||
}
|
||
|
||
/// 格式化符号形式的根
|
||
String _formatSymbolicRoot(
|
||
double b,
|
||
String sqrtExpr,
|
||
double denominator,
|
||
bool isPlus,
|
||
) {
|
||
final sign = isPlus ? '+' : '-';
|
||
|
||
// 处理分母
|
||
final denomStr = denominator == 2 ? '2' : denominator.toString();
|
||
|
||
if (b == 0) {
|
||
// 简化为 ±sqrt(discriminant)/denominator
|
||
if (denominator == 2) {
|
||
return isPlus ? '\\frac{$sqrtExpr}{2}' : '-\\frac{$sqrtExpr}{2}';
|
||
} else {
|
||
return isPlus
|
||
? '\\frac{$sqrtExpr}{$denomStr}'
|
||
: '-\\frac{$sqrtExpr}{$denomStr}';
|
||
}
|
||
} else {
|
||
// 完整的表达式：(-b ± sqrt(discriminant))/denominator
|
||
final bInt = b.toInt();
|
||
|
||
// 检查是否可以简化
|
||
if (bInt % denominator.toInt() == 0) {
|
||
final simplifiedB = bInt ~/ denominator.toInt();
|
||
|
||
if (simplifiedB == 0) {
|
||
return isPlus ? sqrtExpr : '-$sqrtExpr';
|
||
} else if (simplifiedB == 1) {
|
||
return isPlus
|
||
? '1 $sign $sqrtExpr'
|
||
: '1 $sign $sqrtExpr'.replaceAll('+', '-').replaceAll('--', '+');
|
||
} else if (simplifiedB == -1) {
|
||
return isPlus
|
||
? '-1 $sign $sqrtExpr'
|
||
: '-1 $sign $sqrtExpr'.replaceAll('+', '-').replaceAll('--', '+');
|
||
} else if (simplifiedB > 0) {
|
||
return isPlus
|
||
? '$simplifiedB $sign $sqrtExpr'
|
||
: '$simplifiedB $sign $sqrtExpr'
|
||
.replaceAll('+', '-')
|
||
.replaceAll('--', '+');
|
||
} else {
|
||
final absB = (-simplifiedB).toString();
|
||
return isPlus
|
||
? '-$absB $sign $sqrtExpr'
|
||
: '-$absB $sign $sqrtExpr'
|
||
.replaceAll('+', '-')
|
||
.replaceAll('--', '+');
|
||
}
|
||
} else {
|
||
// 无法简化，使用分数形式
|
||
final bStr = b > 0 ? '$bInt' : '($bInt)';
|
||
final numerator = b > 0
|
||
? '-$bStr $sign $sqrtExpr'
|
||
: '($bInt) $sign $sqrtExpr';
|
||
|
||
if (denominator == 2) {
|
||
return '\\frac{$numerator}{2}';
|
||
} else {
|
||
return '\\frac{$numerator}{$denomStr}';
|
||
}
|
||
}
|
||
}
|
||
}
|
||
|
||
/// 测试方法：验证修复效果
|
||
void testParenthesesFix() {
|
||
print('=== 测试括号修复效果 ===');
|
||
|
||
// 测试案例1: 已经标准化的方程
|
||
final test1 = 'x^2+4x-8=0';
|
||
print('测试输入: $test1');
|
||
final result1 = solve(test1);
|
||
print('整理方程步骤:');
|
||
result1.steps.forEach((step) {
|
||
if (step.title == '整理方程') {
|
||
print(' 公式: ${step.formula}');
|
||
}
|
||
});
|
||
print('预期: x^2+4x-8=0 (无括号)');
|
||
print('');
|
||
|
||
// 测试案例2: 需要展开的方程
|
||
final test2 = '(x+2)^2=x^2+4x+4';
|
||
print('测试输入: $test2');
|
||
final result2 = solve(test2);
|
||
print('整理方程步骤:');
|
||
result2.steps.forEach((step) {
|
||
if (step.title == '整理方程') {
|
||
print(' 公式: ${step.formula}');
|
||
}
|
||
});
|
||
print('预期: 展开后无不必要的括号');
|
||
print('');
|
||
|
||
// 测试案例3: 因式分解
|
||
final test3 = '(x+1)(x-1)=x^2-1';
|
||
print('测试输入: $test3');
|
||
final result3 = solve(test3);
|
||
print('整理方程步骤:');
|
||
result3.steps.forEach((step) {
|
||
if (step.title == '整理方程') {
|
||
print(' 公式: ${step.formula}');
|
||
}
|
||
});
|
||
print('预期: 展开后无不必要的括号');
|
||
}
|
||
}
|