🐛 Bug fixes
This commit is contained in:
149
lib/solver.dart
149
lib/solver.dart
@@ -7,13 +7,18 @@ 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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final Rational 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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/// 格式化数字，移除不必要的尾随零
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String _formatNumber(double value, {int precision = 4}) {
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String _formatNumber(dynamic value, {int precision = 4}) {
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if (value is Rational) {
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return _formatRational(value);
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} else if (value is FractionExpr) {
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return '\\frac{${value.numerator}}{${value.denominator}}';
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} else if (value is double) {
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String formatted = value.toStringAsFixed(precision);
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// 移除尾随的零和小数点
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formatted = formatted.replaceAll(RegExp(r'\.0+$'), '');
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@@ -22,6 +27,9 @@ class SolverService {
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formatted = formatted.substring(0, formatted.length - 1);
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}
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return formatted;
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} else {
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return value.toString();
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}
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}
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/// 尝试对二次方程进行因式分解
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@@ -345,8 +353,8 @@ class SolverService {
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final parts = _parseLinearEquation(input, variable);
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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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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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@@ -354,7 +362,15 @@ class SolverService {
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title: '移项',
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explanation: '将所有含 $variable 的项移到等式左边，常数项移到右边。',
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formula:
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'\$\$$a$variable ${c >= 0 ? '-' : '+'} ${c.abs()}$variable = $d ${b >= 0 ? '-' : '+'} ${b.abs()}\$\$',
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'\$\$${_formatNumber(a)}$variable ${_formatNumber(c) == '0'
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? ''
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: c >= Rational.zero
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? '-'
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: '+'} ${_formatNumber(c.abs())}$variable = ${_formatNumber(d)} ${_formatNumber(b) == '0'
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? ''
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: b >= Rational.zero
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? '-'
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: '+'} ${_formatNumber(b.abs())}\$\$',
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),
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);
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@@ -364,7 +380,7 @@ class SolverService {
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title: '合并同类项',
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explanation: '合并等式两边的项。',
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formula:
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'\$\$${_formatNumber(newA.toDouble())}$variable = ${_formatNumber(newD.toDouble())}\$\$',
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'\$\$${_formatNumber(newA)}$variable = ${_formatNumber(newD)}\$\$',
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),
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);
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@@ -380,14 +396,15 @@ class SolverService {
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CalculationStep(
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stepNumber: 3,
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title: '求解 $variable',
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explanation: '两边同时除以 $variable 的系数 ($newA)。',
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formula: '\$\$$variable = \\frac{$newD}{$newA}\$\$',
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explanation: '两边同时除以 $variable 的系数 (${_formatNumber(newA)})。',
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formula:
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'\$\$$variable = \\frac{${_formatNumber(newD)}}{${_formatNumber(newA)}}\$\$',
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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: '\$\$$variable = $x\$\$',
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finalAnswer: '\$\$$variable = ${_formatNumber(x)}\$\$',
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);
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}
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@@ -801,8 +818,12 @@ class SolverService {
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final p1 = _parseTwoVariableLinear(equations[0]);
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final p2 = _parseTwoVariableLinear(equations[1]);
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double a1 = p1[0], b1 = p1[1], c1 = p1[2];
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double a2 = p2[0], b2 = p2[1], c2 = p2[2];
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final a1Rat = _rationalFromDouble(p1[0]);
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final b1Rat = _rationalFromDouble(p1[1]);
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final c1Rat = _rationalFromDouble(p1[2]);
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final a2Rat = _rationalFromDouble(p2[0]);
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final b2Rat = _rationalFromDouble(p2[1]);
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final c2Rat = _rationalFromDouble(p2[2]);
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steps.add(
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CalculationStep(
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@@ -813,43 +834,56 @@ class SolverService {
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'''
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\$\$
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\\begin{cases}
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${a1}x ${b1 >= 0 ? '+' : ''} ${b1}y = $c1 & (1) \\\\
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${a2}x ${b2 >= 0 ? '+' : ''} ${b2}y = $c2 & (2)
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${_formatNumber(a1Rat)}x ${_formatNumber(b1Rat) == '0'
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? ''
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: b1Rat >= Rational.zero
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? '+'
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: ''} ${_formatNumber(b1Rat)}y = ${_formatNumber(c1Rat)} & (1) \\\\
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${_formatNumber(a2Rat)}x ${_formatNumber(b2Rat) == '0'
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? ''
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: b2Rat >= Rational.zero
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? '+'
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: ''} ${_formatNumber(b2Rat)}y = ${_formatNumber(c2Rat)} & (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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final det = a1Rat * b2Rat - a2Rat * b1Rat;
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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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final infiniteCheck = a1Rat * c2Rat - a2Rat * c1Rat;
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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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final newA1 = a1Rat * b2Rat;
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final newC1 = c1Rat * b2Rat;
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final newA2 = a2Rat * b1Rat;
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final newC2 = c2Rat * b1Rat;
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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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explanation:
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'为了消去变量 y，将方程(1)两边乘以 ${_formatNumber(b2Rat)}，方程(2)两边乘以 ${_formatNumber(b1Rat)}。',
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formula:
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'''
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\$\$
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\\begin{cases}
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${_formatNumber(newA1.toDouble(), precision: 2)}x ${b1 * b2 >= 0 ? '+' : ''} ${_formatNumber((b1 * b2), precision: 2)}y = ${_formatNumber(newC1.toDouble(), precision: 2)} & (3) \\\\
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${_formatNumber(newA2.toDouble(), precision: 2)}x ${b1 * b2 >= 0 ? '+' : ''} ${_formatNumber((b1 * b2), precision: 2)}y = ${_formatNumber(newC2.toDouble(), precision: 2)} & (4)
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${_formatNumber(newA1)}x ${_formatNumber(b1Rat * b2Rat) == '0'
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? ''
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: b1Rat * b2Rat >= Rational.zero
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? '+'
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: ''} ${_formatNumber(b1Rat * b2Rat)}y = ${_formatNumber(newC1)} & (3) \\\\
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${_formatNumber(newA2)}x ${_formatNumber(b1Rat * b2Rat) == '0'
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? ''
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: b1Rat * b2Rat >= Rational.zero
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? '+'
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: ''} ${_formatNumber(b1Rat * b2Rat)}y = ${_formatNumber(newC2)} & (4)
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\\end{cases}
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\$\$
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''',
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@@ -865,7 +899,7 @@ ${_formatNumber(newA2.toDouble(), precision: 2)}x ${b1 * b2 >= 0 ? '+' : ''} ${_
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title: '相减',
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explanation: '将方程(3)减去方程(4)，得到一个只含 x 的方程。',
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formula:
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'\$\$(${_formatNumber(newA1.toDouble(), precision: 2)} - ${_formatNumber(newA2.toDouble(), precision: 2)})x = ${_formatNumber(newC1.toDouble(), precision: 2)} - ${_formatNumber(newC2.toDouble(), precision: 2)} \\Rightarrow ${_formatNumber(xCoeff.toDouble(), precision: 2)}x = ${_formatNumber(constCoeff.toDouble(), precision: 2)}\$\$',
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'\$\$(${_formatNumber(newA1)} - ${_formatNumber(newA2)})x = ${_formatNumber(newC1)} - ${_formatNumber(newC2)} \\Rightarrow ${_formatNumber(xCoeff)}x = ${_formatNumber(constCoeff)}\$\$',
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),
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);
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@@ -875,27 +909,27 @@ ${_formatNumber(newA2.toDouble(), precision: 2)}x ${b1 * b2 >= 0 ? '+' : ''} ${_
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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 = ${_formatNumber(x.toDouble())}\$\$',
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formula: '\$\$x = ${_formatNumber(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.toDouble();
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if (b1Rat.abs() < Rational(BigInt.from(1), BigInt.from(1000000000))) {
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final yCoeff = b2Rat;
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final yConst = c2Rat - a2Rat * 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 = ${_formatNumber(x.toDouble())} 代入原方程(2)中。',
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explanation: '将 x = ${_formatNumber(x)} 代入原方程(2)中。',
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formula:
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'''
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\$\$
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\\begin{aligned}
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$a2(${_formatNumber(x.toDouble())}) + ${b2}y &= $c2 \\\\
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${a2 * x.toDouble()} + ${b2}y &= $c2 \\\\
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${b2}y &= $c2 - ${a2 * x.toDouble()} \\\\
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${b2}y &= ${c2 - a2 * x.toDouble()}
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${_formatNumber(a2Rat)}(${_formatNumber(x)}) + ${_formatNumber(b2Rat)}y &= ${_formatNumber(c2Rat)} \\\\
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${_formatNumber(a2Rat * x)} + ${_formatNumber(b2Rat)}y &= ${_formatNumber(c2Rat)} \\\\
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${_formatNumber(b2Rat)}y &= ${_formatNumber(c2Rat)} - ${_formatNumber(a2Rat * x)} \\\\
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${_formatNumber(b2Rat)}y &= ${_formatNumber(c2Rat - a2Rat * x)}
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\\end{aligned}
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\$\$
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''',
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@@ -906,31 +940,31 @@ ${b2}y &= ${c2 - a2 * x.toDouble()}
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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 = ${_formatNumber(y.toDouble())}\$\$',
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formula: '\$\$y = ${_formatNumber(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:
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'\$\$x = ${_formatNumber(x.toDouble())}, \\quad y = ${_formatNumber(y.toDouble())}\$\$',
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'\$\$x = ${_formatNumber(x)}, \\quad y = ${_formatNumber(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.toDouble();
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final yCoeff = b1Rat;
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final yConst = c1Rat - a1Rat * 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 = ${_formatNumber(x.toDouble())} 代入原方程(1)中。',
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explanation: '将 x = ${_formatNumber(x)} 代入原方程(1)中。',
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formula:
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'''
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\$\$
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\\begin{aligned}
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$a1(${_formatNumber(x.toDouble())}) + ${b1}y &= $c1 \\\\
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${a1 * x.toDouble()} + ${b1}y &= $c1 \\\\
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${b1}y &= $c1 - ${a1 * x.toDouble()} \\\\
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${b1}y &= ${c1 - a1 * x.toDouble()}
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${_formatNumber(a1Rat)}(${_formatNumber(x)}) + ${_formatNumber(b1Rat)}y &= ${_formatNumber(c1Rat)} \\\\
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${_formatNumber(a1Rat * x)} + ${_formatNumber(b1Rat)}y &= ${_formatNumber(c1Rat)} \\\\
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${_formatNumber(b1Rat)}y &= ${_formatNumber(c1Rat)} - ${_formatNumber(a1Rat * x)} \\\\
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${_formatNumber(b1Rat)}y &= ${_formatNumber(c1Rat - a1Rat * x)}
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\\end{aligned}
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\$\$
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''',
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@@ -941,13 +975,13 @@ ${b1}y &= ${c1 - a1 * x.toDouble()}
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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 = ${_formatNumber(y.toDouble())}\$\$',
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formula: '\$\$y = ${_formatNumber(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:
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'\$\$x = ${_formatNumber(x.toDouble())}, \\quad y = ${_formatNumber(y.toDouble())}\$\$',
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'\$\$x = ${_formatNumber(x)}, \\quad y = ${_formatNumber(y)}\$\$',
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);
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}
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}
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@@ -1218,10 +1252,10 @@ ${b1}y &= ${c1 - a1 * x.toDouble()}
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final rightCoeffs = _parsePolynomial(parts[1], variable);
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return LinearEquationParts(
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(leftCoeffs[1] ?? 0.0),
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(leftCoeffs[0] ?? 0.0),
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(rightCoeffs[1] ?? 0.0),
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(rightCoeffs[0] ?? 0.0),
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_rationalFromDouble(leftCoeffs[1] ?? 0.0),
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_rationalFromDouble(leftCoeffs[0] ?? 0.0),
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_rationalFromDouble(rightCoeffs[1] ?? 0.0),
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_rationalFromDouble(rightCoeffs[0] ?? 0.0),
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);
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}
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@@ -1702,7 +1736,18 @@ ${b1}y &= ${c1 - a1 * x.toDouble()}
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/// 格式化有理数为 LaTeX 分数形式
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String _formatRational(Rational r) {
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if (r.denominator == BigInt.one) return r.numerator.toString();
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return '\\frac{${r.numerator}}{${r.denominator}}';
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// 简化分数
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final gcd = r.numerator.gcd(r.denominator);
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final num = r.numerator ~/ gcd;
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final den = r.denominator ~/ gcd;
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// 处理负号，将其移到分子
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if (den < BigInt.zero) {
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return '\\frac{${-num}}{${-den}}';
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}
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return '\\frac{$num}{$den}';
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}
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/// 从因式分解形式计算二次方程的根
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