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🐛 Bug fixes

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LittleSheep
2025-09-17 22:41:39 +08:00
parent 2e542a6c23
commit 48c00cc1e6
1 changed files with 104 additions and 59 deletions
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163
lib/solver.dart
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@@ -7,21 +7,29 @@ import 'models/calculation_step.dart';
/// 帮助解析一元一次方程 ax+b=cx+d 的辅助类
class LinearEquationParts {
final double a, b, c, d;
final Rational a, b, c, d;
LinearEquationParts(this.a, this.b, this.c, this.d);
}
class SolverService {
/// 格式化数字，移除不必要的尾随零
String _formatNumber(double value, {int precision = 4}) {
String formatted = value.toStringAsFixed(precision);
// 移除尾随的零和小数点
formatted = formatted.replaceAll(RegExp(r'\.0+$'), '');
// 如果最后是小数点，也移除
if (formatted.endsWith('.')) {
formatted = formatted.substring(0, formatted.length - 1);
String _formatNumber(dynamic value, {int precision = 4}) {
if (value is Rational) {
return _formatRational(value);
} else if (value is FractionExpr) {
return '\\frac{${value.numerator}}{${value.denominator}}';
} else if (value is double) {
String formatted = value.toStringAsFixed(precision);
// 移除尾随的零和小数点
formatted = formatted.replaceAll(RegExp(r'\.0+$'), '');
// 如果最后是小数点，也移除
if (formatted.endsWith('.')) {
formatted = formatted.substring(0, formatted.length - 1);
}
return formatted;
} else {
return value.toString();
}
return formatted;
}
/// 尝试对二次方程进行因式分解
@@ -345,8 +353,8 @@ class SolverService {
final parts = _parseLinearEquation(input, variable);
final a = parts.a, b = parts.b, c = parts.c, d = parts.d;
final newA = _rationalFromDouble(a) - _rationalFromDouble(c);
final newD = _rationalFromDouble(d) - _rationalFromDouble(b);
final newA = a - c;
final newD = d - b;
steps.add(
CalculationStep(
@@ -354,7 +362,15 @@ class SolverService {
title: '移项',
explanation: '将所有含 $variable 的项移到等式左边，常数项移到右边。',
formula:
'\$\$$a$variable ${c >= 0 ? '-' : '+'} ${c.abs()}$variable = $d ${b >= 0 ? '-' : '+'} ${b.abs()}\$\$',
'\$\$${_formatNumber(a)}$variable ${_formatNumber(c) == '0'
? ''
: c >= Rational.zero
? '-'
: '+'} ${_formatNumber(c.abs())}$variable = ${_formatNumber(d)} ${_formatNumber(b) == '0'
? ''
: b >= Rational.zero
? '-'
: '+'} ${_formatNumber(b.abs())}\$\$',
),
);
@@ -364,7 +380,7 @@ class SolverService {
title: '合并同类项',
explanation: '合并等式两边的项。',
formula:
'\$\$${_formatNumber(newA.toDouble())}$variable = ${_formatNumber(newD.toDouble())}\$\$',
'\$\$${_formatNumber(newA)}$variable = ${_formatNumber(newD)}\$\$',
),
);
@@ -380,14 +396,15 @@ class SolverService {
CalculationStep(
stepNumber: 3,
title: '求解 $variable',
explanation: '两边同时除以 $variable 的系数 ($newA)。',
formula: '\$\$$variable = \\frac{$newD}{$newA}\$\$',
explanation: '两边同时除以 $variable 的系数 (${_formatNumber(newA)})。',
formula:
'\$\$$variable = \\frac{${_formatNumber(newD)}}{${_formatNumber(newA)}}\$\$',
),
);
return CalculationResult(
steps: steps,
finalAnswer: '\$\$$variable = $x\$\$',
finalAnswer: '\$\$$variable = ${_formatNumber(x)}\$\$',
);
}
@@ -801,8 +818,12 @@ class SolverService {
final p1 = _parseTwoVariableLinear(equations[0]);
final p2 = _parseTwoVariableLinear(equations[1]);
double a1 = p1[0], b1 = p1[1], c1 = p1[2];
double a2 = p2[0], b2 = p2[1], c2 = p2[2];
final a1Rat = _rationalFromDouble(p1[0]);
final b1Rat = _rationalFromDouble(p1[1]);
final c1Rat = _rationalFromDouble(p1[2]);
final a2Rat = _rationalFromDouble(p2[0]);
final b2Rat = _rationalFromDouble(p2[1]);
final c2Rat = _rationalFromDouble(p2[2]);
steps.add(
CalculationStep(
@@ -813,43 +834,56 @@ class SolverService {
'''
\$\$
\\begin{cases}
${a1}x ${b1 >= 0 ? '+' : ''} ${b1}y = $c1 & (1) \\\\
${a2}x ${b2 >= 0 ? '+' : ''} ${b2}y = $c2 & (2)
${_formatNumber(a1Rat)}x ${_formatNumber(b1Rat) == '0'
? ''
: b1Rat >= Rational.zero
? '+'
: ''} ${_formatNumber(b1Rat)}y = ${_formatNumber(c1Rat)} & (1) \\\\
${_formatNumber(a2Rat)}x ${_formatNumber(b2Rat) == '0'
? ''
: b2Rat >= Rational.zero
? '+'
: ''} ${_formatNumber(b2Rat)}y = ${_formatNumber(c2Rat)} & (2)
\\end{cases}
\$\$
''',
),
);
final det =
_rationalFromDouble(a1) * _rationalFromDouble(b2) -
_rationalFromDouble(a2) * _rationalFromDouble(b1);
final det = a1Rat * b2Rat - a2Rat * b1Rat;
if (det == Rational.zero) {
final infiniteCheck =
_rationalFromDouble(a1) * _rationalFromDouble(c2) -
_rationalFromDouble(a2) * _rationalFromDouble(c1);
final infiniteCheck = a1Rat * c2Rat - a2Rat * c1Rat;
return CalculationResult(
steps: steps,
finalAnswer: infiniteCheck == Rational.zero ? '有无穷多解' : '无解',
);
}
final newA1 = _rationalFromDouble(a1) * _rationalFromDouble(b2);
final newC1 = _rationalFromDouble(c1) * _rationalFromDouble(b2);
final newA2 = _rationalFromDouble(a2) * _rationalFromDouble(b1);
final newC2 = _rationalFromDouble(c2) * _rationalFromDouble(b1);
final newA1 = a1Rat * b2Rat;
final newC1 = c1Rat * b2Rat;
final newA2 = a2Rat * b1Rat;
final newC2 = c2Rat * b1Rat;
steps.add(
CalculationStep(
stepNumber: 1,
title: '消元',
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。',
explanation:
'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)}。',
formula:
'''
\$\$
\\begin{cases}
${_formatNumber(newA1.toDouble(), precision: 2)}x ${b1 * b2 >= 0 ? '+' : ''} ${_formatNumber((b1 * b2), precision: 2)}y = ${_formatNumber(newC1.toDouble(), precision: 2)} & (3) \\\\
${_formatNumber(newA2.toDouble(), precision: 2)}x ${b1 * b2 >= 0 ? '+' : ''} ${_formatNumber((b1 * b2), precision: 2)}y = ${_formatNumber(newC2.toDouble(), precision: 2)} & (4)
${_formatNumber(newA1)}x ${_formatNumber(b1Rat * b2Rat) == '0'
? ''
: b1Rat * b2Rat >= Rational.zero
? '+'
: ''} ${_formatNumber(b1Rat * b2Rat)}y = ${_formatNumber(newC1)} & (3) \\\\
${_formatNumber(newA2)}x ${_formatNumber(b1Rat * b2Rat) == '0'
? ''
: b1Rat * b2Rat >= Rational.zero
? '+'
: ''} ${_formatNumber(b1Rat * b2Rat)}y = ${_formatNumber(newC2)} & (4)
\\end{cases}
\$\$
''',
@@ -865,7 +899,7 @@ ${_formatNumber(newA2.toDouble(), precision: 2)}x ${b1 * b2 >= 0 ? '+' : ''} ${_
title: '相减',
explanation: '将方程(3)减去方程(4)，得到一个只含 x 的方程。',
formula:
'\$\$(${_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 = ${_formatNumber(constCoeff.toDouble(), precision: 2)}\$\$',
'\$\$(${_formatNumber(newA1)} - ${_formatNumber(newA2)})x = ${_formatNumber(newC1)} - ${_formatNumber(newC2)} \\Rightarrow ${_formatNumber(xCoeff)}x = ${_formatNumber(constCoeff)}\$\$',
),
);
@@ -875,27 +909,27 @@ ${_formatNumber(newA2.toDouble(), precision: 2)}x ${b1 * b2 >= 0 ? '+' : ''} ${_
stepNumber: 3,
title: '解出 x',
explanation: '求解上述方程得到 x 的值。',
formula: '\$\$x = ${_formatNumber(x.toDouble())}\$\$',
formula: '\$\$x = ${_formatNumber(x)}\$\$',
),
);
if (b1.abs() < 1e-9) {
final yCoeff = b2;
final yConst = c2 - a2 * x.toDouble();
if (b1Rat.abs() < Rational(BigInt.from(1), BigInt.from(1000000000))) {
final yCoeff = b2Rat;
final yConst = c2Rat - a2Rat * x;
final y = yConst / yCoeff;
steps.add(
CalculationStep(
stepNumber: 4,
title: '回代求解 y',
explanation: '将 x = ${_formatNumber(x.toDouble())} 代入原方程(2)中。',
explanation: '将 x = ${_formatNumber(x)} 代入原方程(2)中。',
formula:
'''
\$\$
\\begin{aligned}
$a2(${_formatNumber(x.toDouble())}) + ${b2}y &= $c2 \\\\
${a2 * x.toDouble()} + ${b2}y &= $c2 \\\\
${b2}y &= $c2 - ${a2 * x.toDouble()} \\\\
${b2}y &= ${c2 - a2 * x.toDouble()}
${_formatNumber(a2Rat)}(${_formatNumber(x)}) + ${_formatNumber(b2Rat)}y &= ${_formatNumber(c2Rat)} \\\\
${_formatNumber(a2Rat * x)} + ${_formatNumber(b2Rat)}y &= ${_formatNumber(c2Rat)} \\\\
${_formatNumber(b2Rat)}y &= ${_formatNumber(c2Rat)} - ${_formatNumber(a2Rat * x)} \\\\
${_formatNumber(b2Rat)}y &= ${_formatNumber(c2Rat - a2Rat * x)}
\\end{aligned}
\$\$
''',
@@ -906,31 +940,31 @@ ${b2}y &= ${c2 - a2 * x.toDouble()}
stepNumber: 5,
title: '解出 y',
explanation: '求解得到 y 的值。',
formula: '\$\$y = ${_formatNumber(y.toDouble())}\$\$',
formula: '\$\$y = ${_formatNumber(y)}\$\$',
),
);
return CalculationResult(
steps: steps,
finalAnswer:
'\$\$x = ${_formatNumber(x.toDouble())}, \\quad y = ${_formatNumber(y.toDouble())}\$\$',
'\$\$x = ${_formatNumber(x)}, \\quad y = ${_formatNumber(y)}\$\$',
);
} else {
final yCoeff = b1;
final yConst = c1 - a1 * x.toDouble();
final yCoeff = b1Rat;
final yConst = c1Rat - a1Rat * x;
final y = yConst / yCoeff;
steps.add(
CalculationStep(
stepNumber: 4,
title: '回代求解 y',
explanation: '将 x = ${_formatNumber(x.toDouble())} 代入原方程(1)中。',
explanation: '将 x = ${_formatNumber(x)} 代入原方程(1)中。',
formula:
'''
\$\$
\\begin{aligned}
$a1(${_formatNumber(x.toDouble())}) + ${b1}y &= $c1 \\\\
${a1 * x.toDouble()} + ${b1}y &= $c1 \\\\
${b1}y &= $c1 - ${a1 * x.toDouble()} \\\\
${b1}y &= ${c1 - a1 * x.toDouble()}
${_formatNumber(a1Rat)}(${_formatNumber(x)}) + ${_formatNumber(b1Rat)}y &= ${_formatNumber(c1Rat)} \\\\
${_formatNumber(a1Rat * x)} + ${_formatNumber(b1Rat)}y &= ${_formatNumber(c1Rat)} \\\\
${_formatNumber(b1Rat)}y &= ${_formatNumber(c1Rat)} - ${_formatNumber(a1Rat * x)} \\\\
${_formatNumber(b1Rat)}y &= ${_formatNumber(c1Rat - a1Rat * x)}
\\end{aligned}
\$\$
''',
@@ -941,13 +975,13 @@ ${b1}y &= ${c1 - a1 * x.toDouble()}
stepNumber: 5,
title: '解出 y',
explanation: '求解得到 y 的值。',
formula: '\$\$y = ${_formatNumber(y.toDouble())}\$\$',
formula: '\$\$y = ${_formatNumber(y)}\$\$',
),
);
return CalculationResult(
steps: steps,
finalAnswer:
'\$\$x = ${_formatNumber(x.toDouble())}, \\quad y = ${_formatNumber(y.toDouble())}\$\$',
'\$\$x = ${_formatNumber(x)}, \\quad y = ${_formatNumber(y)}\$\$',
);
}
}
@@ -1218,10 +1252,10 @@ ${b1}y &= ${c1 - a1 * x.toDouble()}
final rightCoeffs = _parsePolynomial(parts[1], variable);
return LinearEquationParts(
(leftCoeffs[1] ?? 0.0),
(leftCoeffs[0] ?? 0.0),
(rightCoeffs[1] ?? 0.0),
(rightCoeffs[0] ?? 0.0),
_rationalFromDouble(leftCoeffs[1] ?? 0.0),
_rationalFromDouble(leftCoeffs[0] ?? 0.0),
_rationalFromDouble(rightCoeffs[1] ?? 0.0),
_rationalFromDouble(rightCoeffs[0] ?? 0.0),
);
}
@@ -1702,7 +1736,18 @@ ${b1}y &= ${c1 - a1 * x.toDouble()}
/// 格式化有理数为 LaTeX 分数形式
String _formatRational(Rational r) {
if (r.denominator == BigInt.one) return r.numerator.toString();
return '\\frac{${r.numerator}}{${r.denominator}}';
// 简化分数
final gcd = r.numerator.gcd(r.denominator);
final num = r.numerator ~/ gcd;
final den = r.denominator ~/ gcd;
// 处理负号，将其移到分子
if (den < BigInt.zero) {
return '\\frac{${-num}}{${-den}}';
}
return '\\frac{$num}{$den}';
}
/// 从因式分解形式计算二次方程的根