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✨ Add latex generating in steps

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LittleSheep
2025-09-13 00:48:44 +08:00
parent 9c820cf762
commit 9f88a2273c
1 changed files with 148 additions and 155 deletions
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303
lib/solver_service.dart
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@@ -56,13 +56,12 @@ class SolverService {
final steps = <CalculationStep>[];
steps.add(
CalculationStep(
title: "第一步：表达式求值",
explanation: "这是一个标准的数学表达式，我们将直接计算其结果。",
title: '第一步：表达式求值',
explanation: '这是一个标准的数学表达式，我们将直接计算其结果。',
formula: input,
),
);
// **FIXED**: Correct usage of the math_expressions library
Parser p = Parser();
Expression exp = p.parse(input);
ContextModel cm = ContextModel();
@@ -75,134 +74,118 @@ class SolverService {
CalculationResult _solveLinearEquation(String input) {
final steps = <CalculationStep>[];
steps.add(
CalculationStep(title: "原方程", explanation: "这是一元一次方程。", formula: input),
CalculationStep(
title: '原方程',
explanation: '这是一元一次方程。',
formula: '\$\$$input\$\$',
),
);
// 解析方程: ax+b=cx+d
final parts = _parseLinearEquation(input);
final a = parts.a, b = parts.b, c = parts.c, d = parts.d;
// 移项合并
final newA = a - c;
final newD = d - b;
steps.add(
CalculationStep(
title: "第一步：移项",
explanation: "将所有含 x 的项移到等式左边，常数项移到右边。",
title: '第一步：移项',
explanation: '将所有含 x 的项移到等式左边，常数项移到右边。',
formula:
"${a}x ${c >= 0 ? '-' : '+'} ${c.abs()}x = $d ${b >= 0 ? '-' : '+'} ${b.abs()}",
'\$\$${a}x ${c >= 0 ? '-' : '+'} ${c.abs()}x = $d ${b >= 0 ? '-' : '+'} ${b.abs()}\$\$',
),
);
steps.add(
CalculationStep(
title: "第二步：合并同类项",
explanation: "合并等式两边的项。",
formula: "${newA}x = $newD",
title: '第二步：合并同类项',
explanation: '合并等式两边的项。',
formula: '\$\$${newA}x = $newD\$\$',
),
);
// 求解x
if (newA == 0) {
if (newD == 0) {
return CalculationResult(steps: steps, finalAnswer: "有无穷多解");
} else {
return CalculationResult(steps: steps, finalAnswer: "无解");
}
return CalculationResult(
steps: steps,
finalAnswer: newD == 0 ? '有无穷多解' : '无解',
);
}
final x = newD / newA;
steps.add(
CalculationStep(
title: "第三步：求解 x",
explanation: "两边同时除以 x 的系数 ($newA)。",
formula: "x = $newD / $newA",
title: '第三步：求解 x',
explanation: '两边同时除以 x 的系数 ($newA)。',
formula: '\$\$x = \frac{$newD}{$newA}\$\$',
),
);
return CalculationResult(steps: steps, finalAnswer: "x = $x");
return CalculationResult(steps: steps, finalAnswer: '\$\$x = $x\$\$');
}
/// 3. 求解一元二次方程 (升级版)
CalculationResult _solveQuadraticEquation(String input) {
final steps = <CalculationStep>[];
// 整理成 ax^2+bx+c=0 的形式并提取系数
final eqParts = input.split('=');
if (eqParts.length != 2) throw Exception("方程格式错误，应包含一个 '='。");
final leftCoeffs = _parsePolynomial(eqParts[0]);
final rightCoeffs = _parsePolynomial(eqParts[1]);
// 移项合并
final a = (leftCoeffs[2] ?? 0) - (rightCoeffs[2] ?? 0);
final b = (leftCoeffs[1] ?? 0) - (rightCoeffs[1] ?? 0);
final c = (leftCoeffs[0] ?? 0) - (rightCoeffs[0] ?? 0);
if (a == 0) {
// 如果 a=0, 这实际上是一个一元一次方程
return _solveLinearEquation("${b}x+${c}=0");
return _solveLinearEquation('${b}x+$c=0');
}
// **FIXED**: Corrected typo 'ax' to 'a'
steps.add(
CalculationStep(
title: "第一步：整理方程",
explanation: "将方程整理成标准形式 ax^2+bx+c=0。",
title: '第一步：整理方程',
explanation: r'将方程整理成标准形式 $ax^2+bx+c=0$。',
formula:
"${a}x^2 ${b >= 0 ? '+' : ''} ${b}x ${c >= 0 ? '+' : ''} $c = 0",
'\$\$${a}x^2 ${b >= 0 ? '+' : ''} ${b}x ${c >= 0 ? '+' : ''} $c = 0\$\$',
),
);
// 尝试因式分解 (十字相乘法)
if (a == a.round() && b == b.round() && c == c.round()) {
final factors = _tryFactorization(a.toInt(), b.toInt(), c.toInt());
if (factors != null) {
steps.add(
CalculationStep(
title: "第二步：因式分解法 (十字相乘)",
explanation: "我们发现可以将方程分解为两个一次因式的乘积。",
title: '第二步：因式分解法 (十字相乘)',
explanation: '我们发现可以将方程分解为两个一次因式的乘积。',
formula: factors.formula,
),
);
steps.add(
CalculationStep(
title: "第三步：求解",
explanation: "分别令每个因式等于 0，解出 x。",
formula: "解得 ${factors.solution}",
title: '第三步：求解',
explanation: '分别令每个因式等于 0，解出 x。',
formula: '解得 ${factors.solution}',
),
);
return CalculationResult(steps: steps, finalAnswer: factors.solution);
} else {
steps.add(
CalculationStep(
title: "第二步：选择解法",
explanation: "尝试因式分解失败，我们选择使用求根公式法。",
formula: "\$\\Delta = b^2 - 4ac\$",
),
);
}
} else {
steps.add(
CalculationStep(
title: "第二步：选择解法",
explanation: "系数包含小数，我们使用求根公式法。",
formula: "\$\\Delta = b^2 - 4ac\$",
),
);
}
// 使用求根公式法
steps.add(
CalculationStep(
title: '第二步：选择解法',
explanation: '无法进行因式分解，我们选择使用求根公式法。',
formula: r'\$\$\Delta = b^2 - 4ac\$\$',
),
);
final delta = b * b - 4 * a * c;
steps.add(
CalculationStep(
title: "第三步：计算判别式 (Delta)",
explanation: "\$\\Delta = ($b)^2 - 4 \\cdot ($a) \\cdot ($c) = $delta",
// **FIXED**: Removed unnecessary braces for linter warning
formula: "\$\\Delta = $delta",
title: '第三步：计算判别式 (Delta)',
explanation:
'\$\$\Delta = b^2 - 4ac = ($b)^2 - 4 \cdot ($a) \cdot ($c) = $delta\$\$',
formula: '\$\$\Delta = $delta\$\$',
),
);
@@ -211,40 +194,40 @@ class SolverService {
final x2 = (-b - sqrt(delta)) / (2 * a);
steps.add(
CalculationStep(
title: "第四步：应用求根公式",
title: '第四步：应用求根公式',
explanation:
"因为 \$\\Delta > 0\$，方程有两个不相等的实数根。公式: \$x = \\frac{-b \\pm \\sqrt{\\Delta}}{2a}\$",
r'因为 $\Delta > 0$，方程有两个不相等的实数根。公式: $x = \frac{-b \pm \sqrt{\Delta}}{2a}$。',
formula:
"x₁ = ${x1.toStringAsFixed(4)}, x₂ = ${x2.toStringAsFixed(4)}",
'\$\$x_1 = ${x1.toStringAsFixed(4)}, \quad x_2 = ${x2.toStringAsFixed(4)}\$\$',
),
);
return CalculationResult(
steps: steps,
finalAnswer:
"x₁ = ${x1.toStringAsFixed(4)}, x₂ = ${x2.toStringAsFixed(4)}",
'\$\$x_1 = ${x1.toStringAsFixed(4)}, \quad x_2 = ${x2.toStringAsFixed(4)}\$\$',
);
} else if (delta == 0) {
final x = -b / (2 * a);
steps.add(
CalculationStep(
title: "第四步：应用求根公式",
explanation: "因为 \$\Delta = 0\$，方程有两个相等的实数根。",
formula: "x₁ = x₂ = ${x.toStringAsFixed(4)}",
title: '第四步：应用求根公式',
explanation: r'因为 $\Delta = 0$，方程有两个相等的实数根。',
formula: '\$\$x_1 = x_2 = ${x.toStringAsFixed(4)}\$\$',
),
);
return CalculationResult(
steps: steps,
finalAnswer: "x₁ = x₂ = ${x.toStringAsFixed(4)}",
finalAnswer: '\$\$x_1 = x_2 = ${x.toStringAsFixed(4)}\$\$',
);
} else {
steps.add(
CalculationStep(
title: "第四步：判断解",
explanation: "因为 \$\Delta < 0\$，该方程在实数范围内无解。",
formula: "无实数解",
title: '第四步：判断解',
explanation: r'因为 $\Delta < 0$，该方程在实数范围内无解。',
formula: '无实数解',
),
);
return CalculationResult(steps: steps, finalAnswer: "无实数解");
return CalculationResult(steps: steps, finalAnswer: '无实数解');
}
}
@@ -262,32 +245,44 @@ class SolverService {
steps.add(
CalculationStep(
title: "原始方程组",
explanation: "这是一个二元一次方程组，我们将使用加减消元法求解。",
title: '原始方程组',
explanation: '这是一个二元一次方程组，我们将使用加减消元法求解。',
formula:
"${a1}x ${b1 >= 0 ? '+' : ''} ${b1}y = $c1 ---(1)\n${a2}x ${b2 >= 0 ? '+' : ''} ${b2}y = $c2 ---(2)",
'''
egin{cases}
${a1}x ${b1 >= 0 ? '+' : ''} ${b1}y = $c1 & (1) \\
${a2}x ${b2 >= 0 ? '+' : ''} ${b2}y = $c2 & (2)
\\end{cases}
''',
),
);
final det = a1 * b2 - a2 * b1;
if (det == 0) {
if (a1 * c2 - a2 * c1 == 0) {
return CalculationResult(steps: steps, finalAnswer: "有无穷多解");
} else {
return CalculationResult(steps: steps, finalAnswer: "无解");
}
return CalculationResult(
steps: steps,
finalAnswer: a1 * c2 - a2 * c1 == 0 ? '有无穷多解' : '无解',
);
}
final newA1 = a1 * b2, newC1 = c1 * b2;
final newA2 = a2 * b1, newC2 = c2 * b1;
// **FIXED**: Wrapped calculations in braces {} for string interpolation
steps.add(
CalculationStep(
title: "第一步：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\$。",
title: '第一步：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 $b1。',
formula:
"${newA1}x ${b1 * b2 >= 0 ? '+' : ''} ${b1 * b2}y = $newC1 ---(3)\n${newA2}x ${b1 * b2 >= 0 ? '+' : ''} ${b1 * b2}y = $newC2 ---(4)",
'''
egin{cases}
${newA1}x ${b1 * b2 >= 0 ? '+' : ''} ${b1 * b2}y = $newC1 & (3) \\
${newA2}x ${b1 * b2 >= 0 ? '+' : ''} ${b1 * b2}y = $newC2 & (4)
\\end{cases}
''',
),
);
@@ -296,79 +291,96 @@ class SolverService {
steps.add(
CalculationStep(
title: "第二步：相减",
explanation: "将方程(3)减去方程(4)，得到一个只含 x 的方程。",
title: '第二步：相减',
explanation: '将方程(3)减去方程(4)，得到一个只含 x 的方程。',
formula:
"($newA1 - $newA2)x = $newC1 - $newC2 \n=> ${xCoeff}x = $constCoeff",
'\$\$($newA1 - $newA2)x = $newC1 - $newC2 \Rightarrow ${xCoeff}x = $constCoeff\$\$',
),
);
final x = constCoeff / xCoeff;
steps.add(
CalculationStep(
title: "第三步：解出 x",
explanation: "求解上述方程得到 x 的值。",
formula: "x = $x",
title: '第三步：解出 x',
explanation: '求解上述方程得到 x 的值。',
formula: '\$\$x = $x\$\$',
),
);
// **FIXED**: Added a check for b1 to avoid division by zero if we substitute into an equation without y.
if (b1.abs() < 1e-9) {
// If b1 is very close to 0, use the second equation
final yCoeff = b2;
final yConst = c2 - a2 * x;
final y = yConst / yCoeff;
steps.add(
CalculationStep(
title: "第四步：回代求解 y",
explanation: "将 x = $x 代入原方程(2)中。",
// **FIXED**: Corrected string interpolation for calculations and substitutions
title: '第四步：回代求解 y',
explanation: '将 x = $x 代入原方程(2)中。',
formula:
"${a2}(${x}) + ${b2}y = $c2 \n=> ${a2 * x} + ${b2}y = $c2 \n=> ${b2}y = $c2 - ${a2 * x} \n=> ${b2}y = ${c2 - a2 * x}",
'''
\\begin{aligned}
$a2($x) + ${b2}y &= $c2 \\
${a2 * x} + ${b2}y &= $c2 \\
${b2}y &= $c2 - ${a2 * x} \\
${b2}y &= ${c2 - a2 * x}
\\end{aligned}
''',
),
);
steps.add(
CalculationStep(
title: "第五步：解出 y",
explanation: "求解得到 y 的值。",
formula: "y = $y",
title: '第五步：解出 y',
explanation: '求解得到 y 的值。',
formula: '\$\$y = $y\$\$',
),
);
return CalculationResult(steps: steps, finalAnswer: "x = $x, y = $y");
return CalculationResult(
steps: steps,
finalAnswer: '\$\$x = $x, \quad y = $y\$\$',
);
} else {
// Default case, substitute into the first equation
final yCoeff = b1;
final yConst = c1 - a1 * x;
final y = yConst / yCoeff;
steps.add(
CalculationStep(
title: "第四步：回代求解 y",
explanation: "将 x = $x 代入原方程(1)中。",
title: '第四步：回代求解 y',
explanation: '将 x = $x 代入原方程(1)中。',
formula:
"${a1}(${x}) + ${b1}y = $c1 \n=> ${a1 * x} + ${b1}y = $c1 \n=> ${b1}y = $c1 - ${a1 * x} \n=> ${b1}y = ${c1 - a1 * x}",
'''
\\begin{aligned}
$a1($x) + ${b1}y &= $c1 \\
${a1 * x} + ${b1}y &= $c1 \\
${b1}y &= $c1 - ${a1 * x} \\
${b1}y &= ${c1 - a1 * x}
\\end{aligned}
''',
),
);
steps.add(
CalculationStep(
title: "第五步：解出 y",
explanation: "求解得到 y 的值。",
formula: "y = $y",
title: '第五步：解出 y',
explanation: '求解得到 y 的值。',
formula: '\$\$y = $y\$\$',
),
);
return CalculationResult(steps: steps, finalAnswer: "x = $x, y = $y");
return CalculationResult(
steps: steps,
finalAnswer: '\$\$x = $x, \\quad y = $y\$\$',
);
}
}
/// ---- 辅助函数 ----
/// 展开表达式，例如 (x-1)(x+2) -> x^2+x-2
String _expandExpressions(String input) {
String result = input;
// 循环直到没有更多的表达式可以展开
while (true) {
String oldResult = result;
// 模式1: k*(ax+b)^2, 例如 4(x-1)^2 or (x-1)^2
final powerMatch = RegExp(
r'(-?\d*\.?\d*)?\(([^)]+)\)\^2',
).firstMatch(result);
@@ -376,11 +388,7 @@ class SolverService {
final kStr = powerMatch.group(1);
double k = 1.0;
if (kStr != null && kStr.isNotEmpty) {
if (kStr == '-') {
k = -1.0;
} else {
k = double.parse(kStr);
}
k = kStr == '-' ? -1.0 : double.parse(kStr);
}
final factor = powerMatch.group(2)!;
@@ -388,18 +396,16 @@ class SolverService {
final a = coeffs[1] ?? 0;
final b = coeffs[0] ?? 0;
// (ax+b)^2 = a^2*x^2 + 2ab*x + b^2
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)");
continue; // 重新开始循环以处理嵌套表达式
'${newA}x^2${newB >= 0 ? '+' : ''}${newB}x${newC >= 0 ? '+' : ''}$newC';
result = result.replaceFirst(powerMatch.group(0)!, '($expanded)');
continue;
}
// 模式2: (ax+b)(cx+d), 例如 (x-3)(x+2)
final factorMulMatch = RegExp(
r'\(([^)]+)\)\(([^)]+)\)',
).firstMatch(result);
@@ -414,21 +420,17 @@ class SolverService {
final c = coeffs2[1] ?? 0;
final d = coeffs2[0] ?? 0;
// (ax+b)(cx+d) = ac*x^2 + (ad+bc)*x + bd
final newA = a * c;
final newB = a * d + b * c;
final newC = b * d;
final expanded =
"${newA}x^2${newB >= 0 ? '+' : ''}${newB}x${newC >= 0 ? '+' : ''}${newC}";
result = result.replaceFirst(factorMulMatch.group(0)!, "($expanded)");
continue; // 重新开始循环
'${newA}x^2${newB >= 0 ? '+' : ''}${newB}x${newC >= 0 ? '+' : ''}$newC';
result = result.replaceFirst(factorMulMatch.group(0)!, '($expanded)');
continue;
}
// 如果在这次迭代中没有改变，则跳出循环
if (result == oldResult) {
break;
}
if (result == oldResult) break;
}
return result;
}
@@ -440,18 +442,18 @@ class SolverService {
final leftCoeffs = _parsePolynomial(parts[0]);
final rightCoeffs = _parsePolynomial(parts[1]);
final a = leftCoeffs[1] ?? 0.0;
final b = leftCoeffs[0] ?? 0.0;
final c = rightCoeffs[1] ?? 0.0;
final d = rightCoeffs[0] ?? 0.0;
return LinearEquationParts(a, b, c, d);
return LinearEquationParts(
(leftCoeffs[1] ?? 0.0),
(leftCoeffs[0] ?? 0.0),
(rightCoeffs[1] ?? 0.0),
(rightCoeffs[0] ?? 0.0),
);
}
Map<int, double> _parsePolynomial(String side) {
final coeffs = <int, double>{};
final pattern = RegExp(
r'([+-]?(?:\d*\.?\d*))?x(?:\^(\d+))?|([+-]?\d*\.?\d+)',
r'([+-]?(?:\d*\.?\d*)?)x(?:\^(\d+))?|([+-]?\d*\.?\d+)',
);
var s = side.startsWith('+') || side.startsWith('-') ? side : '+$side';
@@ -463,11 +465,7 @@ class SolverService {
String coeffStr = match.group(1) ?? '+';
double coeff = 1.0;
if (coeffStr.isNotEmpty && coeffStr != '+') {
if (coeffStr == '-') {
coeff = -1.0;
} else {
coeff = double.parse(coeffStr);
}
coeff = coeffStr == '-' ? -1.0 : double.parse(coeffStr);
} else if (coeffStr == '-') {
coeff = -1.0;
}
@@ -486,7 +484,7 @@ class SolverService {
final xMatch = RegExp(r'([+-]?\d*\.?\d*)x').firstMatch(parts[0]);
if (xMatch != null) {
final coeff = xMatch.group(1);
if (coeff == null || coeff == '+') {
if (coeff == null || coeff.isEmpty || coeff == '+') {
a = 1.0;
} else if (coeff == '-') {
a = -1.0;
@@ -497,7 +495,7 @@ class SolverService {
final yMatch = RegExp(r'([+-]?\d*\.?\d*)y').firstMatch(parts[0]);
if (yMatch != null) {
final coeff = yMatch.group(1);
if (coeff == null || coeff == '+') {
if (coeff == null || coeff.isEmpty || coeff == '+') {
b = 1.0;
} else if (coeff == '-') {
b = -1.0;
@@ -523,10 +521,7 @@ class SolverService {
return null;
}
// **FIXED**: Simplified check method
bool check(int m, int n, int b) {
return m + n == b;
}
bool check(int m, int n, int b) => m + n == b;
({String formula, String solution}) formatFactor(int m, int n, int a) {
int common = gcd(n.abs(), a.abs());
@@ -538,25 +533,23 @@ class SolverService {
final a2 = a ~/ den;
final c2 = m ~/ a2;
// **FIXED**: Correctly handle coefficients of 1
final f1Part1 = a1 == 1 ? 'x' : '${a1}x';
final f1 = c1 == 0 ? f1Part1 : "$f1Part1 ${c1 >= 0 ? '+' : ''} $c1";
final f1 = c1 == 0 ? f1Part1 : '$f1Part1 ${c1 >= 0 ? '+' : ''} $c1';
final f2Part1 = a2 == 1 ? 'x' : '${a2}x';
final f2 = c2 == 0 ? f2Part1 : "$f2Part1 ${c2 >= 0 ? '+' : ''} $c2";
final f2 = c2 == 0 ? f2Part1 : '$f2Part1 ${c2 >= 0 ? '+' : ''} $c2';
// **FIXED**: Renamed variables to lowerCamelCase
final int x1Num = -c1, x1Den = a1;
final int x2Num = -c2, x2Den = a2;
final sol1 = x1Den == 1 ? "$x1Num" : "$x1Num/$x1Den";
final sol2 = x2Den == 1 ? "$x2Num" : "$x2Num/$x2Den";
final sol1 = x1Den == 1 ? '$x1Num' : '\\frac{$x1Num}{$x1Den}';
final sol2 = x2Den == 1 ? '$x2Num' : '\\frac{$x2Num}{$x2Den}';
final solution = x1Num * x2Den == x2Num * x1Den
? "x₁ = x₂ = $sol1"
: "x₁ = $sol1, x₂ = $sol2";
? 'x_1 = x_2 = $sol1'
: 'x_1 = $sol1, \\quad x_2 = $sol2';
return (formula: "(${f1})(${f2}) = 0", solution: solution);
return (formula: '\$\$($f1)($f2) = 0\$\$', solution: '\$\$$solution\$\$');
}
int gcd(int a, int b) => b == 0 ? a : gcd(b, a % b);