✨ 配方法
This commit is contained in:
325
lib/solver.dart
325
lib/solver.dart
@@ -261,93 +261,129 @@ class SolverService {
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CalculationStep(
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stepNumber: 2,
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title: '选择解法',
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explanation: '无法进行因式分解，我们选择使用求根公式法。',
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formula: '\$\$\\Delta = b^2 - 4ac\$\$',
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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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// Calculate delta symbolically
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final deltaSymbolic = _calculateDeltaSymbolic(
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aSymbolic,
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bSymbolic,
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cSymbolic,
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);
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final delta =
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_rationalFromDouble(b).pow(2) -
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Rational.fromInt(4) * _rationalFromDouble(a) * _rationalFromDouble(c);
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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: '计算判别式 (Delta)',
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explanation: '\$\$\\Delta = b^2 - 4ac = $deltaSymbolic\$\$',
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formula:
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'\$\$\\Delta = $deltaSymbolic = ${delta.toDouble().toStringAsFixed(4)}\$\$',
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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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final deltaDouble = delta.toDouble();
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if (deltaDouble > 0) {
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// Pass delta directly to maintain precision
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final x1Expr = _formatQuadraticRoot(-b, delta, 2 * a, true);
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final x2Expr = _formatQuadraticRoot(-b, delta, 2 * a, false);
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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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final constantStr = constantTerm >= 0
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? '+${constantTerm}'
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: constantTerm.toString();
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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: 4,
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title: '应用求根公式',
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explanation:
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r'因为 $\Delta > 0$，方程有两个不相等的实数根。公式: $x = \frac{-b \pm \sqrt{\Delta}}{2a}$。',
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formula: '\$\$x_1 = $x1Expr, \\quad x_2 = $x2Expr\$\$',
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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(
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steps: steps,
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finalAnswer: '\$\$x_1 = $x1Expr, \\quad x_2 = $x2Expr\$\$',
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);
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} else if (deltaDouble == 0) {
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final x = -b / (2 * a);
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steps.add(
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CalculationStep(
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stepNumber: 4,
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title: '应用求根公式',
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explanation: r'因为 $\Delta = 0$，方程有两个相等的实数根。',
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formula: '\$\$x_1 = x_2 = ${x.toStringAsFixed(4)}\$\$',
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),
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);
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return CalculationResult(
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steps: steps,
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finalAnswer: '\$\$x_1 = x_2 = ${x.toStringAsFixed(4)}\$\$',
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);
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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: 4,
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title: '判断解',
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explanation: r'因为 $\Delta < 0$，该方程在实数范围内无解，但有虚数解。',
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formula: '无实数解，有虚数解',
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),
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);
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// For complex roots, we need to handle -delta
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final negDelta = -delta;
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final sqrtNegDeltaStr = _formatSqrtFromRational(negDelta);
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final realPart = -b / (2 * a);
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final imagPartExpr = _formatImaginaryPart(sqrtNegDeltaStr, 2 * a);
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steps.add(
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CalculationStep(
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stepNumber: 5,
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title: '计算虚数根',
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explanation: '使用求根公式计算虚数根。',
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formula: r'$$x = \frac{-b \pm \sqrt{-\Delta} i}{2a}$$',
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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 = ${realPart.toStringAsFixed(4)} + $imagPartExpr, \\quad x_2 = ${realPart.toStringAsFixed(4)} - $imagPartExpr\$\$',
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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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@@ -1522,6 +1558,169 @@ ${b1}y &= ${c1 - a1 * x.toDouble()}
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return Rational(numerator, denominator);
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}
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/// 检查数值是否可以表示为符号平方根形式
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String? _getSymbolicSquareRoot(double value) {
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if (value <= 0) return null;
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// 对于完全平方数，直接返回整数平方根
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final sqrtValue = sqrt(value);
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final intSqrt = sqrtValue.toInt();
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if ((sqrtValue - intSqrt).abs() < 1e-10) {
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return intSqrt.toString();
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}
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// 检查是否可以表示为 k√m 的形式，其中 m 不是完全平方数
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// 遍历可能的 k 值，从大到小
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for (int k = sqrt(value).toInt(); k >= 2; k--) {
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final kSquared = k * k;
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if (kSquared > value) continue;
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final remaining = value / kSquared;
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final remainingSqrt = sqrt(remaining);
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final intRemainingSqrt = remainingSqrt.toInt();
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// 检查剩余部分是否为完全平方数
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if ((remainingSqrt - intRemainingSqrt).abs() < 1e-10) {
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// 找到匹配：value = k² * m，其中 m 是完全平方数
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if (intRemainingSqrt == 1) {
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return k.toString(); // k√1 = k
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} else {
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return '$k\\sqrt{$intRemainingSqrt}';
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}
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}
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}
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// 特殊情况：检查是否为简单的分数形式，如 48 = 16*3 = 4²*3
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// 对于 value = 48, k = 4, remaining = 48/16 = 3, sqrt(3) ≈ 1.732, intRemainingSqrt = 1
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// 但 1.732 != 1, 所以上面的循环不会匹配
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// 我们需要检查 remaining 是否是整数且不是完全平方数
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final intValue = value.toInt();
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if (value == intValue.toDouble()) {
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// 尝试找到最大的完全平方因子
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int maxK = 1;
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for (int k = 2; k * k <= intValue; k++) {
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if (intValue % (k * k) == 0) {
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maxK = k;
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}
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}
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if (maxK > 1) {
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final remaining = intValue ~/ (maxK * maxK);
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if (remaining > 1) {
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return '$maxK\\sqrt{$remaining}';
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}
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}
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}
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return null; // 无法用简单符号形式表示
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}
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/// 计算符号形式的二次方程根
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({String formula, String finalAnswer}) _calculateSymbolicRoots(
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double a,
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double b,
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double discriminant,
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String? symbolicSqrt,
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) {
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final halfCoeff = b / (2 * a);
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final denominator = 2 * a;
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String formula;
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String finalAnswer;
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if (symbolicSqrt != null) {
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// 使用符号形式
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final sqrtExpr = symbolicSqrt;
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// 计算根：(-b ± sqrt(discriminant)) / (2a)
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final root1Expr = _formatSymbolicRoot(-b, sqrtExpr, denominator, true);
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final root2Expr = _formatSymbolicRoot(-b, sqrtExpr, denominator, false);
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formula = '\$\$x_1 = $root1Expr, \\quad x_2 = $root2Expr\$\$';
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finalAnswer = '\$\$x_1 = $root1Expr, \\quad x_2 = $root2Expr\$\$';
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} else {
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// 回退到数值计算
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final sqrtValue = sqrt(discriminant);
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final x1 = -halfCoeff + sqrtValue;
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final x2 = -halfCoeff - sqrtValue;
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formula =
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'\$\$x_1 = ${-halfCoeff} + $sqrtValue = $x1, \\quad x_2 = ${-halfCoeff} - $sqrtValue = $x2\$\$';
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finalAnswer = '\$\$x_1 = $x1, \\quad x_2 = $x2\$\$';
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}
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return (formula: formula, finalAnswer: finalAnswer);
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}
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/// 格式化符号形式的根
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String _formatSymbolicRoot(
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double b,
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String sqrtExpr,
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double denominator,
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bool isPlus,
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) {
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final sign = isPlus ? '+' : '-';
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// 处理分母
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final denomStr = denominator == 2 ? '2' : denominator.toString();
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if (b == 0) {
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// 简化为 ±sqrt(discriminant)/denominator
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if (denominator == 2) {
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return isPlus ? '\\frac{$sqrtExpr}{2}' : '-\\frac{$sqrtExpr}{2}';
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} else {
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return isPlus
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? '\\frac{$sqrtExpr}{$denomStr}'
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: '-\\frac{$sqrtExpr}{$denomStr}';
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}
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} else {
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// 完整的表达式：(-b ± sqrt(discriminant))/denominator
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final bInt = b.toInt();
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// 检查是否可以简化
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if (bInt % denominator.toInt() == 0) {
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final simplifiedB = bInt ~/ denominator.toInt();
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if (simplifiedB == 0) {
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return isPlus ? sqrtExpr : '-$sqrtExpr';
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} else if (simplifiedB == 1) {
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return isPlus
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? '1 $sign $sqrtExpr'
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: '1 $sign $sqrtExpr'.replaceAll('+', '-').replaceAll('--', '+');
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} else if (simplifiedB == -1) {
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return isPlus
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? '-1 $sign $sqrtExpr'
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: '-1 $sign $sqrtExpr'.replaceAll('+', '-').replaceAll('--', '+');
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} else if (simplifiedB > 0) {
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return isPlus
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? '$simplifiedB $sign $sqrtExpr'
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: '$simplifiedB $sign $sqrtExpr'
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.replaceAll('+', '-')
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.replaceAll('--', '+');
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} else {
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final absB = (-simplifiedB).toString();
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return isPlus
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? '-$absB $sign $sqrtExpr'
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: '-$absB $sign $sqrtExpr'
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.replaceAll('+', '-')
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.replaceAll('--', '+');
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}
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} else {
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// 无法简化，使用分数形式
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final bStr = b > 0 ? '$bInt' : '($bInt)';
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final numerator = b > 0
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? '-$bStr $sign $sqrtExpr'
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: '($bInt) $sign $sqrtExpr';
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if (denominator == 2) {
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return '\\frac{$numerator}{2}';
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} else {
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return '\\frac{$numerator}{$denomStr}';
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}
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}
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}
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}
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/// 测试方法：验证修复效果
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void testParenthesesFix() {
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print('=== 测试括号修复效果 ===');
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Reference in New Issue
Block a user