Math.NET 数值 http://numerics.mathdotnet.com/ can 数值求解任何线性系统 http://numerics.mathdotnet.com/LinearEquations.html,但我想这不是你要找的。
Math.NET 符号 http://symbolics.mathdotnet.com/可以处理符号表达式,尽管这个项目还处于早期阶段,还不理解方程的概念。然而,我们仍然可以用它来解决像这样的简单系统,只需做一些工作 - 通过做我们手工做的事情。
首先,我们定义一个小函数来求解 1 阶以下的单个线性方程:
using Expr = MathNet.Symbolics.Expression;
Expr SolveSimpleRoot(Expr variable, Expr expr)
{
// try to bring expression into polynomial form
Expr simple = Algebraic.Expand(Rational.Numerator(Rational.Simplify(variable,expr)));
// extract coefficients, solve known forms of order up to 1
Expr[] coeff = Polynomial.Coefficients(variable,simple);
switch(coeff.Length)
{
case 1: return Expr.Zero.Equals(coeff[0]) ? variable : Expr.Undefined;
case 2: return Rational.Simplify(variable,Algebraic.Expand(-coeff[0]/coeff[1]));
default: return Expr.Undefined;
}
}
然后我们可以用它来求解系统,如下所示:
// declare variables
var x = Expr.Symbol("x");
var y = Expr.Symbol("y");
// Parse left and right side of both equations
Expr aleft = Infix.ParseOrThrow("(4-x)*2");
Expr aright = Infix.ParseOrThrow("(y-1)*10+2");
Expr bleft = Infix.ParseOrThrow("x");
Expr bright = Infix.ParseOrThrow("y*2+1");
// Solve both equations to x
Expr ax = SolveSimpleRoot(x,aleft-aright); // "8 - 5*y"
Expr bx = SolveSimpleRoot(x,bleft-bright); // "1 + 2*y"
// Equate both terms of x, solve to y
Expr cy = SolveSimpleRoot(y,ax-bx); // "1"
// Substitute term of y into one of the terms of x
Expr cx = Algebraic.Expand(Structure.Substitute(y,cy,ax)); // "3"
// Print expression in Infix notation
Console.WriteLine(Infix.Print(cx)); // x=3
Console.WriteLine(Infix.Print(cy)); // y=1