Solving a simultaneous equation through code

问题

This seems like an incredibly simple and silly question to ask, but everything I've found about it has been too complex for me to understand.

I have these two very basic simultaneous equations:

X = 2x + 2z
Y = z - x

Given that I know both X and Y, how would I go about finding x and z? It's very easy to do it by hand, but I have no idea how this would be done in code.

回答1:

This seems like an incredibly simple and silly question to ask

Not at all. This is a very good question, and it has unfortunately a complex answer. Let's solve

a * x + b * y = u
c * x + d * y = v

I stick to the 2x2 case here. More complex cases will require you to use a library.

The first thing to note is that Cramer formulas are not good to use. When you compute the determinant

a * d - b * c

as soon as you have a * d ~ b * c, then you have catastrophic cancellation. This case is typical, and you must guard against it.

The best tradeoff between simplicity / stability is partial pivoting. Suppose that |a| > |c|. Then the system is equivalent to

a * c/a * x + bc/a * y = uc/a
      c * x +    d * y = v

which is

cx + bc/a * y = uc/a
cx +       dy = v

and now, substracting the first to the second yields

cx +       bc/a * y = uc/a
     (d - bc/a) * y = v - uc/a

which is now straightforward to solve: y = (v - uc/a) / (d - bc/a) and x = (uc/a - bc/a * y) / c. Computing d - bc/a is stabler than ad - bc, because we divide by the biggest number (it is not very obvious, but it holds -- do the computation with very close coefficients, you'll see why it works).

Now, if |c| > |a|, you just swap the rows and proceed similarly.

In code (please check the Python syntax):

def solve(a, b, c, d, u, v):
    if abs(a) > abs(c):
         f = u * c / a
         g = b * c / a
         y = (v - f) / (d - g)
         return ((f - g * y) / c, y)
    else
         f = v * a / c
         g = d * a / c
         x = (u - f) / (b - g)
         return (x, (f - g * x) / a)

You can use full pivoting (requires you to swap x and y so that the first division is always by the largest coefficient), but this is more cumbersome to write, and almost never required for the 2x2 case.

For the n x n case, all the pivoting stuff is encapsulated into the LU decomposition, and you should use a library for this.

回答2:

@Alexandre , you missed one condition.. Here is the final code

void SolveLinearEquations (float a,float b,float c,float d,float u,float v, float &x, float &y)
{
float f;
float g;
if (abs(a) > abs(c))
{
     f = u * c / a;
     g = b * c / a;
     y = (v - f) / (d - g);
     if(c != 0)
        x = (f - g * y) / c;
     else
        x = (u - b * y)/a;
}
else
{
    f = v * a / c;
     g = d * a / c;
     x = (u - f) / (b - g);
     if (a != 0)
        y = (f - g * x) / a ;
     else
        y = (v - d * x)/c;
}
}

回答3:

(1)  ax + by = c    
(2)  dx + dy = f

(3)1*d     adx + bdy  = cd
(4)2*b     abx + bdy  = fb

(3)-(4)   adx - abx  = cd - fb

          x(ad-ab) = cd - fb


          x = (c*d - f*b)/(a*d-a*b) //use this equation for x



ax + by = c
     by = c - ax

     y = (c - a*x)/b    //use this equation for y

回答4:

The following function might be useful for some:

function solve(s1,s2){    //only works if coefficients > 0

   str=s1 + " " +s2
   str=str.replace(/[^0123456789.-]/g, ' ') //eliminate letters
   str=str.replace( /\s\s+/g, ' ' )         //no double spaces


   var n=str.split(" ");            //put into an array
   var a=0,b=1,c=2,d=3,e=4,f=5      //see what were doing
   var x = ( n[c]*n[e] -n[b]*n[f])/(n[a]*n[e] - n[b]*n[d])
   var y= (n[c]-n[a]*x)/n[b]

   return({x:x, y:y})
}

To use:

result=solve("12x +  2y =32", "9x -5y=55")
alert (result.x+" ----- "+result.y)

回答5:

Algorithm to sove: Ax + By = C, Dx + Ey = F

START

1)  PRINT "Enter A"
2)  INPUT A
3)  PRINT "Enter B"
4)  INPUT B
5)  PRINT "Enter C"
6)  INPUT C
7)  PRINT "Enter D"
8)  INPUT D
9)  PRINT "Enter E"
10) INPUT E
11) PRINT "Enter F"
21) INPUT F

22) detS <-- AE - BC
22) detX <-- CE - BF
24) detY <-- AF - CD

25) x <-- detX / detS
26) y <-- detY / detS

27) PRINT x, y

END

来源：https://stackoverflow.com/questions/11609107/solving-a-simultaneous-equation-through-code

标签

algorithm

math

equation