Quality series (1)-mathematical charm (2): Expanding Euclidean Algorithms

Code
///   <Summary>
/// Solve XM + ny = D (D is m, n is the maximum public approx)
///   </Summary>
///   <Param name = "M"> </param>
///   <Param name = "N"> </param>
///   <Param name = "X"> </param>
///   <Param name = "Y"> </param>
///   <Returns> </returns>
Public   Static   Void Oglidextension ( Long M, Long N, Ref   Long X, Ref   Long Y)
{
If (M < N)
{
Oglidextension (n, m, Ref X, Ref Y );
Return ;
}
If (N =   0 )
{
X =   1 ;
Y =   0 ;
Return ;
}

Else
{
Long Q = M % N;
Oglidextension (n, Q, Ref X, Ref Y );
Long Temp = X;
X = Y;
Y = Temp - M / N * Y;
}

}

 

AlgorithmExplanation: (I have no time. I have to go to work now. I will explain it in detail later)

 

Algorithm idea:

 

This is roughly the Euclidean theorem.:

IntegerA, BAndA, BThe maximum number of public approx.Gcd (A, B)

So it must existX, YMakeAx + by = gcd (A, B)

 

Thoughts:Prerequisites,Euclidean Definition

Hypothesis: Ax1 + by1 = gcd (A, B)Founded,IfB = 0SoGcd (a, B) = 0;

In this caseX1 = 1, Y1 = 0You can obtain the correct solution.

IfB! = 0SoGcd (B, A % B) = gcd (B, a-a/B * B)

According to our assumptionsBx2 + (a-a/B * B) y2 = gcd (B, a-a/B * B)

ThereforeAx1 + by1 = bx2 + (a-a/B * B) y2 = ay2 + (x2-a/by2) B

ThereforeX1 = Y2, Y1 = x2-A/by2

While,For anyA, BIfB! = 0Perform the following iteration:A = B, B = A % B

EventuallyB = 0

Then we can carry out iterative solutions based on the above theorem.

A	B	A/B	R	X = 1	Y = 0
15	9	1	6	1	0
9	6	1	3	0	-1
6	3	2	0	-1	2
3	0	End	End	ENE	End