斐波那契 —— 矩阵形式推导

https://blog.csdn.net/lanchunhui/article/details/50569311

1. 矩阵形式的通项

(Fn+2Fn+1)=(1,1,10)⋅(Fn+1Fn)$$\left(\begin{array}{c}{F}_{n+2}\\ F_{n+1}\end{array}\right)=\left(\begin{array}{cc}1,& 1\\ 1,& 0\end{array}\right)\cdot \left(\begin{array}{c}{F}_{n+1}\\ F_n\end{array}\right)$$

不妨令：A=(1,1,10),F1=1,F0=0$A=\left(\begin{array}{cc}1,& 1\\ 1,& 0\end{array}\right),F_1=1,F_0=0$，证明，An=(Fn+1,Fn,FnFn−1)$A^n=\left(\begin{array}{cc}{F}_{n+1},& F_n\\ F_n,& F_{n-1}\end{array}\right)$，采用数学归纳法进行证明，A1=(F2,F1,F1F0)$A^1=\left(\begin{array}{cc}{F}_2,& F_1\\ F_1,& F_0\end{array}\right)$，显然成立，

An+1=An⋅A=(Fn+1,Fn,FnFn−1)⋅(1,1,10)=(Fn+2,Fn+1,Fn+1Fn)$$A^{n+1}=A^n\cdot A=\left(\begin{array}{cc}{F}_{n+1},& F_n\\ F_n,& F_{n-1}\end{array}\right)\cdot \left(\begin{array}{cc}1,& 1\\ 1,& 0\end{array}\right)=\left(\begin{array}{cc}{F}_{n+2},& F_{n+1}\\ F_{n+1},& F_n\end{array}\right)$$

2. 偶数项和奇数项

因为 An=(Fn+1,Fn,FnFn−1)$A^n=\left(\begin{array}{cc}{F}_{n+1},& F_n\\ F_n,& F_{n-1}\end{array}\right)$，则有：

A2m====(F2m+1,F2m,F2mF2m−1)Am⋅Am(Fm+1,Fm,FmFm−1)⋅(Fm+1,Fm,FmFm−1)(F2m+1+F2m,Fm(Fm+2Fm−1),Fm(Fm+2Fm−1)F2m+F2m−1)$$\begin{array}{rl}{A}^{2m}=& \left(\begin{array}{cc}{F}_{2m+1},& F_{2m}\\ F_{2m},& F_{2m-1}\end{array}\right)\\ =& A^m\cdot A^m\\ =& \left(\begin{array}{cc}{F}_{m+1},& F_m\\ F_m,& F_{m-1}\end{array}\right)\cdot \left(\begin{array}{cc}{F}_{m+1},& F_m\\ F_m,& F_{m-1}\end{array}\right)\\ =& \left(\begin{array}{cc}{F}_{m+1}^2+F_m^2,& F_m(F_m+2F_{m-1})\\ F_m(F_m+2F_{m-1}),& F_m^2+F_{m-1}^2\end{array}\right)\end{array}$$

所以有：

F2m+1=F2m+1+F2mF2m=Fm(Fm+2Fm−1)$$\begin{gathered} F_{2m+1}=F_{m+1}^2+F_m^2 \\ {F}_{2m}=F_m(F_m+2F_{m-1}) \end{gathered}$$

3. 矩形形式求解 Fib(n)

因为涉及到矩阵幂次，考虑到数的幂次的递归解法：

• n 为奇数：n=2k+1$n=2k+1$
  
  • Fn=F2k+1=F2k+1+F2k$F_n=F_{2k+1}=F_{k+1}^2+F_k^2$
  • Fn+1=F2k+2=Fk+1(Fk+1+2Fk)$F_{n+1}=F_{2k+2}=F_{k+1}(F_{k+1}+2F_k)$
• n 为偶数：n=2k$n=2k$
  
  • Fn=F2k=Fk(Fk+2Fk−1)=Fk(Fk+2(Fk+1−Fk))$F_n=F_{2k}=F_k(F_k+2F_{k-1})=F_k(F_k+2(F_{k+1}-F_k))$
  • Fn+1=F2k+1=F2k+1+F2k$F_{n+1}=F_{2k+1}=F_{k+1}^2+F_k^2$

4. Python

def fib(n):

    if n > 0:
        f0, f1 = fib(n // 2)
        if n % 2 == 1:
            return f0**2+f1**2, f1*(f1+2*f0)
        return f0*(f0+2*(f1-f0)), f0**2+f1**2
    return 0, 1


if __name__ == '__main__':
    print([fib(i)[0] for i in range(10)])