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)(Fn+2Fn+1)=(1,11,0)⋅(Fn+1Fn)

\begin{pmatrix} F_{n+2}\\ F_{n+1} \end{pmatrix}=\begin{pmatrix} 1,&1\\ 1,&0 \end{pmatrix}\cdot \begin{pmatrix} F_{n+1}\\ F_{n} \end{pmatrix}

不妨令：A=(1,1,10),F1=1,F0=0A=(1,11,0),F1=1,F0=0 A=\begin{pmatrix} 1,&1\\ 1,&0 \end{pmatrix} , F_1=1, F_0=0，证明，An=(Fn+1,Fn,FnFn−1)An=(Fn+1,FnFn,Fn−1)A^n=\begin{pmatrix} F_{n+1},&F_n\\ F_n,&F_{n-1} \end{pmatrix}，采用数学归纳法进行证明，A1=(F2,F1,F1F0)A1=(F2,F1F1,F0)A^1=\begin{pmatrix} F_{2},&F_1\\ F_1,&F_{0} \end{pmatrix}，显然成立，

An+1=An⋅A=(Fn+1,Fn,FnFn−1)⋅(1,1,10)=(Fn+2,Fn+1,Fn+1Fn)An+1=An⋅A=(Fn+1,FnFn,Fn−1)⋅(1,11,0)=(Fn+2,Fn+1Fn+1,Fn)

A^{n+1}=A^n\cdot A=\begin{pmatrix} F_{n+1},&F_n\\ F_n,&F_{n-1} \end{pmatrix}\cdot \begin{pmatrix} 1,&1\\ 1,&0 \end{pmatrix}=\begin{pmatrix} F_{n+2},&F_{n+1}\\ F_{n+1},&F_{n} \end{pmatrix}

2. 偶数项和奇数项

因为 An=(Fn+1,Fn,FnFn−1)An=(Fn+1,FnFn,Fn−1) A^n=\begin{pmatrix} F_{n+1},&F_n\\ F_n,&F_{n-1} \end{pmatrix} ，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)A2m=(F2m+1,F2mF2m,F2m−1)=Am⋅Am=(Fm+1,FmFm,Fm−1)⋅(Fm+1,FmFm,Fm−1)=(Fm+12+Fm2,Fm(Fm+2Fm−1)Fm(Fm+2Fm−1),Fm2+Fm−12)

\begin{split} A^{2m}=&\begin{pmatrix} F_{2m+1},&F_{2m}\\ F_{2m},&F_{2m-1} \end{pmatrix}\\ =&A^m\cdot A^m\\ =&\begin{pmatrix} F_{m+1},&F_m\\ F_m,&F_{m-1} \end{pmatrix}\cdot \begin{pmatrix} F_{m+1},&F_m\\ F_m,&F_{m-1} \end{pmatrix}\\ =&\begin{pmatrix} F^2_{m+1}+F^2_{m},&F_m(F_m+2F_{m-1})\\ F_m(F_m+2F_{m-1}),&F^2_m+F^2_{m-1} \end{pmatrix} \end{split}

所以有：

F2m+1=F2m+1+F2mF2m=Fm(Fm+2Fm−1)F2m+1=Fm+12+Fm2F2m=Fm(Fm+2Fm−1)

F_{2m+1}=F^2_{m+1}+F^2_{m}\\ F_{2m}=F_m(F_m+2F_{m-1})

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+1n=2k+1n=2k+1
- Fn=F2k+1=F2k+1+F2kFn=F2k+1=Fk+12+Fk2F_n =F_{2k+1}= F^2_{k+1}+F^2_{k}
- Fn+1=F2k+2=Fk+1(Fk+1+2Fk)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=2kn=2kn=2k
- Fn=F2k=Fk(Fk+2Fk−1)=Fk(Fk+2(Fk+1−Fk))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+F2kFn+1=F2k+1=Fk+12+Fk2F_{n+1}=F_{2k+1}=F^2_{k+1}+F^2_{k}

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**2return 0, 1if __name__ == '__main__':print([fib(i)[0] for i in range(10)])