斐波那契 —— 矩阵形式推导

 

1. 矩阵形式的通项

 

(Fn+2Fn+1)=(1,1,10)⋅(Fn+1Fn)(Fn+2Fn+1)=(1,11,0)⋅(Fn+1Fn)

 

不妨令：A=(1,1,10),F1=1,F0=0A=(1,11,0),F1=1,F0=0，证明，An=(Fn+1,Fn,FnFn−1)An=(Fn+1,FnFn,Fn−1)，采用数学归纳法进行证明，A1=(F2,F1,F1F0)A1=(F2,F1F1,F0)，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)

 

2. 偶数项和奇数项

因为 An=(Fn+1,Fn,FnFn−1)An=(Fn+1,FnFn,Fn−1)，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)

 

所以有：

 

F2m+1=F2m+1+F2mF2m=Fm(Fm+2Fm−1)F2m+1=Fm+12+Fm2F2m=Fm(Fm+2Fm−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+1
  
  • Fn=F2k+1=F2k+1+F2kFn=F2k+1=Fk+12+Fk2
  • Fn+1=F2k+2=Fk+1(Fk+1+2Fk)Fn+1=F2k+2=Fk+1(Fk+1+2Fk)
• n 为偶数：n=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))
  • Fn+1=F2k+1=F2k+1+F2kFn+1=F2k+1=Fk+12+Fk2

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)])