def f(x):return x*x-4*x+5
a0,b0=1,3
1、均勻搜索
令 δ = ( b 0 ? a 0 ) / N , a i = a 0 + i δ , i = 1 , 2 , 3 \delta=(b_0-a_0)/N,a_i=a_0+i\delta,i=1,2,3 δ=(b0??a0?)/N,ai?=a0?+iδ,i=1,2,3
while b0-a0>0.1:a=np.linspace(a0,b0,5)for i in range(1,4):if f(a[i-1])>f(a[i]) and f(a[i+1])>f(a[i]):a0,b0=a[i-1],a[i+1]break
print((a0+b0)/2)
極小點為2
2、黃金區間法
令 τ = ( 5 ? 1 ) / 2 , λ = a 0 + ( 1 ? τ ) ( b 0 ? a 0 ) , μ = a 0 + τ ( b 0 ? a 0 ) \tau=(\sqrt{5}-1)/2,\lambda=a_0+( 1-\tau)(b_0-a_0),\mu=a_0+\tau(b_0-a_0) τ=(5??1)/2,λ=a0?+(1?τ)(b0??a0?),μ=a0?+τ(b0??a0?)
tao=(pow(5,0.5)-1)/2
lamda=a0+(1-tao)*(b0-a0)
miu=a0+tao*(b0-a0)
while miu-lamda>0.1:if f(lamda)<f(miu):b0=miuelse:a0=lamdalamda=a0+(1-tao)*(b0-a0)miu=a0+tao*(b0-a0)
print((lamda+miu)/2)
極小點約等于2.09
3、導數二分法
記區間中點 λ = ( a 0 + b 0 ) / 2 \lambda=(a_0+b_0)/2 λ=(a0?+b0?)/2
def first_order(x):return 2*x-4while True:lamda=(a0+b0)/2value=first_order(lamda)if value==0:print(lamda)breakelif value<0:a0=lamdaelse:b0=lamda
極小點為2
