情報学部 | 菅沼ホーム | SE目次 | 索引 |
∥∇f(x(k))∥ = ∥[∂f/∂x1,・・・,∂f/∂xn]T∥ = 0
∥f(x(k+1)) - f(x(k))∥ < ε2 ∥x(k+1) - x(k)∥ < ε3 ∥f(x(k))∥ < ε4
x(k+1) = x(k) + α(k)p(k)
f(x(k) + α(k)p(k))
[a(i), b(i)] : ある段階における区間 x1(i), x2(i) : 区間内の点(x1(i) < x2(i))
(x2(i) - a(i)) / (b(i) - a(i)) = (b(i) - x1(i)) / (b(i) - a(i)) = τ (1)
x1(i) - a(i) = b(i) - x2(i) (2)
b(i+1) = x2(i) a(i+1) = a(i)
x2(i+1) = x1(i)
(x2(i+1) - a(i)) / (x2(i) - a(i)) = (x1(i) - a(i)) / (x2(i) - a(i)) = τ (3)
x1(i) - a(i) = b(i) - a(i) - (x2(i) - a(i))
(x1(i) - a(i)) / (x2(i) - a(i)) = 1 / τ - 1 = τ
τ2 + τ - 1 = 0 ∴τ = (50.5 - 1) / 2 ≒ 0.618
a(i+1) = a(i) b(i+1) = x2(i) x1(i+1) = a(i) + (1 - τ)(b(i+1) - a(i)) x2(i+1) = x1(i)
a(i+1) = x1(i) b(i+1) = b(i) x1(i+1) = x2(i) x2(i+1) = b(i) - (1 - τ)(b(i) - a(i+1))
f(x) = ax2 + bx + c
x = -b / (2a)
f(xk+1) > f(xk) で,かつ,k = 0 ならば,
z = f(x, y)
xi = x0 + k * ei i = 1, ・・・, n
H : 関数値が最大になる点 fH = f( xH ) G : 2 番目に関数値が最大になる点 fG = f( xG ) L : 関数値が最小になる点 fL = f( xL )
xR = 2xC - xH
fR ≧ fH
B>xN = ( 1 - λ1)xH + λ1xR, fN = f( xN ) ただし,0 < λ1 < 1, λ1 ≠ 0.5
fR < ( fL + (λ2 - 1 )fH) / λ2 ただし,1 < λ2
xE = λ2xR - (λ2 - 1 )xH, fE = f( xE )
fE ≦ fR
fN ≧ fG
xi = 0.5 * (xi + xL) i = 1, ・・・, n+1
情報学部 | 菅沼ホーム | SE目次 | 索引 |