Алгоритм Хука и Дживса

If f (x(0) h1e1) f (x(0) )

, then a success is

recorded and

x(0) h1e1

h,efrom

the starting point is moved in x(0) h e

1 1

2

2

which an

h2 R

(0)

(0)

analogous check is carried out at point

f

(x

h1e1) f (x

)

x(0) h e

with

.

1 1

If, instead,

e

, then a failure

is recorded and2

a similar check is performed at

. If a

success2

is

x(0)

13

the behavior

To achieve a certain accuracy, the stepi lengths

must be

selected in such

h

a way that the

quantities

(0)

f (x

(0)

hje j ) f (x

)

, j 1,..., n

point,

x(0)

,

y(0)

max h

after at most 2n + 1 functional evaluations.

x(0)

i 1,...n

Only two possibilities may arise:

the method

1.

. In such ai

case, if

h

x(0)

terminates and yields the

14

.

approximate solution

(0)

(0)

max

h

(0)

2. y

x

i 1,....n

Ifi

then the method yterminates

yielding

as

y(0)

an approximate solution, otherwise the advancing step starts.

(0)

(0)

step consists of moving further from

y(0)

The advancingy x

along the direction

, (which is the direction that

x(1)

recorded the maximum decrease of f during the exploration step),

rather then simply setting

2 y(0) x(0)

.

as a new starting point

This new starting point is instead set equal to

. From

y(1)

f ( y

(1)

) f

( y

(0)

x

(0)

)

this point

a new series of exploration moves is started. If this exploration

leads to a

such that

y(1) y(0) x(0)

point

x(1)

, then a new

starting point

15

.

for further explorations is set equal to

The method is now ready to restart from the point

just

is the value of F at the mimimizer X.}

x = z;

x = x0; minf = eval(f); iter

= 0;

end {if z == x}

while h > tol

end {if y == x}

[y] = explore(f,n,h,x);

iter = iter +1;

if y == x

end {while}

h = h/2;

minf = eval(f);

else

return

x = 2*y-x;

[z] = explore(f,n,h,x);

if z == x

x = y;

16

else

x = x0; f0 = eval(f);

if ff < f0

for i=1:n

f0 = ff;

x(i) = x(i) + h(i); ff =

else

eval(f);

x(i) = x0(i);

if ff < f0

end {if ff < f0}

f0 = ff;

end {if ff < f0}

else

end {for}

x(i) = x0(i) - h(i);

return

ff = eval(f);