Information Technology Reference
In-Depth Information
-/?/+??????331040-[ 8]-
{-1.02887, -0.418549, 0.227905, -0.816986, -0.466431}
-+/+-??????240300-[ 9]-
{1.83539, 1.11624, -0.917511, 0.395264, -1.71542}
+*/**??????021420-[10]-
{-0.88263, -0.065887, -0.422607, -1.30252, -1.18634}
++**+??????032403-[11]-
{1.40912, 0.489258, 1.1077, -1.08975, -0.546967}
*+/*+??????132410-[12]-
{-0.243469, -0.438812, -0.882447, -0.906128, -1.05746}
-/-*+??????000234-[13]-
{-1.58066, -0.344665, -0.249664, -0.405212, 0.219116}
/*-?*??????124240-[14]-
{-0.419189, -1.16443, 1.82031, 0.203216, 1.75998}
***//??????010200-[15]-
{-1.87112, -0.15274, -0.238952, -1.12656, -0.200226}
+*+?/??????112430-[16]-
{-1.34305, -0.608338, 0.092956, -1.07773, -1.43314}
+-/-*??????340131-[17]-
{1.5076, -0.039703, 1.36438, 1.03183, -1.09851}
/+/+*??????323432-[18]-
{1.89792, -0.39209, 1.95444, -0.865998, -1.61389}
/*+*/??????021334-[19]-
{-0.041564, 0.254913, 1.37637, -1.80347, -1.7045}
Outputs:
[ 0] = f(0.811676) = 1.2911
[ 1] = f(-3.57447) = --
[ 2] = f(11.664) = --
[ 3] = f(-9.02453) = --
[ 4] = f(-0.914593) = 0.59524
[ 5] = f(0.0665201) = 1.05776
[ 6] = f(-1.18539) = --
[ 7] = f(0.143316) = 0.859833
[ 8] = f(-0.488403) = 1.17403
[ 9] = f(1.69579) = 1.22345
[10] = f(2.11769) = --
[11] = f(-1.8221) = --
[12] = f(0.121492) = 0.924063
[13] = f(-2.12059) = --
[14] = f(-61.8317) = --
[15] = f(2.23794) = --
[16] = f(-1.32433) = --
[17] = f(0.64897) = 1.64863
[18] = f(9.4166) = --
[19] = f(-0.364412) = 0.672306
Note that, in this initial population, 11 out of 20 individuals have zero
fitness but, as the system starts learning, this rate will start to decrease, being
kept to a minimum in later generations.
The expression of the best individual of this generation (chromosome 17)
is shown in Figure 8.4. As you can see, it corresponds to the parameter value
x
0
= 0.64897, which gives
f
(
x
0
) = 1.64863. Note that some numerical con-
stants are used more than once (
c
1
and
c
3
are both used twice) to create new
ones. Note also that, for this small set of just five RNCs per gene, the algo-
rithm is using most of the random constants at its disposal. In this particular
case, only constant
c
2
has no expression whatsoever in the final solution.
