Genetic algorithm (ga) binary and real Vijay Bhaskar Semwal

Introduction to GeneticAlgorithms 2
We toss a fair coin 60 times and get the
following initial population:
s1 = 1111010101 f (s1) = 7
s2 = 0111000101 f (s2) = 5
s3 = 1110110101 f (s3) = 7
s4 = 0100010011 f (s4) = 4
s5 = 1110111101 f (s5) = 8
s6 = 0100110000 f (s6) = 3

Next we apply fitness proportionate selection with the
roulette wheel method:
21
n
3
Area is
Proportional
to fitness
value
Individual i will have a
probability to be chosen
i
if
if
)(
)(
4
We repeat the extraction
as many times as the
number of individuals we
need to have the same
parent population size
(6 in our case)

Suppose that, after performing selection, we get
the following population:
s1` = 1111010101 (s1)
s2` = 1110110101 (s3)
s3` = 1110111101 (s5)
s4` = 0111000101 (s2)
s5` = 0100010011 (s4)
s6` = 1110111101 (s5)

Next we mate strings for crossover. For each
couple we decide according to crossover
probability (for instance 0.6) whether to actually
perform crossover or not
Suppose that we decide to actually perform
crossover only for couples (s1`, s2`) and (s5`, s6`). For
each couple, we randomly extract a crossover
point, for instance 2 for the first and 5 for the
second

s1` = 1111010101 s2` =
1110110101
s5` = 0100010011 s6` =
1110111101
Before crossover:
After crossover:
s1`` = 1110110101 s2`` =
1111010101
s5`` = 0100011101 s6`` =
1110110011

The final step is to apply random mutation: for each bit that we are to copy to
the new population we allow a small probability of error (for instance 0.1)
Before applying mutation:
s1`` = 1110110101
s2`` = 1111010101
s3`` = 1110111101
s4`` = 0111000101
s5`` = 0100011101
s6`` = 1110110011

In one generation, the total population fitness
changed from 34 to 37, thus improved by ~9%
At this point, we go through the same process
all over again, until a stopping criterion is met

A vector v = (i1 i2… in) represents a tour (v is a
permutation of {1,2,…,n})
Fitness f of a solution is the inverse cost of the
corresponding tour
Initialization: use either some heuristics, or a
random sample of permutations of {1,2,…,n}
We shall use the fitness proportionate selection

Next, starting from the second cut point of one parent, the cities from the other
parent are copied in the same order
The sequence of the cities in the second parent is 9 – 3 – 4 – 5 – 2 –
1 – 8 – 7 – 6
After removal of cities from the first offspring we get 9 – 3 – 2 – 1 – 8
This sequence is placed in the first offspring
o1 = (2 1 8 4 5 6 7 9 3), and similarly in the second
o2 = (3 4 5 1 8 7 6 9 2)

The sub-string between two randomly selected
points in the path is reversed
Example:
(1 2 3 4 5 6 7 8 9) is changed into (1 2 7 6 5 4 3 8 9)
Such simple inversion guarantees that the
resulting offspring is a legal tour

 Simple problem: max x2 over {0,1,…,31}
 GA approach:
 Representation: binary code, e.g. 01101  13
 Population size: 4
 1-point xover, bitwise mutation
 Roulette wheel selection
 Random initialisation
 We show one generational cycle done by
hand

 Has been subject of many (early) studies
 still often used as benchmark for novel GAs
 Shows many shortcomings, e.g.
 Representation is too restrictive
 Mutation & crossovers only applicable for bit-string &
integer representations
 Selection mechanism sensitive for converging
populations with close fitness values
 Generational population model (step 5 in SGA repr.
cycle) can be improved with explicit survivor selection

 Performance with 1 Point Crossover depends on the order
that variables occur in the representation
 more likely to keep together genes that are near
each other
 Can never keep together genes from opposite
ends of string
 This is known as Positional Bias
 Can be exploited if we know about the structure
of our problem, but this is not usually the case

 Choose n random crossover points
 Split along those points
 Glue parts, alternating between parents
 Generalisation of 1 point (still some positional bias)

 Assign 'heads' to one parent, 'tails' to the other
 Flip a coin for each gene of the first child
 Make an inverse copy of the gene for the second child
 Inheritance is independent of position

 Decade long debate: which one is better / necessary / main-
background
 Answer (at least, rather wide agreement):
 it depends on the problem, but
 in general, it is good to have both
 both have another role
 mutation-only-EA is possible, xover-only-EA would not work

Exploration: Discovering promising areas in the search space,
i.e. gaining information on the problem
Exploitation: Optimising within a promising area, i.e. using
information
There is co-operationAND competition between them
 Crossover is explorative, it makes a big jump to an area
somewhere “in between” two (parent) areas
 Mutation is exploitative, it creates random small diversions,
thereby staying near (in the area of ) the parent

 Only crossover can combine information from two
parents
 Only mutation can introduce new information (alleles)
 Crossover does not change the allele frequencies of the
population (thought experiment: 50% 0’s on first bit in
the population, ?% after performing n crossovers)
 To hit the optimum you often need a ‘lucky’ mutation

In this section we take an in-depth look at the
working of the standard genetic algorithm,
explaining why GAs constitute an effective
search procedure
For simplicity we discuss binary string
representation of individuals

{0,1,#} is the symbol alphabet, where # is a
special wild card symbol
A schema is a template consisting of a string
composed of these three symbols
Example: the schema [01#1#] matches the
strings: [01010], [01011], [01110] and [01111]

The order of the schema S (denoted by o(S)) is the
number of fixed positions (0 or 1) presented in the
schema
Example: for S1 = [01#1#], o(S1) = 3
for S2 = [##1#1010], o(S2) = 5
The order of a schema is useful to calculate
survival probability of the schema for mutations
There are 2
l-o(S)
different strings that match S

Genetic algorithm (ga) binary and real Vijay Bhaskar Semwal

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