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Artificial Intelligence (CS607)
Lecture No. 11-13
3 Genetic Algorithms
3.1 Discussion on Problem Solving
In the previous chapter we studied problem solving in general and elaborated on
various search strategies that help us solve problems through searching in
problem trees. We kept the information about the tree traversal in memory (in the
queues), thus we know the links that have to be followed to reach the goal. At
times we don't really need to remember the links that were followed. In many
problems where the size of search space grows extremely large we often use
techniques in which we don't need to keep all the history in memory. Similarly, in
problems where requirements are not clearly defined and the problem is ill-
structured, that is, we don't exactly know the initial state, goal state and operators
etc, we might employ such techniques where our objective is to find the solution
not how we got there.
Another thing we have noticed in the previous chapter is that we perform a
sequential search through the search space. In order to speed up the techniques
we can follow a parallel approach where we start from multiple locations (states)
in the solution space and try to search the space in parallel.
3.2 Hill Climbing in Parallel
Suppose we were to climb up a hill. Our goal is to reach the top irrespective of
how we get there. We apply different operators at a given position, and move in
the direction that gives us improvement (more height). What if instead of starting
from one position we start to climb the hill from different positions as indicated by
the diagram below.
In other words, we start with different independent search instances that start
from different locations to climb up the hill.
Further think that we can improve this using a collaborative approach where
these instances interact and evolve by sharing information in order to solve the
problem. You will soon find out that what we mean by interact and evolve.
However, it is possible to implement parallelism in the sense that the instances
can interact and evolve to solve the solution. Such implementations and
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algorithms are motivated from the biological concept of evolution of our genes,
hence the name Genetic Algorithms, commonly terms as GA.
3.3 Comment on Evolution
Before we discuss Genetic Algorithms in detail with examples lets go though
some basic terminology that we will use to explain the technique. The genetic
algorithm technology comes from the concept of human evolution. The following
paragraph gives a brief overview of evolution and introduces some terminologies
to the extent that we will require for further discussion on GA. Individuals (animals
or plants) produce a number of offspring (children) which are almost, but not
entirely, like themselves. Variation may be due to mutation (random changes), or
due to inheritance (offspring/children inherit some characteristics from each
parent). Some of these offspring may survive to produce offspring of their own--
some will not. The "better adapted" individuals are more likely to survive. Over
time, generations become better and better adapted to survive.
3.4 Genetic Algorithm
Genetic Algorithms is a search method in which multiple search paths are
followed in parallel. At each step, current states of different pairs of these paths
are combined to form new paths. This way the search paths don't remain
independent, instead they share information with each other and thus try to
improve the overall performance of the complete search space.
3.5 Basic Genetic Algorithm
A very basic genetic algorithm can be stated as below.
Start with a population of randomly generated, (attempted) solutions to a
Repeatedly do the following:
Evaluate each of the attempted solutions
Keep the "best" solutions
Produce  next  generation  from  these
"inheritance" and "mutation")
Quit when you have a satisfactory solution (or you run out of time)
The two terms introduced here are inheritance and mutation. Inheritance has the
same notion of having something or some attribute from a parent while mutation
refers to a small random change. We will explain these two terms as we discuss
the solution to a few problems through GA.
3.6 Solution to a Few Problems using GA
3.6.1 Problem 1:
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Suppose your "individuals" are 32-bit computer words
You want a string in which all the bits in these words are ones
Here's how you can do it:
 Create 100 randomly generated computer words
 Repeatedly do the following:
 Count the 1 bits in each word
 Exit if any of the words have all 32 bits set to 1
 Keep the ten words that have the most 1s (discard the
 From each word, generate 9 new words as follows:
 Pick a random bit in the word and toggle (change)
Note that this procedure does not guarantee that the next
"generation" will have more 1 bits, but it's likely
As you can observe, the above solution is totally in accordance with the basic
algorithm you saw in the previous section. The table on the next page shows
which steps correspond to what.
Basic GA
a Create
of randomly
computer words
attempted solutions
to a problem
Evaluate each of
Count the 1 bits
in each word.
Exit if any of the
Keep  the  "best"
words have all
32 bits set to 1
Keep  the  ten
words that have
the  most  1s
word, generate
9 new words as
(using "inheritance"
and "mutation")
Pick a random
bit in the word
(change) it
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For the sake of simplicity we only use mutation for now to generate the new
individuals. We will incorporate inheritance later in the example. Let's introduce
the concept of an evaluation function. An evaluation function is the criteria that
check various individuals/ solutions for being better than others in the population.
Notice that mutation can be as simple as just flipping a bit at random or any
number of bits.
We go on repeating the algorithm until we either get our required word that is a
32-bit number with all ones, or we run out of time. If we run out of time, we either
present the best possible solution (the one with most number of 1-bits) as the
answer or we can say that the solution can't be found. Hence GA is at times used
to get optimal solution given some parameters.
3.6.2 Problem 2:
Suppose you have a large number of data points (x, y), e.g., (1, 4), (3,
9), (5, 8), ...
You would like to fit a polynomial (of up to degree 1) through these
data points
 That is, you want a formula y = mx + c that gives you a
reasonably good fit to the actual data
 Here's the usual way to compute goodness of fit of the
polynomial on the data points:
 Compute the sum of (actual y ­ predicted y)2 for all the
data points
 The lowest sum represents the best fit
You can use a genetic algorithm to find a "pretty good" solution
By a pretty good solution we simply mean that you can get reasonably good
polynomial that best fits the given data.
Your formula is y = mx + c
Your unknowns are m and c; where m and c are integers
Your representation is the array [m, c]
Your evaluation function for one array is:
 For every actual data point (x, y)
 Compute = mx + c
 Find the sum of (y ­ )2 over all x
 The sum is your measure of "badness" (larger numbers
are worse)
 Example: For [5, 7] and the data points (1, 10) and (2, 13):
  = 5x + 7 = 12 when x is 1
  = 5x + 7 = 17 when x is 2
 (10 - 12)2 + (13 ­ 17)2 = 22 + 42 = 20
 If these are the only two data points, the "badness" of [5,
7] is 20
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Your algorithm might be as follows:
 Create two-element arrays of random numbers
 Repeat 50 times (or any other number):
 For each of the arrays, compute its badness (using all
data points)
 Keep the best arrays (with low badness)
 From the arrays you keep, generate new arrays as
 Convert the numbers in the array to binary, toggle
one of the bits at random
 Quit if the badness of any of the solution is zero
 After all 50 trials, pick the best array as your final answer
Let us solve this problem in detail. Consider that the given points are as follows.
(x, y) : {(1,5) (3, 9)}
We start will the following initial population which are the arrays representing the
solutions (m and c).
[2 7][1 3]
Compute badness for [2 7]
= 2x + 7 = 9 when x is 1
= 2x + 7 = 13 when x is 3
(5 ­ 9)2 + (9 ­ 13)2 = 42 + 42 = 32
= 1x + 3 = 4 when x is 1
= 1x + 3 = 6 when x is 3
(5 ­ 4)2 + (9 ­ 6)2 = 12 + 32 = 10
Lets keep the one with low "badness" [1 3]
Representation [001 011]
Apply mutation to generate new arrays [011 011]
Now we have [1 3] [3 3] as the new population considering that we
keep the two best individuals
Second iteration
(x, y) : {(1,5) (3, 9)}
[1 3][3 3]
  = 1x + 3 = 4 when x is 1
  = 1x + 3 = 6 when x is 3
 (5 ­ 4)2 + (9 ­ 6)2 = 12 + 32 = 10
= 3x + 3 = 6 when x is 1
= 3x + 3 = 12 when x is 3
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(5 ­ 6)2 + (9 ­ 12)2 = 1 + 9 = 10
Lets keep the [3 3]
Representation [011 011]
Apply mutation to generate new arrays [010 011]
Now we have [3 3] [2 3] as the new population
Third Iteration
(x, y) : {(1,5) (3, 9)}
[3 3][2 3]
  = 3x + 3 = 6 when x is 1
  = 3x + 3 = 12 when x is 3
 (5 ­ 6)2 + (9 ­ 12)2 = 1 + 9 = 10
= 2x + 3 = 5 when x is 1
= 2x + 3 = 9 when x is 3
(5 ­ 5)2 + (9 ­ 9)2 = 02 + 02 = 0
Solution found [2 3]
y = 2x+3
So you see that how by going though the iteration of a GA one can find a solution
to the given problem. It is not necessary in the above example that you get a
solution that gives 0 badness. In case we go on doing iterations and we run out of
time, we might just present the solution that has the least badness as the most
optimal solution given these number of iterations on this data.
In the examples so far, each "Individual" (or "solution") had only one parent. The
only way to introduce variation was through mutation (random changes). In
Inheritance or Crossover, each "Individual" (or "solution") has two parents.
Assuming that each organism has just one chromosome, new offspring are
produced by forming a new chromosome from parts of the chromosomes of each
Let us repeat the 32-bit word example again but this time using crossover instead
of mutation.
Suppose your "organisms" are 32-bit computer words, and you want
a string in which all the bits are ones
Here's how you can do it:
 Create 100 randomly generated computer words
 Repeatedly do the following:
 Count the 1 bits in each word
 Exit if any of the words have all 32 bits set to 1
 Keep the ten words that have the most 1s (discard the
 From each word, generate 9 new words as follows:
 Choose one of the other words
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Take the first half of this word and combine it with
the second half of the other word
Notice that we are generating new individuals from the best ones by using
crossover. The simplest way to perform this crossover is to combine the head of
one individual to the tail of the other, as shown in the diagram below.
In the 32-bit word problem, the (two-parent, no mutation) approach, if it succeeds,
is likely to succeed much faster because up to half of the bits change each time,
not just one bit. However, with no mutation, it may not succeed at all. By pure bad
luck, maybe none of the first (randomly generated) words have (say) bit 17 set to
1. Then there is no way a 1 could ever occur in this position. Another problem is
lack of genetic diversity. Maybe some of the first generation did have bit 17 set to
1, but none of them were selected for the second generation. The best technique
in general turns out to be a combination of both, i.e., crossover with mutation.
3.7 Eight Queens Problem
Let us now solve a famous problem which will be discussed under GA in many
famous books in AI. Its called the Eight Queens Problem.
The problem is to place 8 queens on a chess board so that none of them can
attack the other. A chess board can be considered a plain board with eight
columns and eight rows as shown below.
The possible cells that the Queen can move to when placed in a particular square
are shown (in black shading)
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We now have to come up with a representation of an individual/ candidate
solution representing the board configuration which can be used as individuals in
the GA.
We will use the representation as shown in the figure below.
Where the 8 digits for eight columns specify the index of the row where the queen
is placed. For example, the sequence 2 6 8 3 4 5 3 1 tells us that in first column
the queen is placed in the second row, in the second column the queen is in the
6th row so on till in the 8th column the queen is in the 1st row.
Now we need a fitness function, a function by which we can tell which board
position is nearer to our goal. Since we are going to select best individuals at
every step, we need to define a method to rate these board positions or
individuals. One fitness function can be to count the number of pairs of Queens
that are not attacking each other. An example of how to compute the fitness of a
board configuration is given in the diagram on the next page.
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So once representation and fitness function is decided, the solution to the
problem is simple.
Choose initial population
Evaluate the fitness of each individual
Choose the best individuals from the population for crossover
Let us quickly go though an example of how to solve this problem using GA.
Suppose individuals (board positions) chosen for crossover are:
Where the numbers 2 and 3 in the boxes to the left and right show the fitness of
each board configuration and green arrows denote the queens that can attack
The following diagram shows how we apply crossover:
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The individuals in the initial population are shown on the left and the children
generated by swapping their tails are shown on the right. Hence we now have a
total of 4 candidate solutions. Depending on their fitness we will select the best
The diagram below shows where we select the best two on the bases of their
fitness. The vertical over shows the children and the horizontal oval shows the
selected individuals which are the fittest ones according to the fitness function.
Similarly, the mutation step can be done as under.
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That is, we represent the individual in binary and we flip at random a certain
number of bits. You might as well decide to flip 1, 2, 3 or k number of bits, at
random position. Hence GA is totally a random technique.
This process is repeated until an individual with required fitness level is found. If
no such individual is found, then the process is repeated till the overall fitness of
the population or any of its individuals gets very close to the required fitness
level. An upper limit on the number of iterations is usually used to end the
process in finite time.
One of the solutions to the problem is shown as under whose fitness value is 8.
The following flow chart summarizes the Genetic Algorithm.
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Evaluate Fitness
Apply mutation
of Population
individuals in
You are encouraged to explore the internet and other books to find more
applications of GA in various fields like:
 Genetic Programming
 Evolvable Systems
 Composing Music
 Market Strategies
 Industrial Optimization
and many more.
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3.8 Problems
Q1 what type of problems can be solved using GA. Give examples of at least 3
problems from different fields of life. Clearly identify the initial population,
representation, evaluation function, mutation and cross over procedure and exit
Q2 Given pairs of (x, y) coordinates, find the best possible m, c parameters of the
line y = mx + c that generates them. Use mutation only. Present the best possible
solution given the data after at least three iterations of GA or exit if you find the
solution earlier.
(x, y) : {(1,2.5) (2, 3.75)}
Initial population [2 0][3 1]
Q3 Solve the 8 Queens Problem on paper. Use the representations and strategy
as discussed in the chapter.