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CS621: Artificial Intelligence
Pushpak Bhattacharyya
CSE Dept.,
IIT Bombay
Lecture 5: Power of Heuristic; nonconventional search
A*: Definitions and Properties
A* Algorithm – Definition and
Properties
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f(n) = g(n) + h(n)
The node with the least
value of f is chosen from the
OL.
S s
g(n)
f*(n) = g*(n) + h*(n),
where,
g*(n) = actual cost of
the optimal path (s, n)
h*(n) = actual cost of
optimal path (n, g)
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g(n) ≥ g*(n)
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By definition, h(n) ≤ h*(n)
n
h(n)
goal
State space graph G
8-puzzle: heuristics
Example: 8 puzzle
s
2
1
4
1
6
7
1
2
3
7
8
3
4
3
2
4
5
6
5
6
8
7
8
g
5
n
h*(n) = actual no. of moves to transform n to g
1.
2.
h1(n) = no. of tiles displaced from their destined
position.
h2(n) = sum of Manhattan distances of tiles from
their destined position.
h1(n) ≤ h*(n) and h1(n) ≤ h*(n)
h*
h2
h1
Comparison
A* Algorithm- Properties
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Admissibility: An algorithm is called admissible if it
always terminates and terminates in optimal path
Theorem: A* is admissible.
Lemma: Any time before A* terminates there exists
on OL a node n such that f(n) <= f*(s)
Observation: For optimal path s → n1 → n2 → … →
g,
1. h*(g) = 0, g*(s)=0 and
2. f*(s) = f*(n1) = f*(n2) = f*(n3)… = f*(g)
A* Properties (contd.)
f*(ni) = f*(s),
ni ≠ s and ni ≠ g
Following set of equations show the above equality:
f*(ni) = g*(ni) + h*(ni)
f*(ni+1) = g*(ni+1) + h*(ni+1)
g*(ni+1) = g*(ni) + c(ni , ni+1)
h*(ni+1) = h*(ni) - c(ni , ni+1)
Above equations hold since the path is optimal.
Admissibility of A*
A* always terminates finding an optimal path to the goal if such a
path exists.
Intuition
S
(1) In the open list there always exists a node
n such that f(n) <= f*(S) .
g(n)
n
(2) If A* does not terminate, the f value of the
nodes expanded become unbounded.
h(n)
G
1) and 2) are together inconsistent
Hence A* must terminate
Lemma
Any time before A* terminates there exists in the open list a node n'
such that f(n') <= f*(S)
Optimal path
S
n1
n2
G
For any node ni on optimal path,
f(ni) = g(ni) + h(ni)
<= g*(ni) + h*(ni)
Also f*(ni) = f*(S)
Let n' be the fist node in the optimal path that
is in OL. Since all parents of n' have gone to
CL,
g(n') = g*(n') and h(n') <= h*(n')
=> f(n') <= f*(S)
If A* does not terminate
Let e be the least cost of all arcs in the search graph.
Then g(n) >= e.l(n) where l(n) = # of arcs in the path from S to
n found so far. If A* does not terminate, g(n) and hence
f(n) = g(n) + h(n) [h(n) >= 0] will become unbounded.
This is not consistent with the lemma. So A* has to terminate.
2nd part of admissibility of A*
The path formed by A* is optimal when it has terminated
Proof
Suppose the path formed is not optimal
Let G be expanded in a non-optimal path.
At the point of expansion of G,
f(G) = g(G) + h(G)
= g(G) + 0
> g*(G) = g*(S) + h*(S)
= f*(S) [f*(S) = cost of optimal path]
This is a contradiction
So path should be optimal
Better Heuristic Performs
Better
Theorem
A version A2* of A* that has a “better” heuristic than another version
A1* of A* performs at least “as well as” A1*
Meaning of “better”
h2(n) > h1(n) for all n
Meaning of “as well as”
A1* expands at least all the nodes of A2*
h*(n)
h2*(n)
h1*(n)
For all nodes n,
except the goal
node
Proof by induction on the search tree of A2*.
A* on termination carves out a tree out of G
Induction
on the depth k of the search tree of A2*. A1* before termination
expands all the nodes of depth k in the search tree of A2*.
k=0. True since start node S is expanded by both
Suppose A1* terminates without expanding a node n at depth (k+1) of
A2* search tree.
Since A1* has seen all the parents of n seen by A2*
g1(n) <= g2(n)
(1)
Since A1* has terminated without
expanding n,
f1(n) >= f*(S) (2)
S
k+1
G
Any node whose f value is strictly less
than f*(S) has to be expanded.
Since A2* has expanded n
f2(n) <= f*(S)
(3)
From (1), (2), and (3)
h1(n) >= h2(n) which is a contradiction. Therefore, A1* has to expand
all nodes that A2* has expanded.
Exercise
If better means h2(n) > h1(n) for some n and h2(n) = h1(n) for others,
then Can you prove the result ?
A list of AI Search Algorithms
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A*
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AO*
IDA* (Iterative Deepening)
Minimax Search on Game Trees
Viterbi Search on Probabilistic FSA
Hill Climbing
Simulated Annealing
Gradient Descent
Stack Based Search
Genetic Algorithms
Memetic Algorithms
Evolutionary search: Genetic
and Memetic Algoritms
Evolution – Fundamental Laws
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Survival of the fittest.
Change in species is due to change in genes
over reproduction or/and due to mutation.
An Example showing the concept of survival of the fittest and reproduction over
generations.
Evolutionary Computation
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Evolutionary Computation (EC) refers to
computer-based problem solving systems that
use computational models of evolutionary
process.
Terminology:
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Chromosome – It is an individual representing a
candidate solution of the optimization problem.
Population – A set of chromosomes.
gene – It is the fundamental building block of the
chromosome, each gene in a chromosome represents
each variable to be optimized. It is the smallest unit of
information.
Objective: To find a best possible chromosome
to a given optimization problem.
Evolutionary Algorithm:
A meta-heuristic
Let t = 0 be the generation counter;
create and initialize a population P(0);
repeat
Evaluate the fitness, f(xi), for all xi belonging to P(t);
Perform cross-over to produce offspring;
Perform mutation on offspring;
Select population P(t+1) of new generation;
Advance to the new generation, i.e., t = t+1;
until stopping condition is true;
On Overview of GAs
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GA emulate genetic evolution.
A GA has distinct features:
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A string representation of chromosomes.
A selection procedure for initial population and for offspring creation.
A cross-over method and a mutation method.
A fitness function be to minimized.
A replacement procedure.
Parameters that affect GA are initial population,
size of the population, selection process and
fitness function.
Anatomy of GA
Selection
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Selection is a procedure of picking parent
chromosome to produce off-spring.
Types of selection:
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Random Selection – Parents are selected randomly
from the population.
Proportional Selection – probabilities for picking each
chromosome is calculated as:
P(xi) = f(xi)

/Σf(x ) for all j
j
Rank Based Selection – This method uses ranks
instead of absolute fitness values.
P(xi) = (1/β)(1 – er(xi))
Roulette Wheel Selection
Let i = 1, where i denotes chromosome index;
Calculate P(xi) using proportional selection;
sum = P(xi);
choose r ~
U(0,1);
while sum < r do
i = i + 1; i.e. next chromosome
sum = sum + P(xi);
end
return xi as one of the selected parent;
repeat until all parents are selected
Reproduction
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Reproduction is a processes of creating new
chromosomes out of chromosomes in the
population.
Parents are put back into population after
reproduction.
Cross-over and Mutation are two parts in
reproduction of an off-spring.
Cross-over : It is a process of creating one or
more new individuals through the combination
of genetic material randomly selected from two
or parents.
Cross-over
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Uniform cross-over : where corresponding bit
positions are randomly exchanged between two
parents.
One point : random bit is selected and entire
sub-string after the bit is swapped.
Two point : two bits are selected and the substring between the bits is swapped.
Uniform
Cross-over
One point
Cross-over
Two point
Cross-over
Parent1
Parent2
00110110
11011011
00110110
11011011
00110110
11011011
Off-spring1
Off-spring2
01110111
10011010
00111011
11010110
01011010
10110111
Mutation
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Mutation procedures depend upon the
representation schema of the chromosomes.
This is to prevent falling all solutions in
population into a local optimum.
For a bit-vector representation:
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random mutation : randomly negates bits
in-order mutation : performs random mutation
between two randomly selected bit position.
Random
Mutation
In-order
Mutation
Before mutation
1110010011
1110010011
After mutation
1100010111
1110011010