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Artificial Intelligence

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Artificial Intelligence (CS607)
Lecture No. 4 -10
2 Problem Solving
In chapter one, we discussed a few factors that demonstrate intelligence.
Problem solving was one of them when we referred to it using the examples of a
mouse searching a maze and the next number in the sequence problem.
Historically people viewed the phenomena of intelligence as strongly related to
problem solving. They used to think that the person who is able to solve more
and more problems is more intelligent than others.
In order to understand how exactly problem solving contributes to intelligence, we
need to find out how intelligent species solve problems.
2.1 Classical Approach
The classical approach to solving a problem is pretty simple. Given a problem at
hand use hit and trial method to check for various solutions to that problem.
This hit and trial approach usually works well for trivial problems and is referred to
as the classical approach to problem solving.
Consider the maze searching problem. The mouse travels though one path and
finds that the path leads to a dead end, it then back tracks somewhat and goes
along some other path and again finds that there is no way to proceed. It goes on
performing such search, trying different solutions to solve the problem until a
sequence of turns in the maze takes it to the cheese. Hence, of all the solutions
the mouse tries, the one that reached the cheese was the one that solved the
problem.
Consider that a toddler is to switch on the light in a dark room. He sees the
switchboard having a number of buttons on it. He presses one, nothing happens,
he presses the second one, the fan gets on, he goes on trying different buttons till
at last the room gets lighted and his problem gets solved.
Consider another situation when we have to open a combinational lock of a
briefcase. It is a lock which probably most of you would have seen where we
have different numbers and we adjust the individual dials/digits to obtain a
combination that opens the lock. However, if we don't know the correct
combination of digits that open the lock, we usually try 0-0-0, 7-7-7, 7-8-6 or any
such combination for opening the lock. We are solving this problem in the same
manner as the toddler did in the light switch example.
All this discussion has one thing in common. That different intelligent species use
a similar approach to solve the problem at hand. This approach is essentially the
classical way in which intelligent species solve problems. Technically we call this
hit and trial approach the "Generate and Test" approach.
2.2 Generate and Test
This is a technical name given to the classical way of solving problems where we
generate different combinations to solve our problem, and the one which solves
the problem is taken as the correct solution. The rest of the combinations that we
try are considered as incorrect solutions and hence are destroyed.
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Possible
Tester
Correct
Solutions
Solutions
Solution
Generator
Incorrect
Solutions
The diagram above shows a simple arrangement of a Generate and Test
procedure. The box on the left labeled "Solution Generator" generates different
solutions to a problem at hand, e.g. in the case of maze searching problem, the
solution generator can be thought of as a machine that generates different paths
inside a maze. The "Tester" actually checks that either a possible solution from
the solution generates solves out problem or not. Again in case of maze
searching the tester can be thought of as a device that checks that a path is a
valid path for the mouse to reach the cheese. In case the tester verifies the
solution to be a valid path, the solution is taken to be the "Correct Solution". On
the other hand if the solution was incorrect, it is discarded as being an "Incorrect
Solution".
2.3 Problem Representation
All the problems that we have seen till now were trivial in nature. When the
magnitude of the problem increases and more parameters are added, e.g. the
problem of developing a time table, then we have to come up with procedures
better than simple Generate and Test approach.
Before even thinking of developing techniques to systematically solve the
problem, we need to know one more thing that is true about problem solving
namely problem representation. The key to problem solving is actually good
representation of a problem. Natural representation of problems is usually done
using graphics and diagrams to develop a clear picture of the problem in your
mind. As an example to our comment consider the diagram below.
OFF | OFF | OFF
OFF
ON | OFF | OFF
OFF | ON | OFF
OFF | OFF | ON
ON | ON | OFF
ON | OFF | ON
OFF | ON | ON
OFF | ON | ON
ON | ON | OFF
ON | OFF | ON
It shows the problem of switching on the light by a toddler in a graphical form.
Each rectangle represents the state of the switch board. OFF | OFF| OFF means
that all the three switches are OFF. Similarly OFF| ON | OFF means that the first
and the last switch is OFF and the middle one is ON. Starting from the state
when all the switches are OFF the child can proceed in any of the three ways by
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switching either one of the switch ON. This brings the toddler to the next level in
the tree. Now from here he can explore the other options, till he gets to a state
where the switch corresponding to the light is ON. Hence our problem was
reduced to finding a node in the tree which ON is the place corresponding to the
light switch. Observe how representing a problem in a nice manner clarifies the
approach to be taken in order to solve it.
2.4 Components of Problem Solving
Let us now be a bit more formal in dealing with problem solving and take a look at
the topic with reference to some components that constitute problem solving.
They are namely: Problem Statement, Goal State, Solution Space and Operators.
We will discuss each one of them in detail.
2.4.1 Problem Statement
This is the very essential component where by we get to know what exactly the
problem at hand is. The two major things that we get to know about the problem
is the Information about what is to be done and constraints to which our solution
should comply. For example we might just say that given infinite amount of time,
one will be able to solve any problem he wishes to solve. But the constraint
"infinite amount of time" is not a practical one. Hence whenever addressing a
problem we have to see that how much time shall out solution take at max. Time
is not the only constraint. Availability of resources, and all the other parameters
laid down in the problem statement actually tells us about all the rules that have
to be followed while solving a problem. For example, taking the same example of
the mouse, are problem statement will tell us things like, the mouse has to reach
the cheese as soon as possible and in case it is unable to find a path within an
hour, it might die of hunger. The statement might as well tell us that the mouse is
located in the lower left corner of the maze and the cheese in the top left corner,
the mouse can turn left, right and might or might not be allowed to move
backward and things like that. Thus it is the problem statement that gives us a
feel of what exactly to do and helps us start thinking of how exactly things will
work in the solution.
2.4.2 Problem Solution
While solving a problem, this should be known that what will be out ultimate aim.
That is what should be the output of our procedure in order to solve the problem.
For example in the case of mouse, the ultimate aim is to reach the cheese. The
state of world when mouse will be beside the cheese and probably eating it
defines the aim. This state of world is also referred to as the Goal State or the
state that represents the solution of the problem.
2.4.3 Solution space
In order to reach the solution we need to check various strategies. We might or
might not follow a systematic strategy in all the cases. Whatever we follow, we
have to go though a certain amount of states of nature to reach the solution. For
example when the mouse was in the lower left corner of the maze, represents a
state i.e. the start state. When it was stuck in some corner of the maze
represents a state. When it was stuck somewhere else represents another state.
When it was traveling on a path represents some other state and finally when it
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reaches the cheese represents a state called the goal state. The set of the start
state, the goal state and all the intermediate states constitutes something which
is called a solution space.
2.4.4 Traveling in the solution space
We have to travel inside this solution space in order to find a solution to our
problem. The traveling inside a solution space requires something called as
"operators". In case of the mouse example, turn left, turn right, go straight are
the operators which help us travel inside the solution space. In short the action
that takes us from one state to the other is referred to as an operator. So while
solving a problem we should clearly know that what are the operators that we can
use in order to reach the goal state from the starting state. The sequence of
these operators is actually the solution to our problem.
2.5 The Two-One Problem
In order to explain the four components of problem solving in a better way we
have chosen a simple but interesting problem to help you grasp the concepts.
The diagram below shows the setting of our problem called the Two-One
Problem.
Start
Goal
11?22
22?11
Legal Moves:
· Slide
Rules:
· 1s' move right
· 2s' move left
· Hop
· Only one move at a time
· No backing up
A simple problem statement to the problem at hand is as under.
You are given a rectangular container that has 5 slots in it. Each slot can hold
only one coin at a time. Place Rs.1 coins in the two left slots; keep the center slot
empty and put Rs.2 coins in the two right slots. A simple representation can be
seen in the diagram above where the top left container represents the Start State
in which the coined are placed as just described. Our aim is to reach a state of
the container where the left two slots should contain Rs.2 coins, the center slot
should be empty and the right two slots should contain Rs.1 coin as shown in the
Goal State. There are certain simple rules to play this game. The rules are
mentioned clearly in the diagram under the heading of "Rules". The rules actually
define the constraints under which the problem has to be solved. The legal
moves are the Operators that we can use to get from one state to the other. For
example we can slide a coin to its left or right if the left or right slot is empty, or
we can hop the coin over a single slot. The rules say that Rs.1 coins can slide or
hop only towards right. Similarly the Rs.2 coins can slide or hop only towards the
left. You can only move one coin at a time.
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Now let us try to solve the problem in a trivial manner just by using a hit and trial
method without addressing the problem in a systematic manner.
Trial 1
Start State
Move 1
Move 2
Move 3
Move 4
Move 5
In Move 1 we slide a 2 to the left, then we hop a 1 to the right, then we slide the 2
to the left again and then we hop the 2 to the left, then slide the one to the right
hence at least one 2 and one 1 are at the desired positions as required in the
goal state but then we are stuck. There is no other valid move which takes us out
of this state. Let us consider another trial.
Trial 2
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Starting from the start state we first hop a 1 to the right, then we slide the other 1
to the right and then suddenly we get STUCK!! Hence solving the problem
through a hit and trial might not give us the solution.
Let us now try to address the problem in a systematic manner. Consider the
diagram below.
Starting from the goal state if we hop, we get stuck. If we slide we can further
carry on. Keeping this observation in mind let us now try to develop all the
possible combinations that can happen after we slide.
H
H
?1122
11?22
1122 ?
S
S
1?122
1 1 2? 2
S
S
H
H
?1122
1 2 1? 2
1?212
1122 ?
S
S
S
S
1212 ?
12 ?12
?1212
H
H
H
H
12?21
?2112
1 2 2 1?
21?12
S
H
H
S
S
S
1 2 2? 1
?2121
2?112
1 2 2 ?1
2112 ?
2?112
S
S
2?121
212 ?1
H
H
2?211
2 2 1? 1
S
S
1 1? 2 2
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The diagram above shows a tree sort of structure enumerating all the possible
states and moves. Looking at this diagram we can easily figure out the solution to
our problem. This tree like structure actually represents the "Solution Space" of
this problem. The labels on the links are H and S representing hop and slide
operators respectively. Hence H and S are the operators that help us travel
through this solution space in order to reach the goal state from the start state.
We hope that this example actually clarifies the terms problem statement, start
state, goal state, solution space and operators in your mind. It will be a nice
exercise to design your own simple problems and try to identify these
components in them in order to develop a better understanding.
2.6 Searching
All the problems that we have looked at can be converted to a form where we
have to start from a start state and search for a goal state by traveling through a
solution space. Searching is a formal mechanism to explore alternatives.
Most of the solution spaces for problems can be represented in a graph where
nodes represent different states and edges represent the operator which takes us
from one state to the other. If we can get our grips on algorithms that deal with
searching techniques in graphs and trees, we'll be all set to perform problem
solving in an efficient manner.
2.7 Tree and Graphs Terminology
Before studying the searching techniques defined on trees and graphs let us
briefly review some underlying terminology.
A
B
C
D
E
F
G
H
·"A" is the "root node"
I
J
·"A, B, C .... J" are "nodes"
·"B" is a "child" of "A"
·"A" is ancestor of "D"
·"D" is a descendant of "A"
·"D, E, F, G, I, J" are "leaf nodes"
·Arrows represent "edges" or "links"
The diagram above is just to refresh your memories on the terminology of a tree.
As for graphs, there are undirected and directed graphs which can be seen in the
diagram below.
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A
A
B
C
B
C
D
E
F
G
D
E
F
G
H
I
H
I
Directed Graph
Undirected Graph
Let us first consider a couple of examples to learn how graphs can represent
important information by the help of nodes and edges.
Graphs can be used to represent city routes.
A
A
B
C
B
C
D
E
F
G
D
E
F
G
H
I
H
I
Directed Graph
Undirected Graph
Graphs can be used to plan actions.
We will use graphs to represent problems and their solution spaces. One thing to
be noted is that every graph can be converted into a tree, by replicating the
nodes. Consider the following example.
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S
3
3
A
A
D
B
C
B
D
A
E
G
S
S
C
E
E
B
B
F
3
2
D
F
B
F
C
E
A
C
G
F
E
D
1
3
F
G
C
G
G
The graph in the figure represents a city map with cities labeled as S, A, B, C, D,
E, F and G. Just by following a simple procedure we can convert this graph to a
tree.
Start from node S and make it the root of your tree, check how many nodes are
adjacent to it. In this case A and D are adjacent to it. Hence in the tree make A
and D, children of S. Now go on proceeding in this manner and you'll get a tree
with a few nodes replicated. In this manner depending on a starting node you can
get a different tree too. But just recall that when solving a problem; we usually
know the start state and the end state. So we will be able to transform our
problem graphs in problem trees. Now if we can develop understanding of
algorithms that are defined for tree searching and tree traversals then we will be
in a better shape to solve problems efficiently.
We know that problems can be represented in graphs, and are well familiar with
the components of problem solving, let us now address problem solving in a
more formal manner and study the searching techniques in detail so that we can
systematically approach the solution to a given problem.
2.8 Search Strategies
Search strategies and algorithms that we will study are primarily of four types,
blind/uninformed, informed/heuristic, any path/non-optimal and optimal path
search algorithms. We will discuss each of these using the same mouse
example.
Suppose the mouse does not know where and how far is the cheese and is
totally blind to the configuration of the maze. The mouse would blindly search the
maze without any hints that will help it turning left or right at any junction. The
mouse will purely use a hit and trial approach and will check all combinations till
one takes it to the cheese. Such searching is called blind or uninformed
searching.
Consider now that the cheese is fresh and the smell of cheese is spread through
the maze. The mouse will now use this smell as a guide, or heuristic (we will
comment on this word in detail later) to guess the position of the cheese and
choose the best from the alternative choices. As the smell gets stronger, the
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mouse knows that the cheese is closer. Hence the mouse is informed about the
cheese through the smell and thus performs an informed search in the maze.
For now you might think that the informed search will always give us a better
solution and will always solve our problem. This might not be true as you will find
out when we discuss the word heuristic in detail later.
When solving the maze search problem, we saw that the mouse can reach the
cheese from different paths. In the diagram above two possible paths are shown.
In any-path/non optimal searches we are concerned with finding any one solution
to our problem. As soon as we find a solution, we stop, without thinking that there
might as well be a better way to solve the problem which might take lesser time
or fewer operators.
Contrary to this, in optimal path searches we try to find the best solution. For
example, in the diagram above the optimal path is the blue one because it is
smaller and requires lesser operators. Hence in optimal searches we find
solutions that are least costly, where cost of the solution may be different for
each problem.
2.9 Simple Search Algorithm
Let us now state a simple search algorithm that will try to give you an idea about
the sort of data structures that will be used while searching, and the stop criteria
for your search. The strength of the algorithm is such that we will be able to use
this algorithm for both Depth First Search (DFS) and Breadth First Search (BFS).
Let S be the start state
1. Initialize Q with the start node Q=(S) as the only entry; set Visited =
(S)
2. If Q is empty, fail. Else pick node X from Q
3. If X is a goal, return X, we've reached the goal
4. (Otherwise) Remove X from Q
5. Find all the children of state X not in Visited
6. Add these to Q; Add Children of X to Visited
7. Go to Step 2
Here Q represents a priority queue. The algorithm is simple and doesn't need
much explanation. We will use this algorithm to implement blind and uninformed
searches. The algorithm however can be used to implement informed searches
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as well. The critical step in the Simple Search Algorithm is picking of a node X
from Q according to a priority function. Let us call this function P(n). While using
this algorithm for any of the techniques, our priority will be to reduce the value of
P(n) as much as we can. In other words, the node with the highest priority will
have the smallest value of the function P(n) where n is the node referred to as X
in the algorithm.
2.10 Simple Search Algorithm Applied to Depth First Search
Depth First Search dives into a tree deeper and deeper to fine the goal state. We
will use the same Simple Search Algorithm to implement DFS by keeping our
priority function as
1
P(n) =
height (n)
As mentioned previously we will give priority to the element with minimum P(n)
hence the node with the largest value of height will be at the maximum priority to
be picked from Q. The following sequence of diagrams will show you how DFS
works on a tree using the Simple Search Algorithm.
We start with a tree containing nodes S, A, B, C, D, E, F, G, and H, with H as the
goal node. In the bottom left table we show the two queues Q and Visited.
According to the Simple Search Algorithm, we initialize Q with the start node S,
shown below.
If Q is not empty, pick the node X with the minimum P(n) (in this case S), as it is
the only node in Q. Check if X is goal, (in this case X is not the goal). Hence find
all the children of X not in Visited and add them to Q and Visited. Goto Step 2.
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Again check if Q is not empty, pick the node X with the minimum P(n) (in this
case either A or B), as both of them have the same value for P(n). Check if X is
goal, (in this case A is not the goal). Hence, find all the children of A not in Visited
and add them to Q and Visited. Go to Step 2.
Go on following the steps in the Simple Search Algorithm till you find a goal node.
The diagrams below show you how the algorithm proceeds.
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Here, from the 5th row of the table when we remove H and check if it's the goal,
the algorithm says YES and hence we return H as we have reached the goal
state. The path followed by the DFS is shown by green arrows at each step. The
diagram below also shows that DFS didn't have to search the entire search
space, rather only by traveling in half the tree, the algorithm was able to search
the solution.
Hence simply by selecting a specific P(n) our Simple Search Algorithm was
converted to a DFS procedure.
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2.11 Simple Search Algorithm Applied to Breadth First Search
Breadth First Search explores the breadth of the tree first and progresses
downward level by level. Now, we will use the same Simple Search Algorithm to
implement BFS by keeping our priority function as
P(n) = height (n)
As mentioned previously, we will give priority to the element with minimum P(n)
hence the node with the largest value of height will be at the maximum priority to
be picked from Q. In other words, greater the depth/height greater the priority.
The following sequence of diagrams will show you how BFS works on a tree
using the Simple Search Algorithm.
We start with a tree containing nodes S, A, B, C, D, E, F, G, and H, with H as the
goal node. In the bottom left table we show the two queues Q and Visited.
According to the Simple Search Algorithm, we initialize Q with the start node S.
If Q is not empty, pick the node X with the minimum P(n) (in this case S), as it is
the only node in Q. Check if X is goal, (in this case X is not the goal). Hence find
all the children of X not in Visited and add them to Q and Visited. Goto Step 2.
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Again, check if Q is not empty, pick the node X with the minimum P(n) (in this
case either A or B), as both of them have the same value for P(n). Remember, n
refers to the node X. Check if X is goal, (in this case A is not the goal). Hence find
all the children of A not in Visited and add them to Q and Visited. Go to Step 2.
Now, we have B, C and D in the list Q. B has height 1 while C and D are at a
height 2. As we are to select the node with the minimum P(n) hence we will select
B and repeat. The following sequence of diagram tells you how the algorithm
proceeds till it reach the goal state.
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When we remove H from the 9th row of the table and check if it's the goal, the
algorithm says YES and hence we return H since we have reached the goal
state. The path followed by the BFS is shown by green arrows at each step. The
diagram below also shows that BFS travels a significant area of the search space
if the solution is located somewhere deep inside the tree.
Hence, simply by selecting a specific P(n) our Simple Search Algorithm was
converted to a BFS procedure.
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2.12 Problems with DFS and BFS
Though DFS and BFS are simple searching techniques which can get us to the
goal state very easily yet both of them have their own problems.
DFS has small space requirements (linear in depth) but has major problems:
DFS can run forever in search spaces with infinite length paths
DFS does not guarantee finding the shallowest goal
BFS guarantees finding the shallowest path even in presence of infinite paths,
but it has one great problem
BFS requires a great deal of space (exponential in depth)
We can still come up with a better technique which caters for the drawbacks of
both these techniques. One such technique is progressive deepening.
2.13 Progressive Deepening
Progressive deepening actually emulates BFS using DFS. The idea is to simply
apply DFS to a specific level. If you find the goal, exit, other wise repeat DFS to
the next lower level. Go on doing this until you either reach the goal node or the
full height of the tree is explored. For example, apply a DFS to level 2 in the tree,
if it reaches the goal state, exit, otherwise increase the level of DFS and apply it
again until you reach level 4. You can increase the level of DFS by any factor. An
example will further clarify your understanding.
Consider the tree on the previous page with nodes from S ... to N, where I is the
goal node.
Apply DFS to level 2 in the tree. The green arrows in the diagrams below show
how DFS will proceed to level 2.
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After exploring to level 2, the progressive deepening procedure will find out that
the goal state has still not been reached. Hence, it will increment the level by a
factor of, say 2, and will now perform a DFS in the tree to depth 4. The blue
arrows in the diagrams below show how DFS will proceed to level 4.
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As soon as the procedure finds the goal state it will quit. Notice that it guarantees
to find the solution at a minimum depth like BFS. Imagine that there are a number
of solutions below level 4 in the tree. The procedure would only travel a small
portion of the search space and without large memory requirements, will find out
the solution.
2.14 Heuristically Informed Searches
So far we have looked into procedures that search the solution space in an
uninformed manner. Such procedures are usually costly with respect to either
time, space or both. We now focus on a few techniques that search the solution
space in an informed manner using something which is called a heuristic. Such
techniques are called heuristic searches. The basic idea of a heuristic search is
that rather than trying all possible search paths, you try and focus on paths that
seem to be getting you closer to your goal state using some kind of a "guide". Of
course, you generally can't be sure that you are really near your goal state.
However, we might be able to use a good guess for the purpose. Heuristics are
used to help us make that guess. It must be noted that heuristics don't always
give us the right guess, and hence the correct solutions. In other words educated
guesses are not always correct.
Recall the example of the mouse searching for cheese. The smell of cheese
guides the mouse in the maze, in other words the strength of the smell informs
the mouse that how far is it from the goal state. Here the smell of cheese is the
heuristic and it is quite accurate.
Similarly, consider the diagram below. The graph shows a map in which the
numbers on the edges are the distances between cities, for example, the
distance between city S and city D is 3 and between B and E is 4.
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Suppose our goal is to reach city G starting from S. There can be many choices,
we might take S, A, D, E, F, G or travel from S, to A, to E, to F, and to G. At each
city, if we were to decide which city to go next, we might be interested in some
sort of information which will guide us to travel to the city from which the distance
of goal is minimum.
If someone can tell us the straight-line distance of G from each city then it might
help us as a heuristic in order to decide our route map. Consider the graph
below.
It shows the straight line distances from every city to the goal. Now, cities that are
closer to the goal should be our preference. These straight line distances also
known as "as the crow flies distance" shall be our heuristic.
It is important to note that heuristics can sometimes misguide us. In the example
we have just discussed, one might try to reach city C as it is closest from the goal
according to our heuristic, but in the original map you can see that there is no
direct link between city C and city G. Even if someone reaches city C using the
heuristic, he won't be able to travel to G from C directly, hence the heuristic can
misguide. The catch here is that crow-flight distances do not tell us that the two
cities are directly connected.
Similarly, in the example of mouse and cheese, consider that the maze has
fences fixed along some of the paths through which the smell can pass. Our
heuristic might guide us on a path which is blocked by a fence, hence again the
heuristic is misguiding us.
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The conclusion then is that heuristics do help us reduce the search space, but it
is not at all guaranteed that we'll always find a solution. Still many people use
them as most of the time they are helpful. The key lies in the fact that how do we
use the heuristic. Consider the notion of a heuristic function.
Whenever we choose a heuristic, we come up with a heuristic function which
takes as input the heuristic and gives us out a number corresponding to that
heuristic. The search will now be guided by the output of the heuristic function.
Depending on our application we might give priority to either larger numbers or
smaller numbers.
Hence to every node/ state in our graph we will assign a heuristic value,
calculated by the heuristic function. We will start with a basic heuristically
informed search which is called Hill Climbing.
2.15 Hill Climbing
Hill Climbing is basically a depth first search with a measure of quality that is
assigned to each node in the tree. The basic idea is: Proceed as you would in
DFS except that you order your choices according to some heuristic
measurement of the remaining distance to the goal. We will discuss the Hill
climbing with an example.
Before going to the actual example, let us give another analogy for which the
name Hill Climbing has been given to this procedure. Consider a blind person
climbing a hill. He can not see the peak of the hill. The best he can do is that from
a given point he takes steps in all possible directions and wherever he finds that
a step takes him higher he takes that step and reaches a new, higher point. He
goes on doing this until all possible steps in any direction will take him higher and
this would be the peak, hence the name hill climbing. Notice that each step that
we take, gets us closer to our goal which in this example is the peak of a hill.
Such a procedure might as well have some problems.
Foothill Problem: Consider the diagram of a mountain below. Before reaching
the global maxima, that is the highest peak, the blind man will encounter local
maxima that are the intermediate peaks and before reaching the maximum
height. At each of these local maxima, the blind man gets the perception of
having reached the global maxima as none of the steps takes him to a higher
point. Hence he might just reach local maxima and think that he has reached the
global maxima. Thus getting stuck in the middle of searching the solution space.
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Plateau Problem: Similarly, consider another problem as depicted in the
diagram below. Mountains where flat areas called plateaus are frequently
encountered the blind person might again get stuck.
When he reaches the portion of a mountain which is totally flat, whatever step he
takes gives him no improvement in height hence he gets stuck.
Ridge Problem: Consider another problem; you are standing on what seems like
a knife edge contour running generally from northeast to southwest. If you take
step in one direction it takes you lower, on the other hand when you step in some
other direction it gives you no improvement.
All these problems can be mapped to situations in our solution space searching.
If we are at a state and the heuristics of all the available options take us to a
lower value, we might be at local maxima. Similarly, if all the available heuristics
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take us to no improvement we might be at a plateau. Same is the case with ridge
as we can encounter such states in our search tree.
The solution to all these problems is randomness. Try taking random steps in
random direction of random length and you might get out of the place where you
are stuck.
Example
Let us now take you through an example of searching a tree using hill climbing to
end out discussion on hill climbing.
Consider the diagram below. The tree corresponds to our problem of reaching
city M starting from city S. In other words our aim is to find a path from S to M.
We now associate heuristics with every node, that is the straight line distance
from the path-terminating city to the goal city.
When we start at S we see that if we move to A we will be left with 9 units to
travel. As moving on A has given us an improvement in reaching our goal hence
we move to A. Exactly in the same manner as the blind man moves up a step
that gives him more height.
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Standing on A we see that C takes us closer to the goal hence we move to C.
From C we see that city I give us more improvement hence we move to I and
then finally to M.
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Notice that we only traveled a small portion of the search space and reached our
goal. Hence the informed nature of the search can help reduce space and time.
2.16 Beam Search
You just saw how hill climbing procedure works through the search space of a
tree. Another procedure called beam search proceeds in a similar manner. Out of
n possible choices at any level, beam search follows only the best k of them; k is
the parameter which we set and the procedure considers only those many nodes
at each level.
The following sequence of diagrams will show you how Beam Search works in a
search tree.
We start with a search tree with L as goal state and k=2, that is at every level we
will only consider the best 2 nodes. When standing on S we observe that the only
two nodes available are A and B so we explore both of them as shown below.
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From here we have C, D, E and F as the available options to go. Again, we select
the two best of them and we explore C and E as shown in the diagram below.
From C and E we have G, H, I and J as the available options so we select H and
J and similarly at the last level we select L and N of which L is the goal.
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2.17 Best First Search
Just as beam search considers best k nodes at every level, best first search
considers all the open nodes so far and selects the best amongst them. The
following sequence of diagrams will show you how a best first search procedure
works in a search tree.
We start with a search tree as shown above. From S we observe that A is the
best option so we explore A.
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At A we now have C, G, D and B as the options. We select the best of them
which is D.
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At D we have S, G, B, H, M and J as the options. We select H which is the best of
them.
At last from H we find L as the best. Hence best first search is a greedy approach
will looks for the best amongst the available options and hence can sometimes
reduce the searching time.  All these heuristically informed procedures are
considered better but they do not guarantee the optimal solution, as they are
dependent on the quality of heuristic being used.
2.18 Optimal Searches
So far we have looked at uninformed and informed searches. Both have their
advantages and disadvantages. But one thing that lacks in both is that whenever
they find a solution they immediately stop. They never consider that their might
be more than one solution to the problem and the solution that they have ignored
might be the optimal one.
A simplest approach to find the optimal solution is this; find all the possible
solutions using either an uninformed search or informed search and once you
have searched the whole search space and no other solution exists, then choose
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the most optimal amongst the solutions found. This approach is analogous to the
brute force method and is also called the British museum procedure.
But in reality, exploring the entire search space is never feasible and at times is
not even possible, for instance, if we just consider the tree corresponding to a
game of chess (we will learn about game trees later), the effective branching
factor is 16 and the effective depth is 100. The number of branches in an
exhaustive survey would be on the order of 10120. Hence a huge amount of
computation power and time is required in solving the optimal search problems in
a brute force manner.
2.19 Branch and Bound
In order to solve our problem of optimal search without using a brute force
technique, people have come up with different procedures. One such procedure
is called branch-and-bound method.
The simple idea of branch and bound is the following:
The length of the complete path from S to G is 9. Also note that while traveling
from S to B we have already covered a distance of 9 units. So traveling further
from S D A B to some other node will make the path longer. So we ignore any
further paths ahead of the path S D A B.
We will show this with a simple example.
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The diagram above shows the same city road map with distance between the
cities labels on the edges. We convert the map to a tree as shown below.
We proceed in a Best First Search manner. Starting at S we see that A is the
best option so we explore A.
From S the options to travel are B and D, the children of A and D the child of S.
Among these, D the child of S is the best option. So we explore D.
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From here the best option is E so we go there,
then B,
then D,
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Here we have E, F and A as equally good options so we select arbitrarily and
move to say A,
then E.
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When we explore E we find out that if we follow this path further, our path length
will increase beyond 9 which is the distance of S to G. Hence we block all the
further sub-trees along this path, as shown in the diagram below.
We then move to F as that is the best option at this point with a value 7.
then C,
We see that C is a leaf node so we bind C too as shown in the next diagram.
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Then we move to B on the right hand side of the tree and bind the sub trees
ahead of B as they also exceed the path length 9.
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We go on proceeding in this fashion, binding the paths that exceed 9 and hence
we are saved from traversing a considerable portion of the tree. The subsequent
diagrams complete the search until it has found all the optimal solution, that is
along the right hand branch of the tree.
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Notice that we have saved ourselves from traversing a considerable portion of
the tree and still have found the optimal solution. The basic idea was to reduce
the search space by binding the paths that exceed the path length from S to G.
2.20 Improvements in Branch and Bound
The above procedure can be improved in many different ways. We will discuss
the two most famous ways to improve it.
1. Estimates
2. Dynamic Programming
The idea of estimates is that we can travel in the solution space using a heuristic
estimate. By using "guesses" about remaining distance as well as facts about
distance already accumulated we will be able to travel in the solution space more
efficiently. Hence we use the estimates of the remaining distance. A problem
here is that if we go with an overestimate of the remaining distance then we might
loose a solution that is somewhere nearby. Hence we always travel with
underestimates of the remaining distance. We will demonstrate this improvement
with an example.
The second improvement is dynamic programming. The simple idea behind
dynamic programming is that if we can reach a specific node through more than
one different path then we shall take the path with the minimum cost.
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In the diagram you can see that we can reach node D directly from S with a cost
of 3 and via S A D with a cost of 6 hence we will never expand the path with the
larger cost of reaching the same node.
When we include these two improvements in branch and bound then we name it
as a different technique known as A* Procedure.
2.21 A* Procedure
This is actually branch and bound technique with the improvement of
underestimates and dynamic programming.
We will discuss the technique with the same example as that in branch-and-
bound.
The values on the nodes shown in yellow are the underestimates of the distance
of a specific node from G. The values on the edges are the distance between two
adjacent cities.
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Our measure of goodness and badness of a node will now be decided by a
combination of values that is the distance traveled so far and the estimate of the
remaining distance. We construct the tree corresponding to the graph above.
We start with a tree with goodness of every node mentioned on it.
Standing at S we observe that the best node is A with a value of 4 so we move to
4.
Then B. As all the sub-trees emerging from B make our path length more than 9
units so we bound this path, as shown in the next diagram.
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Now observe that to reach node D that is the child of A we can reach it either with
a cost of 12 or we can directly reach D from S with a cost of 9. Hence using
dynamic programming we will ignore the whole sub-tree beneath D (the child of
A) as shown in the next diagram.
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Now we move to D from S.
Now A and E are equally good nodes so we arbitrarily choose amongst them,
and we move to A.
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As the sub-tree beneath A expands the path length is beyond 9 so we bind it.
We proceed in this manner. Next we visit E, then we visit B the child of E, we
bound the sub-tree below B. We visit F and finally we reach G as shown in the
subsequent diagrams.
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Notice that by using underestimates and dynamic programming the search space
was further reduced and our optimal solution was found efficiently.
2.22 Adversarial Search
Up until now all the searches that we have studied there was only one person or
agent searching the solution space to find the goal or the solution. In many
applications there might be multiple agents or persons searching for solutions in
the same solution space.
Such scenarios usually occur in game playing where two opponents also called
adversaries are searching for a goal. Their goals are usually contrary to each
other. For example, in a game of tic-tac-toe player one might want that he should
complete a line with crosses while at the same time player two wants to complete
a line of zeros. Hence both have different goals. Notice further that if player one
puts a cross in any box, player-two will intelligently try to make a move that would
leave player-one with minimum chance to win, that is, he will try to stop player-
one from completing a line of crosses and at the same time will try to complete
his line of zeros.
Many games can be modeled as trees as shown below. We will focus on board
games for simplicity.
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Searches in which two or more players with contrary goals are trying to explore
the same solution space in search of the solution are called adversarial searches.
2.23 Minimax Procedure
In adversarial searches one player tries to cater for the opponent's moves by
intelligently deciding that what will be the impact of his own move on the over all
configuration of the game. To develop this stance he uses a look ahead thinking
strategy. That is, before making a move he looks a few levels down the game
tree to see that what can be the impact of his move and what options will be open
to the opponent once he has made this move.
To clarify the concept of adversarial search let us discuss a procedure called the
minimax procedure.
Here we assume that we have a situation analyzer that converts all judgments
about board situations into a single, over all quality number. This situation
analyzer is also called a static evaluator and the score/ number calculated by the
evaluator is called the static evaluation of that node. Positive numbers, by
convention indicate favor to one player. Negative numbers indicate favor to the
other player. The player hoping for positive numbers is called maximizing player
or maximizer. The other player is called minimizing player or minimizer. The
maximizer has to keep in view that what choices will be available to the minimizer
on the next step. The minimizer has to keep in view that what choices will be
available to the maximizer on the next step.
Consider the following diagram.
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Standing at node A the maximizer wants to decide which node to visit next, that
is, choose between B or C. The maximizer wishes to maximize the score so
apparently 7 being the maximum score, the maximizer should go to C and then to
G. But when the maximizer will reach C the next turn to select the node will be of
the minimizer, which will force the game to reach configuration/node F with a
score of 2. Hence maximizer will end up with a score of 2 if he goes to C from A.
On the other hand, if the maximizer goes to B from A the worst which the
minimizer can do is that he will force the maximizer to a score of 3. Now, since
the choice is between scores of 3 or 2, the maximizer will go to node B from A.
2.24 Alpha Beta Pruning
In Minimax Procedure, it seems as if the static evaluator must be used on each
leaf node. Fortunately there is a procedure that reduces both the tree branches
that must be generated and the number of evaluations. This procedure is called
Alpha Beta pruning which "prunes" the tree branches thus reducing the number
of static evaluations.
We use the following example to explain the notion of Alpha Beta Pruning.
Suppose we start of with a game tree in the diagram below. Notice that all
nodes/situations have not yet been previously evaluated for their static evaluation
score. Only two leaf nodes have been evaluated so far.
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Sitting at A, the player-one will observe that if he moves to B the best he can get
is 3.
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So the value three travels to the root A. Now after observing the other side of the
tree, this score will either increase or will remain the same as this level is for the
maximizer.
When he evaluates the first leaf node on the other side of the tree, he will see
that the minimizer can force him to a score of less than 3 hence there is no need
to fully explore the tree from that side. Hence the right most branch of the tree will
be pruned and won't be evaluated for static evaluation.
We have discussed a detailed example on Alpha Beta Pruning in the lectures.
We have shown the sequence of steps in the diagrams below. The readers are
required to go though the last portion of Lecture 10 for the explanation of this
example, if required.
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2.25 Summary
People used to think that one who can solve more problems is more intelligent
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Generate and test is the classical approach to solving problems
Problem representation plays a key role in problem solving
The components of problem solving include
o Problem Statement
o Operators
o Goal State
o Solution Space
Searching is a formal mechanism to explore alternatives
Searches can be blind or uninformed, informed, heuristic, non-optimal and
optional.
Different procedures to implement different search strategies form the major
content of this chapter
2.26 Problems
Q1 Consider that a person has never been to the city air port. Its early in the
morning and assume that no other person is awake in the town who can guide
him on the way. He has to drive on his car but doesn't know the way to air port.
Clearly identify the four components of problem solving in the above statement,
i.e. problem statement, operators, solution space, and goal state. Should he
follow blind or heuristic search strategy? Try to model the problem in a graphical
representation.
Q2 Clearly identify the difference between WSP (Well-Structured Problems) and
ISP (Ill- Structured) problems as discussed in the lecture. Give relevant
examples.
Q3 Given the following tree. Apply DFS and BFS as studied in the chapter. Show
the state of the data structure Q and the visited list clearly at every step. S is the
initial state and D is the goal state.
S
A
B
C
D
E
F
G
H
I
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Q4 Discuss how progressive deepening uses a mixture of DFS and BFS to
eliminate the disadvantages of both and at the same time finds the solution is a
given tree. Support your answer with examples of a few trees.
Q5 Discuss the problems in Hill Climbing. Suggest solutions to the commonly
encountered problems that are local maxima, plateau problem and ridge problem.
Given the following tree, use the hill climbing procedure to climb up the tree. Use
your suggested solutions to the above mention problems if any of them are
encountered.  K is the goal state and numbers written on each node is the
estimate of remaining distance to the goal.
10
S
9
11
A
B
7
9
12
C
D
E
7
7
7
G
H
F
7
5
I
J
0
K
Q6 Discuss how best first search works in a tree. Support your answer with an
example tree. Is best first search always the best strategy? Will it always
guarantee the best solution?
Q7 Discuss how beam search with degree of the search = 3 propagates in the
given search tree. Is it equal to best first search when the degree = 1.
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Q8 Discuss the main concept behind branch and bound search strategy. Suggest
Improvements in the Algorithm. Simulate the algorithm on the given graph below.
The values on the links are the distances between the cities. The numbers on the
nodes are the estimated distance on the node from the goal state.
Q9. Run the MiniMax procedure on the given tree. The static evaluation scores
for each leaf node are written under it. For example the static evaluation scores
for the left most leaf node is 80.
80
10
55
45
65
100
20
35
70
Q10 Discuss how Alpha Beta Pruning minimizes the number of static evaluations
at the leaf nodes by pruning branches. Support your answer with small examples
of a few trees.
Q11 Simulate the Minimax procedure with Alpha Beta Pruning algorithm on the
following search tree.
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Maximizing Level
Minimizing Level
Maximizing Level
30
30
50
40
70
60
20
80
90
10
90
70
60
Adapted from: Artificial Intelligence, Third Edition by Patrick Henry Winston
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