AI Lesson 04

Module
2
Problem Solving
using Search-
(Single agent search)
Version 2 CSE IIT, Kharagpur

Lesson
4
Uninformed Search

2.4 Search
Searching through a state space involves the following:
• A set of states
• Operators and their costs
• Start state
• A test to check for goal state

We will now outline the basic search algorithm, and then consider various variations of
this algorithm.

2.4.1 The basic search algorithm

Let L be a list containing the initial state (L= the fringe)
Loop
if L is empty return failure
Node select (L)
if Node is a goal
then return Node
(the path from initial state to Node)
else generate all successors of Node, and
merge the newly generated states into L
End Loop

We need to denote the states that have been generated. We will call these as nodes. The
data structure for a node will keep track of not only the state, but also the parent state or
the operator that was applied to get this state. In addition the search algorithm maintains a
list of nodes called the fringe. The fringe keeps track of the nodes that have been
generated but are yet to be explored. The fringe represents the frontier of the search tree
generated. The basic search algorithm has been described above.

Initially, the fringe contains a single node corresponding to the start state. In this version
we use only the OPEN list or fringe. The algorithm always picks the first node from
fringe for expansion. If the node contains a goal state, the path to the goal is returned. The
path corresponding to a goal node can be found by following the parent pointers.
Otherwise all the successor nodes are generated and they are added to the fringe.

The successors of the current expanded node are put in fringe. We will soon see that the
order in which the successors are put in fringe will determine the property of the search
algorithm.


2.4.2 Search algorithm: Key issues
Corresponding to a search algorithm, we get a search tree which contains the generated
and the explored nodes. The search tree may be unbounded. This may happen if the state
space is infinite. This can also happen if there are loops in the search space. How can we
handle loops?

Corresponding to a search algorithm, should we return a path or a node? The answer to
this depends on the problem. For problems like N-queens we are only interested in the
goal state. For problems like 15-puzzle, we are interested in the solution path.

We see that in the basic search algorithm, we have to select a node for expansion. Which
node should we select? Alternatively, how would we place the newly generated nodes in
the fringe? We will subsequently explore various search strategies and discuss their
properties,

Depending on the search problem, we will have different cases. The search graph may be
weighted or unweighted. In some cases we may have some knowledge about the quality
of intermediate states and this can perhaps be exploited by the search algorithm. Also
depending on the problem, our aim may be to find a minimal cost path or any to find path
as soon as possible.

Which path to find?
The objective of a search problem is to find a path from the initial state to a goal state. If
there are several paths which path should be chosen? Our objective could be to find any
path, or we may need to find the shortest path or least cost path.

2.4.3 Evaluating Search strategies
We will look at various search strategies and evaluate their problem solving performance.
What are the characteristics of the different search algorithms and what is their
efficiency? We will look at the following three factors to measure this.

1. Completeness: Is the strategy guaranteed to find a solution if one exists?

2. Optimality: Does the solution have low cost or the minimal cost?

3. What is the search cost associated with the time and memory required to find
a solution?
a. Time complexity: Time taken (number of nodes expanded) (worst or
average case) to find a solution.
b. Space complexity: Space used by the algorithm measured in terms of
the maximum size of fringe


The different search strategies that we will consider include the following:
1. Blind Search strategies or Uninformed search
a. Depth first search
b. Breadth first search
c. Iterative deepening search
d. Iterative broadening search
2. Informed Search
3. Constraint Satisfaction Search
4. Adversary Search

Blind Search
In this lesson we will talk about blind search or uninformed search that does not use any
extra information about the problem domain. The two common methods of blind search
are:

• BFS or Breadth First Search
• DFS or Depth First Search

2.4.4 Search Tree
Consider the explicit state space graph shown in the figure.

One may list all possible paths, eliminating cycles from the paths, and we would get the
complete search tree from a state space graph. Let us examine certain terminology
associated with a search tree. A search tree is a data structure containing a root node,
from where the search starts. Every node may have 0 or more children. If a node X is a
child of node Y, node Y is said to be a parent of node X.

B E

A D F H
C
G

Figure 1: A State Space Graph


A

B C

D E D G

C F B F

G G H E G H

Figure 2: Search tree for the state space graph in Figure 25

Consider the state space given by the graph in Figure 25. Assume that the arcs are
bidirectional. Starting the search from state A the search tree obtained is shown in Figure
26.

Search Tree – Terminology
• Root Node: The node from which the search starts.
• Leaf Node: A node in the search tree having no children.
• Ancestor/Descendant: X is an ancestor of Y is either X is Y’s parent or X is an
ancestor of the parent of Y. If S is an ancestor of Y, Y is said to be a descendant
of X.
• Branching factor: the maximum number of children of a non-leaf node in the
search tree
• Path: A path in the search tree is a complete path if it begins with the start node
and ends with a goal node. Otherwise it is a partial path.

We also need to introduce some data structures that will be used in the search algorithms.

Node data structure
A node used in the search algorithm is a data structure which contains the following:
1. A state description
2. A pointer to the parent of the node
3. Depth of the node
4. The operator that generated this node
5. Cost of this path (sum of operator costs) from the start state

The nodes that the algorithm has generated are kept in a data structure called OPEN or
fringe. Initially only the start node is in OPEN.

The search starts with the root node. The algorithm picks a node from OPEN for
expanding and generates all the children of the node. Expanding a node from OPEN
results in a closed node. Some search algorithms keep track of the closed nodes in a data
structure called CLOSED.

A solution to the search problem is a sequence of operators that is associated with a path
from a start node to a goal node. The cost of a solution is the sum of the arc costs on the
solution path. For large state spaces, it is not practical to represent the whole space. State
space search makes explicit a sufficient portion of an implicit state space graph to find a
goal node. Each node represents a partial solution path from the start node to the given
node. In general, from this node there are many possible paths that have this partial path
as a prefix.

The search process constructs a search tree, where
• root is the initial state and
• leaf nodes are nodes
• not yet expanded (i.e., in fringe) or
• having no successors (i.e., “dead-ends”)
Search tree may be infinite because of loops even if state space is small

The search problem will return as a solution a path to a goal node. Finding a path is
important in problems like path finding, solving 15-puzzle, and such other problems.
There are also problems like the N-queens problem for which the path to the solution is
not important. For such problems the search problem needs to return the goal state only.

2.5 Breadth First Search
2.5.1 Algorithm

Breadth first search
Let fringe be a list containing the initial state
Loop
if fringe is empty return failure
Node remove-first (fringe)
if Node is a goal
then return the path from initial state to Node
(merge the newly generated nodes into fringe)
add generated nodes to the back of fringe
End Loop

Note that in breadth first search the newly generated nodes are put at the back of fringe or
the OPEN list. What this implies is that the nodes will be expanded in a FIFO (First In


First Out) order. The node that enters OPEN earlier will be expanded earlier. This
amounts to expanding the shallowest nodes first.

2.5.2 BFS illustrated
We will now consider the search space in Figure 1, and show how breadth first search
works on this graph.

Step 1: Initially fringe contains only one node corresponding to the source state A.

A

B C

D E D G

C F B F

G G H E G H

Figure 3

FRINGE: A

Step 2: A is removed from fringe. The node is expanded, and its children B and C are
generated. They are placed at the back of fringe.

A

B C

D E D G

C F B F

G G H E G H

Figure 4

FRINGE: B C

Step 3: Node B is removed from fringe and is expanded. Its children D, E are generated
and put at the back of fringe.


A

B C

D E D G

C F B F

G G H E G H

Figure 5

FRINGE: C D E

Step 4: Node C is removed from fringe and is expanded. Its children D and G are added
to the back of fringe.

A

B C

D E D G

C F B F

G G H E G H

Figure 6

FRINGE: D E D G

Step 5: Node D is removed from fringe. Its children C and F are generated and added to
the back of fringe.


A

B C

D E D G

C F B F

G G H E G H

Figure 7

FRINGE: E D G C F

Step 6: Node E is removed from fringe. It has no children.
A

B C

D E D G

C F B F

G G H E G H

FRINGE: D G C F

Step 7: D is expanded, B and F are put in OPEN.

A

B C

D E D G Goal!

C F B F

G G H E G H

Figure 8


FRINGE: G C F B F

Step 8: G is selected for expansion. It is found to be a goal node. So the algorithm returns
the path A C G by following the parent pointers of the node corresponding to G. The
algorithm terminates.

2.5.3 Properties of Breadth-First Search
We will now explore certain properties of breadth first search. Let us consider a model of
the search tree as shown in Figure 3. We assume that every non-leaf node has b children.
Suppose that d is the depth o the shallowest goal node, and m is the depth of the node
found first.

…. b children

….
Figure 9: Model of a search tree with uniform branching factor b

Breadth first search is:
• Complete.
• The algorithm is optimal (i.e., admissible) if all operators have the same cost.
Otherwise, breadth first search finds a solution with the shortest path length.
• The algorithm has exponential time and space complexity.
Suppose the search tree can be modeled as a b-ary tree as shown in Figure 3. Then the
time and space complexity of the algorithm is O(bd) where d is the depth of the
solution and b is the branching factor (i.e., number of children) at each node.

A complete search tree of depth d where each non-leaf node has b children, has a total of
1 + b + b2 + ... + bd = (b(d+1) - 1)/(b-1) nodes

Consider a complete search tree of depth 15, where every node at depths 0 to14 has 10
children and every node at depth 15 is a leaf node. The complete search tree in this case
will have O(1015) nodes. If BFS expands 10000 nodes per second and each node uses 100
bytes of storage, then BFS will take 3500 years to run in the worst case, and it will use
11100 terabytes of memory. So you can see that the breadth first search algorithm cannot
be effectively used unless the search space is quite small. You may also observe that even
if you have all the time at your disposal, the search algorithm cannot run because it will
run out of memory very soon.


Advantages of Breadth First Search
Finds the path of minimal length to the goal.

Disadvantages of Breadth First Search
Requires the generation and storage of a tree whose size is exponential the the depth of
the shallowest goal node

2.6 Uniform-cost search
This algorithm is by Dijkstra [1959]. The algorithm expands nodes in the order of their
cost from the source.

We have discussed that operators are associated with costs. The path cost is usually taken
to be the sum of the step costs.

In uniform cost search the newly generated nodes are put in OPEN according to their
path costs. This ensures that when a node is selected for expansion it is a node with the
cheapest cost among the nodes in OPEN.
Let g(n) = cost of the path from the start node to the current node n. Sort nodes by
increasing value of g.
Some properties of this search algorithm are:
• Complete
• Optimal/Admissible
• Exponential time and space complexity, O(bd)

2.7 Depth first Search
2.7.1 Algorithm
Depth First Search
Loop
if Node is a goal
merge the newly generated nodes into fringe
add generated nodes to the front of fringe
End Loop

The depth first search algorithm puts newly generated nodes in the front of OPEN. This
results in expanding the deepest node first. Thus the nodes in OPEN follow a LIFO order
(Last In First Out). OPEN is thus implemented using a stack data structure.


2.7.2 DFS illustrated

B E

A D F H
C
G
Figure 10

A

B C

D E D G

C F B F

G G H E G H
Figure 11: Search tree for the state space graph in Figure 34

Let us now run Depth First Search on the search space given in Figure 34, and trace its
progress.

Step 1: Initially fringe contains only the node for A.
A

B C

D E D G

C F B F

G G H E G H

FRINGE: A

Figure 12


Step 2: A is removed from fringe. A is expanded and its children B and C are put in front
of fringe.

A

B C

D E D G

C F B F

G G H E G H

FRINGE: B C

Figure 13

Step 3: Node B is removed from fringe, and its children D and E are pushed in front of
fringe.
A

B C

D E D G

C F B F

G G H E G H

FRINGE: D E C

Figure 14

Step 4: Node D is removed from fringe. C and F are pushed in front of fringe.


A

B C

D E D G

C F B F

G G H E G H

FRINGE: C F E C

Figure 15

Step 5: Node C is removed from fringe. Its child G is pushed in front of fringe.

A

B C

D E D G

C F B F

G G H E G H

FRINGE: G F E C

Figure 16

Step 6: Node G is expanded and found to be a goal node. The solution path A-B-D-C-G
is returned and the algorithm terminates.


A

B C

D E D G

C F B F

Goal! G G H E G H

FRINGE: G F E C

Figure 17

2.7.3 Properties of Depth First Search
Let us now examine some properties of the DFS algorithm. The algorithm takes
exponential time. If N is the maximum depth of a node in the search space, in the worst
case the algorithm will take time O(bd). However the space taken is linear in the depth of
the search tree, O(bN).
Note that the time taken by the algorithm is related to the maximum depth of the search
tree. If the search tree has infinite depth, the algorithm may not terminate. This can
happen if the search space is infinite. It can also happen if the search space contains
cycles. The latter case can be handled by checking for cycles in the algorithm. Thus
Depth First Search is not complete.

2.7.4 Depth Limited Search
A variation of Depth First Search circumvents the above problem by keeping a depth
bound. Nodes are only expanded if they have depth less than the bound. This algorithm is
known as depth-limited search.

Depth limited search (limit)
Loop
if Node is a goal
else if depth of Node = limit return cutoff
else add generated nodes to the front of fringe
End Loop
.


2.7.5 Depth-First Iterative Deepening (DFID)
First do DFS to depth 0 (i.e., treat start node as having no successors), then, if no solution
found, do DFS to depth 1, etc.

DFID
until solution found do
DFS with depth cutoff c
c = c+1

Advantage
• Linear memory requirements of depth-first search
• Guarantee for goal node of minimal depth

Procedure
Successive depth-first searches are conducted – each with depth bounds increasing by 1

Figure 18: Depth First Iterative Deepening

Properties
For large d the ratio of the number of nodes expanded by DFID compared to that of DFS
is given by b/(b-1).
For a branching factor of 10 and deep goals, 11% more nodes expansion in iterative-
deepening search than breadth-first search

The algorithm is
• Complete
• Optimal/Admissible if all operators have the same cost. Otherwise, not optimal
but guarantees finding solution of shortest length (like BFS).
• Time complexity is a little worse than BFS or DFS because nodes near the top of
the search tree are generated multiple times, but because almost all of the nodes
are near the bottom of a tree, the worst case time complexity is still exponential,
O(bd)


If branching factor is b and solution is at depth d, then nodes at depth d are
generated once, nodes at depth d-1 are generated twice, etc.
Hence bd + 2b(d-1) + ... + db <= bd / (1 - 1/b)2 = O(bd).
• Linear space complexity, O(bd), like DFS

Depth First Iterative Deepening combines the advantage of BFS (i.e., completeness) with
the advantages of DFS (i.e., limited space and finds longer paths more quickly)
This algorithm is generally preferred for large state spaces where the solution depth is
unknown.

There is a related technique called iterative broadening is useful when there are many
goal nodes. This algorithm works by first constructing a search tree by expanding only
one child per node. In the 2nd iteration, two children are expanded, and in the ith iteration
I children are expanded.

Bi-directional search
Suppose that the search problem is such that the arcs are bidirectional. That is, if there is
an operator that maps from state A to state B, there is another operator that maps from
state B to state A. Many search problems have reversible arcs. 8-puzzle, 15-puzzle, path
planning etc are examples of search problems. However there are other state space search
formulations which do not have this property. The water jug problem is a problem that
does not have this property. But if the arcs are reversible, you can see that instead of
starting from the start state and searching for the goal, one may start from a goal state and
try reaching the start state. If there is a single state that satisfies the goal property, the
search problems are identical.
How do we search backwards from goal? One should be able to generate predecessor
states. Predecessors of node n are all the nodes that have n as successor. This is the
motivation to consider bidirectional search.

Algorithm: Bidirectional search involves alternate searching from the start state toward
the goal and from the goal state toward the start. The algorithm stops when the frontiers
intersect.


A search algorithm has to be selected for each half. How does the algorithm know when
the frontiers of the search tree intersect? For bidirectional search to work well, there must
be an efficient way to check whether a given node belongs to the other search tree.

Bidirectional search can sometimes lead to finding a solution more quickly. The reason
can be seen from inspecting the following figure.

Also note that the algorithm works well only when there are unique start and goal states.
Question: How can you make bidirectional search work if you have 2 possible goal
states?

Time and Space Complexities
Consider a search space with branching factor b. Suppose that the goal is d steps away
from the start state. Breadth first search will expand O(bd) nodes.
If we carry out bidirectional search, the frontiers may meet when both the forward and
the backward search trees have depth = d/2. Suppose we have a good hash function to
check for nodes in the fringe. IN this case the time for bidirectional search will be
O((bd/2). Also note that for at least one of the searches the frontier has to be stored. So the
space complexity is also O((bd/2).

Comparing Search Strategies


Breadth Depth first Iterative Bidirectional
first deepening (if applicable)

Time bd bd bd bd/2
Space bd bm bd bd/2
Optimum? Yes No Yes Yes
Complete? Yes No Yes Yes

Search Graphs
If the search space is not a tree, but a graph, the search tree may contain different nodes
corresponding to the same state. It is easy to consider a pathological example to see that
the search space size may be exponential in the total number of states.

In many cases we can modify the search algorithm to avoid repeated state expansions.
The way to avoid generating the same state again when not required, the search algorithm
can be modified to check a node when it is being generated. If another node
corresponding to the state is already in OPEN, the new node should be discarded. But
what if the state was in OPEN earlier but has been removed and expanded? To keep track
of this phenomenon, we use another list called CLOSED, which records all the expanded
nodes. The newly generated node is checked with the nodes in CLOSED too, and it is put
in OPEN if the corresponding state cannot be found in CLOSED. This algorithm is
outlined below:

Graph search algorithm
Let closed be initially empty
Loop
if Node is a goal
else put Node in closed
generate all successors of Node S
for all nodes m in S
if m is not in fringe or closed
merge m into fringe
End Loop

But this algorithm is quite expensive. Apart from the fact that the CLOSED list has to be
maintained, the algorithm is required to check every generated node to see if it is already
there in OPEN or CLOSED. Unless there is a very efficient way to index nodes, this will
require additional overhead for every node.


In many search problems, we can adopt some less computationally intense strategies.
Such strategies do not stop duplicate states from being generated, but are able to reduce
many of such cases.

The simplest strategy is to not return to the state the algorithm just came from. This
simple strategy avoids many node re-expansions in 15-puzzle like problems.

A second strategy is to check that you do not create paths with cycles in them. This
algorithm only needs to check the nodes in the current path so is much more efficient
than the full checking algorithm. Besides this strategy can be employed successfully with
depth first search and not require additional storage.

The third strategy is as outlined in the table. Do not generate any state that was ever
created before.

Which strategy one should employ must be determined by considering the frequency of
“loops” in the state space.

Questions for Module 2
1. Give the initial state, goal test, sucessor function, and cost function for each of the
following.
Choose a formulation that is precise enough to be implemented.
a) You have to colour a planar map using only four colours, in such a way that
no two adjacent regions have the same colour.
b) In the travelling salesperson problem (TSP) there is a map involving N cities
some of which are connected by roads. The aim is to find the shortest tour that
starts from a city, visits all the cities exactly once and comes back to the
starting city.
c) Missionaries & Cannibals problem: 3 missionaries & 3 cannibals are on one
side of the river. 1 boat carries 2. Missionaries must never be outnumbered by
cannibals. Give a plan for all to cross the river.

2. Given a full 5-gallon jug and an empty 2-gallon jug, the goal is to fill the 2-gallon
jug with exactly one gallon of water. You may use the following state space
formulation.
State = (x,y), where x is the number of gallons of water in the 5-gallon jug and y
is # of gallons in the 2-gallon jug
Initial State = (5,0)
Goal State = (*,1), where * means any amount
Create the search tree. Discuss which search strategy is appropriate for this
problem.

3. Consider the following graph.


B E

A D F H
C
G

Starting from state A, execute DFS. The goal node is G. Show the order in which the
nodes are expanded. Assume that the alphabetically smaller node is expanded first to
break ties.

4. Suppose you have the following search space:

State next cost
A B 4
A C 1
B D 3
B E 8
C C 0
C D 2
C F 6
D C 2
D E 4
E G 2
F G 8

a) Draw the state space of this problem.
b) Assume that the initial state is A and the goal state is G. Show how each of
the following search strategies would create a search tree to find a path from
the initial state to the goal state:

I. Breadth-first search
II. Depth-first search
III. Uniform cost search
IV. Iterative deepening search
At each step of the search algorithm, show which node is being expanded, and the
content of fringe. Also report the eventual solution found by each algorithm, and the


solution cost.

5. Suppose that breadth first search expands N nodes for a particular graph. What will
be the maximum number of nodes expanded by Iterative Deepening search ?

Solutions
1. You have to colour a planar map using only four colours, in such a way that no two
adjacent regions have the same colour.

The map is represented by a graph. Each region corresponds to a vertex of the graph.
If two regions are adjacent, there is an edge connecting the corresponding vertices.

The vertices are named <v1, v2, … , vN>.
The colours are represented by c1, c2, c3, c4.
A state is represented as a N-tuple representing the colours of the vertices. A vertes
has colour x if its colour has not yet been assigned. An example state is:
{c1, x, c1, c3, x, x, x …}
colour(i) denotes the colour of si.

Consider the map below consisting of 5 regions namely A, B, C, D and E. The adjacency
information is represented by the corresponding graph shown.

A
A
B C
D
B
D E
E
C

A state of this problem is shown below.


A
D
B

E
C

This state is represented as {blue, green, x, x, blue}.

The initial state for this problem is given as {x, x, x, x, x}
The goal test is as follows. For every pair of states si and sj that are adjacent, colour(i)
must be different from colour(j).

The successor functions are of the form:
• Change (i, c): Change the colour of a state i to c.

2. In the travelling salesperson problem (TSP) there is a map involving N cities some of
which are connected by roads. The aim is to find the shortest tour that starts from a
city, visits all the cities exactly once and comes back to the starting city.

A
41
94
4

190
38

65
B C
100

302
4
15

7
35
D 204 E

Y: set of N cities
d(x,y) : distance between cities x and y. x,y∈Y
A state is a Hamiltonian path (does not visit any city twice)
X: set of states

X: set of states. X =
{(x1, x2, …, xn)|
n=1, …, N+1,
xi ∈Y for all I,


xi ≠ xj unless i=1, j=N+1}
Successors of state
(x1, x2, …, xn):

δ (x1, x2, …, xn) = {(x1, x2, …, xn, xn+1) | xn+1∈ Y
xn+1 ≠ xi for all 1≤ i≤ n }
The set of goal states include all states of length N+1

3. Missionaries & Cannibals problem: 3 missionaries & 3 cannibals are on one side of
the river. 1 boat carries 2. Missionaries must never be outnumbered by cannibals.
Give a plan for all to cross the river.
State: <M, C, B>
♦ M: no of missionaries on the left bank
♦ C: no of cannibals on the left bank
♦ B: position of the boat: L or R
4. Initial state: <3, 3, L>
5. Goal state: <0, 0, R>
6. Operators: <M,C>
► M: No of missionaries on the boat
► C: No of cannibals on the boat
Valid operators: <1,0> <2,0>, <1,1>, <0,1> <0,2>

<3, 3, L> <3, 3, R>
<2, 3, L> <2, 3, R>
<1, 3, L> <1, 3, R>
<3, 2, L> <3, 2, R>
<2, 2, L> <2, 2, R>
<1, 2, L> <1, 2, R>
<3, 1, L>
<3, 1, R>
<2, 1, L> <2, 1, R>
<1, 1, L> <1, 1, R>
<0, 2, L> <0, 2, R>
<0, 1, L> <0, 1, R>
<0, 3, L> <0, 3, R>
<3, 0, L> <3, 0, R>
<2, 0, L> <2, 0, R>
<1, 0, L> <1, 0, R>
<0, 0, L> <0, 0, R>


<3, 3, L> <3, 3, R>

<3, 2, L> <3, 2, R>

<2, 2, L> <2, 2, R>

<3, 1, R>
<3, 1, L>
<1, 1, R>
<1, 1, L>
<0, 2, R>
<0, 2, L>
<0, 1, R>
<0, 1, L>
<0, 3, R>
<0, 3, L>
<3, 0, R>
<3, 0, L>
<0, 0, R>
<0, 0, L>

2. Given a full 5-gallon jug and an empty 2-gallon jug, the goal is to fill the 2-gallon jug
with exactly one gallon of water. You may use the following state space formulation.
• State = (x,y), where x is the number of gallons of water in the 5-gallon jug and y
is # of gallons in the 2-gallon jug
• Initial State = (5,0)
• Goal State = (*,1), where * means any amount
Create the search tree. Discuss which search strategy is appropriate for this problem.

Solution:
The table below shows the different operators and their effects.

Name Cond. Transition Effect

Empty5 – (x,y)→(0,y) Empty 5-gal. jug

Empty2 – (x,y)→(x,0) Empty 2-gal. jug

2to5 x≤3 (x,2)→(x+2,0) Pour 2-gal. into 5-gal.

5to2 x≥2 (x,0)→(x-2,2) Pour 5-gal. into 2-gal.

5to2part y<2 (1,y)→(0,y+1) Pour partial 5-gal. into 2-gal.


The figure below shows the different states and the transitions between the states using
the operators above. The transitions corresponding to Empty2 have not been marked to
keep the figure clean.

Empty5 5, 2 5, 1 5, 0

Empty2 4, 2 4, 1 4, 0

3, 2 3, 1 3, 0
2to5

2, 2 2, 1 2, 0
5to2
1, 2 1, 1 1, 0
5to2part
0, 2 0, 1 0, 0

(5,0) is the initial state.
(0,1) is the goal state.
A solution to this problem is given by the path
(5,0) – (3,2) – (3,0) – (1,2) – (1,0) – (0,1).
using the operators
5to2, Empty2, 5to2, Empty2, 5to2.

Depth search is not appropriate
The search space is a graph.
The search tree has many repeated states
Breadth first search is appropriate

3. Consider the following graph.


B E

A D F H
C
G

Starting from state A, execute DFS. The goal node is G. Show the order in which the
nodes are expanded. Assume that the alphabetically smaller node is expanded first to
break ties.

Solution:

Step Fringe Node Expanded Comments
1 A
2 BC A
3 DEC B
4 FEC D
5 HEC F
6 EC H
7 C E
8 DG C
9 FG D
10 HG F
11 G H
12 G Goal reached!


AI Lesson 04

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AI Lesson 04