INFO 2950
Prof. Carla Gomes
gomes@cs.cornell.edu
Module
Modeling Computation:
Finite State Machines without Output
Rosen, Chapter 12.2 and 12.3
1
Finite-State Machines with No Output
Definition:
Concatenation of A and B
A and B are subsets of V*, where V is a vocabulary
The concatenation of A and B is
AB={xy: x string in A and y string in B}
Example: A={0, 11} and B={1, 10, 110}
AB={01,010,0110,111,1110,11110}
What is BA?
A0={λ}
An+1=AnA for n=0,1,2,…
Definition:
Concatenation of A and B
Example: A={1,00}
An ? n=1, 2, 3,…
A0
A0={λ}
A1=A0A1={λ} A ?
A1=A0A1={λ} A={1,00}
A2=A1A ={1,00} {1,00}
A2=A1A ={1,00} {1,00} = {11,100,001,0000}
A3=A2A ={11,100,001,0000}{1,00}
A3=A2A ={11,100,001,0000}{1,00} =
{111,1100,1001,10000,001111,00100,00001,000000}
Kleene Closure
Let A be any subset of V*.
Kleene closure of A, denoted by A*, is

A A
*
k 0
k
The set consisting of concatenations of
arbitrarily many strings from A
If B={0,1}, B*=V*.
Kleene Closure
What are the Kleen closures of the set A={0}; B={0,1}, and C={11}?
A*={,0,00,000,…} ={0n |n=0,1,2,…}
If B={0,1}, B*=V*. (The set of all strings over the alphabet V)
C*={,11,1111,111111,…} ={(11)n |n=0,1,2,…} ={12n |n=0,1,2,…}
Finite State Automata
Representations
As with finite-state machines, finite-state
automata have the following common
representations
– A state table is used to represent a finite-state automaton by giving
the values of the function f.
• Just like for finite-state machines, except the second half of the
columns are omitted, since there is no output.
– A state diagram is a directed graph representation of a finite-state
automaton.
• Final states are usually denoted by double-circles.
• Values separated by commas denote several possible inputs, not inputs
and outputs as with finite state machines.
Finite State Automata
A finite state automation over an alphabet is illustrated by a state
diagram:
– a directed graph
– edges are labeled with elements of alphabet,
– some nodes (or states), marked as final
of “accepting”.
– one node marked as start state
Dealing with Input Strings
Let x=x1x2…xnI * (That is, x is a string over I)
Then we can extend the transition function f to all state-input
string pairs (rather than simply state-input pairs) in the
obvious way
–
–
–
–
–
Assume the machine is in state S0,
Compute f(S0, x1) = Si2
Next compute f(Si2, x2) = Si3
Continue until you get f(Sin, xn) = Sin+1
We define f(S0, x) = Sin+1
From now on, we will speak of the transition function f being
applied to input strings, not just single inputs.
Finite Automaton
Input
String
Output
Finite
Automaton
“Accept”
or
“Reject”
Transition Graph
a, b
q5
b
q0 a
a
a
b
q1 b q2 b q3 a
initial
state
transition
state
a, b
q4
accepting
state
Initial Configuration
Input String
a b b a
a, b
q5
b
q0 a
a
a
b
q1 b q2 b q3 a
a, b
q4
Reading the Input
a b b a
a, b
q5
b
q0 a
a
a
b
q1 b q2 b q3 a
a, b
q4
a b b a
a, b
q5
b
q0 a
a
a
b
q1 b q2 b q3 a
a, b
q4
a b b a
a, b
q5
b
q0 a
a
a
b
q1 b q2 b q3 a
a, b
q4
a b b a
a, b
q5
b
q0 a
a
a
b
q1 b q2 b q3 a
a, b
q4
Input finished
a b b a
a, b
q5
b
q0 a
a
a
b
q1 b q2 b q3 a
a, b
q4
accept
Rejection
a b a
a, b
q5
b
q0 a
a
a
b
q1 b q2 b q3 a
a, b
q4
a b a
a, b
q5
b
q0 a
a
a
b
q1 b q2 b q3 a
a, b
q4
a b a
a, b
q5
b
q0 a
a
a
b
q1 b q2 b q3 a
a, b
q4
a b a
a, b
q5
b
q0 a
a
a
b
q1 b q2 b q3 a
a, b
q4
Input finished
a b a
a, b
q5
b
q0 a
a
a
b
q1 b q2 b q3 a
reject
a, b
q4
Another Rejection

a, b
q5
b
q0 a
a
a
b
q1 b q2 b q3 a
a, b
q4

a, b
q5
b
q0 a
reject
a
a
b
q1 b q2 b q3 a
a, b
q4
Another Example
a a b
a, b
a
q0
b
q1
a, b
q2
a a b
a, b
a
q0
b
q1
a, b
q2
a a b
a, b
a
q0
b
q1
a, b
q2
a a b
a, b
a
q0
b
q1
a, b
q2
Input finished
a a b
a
q0
a, b
accept
b
q1
a, b
q2
Rejection Example
b a b
a, b
a
q0
b
q1
a, b
q2
b a b
a, b
a
q0
b
q1
a, b
q2
b a b
a, b
a
q0
b
q1
a, b
q2
b a b
a, b
a
q0
b
q1
a, b
q2
Input finished
b a b
a, b
a
q0
b
q1
a, b
q2
reject
Finite-State Automata
Definition: A finite-state automaton is a 5-tuple
M=(S, I, f, S0, F) where
– S is a finite set of states
– I is a finite input alphabet
– f:SIS is a transition function from each state-input
pair to a state
– S0 is the initial state
– FS is a set of final states
Note: automaton is the singular of automata.
Finite Automata
’
If end of input
– If in accepting state => accept
– Otherwise => reject
If no transition possible (got stuck) => reject
Language Recognition
Definition: A finite-state automaton accepts (or
recognizes) a string x if f(S0, x)F. That is, the
finite state automaton ends up in a final state.
Definition: The language accepted (or recognized)
by a finite-state automaton M, denoted by L(M), is
the set of all strings recognized by M.
Definition: Two finite-state automata are equivalent
if they recognize the same language.
LM   abba
Example
M
a, b
q5
b
q0 a
a
a
b
q1 b q2 b q3 a
a, b
q4
accept
Example
LM    , ab, abba
M
a, b
q5
b
q0 a
accept
a
a
b
q1 b q2 b q3 a
accept
a, b
q4
accept
Example
LM   {a b : n  0}
n
a, b
a
q0
b
q1
accept
a, b
q2
trap state
What is the Language?
Example: What language is recognized by the
following finite-state automaton?
Input 0 or 1
• Solution: Since the only final state is the start state, and
only an input of 1 will leave the machine in the start state,
it is easy to see that
L(M)={1n: n=0,1,2,…}
What is the Language?
Example: What language is recognized by the following
finite-state automaton?
• Solution: Notice that
Any input that does not start with 0 cannot go to a final state
The final state can only be arrived at if the last input is a 1
Any string can occur in between the first 0 and last 1.
• Thus, we can see that
L(M)={binary strings beginning with 0 and ending with 1}
FSM Examples
Example: What language is recognized by the
following finite-state automaton?
0
1
S0
0
1
S1
Accepts strings
over alphabet
{0,1} that end in 1
44
FSM Examples
Example: What language is recognized by the
following finite-state automaton?
S0
a
a
b
S1
b
S2
a
4
b
b
b
a
5
a
Accepts strings
over alphabet {a,b}
that begin and end
with same symbol
45
Example: What language is recognized by the
following finite-state automaton?
0
S1
Start
0
2
S0
FSM Examples
Accepts strings over
{0,1,2} such that sum of
digits is a multiple of 3
1
1
2
S2
0
2
1
46
FSM Examples
0
0
1
Even
Odd
1
Accepts strings over {0,1}
that have an odd number
of ones
47
FSM Examples
1
0
1
1
0
'0'
0,1
'00'
'001'
0
Accepts strings over
{0,1} that contain
the substring 001
48
Examples
Design a FSM to recognize strings with an equal number of ones and
zeros.
– Not possible
Design a FSM to recognize strings with an equal number of substrings
"01" and "10".
– Perhaps surprisingly, this is possible
49
FSM Examples
0
1
0
1
Accepts strings with
an equal number
of substrings "01" and
"10"
0
1
1
1
0
0
50
Non-deterministic Finite
Automaton (NDFA)
a non-deterministic finite state machine or non-deterministic finite
automaton (NDFA) is a finite state machine where for each pair of
state and input symbol it assigns a set of states i.e., from a pair of state
and input symbol there are possibly two or more transitions to a next
state.
Every NDFA can be converted to a DFA
Nondeterministic
Finite-State Automata
Definition: A Nondeterministic fnite-state
automaton is a 5-tuple M=(S, I, f, S0, F) where
– S is a finite set of states
– I is a finite input alphabet
– f:SI P(S) is a transition function from each stateinput pair to a set of states
– S0 is the initial state
– FS is a set of final states
0
0
S1
1
Start
S3
0,1
S0
1
0
0
State
F
0
F
1
S0
S0,S2
S1
S1
S3
S4
S2
S2
1
S4
S4
S3
S3
S4
S3
S3
53
S1
1
Start
0
0
S3
0,1
S0
1
0
0
S2
1
S4
What does it mean for a nondeterministic FSA (NFSA)
to recognize a string?
T he NDFSA recognizes or accepts a string x if there is a final
state in the set of all states that can be obtained from S0, using x.
54
What language does it accept?
0n 11 |n≥0
1
0
S
0
S
3
0
0,1
1
0
Start
0
S
1
S
2
0n |n≥0
1
S
4
0n 01 |n≥0
55
Equivalence of DFSM and NFSM
Theorem:
– For each non-deterministic finite state machine N, we can construct a
deterministic finite state machine D such that N and D accept the same
language.
– [proof omitted]
Theorem:
– Every deterministic finite state machine can be regarded as a non–
deterministic finite state machine that just doesn’t use the extra non–
deterministic capabilities.
56
F
0
F
1
S0
S0,S2
S1
S1
S3
S4
S2
1
S
0
Start
0
S3
S4
S3
S3
{S0}
S
2
{S0,
S2}
0,1
S
4
{S1,
S4}
0
0,1
Ø
1
0
1
{S3}
0,1
0
1
0n |n≥0
0n 01 |n≥0
1
0
1
0
start
S
3
1
0
S4
S3
0
0
State
S
1
{S3,
S4}
0,1
{S1}
{S4}
0n 11 |n≥0
1
57