Finite-state Recognizers
1
Zvi Kohavi and Niraj K. Jha
Deterministic Recognizers
Treat FSM as a recognizer that classifies input strings into two classes:
strings it accepts and strings it rejects
Tape
1
0
0
1
0
0
1
1
Head
Finite
control
Finite-state recognizer:
• Equivalent to a string of input symbols that enter the machine at
successive times
• Finite-state control: Moore FSM
• States in which output symbol is 1 (0): accepting (rejecting) states
• A string is accepted by an FSM: if and only if the state the FSM enters
after having read the rightmost symbol is an accepting state
• Set of strings recognized by an FSM: all input strings that take the FSM
from its starting state to an accepting state
2
Transition Graph
Example: a machine that accepts a string if and only if the string begins
and ends with a 1, and every 0 in the string is preceded and
followed by at least a single 1
1
A
1
A
1
0
0
0
0,1
1
C
B
(a) Deterministic state diagram.
B
(b) Transition graph.
Transition graph: consists of a set of vertices and various directed arcs
connecting them
•
•
•
•
At least one of the vertices is specified as a starting vertex
Arcs are labeled with symbols from the input alphabet
A vertex may have one or more Ii-successors or none
It accepts a string if the string is described by at least one path emanating
from a starting vertex and terminating at an accepting vertex
• It may be deterministic or non-deterministic
3
Example
Example: 1110 and 11011 accepted by the transition graph below, but 100
rejected
A
D
0
1
1
1
1
0
B
0
C
Equivalent transition graphs: two or more graphs that recognize the same
set of strings
• Each graph below accepts a string: if and only if each 1 is preceded by at
least two 0’s
0
0
0
A
1
B
0
C
0
A
B
0
C
1
4
Graphs Containing  -Transitions
 -transitions: when no input symbol is used to make the transition
Example: Graph that recognizes a set of strings that start with an even
number of 1’s, followed by an even number of 0’s, and end with
substring 101
B
D
0
1
1
0
1
C
A
E
0
F
1
G
(a) A graph containing a -transition.
B
D
1
0
0
1
A
0
C
1
E
0
F
1
G
1
(b) An equivalent graph without -transitions.
5
Converting Nondeterministic into
Deterministic Graphs
Example: Transition graph and its transition table
0,1
A
0
1
0
C
0
C
1
A
B
AC
C AB A
1
B
(a) Transition graph.
•
(b) Transition table.
Successor table and deterministic graph:
0
AB
AB
0
1
C
AC
C
0
1
1
C
AB
A
AC
ABC
A
A
C
0
1
AC
A
1
0
ABC
ABC
AC
ABC
0
(a) Successor table.
0,1
(b) State diagram of an equivalent
deterministic machine.
6
Theorem
Theorem: Let S be a set of strings that can be recognized by a
nondeterministic transition graph Gn. Then S can also be recognized by
an equivalent deterministic graph Gd. Moreover, if Gn has p vertices,
Gd will have at most 2p vertices
7
Regular Expressions
Example: Sets of strings and the corresponding expression
•
•
•
•
Graph (a) recognizes set {101}: expression denoted as 101
Graph (b) recognizes set {01,10}: expression = 01 + 10
Graph (c) recognizes {0111,1011}: expression = 0111 + 1011
– Concatenation of 01 + 10 and 11
Graph (d) recognizes set {  ,1,11,111,1111,…}: expression = 1*
1
0
1
(a)
0
1
1
0
1
0
1
0
(b)
1
1
1
(d)
(c)
8
Regular Expressions (Contd.)
Example: 01(01)* = 01 + 0101 + 010101 + 01010101 + …
• R* =  + R + R2 + R3 + …
Example: Set of strings on {0,1} beginning with a 0 and followed only by 1’s:
01*
Example: Set of strings on {0,1} containing exactly two 1’s: 0*10*10*
Example: Set of all strings on {0,1}:
(0+1)* =  + 0 + 1 + 00 + 01 + 10 + 11 + 000 + …
Example: Set of strings on {0,1} that begin with substring 11: 11(0+1)*
Example: Transition graphs and the sets of strings they recognize
B
1
0
0
1
A
D
1
1
E
A
B
1
0
C
9
(a) (01 + 10)*11.
(b) (10*)*.
Definition and Basic Properties
Let A = {a1,a2,…,ap} be a finite alphabet: then the class of regular
expressions over alphabet A is defined recursively as follows:
• Any symbol, a1, a2, …, ap alone is a regular expression: as are null string 
and empty set 
• If P and Q are regular expressions: then so is their concatenation PQ and
their union P+Q
– If P is a regular expression: then so is its closure P*
• No other expressions are regular: unless they can be generated in a finite
number of applications of the above rules
Recognizers for  and :
A
(a) A graph accepting .
B
(b) A graph accepting
.
10
Identities
+R=R
 R = R = 
R = R = R
* = 
* = 
Set of strings that can be described by a regular expression: regular set
• Not every set of strings is regular
• Set over {0,1}, which consists of k 0’s (for all k), followed by a 1, followed
in turn by k 0’s, is not regular: 010 + 00100 + 0001000 + … + 0k10k + …
– Requires an infinite number of applications of the union operation
• However, certain infinite sums are regular
– Set consisting of alternating 0’s and 1’s, starting and ending with a 1:
1(01)*
11
Manipulating Regular Expressions
A regular set may be described by more than one regular expression
•
Such expressions are called equivalent
Example: Alternating 0’s and 1’s, starting and ending with 1
•
1(01)* or (10)*1
Let P, Q, and R be regular expressions: then
R+R=R
PQ + PR = P(Q+R); PQ + RQ = (P + R)Q
R*R* = R*
RR* = R*R
(R*)* = R*
 + RR* = R*
(PQ)*P = P(QP)*
(P + Q)* = (P*Q*)* = (P* + Q*)* = P*(QP*)* = (P*Q)*P*
 + (P + Q)*Q = (P*Q)*
12
Examples
Example: Prove that the set of strings in which every 0 is immediately
followed by at least two 1’s can be described by both R1 and R2,
where
R1 =  + 1*(011)*(1*(011)*)*
R2 = (1 + 011)*
Proof: R1 =  + 1*(011)*(1*(011)*)*
= (1*(011)*)*
= (1 + 011)* = R2
Example: Prove the identity
(1 + 00*1) + (1 + 00*1)(0 +10*1)*(0 + 10*1) = 0*1(0 + 10*1)*
Proof: LHS = (1 + 00*1)[+ (0 + 10*1)*(0 + 10*1)]
= [(+ 00*)1][+ (0 + 10*1)*(0 + 10*1)]
= 0*1(0 + 10*1)*
13
Transition Graphs Recognizing Regular
Sets
Theorem: Every regular expression R can be recognized by a transition
graph
i
Proof:
(a) R = .
(b) R = .
(c) R =
G
G
H
H
(a) Graphs recognizing P and Q.
(b) A graph recognizing P+Q.
G
i.
H
(c) A graph recognizing PQ.
G
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(d) A graph recognizing P*.
Example
Example: Construct a transition graph recognizing R = (0 + 1(01)*)*
0
P
A
A
B
B
C
T
1
C
D
(a) R = P*; P = 0 + 1(01)*.
(c) Q = 1T; T = (01)*.
0
A
B
0
C
A
B
C
1
Q
(b) P = 0 + Q; Q = 1(01)*.
1
D
E
(d) Final step.
F
0
15
Example (Contd.)
Example: Prove that (P + Q)* = P*(QP*)*
P,Q
P
Q
Q
Q
P
P
(a) Graph recognizing P*(QP*)*.
P
P
(b) Equivalent graph with no
-transitions.
P,Q
(a) Equivalent deterministic
graph recognizing (P + Q)*.
16
Informal Techniques
Example: Construct a graph that recognizes P = (01 + (11 + 0)1*0)*11
1
Graph for Q = (11 + 0)1*0
B
1
1
A
C
0
0
D
Graph for P
0
D
1
A
1
1
B
1
1
0
C
0
E
1
F
17
Example
Example: Construct a graph that recognizes R = (1(00)*1 + 01*0)*
1
1
0
0
B
0
C
B
0
C
0
A
0
A
1
1
F
D
1
F
D
1
0
1
1
0
0
E
(a) Partial graph.
0
E
(b) Complete graph.
18
Regular Sets Corresponding to Transition
Graphs
The set of strings that can be recognized by a transition graph (hence, an
FSM) is a regular set
Theorem: Let Q, P, and R be regular expressions on a finite alphabet.
Then, if P does not contain  :
• Equation R = Q + RP has a unique solution given by R = QP*
• Equation R = Q + PR has a unique solution given by R = P*Q
19
Systems of Equations
Example: Derive the set of strings derived by the following transition graph
0
1
A
1
0
C
B
A = + A0 + B1
(1)
B = A0 + B1 + C0 (2)
C = B0
(3)
Substituting (3) into (2):
B = A0 + B1 + B00 = A0 + B(1 + 00)
(4)
From the theorem:
B = A0(1 + 00)*
(5)
Substituting (5) into (1):
A = + A0 + A0(1 + 00)*1 =  + A(0 + 0(1 + 00)*1)
From the theorem:
A =  (0 + 0(1 + 00)*1)* = (0 + 0(1 + 00)*1)* (7)
Hence, solution C from (7), (5) and (3):
C = (0 + 0(1 + 00)*1)*0(1 + 00)*0
0
0
(6)
20
Theorem
Theorem: The set of strings that take an FSM M from an arbitrary state Si
to another state Sj is a regular set
• Combining the two theorems:
– An FSM recognizes a set of strings if and only if it is a regular set
Applications: the correspondence between regular sets and FSMs enables
us to determine whether certain sets are regular
Example: Let R denote a regular set on alphabet A that can be recognized
by machine M1
• Complement R’: set containing all the strings on A that are not contained
in R
• R’ describes a regular set: since it can be recognized by a machine M2,
which is obtained from M1 by complementing the output values associated
with the states of M1
21
Examples
Example: Let P&Q represent the intersection of sets P and Q
•
•
Prove P&Q is regular
Since P’ and Q’ are regular:
– P’ + Q’ is regular
– Hence, (P’ + Q’)’ is regular
– Since P&Q = (P’ + Q’)’: P&Q is regular
Regular expressions containing complementation, intersection, union,
concatenation, closure: extended regular expressions
Example: Consider the set of strings on {0,1} s.t. no string in the set
contains three consecutive 0’s
•
•
Set can be described by: [(0 + 1)*000(0 + 1)*]’
More complicated expression if complementation not used:
(1 + 01 + 001)*( + 0 + 00)
22
Example
Example: Let M be an FSM whose input/output alphabet is {0,1}. Assume
the machine has a designated starting state. Let z1z2…zn denote
the output sequence produced by M in response to input sequence
x1x2…xn. Define a set SM, which consists of all the strings w s.t.
w = z1x1z2x2…znxn for any x1x2…xn in (0 + 1)*. Prove that SM is
regular.
•
•
•
Given the state diagram of M: replace each directed arc with two directed
arcs and a new state, as shown in the figure
Retain the original starting state: designate all the original states as
accepting states
The resulting nondeterministic graph recognizes SM: thus SM is regular
Replace
A
x/z
B
with
A
z
x
B
23
Example (Contd.)
Example (contd.): Derive SN for machine N shown below
1/0
0/1
A
0/0
B
1/1
1
E
A
0
1
1
1
C
DF
0
A
B
0
D
1
0
0
0,1
0
1
0
1
F
(a) Transition graph.
0
CE
1
B
(b) Equivalent deterministic form.
24
Two-way Recognizers
Two-way recognizer (or two-way machine): consists of a finite-state control
coupled through a head to a tape
•
•
•
•
•
•
•
Initially: the finite-state control is in its designated starting state, with its
head scanning the leftmost square of the tape
The machine then proceeds to read the symbols of the tape: one at a time
In each cycle of computation: the machine examines the symbol currently
scanned by the head, shifts the head one square to the right or left, and
then enters a new (not necessarily distinct) state
If the machine eventually moves off the tape on the right end entering an
accepting state: the tape is accepted by the machine
A machine can reject a tape: either by moving off its right end while
entering a rejecting state or by looping within the tape
Null string  can be represented either by: the absence of an input tape or
by a completely blank tape
A machine accepts  if and only if: its starting state is an accepting state
25
Example
Example: A two-way machine recognizing set 100*
c
0
c
1
1
A
A
A
A
B
D
D
D
D
B
A
B
(a) A loop.
(b) Rejection of a tape.
26
Convenience of Using Two-way Machines
Two-way machines are as powerful as one-way machines w.r.t the class of
tapes they can recognize
• However, for some computations: it is convenient to use two-way
machines since they may require fewer states
Example: Consider the two-way machine shown in the table, which
accepts a tape if and only if it contains at least three 1’s and at
least two 0’s
• The minimal one-way machine that is equivalent to the two-way machine
has 12 states: since it must examine the tapes for the appropriate number
of 0’s and 1’s simultaneously
c
1
0
0
1
0
0
A
A
B
B
B
C
C
(a) Rejecting a tape.
C
c
1
0
1
1
A
A
B
B
C
D
D
D
D
E
E
F
F
0
0
F
G
(b) Accepting a tape.
27
G