zvi kohavi and niraj k. jha 1 finite-state recognizers
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Zvi Kohavi and Niraj K. Jha
Finite-state RecognizersFinite-state Recognizers
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Deterministic RecognizersDeterministic Recognizers
Treat FSM as a recognizer that classifies input strings into two classes:
strings it accepts and strings it rejects
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
1 01
Finitecontrol
0
Head
Tape
0 0 11
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Transition GraphTransition GraphExample: 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
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
A
C B
(a) Deterministic state diagram.
01
01
0,1
A
B
1 0
1
(b) Transition graph.
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ExampleExample
Example: 1110 and 11011 accepted by the transition graph below, but 100 rejected
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
A D
B0
1 0
1
11
C0
A B C0 0
0
1
A B C0 0
0
1
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Graphs Containing -TransitionsGraphs 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
A
0
1
01
0
C1
E F G1
B D
A
0
1
01
0
C1
E F G1
(a) A graph containing a -transition.
(b) An equivalent graph without -transitions.
0
1
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Converting Nondeterministic into Converting Nondeterministic into Deterministic GraphsDeterministic Graphs
Example: Transition graph and its transition table
• Successor table and deterministic graph:
B
A C
(a) Transition graph.
0
101
0,1
0 1
A
B
C
C
AB
AC
A
(b) Transition table.
(a) Successor table.
0 1
AB
C
C
AB
AC
A
AB C
AC
0
0
0
1
1
A
(b) State diagram of an equivalentdeterministic machine.
AC
ABC
ABC A
C
ACABC
A
ABC
0
1
1
0 0,1
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TheoremTheorem
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
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Regular ExpressionsRegular 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*
0 11
(a)
1
1 0
0
(c)
1
1
1
1 0
0
(b)
(d)
1
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Regular Expressions (Contd.)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
C
1
B
A
0
1
1
A1
D E
(a) (01 + 10)*11.
0
1
(b) (10*)*.
0
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Definition and Basic PropertiesDefinition 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 :
BA
(a) A graph accepting . (b) A graph accepting .
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IdentitiesIdentities
+ 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)*
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Manipulating Regular ExpressionsManipulating 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: thenR + 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)*
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ExamplesExamples
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)*
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Transition Graphs Recognizing Regular Transition Graphs Recognizing Regular SetsSets
Theorem: Every regular expression R can be recognized by a transition graph
Proof:(a) R = . (b) R = . (c) R = .i
i
(a) Graphs recognizing P and Q.
G
H
(c) A graph recognizing PQ.
G H
(b) A graph recognizing P+Q.
G
H
G
(d) A graph recognizing P*.
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ExampleExample
Example: Construct a transition graph recognizing R = (0 + 1(01)*)*
0
P
(a) R = P*; P = 0 + 1(01)*.
A B C
(b) P = 0 + Q; Q = 1(01)*.
A B C
Q
0
(c) Q = 1T; T = (01)*.
A B C
T
D
1
0
(d) Final step.
A B C
D E
1
1
0
F
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Example (Contd.)Example (Contd.)
Example: Prove that (P + Q)* = P*(QP*)*
P
(a) Graph recognizing P*(QP*)*.
Q
(b) Equivalent graph with no-transitions.
P
P
Q
P,Q
PP
Q
P,Q
(a) Equivalent deterministicgraph recognizing (P + Q)*.
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Informal TechniquesInformal Techniques
Example: Construct a graph that recognizes P = (01 + (11 + 0)1*0)*11
Graph for Q = (11 + 0)1*0
Graph for P
A
0
B 1
C
D
11
0
A0
B 1
C
F
11
0
D
1
E
0
1
1
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ExampleExample
Example: Construct a graph that recognizes R = (1(00)*1 + 01*0)*
E
B0
0
1 D
1
F
C
(a) Partial graph.
0
1
A
0
E
B0
0
1 D
1
F
C
(b) Complete graph.
0
1
A
0
0
11
0
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Regular Sets Corresponding to Transition Regular Sets Corresponding to Transition GraphsGraphs
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
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Systems of EquationsSystems of Equations
Example: Derive the set of strings derived by the following transition graph
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) (6)
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
A B C
0
0 01
0
1
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TheoremTheorem
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
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ExamplesExamples
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)
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ExampleExample
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
xzReplace A Bx/z
A B with
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Example (Contd.)Example (Contd.)
Example (contd.): Derive SN for machine N shown below
A B0/1
1/0
1/1
0/0
A B
F
1 0
1
0
D
0
C
1
0
E
1
A
1
CE
(a) Transition graph.
DF
B
1
1
1
000
0
(b) Equivalent deterministic form.
0,1
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Two-way RecognizersTwo-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
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ExampleExample
Example: A two-way machine recognizing set 100*
1
A
0c
A
B
A
B
1
A
c
A B D D D D
(a) A loop. (b) Rejection of a tape.
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Convenience of Using Two-way MachinesConvenience 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
01
A
c
A B B B C C
(a) Rejecting a tape. (b) Accepting a tape.
0 00 1
C
01
A
c
A B B C
DD
0 01 1
DD
E E F F F G G