sequence detection : regular expressions conversion to finite state machines

8/6/2019 Sequence Detection : Regular Expressions conversion to Finite State Machines

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State1 State2a

a

Figure1: Representation of a+

since a+ indicates at least one occurrence of a

from a current state state1, the fsm has to transit

to another state and consume as many 'a's as possible

and on occurrence of any other symbol can traverse to

the next state


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State2

a

State3

aState1

a'

Figure2: Representation of a*

since a* indicates zero or one or more occurrence of a

the fsm will have a bypass path a' which is used when any

other symbol is consumed instead of a letter a. if there

are letters a being consumed they move through the

intermediate State2 as shown.


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State2 State3a

State1

a'

a'

Figure3: Representation of a?

since a? indicates zero or one occurrence of a

the fsm reaches its final state State3 either via the bypass

path which indicates a zero occurrence of a OR via

the State2 which shall only consume a single occurrence of a


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Problem statement : To convert regex 1+0(10)+11 into a statemachine

Solution Approach:

Let's denote the inner 10 as an X, so the regex is 1+0X+11 We split it into smaller parts of the FSM

First part as 1+ Followed by 0 Followed by X+ Followed by two consecutive 1s


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1. From the atomic regex representations mentioned in

previous slides, 1+ is represented as in Figure4

State1 State21

1

2. To consume a 0, the representation of the state diagram

would be as in Figure5

State30

Figure4: Representation of 1+

3. To consume a (X)+, the representation of the state

diagram would be in Figure6

Figure5: Representation of 0

StateX StateYX

Figure6: Representation of X+

X


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4. To consume two consecutive 1s, the representation of thestate diagram would be as in Figure7

State9 State101

Figure6: Representation of 11

1


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LEVEL 2 Combine the individual state machines graphs

(1. and 2. for now!)

State1 State21

State30

1

In case of binary stream sequence match, the only possible

values of the symbols are 0 and 1. Hence there can be only

two path in the deterministic finite automaton. A path for

consuming the 0 symbol and another for consuming the 1

symbol.

Any path that is not defined above, for instance the 0 path

from State1 leads to the initial state. So State1's 0 pathleads to itself.

State1 State21 State30

1

0

Graph1


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Expanding 3. Consumption of the symbol X where X is 10

StateX StateYX

StateX StateY10

10

StateX1 StateX21

StateX30

StateX4

0

Graph2

0

1

1


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Joining Graph1 and Graph2

State1 State21

State30

1

0

StateX1 StateX21

StateX30 0

Note the transition from State3 to StateX1 is on

an empty symbol which means reaching State3 is

same as reaching StateX1

StateX4

0

1

1


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Simplifying the Graph by removing the empty symbol

State1 State21 0

1

0

StateX1 StateX21

StateX30 0

StateX4

0

1

1

StateX1 s 0 path shall transit to State1 since an

occurrence of 0 breaks the path

StateX2 s 1 path shall transit to State2 since an

occurrence of 1 breaks the path but might be a

starting of another valid sequence of 1+

0 1


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Adding the last graph to this final graph we get the

final finite state machine

State1 State21 0

1

0

State

X1

State

X2

1 State

X3

0 0

State9

0

1

10 1

State10

Note that StateX4 has become State9, because StateX4

already detects a 1 following a 10, hence is

equivalent to the first state of the 4. and so can be

merged with State9


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State1

Out

sequence detection : regular expressions conversion to finite state machines

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