Copyright © 2014 - Curt Hill Finite State Automata Again This Time No Output

Copyright © 2014 - Curt Hill

Finite State Automata Again

This Time No Output


Recall…• We last considered FSA • Each has

• An input alphabet• A finite number of states• State transition function• Output function• Initial node

• They are often represented as– Graphs with nodes and arcs– State charts

• Now we are considering FSA without output

What is new?• Now our FSMs do not produce

output, so we do not need the Output function

• What we add instead is the requirement of a final node

• To be legal input we must start at an initial state and end at a final state

• Thus their output is based on final state– Legal or illegalCopyright © 2014 - Curt Hill

Why?• FSA with output are usually used to

model a simple machine and display intermediate results – These intermediate results may be all

we are interested in• The no output FSA are usually

language recognizers– Does the input string conform to

some predefined rule or not


Definition• An output-less finite state machine

is 5 tupleM=(S,I,f,s0,F)

• S is a set of states• I is the set of inputs• f is the transition function• s0 is the initial state, s0S• F is the set of final states, FS

Copyright © 2011-2014 - Curt Hill

Error States• There are two possible errors that

can occur with these• If the state does not allow for this

particular input this puts the machine in an error state– The book does not allow this one yet

• If the machine does not end in a designated final state this is also an error– There may be multiple final states


Exercise• Recall the FSM from previous

presentation that we saw– I will show again on next slide

• Lets assume that the initial and terminal state is s0

• Find three strings that this automaton accepts

• What is shortest string it may accept?

• What is the longest string it may accept? Copyright © 2011-2014 - Curt Hill

Exercise


s0s1

s2

a b

b

a

a

b

s0

s1

s2

ba

s0

s0

s1

s1 s2

s2

Which are not legal?


s0s1

s2

a b

b

a

a

b

s0

s1

s2

ba

s0

s0

s1

s1 s2

s2

• aaabb• bbaa• abaab• bababba• abababa• bbbbabab

Strings• These FSA are language

recognizers– That is, given an input string they

pronounce it legal or illegal based on the state on which they stopped

• I* is the set of all strings, whose characters are composed solely from the input alphabet, I

• The language they accept is a subset of the set of all possible strings, I*


Definitions• Suppose we have two sets of

strings A,B both of which are subsets of I*

• The concatenation of A and B is AB– This is the set of every element of A

concatenated with each element of B– Something like a Cartesian product

but we end up with strings rather than ordered pairs


Example• Suppose that A = {0, 1} and

B={a,b,c}• AB = {0a, 0b, 0c, 1a, 1b, 1c} • BA = {a0, a1, b0, b1, c0, c1}• AA = {00,01,10,11}• BB = {aa, ab, ac, ba, bb, bc, ca,

cb, cc}


Kleene Closure• The Kleene Closure is the set

consisting of all strings made up by concatenating of strings from a set of characters– If the set of characters is A then the

Kleene Closure is A*

• This closure is all possible strings from that set:


Example• Suppose that A={0,1}, B={a,b,c}• A* = {0,1,00,01,10,11,000,001,…}

– This is the set of all bit strings• B* = {a,b,c,aa,ab,ac,ba,bb,bc,…}

– This is the set of all strings composed of the three letters abc

• If C = {ab} what is C*?


Language Recognition• Each FSM with starting and final

states recognizes a language– A type 3 grammar to be exact

• If the machine ends on a final state the string is correct– If not the string is not in the language

• If this machine is denoted by M then the language is denoted by L(M)


Example

• Suppose that S0 is the start state and S1 and S3 are final states

• What language does this accept?• What productions?


S0 S1

0S2

1

1 0

S3

0,1

0,1

Other Way• Can we construct a DFM that

recognizes all bit strings where there are not two successive ones?


Result• What we want is a DFM with node

that is obtained with a one but only accepts zeros from there

• Start is S0 and final is S1 and S2


S0

S11

S2

0

1

0

S30,1

1

0

Determinism• All the FSMs we have looked at are

deterministic– For each pair of state and input there

is a unique new state• The alternative is a non-

deterministic finite state automaton– For any state-input pair there may be

zero or more states• A NDFA accepts a string if any

choice of transitions accepts the string Copyright © 2014 - Curt Hill

Example

• State S0 is the start state and S1 is the final state– S0 has no two 1 transitions

– S1 has no transitions

– S2 has two 1 transitions

– S3 has no 1 transitionCopyright © 2014 - Curt Hill

S0 S1

0,1S2

0

1

1

S3

1

0

Commentary• The understanding problem is the

states that have multiple transitions or no transitions

• A NDFA can be viewed in several ways:– If any of the alternatives work the

machine accepts the language– Multiple alternatives are pursued

concurrently– Failed alternatives do not matter as

long as one worksCopyright © 2014 - Curt Hill

NDFA• Non-Deterministic Finite Automata

are useful in variety of theoretical contexts

• Often easier to define a NDFA for a language

• Many proofs are easier with NDFA• Closures are also easier to

compute


Example

• Suppose that S0 is the start state and S1 is the final state

• What language does this accept?• What productions?• What errors are possible


S0 S1

1S2

0

0,1

Equivalence• Although NDFA have more

flexibility than DFA the two have equivalent strength in regards to language recognition

• In particular for any NDFA and equivalent DFA may be constructed

• How is this done?– It is called the Powerset or subset

constructionCopyright © 2014 - Curt Hill

Construction• Suppose we have a NDFA, N • Converting it to a DFA, D, will have

several characteristics• The worst case is that if the N has

n nodes then D will have 2n nodes• Each node in D will be labeled by a

subset of the nodes from N


Building the States• Add {s0} to SD – this is the start node• While there are still states to be defined

– Choose s in SD – For each a in the alphabet determine the set

of states that are reachable from q on input a• If there is no state in SD representing the above

add it• Define a new state corresponding to this

• Define fD as any state in SD that corresponds to a subset containing any final state in N


Easy Example

• There are two states that have incomplete specification, S0 and S1

• All we need to do is generate an error state that captures


S0 S1

1S2

0

0,1

Easy Example


S0 S1

1S2

0

0,1

S0 S1

1S2

0

0,1

Becomes:

Serr

0 1

0,1

Does not matter if we have 1 or 2 error states

Harder


S0 S1

0,1S2

0

1

1

• No missing transitions, but two 1’s coming from S0

• State S0 is start, S1 is final

Harder


S0 S1

0,1S2

0

1

1

S0 S1

0

S2

1

1

S3

Serr

0,1

0,1

S3

1

0

0 0,1

Six states and three final states.S1 and S3 could have been merged

Exercises• 13.3

– 1, 5, 9, 13, 17, 21, 45


Documents

Copyright © 2014 - Curt Hill Finite State Automata Again This Time No Output