Finite Automata

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Finite Automata

• A simple computing model

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Outline (this and next lecture)

• Introduction
• Deterministic finite automata (DFAs)
• Non-deterministic finite automata (NFAs)
• NFAs to DFAs
• Simplifying DFAs
• Regular expressions ? finite automata
• Only attempt up to question 5 of tutorial 4, the
  rest will be covered next lecture

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An automatic one way door
Consider the control system for a one-way
swinging door There are two states Open and
Closed It has two inputs, person detected at
position A and person detected at position B If
the door is closed, it should open only if a
person is detected at A but not B Door should
close only if no one is detected
A
B
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Control schematic
Open
Closed
A, no B
No A or B
A and B B, no A No A or B
A and B A, no B B, no A
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This is a finite automaton!

• A finite automaton is usually represented like
  this as a directed graph
• Two parts of a directed graph
• The states (also called nodes or vertices)
• The edges with arrows which represent the
  allowed transitions
• One state is usually picked out as the starting
  point
• For so-called accepting automata, some states
  are chosen to be final states

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Strings and automata
The input data are represented by a string over
some alphabet and it determines how the machine
progresses from state to state. Beginning in the
start state, the characters of the input string
cause the machine to change from one state to
another. Accepting automata give only yes or no
answers, depending on whether they end up in a
final state. Strings which end in a final
state are accepted by the automaton.
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The labeled graph in the figure above represents
a FA over the alphabet A a, b with start
state 0 and final state 3. Final states are
denoted by a double circle.
Example
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Deterministic finite automata (DFAs)

• The previous graph was an example of a
  deterministic finite automaton every node had
  two edges (a and b) coming out
• A DFA over a finite alphabet A is a finite
  directed graph with the property that each node
  emits one labeled edge for each distinct element
  of A.

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More formally

• A DFA accepts a string w in A if there is a path
  from the start state to some final state such
  that w is the concatenation of the labels on the
  edges of the path.
• Otherwise, the DFA rejects w .
• The set of all strings accepted by a DFA M is
  called the language of M and is denoted by
• L(M)

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EXAMPLE (ab)

• Construct a DFA to recognize the regular
  languages represented by the regular expression
  (a b) over alphabet A a, b.
• This is the set a, b of all strings over a,
  b. This can be recognised by

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EXAMPLE a(ab)

• Find a DFA to recognize the language represented
  by the regular expression a(a b) over the
  alphabet A a, b.
• This is the set of all strings in A which begin
  with a. One possible DFA is

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EXAMPLE Pattern recognition

• Build a DFA to recognize the regular language
  represented by the regular expression (a
  b)abb over the alphabet A a, b.
• The language is the set of strings that begin
  with anything, but must end with the string abb.
• Effectively, were looking for strings which have
  a particular pattern to them

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Solution - (a b)abb
The diagram below shows a DFA to recognize this
language.
If in state 1 the last character was a If in
state 2 the last two symbols were ab If in
state 3 the last three were abb
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State transition function
We can also represent a DFA by a state
transition function, which we'll denote by T,
where any state transition of the form is
represented by T(i,a) j To describe a full
DFA we need to know what states there are,
which are the start and final ones, the set
of transitions between them.
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Regular languages

• The class of regular languages is exactly the
  same as the class of languages accepted by DFAs!
• Kleene (1956)
• For any regular language, we can find a DFA which
  recognizes it!

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Applications of DFAs

• DFAs are very often used for pattern matching,
  e.g. searching for words/structures in strings
• This is used often in UNIX, particularly by the
  grep command, which searches for combinations of
  strings and wildcards (, ?)
• grep stands for Global (search for) Regular
  Expressions Parser
• DFAs are also used to design and check simple
  circuits, verifying protocols, etc.
• They are of use whenever significant memory is
  not required

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Non-deterministic Finite Automata

• DFAs are called deterministic because following
  any input string, we know exactly which state its
  in and the path it took to get there
• For NFAs, sometimes there is more than one
  direction we can go with the same input character
  
• Non-determinism can occur, because following a
  particular string, one could be in many possible
  states, or taken different paths to end at the
  same state!

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NFAs

• A non-deterministic finite automaton (NFA) over
  an alphabet A is a finite directed graph with
  each node having zero or more edges,
• Each edge is labelled either with a letter from
  A or with ?.
• Multiple edges may be emitted from the same node
  with the same label.
• Some letters may not have an edge associated
  with them. Strings following such paths are not
  recognised.

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Non-determinism

• If an edge is labelled with the empty string ?,
  then we can travel the edge without consuming an
  input letter. Effectively we could be in either
  state, and so the possible paths could branch.
• If there are two edges with the same label, we
  can take either path.
• NFAs recognise a string if any one of its many
  possible states following it is a final state
• Otherwise, it rejects it.

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NFAs versus DFAs
NFA for aa
DFA for aa
Why is the top an NFA while the bottom is a DFA?
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EXAMPLE

• Draw two NFAs to recognize the language of the
  regular expression ab aa.
• This NFA has a ? edge, which allows us to travel
  to state 2 without consuming an input letter.
• The upper path corresponds to ab and the lower
  one to aa

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An equivalent NFA
This NFA also recognizes the same language.
Perhaps it's easier to see this by considering
the equality ab aa ab aa.
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NFA transition functions

• Since there may be nondeterminism, we'll let the
  values of this function be sets of states.
• For example, if there are no edges from state k
  labelled with a, we'll write
• T(k, a) ?.
• If there are three edges from state k, all
  labelled with a, going to states i, j and k,
  we'll write 
• T(k, a) i, j, k.

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Comments on non-determinism

• All digital computers are deterministic quantum
  computers may be another story!
• The usual mechanism for deterministic computers
  is to try one particular path and to backtrack to
  the last decision point if that path proves poor
  
• Parallel computers make non-determinism almost
  realizable. We can let each process make a random
  choice at each branch point, thereby exploring
  many possible trees.

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Perhaps surprisingly

• The class of regular languages is exactly the
  same as the class of languages accepted by NFAs!
• Rabin and Scott (1959)
• Just like for DFAs!
• Every NFA has an equivalent DFA which recognises
  the same language.

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NFAs to DFAs

• We prove the equivalence of NFAs and DFAs by
  showing how, for any NFA, to construct a DFA
  which recognises the same language
• Generally the DFA will have more possible states
  than the NFA. If the NFA has n states, then the
  DFA could have as many as 2n states!
• Example NFA has three states A, B, C
• the DFA could have eight ?, A, B, C,
• A, B, A, C, B, C, A, B, C
• These correspond to the possible states the NFA
  could be in after any string

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DFA construction

• Begin in the NFA start state, which could be a
  multiple state if its connected to any by ?
• Determine the set of possible NFA states you
  could be in after receiving each character. Each
  set is a new DFA state, and is connected to the
  start by that character.
• Repeat for each new DFA state, exploring the
  possible results for each character until the
  system is closed
• DFA final states are any that contain a NFA
  final state

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Example (a b) ab
C
A
B
b
a
NFA
a,b

• The start state is A, but following an a you
  could be in A or B following a b you could only
  be in state A

A
A,B
A,C
b
a
DFA
b
a
a
b
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Summary

• Regular expressions represent the regular
  languages.
• DFAs recognize the regular languages.
• NFAs also recognize the regular languages.

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Finite automata

• So far, weve introduced two kinds of automata
  deterministic and non-deterministic.
• Weve shown that we can find a DFA to recognise
  anything language that a given NFA recognises.
• Weve asserted that both DFAs and NFAs
  recognise the regular languages, which themselves
  are represented by regular expressions
• We prove this by construction, by showing how any
  regular expression can be made into a NFA and
  vice versa

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Regular Expressions ? Finite Automata

• Given a regular expression, we can find an
  automata which recognises its language
• Start the algorithm with a machine that has a
  start state, a single final state, and an edge
  labelled with the given regular expression as
  follows

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Four step algorithm

• 1. If an edge is labelled with ?, then erase the
  edge.
• 2. Transform any diagram like

into the diagram
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• 3. Transform any diagram like

into the diagram
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• 4. Transform any diagram like

into the diagram
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Example a ab

• Construct a NFA for the regular expression,
  a ab
• Start with
• Apply rule 2

a ab
a
ab
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Example a ab
a
Apply rule 4 to a
?
?
ab
a
Apply rule 3 to ab
?
?
a
b
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Finite Automaton ? Regular Expression

• 1.Create a new start state s, and draw a new edge
  labelled with ? from s to the original start
  state.
• 2. Create a new final state f, and draw new edges
  labelled with ? from all the original final
  states to f

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• 3. For each pair of states i and j that have more
  than one edge from i to j, replace all the edges
  from i to j by a single edge labelled with the
  regular expression formed by the sum of the
  labels on each of the edges from i to j.
• 4. Construct a sequence of new machines by
  eliminating one state at a time until the only
  states remaining are s and the f.

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Eliminating states

• As each state is eliminated, a new machine is
  constructed from the previous machine as follows
• Let old(i,j) denote the label on edge ?i,j? of
  the current machine. If no edge exists, label it
  ?.
• Assume that we wish to eliminate state k. For
  each pair of edges ?i,k? (incoming edge) and
  ?k,j? (outgoing edge) we create a new edge label
  new(i, j)

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Eliminate state k

• The label of this new edge is given by
• new(i,j) old(i,j) old(i, k) old(k, k)
  old(k,j)
• All other edges, not involving state k, remain
  the same
• new(i, j) old(i, j).
• After eliminating all states except s and f, we
  wind up with a two-state machine with the single
  edge ?s, f? labelled with the desired regular
  expression new(s, f)

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Example
Initial DFA
Steps 1 and 2 Add start and final states
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Eliminate state 2 (No path to f)
Eliminate state 0
Eliminate state 1
Final regular expression
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Finding simpler automata

• Sometimes our constructions lead to more
  complicated automata than we need, having more
  states than are really necessary
• Next, we look for ways of making DFAs with a
  minimum number of states
• Myhill-Nerode theorem
• Every regular expression has a unique minimum
  state DFA
• up to a simple renaming of the states

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Finding minimum state DFA

• Two steps to minimizing DFA
• 1) Discover which, if any, pairs of states are
  indistinguishable. Two states, s and t, are
  equivalent if for all possible strings w , T(s,w)
  and T(t,w) are both either final or non-final.
• 2) Combine all equivalent states into a single
  state, modifying the transition functions
  appropriately.

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Consider the DFA
b
a,b
a
1
a
b
a
2
b

• States 1 and 2 are indistinguishable! Starting in
  either, b is rejected and anything with a in it
  is accepted.

b
a,b
a
a,b
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Part 1 finding indistinguishable pairs

• Remove all inaccessible states, where no path
  exists to them from start.
• Construct a grid of pairs of states.
• Begin by marking those pairs which are clearly
  distinguishable, where one is final and the other
  non-final.
• Next eliminate all pairs, which on the same
  input, lead to a distinguishable pair of states.
  Repeat until you have considered all pairs.
• The remaining pairs are indistinguishable.

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Part 2 construct minimum DFA

• Construct a new DFA where any pairs of
  indistinguishable states form a single state in
  the new DFA
• The start state will be the state containing the
  original start state.
• The final states will be those which contain
  original final states
• The transitions will be the full set of
  transitions from the original states (these
  should all be consistent.)

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Example
1
a
b
b
b
a,b
a
0
2
4
b
a
a
3
What are the distinguishable pairs of
states? Clearly, 0, 4 1, 4 2, 4 3, 4 are
all distinguishable because 4 is final but none
of the others are.
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Grid of pairs of states

• We eliminate these as possible indistinguishable
  pairs
• Next consider 0, 1. With input a, this becomes
  3, 4 which is distinguishable, so 0, 1 is as
  well.
• Similarly, we can show 0, 2 and 0, 3 are also
  distinguishable, leading to the modified grid

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Remaining pairs

• We are left with
• 1, 2 given a 4, 4
• given b 2, 1
• 2, 3 given a 4, 4
• given b 1, 2
• 1, 3 given a 4, 4
• given b 2, 2
• These do not lead to pairs we know to be
  distinguishable, and are therefore
  indistinguishable!

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Construct minimal DFA

• States 1, 2, and 3 are all indistinguishable,
  thus the minimal DFA will have three states
• 0 1, 2, 3 4
• Since originally T(0, a) 3 and T(0, b) 1, the
  new transitions are T(0, a) T(0, b) 1,2,3
• Similarly,
• T(1,2,3, a) 4 and T(1,2,3, b) 1,2,3
• Finally, as before,
• T(4, a) 4 and T(4, b) 4

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Resulting minimal DFA

• The resulting DFA is much simpler

b
a,b
a
a,b
0
1, 2, 3
4
This recognises regular expressions of the form,
(a b) b a (a b) This is the simplest DFA
which will recognise this language!
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Conclusions

• We now have many equivalent ways of representing
  regular languages DFAs, NFAs, regular
  expressions and regular grammars
• We can also now simply(?!) move between these
  various representations.
• Well see next lecture that the automata
  representation leads to a simple way of
  recognising some languages which are not regular.
  
• Well also begin to consider more powerful
  language types and correspondingly more powerful
  computing models!