finite state automata and determinisationncss.edu.au/ckeditor_assets/attachments/26/fsa.pdf5...

Finite State Automata and Determinisation

Tim Dawborn

January, 2017

Languages fsa nfa re dfa Determinisation 2

Outline

1 Languages

2 Finite State Automata (fsa)

3 Non-deterministic Finite State Automata (nfa)

4 Regular Expressions

5 Deterministic Finite State Automata (dfa)

6 Determinisation

Tim Dawborn Finite State Automata and Determinisation January, 2017


Alphabets

• An alphabet Σ is a set of tokens used by a language• Σ = {0, 1} is the alphabet for binary• Σ = {., 0, 1, . . . , 9} is the alphabet for a decimal number• Σ = {marco, polo} is the alphabet for the communications

protocol in the game Marco Polo1

• What language(s) is this the alphabet for?Σ = {x, 0, 1, . . . , 9, A, B, . . . , F}

1A game involving calling out “Marco!” and “Polo!”,with an optional swimming pool



Strings

• A combination2 of elements from Σ is called a string• “0”, “101101”, “111” are all strings of Σ = {0, 1}

• The set Σ∗ is the infinite set of all combinations of theelements of Σ

• For the binary alphabet, Σ = {0, 1}, Σ∗ is3

Σ∗ = {∅, (0), (1), (00), (01), (10), (11), (000), (001), . . .}

2Obviously, a contiguous ordered tuple of elements, not a set:“01” is a different string from “10”

3Here the empty set will do to represent a null or empty string



Languages

• A language L is a set of items from Σ∗ which are deemed tobe valid (L ⊆ Σ∗)

• For hexadecimal, only the strings which begin with “0x” andhave at least one more non-“x” token are deemed valid4 “0xF9”8 “1234”4 “0x0”8 “0x”8 “1xA”8 “” (the empty string)

• A language in this context is not a set of rules, syntax,grammar and all: it’s just a set of valid strings



Finite State Automata

• Finite State Automata are representations of a self-containedfinite set of states, with rules that govern movement amongthose states.

• Finite State Automata (fsas) exist everywhere

• They are used to illustrate• the possible states some process can be in• how the process moves between these states




• E.g. a state machine for PIN code entry on your mobile4:

waiting

waiting0-9

cancel

waiting

0-9

cancel

waiting

0-9cancel

check password!

0-9

cancel

4How many people do you know who say “PIN Number”?




waiting

waiting0-9

cancel

waiting

0-9

cancel

waiting

0-9cancel

check password!

0-9

cancel

• fsas have nodes

• fsas have edges

• fsa edges (transitions) have labels

• fsas have a start node (entry point)

• fsas have accepting nodes, denoted with the double outline




waiting

waiting0-9

cancel

waiting

0-9

cancel

waiting

0-9cancel

check password!

0-9

cancel

• You are always “on” one of the nodes, i.e., in one of thestates.

• You can travel between the nodes following the directed edges

• You can only travel along an edge if you see the label that ison the edge



Pattern Recognition using fsas

q0 q3

0-9,A-F

q10 q2x 0-9,A-F

Walk the following strings through the fsa. Are they accepted?String Path Accepts?“0xF9”

q0 → q1 → q2 → q3 → q3 4

“1234” q0 8

“0x0” q0 → q1 → q2 → q3 4

“” q0 8




q0 q3

0-9,A-F

q10 q2x 0-9,A-F

Walk the following strings through the fsa. Are they accepted?String Path Accepts?“0xF9” q0 → q1 → q2 → q3 → q3

4

“1234” q0 8

“0x0” q0 → q1 → q2 → q3 4

“” q0 8




q0 q3

0-9,A-F

q10 q2x 0-9,A-F

Walk the following strings through the fsa. Are they accepted?String Path Accepts?“0xF9” q0 → q1 → q2 → q3 → q3 4

“1234” q0 8

“0x0” q0 → q1 → q2 → q3 4

“” q0 8




q0 q3

0-9,A-F

q10 q2x 0-9,A-F


“1234”

q0 8

“0x0” q0 → q1 → q2 → q3 4

“” q0 8




q0 q3

0-9,A-F

q10 q2x 0-9,A-F


“1234” q0

8

“0x0” q0 → q1 → q2 → q3 4

“” q0 8




q0 q3

0-9,A-F

q10 q2x 0-9,A-F


“1234” q0 8

“0x0” q0 → q1 → q2 → q3 4

“” q0 8




q0 q3

0-9,A-F

q10 q2x 0-9,A-F


“1234” q0 8

“0x0”

q0 → q1 → q2 → q3 4

“” q0 8




q0 q3

0-9,A-F

q10 q2x 0-9,A-F


“1234” q0 8

“0x0” q0 → q1 → q2 → q3

4

“” q0 8




q0 q3

0-9,A-F

q10 q2x 0-9,A-F


“1234” q0 8

“0x0” q0 → q1 → q2 → q3 4

“” q0 8




q0 q3

0-9,A-F

q10 q2x 0-9,A-F


“1234” q0 8

“0x0” q0 → q1 → q2 → q3 4

“”

q0 8




q0 q3

0-9,A-F

q10 q2x 0-9,A-F


“1234” q0 8

“0x0” q0 → q1 → q2 → q3 4

“” q0

8




q0 q3

0-9,A-F

q10 q2x 0-9,A-F


“1234” q0 8

“0x0” q0 → q1 → q2 → q3 4

“” q0 8



Non-deterministic Finite State Automata

• There are two main types of fsas• Deterministic• Non-deterministic

• What we have been looking at so far are examples ofnon-deterministic finite [state] automata (nfa)

• E.g. nfa for recognising a hexadecimal number:

q0 q3

0-9,A-F

q10 q2x 0-9,A-F



Properties of nfas

• Must have a start node

• No restrictions on the edge labels

• Allowed special ε edge labels (we’ll come back to this)

• Allowed to be “on” multiple states simultaneously• non-determinism

Basically an NFA is a FSA that isn’t fully defined: in a DFA(Deterministic Finite (State) Automaton of course),everything has to be determined: if there’s an error conditionthat has to be explained, there can’t be any ambiguity, and soon.



Non-determinism example

Is the string “ca” accepted by the nfa?Path:

{q0}→ {q1}→ {q3}

Accepted:

Yes

q0 q3a,b

q1

c

q2

a

a

b,c

c

b

a,b




Is the string “ca” accepted by the nfa?Path: {q0}

→ {q1}→ {q3}

Accepted:

Yes

q0 q3a,b

q1

c

q2

a

a

b,c

c

b

a,b




Is the string “ca” accepted by the nfa?Path: {q0}→ {q1}

→ {q3}

Accepted:

Yes

q0 q3a,b

q1

c

q2

a

a

b,c

c

b

a,b




Is the string “ca” accepted by the nfa?Path: {q0}→ {q1}→ {q3}Accepted:

Yes

q0 q3a,b

q1

c

q2

a

a

b,c

c

b

a,b




Is the string “ca” accepted by the nfa?Path: {q0}→ {q1}→ {q3}Accepted: Yes

q0 q3a,b

q1

c

q2

a

a

b,c

c

b

a,b




Is the string “ac” accepted by the nfa?Path:

{q0}→ {q0, q2}→ {q1, q3}

Accepted:

Yes

q0 q3a,b

q1

c

q2

a

a

b,c

c

b

a,b




Is the string “ac” accepted by the nfa?Path: {q0}

→ {q0, q2}→ {q1, q3}

Accepted:

Yes

q0 q3a,b

q1

c

q2

a

a

b,c

c

b

a,b




Is the string “ac” accepted by the nfa?Path: {q0}→ {q0, q2}

→ {q1, q3}

Accepted:

Yes

q0 q3a,b

q1

c

q2

a

a

b,c

c

b

a,b




Is the string “ac” accepted by the nfa?Path: {q0}→ {q0, q2}→ {q1, q3}Accepted:

Yes

q0 q3a,b

q1

c

q2

a

a

b,c

c

b

a,b




Is the string “ac” accepted by the nfa?Path: {q0}→ {q0, q2}→ {q1, q3}Accepted: Yes

q0 q3a,b

q1

c

q2

a

a

b,c

c

b

a,b




Is the string “aabc” accepted by the nfa?Path:

{q0}→ {q0, q2}→ {q0, q2}→ {q0, q1, q2}→ {q1, q3}

Accepted:

Yes

q0 q3a,b

q1

c

q2

a

a

b,c

c

b

a,b




Is the string “aabc” accepted by the nfa?Path: {q0}

→ {q0, q2}→ {q0, q2}→ {q0, q1, q2}→ {q1, q3}

Accepted:

Yes

q0 q3a,b

q1

c

q2

a

a

b,c

c

b

a,b




Is the string “aabc” accepted by the nfa?Path: {q0}→ {q0, q2}

→ {q0, q2}→ {q0, q1, q2}→ {q1, q3}

Accepted:

Yes

q0 q3a,b

q1

c

q2

a

a

b,c

c

b

a,b




Is the string “aabc” accepted by the nfa?Path: {q0}→ {q0, q2}→ {q0, q2}

→ {q0, q1, q2}→ {q1, q3}

Accepted:

Yes

q0 q3a,b

q1

c

q2

a

a

b,c

c

b

a,b




Is the string “aabc” accepted by the nfa?Path: {q0}→ {q0, q2}→ {q0, q2}→ {q0, q1, q2}

→ {q1, q3}

Accepted:

Yes

q0 q3a,b

q1

c

q2

a

a

b,c

c

b

a,b




Is the string “aabc” accepted by the nfa?Path: {q0}→ {q0, q2}→ {q0, q2}→ {q0, q1, q2}→ {q1, q3}Accepted:

Yes

q0 q3a,b

q1

c

q2

a

a

b,c

c

b

a,b




Is the string “aabc” accepted by the nfa?Path: {q0}→ {q0, q2}→ {q0, q2}→ {q0, q1, q2}→ {q1, q3}Accepted: Yes

q0 q3a,b

q1

c

q2

a

a

b,c

c

b

a,b



ε transitions

• nfas can have a special transition label ε

• ε allows a transition without consuming any input tokens

q0 q4q1cq2a

q3b

cϵ

• In English, what pattern does this nfa therefore allow us tomatch?

• Any string starting with a “c”, followed by one or moreinstances of “ab”, followed by a “c”



Your turn – Woo!

Construct a nfa which accepts only binary numbers with an evennumber of zero’s.

q0

1

q100

1

As above, except for one’s instead of zero’s.

q0

0

q111

0



Your turn

Construct a nfa which accepts only binary numbers with an evennumber of zero’s OR an even number of one’s.

q0

q1

1

q20

q3

0

q41

ϵ

ϵ

0

1

1

0



Regular Expressions (re)

You remember thepower of regularexpressions? Thesethings can save theworld!regex’s rule.

. . . and it turns out they’re much better interpreted and dealt withby using FSAs.



Regular Expressions as nfas

• Regular expressions can easily berepresented using nfas

• We can group regular expressionsinto 4 different components

Character a single character:/a/

Concatenation two adjacentexpressions: /ab/

Union two OR’dexpressions: /a|b/

Kleene star zero or morerepetitions: /a*/

/a(bc|d*)*e/ can beviewed as

a

b c d

e

concat star

union

star

concat

concat



And it really does matter

Knowing the right way to do regular expressions, using goodcomputer science, means that you don’t have to make major errorsleading to really really bad performance, in a hardly-used bit ofsoftware, like say, Perl.

This is the time taken to match the sequence a?nan against an where thesuperscript represents string repeats, so a?2a2 = a?a?aa

Source: Russ Cox / [email protected] a great discussion of this, see

http://swtch.com/~rsc/regexp/regexp1.html



re to nfa: Character

/a/

q0 q1a



re to nfa: Concatenation

/ab/

q0 q1a + q2 q3b

becomes

q0 q3q1a q2ϵ b



re to nfa: Union

/a|b/

q0 q1a + q2 q3b

becomes

q4 q5

q0ϵ

q2ϵ

q1aϵ

q3bϵ



re to nfa: Kleene Star

/a*/

q0 q1a

becomes

q2 q3ϵ

q0ϵ

q1aϵ

ϵ



An example conversion

/a(bc|d*)*e/

becomes

q0

q14

q1a q2ϵq3ϵ

q4ϵ

q12

ϵ

q5b

q6

ϵ q10ϵ

q13ϵ

q7ϵ

q8d

q11ϵ

q9c

ϵ

ϵ

ϵ

ϵ

ϵ

e



DFAs vs NFAs

• How do dfas differ from nfas?• Deterministic (not non-deterministic)• No ε edges• Every edge from a node must have a unique label (can’t be in

multiple states)• Every node must have an outward edge for each token of the

alphabet (it is completely described)



dfas vs. nfas

• E.g. dfa for recognising a hexadecimal number:

q1

q4

0-9,A-F

q5

x

q20

1-9,A-F,x

q3x 0-9,A-F

0-9,A-F

x0-9,A-F,x



Why do we need dfas? What’s wrong with nfas?

• The non-deterministic aspect of nfas makes them bad formodern-day computation

• Computers cannot efficiently perform the “be in multiplestates at once”

• In a dfa, you can only be in one state at a time



Your turn

dfas can be harder to manually construct than nfas, due to theirrestrictions.

Construct a dfa over Σ = {a, b} accepting strings ending in “bb”.

q0 q2

b

a

a

q1b ba



nfa?= dfa

• At first glance, it would appear that nfas are more powerfulthan dfas. Why?

• multiple edges with the same label coming off a node• ε edges• ability to be in multiple states at once

• In 1959, Rabin and Scott proved that nfas and dfas have thesame expressive power

• The proof is surprisingly simple: we need to show that foreach NFA there is a DFA, and vice-versa, that accept preciselythe same languages.



Determinisation Algorithm

• This algorithm converts any nfa into an equivalent dfa

q1

q3

q4

0

q2

ϵ

0ϵ

0 1

1

into

{q1,q2,q3} {q2,q4}0,1

1

{q2,q3}01

{q4}00

{}1

0,1



Move and Epsilon-Closure

Move M(states, token)Given a set of states and an input token, what set ofstates do you end up at

Epsilon-Closure EC (states)Given a set of states, what set of states do you getby expanding all ε transitions

q0q3

b

q1c

q2

aa

b,c

cϵ

ba,b

M({q0, q1},“a”) = {q2, q3} EC({q2, q3}) = {q0, q2, q3}



Algorithm Idea

• We need to remove the concept of non-determinism in thedfa

• Achieved by creating a state in the dfa for every possible setof states in the nfa

• Q: If the nfa has n states, what’s the maximum number ofstates its corresponding dfa could have?

• A: |P(n)| = 2n



PseudocodeRequire: nfa

dfa ← a new DFA objectdfa.start ← EC (nfa.start)todo ← [dfa.start]

while todo 6= ∅ dostates = pop the next item off todo

for all σ ∈ ΣNFA \ {ε} dos = EC (M(states, σ))if s 6= ∅ then

Add the edge statesσ−→ s to dfa

Add s to todo if it’s newend if

end forend whilereturn dfa



Example

q0q3

b

q1c

q2

aa

b,c

cϵ

ba,b

todo{0}

{0, 2}{1}{0, 1, 2}{1, 3}{3}{0, 2, 3}



Example

q0q3

b

q1c

q2

aa

b,c

cϵ

ba,b

todo{0}

{0, 2}{1}{0, 1, 2}{1, 3}{3}{0, 2, 3}

{q0}



Example

q0q3

b

q1c

q2

aa

b,c

cϵ

ba,b

todo{0}{0, 2}{1}

{0, 1, 2}{1, 3}{3}{0, 2, 3}

{q0}

b {q0,q2}a

{q1}c



Example

q0q3

b

q1c

q2

aa

b,c

cϵ

ba,b

todo{0}{0, 2}{1}{0, 1, 2}{1, 3}

{3}{0, 2, 3}

{q0}

b {q0,q2}a

{q1}c

a{q0,q1,q2}b {q1,q3}

c



Example

q0q3

b

q1c

q2

aa

b,c

cϵ

ba,b

todo{0}{0, 2}{1}{0, 1, 2}{1, 3}{3}

{0, 2, 3}

{q0}

b {q0,q2}a

{q1}c

a{q0,q1,q2}b {q1,q3}

cb,c

{q3}a



Example

q0q3

b

q1c

q2

aa

b,c

cϵ

ba,b

todo{0}{0, 2}{1}{0, 1, 2}{1, 3}{3}{0, 2, 3}

{q0}

b {q0,q2}a

{q1}c

a{q0,q1,q2}b {q1,q3}

c

b{q0,q2,q3}

a

c

b,c{q3}

a



Example

q0q3

b

q1c

q2

aa

b,c

cϵ

ba,b

todo{0}{0, 2}{1}{0, 1, 2}{1, 3}{3}{0, 2, 3}

{q0}

b {q0,q2}a

{q1}c

a{q0,q1,q2}b {q1,q3}

c

b{q0,q2,q3}

a

c

b,c{q3}

ab,c

a



Example

q0q3

b

q1c

q2

aa

b,c

cϵ

ba,b

todo{0}{0, 2}{1}{0, 1, 2}{1, 3}{3}{0, 2, 3}

{q0}

b {q0,q2}a

{q1}c

a{q0,q1,q2}b {q1,q3}

c

b{q0,q2,q3}

a

c

b,c{q3}

ab,c

a{}

a,b,c

a,b,c



Example

q0q3

b

q1c

q2

aa

b,c

cϵ

ba,b

todo{0}{0, 2}{1}{0, 1, 2}{1, 3}{3}{0, 2, 3}

{q0}

b {q0,q2}a

{q1}c

a{q0,q1,q2}b {q1,q3}

c

b{q0,q2,q3}

a

c

a,b

c

b,c{q3}

ab,c

a{}

a,b,c

a,b,c



Finish

End of presentation.


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