1. Introduction to Finite Automata
Fundamentals
Deterministic finite automata
Representations of automata
Definition of a language
Proof of language content

2. Review:
What is the difference between
S not equal to T and S  T = null?
Recursive definitions use that thing being defined
as part of the definition.
Example: recursive definition of a tree
Basis: single node is a tree
IH: assume that T1 … Tk are trees
Induction: connect T1 … Tk to the node that is
the root of the recursively defined tree

3. Review: FA Recognizing Strings Ending in “ing”
We can get the next character in any state
Where we go depends on which character
empty
Saw i i
Not i
Saw ing g
Not i or g
Saw in n
Not i or n
Start
Double circle denotes
“finish” or “accepting”
state
What is wrong with this FA?

4. Rules for deterministic finite automata
Ready
Sending
data in
done
timeout
Start
Finite automata move from state to state in response to input controls
Deterministic finite automata (DFA) are always in a unique state
Each input determines one and only one state to which the automaton will transition from its current state

5. Review: FA Recognizing Strings Ending in “ing”
Does this example obey the rules?
empty
Saw i i
Not i
Saw ing g
Not i or g
Saw in n
Not i or n
Start
Any letter
How many string ending in “ing” will this FA find in
the input?

6. Review: FA Recognizing Strings Ending in “ing”
How do I change this FA so that it only accepts
input string ending in “ing”
empty
Saw i i
Not i
Saw ing g
Not i or g
Saw in n
Not i or n
Start
Any letter

7. Central concepts of automata theory
Alphabets: sets of symbols
Strings: lists of symbols from an alphabet
Languages: sets of string from the same alphabet

8. Alphabets Any finite set of symbols Usually denoted by S Examples:
S=set of all ASCII characters
S={0,1} binary alphabet
S={a,b,…,z} lower case letters

9. Sets of Strings
A set of strings over alphabet Σ is a set of lists, each element of which is a member of Σ
e denotes the empty string (length=0)
Σ* denotes the set of all strings over S
{0,1}*={ε,0,1,00,01,10,11,000,…}
Sk denotes strings of length k over a specified alphabet
S0={e} for any alphabet
S+ denotes non-empty strings

10. More about Strings S* = S0  S1  S2 …
S and S1 give different meaning to the same entity
S denotes an alphabet
S1 denotes strings of unit length over alphabet
If x and y are strings, then xy is a string formed by their concatenation

11. Languages A language is a subset of strings for some alphabet
L=Σ* language of all strings from S
L can be empty
L={e} is not empty. It contains the empty string
See text p31 for examples from binary alphabet

12. Using DFAs to define languages
Elements in the formalism for defining languages:
A finite set of states (Q, typically).
An input alphabet (Σ, typically).
A transition function (δ, typically).
A start state in Q (q0, typically).
A set of final states ⊆ Q (F, typically).
“Final” and “accepting” are synonyms.
Language A=(Q,S,d,q0,F) is a “five tuple”

13. The Transition Function
Takes two arguments: a state and an input symbol.
δ(q, a) = the state that the DFA goes to when it is in state q and input a is received.
In graph representation δ(q, a) = p is shown by arc from state q to state p labeled by a

14. The set of strings “accepted” by A=(Q,S,d,q0,F) its language
How do we determine if A=(Q,S,d,q0,F) accepts a string?
Let a1,a2,…an denote a finite string
Locate q0 in A using “start”
Find states q1,q2,…qn such that d(qi-1,ai)=qi
If qn  F then string accepted, otherwise string is rejected

15. DFA that accepts all strings without two consecutive 1’s
L=({A,B,C}, {0,1}, d, A, {A,B})
Start 1 A C B
0,1
3 possible
terminations
Consecutive
1’s have
been seen.
No 11’s
and ends
in 0
No 11’s
and ends
In 1

16. Transition Table: δ(q, a) is the element in row q of columns a
Final states
starred * 1
A A B
B A C
C C C
Arrow for
start state
Start 1 A C B
0,1

17. Using graph to answer Is string in language?
Is string 101 in the language of the DFA below?
Start at A.
Start 1 A C B
0,1

18. Is string in language 2
Is string 101 in the language of the DFA below?
Follow arc labeled 1.
0,1 1 1 A B C
Start

19. Is string in language 3
Is string 101 in the language of the DFA below?
Then arc labeled 0 from current state B.
0,1 1 1 A B C
Start

20. Is string in language 4
Is string 101 in the language of the DFA below?
Finally arc labeled 1 from current state A. Result
is an accepting state, so 101 is in the language.
0,1 1 1 A B C
Start

21. Extended Transition Function Delta-hat
We defined δ(q, a) as the state that the DFA goes to when it is in state q and input a is received.
We want an extended transition function d(q, w) defined as the state a DFA is in after it processes string w starting from state q
This extended transition function allows a language to be defined by L={w|d(q0,w)F} ˄ ˄ ˄

22. Recursive definition of Delta-hat
Write w=xa where a is last symbol in w
Basis: δ(q, ε) = q, where e is empty string
Basis: δ(q, a) = δ(q, a)
Induction: δ(q,xa) = δ(δ(q,x),a)
If δ(q,x)=p then δ(q,w) = δ(p,a) ˄ ˄ ˄ ˄ ˄ ˄ ˄

23. Use δ(q,xa) = δ(δ(q,x),a) to decompose delta-hat to nested delta’s˄ ˄
Use δ(q,xa) = δ(δ(q,x),a) to decompose delta-hat to nested delta’s
For convenience of slide
preparation, I will use
delta-underline to mean
delta-hat 1
A* A B
B* A C
C C C
δ(B,011) = δ(δ(B,01),1) = δ(δ(δ(B,0),1),1) =
δ(δ(A,1),1) = δ(B,1) = C
011 not accepted

24. Quiz #2 Monday
on material in lecture “finite automata 2”
and text pp28-52

25. Prove
j=1 to n j = n(n+1)/2
Base case n=1
j=1 to 1 j = 1(1+1)/2
Review: What method
is used in this proof?
If S(?) then S(?)

26. Review: Induction on integers
Methods if S(n) then S(n+1) and if S(n-1) then S(n)
are equivalent but if S(n+1) then S(n) is inconsistent
with induction principal.
What you assume is your IH. Make sure it is true.
I’m not accepting the “identity” method (text pp20-21)
because it does not generalize to other forms of induction.

27. Review: Language of a DFA
Automata of all kinds define languages.
If A=(Q,S,d,q0,F) is an automaton, L(A) is its language.
L(A) = the set of strings w such that
δ(q0, w) is in F. ˄

28. Formal proof of language content1
A* A B
B* A C
C C C
Delta_hat and transition table make it easy to test if a string is in L(DFA)
Formal proof of language content is a problem in equality of sets ˄ ˄

29. Proof of language content
Start 1 A C B
0,1
Proof of language content
S = L(DFA)
T = strings of 0’s and 1’s with no consecutive 1’s
To prove S=T, we need to prove
If w is in S, then w is in T
If w is in T, then w is in S
Proof of “If w is in S, then w is in T” is by induction on length of the string |w|

30. Since DFA has 2 accepting states, if w is in S=L(DFA) has 2 cases
If δ(A, w) = A (string ends with 0)
If δ(A, w) = B (string ends with 1)
Start 1 A C B
0,1
No 11’s
and ends
in 0
in 1

31. Prove: if w in L(DFA) then w no 11’s
Basis: |w| = 0; i.e., w = ε
δ(A, e)=A by definition (e in S=L(DFA))
since ε has no 1’s, e is in T
Basis is true
Induction: |w| >0, write w = xa, where a is the last symbol of w.
inductive hypothesis: if x in L(DFA) then x no 11’s

32. IH: if x in L(DFA) then x no 11’s has 2 cases
Start 1 A C B
0,1
Inductive hypothesis:
if δ(A, x) = A, then x has no consecutive 1’s and ends in 0.
if δ(A, x) = B, then x has no consecutive 1’s and ends in a single 1.

33. Start 1 A C B
0,1
Induction w=xa
If δ(A, xa) = A then δ(A, x) must be A or B and a must be 0 (see DFA above).
By the IH, x has no 11’s and ends in 0 or 1.
Thus, w has no 11’s and end in 0.
If δ(A, xa) = B then δ(A, x) must be A and a must be 1 (see DFA above).
By IH (case 1), x has no 11’s and ends in 0.
Thus, w has no 11’s and ends in 1

34. Prove: if w no 11’s then w in L(DFA)
Contrapositive : If w is not accepted by DFA (i.e. d(A,w)=C) then w has 11’s.
If δ(A,w)=C then w = x1y, where δ(A,x)=B and y is the tail of w that follows what gets to C the first time.
If δ(A,x)=B then x = z1 for some z.
Thus w = z11y and has 11.
Start 1 A C B
0,1

35. prove d(q, xy)=d(d(q, x), y)
Exercise p53:
prove d(q, xy)=d(d(q, x), y)
Recall: xy is the concatenation of strings x and y
When a is last character in string, true by definition of delta_hat: d(q, xa)=d(d(q, x), a)
Base case is true
Complete proof by induction on |y|

36. prove d(q,xy)=d(d(q,x),y) by induction on |y|
Let y=za
d(d(q,x),y)=d(d(q,x),za) let d(q,x)=p
=d(p,za)
=d(d(p,z),a) def of d txt p49
=d(d(d(q,x),z)a) substitution for p
Assume d(q,xz)=d(d(q,x),z) for strings shorter than y
Since |z|<|y|
d(d(q,x),y)=d(d(d(q,x),z)a)= d(d(q,xz),a)
=d(q,xza) def of d txt p49
=d(q,xy) substitution for za

37. Review: Assignment 2
Due
Exercise 2.2.9a, text p 54
Given: A=(Q,S,d,q0,{qf}) and d(q0, a)=d(qf, a) for all a in S
Prove: d(q0, w) = d(qf, w) for all w with |w| > 0 using induction on the length of the string.   
Your proof must include the following:
(1) truth of base case
(2) statement of inductive hypothesis
(3) application of inductive hypothesis
Hint: let w=xa and used definition of delta_hat 

38. Exercise 2.2.9b, text p54
Given: A = (Q, S, d, q0, {qf}) and d(q0, w) = d(qf, w) iff w in L(A) with |w| > 0
Prove by induction on integers that if |x| > 0 and x is in L(A)
then xk (k > 0 concatenations of x) is also in L(A)
Your proof must include the following:
(1) truth of base cases k=1 and k=2
(2) statement of inductive hypothesis k>3
(3) application of inductive hypothesis to show that if S(k-1) then S(k)
Note: the proof is about membership in L(A) 
Hint: use the result from exercise in text p53

39. Review: designing finite automata
Exercise 2.2.4a: A=({A,B,C},{0,1},d,A,{C})
L(A)=set of input strings ending in 00
Graph A
Exercise : A= >A A B
*B B A
L(A)=set input strings with an odd number of 1’s
Prove by induction on length of string that an input string with an odd number of 1’s is accepted by A

40. Assignment #3 due 9-19-14 Exercise 2.2.10: DFA= 0 1 ->A A B *B B A
L(DFA)=set input strings with an odd number of 1’s
Prove by induction on length of string that
d(A,w)=A if w contains an even number of 1’s
d(A,w)=B if w contains an odd number of 1’s
Hints: Basis includes |w|=0 and |w|=1
if w=xa, then a can be zero or one
if w=xa and a=0 then x and w contain the same number of 1’s

41. More hints on Assignment #3
Exercise : DFA= >A A B
*B B A
Define parity(w) as whether w contains and even or odd number of ones.
Prove by induction on |w|
If parity(w)=even, then d(A,w)=A
If parity(w)=odd, then d(A,w)=B
Basis includes w={e,0,1}. For each case, your proof must ask “what is parity(w)?” and “what is d(A,w)?”.
Induction with w=xa has 4 cases: party(x)=even, a={0,1} and party(x)=odd, a={0,1}. For each case, your proof must ask the same questions as in the Basis.

42. Regular Languages
A language L is regular if it consist entirely of strings accepted by some DFA.
The DFA must accept only the strings in L, no others.
Some languages LNR are not regular.
No DFA exist that accepts all the string in LNR

43. Example of LNR L1 = {0n1n | n ≥ 1}
L1 is the set of strings consisting of n 0’s followed by n 1’s, such that n is at least 1.
Thus, L1 = {01, 0011, ,…}

44. Another LNR L2 = {w | w in {(, )}* and w is balanced }
alphabet consists of the parenthesis symbols ’(’ and ’)’.
Balanced parentheses are those that can appear in an arithmetic expression. (), ()(), (()), (()()), etc

45. Many Languages are Regular
We will discuss 4 ways to describe a Regular Language
DFA’s
Non-deterministic FA’s
Non-deterministic FA’s with e-transtions
Regular expressions
Will show at all are equivalent