- Introduction
- String in automata
- Power of Alphabet ( ∑ )
- Introduction of Automata
- Every DFA is NFA
- Design a DFA for Language: “Starting with ‘a’ "
- Design a DFA for Language: “Ending with ‘a’ "
- Design a DFA for Language: “Start with a and end with b”
- Design a DFA for Language: “Not start with ‘a’ or not end with ‘b’ “
- Design a DFA for L = “Start with ‘a’ and end with ‘a’”
- Design a DFA for L = “Start with ‘001’ ”
- Design a DFA for L = “Ending with ‘001’ ”.
- Design a DFA for L = “At least one ‘a’”
- Design a DFA for L = “At most one ‘a’ ”
- Design a DFA for L = “At least two ‘a’ ”
- Design a DFA for L = “Exactly one ‘a’ ”
- Design a DFA for L = “At most two ‘a’ ”
- Design a Machine for L = “Containing substring ‘ab’”.
- Design a Machine for L = “Not containing substring ‘001’ ”
- Difference between DFA and NFA: Minimum number of states
- Design a machine of L = Containing even a where ∑ = {a, b}
- Design a machine of L = Containing odd b where ∑ = {a, b}
- Machine with Unary Alphabet
- Design a machine of L = Even number of a’s AND odd number of b’s
- Design a machine of L = Even number of a’s OR odd number of b’s
- Some exercise problems with same pattern
- Q. UGCNET-July-2018-II-32
- (Number of a) mod 2 = 0 and (Number of b) mod 3 =0
- (Number of a) mod 2 =0 and (Number of b) mod 4 =0
- Design a machine of L= (Number of a) mod 4 <= 2
- Design a machine of L = { ω | |ω|=3}, ∑ = {a, b}
- Solve the related Question
- Design a machine L= {Every substrings of 001}
- Design a machine of L = a* b* where ∑ = {a, b}
- Set of all string no consecutive 1’s where ∑ = {0, 1}
- Design a machine of L = Exactly two a’s where ∑ = {a,b}
- Design a machine of L = At most two a’s where ∑ = {a, b}
- Design a machine of L = Exactly two a’s and two b’s ∑ = {a, b}
- At least two a’s and at least two b’s ∑ = {a, b}
- Design a machine of L = (111)*+ (11111)* , ∑ = {1}
- Design a machine of L = (111+ 11111)* , ∑ = {1}
- NFA to DFA Conversion
- Minimization of DFA
- Null removal or λ move removed
- Some more DFA machine design examples
- Application of DFA and NFA
- How to convert DFA/NFA to Higher Machine
- DFA/NFA to Regular Expression Calculation
- Write a Regular Expression: Example1
- Write Regular Expression: Example2
- Write Regular Expression: Example3
- Example1: find Regular Expression
- Example2: find regular expression
- Find Regular Expression of the given machine
- Example3: Find Regular Expression
- Example4: Find regular expression
- Example5: Find regular expression
- Example6: Find Regular Expression
- Here we calculate Regular Expression
- Q. Select the correct Regular Expression
- Simplification Regular Expressions
- Definition of Mealy and Moore
- Example 1: Machine for Same Input
- Example 2: Machine for 1’s complement
- Example 3: Machine for 2’s complement
- Example 4: Machine for change Sign Bit
- Example 5: Machine for Increment by one
- Example 6: Machine for Full adder
- Example 7: Machine for Mod 3
- How to find minimum no. of state in mod N
- Mealy machine to Moore machine conversion
- Moore machine to Mealy machine conversion
- Power of Language and Machine
- How to identify language
- Example 1: Identification of Language
- Example 2: Identification of Language
- Example 3: Identification of Language
- Example 4: Identification of Language
- Language Identification: Important points
- Language Identify: Examples
- Language Identify: Examples
- Some Exercise and Solve
- Some Exercise and Solve
- Some Examples
- Some More Examples
- Important Exercise and Solutions
- Special Cases: Examples
- Special Cases: Examples
- Special Cases: Examples
- Special Cases: Examples
- Special Cases: Examples
- Special Cases: Examples
- Special Cases: Examples
- Special Cases of CSL: Example
- Important Examples: Identification of language
- Example CSL: Identification of language
- Important Examples: Identification of language
- Important Examples: Identification of language
- Important Examples: Identification of language
- Important Examples: Identification of language
- Important Examples: Identification of language
- Important Examples: Identification of language
- Exercise: Identification of language
- Exercise: Identification of language
- Introduction to Pumping Lemma
- Pumping Lemma Theorem for Regular Language
- Q .UGC-NET 2017
- Example: L= a^n b^n
- Example: L={a^i^2, i≥1}
- Example: L={a^p, p is prime}
- Example: L={a^p, p is not prime}
- Example: L={ωω^R, ωϵ∑*}
- Example: L= Balance parenthesis
- Example: L={ωω, ωϵ∑*}
- Example: L={0^n 1^n, n≥0}
- Example: L={a^n ; n is perfect square}
- Example: L={ω|n_a (ω) = n_b (ω)}, where ∑={a, b}, ω ϵ∑*}
- Exercise
- Shortcut Prove: One Language is Not Regular
- Pumping Lemma Theorem for Context Free Language
- Example: L={a^p ; p is prime}
- Example: L={a^i^2, i≥1}
- Example: L={ωω, ωϵ∑*}
- Exercise
- Ogden’s Lemma for CFL
- Myhill-Nerode Theorem
- Introduction: Push Down Automata
- DPDA and NPDA
- PDA Design Method
- Design PDA of L = a^n b^n
- Design PDA of L = a^m b^n, m<n
- Design PDA of L = a^m b^n, m>n
- Design PDA of L = a^m b^n, m ≤ n
- Design PDA of L = a^m b^n, m ≥ n
- Design PDA of L = a^m b^n, m ≠ n
- Design PDA of L = a^n b^2n
- Design PDA of L = a^2n b^n
- Design PDA of L = n_a (ω) = n_b (ω)
- Design PDA of L = ωω^R
- Subject Mock

Please see __Special Cases: Examples__ ** and **__Special Cases: Examples__** **before start this section.

** **

**Answer:**

This given language is the same as

because x is not present in that case.

So, this same problem already discussed in __Example 2__** ****please see for details**.

And because there is no special character so, it needs multiple copies of the stack.

**For details see** ** Example 2**.

**Answer:**

Like previous question __Example 1__

Now, we can try to write regular expression initially

But we can’t derive all the strings from the above RE.

**Example:**

We can’t derive this string from above,

because it is always ending with either **aa** or **bb**, it never ends with **ba** or **ab**.

So, as per above example for all string is not accepted in this,

Now,

**Answer:**

This language same as above

The only position of x is different but the logic is the same.

So, the same way as above us cannot write **Regular Expression** for this given language

So, same as **Example 2**,

Now we first see that we can write a regular expression for it or not.

Here first and the last letter is the same as either a or b.

In this expression, every string has the same starting and ending symbol, and a minimum three-length string.

**Example: **

Now we have to check this **Regular Expression (RE)** can cover all the strings of the given language.

**Take an example **

Let,

So, we can visualize like that

and starting and ending symbol is the same.

**Take another example **

Let,

Now,

Here we also can write the expression as

and starting and ending symbol is the same.

Now, we can cover all string where ending and starting string is same.

So, we can form a regular expression for this language.