Minimization of DFA

Construct a minimum state automata equivalent to given automata ?

(RGPV 2008)

Solution:
Transition table for above automata.

State

Input = a

Input = b

->q0 Initial state

q1

q3

q1

q2

q4

q2

q1

q1

q3

q2

q4

q4 Final state

q4

q4


Step 01: Remove steps which are unreachable from initial states.
Step 02: Split final states and non final states.

A0 = {q4}
A1 = {q0,q1,q2,q3}

π0 = {q4}, {q0,q1,q2,q3}

A0 cannot be partition further.

In A1,
q0 is 1 equivalent to q2 for input a, but not equivalent to q1 and q3.
q1 is 1 equivalent to q3 for input a and b, but not to q0 and q2.

So, A1 can be partitioned as,
B0 = {q0, q2}
B1 = {q1, q3}

π1 = {q4}, {q0,q2}, {q1,q3}

Now, B0 and B1 can not be partitioned further.

π2 = {q4}, {q0,q2}, {q1,q3}

π2 = π1

In minimized DFA, we have three states,
  1. {q4}, 
  2. {q0,q2}, 
  3. {q1,q3}

State

Input = a

Input = b

->{q0,q2} Initial state

{q1,q3}

{q1,q3}

{q1,q3}

{q0,q2}

{q4}

{q4} Final state

{q4}

{q4}



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NFA to DFA using Indirect method

NFA to DFA using Indirect method

Step 01 : First convert NFA with ∈ moves to NFA without ∈ moves.
Step 01 :Then convert NFA without ∈ to the DFA.

Let's take an example,

We have a NFA with epsilon moves, now convert it to the NFA without epsilon moves.

NFA with epsilon moves


First convert above NFA to without epsilon moves,

Find ∈-closure of (q0), (q1) and (q2).
  1. ∈-closure of (q0) = {q0, q1, q2} 
  2. ∈-closure of (q1) = {q1, q2}
  3. ∈-closure of (q2) = {q2}

State

0

1

2

->q1

{q1,q2,q3}

{q2,q3}

{q3}

q2

φ

{q2,q3}

{q3}

q3

φ

φ

{q3}


From the above transition table, draw the transition diagram,

NFA without  epsilon

From the question diagram, it is clear that only with ∈ input q1 and q2 state can reach to the final state.
So, now without ∈ input, q1 and q2 is also treated as final states.
As shown in diagram beabove.

Now, NFA without epsilon to DFA

Using Subset construction method.

Again I am going to use information from this transition table.

State

0

1

2

->q1

{q1,q2,q3}

{q2,q3}

{q3}

q2

φ

{q2,q3}

{q3}

q3

φ

φ

{q3}


Subsets are,
  1. {q1}
  2. {q2}
  3. {q3}
  4. {q1,q2,q3}
  5. {q2,q3}
Treat each subset as state in DFA.
If there is no transition use φ symbol.

State

Input = 0

Input = 1

Input = 2

{q1}

{q1,q2,q3}

{q2,q3}

{q3}

{q2}

φ

{q2,q3}

{q3}

{q3}

φ

φ

{q3}

{q1,q2,q3}

{q1,q2,q3}

{q2,q3}

{q3}

{q2,q3}

φ

{q2,q3}

{q3}

φ

φ

φ

φ


In NFA withoud epsilon, q1, q2, q3 was final states. In subset {q1,q2,q3} and {q2,q3} final states exist.
So these subsets will also treated as final states.

DFA from NFA

Now from the transition table of above DFA,

State

Input = 0

Input = 1

Input = 2

{q1}

{q1,q2,q3}

{q2,q3}

{q3}

{q2}

φ

{q2,q3}

{q3}

{q3}

φ

φ

{q3}

{q1,q2,q3}

{q1,q2,q3}

{q2,q3}

{q3}

{q2,q3}

φ

{q2,q3}

{q3}

φ

φ

φ

φ


We find state {q1,q2,q3} is equivalent to state {q1}, so replce state {q1,q2,q3} by state {q1}.
Simillary, replace state {q2,q3} by state {q2}.

So, new DFA transition table become,

State

Input = 0

Input = 1

Input = 2

{q1}

{q1}

{q2}

{q3}

{q2}

φ

{q2}

{q3}

{q3}

φ

φ

{q3}

φ

φ

φ

φ


And DFA transition diagram,


Practice problems:

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Construct NFA without ∈

Construct NFA without ∈ transitions


NFA with ∈

Sol.

Step 01: Find ∈-closure of (q1), (q2) and (q3).
∈-closure of (q1) = {q1, q2, q3} 
∈-closure of (q2) = {q2, q3}
∈-closure of (q3) = {q3}

For each state find the next state for each input. See the table below,

State

0

1

2

->q1

{q1,q2,q3}

{q2,q3}

{q3}

q2

φ

{q2,q3}

{q3}

q3

φ

φ

{q3}


From the question diagram, it is clear that only with ∈ input q1 and q2 state can reach to the final state.
So, now without ∈ input, q1 and q2 is also treated as final states.
As shown in diagram below.

NFA without ∈


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Suppose a system has 12 instances | UGC NET Dec 2018

Q2. UGC NET Dec 2018 : 

Suppose a system has 12 instances of some resource with n processes competing for that resource. Each process may require 4 instances of the resources. The maximum value of n for which the system never enters into deadlock is
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Q1. UGC NET Dec 2018 

​​​​​​​Data warehouse contains _____ data that is never found in operational environment.
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Remove Windows not activated from windows 10

How to remove Windows not activated, from Windows 10.

In a blank text document, paste the code.

@echo off
taskkill /F /IM explorer.exe
explorer.exe
exit


Save the file as, 
File name: Activation.bat
Save as type: All Files


File will look like this,
Run it as administration.

And restart the computer.
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Divide the screen into two halfs

Divide the web page screen in to equal parts


<div style="width: 50%; float:left">
   #left content here
</div>

<div style="width: 50%; float:right">
   #right content here
</div>

Similar code, as above can be used to divide in 3, 4, 5 or more parts.



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Grammar is ambiguous. S → aSbS|bSaS|∈

RGPV 2020
Show that the following grammar is ambiguous. 
S → aSbS|bSaS|∈

Ans. For grammar to be ambiguous, there should be more than one parse tree for same string.

Above grammar can be written as

S → aSbS
S → bSaS
S → ∈

Lets generate a string 'abab'.

So, now parse tree for 'abab'.

Left most derivative parse tree 01

S → aSbS
S → a∈bS
S → a∈baSbS
S → a∈ba∈b∈
S → abab

Parse Tree 01

Left most derivative parse tree 02

S → aSbS
S → abSaSbS
S → ab∈aSbS
S → ab∈a∈bS
S → ab∈a∈b∈
S → abab

Parse tree 02

So there are more than 1 parse tree for same string, that means grammar is ambiguous.
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Construct Moore machine for Mealy machine

RGPV 2020
Construct Moore machine for the following Mealy machine.



Sol.
Transition table for above Mealy machine.

Present State

Next State

Input = 0

Input = 1

State

Output

State

Output

q0

q0

0

q1

1

q1

q2

2

q0

0

q2

q1

1

q2

2


Transition table for Moore machine from above Mealy machine transition table.

Present State

Next State

Output

Input = 0

Input = 1

q0

q0

q1

0

q1

q2

q0

1

q2

q1

q2

2


Transition diagram for Moore machine fram above Moore machine transition table.


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Practicals:
Previous years solved papers:
A list of Video lectures References:
  1. Sebesta,”Concept of programming Language”, Pearson Edu 
  2. Louden, “Programming Languages: Principles & Practices” , Cengage Learning 
  3. Tucker, “Programming Languages: Principles and paradigms “, Tata McGraw –Hill. 
  4. E Horowitz, "Programming Languages", 2nd Edition, Addison Wesley

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