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salam bro, hope all are good...

did any do some thing regarding Diophantine equation
i tried but the result is not solvable

anyone please guide

I think this equation has no solution

CS701 Assignment No 1 Solution Idea

Describe and Design a Turing machine in following three ways description that decide the following language

 

  1. Formal description of Turing machine
  2. Implementation level descriptions of Turing machine
  3. High level description of Turing machine

 

  1. a) L = {13n+1 : n≥0}, the language consisting of all strings of 1s in given exponential function. Some example words of the language are as follows:

 

Solution

 

if we put n=0 then  in put string will be

  1. 000

if we put n= 2 then the input string will be

2. 000000000

if we put n= 3 then the input string will be

3. 000000000000000000000000000

Formal Description of Turing Machine 0^3^n+1

  1.  

A Turing machine M1 that decides the language given in question above is defined by the following 7-Tuple.

M = (Q, Σ, Γ, δ, q­1, q­accept , qreject)

Q = (q1, ­q2, q3 … q7,  q­accept , qreject)

qis start state, q­accept is the accept state and  qreject  is the reject state.

Σ = {0} // Input alphabet

Γ = {0, x, []} // tape or work alphabet where Σ⊆Γ and     ∈ Γ.

δ : (Q × Γ) → (Q × Γ × {L, R}) // The transition function is described by the following state diagram.

  1. Implementation level description of Turing Machine(0^3^n+1) 03n+1 
  • A Turing machine M1 that decides L = {03n+1: n≥0}
  • M1 = on input string w
    1. Sweep left to right across the tape, crossing of every other 0.
    2. If in stage 1 the tape contained a single 0, accept.
    3. If in stage 1 the tape contained more than single 0 and the number of 0’s are even, reject.
    4. Return the head to the left-hand end of the tape.
    5. Go to stage 1.
  • Basically every sweep cuts the number of 0’s by three.
  • At the end only one should remain and if so the original number of zeroes was a power of three.
  1. High Level Description
  2. Sweep from left to right across the tape, crossing of every other 0.
  3. If there is a single 0 on the tape, accept.
  4. If there are more than one 0’s and the number of 0’s is even, reject.
  5. Return the head to the left end and repeat.

Note: You may verify.. up to me its correct version

Question No 2

CS701 Assignment Qustion No 1 Solution – by- Ghulam Shabbir
Sol.
Ax+By=C
Coefficient ko comapre karain with given equation 33x+15y=14

A=33
B=15
C=14
The greatest common factor (GCR) of A and B must be divisible by C
GCF of A and B is 3
33/3=11
And
15/3=5

But this 3 is not not divisible by 14
As
14/3===not divisible
hence the given statement has no integer solution

Thanks for Guiding

Did Any one write the answer of second part of this assignment .....

1) How can you differentiate and analysis between Turing Machine and Fuzzy Turing Machine as discussed in the paper? Elaborate it critically in your own words.
2) What functionalities have been expressed in extended Church thesis? Elaborate it critically in your own words.

aoa
Dear Students you can view this link for installing jflap in your windows.
https://www.youtube.com/watch?v=-KNnBBUPX2k
Regards

Dear Students you can also consider this video before using jflap for make a turing machine.
https://www.youtube.com/watch?v=lTGl9EAMIy0
Regards

Salam to all
did any one of you attempt the second question of this assignment.
please share some valuable knowledge...

Turing machine example is same like as ....

Attachments:

L = {03n+1: n≥0}={000,000000,000000000,000000000000.....}

so you follow all of my sharing from first to end
regards

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