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The reference polynomial used in a CRC scheme  is x4 + x3 + 1. A data sequence 1010101010 is to be sent, Determine the actual bit string that is transmitted.
  • a)
    10101010100011
  • b)
    10101010101110
  • c)
    10101010100110
  • d)
    10101010100010
Correct answer is option 'D'. Can you explain this answer?
Verified Answer
The reference polynomial used in a CRC scheme is x4 + x3 + 1. A data s...
Reference polynomial x4 + r3 + 1 =11001
Datasequence = 1010101010
On dividing 1010101010 by 11001,
We get CRC (i.e. remainder) = 0010
On appending this CRC on data sequence, we get the actual message transmitted i.e. 10101010100010.
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Most Upvoted Answer
The reference polynomial used in a CRC scheme is x4 + x3 + 1. A data s...
Given: reference polynomial x^4 + x^3 + 1, data sequence 1010101010

Steps to calculate actual bit string:

1. Append 4 zeros to the data sequence: 10101010100000
2. Divide the extended data sequence by the reference polynomial using modulo-2 division:
- First, perform XOR operation between the first 5 bits (10101) and the reference polynomial (x^4 + x^3 + 1) to get the remainder 00110. This is the first 5 bits of the actual bit string.
- Next, append the next bit from the data sequence (0) to the remainder (00110) to get 001100. Perform XOR operation between 001100 and the reference polynomial to get the new remainder 000010. This is the next bit of the actual bit string.
- Repeat the above step for all the remaining bits of the data sequence.
3. The final remainder obtained after dividing the extended data sequence is appended to the original data sequence to get the actual bit string.

Therefore, the actual bit string transmitted is 10101010100010.
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Understanding the Complexity ClassesIn computational theory, problems are categorized into complexity classes based on the resources required to solve them. Here, we focus on four specific problems: PATH, HAMPATH, SAT, and 3SAT, to determine which belong to class P.What is Class P?- Class P consists of decision problems that can be solved by a deterministic Turing machine in polynomial time.- If a problem is in P, it means there exists an algorithm that can solve it efficiently for all input sizes.Analysis of Each Problem- PATH: - The problem asks whether there exists a path between two vertices in a graph. - This can be solved using Depth-First Search (DFS) or Breadth-First Search (BFS), both of which run in polynomial time. - Conclusion: PATH is in P.- HAMPATH: - This problem involves determining whether there is a Hamiltonian path in a graph (a path that visits each vertex exactly onc e). - HAMPATH is NP-complete, meaning it is not known to be solvable in polynomial time. - Conclusion: HAMPATH is not in P.- SAT (Satisfiability): - SAT asks whether a boolean formula can be satisfied by some assignment of truth values. - While SAT is NP-complete, it was proven to be in NP, and due to advancements, it can be solved in polynomial time for specific cases. - Conclusion: SAT is not in P generally.- 3SAT: - This is a specific case of SAT where the formula is in conjunctive normal form with exactly three literals per clause. - 3SAT is also NP-complete and does not have known polynomial-time solutions. - Conclusion: 3SAT is not in P.Final ConclusionBased on the analyses, the problems that belong to class P are:- Option A: SAT - Not generally in P.- Option C: PATH - In P.Thus, only PATH is confirmed to be in P.

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The reference polynomial used in a CRC scheme is x4 + x3 + 1. A data sequence 1010101010 is to be sent, Determine the actual bit string that is transmitted.a)10101010100011b)10101010101110c)10101010100110d)10101010100010Correct answer is option 'D'. Can you explain this answer?
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