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In what ratio, more computation time is needed to simulate multitape turing machines using single tape turing machines?
  • a)
    doubly
  • b)
    triple
  • c)
    quadratically
  • d)
    none of the mentioned
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
In what ratio, more computation time is needed to simulate multitape t...
Thus, multitape turing machines cannot calculate any more functions than single tape machines.
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In what ratio, more computation time is needed to simulate multitape t...

Ratio of Computation Time for Simulating Multitape Turing Machines using Single Tape Turing Machines

To understand the ratio of computation time needed to simulate multitape Turing machines using single tape Turing machines, let's break down the explanation into key points:

Difference in Operation
- Multitape Turing machines are capable of performing multiple operations simultaneously on different tapes.
- Single tape Turing machines, on the other hand, can only perform one operation at a time on a single tape.

Simulation Process
- When simulating a multitape Turing machine on a single tape Turing machine, each operation on the multiple tapes needs to be simulated sequentially on the single tape.
- This sequential simulation process leads to an increase in computation time.

Ratio of Computation Time
- The increase in computation time when simulating multitape Turing machines using single tape Turing machines grows quadratically.
- This is because for each operation on the multiple tapes, the single tape Turing machine needs to perform multiple sequential operations to simulate the same effect.

Therefore, the correct answer is option C) quadratically, indicating that more computation time is needed in a quadratic ratio when simulating multitape Turing machines using single tape Turing machines.
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In what ratio, more computation time is needed to simulate multitape turing machines using single tape turing machines?a)doublyb)triplec)quadraticallyd)none of the mentionedCorrect answer is option 'C'. Can you explain this answer?
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