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I'm trying to prove that for every multihead Turing machine X, there is a multihead Turing machine y such that for any input string z, we have volume(X, z) = Θ(Y(z)) and volume(Y,z) = Θ(Y(z)). In other words, the three values are all equal up to a constant. The volume is defined as the summation of all head movements during the computation. The tape is one-way infinite to the right.

So far, I have a representation that goes as the follow. Let Y have 3 areas for info. The areas do as follows:

  1. Simulates X
  2. Counts the current number of cells in use by X:
    • Binary counter where the least significant bit is read first
    • Initially every other symbol is a 1 i.e | |1| |1| |....
    • The * symbol will indicate where the head currently is
    • For each move right in (1) a 1 gets written at the beginning |1*|1| |1|...,
    • For each move left in (1) a 0 gets written at the beginning |0*|1| |1|
    • If there is no shift in (1) after that step, don't do anything, continue to next transition
    • Adding, subtracting, or no change, add the current amount to (3)
  3. Accumulation of volume (#heads used throughout X's computation):
    • From the head's position in (2) add a 1 in the same position in (3)
    • When the most significant bit in (2) is to be added (as like above), transition the head to (3), mark that position with a 1, transition back to (2), then add the 1 (namely, place the counter head in that spot as the 1 is already there)

At the end of (1)'s computation, (when the machine halts and all heads have dropped off the tape), the output of Y should be computed volume for X.

I am currently trying to figure out a relation between X, Y, and there runtime/volumes to prove my observation from above.

Telo Springs
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  • I think it's more suitable for http://cstheory.stackexchange.com/ – lared Feb 09 '15 at 22:42
  • Perhaps you can clarify part of the question. The statement provided is trivially true by allowing the machines Y and X to be identical. Is there an expected restriction that Y must meet? – Bill Province Feb 10 '15 at 23:21
  • @BillProvince The only restriction is just that, X is not the same a Y. Otherwise, the way Y is setup as described above is as. – Telo Springs Feb 11 '15 at 23:20
  • Sorry, still not clear on the requirements for Y. Again, I can make Y trivially different from X by just adding a new state to Y that is never entered. Or, I could add a 'do-nothing' pair of states to X in Y that merely move the tape head left and right a single time then continue on the regular logic of machine X. – Bill Province Feb 12 '15 at 22:48
  • A Google search for J.F. Costa's paper **"Turing Machines as Clocks, Rulers, and Randomizers"** is a good place to start for some existing work in this area. – Bill Province Feb 12 '15 at 23:27

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