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Problem 1

Show that for any function \(f: \mathcal{N} \longrightarrow \mathcal{R}^{+}\), where \(f(n) \geq n\), the space complexity class \(\operatorname{SPACE}(f(n))\) is the same whether you define the class by using the singletape TM model or the two-tape read-only input TM model.

Short Answer

Expert verified
In conclusion, for any function \(f(n) \geq n\), the space complexity class \(\operatorname{SPACE}(f(n))\) is the same for both the single-tape Turing Machine and the two-tape read-only input Turing Machine models. This is demonstrated by showing that for any algorithm computed by a two-tape Turing Machine in \(k\) space, an equivalent algorithm can be computed by a single-tape Turing Machine in the same space \(f(n)\), thereby proving that the space complexity classes under both models are equivalent.
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Step 1: Understand Space Complexity

Space complexity describes the amount of memory an algorithm uses in relation to its input size. In computational complexity theory, it is represented as the function \( f(n) \), which denotes the maximum amount of cells a Turing Machine needs to solve a problem of size \( n \).

Step 2: Understand the Single-Tape Turing Machine

A single-tape Turing Machine consists of a tape- an infinite sequence of cells, a read/write head and a state register. For any computation on an input \( n \), the tape head moves left and right reading or writing symbols to the cells and changes its state according to the instructions programmed.

Step 3: Understand the Two-Tape Read-Only Input Turing Machine

A two-tape read-only input Turing Machine is similar to the single-tape Turing Machine but with a slight difference that it has two tapes instead of one. The first tape is a read-only input tape which has the input written on it, and the second tape is a working tape similar to the tape in a single-tape Turing Machine.

Step 4: Prove the Assertion

Now, the goal is to prove that for any function \( f(n) \geq n \) the space complexity class \(\operatorname{SPACE}(f(n))\) is the same using single-tape TM or the two-tape read-only input TM model. Assume a two-tape Turing Machine \( M' \), with read-only input tape and working tape, uses \( k \) space to solve a problem. To show that this same space complexity can be achieved by a Single-Tape Turing Machine \( M \), the idea is to let \( M \) simulate the operations of \( M' \) by storing the content of both the tapes of \( M' \) into its single tape. Since \( M' \)'s input is read-only and working space is \( k \), \( M \) can encode both tapes in the space \( f(n) = k \geq n \) (since \( f(n) \geq n\) by given condition). So, for any algorithm computed by \( M' \) in \( k \) space, you can find an equivalent algorithm computed by \( M \) in the same space \( f(n) \). Hence, the space complexity class under both models are equivalent, thereby proving the assertion.

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Most popular questions from this chapter

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