 Suggested languages for you:

Europe

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.

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.
See the step by step solution

## 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.

We value your feedback to improve our textbook solutions.

## Access millions of textbook solutions in one place

• Access over 3 million high quality textbook solutions
• Access our popular flashcard, quiz, mock-exam and notes features ## Join over 22 million students in learning with our Vaia App

The first learning app that truly has everything you need to ace your exams in one place.

• Flashcards & Quizzes
• AI Study Assistant
• Smart Note-Taking
• Mock-Exams
• Study Planner 