Question

Analyze the time and space efficiency of the following Turing machine operating on a unary string of length \(n\). \((1,1,1,1, R)\) \((1, b, b, 2, L)\) \((2,1,0,2, L)\) \((2, b, b, 3, R)\) \((3,0,1,3, R)\)

Short answer

The Turing machine has a time complexity of \(O(n)\) and a space complexity of \(O(n)\).

Step-by-step solution

Understand the Turing Machine Transitions

The Turing machine has several transitions. Transition (1,1,1,1, R) means if in state 1 and reading a '1', remain in state 1, write a '1', and move right. Transition (1, b, b, 2, L) moves to state 2 when reading 'b' from state 1, writing 'b' and moving left. The transitions continue similarly, outlining how the machine processes the input string.

Initial Analysis of State Transitions

The machine begins in state 1, traversing right across all '1's until it encounters a blank ('b'), then it moves to state 2 and shifts left. In state 2, it changes '1's to '0's until reaching another blank at the start, converting back to state 3 and shifting right. In state 3, all '0's are transformed back to '1's.

Determine Time Complexity

The machine in state 1 moves right across the entire tape of length n, requiring exactly n steps. In state 2, it moves back across the entire tape from the blank space, which involves n additional steps to change every '1' to '0'. Transition to state 3 involves reading n '0's to change them back to '1's. Therefore, each phase requires a full traversal of n steps, resulting in a total time complexity of \(O(n)\).

Determine Space Complexity

Space complexity is determined by the maximum amount of tape used. Given that the unary string already occupies n spaces, and no additional space is required beyond the input length, the space complexity is \(O(n)\), corresponding to the length of the input.

Key concepts

Time Complexity

Understanding the time complexity of a Turing machine involves analyzing how many steps are required for the machine to accomplish its task relative to input size. In our given machine, the task is executed by a finite set of state transitions across a unary string of length \( n \).

In state 1, the Turing machine traverses right across the tape of length \( n \), effectively examining each '1' and moving one cell to the right at each step. This process alone requires exactly \( n \) steps. When moving to state 2, the machine reverses direction, moving left across the tape.

Again, it requires \( n \) steps to transform all '1's into '0's. In state 3, the Turing machine moves right again, requiring another \( n \) steps to transform '0's back to '1's. Therefore, each phase of traversal – one in each state – requires \( n \) steps, summing up to a total time complexity of \( O(n) \). This linear complexity indicates that the time taken increases directly with the size of the input string.

Space Complexity

Space complexity refers to the amount of tape space utilized by the Turing machine relative to the input size. In the given problem, the input itself is a unary string of length \( n \).

The Turing machine does not require any additional space beyond this input length, making the total space allocated equivalent to \( n \). Each transition writes data within the existing space but does not extend past the initial input. This means that there are no additional tape cells being used.

As such, the space complexity is classified as \( O(n) \), which means that the space requirement grows linearly with the size of the input tape. Therefore, the space complexity is efficient, as it directly corresponds to the input size without adding unnecessary overhead.

State Transitions

State transitions in a Turing machine describe the series of operations performed on the input. These transitions define how the machine reads, writes, and shifts direction based on current states and tape symbols.

In the provided Turing machine, several state transitions guide its computation. Initially, the machine resides in state 1, moving right across any sequence of '1's until encountering a blank ('b'). This provides a straightforward mechanism for processing the input in one direction. Upon reaching 'b', the machine transitions to state 2, starts moving left, and changes '1's to '0's.

This leftward movement continues until another 'b' is identified, prompting a transition to state 3. In state 3, the process involves transforming '0's back into '1's while moving right again. These clearly defined transitions ensure that the machine effectively processes the tape, enabling the desired computation to reach its final state.