91Ó°ÊÓ

Suppose that \(L\) is a subset of \(I^{*},\) where \(I\) is a nonempty set of symbols. If \(x \in I^{*},\) we let \(L / x=\left\\{z \in I^{*} | x z \in L\right\\} .\) We say that the strings \(x \in I^{*}\) and \(y \in I^{*}\) are distinguishable with respect to \(L\) if \(L / x \neq L / y .\) A string \(z\) for which \(x z \in L\) but \(y z \notin L,\) or \(x z \notin L,\) but \(y z \in L\) is said to distinguish \(x\) and \(y\) with respect to \(L .\) When \(L / x=L / y,\) we say that \(x\) and \(y\) are indistinguishable with respect to \(L .\) $$ \begin{array}{l}{\text { Let } L \text { be the set of all bit strings that end with } 01 . \text { Show }} \\ {\text { that } 11 \text { and } 10 \text { are distinguishable with respect to } L \text { and }} \\ {\text { that the strings } 1 \text { and } 11 \text { are indistinguishable with re- }} \\\ {\text { spect to } L .}\end{array} $$

Step by step solution

01

Define the Set L

The set \( L \) consists of all bit strings that end with '01', i.e., \( L = \{ x \, | \, x \, \text{ends with} \, \text{01} \} \).

02

Consider Subset L / 11

Determine \( L / 11 \). This consists of all strings \( z \) such that \( 11z \) ends with '01'. For example, if \( z = 01 \), then \( 1101 \in L \). Hence, \( L / 11 = \{ 01, 101, 00101, \ldots \} \).

03

Consider Subset L / 10

Determine \( L / 10 \). This consists of all strings \( z \) such that \( 10z \) ends with '01'. For example, if \( z = 1 \), then \( 101 \in L \). Hence, \( L / 10 = \{ 1, 001, 101, 0001, \ldots \} \).

04

Check for Distinguishability Between 11 and 10

Since \( L / 11 eq L / 10 \), 11 and 10 are distinguishable with respect to \( L \). For instance, the string '1' distinguishes them because '1101' is in \( L \) but '10101' is not.

05

Consider Subset L / 1

Determine \( L / 1 \). This consists of all strings \( z \) such that \( 1z \) ends with '01'. For \( z = 01 \), \( 101 \in L \). Hence, \( L / 1 = \{ 01, 001, 101, \ldots \} \).

06

Consider Subset L / 11 Again

From the earlier step, we know \( L / 11 = \{ 01, 101, 00101, \ldots \} \).

07

Check for Indistinguishability Between 1 and 11

Since \( L / 1 = L / 11 \), 1 and 11 are indistinguishable with respect to \( L \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

subset

A subset is a set which consists of elements from another set, known as the superset. In other words, if every element of a set A is also an element of a set B, then A is a subset of B, denoted as \( A \subseteq B \). For example, if we have a set of all even numbers, then the set containing {2, 4, 6} is a subset of that set. If we consider the problem where L consists of bit strings ending in '01', a subset L/x would be strings z where the concatenation xz ends in '01'. This is how we determine whether bit strings are distinguishable or indistinguishable in the problem context. This concept relies on forming subsets that match specific criteria (ending in '01' in this case). It's important to understand this foundation, as it underpins the logical steps in solving distinguishability problems.

bit strings

A bit string is a sequence made up of bits, where each bit is either 0 or 1. For instance, 101 and 0110 are bit strings. These strings are fundamental in computer science and digital logic.
Consider the set L in the given problem, which includes all bit strings ending with 01. Analyzing bit strings involves checking their ending bits and understanding how they transform when other bits are added or manipulated.
In the context of the exercise, different bit strings like '11' and '10' or '1' and '11' are checked for their ending properties relative to set L. When augmented by additional bits, we create new potential strings like 1101 or 101. The key skill here is recognizing how the bit arrangements impact whether a string belongs to L or not.

distinguishability

Distinguishability between two strings x and y with respect to a set L determines whether there exists some string z where only one of xz or yz belongs to L. If such a z exists, x and y are distinguishable; if not, they are indistinguishable.
For instance, in the problem above, we see that the bit strings '11' and '10' are distinguishable because adding '1' results in one string fitting the set L and the other not.
On the other hand, '1' and '11' are indistinguishable because, when we check L/1 and L/11, both subsets contain strings ending in '01,' making their memberships identical in L.
Understanding distinguishability helps in predicting and contrasting behavior of different string operations and comparing outcomes in various substring contexts.