Understanding the Language L

The language L is given by \(L = \{2^{nk} \mid k > 0, \text{ and } n \text{ is non − negative integer number}\}\). The alphabet is \(\Sigma = \{2\}\), meaning strings are sequences of the symbol '2'. The notation \(2^m\) represents the string consisting of '2' repeated \(m\) times. For example, \(2^3 = 222\). The length of the string \(2^m\) is \(m\).

The definition of the language involves exponents of the form \(nk\), where \(k\) is a positive integer (\(k \in \{1, 2, 3, \dots\}\)) and \(n\) is a non-negative integer (\(n \in \{0, 1, 2, 3, \dots\}\)).

Interpreting the Role of 'n'

The presence of 'n' in the options suggests that 'n' is a fixed parameter defining the specific language we are considering, rather than a variable that ranges over all non-negative integers within the set definition itself. If 'n' were a variable in the set definition, the language would be the union of languages defined by each possible value of 'n'. Let's explore this possibility briefly:

• If \(n=0\), the exponents are \(0 \cdot k = 0\) for all \(k > 0\). The only exponent is 0. The language would be \(\{2^0\} = \{\epsilon\}\).
• If \(n=1\), the exponents are \(1 \cdot k = k\) for all \(k > 0\). The exponents are \(\{1, 2, 3, \dots\}\). The language would be \(\{2^1, 2^2, 2^3, \dots\} = \{2, 22, 222, \dots\}\).
• If \(n=2\), the exponents are \(2 \cdot k\) for all \(k > 0\). The exponents are \(\{2, 4, 6, \dots\}\). The language would be \(\{2^2, 2^4, 2^6, \dots\}\).
• If \(n=3\), the exponents are \(3 \cdot k\) for all \(k > 0\). The exponents are \(\{3, 6, 9, \dots\}\). The language would be \(\{2^3, 2^6, 2^9, \dots\}\).

If 'n' in the language definition was a variable, the language L would be the union of all these sets: \(\{\epsilon\} \cup \{2^1, 2^2, \dots\} \cup \{2^2, 2^4, \dots\} \cup \{2^3, 2^6, \dots\} \cup \dots\). The set of all exponents would be \(\{0, 1, 2, 3, \dots\}\). The language would be \(L = \{2^m \mid m \ge 0\} = \{2\}^* = \{\epsilon, 2, 22, 222, \dots\}\). The minimum DFA for \(2^*\) has 1 state. However, none of the options evaluate to 1 for arbitrary non-negative integer \(n\).

Therefore, the most likely interpretation, consistent with the options provided, is that \(n\) represents a fixed non-negative integer parameter for which we need to find the minimum number of DFA states.

Analyzing the Language for a Fixed 'n'

Let's assume \(n\) is a fixed non-negative integer, and \(L_n = \{2^{nk} \mid k > 0\}\).

• Case n = 0: \(L_0 = \{2^{0 \cdot k} \mid k > 0\} = \{2^0\} = \{\epsilon\}\). The language contains only the empty string. The minimum DFA for \(\{\epsilon\}\) requires 2 states: a start state that is final, and a non-final state for any input symbol '2'. This doesn't match any of the options if we substitute \(n=0\) into them (options become 0, 1, 1, 0).
• Case n ≥ 1: \(L_n = \{2^{nk} \mid k > 0\}\). The exponents are positive multiples of \(n\): \(n, 2n, 3n, \dots\). The language is the set of strings of '2's whose length is a positive multiple of \(n\). \(L_n = \{w \in \{2\}^* \mid |w| > 0 \text{ and } |w| \pmod n = 0\}\).

We need to construct a minimal DFA for \(L_n = \{w \mid |w| > 0 \text{ and } |w| \pmod n = 0\}\) for \(n \ge 1\).

Constructing the Minimal DFA for \(L_n\) (\(n \ge 1\))

To accept strings whose length is a positive multiple of \(n\), a DFA needs to keep track of the length of the input string modulo \(n\). However, it also needs to distinguish the case of length 0 from positive lengths that are multiples of \(n\).

Let's define states based on the length modulo \(n\), but separating the initial state representing length 0.

• Let \(q_{start}\) be the initial state. It represents having read a string of length 0. Since the empty string \(\epsilon\) (length 0) is not in \(L_n\) for \(n \ge 1\), \(q_{start}\) is not a final state.
• Let \(q_i\) represent the state reached after reading a non-empty string whose length is congruent to \(i \pmod n\), for \(i \in \{0, 1, \dots, n-1\}\).

We need \(1 + n = n+1\) states: \(q_{start}, q_0, q_1, \dots, q_{n-1}\).

Transitions:

• From \(q_{start}\) on reading '2', we have read 1 symbol. Length is 1, and \(1 \pmod n = 1\) (assuming \(n \ge 1\)). So, \(\delta(q_{start}, '2') = q_1\).
• From a state \(q_i\) (representing non-empty length \(\equiv i \pmod n\)) on reading '2', the length becomes congruent to \((i+1) \pmod n\). So, \(\delta(q_i, '2') = q_{(i+1) \pmod n}\) for \(i \in \{0, 1, \dots, n-1\}\).

Final states:

A string \(w\) is accepted if it is non-empty and its length is a multiple of \(n\). A non-empty string \(w\) of length \(m\) ends up in state \(q_{m \pmod n}\).

If \(|w| \pmod n = 0\), the string ends in state \(q_0\).

Since strings in \(L_n\) have positive length that is a multiple of \(n\), they end up in state \(q_0\), and they should be accepted. Therefore, \(q_0\) is a final state.

Strings with length \(\equiv i \pmod n\) where \(i \in \{1, \dots, n-1\}\) are not in \(L_n\). These end up in states \(q_1, \dots, q_{n-1}\). These states are not final.

Summary of the DFA:

• States: \(Q = \{q_{start}, q_0, q_1, \dots, q_{n-1}\}\)
• Alphabet: \(\Sigma = \{2\}\)
• Start state: \(q_{start}\)
• Final states: \(F = \{q_0\}\)
• Transitions:
  • \(\delta(q_{start}, '2') = q_1\)
  • \(\delta(q_i, '2') = q_{i+1}\) for \(0 \le i < n-1\)
  • \(\delta(q_{n-1}, '2') = q_0\)

This DFA has \(1 + n = n+1\) states.

Minimality of the DFA

To show this is the minimum number of states, we can argue that all \(n+1\) states are distinguishable.

• \(q_{start}\) is not final, while \(q_0\) is final. So \(q_{start}\) and \(q_0\) are distinguishable.
• For \(i \in \{1, \dots, n-1\}\), \(q_i\) is not final. Can \(q_i\) be distinguished from \(q_{start}\)? From \(q_i\), reading \(n-i\) '2's leads to state \(q_{(i + n-i) \pmod n} = q_0\), which is final. From \(q_{start}\), reading \(n-i\) '2's leads to state \(q_{n-i}\) (since \(n-i \ge 1\)), which is not final (as \(n-i \ne 0\)). So \(q_i\) is distinguishable from \(q_{start}\).
• For \(i, j \in \{0, 1, \dots, n-1\}\) with \(i \ne j\). From \(q_i\), reading \(n-i \pmod n\) '2's leads to state \(q_0\), which is final. From \(q_j\), reading \(n-i \pmod n\) '2's leads to state \(q_{(j + (n-i \pmod n)) \pmod n}\). Since \(i \ne j\), \((j + (n-i \pmod n)) \pmod n \ne 0\). So it leads to a non-final state. Thus, \(q_i\) and \(q_j\) are distinguishable.

All \(n+1\) states are distinguishable, so the minimum number of states required is \(n+1\).

This result matches option 2.

Conclusion

Based on the analysis where 'n' is a fixed positive integer parameter, the language consists of strings of '2's whose length is a positive multiple of 'n'. The minimum number of states in a DFA accepting this language is \(n+1\).

The interpretation for \(n=0\) leads to 2 states, which is not \(n+1=1\), suggesting the problem intends \(n \ge 1\) or there is an inconsistency for \(n=0\). Given the options, the result \(n+1\) for \(n \ge 1\) is the intended answer.

Revision Table: Finite Automata and Language {2^nk}

Additional Information: Regular Languages and Minimization

The language \(L_n = \{2^{nk} \mid k > 0\}\) for a fixed \(n \ge 1\) is a regular language. Regular languages are precisely those that can be accepted by finite automata. The set of regular languages over a single-symbol alphabet \(\Sigma=\{a\}\) is the set of languages of the form \(a^S\) where \(S\) is a semilinear set of non-negative integers.

In our case, for \(n \ge 1\), the set of lengths is \(\{n, 2n, 3n, \dots\} = \{m \mid m = nk, k \ge 1\}\). This set is arithmetic progression, which is a simple form of a semilinear set.

For \(n=0\), the set of lengths is \(\{0\}\), also a semilinear set.

Every regular language has a unique minimum-state DFA (up to isomorphism). Algorithms like the Myhill-Nerode theorem or partitioning algorithms (like Hopcroft's algorithm) are used for DFA minimization.

The Myhill-Nerode theorem states that the minimum number of states in a DFA for a language L is equal to the number of equivalence classes of the Myhill-Nerode relation (\(u \sim_L v\) if and only if for all strings \(w\), \(uw \in L \Leftrightarrow vw \in L\)). The distinguishable states identified in our construction correspond directly to these equivalence classes.