Theory of Computation

Question

Manu · Accepted Answer

Step 1: (a) Define M formally with a clear definition of the 5-tuple. A nondeterministic finite automaton (NFA) M can be defined as a 5-tuple (Q, Σ, Δ, q₀, F), where: Q is a finite set of states, Σ is a finite set of input symbols (alphabet), Δ: Q × Σ → P(Q) is the transition function (where P(Q) represents the power set of Q), q₀ ∈ Q is the start state, F ⊆ Q is the set of accept states. Explanation:  For language A which is the set of all binary strings that either have an odd number of 0's, or the number of 1's is not a multiple of 3, or both, we need to construct an NFA where Q, Σ, Δ, q₀, and F are designed to satisfy the described properties. Step 2: (b) Draw the state diagram of M. To draw the state diagram, we must determine the states and transitions that will allow the automaton to accept the language A. Let 𝑀 = ( 𝑄 , Σ , Δ , 𝑞 0 , 𝐹 ) M = ( Q , Σ , Δ , q 0 ​ , F ) where: 𝑄 = { 𝑞 0 , 𝑞 1 , 𝑞 2 , 𝑞 3 , 𝑞 4 , 𝑞 5 , 𝑞 6 , 𝑞 7 } Q = { q 0 ​ , q 1 ​ , q 2 ​ , q 3 ​ , q 4 ​ , q 5 ​ , q 6 ​ , q 7 ​ } Σ = { 0 , 1 } Σ = { 0 , 1 } Δ Δ is defined by the transition table we will design next. 𝑞 0 q 0 ​ is the start state. 𝐹 = { 𝑞 1 , 𝑞 3 , 𝑞 5 , 𝑞 6 , 𝑞 7 } F = { q 1 ​ , q 3 ​ , q 5 ​ , q 6 ​ , q 7 ​ } (accept states which correspond to strings with an odd number of 0's, a number of 1's that is not a multiple of 3, or both). The transition function Δ Δ needs to account for the two conditions of language A. We'll need states that keep track of whether we've seen an odd number of 0's and states that keep track of the count of 1's mod 3. The state diagram will depict the states and transitions based on the 0's and 1's in the input string. Here is the description of the state transitions: From the start state 𝑞 0 q 0 ​ , upon reading a '0', we should go to an accept state for an odd number of 0's, which is 𝑞 1 q 1 ​ . Upon reading a '1', we should go to a non-accept state 𝑞 2 q 2 ​ since we have a count of 1's that is 1 mod 3. From 𝑞 1 q 1 ​ (odd 0's), reading a '0' should transition back to 𝑞 0 q 0 ​ , and reading a '1' should go to 𝑞 3 q 3 ​ which is also an accept state because we still have an odd number of 0's. From 𝑞 2 q 2 ​ , reading a '0' should transition to 𝑞 3 q 3 ​ (accept state), and reading a '1' should go to 𝑞 4 q 4 ​ . From 𝑞 3 q 3 ​ , reading a '0' should transition to 𝑞 2 q 2 ​ , and reading a '1' should go to 𝑞 5 q 5 ​ (accept state). From 𝑞 4 q 4 ​ , reading a '0' should transition to 𝑞 5 q 5 ​ (accept state), and reading a '1' should go back to 𝑞 0 q 0 ​ . From 𝑞 5 q 5 ​ , reading a '0' should transition to 𝑞 4 q 4 ​ , and reading a '1' should go to 𝑞 6 q 6 ​ . 𝑞 6 q 6 ​ is a "trap" state where any input will keep the state within 𝑞 6 q 6 ​ or 𝑞 7 q 7 ​ , and since these are non-multiple of 3 states for 1's, we stay in them upon reading a '1'. 𝑞 7 q 7 ​ transitions similarly to 𝑞 6 q 6 ​ , trapping the automaton in states that signify an odd number of 0's and non-multiple of 3 number of 1's. For simplicity, we can merge 𝑞 6 q 6 ​ and 𝑞 7 q 7 ​ since they represent similar concepts, but for this exercise, we will keep them separate to clearly differentiate the odd number of 0's scenario. With these transitions in mind, I'll create the state diagram to illustrate this NFA. Explanation : The transitions between states occur based on the input symbol (0 or 1), and certain states are designed to "trap" the automaton in an accepting condition once it's reached (for instance, when we have an odd number of 0's or a count of 1's that isn't a multiple of 3). Step 3: (c) Is A the only language M recognizes? No, A is not the only language that M recognizes. A nondeterministic finite automaton (NFA) can recognize several different languages depending on its configuration and the definition of its accept states. The language A described in the problem is the primary language for which M is designed, but there could be other strings not in A that M could accept due to its nondeterministic nature, especially if M has multiple accept states or nondeterministic transitions that could lead to an accept state by different paths. Explanation : An NFA can accept multiple languages because it can have nondeterministic transitions leading to accept states for different subsets of strings. Step 4: (d) Give an example string from A and show M accepts that string. Let's consider the example string "0101" which belongs to the language A because it has an odd number of 0's (two 0's) and the number of 1's is not a multiple of 3 (two 1's). Here's how the NFA M accepts the string "0101": Start at state q0 : Initially, we are in the starting state 𝑞 0 q 0 ​ , which represents an even number of 0's and a count of 1's that is a multiple of 3. Read the first symbol, '0' : Transition from 𝑞 0 q 0 ​ to 𝑞 1 q 1 ​ . State 𝑞 1 q 1 ​ represents an odd number of 0's (one 0). Read the second symbol, '1' : Transition from 𝑞 1 q 1 ​ to 𝑞 3 q 3 ​ . State 𝑞 3 q 3 ​ represents an odd number of 0's and a count of 1's that is 1 mod 3 (one

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