Consider the language L = {w ∈ {a,b,c}∗ | nw(a) = nw(b) = nw(c)}, where nw(z) is the number of occurrences of the symbol z in string w. In other words, L contains all strings that have an equal number of a’s, b’s, and c’s. The symbols may be in any order.
Describe a TM T that decides L. You may assume that a ⊔ symbol has been placed at the beginning of the tape.
Draw the state diagram of T. You do not need to include the rejecting state.
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Consider the language L = {w ∈ {a,b,c}∗ | nw(a) = nw(b) = nw(c)}, where nw(z) is the number of oc...
For a string s ∈ {0, 1} let denote the number represented by in the binary * s2 s numeral system. For example 1110 in binary has a value of 14 . Consider the language: L = {u#w | u,w ∈ {0, 1} , u } , * 2 + 1 = w2 meaning it contains all strings u#w such that u + 1 = w holds true in the binary system. For example, 1010#1011 ∈ L and 0011#100 ∈...
It wants a diagram Construct a TM M4 which accepts the language [w2w | w in (0,1)). This means, any input that comprises two copies of a binary string w separated by a single 2 symbol should be accepted. Any other input should be rejected. The input alphabet is (0,1,2). The tape alphabet contains 0, 1,2,_and you may use additional symbols 3, 4, , 9 if you wish (M4 can be constructed comfortably with just one additional symbol 3). The...
John Doe claims that the language L, of all strings over the alphabet Σ = { a, b } that contain an even number of occurrences of the letter ‘a’, is not a regular language. He offers the following “pumping lemma proof”. Explain what is wrong with the “proof” given below. “Pumping Lemma Proof” We assume that L is regular. Then, according to the pumping lemma, every long string in L (of length m or more) must be “pumpable”. We...
1. For a string s e 0, î, 2;" and a symbol d e { 0,1,2} let #(s, d) denote the number of times d appears in s. For example, #(0120012, 0)-3. Consider the language: {0, 1, 2. #(11,0) L- #(w, 1), #(11,2) #(w, 2) } . {utfw #(w, 0), #(11, 1) u, w, e For example, 2021 02#0011222 Construct a TM that decides this language. Provide a formal definition of your TM 1. For a string s e 0,...
Question 1. Let Σ = {a, b}, and consider the language L = {w ∈ Σ ∗ : w contains at least one b and an even number of a’s}. Draw a graph representing a DFA (not NFA) that accepts this language.
Question 1. Let Σ = {a, b}, and consider the language L = {w ∈ Σ ∗ : w contains at least one b and an even number of a’s}. Draw a graph representing a DFA (not NFA) that accepts this language. Question 2. Let L be the language given below. L = {a n b 2n : n ≥ 0} = {λ, abb, aabbbb, aaabbbbbb, . . .} Find production rules for a grammar that generates L.
. Let Σ = { a, b } , and consider the language L = { w ∈ Σ ∗ : w contains at least one b and an even number of a’s } . Draw a graph representing a DFA (not NFA) that accepts this language.
1. Construct a DFSM to accept the language: L = {w € {a,b}*: w contains at least 3 a's and no more than 3 b's} 2. Let acgt} and let L be the language of strings consisting of repeated copies of the pairs at, ta, cg, gc. Construct both a DFSM to accept the language and a regular expression that represents the language 3. Let a,b. For a string w E ', let W denote the string w with the...
Let ?= (a, b). The Language L = {w E ?. : na(w) < na(w)) is not regular. (Note: na(w) and nu(w) are the number of a's and 's in tw, respectively.) To show this language is not regular, suppose you are given p. You now have complete choice of w. So choose wa+1, Of course you see how this satisfies the requirements of words in the language. Now, answer the following: (a) What is the largest value of lryl?...
State diagrams for Turing Machines. Suppose you are given a string w ∈ {0,1}* placed on a Turing Machine tape. Give the state diagram for the Turing Machine required to take the initial string, w, and replace it on the tape with a new string, w′. The new string, w′, is formed by shifting the entire input string one cell to the right. Suppose you are given a string w ∈ {0, 1}* placed on a Turing Machine tape. Give...