This documentation contains a part about expressions in general, a part about the regular. Explain.c. convert your NFA to DFA and explain what NFA states each DFA state represents. Question: Consider the RE (ab + ba + a)*a. convert this RE to an NFA. explain the steps.b. simplify the NFA as much as possible. "/>
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\$\begingroup\$ I think I have managed to produce one that accepts all words starting ab. I am having trouble with implementing the ending with ab. I have a ton of these questions in exercises and wanted to example to go by. \$\endgroup\$ –.

The r.e. is as follows:. Here are a DFA accepting strings ending with ab. Homework 6 Homework 6 307 1. Prove that the following CFG G in Greibach normal form generates exactly the set of nonnull strings over {a, b} with equally many a's and b's: S-aB IbA, A-aSlbAAla, B -bS I aBB I b..

The set {∧, ab } is represented by the regular expression ∧ + ab b. Nfa ending with ab or ba My question is Accept all strings containing " 011 " or " 001 " as a substring and should not contain " 010 " as substring for the following languages over the alphabet {0,1} i have solve it. I am trying to understand the answer here for FA that accepts only the words baa, ab , ... (Don't think of q2 as an ending state --- the program does NOT terminate once it enters q2. It is merely that when the program is in q2, it can accept the input string so far. Think of q2 as an "accepting state".) \$\endgroup\$. .. I am trying to understand the answer here for FA that accepts only the words baa, ab , ... (Don't think of q2 as an ending state --- the program does NOT terminate once it enters q2. It is merely that when the program is in q2, it can accept the input string so far. Think of q2 as an "accepting state".) \$\endgroup\$. ..

Jun 24, 2019 · L1 = {ab, abbab, abaab, .....} Here as we can see that each string of the above language ends with ‘ab’ but the below language is not accepted by this NFA because some of the string of below language does not end with ‘ab’. L2 = {bba, abb, aaabbbb, .....} The state transition diagram of the desired language will be like below:.

How to get each of the five endings for Barbara's Hangout date in Genshin Impact with a list of all the right choices. How to Get All Barbara Hangout Endings.In Barbara's hangout event there are five total endings.Two are definitely bad, two are definitively. L=L1 U L2 Where L1=(ab)*(ba)* L2=aa* Er.Deepinder Kaur 34. NFA for (ab)*(ba)* q1 q2 ba ab ba start q3 a a q4 NFA for aa*.

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An NFA that accepts all strings over {0,1} that contain a 1 either at the third position from the end or at the second position from the end. 0,1 q 1 0,1,ε 0,1 1 q 2 q 4 q 3 • There are two edges labeled 1 coming out of q1. then we expect to see at least one more aafter the sequence of b’s, and so on. The NFA below captures strings that start and end with the same symbol (aor b). q0 q1 q3 q4 q2 a b a , b a a b b a ,b b a 4. Closure Properties of Regular Languages: Prove that if Lis a regular language over alphabet , then so is L0= fxyzjx;y;z2 ^x;z62L ^y2Lg.

NFA stands for non-deterministic finite automata. ... {ε, a, aa, b, bb, ab, ba, aba, bab, .....}, any combination of a and b. The (a + b)* shows any combination with a and b even a null string. Example 1: Write the regular expression for the language accepting all the string which are starting with 1 and ending with 0, over. 41. Build an NFA Ml that accepts (ab)* and an NFA M2 that accepts (ba)*. Use transitions to obtain a machine M that accepts (ab)*(ba)*. Give the input transition function of M. Use Algorithm 5.6.3 to construct the state diagram of a DFA that accepts 42. Build an NFA Ml that accepts (aba)+ and an NFA M2 that accepts (ab)*. Use.

Names Ending with Nfa . Baby name ideas for boys and ... Obtain a DFA to accept strings of a's and b's ending with ab or ba. 14. Design NFA to accept the strings abc, acd and abcd. 15. Construct DFA for the regular expression ab* + b. 16. Prove that the language L = {WW¶We (a + b)*} is not regular. 6. Design an NFA for each of language given in Question 5. Answers (a) Figure 8: NFA for a(ba)b(credited: Connor Dooley) (b) Figure 9: NFA for ( [b)a(credited: Connor Dooley) 7. Give regular expressions for the following languages, where = f0;1g (a) fw: wcontains exactly two 0’s g (b) fw: wcontains at least two 0’s and at most one 1 g.

A string length n, ending in aab can be accepted by the above NFA if: The machine remains in q0 for the first n-3 inputs. On the last three inputs aab, the machine makes the transition as shown below: To reach the final state, and to remain there, it is necessary that the two character of the string should be 10.

2. For the alphabet (0, b) give regular expression and draw finite automata for the following languages: a) 1: = All strings. e) is = All strings that contain the substring ab. b) L2 = All strings starting with ba. 1) L. = All strings that contain the substring ab or ba. c) L = All strings ending with ba. 8) L = All strings containing exactly.

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chevy truck wont go into gear. house of time clock repair. truck parking usa. The state transition diagram of the language containing 'a' just followed by 'bb' will be like: In the above DFA , state 'v' is the initial and final state too which on getting 'b' as the input it remains in the state of itself and on getting 'a' as the input it transit to a.

Feb 22, 2022 · \$+ab(a+b)*(ab+bb+ba) the empty string indicated with \$ is accepted as it says all the words, then the compulsory ab concatenated with any word (a+b)* then also concatenated with all possible ending that are not aa. However this does not accept ab and or abb which should be accepted. Maybe someone has better suggestions?. The NFA that accepts strings ending with "ab" is given by Subproblem 2. The NFA that accepts strings ending with "ba" is given by These two machines can be combined as follows: Add a new start state q 0, with two epsilon transitions to the start states of the two NFAs q 1 and q 4. Now q 0 is the new start state of the combined ε-NFA. 3. NFA that accepts strings that starts with two a's and ends with two a's. Σ = {a, b} Try on your own first. Check here for the NFA. 4. NFA that accepts strings that end with "ab" or "ba". Σ = {a, b} This is left as exercise. 5. NFA that accepts strings that starts and ends with different symbols Σ = {a, b} This is left as an exercise. ε ....

A regular expression for ending with abb; A regular expression for all strings having 010 or 101. Option 4: (ba*a + ab*b)* (ab* + ba*) ... Here w is a string of length n, so it requires a minimum of n+1 states accepted by NFA. Hence the correct answer is n+1. ... This is not a correct combination of {ab,aa,baa}. dfa construction problems.

In this video solution is explained for following problem : Construct a DFA for all Strings ending with either ab or ba >.#DFA #FA #FiniteAutomata #Determinist.. Design a Machine for L = “Not containing substring ‘001’ ” Difference between DFA and NFA : Minimum number of states. Nfa ending with ab or ba willoughby municipal court active warrant list. a( ba )* Regular expression for every odd position is b defined over {a,b} b( ab )* More Examples of Regular Expression Regular Expression for no 0 or many triples of 0’s and many 1 in the strings. RegExp for strings of one or many 11 or no 11. A regular expression for ending with abb; A regular expression for all strings having 010 or 101. Wiki formatting help page on yuba city shooting 2022.

NFA that accepts strings that starts with two a's and ends with two a's. Σ = {a, b} Try on your own first. Check here for the NFA. 4. NFA that accepts strings that end with "ab" or "ba". Σ = {a, b} This is left as exercise. 5. NFA that accepts strings that starts and ends with different symbols Σ = {a, b} This is left as an exercise. ε.

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Regular Expressions . A Regular Expression can be recursively defined as follows −. ε is a Regular Expression indicates the language containing an empty string.. Jun 14, 2021 · Construct NFA with epsilon for a given language L= {ab, ba}. Follow the steps given below −. Step 1 − NFA with epsilon for (a+b) is given below −. It accepts either a or b as an input, and both go to the final state. Step 2 − NFA with epsilon for ab is as follows −. Concatenating a and b with epsilon, and a must followed by b then ....

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The NFA below captures strings that start and end with the same symbol (aor b). q0 q1 q3 q4 q2 a b a , b a a b b a ,b b a 4. Closure Properties of Regular Languages: Prove that if Lis a regular language over alphabet , then so is L0= fxyzjx;y;z2 ^x;z62L ^y2Lg.. The textbook example is also quite good: (ab∪aba)*. A DFA is presented in Figure 2-4 and an NFA for it in Figures 2-5 and 2.6. You can see how much more simple and intuitive the NFAs are. Example 3 As a final addition to our NFA's structure we consider this example: (ab)*∪(aba)*. This is, of course, different than the former example (ab∪. Solution. Step 1 −. Here the ε transition is between q1 and q2, so let q1 is X and qf is Y. Here the outgoing edges from q f is to q f for inputs 0 and 1. Step 2 −. Now we will Copy all these edges from q 1 without changing the edges from q f and get the following FA −. Step 3 −.

Design an NFA with ∑ = {0, 1} accepts all string ending with 01. Solution: Hence, NFA would be: Example 3: Design an NFA with ∑ = {0, 1} in which double '1' is followed by double '0'. Solution: The FA with double 1 is as follows: It should be immediately followed by double 0. Then, Now before double 1, there can be any string of 0 and 1.. Nov 11, 2020 · Both structures are allowed here but as it is ∈-NFA so the second structure is recommended. ∈-NFA for L = {ab, ba} : Following the above-mentioned rules, ∈-NFA of Regular Language L = {ab, ba} is to be constructed. The language consists of ab or ba, which can also be written as (ab + ba).. In this video solution is explained for following problem : Construct a DFA for all Strings ending with either ab or ba .#DFA #FA #FiniteAutomata #Determinist.

I am trying to understand the answer here for FA that accepts only the words baa, ab , ... (Don't think of q2 as an ending state --- the program does NOT terminate once it enters q2. It is merely that when the program is in q2, it can accept the input string so far. Think of q2 as an "accepting state".) \$\endgroup\$. .. c. Convert all NFA that are generated in part a into GTG? Question No. 2 a. Convert following RE into NFA? (ab+ba)* (aa+bb) b. Convert all NFA that are generated in part a into DFA?. 04-29: NFA Examples Create an NFA for the language Give an NFA for the language L = All strings over {0,1} that contain two pairs of adjacent 0's separated by an. In this video solution is explained for following problem : Construct a DFA for all Strings ending with either ab or ba >.#DFA #FA #FiniteAutomata #Determinist..

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COS3701 - SUMMARIZED EXAM NOTES 2021 . Theoretical Computer Science III. Theory of Computation 4 Mustansiriya University – college of sciences – computer science department – second class Language language is the set of all strings of terminal symbols derivable from alphabet. alphabet is a finite set of symbols. cobra crossbow pistol mods. mce logistics inc. esx to qbus script. xmlworkerhelper java example. 3. NFA that accepts strings that starts with two a's and ends with two a's. Σ = {a, b} Try on your own first. Check here for the NFA.4. NFA that accepts strings that end with "ab" or "ba".Σ = {a, b} This is left as exercise. 5. NFA that accepts strings that starts and ends with different symbols.

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I am trying to understand the answer here for FA that accepts only the words baa, ab , ... (Don't think of q2 as an ending state --- the program does NOT terminate once it enters q2. It is merely that when the program is in q2, it can accept the input string so far. Think of q2 as an "accepting state".) \$\endgroup\$.. Apr 06, 2021 · Here, state names are in brackets ([..]) and state names that end with "#" are terminal. Generally it's considered easier to make an NFA than a DFA, so the usual method is to first make an NFA and then make the DFA by modifying the NFA by changing multiple output states to a single intermediate state..

P : A → aB, a, ε (where ε is the empty word) B → bA, b For this regular grammar, create an equivalent NFA. A regular grammar is a 4 tuple G = (N, Σ, P, S). So our NFA is made up of 2 states, A and B. Names Ending with Nfa.Baby name ideas for boys and girls. Discover names that end with a. We have many different letter combinations like Nfa for names. ∈-NFA for ab: For concatenation, a must be followed by b.Only then it can reach the final state. Both structures are allowed here but as it is ∈-NFA so the second structure is recommended. ∈-NFA for L = {ab, ba} : Following the.

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( ab ∪ a)*. This is so because aab can be formed by one application a, followed by one of ab . So it is redundant inside a Kleene star. Now we can write a two state machine: a a b If you put the loop on a on the start state, either in place of where we have it, or in addition to it, it's also right.. In this video solution is explained for following problem : Construct a DFA for all Strings ending with either ab or ba >.#DFA #FA #FiniteAutomata #Determinist.. This documentation contains a part about expressions in general, a part about the regular. Explain.c. convert your NFA to DFA and explain what NFA states each DFA state represents. Question: Consider the RE (ab + ba + a)*a. convert this RE to an NFA. explain the steps.b. simplify the NFA as much as possible.

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13. Obtain a DFA to accept strings of a's and b's ending with ab or ba. 14. Design NFA to accept the strings abc, acd and abcd. 15. Construct DFA for the regular expression ab* + b. 16. Prove that the language L = {WW¶We (a + b)*} is not regular. Wr is the reverse of the string W. 17. Obtain grammar for the following DFA. 18. c. { ab , a, b, bb} Solution a. The set {∧, ab } is represented by the regular expression ∧ + ab b. (b)Draw the state diagram of the NFA of the following languages: (A)* B For full credit your NFA should have no more than six states and the minimal number of transitions in the diagram.

\$\begingroup\$ I think I have managed to produce one that accepts all words starting ab. I am having trouble with implementing the ending with ab. I have a ton of these questions in exercises and wanted to example to go by. \$\endgroup\$ –. Apr 06, 2021 · NFA Nfa ending with ab or ba baston edge french bulldogs melbourne vic python card game source code 3 inch iso board mercy lab durango DFA for the language of all those strings starting and ending with b. DFA for ending with b. DFA for the string of even A's and even b's.

6. Design an NFA for each of language given in Question 5. Answers (a) Figure 8: NFA for a(ba)b(credited: Connor Dooley) (b) Figure 9: NFA for ( [b)a(credited: Connor Dooley) 7. Give regular expressions for the following languages, where = f0;1g (a) fw: wcontains exactly two 0’s g (b) fw: wcontains at least two 0’s and at most one 1 g.

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There are 490 words that end with AB. Click on a word ending with AB to see its definition. → 9 2-letter words in ab - ending in ab:. ab-. Draw a DFA for the language accepting strings starting with 'ab' over input alphabets ∑ = {a, b} Solution- Regular expression for the given language = ab(a + b)* Step-01: All strings of the language.

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2019. 6. 24. · Problem-1: Construction of a minimal NFA accepting a set of strings over {a, b} in which each string of the language ends with ‘ab’. Explanation: The desired language will be like: L1 = {ab, abbab, abaab, .....} Here as we can see that each string of the above language ends with ‘ab’ but the below language is not accepted by this NFA because some of the string of below. 13. Obtain a DFA to accept strings of a's and b's ending with ab or ba. 14. Design NFA to accept the strings abc, acd and abcd. 15. Construct DFA for the regular expression ab +b. 16. Prove that the language L = {WWWE(a + b)"} is not regular. Wis the reverse of the string W. 17. Obtain grammar for the following DFA. )a,b a 18. Define Turing ....

41. Build an NFA Ml that accepts (ab)* and an NFA M2 that accepts (ba)*. Use transitions to obtain a machine M that accepts (ab)*(ba)*. Give the input transition function of M. Use Algorithm 5.6.3 to construct the state diagram of a DFA that accepts 42. Build an NFA Ml that accepts (aba)+ and an NFA M2 that accepts (ab)*. Use. a( ba )* Regular expression for every odd position is b defined over {a,b} b( ab )* More Examples of Regular Expression Regular Expression for no 0 or many triples of 0’s and many 1 in the strings. RegExp for strings of one or many 11 or no 11. A regular expression for ending with abb; A regular expression for all strings having 010 or 101.

I will explain the entire answer through a series of images. Firstly I would like you to carefully examine the final solution below- We will try to obtain the above solution step by step as follows- Step 1 . Take the smallest valid strings 'ab' and 'ba' and draw the appropriate DFA:- Step 2. We want all our st Continue Reading More answers below. I am trying to understand the answer here for FA that accepts only the words baa, ab , ... (Don't think of q2 as an ending state --- the program does NOT terminate once it enters q2. It is merely that when the program is in q2, it can accept the input string so far. Think of q2 as an "accepting state".) \$\endgroup\$. .. 41. Build an NFA Ml that accepts (ab)* and an NFA M2 that accepts (ba)*. Use transitions to obtain a machine M that accepts (ab)*(ba)*. Give the input transition function of M. Use Algorithm 5.6.3 to construct the state diagram of a DFA that accepts 42. Build an NFA Ml that accepts (aba)+ and an NFA M2 that accepts (ab)*. Use. a( ba )* Regular expression for every odd position is b defined over {a,b} b( ab )* More Examples of Regular Expression Regular Expression for no 0 or many triples of 0’s and many 1 in the strings. RegExp for strings of one or many 11 or no 11. A regular expression for ending with abb; A regular expression for all strings having 010 or 101. 41. Build an NFA Ml that accepts (ab)* and an NFA M2 that accepts (ba)*. Use transitions to obtain a machine M that accepts (ab)*(ba)*.Give the input transition function of M. Use Algorithm 5.6.3 to construct the state diagram of a DFA that accepts 42.

Regular Expressions . A Regular Expression can be recursively defined as follows −. ε is a Regular Expression indicates the language containing an empty string. (L (ε) = {ε}) φ is a Regular Expression denoting an empty language. (L (φ) = { }) If X is a Regular Expression denoting the language L (X) and Y is a Regular Expression denoting. In this video solution is explained for following problem : Construct a DFA for all Strings ending with either ab or ba .#DFA #FA #FiniteAutomata #Determinist.

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In the 3rd edition of Sipser's Introduction to the Theory of Computation (example 1.56, p.68), there is a step-by-step procedure to transform ( ab U a)* into a NFA .And then the text ends with: "In this example, the procedure gives an NFA with eight states, but the smallest equivalent NFA has. 2 NFA for (a|b)*abb ba • Has paths to either S0. A string length n, ending in aab can be accepted by the above NFA if: The machine remains in q0 for the first n-3 inputs. On the last three inputs aab, the machine makes the transition as shown below: To reach the final state, and to remain there, it is necessary that the two character of the string should be 10..

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Apr 06, 2021 · Here, state names are in brackets ([..]) and state names that end with "#" are terminal. Generally it's considered easier to make an NFA than a DFA, so the usual method is to first make an NFA and then make the DFA by modifying the NFA by changing multiple output states to a single intermediate state.. 2019. 6. 24. · Problem-1: Construction of a minimal NFA accepting a set of strings over {a, b} in which each string of the language ends with ‘ab’. Explanation: The desired language will be like: L1 = {ab, abbab, abaab, .....} Here as we can see that each string of the above language ends with ‘ab’ but the below language is not accepted by this NFA because some of the string of below. This documentation contains a part about expressions in general, a part about the regular. Explain.c. convert your NFA to DFA and explain what NFA states each DFA state represents. Question: Consider the RE (ab + ba + a)*a. convert this RE to an NFA. explain the steps.b. simplify the NFA as much as possible.

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Construct the NFA for (ba)*: star of (1) Construct the NFA for a(ba)*: concatenation of a and (2) Construct the NFA for a(ba)*b: concatenation of (3) and b; Construct the NFA for ba: concatenation of b and a; Construct the NFA for bab: concatenation of 5 and b; Construct the NFA for a(ba)*b ∪ bab: union of (4) and (6) More Text Exercises. 1. ....

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DFA’S, NFA ’S, REGULAR LANGUAGES We prove the equivalence of these deﬁnitions, often by providing an algorithm for converting one formulation. Nfa ending with. ( ab ∪ a)*. This is so because aab can be formed by one application a, followed by one of ab . So it is redundant inside a Kleene star. Now we can write a two state machine: a a b If you put the loop on a on the start state, either in place of where we have it, or in addition to it, it's also right..

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Regular Expressions . A Regular Expression can be recursively defined as follows −. ε is a Regular Expression indicates the language containing an empty string..

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NFA stands for non-deterministic finite automata. ... {ε, a, aa, b, bb, ab, ba, aba, bab, .....}, any combination of a and b. The (a + b)* shows any combination with a and b even a null string. Example 1: Write the regular expression for the language accepting all the string which are starting with 1 and ending with 0, over.

NFA stands for non-deterministic finite automata. ... {ε, a, aa, b, bb, ab, ba, aba, bab, .....}, any combination of a and b. The (a + b)* shows any combination with a and b even a null string. Example 1: Write the regular expression for the language accepting all the string which are starting with 1 and ending with 0, over. mpc 500 sample pack. Automata theory -- NFA and DFA construction 1. • Draw a DFA for the language accepting strings ending with '01' over input alphabets ∑ = {0, 1} • Draw a DFA for the language accepting strings ending with 'abb' over input alphabets ∑ = {a, b} • Draw a DFA for the language accepting strings ending with 'abba' over input alphabets ∑ = {a, b} minimum.

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13. Obtain a DFA to accept strings of a's and b's ending with ab or ba. 14. Design NFA to accept the strings abc, acd and abcd. 15. Construct DFA for the regular expression ab +b. 16. Prove that the language L = {WWWE(a + b)"} is not regular. Wis the reverse of the string W. 17. Obtain grammar for the following DFA. )a,b a 18. Define Turing.

Apr 06, 2021 · Here, state names are in brackets ([..]) and state names that end with "#" are terminal. Generally it's considered easier to make an NFA than a DFA, so the usual method is to first make an NFA and then make the DFA by modifying the NFA by changing multiple output states to a single intermediate state..

You have to just perform if-elseif type conditions to draw the DFA from table. DFA: i.e Deterministic Finite Automata in which decision for any input should be deterministic i.e. there should be confirmation about output on particular input. So it has only one state as output for particular input. NFA : i.e Non-Deterministic Finite Automata in which decision for any input may.

I am trying to understand the answer here for FA that accepts only the words baa, ab , ... (Don't think of q2 as an ending state --- the program does NOT terminate once it enters q2. It is merely that when the program is in q2, it can accept the input string so far. Think of q2 as an "accepting state".) \$\endgroup\$. ..

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An NFA can have zero, one or more than one move from a given state on a given input symbol. states for each input symbol using transition function of NFA . If this set of states is not in Q', add it to Q'. Step 4: Final state of DFA will be all states with contain F (final states of NFA ).

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baseball tablecloth   • a( ba )* Regular expression for every odd position is b defined over {a,b} b( ab )* More Examples of Regular Expression Regular Expression for no 0 or many triples of 0’s and many 1 in the strings. RegExp for strings of one or many 11 or no 11. A regular expression for ending with abb; A regular expression for all strings having 010 or 101.
• In the 3rd edition of Sipser's Introduction to the Theory of Computation (example 1.56, p.68), there is a step-by-step procedure to transform ( ab U a)* into a NFA .And then the text ends with: "In this example, the procedure gives an NFA with eight states, but the smallest equivalent NFA has. 2 NFA for (a|b)*abb ba • Has paths to either S0 ...
• I am trying to understand the answer here for FA that accepts only the words baa, ab , ... (Don't think of q2 as an ending state --- the program does NOT terminate once it enters q2. It is merely that when the program is in q2, it can accept the input string so far. Think of q2 as an "accepting state".) \$\endgroup\$. .
• An NFA can have zero, one or more than one move from a given state on a given input symbol. states for each input symbol using transition function of NFA . If this set of states is not in Q', add it to Q'. Step 4: Final state of DFA will be all states with contain F (final states of NFA ).
• random person sent me money on cash app reddit. An NFA can have zero, one or more than one move from a given state on a given input symbol. states for each input symbol using transition function of NFA.If this set of states is not in Q', add it to Q'. Step 4: Final state of DFA will be all states with contain F (final states of NFA).A string ending with a substring starting with ch or th ...