Drawing a dfsa where L is a set of strings that contains at most 4 zeros












0














For each of the following languages over alphabet $Σ = {0, 1}$, construct a DFSA that accepts it and a regular expression that denotes it. Prove that your automata and regular expressions are correct.
Use as few states as possible in your DFSA.



(a) $L_1 = {x: text{x is a set of string that contains at most 4 zeros} }$



The regex is



$$R_1 = 1^{∗} + 1^{∗}01^{∗} + 1^{∗}01^{∗}01^{∗} + 1^*01^*01^*01^* + 1^*01^*01^*01^*01^*$$



How would I draw the dfsa for it?






computer-science automata regular-language







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    0














    For each of the following languages over alphabet $Σ = {0, 1}$, construct a DFSA that accepts it and a regular expression that denotes it. Prove that your automata and regular expressions are correct.
    Use as few states as possible in your DFSA.



    (a) $L_1 = {x: text{x is a set of string that contains at most 4 zeros} }$



    The regex is



    $$R_1 = 1^{∗} + 1^{∗}01^{∗} + 1^{∗}01^{∗}01^{∗} + 1^*01^*01^*01^* + 1^*01^*01^*01^*01^*$$



    How would I draw the dfsa for it?






    computer-science automata regular-language







    share|cite|improve this question



























      0












      0








      0







      For each of the following languages over alphabet $Σ = {0, 1}$, construct a DFSA that accepts it and a regular expression that denotes it. Prove that your automata and regular expressions are correct.
      Use as few states as possible in your DFSA.



      (a) $L_1 = {x: text{x is a set of string that contains at most 4 zeros} }$



      The regex is



      $$R_1 = 1^{∗} + 1^{∗}01^{∗} + 1^{∗}01^{∗}01^{∗} + 1^*01^*01^*01^* + 1^*01^*01^*01^*01^*$$



      How would I draw the dfsa for it?






      computer-science automata regular-language







      share|cite|improve this question















      For each of the following languages over alphabet $Σ = {0, 1}$, construct a DFSA that accepts it and a regular expression that denotes it. Prove that your automata and regular expressions are correct.
      Use as few states as possible in your DFSA.



      (a) $L_1 = {x: text{x is a set of string that contains at most 4 zeros} }$



      The regex is



      $$R_1 = 1^{∗} + 1^{∗}01^{∗} + 1^{∗}01^{∗}01^{∗} + 1^*01^*01^*01^* + 1^*01^*01^*01^*01^*$$



      How would I draw the dfsa for it?






      computer-science automata regular-language




      computer-science automata regular-language






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 27 '18 at 10:17









      dkaeae

      304311




      304311










      asked Nov 5 '18 at 10:20









      shah

      384




      384






















          1 Answer
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          oldest

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          1














          Here are the state transitions:



          $s_0rightarrow_1 s_0$,
          $s_0rightarrow_0 s_1$,
          $s_1rightarrow_1 s_1$,
          $s_1rightarrow_0 s_2$,
          $s_2rightarrow_1 s_2$,
          $s_2rightarrow_0 s_3$,
          $s_3rightarrow_1 s_3$,
          $s_3rightarrow_0 s_4$,
          $s_4rightarrow_1 s_4$.



          $s_0$ is the starting state and all states are final states.



          Start in state $s_0$. Produce any number of 1's and then exit or read 0 and move to state $s_1$. In state $s_1$ produce any number of 1's and then exit or read 0 and move to state $s_2$, and so on.






          share|cite|improve this answer























          • Oh wait, this cant be done by dfsa because since it can only have one final state? If this question was changed to "x is a set of string that contains at most 1 zeros" would it be a dfsa and the regex will be $1^{*}$
            – shah
            Nov 5 '18 at 10:38












          • Depends on the definition. Are you allowed to have $epsilon$ transitions (empty word)?
            – Wuestenfux
            Nov 5 '18 at 10:41










          • A DFSA is a 5-tuple $M = (Q, Sigma, delta, s, f)$. Q is the set of states, $Sigma$ is the input alphabet, $delta$ is the transition, $s$ is the initial state, and $f$ is the set of accepting states. So yeah no $epsilon$ transition
            – shah
            Nov 5 '18 at 10:44








          • 1




            So you can have a set of accepting states, not just one.
            – Wuestenfux
            Nov 5 '18 at 10:46










          • Do you know how to explain each state? For example: $s_0: $ x has an even number of 1's. Not sure how too
            – shah
            Nov 5 '18 at 10:51













          Your Answer





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          1 Answer
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          1 Answer
          1






          active

          oldest

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          active

          oldest

          votes






          active

          oldest

          votes









          1














          Here are the state transitions:



          $s_0rightarrow_1 s_0$,
          $s_0rightarrow_0 s_1$,
          $s_1rightarrow_1 s_1$,
          $s_1rightarrow_0 s_2$,
          $s_2rightarrow_1 s_2$,
          $s_2rightarrow_0 s_3$,
          $s_3rightarrow_1 s_3$,
          $s_3rightarrow_0 s_4$,
          $s_4rightarrow_1 s_4$.



          $s_0$ is the starting state and all states are final states.



          Start in state $s_0$. Produce any number of 1's and then exit or read 0 and move to state $s_1$. In state $s_1$ produce any number of 1's and then exit or read 0 and move to state $s_2$, and so on.






          share|cite|improve this answer























          • Oh wait, this cant be done by dfsa because since it can only have one final state? If this question was changed to "x is a set of string that contains at most 1 zeros" would it be a dfsa and the regex will be $1^{*}$
            – shah
            Nov 5 '18 at 10:38












          • Depends on the definition. Are you allowed to have $epsilon$ transitions (empty word)?
            – Wuestenfux
            Nov 5 '18 at 10:41










          • A DFSA is a 5-tuple $M = (Q, Sigma, delta, s, f)$. Q is the set of states, $Sigma$ is the input alphabet, $delta$ is the transition, $s$ is the initial state, and $f$ is the set of accepting states. So yeah no $epsilon$ transition
            – shah
            Nov 5 '18 at 10:44








          • 1




            So you can have a set of accepting states, not just one.
            – Wuestenfux
            Nov 5 '18 at 10:46










          • Do you know how to explain each state? For example: $s_0: $ x has an even number of 1's. Not sure how too
            – shah
            Nov 5 '18 at 10:51


















          1














          Here are the state transitions:



          $s_0rightarrow_1 s_0$,
          $s_0rightarrow_0 s_1$,
          $s_1rightarrow_1 s_1$,
          $s_1rightarrow_0 s_2$,
          $s_2rightarrow_1 s_2$,
          $s_2rightarrow_0 s_3$,
          $s_3rightarrow_1 s_3$,
          $s_3rightarrow_0 s_4$,
          $s_4rightarrow_1 s_4$.



          $s_0$ is the starting state and all states are final states.



          Start in state $s_0$. Produce any number of 1's and then exit or read 0 and move to state $s_1$. In state $s_1$ produce any number of 1's and then exit or read 0 and move to state $s_2$, and so on.






          share|cite|improve this answer























          • Oh wait, this cant be done by dfsa because since it can only have one final state? If this question was changed to "x is a set of string that contains at most 1 zeros" would it be a dfsa and the regex will be $1^{*}$
            – shah
            Nov 5 '18 at 10:38












          • Depends on the definition. Are you allowed to have $epsilon$ transitions (empty word)?
            – Wuestenfux
            Nov 5 '18 at 10:41










          • A DFSA is a 5-tuple $M = (Q, Sigma, delta, s, f)$. Q is the set of states, $Sigma$ is the input alphabet, $delta$ is the transition, $s$ is the initial state, and $f$ is the set of accepting states. So yeah no $epsilon$ transition
            – shah
            Nov 5 '18 at 10:44








          • 1




            So you can have a set of accepting states, not just one.
            – Wuestenfux
            Nov 5 '18 at 10:46










          • Do you know how to explain each state? For example: $s_0: $ x has an even number of 1's. Not sure how too
            – shah
            Nov 5 '18 at 10:51
















          1












          1








          1






          Here are the state transitions:



          $s_0rightarrow_1 s_0$,
          $s_0rightarrow_0 s_1$,
          $s_1rightarrow_1 s_1$,
          $s_1rightarrow_0 s_2$,
          $s_2rightarrow_1 s_2$,
          $s_2rightarrow_0 s_3$,
          $s_3rightarrow_1 s_3$,
          $s_3rightarrow_0 s_4$,
          $s_4rightarrow_1 s_4$.



          $s_0$ is the starting state and all states are final states.



          Start in state $s_0$. Produce any number of 1's and then exit or read 0 and move to state $s_1$. In state $s_1$ produce any number of 1's and then exit or read 0 and move to state $s_2$, and so on.






          share|cite|improve this answer














          Here are the state transitions:



          $s_0rightarrow_1 s_0$,
          $s_0rightarrow_0 s_1$,
          $s_1rightarrow_1 s_1$,
          $s_1rightarrow_0 s_2$,
          $s_2rightarrow_1 s_2$,
          $s_2rightarrow_0 s_3$,
          $s_3rightarrow_1 s_3$,
          $s_3rightarrow_0 s_4$,
          $s_4rightarrow_1 s_4$.



          $s_0$ is the starting state and all states are final states.



          Start in state $s_0$. Produce any number of 1's and then exit or read 0 and move to state $s_1$. In state $s_1$ produce any number of 1's and then exit or read 0 and move to state $s_2$, and so on.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 5 '18 at 11:12

























          answered Nov 5 '18 at 10:25









          Wuestenfux

          3,5791411




          3,5791411












          • Oh wait, this cant be done by dfsa because since it can only have one final state? If this question was changed to "x is a set of string that contains at most 1 zeros" would it be a dfsa and the regex will be $1^{*}$
            – shah
            Nov 5 '18 at 10:38












          • Depends on the definition. Are you allowed to have $epsilon$ transitions (empty word)?
            – Wuestenfux
            Nov 5 '18 at 10:41










          • A DFSA is a 5-tuple $M = (Q, Sigma, delta, s, f)$. Q is the set of states, $Sigma$ is the input alphabet, $delta$ is the transition, $s$ is the initial state, and $f$ is the set of accepting states. So yeah no $epsilon$ transition
            – shah
            Nov 5 '18 at 10:44








          • 1




            So you can have a set of accepting states, not just one.
            – Wuestenfux
            Nov 5 '18 at 10:46










          • Do you know how to explain each state? For example: $s_0: $ x has an even number of 1's. Not sure how too
            – shah
            Nov 5 '18 at 10:51




















          • Oh wait, this cant be done by dfsa because since it can only have one final state? If this question was changed to "x is a set of string that contains at most 1 zeros" would it be a dfsa and the regex will be $1^{*}$
            – shah
            Nov 5 '18 at 10:38












          • Depends on the definition. Are you allowed to have $epsilon$ transitions (empty word)?
            – Wuestenfux
            Nov 5 '18 at 10:41










          • A DFSA is a 5-tuple $M = (Q, Sigma, delta, s, f)$. Q is the set of states, $Sigma$ is the input alphabet, $delta$ is the transition, $s$ is the initial state, and $f$ is the set of accepting states. So yeah no $epsilon$ transition
            – shah
            Nov 5 '18 at 10:44








          • 1




            So you can have a set of accepting states, not just one.
            – Wuestenfux
            Nov 5 '18 at 10:46










          • Do you know how to explain each state? For example: $s_0: $ x has an even number of 1's. Not sure how too
            – shah
            Nov 5 '18 at 10:51


















          Oh wait, this cant be done by dfsa because since it can only have one final state? If this question was changed to "x is a set of string that contains at most 1 zeros" would it be a dfsa and the regex will be $1^{*}$
          – shah
          Nov 5 '18 at 10:38






          Oh wait, this cant be done by dfsa because since it can only have one final state? If this question was changed to "x is a set of string that contains at most 1 zeros" would it be a dfsa and the regex will be $1^{*}$
          – shah
          Nov 5 '18 at 10:38














          Depends on the definition. Are you allowed to have $epsilon$ transitions (empty word)?
          – Wuestenfux
          Nov 5 '18 at 10:41




          Depends on the definition. Are you allowed to have $epsilon$ transitions (empty word)?
          – Wuestenfux
          Nov 5 '18 at 10:41












          A DFSA is a 5-tuple $M = (Q, Sigma, delta, s, f)$. Q is the set of states, $Sigma$ is the input alphabet, $delta$ is the transition, $s$ is the initial state, and $f$ is the set of accepting states. So yeah no $epsilon$ transition
          – shah
          Nov 5 '18 at 10:44






          A DFSA is a 5-tuple $M = (Q, Sigma, delta, s, f)$. Q is the set of states, $Sigma$ is the input alphabet, $delta$ is the transition, $s$ is the initial state, and $f$ is the set of accepting states. So yeah no $epsilon$ transition
          – shah
          Nov 5 '18 at 10:44






          1




          1




          So you can have a set of accepting states, not just one.
          – Wuestenfux
          Nov 5 '18 at 10:46




          So you can have a set of accepting states, not just one.
          – Wuestenfux
          Nov 5 '18 at 10:46












          Do you know how to explain each state? For example: $s_0: $ x has an even number of 1's. Not sure how too
          – shah
          Nov 5 '18 at 10:51






          Do you know how to explain each state? For example: $s_0: $ x has an even number of 1's. Not sure how too
          – shah
          Nov 5 '18 at 10:51




















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