How to find a DFA for combinations of even and odd occurrences $0,1$?












1















Let $L$ be a language over ${0,1}$ whose Nerode equivalence classes are:
$$
{w|#_0(w)mod2=0quadlandquad #_1(w)mod2=0}\
{w|#_0(w)mod2=0quadlandquad #_1(w)mod2=1}\
{w|#_0(w)mod2=1quadlandquad #_1(w)mod2=1}\
{w|#_0(w)mod2=1quadlandquad #_1(w)mod2=0}
$$

($#_0(w)mod2=0$ means that the number of zeroes in word $w$ is even).



Also $epsilon in L, 0,1,1110notin L$. Find DFA for the language.




First because there's a finite number of eq. classes by Nerode theorem we know that $L$ is regular so DFA exists for the language.



The solution to the problem is:



enter image description here



But I don't understand how the DFA accommodates all possible words in $L$.



1) For example what about $111$? It had even number of $0$'s and uneven number of $1$'s so it belongs to the second eq.class. But I don't see how we can arrive to this word using the DFA above.



2) What about the word $00011$? It belongs to the fourth equivalence class but I don't see how to arrive at this word using the DFA.










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    1















    Let $L$ be a language over ${0,1}$ whose Nerode equivalence classes are:
    $$
    {w|#_0(w)mod2=0quadlandquad #_1(w)mod2=0}\
    {w|#_0(w)mod2=0quadlandquad #_1(w)mod2=1}\
    {w|#_0(w)mod2=1quadlandquad #_1(w)mod2=1}\
    {w|#_0(w)mod2=1quadlandquad #_1(w)mod2=0}
    $$

    ($#_0(w)mod2=0$ means that the number of zeroes in word $w$ is even).



    Also $epsilon in L, 0,1,1110notin L$. Find DFA for the language.




    First because there's a finite number of eq. classes by Nerode theorem we know that $L$ is regular so DFA exists for the language.



    The solution to the problem is:



    enter image description here



    But I don't understand how the DFA accommodates all possible words in $L$.



    1) For example what about $111$? It had even number of $0$'s and uneven number of $1$'s so it belongs to the second eq.class. But I don't see how we can arrive to this word using the DFA above.



    2) What about the word $00011$? It belongs to the fourth equivalence class but I don't see how to arrive at this word using the DFA.










    share|cite|improve this question

























      1












      1








      1








      Let $L$ be a language over ${0,1}$ whose Nerode equivalence classes are:
      $$
      {w|#_0(w)mod2=0quadlandquad #_1(w)mod2=0}\
      {w|#_0(w)mod2=0quadlandquad #_1(w)mod2=1}\
      {w|#_0(w)mod2=1quadlandquad #_1(w)mod2=1}\
      {w|#_0(w)mod2=1quadlandquad #_1(w)mod2=0}
      $$

      ($#_0(w)mod2=0$ means that the number of zeroes in word $w$ is even).



      Also $epsilon in L, 0,1,1110notin L$. Find DFA for the language.




      First because there's a finite number of eq. classes by Nerode theorem we know that $L$ is regular so DFA exists for the language.



      The solution to the problem is:



      enter image description here



      But I don't understand how the DFA accommodates all possible words in $L$.



      1) For example what about $111$? It had even number of $0$'s and uneven number of $1$'s so it belongs to the second eq.class. But I don't see how we can arrive to this word using the DFA above.



      2) What about the word $00011$? It belongs to the fourth equivalence class but I don't see how to arrive at this word using the DFA.










      share|cite|improve this question














      Let $L$ be a language over ${0,1}$ whose Nerode equivalence classes are:
      $$
      {w|#_0(w)mod2=0quadlandquad #_1(w)mod2=0}\
      {w|#_0(w)mod2=0quadlandquad #_1(w)mod2=1}\
      {w|#_0(w)mod2=1quadlandquad #_1(w)mod2=1}\
      {w|#_0(w)mod2=1quadlandquad #_1(w)mod2=0}
      $$

      ($#_0(w)mod2=0$ means that the number of zeroes in word $w$ is even).



      Also $epsilon in L, 0,1,1110notin L$. Find DFA for the language.




      First because there's a finite number of eq. classes by Nerode theorem we know that $L$ is regular so DFA exists for the language.



      The solution to the problem is:



      enter image description here



      But I don't understand how the DFA accommodates all possible words in $L$.



      1) For example what about $111$? It had even number of $0$'s and uneven number of $1$'s so it belongs to the second eq.class. But I don't see how we can arrive to this word using the DFA above.



      2) What about the word $00011$? It belongs to the fourth equivalence class but I don't see how to arrive at this word using the DFA.







      proof-explanation equivalence-relations formal-languages automata






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      asked Nov 24 at 9:13









      Yos

      1,137723




      1,137723






















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          Not every equivalence class belongs to the language, otherwise the language would always be $Sigma^*$. Since $varepsilon in L$ and $0,1,1110 notin L$, we know that $L$ equals the first equivalence class. Therefore your examples are not counterexamples.






          share|cite|improve this answer





















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            1 Answer
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            Not every equivalence class belongs to the language, otherwise the language would always be $Sigma^*$. Since $varepsilon in L$ and $0,1,1110 notin L$, we know that $L$ equals the first equivalence class. Therefore your examples are not counterexamples.






            share|cite|improve this answer


























              1














              Not every equivalence class belongs to the language, otherwise the language would always be $Sigma^*$. Since $varepsilon in L$ and $0,1,1110 notin L$, we know that $L$ equals the first equivalence class. Therefore your examples are not counterexamples.






              share|cite|improve this answer
























                1












                1








                1






                Not every equivalence class belongs to the language, otherwise the language would always be $Sigma^*$. Since $varepsilon in L$ and $0,1,1110 notin L$, we know that $L$ equals the first equivalence class. Therefore your examples are not counterexamples.






                share|cite|improve this answer












                Not every equivalence class belongs to the language, otherwise the language would always be $Sigma^*$. Since $varepsilon in L$ and $0,1,1110 notin L$, we know that $L$ equals the first equivalence class. Therefore your examples are not counterexamples.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 24 at 9:24









                mrp

                3,77251537




                3,77251537






























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