Nondeterministic finite automata (NFA) correction

Question

I am trying to solve a question about NFA. The instructions are as following: Alphabet {a, b, c}. • L1 is all strings that their last character is the same as their fifth-last character. For instance, the string aaacbacbca should be accepted, because the fifth-last character and the last character are both a. The string ccaab should be rejected because the fifth-last character is c and the last character is b. Here is with what I came up, but I am really beginner in this topic and I am not sure whether is correct or not:

Answer 1

The automata you have at the moment accepts only the strings that end with acbca . Here are steps to arrive to the solution:

• change the automata you have such that it would accept all strings which have last and five-to-last symbol a
• do the same for the symbols b , c
• combine the 3 automata

Answer 2

You were almost right, but the automata you drew was only to accept strings ending in acbca . This one will accept the strings you want

a,b,c     a      a,b,c   a,b,c   a,b,c    a     a,b,c
,--->[q0]--->[q1]--->[q2]--->[q3]--->[q4]--->{q5}----+>[q16]-----.
|    /|   b      a,b,c   a,b,c   a,b,c    b     a,b,c|   ^ a,b,c |
`---´ +----->[q6]--->[q7]--->[q8]--->[q9]--->{q10}---+   `-------´
      |   c      a,b,c   a,b,c   a,b,c    c     a,b,c|
      `----->[q11]-->[q12]-->[q13]-->[q14]-->{q15}---´

States like {q5} are accepting states, while states like [q0] are not accepting. The meaning of q16 is to make sure that strings that have two equal letters distant 4 characters but not ending there, sink to a non accepting state. Same might be for letters b,c in state [q4] , for letters a,c in state [q9] , and for letters a,b in state [q14] , but for clarity of the drawing, I have omitted them.