Creating DFA's for $\Sigma=\{0,1\}$ Languages

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mathmari
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Hey! :o

I have to draw the DFA's for the following languages:
$\Sigma=\{0, 1\}$

  1. $$\{w \mid w \text{ starts with } 1 \text{ and ends with } 0 \}$$
  2. $$\{w \mid w \text{ contains at least 3 } 1\}$$
  3. $$\{w \mid w \text{ contains the substring } 0101\}$$
  4. $$\{w \mid w \text{ has length at least } 3 \text{ and the third symbol is } 0\}$$
  5. $$\{w \mid w \text{ starts with } 0 \text{ and has an odd length, or with } 1 \text{ and has an even length} \}$$
  6. $$\{w \mid w \text{ doesn't contain the substring } 110\}$$
  7. $$\{w \mid \text{ the length of } w \text{ is smaller than or equal to } 5\}$$
  8. $$\{w \mid w \text{ is any string besides } 11 \text{ and } 111\}$$
  9. $$\{w \mid \text{ in each odd position of } w \text{ there is a } 1\}$$
  10. $$\{w \mid w \text{ contains at least two } 0 \text{ and at most one } 1\}$$
  11. $$\{\epsilon , 0\}$$
  12. $$\{w \mid w \text{ contains an even number of } 0, \text{ or exactly two } 1\}$$
  13. the empty set
  14. all strings besides the empty one.
I have done the following:

View attachment 4142

Could you tell me if it is correct?? (Wondering)

Could you give me some hints how we could find the DFA of the last three languages ?? (Wondering)
 
Attachments
  • Eleven numbered state diagrams of DFAs over the alphabet {0,1}, labelled 1) to 11), with states q0 onward, accepting states drawn as double circles, and transitions labelled 0, 1 or 0,1.
    DFA's.png
    30.5 KB · Views: 144
Last edited by a moderator:
Hey mathmari! (Wave)

I'll just pick up number 1 for now: w starts with 1 and ends with 0.
Hmm... so 1010 would be part of the language wouldn't it? (Thinking)
Does your DFA accept 1010? (Wondering)
 
I like Serena said:
I'll just pick up number 1 for now: w starts with 1 and ends with 0.
Hmm... so 1010 would be part of the language wouldn't it? (Thinking)
Does your DFA accept 1010? (Wondering)

It doesn't accept 1010... (Headbang)

Should it be maybe as followed?? (Wondering)

View attachment 4151
 
Attachments
  • Handwritten state diagram of a DFA with states q0, q1, q2 (accepting, double circle), and q3. Start arrow enters q0. q0 on 1 goes to q1, on 0 goes to q3. q1 loops on 1 and goes on 0 to q2. q2 loops on 0 and goes on 1 back to q1. q3 loops on 0,1.
    DFA_1.png
    3.1 KB · Views: 153
Or have I done something wrong?? (Wondering)
 
mathmari said:
It doesn't accept 1010... (Headbang)

Should it be maybe as followed?? (Wondering)

mathmari said:
Or have I done something wrong?? (Wondering)

No worries. That's it. (Nod)

Number 2 looks fine to me. (Happy)

So let me randomly pick up number 10.
Logically we have to build up a tree that starts as a binary tree where we either accept a 0 to the left and a 1 to the right.
That is, until we can decide what is acceptable or not. (Thinking)