theory of computation

Question

Let sigma = { 0,1 } . Construct a minimal DFA which accepts set of all strings in which

"Every substring of four symbols has at most two 0’s". For example, 001110 and 011001 are
in the language, but 10010 is not since one of its substrings, 0010, contains three zeros.

Answer 1

We add the substring $1111$ as prefix to every $w\,\epsilon \, L$ (to fullfill the below assumption for characters at index 0,1,2)

The state of an automata which checks if a word $w\, \epsilon \,L$,  denotes a substring of length $4$ ending at the current character. So, if we are at index $i$ of the string it denotes that the string upto this index is valid and when we make a transition for index $i+1$ there is only one substring that we have to check for this string to be valid $i.e$ the substring of length $4$ ending at the $\left ( i+1 \right )$^th index.

$q$_d is the dead state for the automata and all the other states are final states, $q$₀ is the start state.

$q$₀ $= \, 1111$  , $q$₁ $= \, 1110$ , $q$₂ $= \, 1100$ , $q$₃ $= \, 1001$ , $q$₄ $= \, 0011$ , $q$₅ $= \, 1101$ , $q$₆ $= \, 1010$

$q$₇ $= \, 1011$ , $q$₈ $= \, 0110$ , $q$₉ $= \, 0101$ , $q$₁₀ $= \, 0111$ , $q$_d $= \, \left \{ 0010,1000, 0100, 1001\right \}$

The transition table for the automata is 

State	0	1
$q$0	$q$1	$q$0
$q$1	$q$2	$q$5
$q$2	$q$d	$q$3
$q$3	$q$d	$q$4
$q$4	$q$8	$q$10
$q$5	$q$6	$q$7
$q$6	$q$d	$q$9
$q$7	$q$8	$q$10
$q$8	$q$2	$q$5
$q$9	$q$6	$q$7
$q$10	$q$1	$q$0
$q$d	$q$d	$q$d

For Example : 

0010 : $q$₀ $\rightarrow$ $q$₁ $\rightarrow$ $q$₂ $\rightarrow$ $q$₃ $\rightarrow$ $q$_d

101001 : $q$₀ $\rightarrow$ $q$₀ $\rightarrow$ $q$₁ $\rightarrow$ $q$₅ $\rightarrow$ $q$₆ $\rightarrow$ $q$_d

0011 : $q$₀ $\rightarrow$ $q$₁ $\rightarrow$ $q$₂ $\rightarrow$ $q$₃ $\rightarrow$ $q$4

 

Correct me if I'm wrong

Answer 2

The above dfa will always go to  dead state if there are more than 2 0s in any substring of length 4.

Answer 3

Except trap state all other states are final states

Comments

For 0010, it still reaches the final state.
q1 -> q3 -> q7 -> q8

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Here q0 is the staring state but not q1.

0010

q0 -> q1 -> q3 -> Trap

So it's not accepting.