(12 points) Given the following regular expression over the alphabet {a,b}, (ba)(a+b) to transform it to a regular grammar, we first transform it to an NFA as the one shown below. Instead of converting this NFA to a regular grammar directly, we convert it to a DFA first and then convert the DFA to a regular grammar (why?). So we construct -closures of the above NFA, N(0)={0,1,3}N(1)={1,3}N(2)={2}N(3)={3}N(4)={4}N()= and build the following tree, ( 3 points) so that the following nodes of the tree can be used to build a DFA. Fill out the following blanks and blanks in the above tree. {0,1,3} (0.75 points) With the selected nodes from the above tree, we build a DFA by constructing the following transition table (the table on the left), and renaming the states as S (start state), I, J, K and L : ( 3.25 points) Then convert the table representation to a digraph representation as follows. Fill out the blanks in the following digraph. ( 1.25 points) However, we don't need the portion circled in the above digraph representation of the DFA (why?). By removing that portion, the FA is like the one shown on the left side of the following figure (no points for this FA). By constructing production rules from this FA, we get the production set of the regular grammar on the right side of the above figure. Fill out the blanks in the above figure. (1.75 points for the production set) Regular expression of the above FA is (2 points) which is equal to the given regular expression (ba)*(a+b). (Why?)