Homework #1 Solution Sketches:

1)

[^aeiou]*a[^aeiou]*e[^aeiou]*i[^aeiou]o*[^aeiou]*u[^aeiou]*,
where [^aeiou] represents any alphabetic character other than 
a e i o u (i.e., a consonant).

<start> -> <consonants>a<consonants>e<consonants>i<consonants>o<consonants>u<consonants>
<consonants> -> <consonant><consonants> | epsilon
<consonant> -> b | c | d ....

2) 

The grammar generates strings of even lengths containing the same
number of a's and b's (including the null string)
(e.g., ~, ab, ba, aaabbb, aabbabaaabbb).

3)

The following statement has two distinct parse trees which illustrate
that the grammar which it was generated from is ambiguous.

if Cond1 then if Cond2 then S1 else if Cond3 then S2 else S3

Parse 1:

<stmt> -> <matched_stmt>
       -> if <expr> then <matched_stmt> else <stmt>
       -> if <expr> then if <expr> then <matched_stmt> else <stmt> else <stmt>
       -> if <expr> then if <expr> then <other> else <stmt> else <stmt>
       -> if <expr> then if <expr> then <other> else <stmt> else <matched_stmt>
       -> if <expr> then if <expr> then <other> else <stmt> else <other>
       -> if <expr> then if <expr> then <other> else if <expr> then <stmt> else <other>
       -> if <expr> then if <expr> then <other> else if <expr> then <matched_stmt> else <other>
       -> if <expr> then if <expr> then <other> else if <expr> then <other> else <other>

Parse 2:

<stmt> -> if <expr> then <stmt>
       -> if <expr> then <matched_stmt>
       -> if <expr> then if <expr> then <matched_stmt> else <stmt>
       -> if <expr> then if <expr> then <other> else <stmt>
       -> if <expr> then if <expr> then <other> else <matched_stmt>
       -> if <expr> then if <expr> then <other> else if <expr> then <matched_stmt> else <stmt>
       -> if <expr> then if <expr> then <other> else if <expr> then <other> else <stmt>
       -> if <expr> then if <expr> then <other> else if <expr> then <other> else <matched_stmt>
       -> if <expr> then if <expr> then <other> else if <expr> then <other> else <other>

4)

No, because the grammar involves the concept of memory.  
Regular expressions cannot describe constructs involving remembering
something.

5)

Trivial.

Consider the sentence: 3 + 4 + 5

Parse 1: (leftmost derivation)

<S> -> <A>
    -> <A> + <A>
    -> <A> + <A> + <A> (where first <A> in line 2 is expanded to <A> + <A>)
    -> <id> + <A> + <A>
    -> <id> + <id> + <A>
    -> <id> + <id> + <id>
    -> 3 + <id> + <id>
    -> 3 + 4 + <id>
    -> 3 + 4 + 5

Parse 2: (rightmost derivation)

<S> -> <A>
    -> <A> + <A>
    -> <A> + <A> + <A> (where second <A> in line 2 is expanded to <A> + <A>)
    -> <A> + <A> + <id>
    -> <A> + <id> + <id>
    -> <id> + <id> + <id>
    -> <id> + <id> + <5>
    -> <id> + 4 + 5
    -> 3 + 4 + 5

6)

a)

{b < 1}
a = 2*b + 1
{a < 3}
b = a - 3
{b < 0}

b)

{2b + a > 1} = {b > (1-a)/2}
a = 3*(2*b + a)
{a > 3}
b = 2*a -1
{b > 5}