 ^{(s)}
Give unambiguous contextfree grammars for each
of the following languages:

The language of identifiers, whose elements are sequences of letters or
digits, starting with a letter

The language of natural numbers, whose elements are sequences of digits,
starting with a digit different from 0

The language of real numbers,
in which either the integer part or the fractional
part can be empty, but not both.
Thus the grammar must allow 12., 1.2, .12,
but not a decimal point by itself.
 ^{(s)}
The set of formulas (FirstOrder Logic Formulas) can be defined
inductively as follows:

An atomic formula is a formula

if "A" and "B"
are formulas, then also "not A", "A and B",
"A or B",
and "A implies B" are formulas

If "A" is a formula and "x" is an identifier, then also
"exists x. A" and "forall x. A" are formulas
We want the following properties to hold:

The precedence order is as follows (from highest to lowest priority):
(1) not, (2) and, (3) or, (4) implies,
(5) exists, forall.
 "and" and "or" are
leftassociative, "implies" is rightassociative.

Formulas can be precedeed by an arbitrary number of negations
(i.e. "not not A", "not not not A", etc. are
allowed)
Give an unambiguous contextfree grammar generating the
language of formulas and respecting these
properties. You may assume given productions for generating identifiers
and atomic formulas starting from the nonterminals Ide
and Atom respectively.
 ^{(s)}
Consider the following productions for commands
Cmd ::= print(Exp)  Cmd ; Cmd  if Exp then Cmd else Cmd
Show that this grammar is ambiguous and explain all possible causes of
ambiguity,
distinguishing between those which are not relevant at the semantical
level and those which are relevant. Note: the sequentialization operator
";" is associative.
 ^{(s)}
Exercise 2.6 in Sethi's book. Note: by "parse tree" Sethi means
"derivation tree"
 Consider the following grammar defining the language of
regular expression on the alphabet {a,b,c}
E ::= \  0  a  b  c  E + E  EE  E*  (E)
Where "\", "0", "a", "b",
"c", "+", "*", "(" and ")"
are the terminal symbols. (Remark: "\" represent
the regular expression "lambda" and "0" represents the
regular expression "emptyset". Here in this grammar
they are just terminals, they should not be confused
with the production of an empty string.)
 Show that the grammar is ambiguous.
 Give an unambiguous grammar which generates exactly the same set
of strings as this one.
 State what associativity and precedence rules your grammar enforces.
 Consider the following grammar generating the language of the
untyped lambdacalculus:
M ::= Var  M M  \Var. M  (M)
in which "\", ".", "(", and ")" are terminal symbols
and Var is a nonterminal which unambiguously generates identifiers.
The productions for Var are not represented here.
 Show that the grammar is ambiguous,
by giving two distinct parse trees for a single string.
 Give an unambiguous grammar which generates exactly
the same set of strings as the one above.

The following grammar generates the language of Types of mini ML:
Type ::= TVar  Type > Type  Type * Type
Note: Type and TVar are nonterminals. ">" and
"*" are terminals.
TVar is assumed to generate a set of names (the Type variables)
"alpha", "beta", "gamma", etc. The
productions for TVar are not represented here.
Show that this grammar is ambiguous by giving two derivation trees for
each of the following type expressions:
alpha > beta > gamma
alpha > beta * gamma
alpha * beta * gamma
 ^{(s)}
Modify the grammar of previous exercise so to obtain an unambiguous grammar,
generating the same language,
by enforcing the following properties:
 > is right associative
 * has higher priority than >
 * is left associative
 Consider the following grammar, generating a subset of the
language of types of a generic imperative language:
Type ::= int  bool  ^Type  Type * Type  (Type)
where "int", "bool", "^",
"(" and ")" are terminal symbols.
 Show that the grammar is ambiguous.
 Give an unambiguous grammar which generates exactly the same set
of strings as this one.
 State what associativity and precedence rules your grammar enforces.
 ^{(s)}
Exercise 2.15 in Sethi's book. Remember that by "parse tree" Sethi means
"derivation tree".
Note1: in this exercise, the tokens are the words in boldface and the
symbols between quotes.
The words in italics are the syntactical categories.
Note 2: In Question (c), the sentence
"Suppose that the first production
for Declarator is changed so to make * a postfix operator"
means: "suppose
that the first production for Declarator
is changed into the following
production:
Declarator ::= Declarator *

Consider the following grammar, which generates all the nonempty
strings of balanced parentheses:
P ::= ()  (P)  PP
 Prove that this grammar is ambiguous.
 Modify the grammar so to obtain a
nonambiguous grammar generating the same language.
 ^{(d)} Show that the new grammar is not ambiguous.
 ^{(s)}
Define a grammar which generates all palindromes
over the English alphabet.
Note: a palindrome is any string
which reads the same backward as forward.
For example, the empty string, "a", "aa", "aba", and "abba" are
palindromes, whereas "ab", "abb", and "abc" are not.

Define a grammar which generates all the strings over the alphabet {a,b}
which have the same number of a's and b's.
For instance, the empty string, "ab", "ba", "aabb" and "abab"
belong to the language,
while "a", "bba", and "aaa" don't.

Define a grammar over the alphabet {a,b,c} generating all the strings
of the form a^{k}b^{m}c^{n}
with n = m + k.
For instance, the empty string, "ac", "bc", "abcc" and "aabccc"
belong to the language,
while "a", "bacc", and "aac" don't.