CS 236 Midterm Example Problems


















For each string and regular expression (regex) pair given in parts a-d,
answer 'yes' if the string matches the pattern defined by the regex,
or 'no' if the string doesn't match.

a.
string: abab
regex:  a(bb)*(ab)*

b.
string: abaa
regex:  a*ba(a|b)*

c.
string: abab
regex:  (ab)*(bb)*

d.
string: abaa
regex:  a*(ba|bb)*


















Consider the following grammar and input string.
The nonterminals in the grammar are { E }.
The terminals are { (,),+,-,*,/,x,y,2 }.
The start symbol is E.
Follow parts a-d below to show the first four steps of
a Derivation of the string using the grammar.

grammar:
	E -> (E + E) | (E - E) | (E * E) | (E / E)
	E -> x | y | 2

string:
	((2 * x) + (y / 2)) 

The initial string of grammar symbols in the Derivation is: E

a. Give the string of grammar symbols that results
from applying one grammar rule to the initial string.

b. Give the string of grammar symbols that results
from applying one grammar rule to the leftmost
nonterminal in the answer for part a.

c. Give the string of grammar symbols that results
from applying one grammar rule to the leftmost
nonterminal in the answer for part b.

d. Give the string of grammar symbols that results
from applying one grammar rule to the leftmost
nonterminal in the answer for part c.


















The following grammar is an ambiguous grammar for expressions.
The terminals in the grammar are enclosed in single quotes.
Write a new grammar for the same language that is not ambiguous.
Construct the grammar so that it enforces
correct operator precedence and associativity.
The precedence of the operators is given by the precedence table.
(Each row in the table is a different level of precedence.)
All binary operators are left-associative
except for '=' which is right-associative.

	E -> E '=' E
        E -> E '--'
	E -> E '++'
	E -> E '<' E
	E -> E '>' E
        E -> '(' E ')'
        E -> '4'

Precedence Table
highest	++ --
	< >
lowest	=


















Consider the following grammar.
The terminals in the grammar are enclosed in single quotes.
Follow parts a-d below to give the contents of some of the positions
in the parse table for a table-driven parser for the grammar.

E -> A X
X -> '=' E 
X -> epsilon
A -> B Y
Y -> '<' B Y
Y -> '>' B Y
Y -> epsilon
B -> C Z
Z -> '++' Z
Z -> epsilon
C -> '(' E ')'
C -> '4'

a. Give the content of the parse table at the single position on the row
indexed by the nonterminal Y and the column indexed by the terminal ')'

b. Give the content of the parse table at the single position on the row
indexed by the nonterminal A and the column indexed by the terminal '4'

c. Give the content of the parse table at the single position on the row
indexed by the nonterminal B and the column indexed by the endmarker '#'

d. Give the content of the parse table at the single position on the row
indexed by the nonterminal X and the column indexed by the terminal '='


















Consider the following grammar and input string.
The terminals in the grammar are enclosed in single quotes.
Follow parts a-d below to show the first four steps of a
stack-based parser when reading the string using the grammar.

Grammar:
	E -> T A
	A -> '+' T A | '-' T A | epsilon
	T -> '(' E ')' | '2'

String:
	2 + (2 - 2)

The initial contents of the stack and the input are:
   Stack: E#    Input: 	2+(2-2)#

a. Give the contents of the stack and the input after one parser step.

b. Give the contents of the stack and the input after a second parser step.

c. Give the contents of the stack and the input after a third parser step.

d. Give the contents of the stack and the input after a fourth parser step.


















Follow parts a-d below to show the steps of algebraically
transforming the initial expression into the final expression.

Initial expression:
	Q and ( not(not P or Q) or not (Q or not P))

Final expression:
	False

In each step below, use only one application of one law
(not including double negation, commutative, or associative laws)
from the Table of Algebraic Laws.
Do not use any laws not provided in the table.
You may use double negation, commutative, and associative
laws as an added part of any step without stating it's use.
In each step, the application of a law needs to
lead toward the final expression.
Note that in each step you need to use one of the laws
other than double negation, commutative, and associative.
In other words, if a step only uses double negation, commutative,
and associative laws, you will not earn points for the step.

a. Starting with the initial expression,
give the result of applying one law
and identify the law that you used.

b. Give the result of applying one law
to the expression given in part a
and identify the law that you used.

c. Give the result of applying one law
to the expression given in part b
and identify the law that you used.

d. Give the result of applying one law
to the expression given in part c
and identify the law that you used.

Table of Algebraic Laws

P and True	<=>  P			Identity
P or False	<=>  P			Identity
P or True	<=>  True		Domination
P and False	<=>  False		Domination
P or P		<=>  P			Idempotent
P and P		<=>  P			Idempotent
not (not P)	<=>  P			Double Negation
P or Q	 	<=>  Q or P		Commutative
P and Q		<=>  Q and P		Commutative
(P or Q) or R	<=>  P or (Q or R)	Associative
(P and Q) and R	<=>  P and (Q and R)	Associative
P or (Q and R)	<=>  (PvQ) and (PvR)	Distributive
P and (Q or R)	<=>  (P^Q) or (P^R)	Distributive
not (P and Q)	<=>  not P or not Q	DeMorgan's
not (P or Q) 	<=>  not P and not Q	DeMorgan's
P or not P	<=>  True		Negation
P and not P	<=>  False		Negation
P -> Q	 	<=>  not P or Q		Implication


















Let the following propositional variables represent
the associated propositions.

P - John brings his lunch
R - John eats lunch with Sue 
S - Sue has a cold
T - John gets a cold

Given these premises:

If John brings his lunch he eats lunch with Sue.
If John eats lunch with Sue and she has a cold then John gets a cold.
John brings his lunch.
Sue has a cold.

And this conclusion:

John gets a cold.

a. Write the premises and the conclusion as logic expressions.

b. Write the premises as clauses (disjunctions of literals).

c. Prove the conclusion using resolution and proof-by-contradiction.
Give a justification for each line in the proof.
Each line in the proof must be either a premise or
the result of applying the resolution rule.
Do not use any other inference rules in the proof.


















The domain of possible values for variables X and Y is
{Jim, Ann, Sal, Pat, Tom}.
The following facts define the values for which the child predicate is true.
The child predicate is false for all other cases.

child (Ann,Jim)
child (Sal,Jim)
child (Pat,Ann)
child (Tom,Sal)

Evaluate each expression using the domain values and predicates as
defined and indicate if the expression is true or false.

a. (all X) child(X,Jim)

b. (exists X) not child(X,Jim)

c. (all X)(exists Y)(child(X,Jim) -> child(Y,X))

d. (exists Y)(all X)(child(X,Jim) -> child(Y,X))

e. (exists X)(all Y)(child(X,Jim) -> child(Y,X))


















Consider the following facts, rules, and queries.

Facts:
   child('Tim','Bea').
   child('Jill','Ned').
   child('Ned','Bea').
   child('Ann','Jill').

Rules:
   sibling(C1,C2) :- parent(P,C1), parent(P,C2).
   parent(P,C) :- child(C,P).

Queries:
   sibling('Tim','Ned')?

a. Write the rules as clauses (disjunctions of literals). 

b. Write one formal proof of the query using
contradiction, instantiation, and resolution.
Give a justification for each line in your proof.
Each step must be a premise or the result of
instantiation or resolution. 


















In this problem you will use mathematical induction
to prove (for all integers n >= 1):

P(n): 1/2^1 + 1/2^2 + ... + 1/2^n = (2^n - 1)/2^n

(We will complete the base case for you as follows.)
Consider the base case when n = 1.
In this case, the summation on the left side of P(n)
contains just the first term (1/2^1) and has the value 1/2.
When n has value 1, the right side of P(n)
becomes (2^1 - 1)/(2^1) and also has the value 1/2.
Thus, the statement is true for the base case.

Complete parts a-d below to show that
the statement works for the inductive case.

a. State the inductive hypothesis.

b. State what is to be proven in the inductive step.

c. Give the result of using the inductive hypothesis (from part a)
to make a substitution into the summation side of part b.

d. Use algebra to show that the result for part c
is equivalent to the right side of part b.