Solution to CS 390 Test

July 23, 2003

1. Find a regular expression for each of the following languages over the alphabet {0, 1}:

(a) The language of all strings containing exactly two 1's.
Ans:   0^*10^*10^*

(b) The language of all strings that do not end in 01. [10 each]
Ans:   (0 + 1)^*(0 + 11) + 1^*

2. For the language represented by the regular expression (0^*1 + 01^*)^*, answer the following questions:

(a) Find a string of the shortest length.
Ans:  

(b) Find a string of length 4.
Ans:   Any string of 0's and 1's of length 4.

(c) Simplify the regular expression.
Ans:   (0 + 1)^*

(d) Describe the language in English as simply as possible. [5 each]
Ans:   The set of all strings of 0's and 1' including .

3. Prove that   |ax| = |xa|   by structural (general) induction, where a is an arbitrary symbol in the alphabet {0, 1}, x is an arbitrary string over the alphabet and |w| denotes the length of the string w.

Note that |w| is defined recursively as follows:
Basis Clause: || = 0
Inductive Clause: |ax| = |x| + 1 for any string x and a symbol a. [20]

Proof:
Basis Step: x = .
Since |ax| = |x| = |x| = |xa|, |ax| = |xa|.

Inductive Step: Assume that |ax| = |xa| for an arbitrary string x.
We show that for an arbitrart symbol b of the alphabet, |abx| = |bxa|.
By the definition of |w|, |abx| = |bx| + 1 = |x| + 1 + 1.
Also by the definition of |w|, |bxa| = |xa| + 1.
Since by the induction hypothesis |ax| = |xa|, |bxa| = |xa| + 1 = |ax| + 1 = |x| + 1 + 1.
Hence |abx| = |bxa|.

4 (a) Find an FA that recognizes the same language as the following NFA and list the accepting states of the FA. Note that { } below denotes the empty set. [20]

State	a	b
0	{1, 2}	{ 2 }
1	{ 3 }	{2, 3}
2	{ 3 }	{ }
3	{ 1 }	{ 1 }

The initial state is state 0 and the accepting state is state 1.

Ans:  

State	a	b	State	a	b
{ 0 }	{1, 2}	{ 2 }	{ 1 }	{ 3 }	{2, 3}
{1, 2}	{ 3 }	{2, 3}	{1, 3}	{1, 3}	{1, 2, 3}
{ 2 }	{ 3 }	{ }	{1, 2, 3}	{1, 3}	{1, 2, 3}
{ 3 }	{ 1 }	{ 1 }	{ }	{ }	{ }
{2, 3}	{ 1, 3 }	{ 1 }

The accepting states are { 1 }, {1, 2}, {1, 3}, {1, 2, 3}.

(b) Find *(0, abb), where is the transition function of the NFA. [5]

Ans:   { 1 }.

5. Prove that for an arbitrary string x over the alphabet {a, b}, (x^r)^r = x, where x^r denotes the reversal of string x.

Note that w^r for a string w is defined recursively as follows:
Basis Clause: ^r =
Inductive Clause: For any arbitrary string x and symbol a, (ax)^r = x^ra. [15]

Proof:
Basis Step: x = .
Since ^r = , (^r)^r = ^r = .
Hence (x^r)^r = x for x = .

Inductive Step: Assume that (x^r)^r = x for an arbitrary string x.
We show that for an arbitrary symbol a, ((ax)^r)^r = ax .
Since by the definition of the reversal (ax)^r = x^ra, ((ax)^r)^r = (x^ra)^r.
Since (xy)^r = y^rx^r for arbitrary strings x and y, (x^ra)^r = a((x)^r)^r
(You can also use (wa)^r = aw^r here).
Since (x^r)^r = x by the induction hypothesis, a((x)^r)^r = ax.
Hence ((ax)^r)^r = ax .