Solutions to Ass. 2 and few more questions from textbook

Chapter 5, Exercise 12, pg 72
i)
-
a
b
+
b
a
University of Ottawa
School of Information Technology and Engineering
a,b
FIN Not a FA
+
CSI 3104/3504, Winter 2006
Solution of Assignment #2
a
b
The changed FIN is not a FA because the start state has more than one path for a.
ii)
b
a
b
b
-
a
+
FIN NIF
-
+
b
a
a,b
iii) The language accepted by NIF is the reverse of every word in the language accepted by FIN. If there
was a path from one state to another on FIN by a sequence of letters, then there must be a path on NIF
using the sequence backwards.
iv)
-
+
FIN
a,b
The words accepted by this FA are not limited to palindromic words. It accepts the non-palindrome words
ab and FIN = NIF.
a
a
-
b

Chapter 5, Exercise 18, pg 74
We start in state 1 and remain there until we encounter an a. State 2 = we have just read
an a. Scan any a’s and return to state 1 on reading c. State 3 = we have read a b following an a. Reading
an a puts us back to state 2 and reading a b sets us back to state 1. However, state 4 = we have just found a
substring abc, and if the whole sequence was read the string is accepted.
Now states 4, 5 and 6 exactly mirror states 1, 2 and 3. Returning to state 1 indicates that we just found
another occurrence of the substring abc. Being in one of the first three states means that we have read an
even number of abc substrings (if any) and are in the midst of finding another one. Ending in an accepting
state, 4. 5 or 6, means that we have read an odd number of abc's.
Chapter 6, Exercise 20, pg 91
i) In a machine without Λ-edges, every transition consumes at least one letter of input. Therefore, we know
that Step 2 is true. The list in Step 3 is exhaustive; it includes all possible paths through the machine
(where a path is represented by the sequence of integers that number its consecutive edges) and also
sequences of integers that cannot represent paths because the edges are discontinuous in the graph. There is
an upper bound on the magnitude of the integers used as edge numbers and on the length of the sequence.
Since the list is finite , we can actually check it for the two required characteristics: 1) Does the sequence
represent a path from a start state to a final state, and 2) when we concatenate the edge labels, does the
result match the word we are testing? Since all possible paths are listed, if our test word has a path we must
find it. Since we test paths for validity at Step 4, we are not deceived by matching edge labels on an
invalid path. The list and the procedures we apply to it are finite and so this algorithm must terminate.
(ii) In a machine with Λ-edges, it is not true that every transition consumes at least one letter of input. We can use
infinitely many Λ-transitions without consuming any letters at all, and therefore we cannot set an upper
bound on the length of the path that gives us the test word.

Chapter 7, Exercise 1(v), pg 142
a
bb
-
-
a
a
ab
-
!
abb
ba
bb+abb+abba
+
a*(bb+abb+abba)(a+b)*
!
+ a,b
a
bb
a
bb
-
-
abba
The regular expression : a*(bb+abb+abba)(a+b)*
= a*bb(a+b)*
a,b
+
!
abb
!
+
abb+abba
+
!
a,b
a,b

Chapter 7, Exercise 3(ii), pg 143
Old state New State a New State b New State
W1, y1 Z1- W2, y2 Z2+ W1,y3 Z3
W2,y2 Z2+ W2,y4 Z4+ W1,y3 Z3
W1,y3 Z3 W2,y2 Z2+ W1,y4 Z5+
W2,y4 Z4+ W2,y4 Z4+ W1,y4 Z5+
W1,y4 Z5+ W2,y4 Z4+ W1,y4 Z5+
w1 or y1
-
a
b
w2 or y2
+
w1 or y3
Chapter 7, Exercise 5(i), pg 143
b
b
w1 w2 or x1
-
a
a
a
b
a
w2 or y4
+
+
w1 or y4
w2 or x1 or x2
a
b
a
b
w1 or x1
a
b
a
b
b
w1 or x3 or x1
+
b
a
b
w2 or x1
or x2 or x3
+
a

Chapter 7, Exercise 6(ii), pg 144
Old state New state a New state b New state
Z0+/- X2 Z2 X1 Z1
X1 Z1 X2 Z2 X1 Z1
X2 Z2 X2 Z2 X3,x1 Z3+
X3,x1 Z3+ X3,x1,x2 Z4+ X3,x1 Z3+
X3,x1,x2 Z4+ X3,x1,x2 Z4+ X3,x1 Z3+
+
-
a
a
x2
b
a
x1
b
b
+
b
x3 or x1
Chapter 8, Exercise 13, pg 167
i)
a
b
+
a
x1 or x2 or x3
ii) There is no way of knowing when the last letter is read and its time to print the letter that was read
first.

Chapter 9, Exercise 5, pg 185
L1: (a+b)b(a+b)*, it means all strings that has the second letter to be b;
L2: (a+b)*aa(a+b)*, it means all the strings that contain the substring aa;
x1
a,b
x2 x3
- +
FA1
b
a,b
a x4
a,b
w1 w2 w3
a
- +
L1 L2
x1 x2 x3
-
+
a,b
+
FA' 1
The FA of L'1 + L'2:
-
+
a
b
x2 or w2
+
x2 or w1
+
a
b
b
a
b
a
x4 or w3
+
x3 or w1
+
b
x4 or w2
+
a
+
x4
a
a,b
b
b
a,b
a,b
x3 or w2
+
a
x4 or w1
+
a
b
-
+
b
b
a
b
b
+
a
FA'2
a
FA2
w1 w2 w3
x3 or w3
a,b
a,b
a,b

The FA of L1 ∩ L2 = (L'1 + L'2)' is:
x1 or w1
-
a
b
x2 or w2
x2 or w1
a
b
a
x4 or w3
x3 or w1
b
x4 or w2
Simplifying the FA of L1 ∩ L2,
Regular Expression:
(a+b)b(a+b)*aa(a+b)*
b
z1
-
a
a,b
x3 or w2
a
a +
a
b
b
b
a
x4 or w1
z3
z2
a
a,b
b
x3 or w3
b
a,b
a
z4 z5
b
b
a
z6
+
a,b