CSE396
Problem Set 4 Answer Key
Spring 2016
(1) Write a regular expression R over the alphabet {0, 1, 2} such that L(R) equals the
language A of numbers in base-3 that are congruent to either 0 or 3 modulo 4. Count the
empty string as belonging to L(R) (i.e., it counts as zero), and allow leading 0s in the numbers.
First design a DFA M such that L(M ) = A. Then show the conversion to a regular
expression. For good measure, give s PD set of size 4 for this language based on your M .
As a hint for designing M , use four states for the possible congruences. The idea of an
“even bank” and “odd bank” like in 1(g) of PS2 may help too. (24 pts.)
Answer: For the DFA M and the algorithm to find a regular expression R see the separate
hand-drawn sheet. Briefly, if you draw the M with states q0 , q2 for the even numbers mod 4
on the left and q1 , q3 for teh odd congruences on the right, it makes a nice square with arcs on
1 going straight across between the “banks,” arcs on 2 going up and down the “even bank”
with 0 as self-loops, and the roles of 0 and 2 reversed on the “odd bank.” The sheet shows
eliminating the state q2 first; elimiating q1 first comes downto the same 2-state GNFA with
both states q0 , q3 accepting and arcs as follows:
α
β
γ
η
=
=
=
=
T (0, 0) = 0 + 20∗ 2 + 12∗ 1
T (0, 3) = 20∗ 1 + 12∗ 0
T (3, 3) = 2 + 10∗ 1 + 02∗ 0
T (3, 0) = 10∗ 2 + 02∗ 1
Then L(M ) = L0,0 ∪ L0,3 . From the notes (which wrote “1,2” in place of “0,3”) the formulas
are:
L0,0 = (α ∪ βγ ∗ η)∗
L0,3 = α∗ β(γ ∪ ηα∗ β)∗
= (α ∪ βγ ∗ η)∗ βγ ∗ .
The latter form gives L0,3 = L0,0 βγ ∗ . Then for the final union you can factor out L0,0 to
give L(M ) = R = L0,0 ( ∪ βγ ∗ ). I don’t know whether this particular answer can be usefully
simplified further—so long as you made clear you knew L(M ) = L0,0 ∪ L0,3 and gave one
or other of the above labeled forms that was OK. It was not required to do a final yucky
substitution to get a mess of 0s, 1s, 2s, and *s—though if you followed the text uncritically
by adding a new final state f with epsilon-arcs (q0 , , f ) and (q3 , , f ) and eliminated state q3
too you’d be locked into the yuck.
(2) Prove that the following two languages are non-regular via the Myhill-Nerode Theorem.
Both use alphabet Σ = {0, 1}. Here xR stands for x reversed, e.g. 0111R = 1110. The condition
x = xR means that x is a palindrome, but the members of La are a little more restrictive,
including 1001 but not 10001. (2 × 12 = 24 pts.)
(a) La = {xxR : x ∈ Σ∗ }.
(b) Lb = {x ∈ Σ∗ : #0(x) > #1(x)}.
Answer to (a): The idea is that “critical first halves” can be 0n 1 where the final 1 makes
a definite boundary to the rest of the string that ahs to balance it. So take S = 0∗ 1. Clearly
S is infinite. Let any x, y ∈ S, 6= y, be given. Then there are distinct numbers m, n ∈ N such
that x = 0m 1 and y = 0n 1. Take z = 10m . Then xz = 0m 110m ∈ La since the halves balance
to make an even-length palindrome, but yz = 0n 110m ∈
/ La since it is not a palindrome at
all—the ‘1’s prevent any way of trying to break the parts with ‘0’s differently to balance them.
So L(xz) 6= L(yz). Since x, y ∈ S are arbitrary distinct members, this shows S is PD for La ,
and since S is infinite, La is not regular by the Myhill-Nerode Theorem. (It is fine to take
just S = 0∗ , but then you have to remember later that z needs to be 110m with two ‘1’s, not
one or none. Another mistake was S = 10∗ and later z = 0m 1 which is mirror-image wrong:
if m = 3 and n = 5 then yz = 105 03 1 = 1000000001 = 104 04 1 which belongs to La after all.)
Answer to (b): Take S = 0∗ . Clearly S is infinite. Let any x, y ∈ S, 6= y, be given. Then
we can helpfully write x = 0m and y = 0n where m, n are natural numbers such that without
loss of generality m < n. Take z = 1m . Then xz = 0m 1m ∈
/ Lb since #0(xz) = m is not
greater than #1(xz) = m, but yz = 0n 1m ∈ Lb since n > m by the “wlog.” provision. Hence
L(xz) 6= L(yz). Since x, y ∈ S are arbitrary, this shows S is PD for Lb , and since S is infinite,
Lb is not regular by the Myhill-Nerode Theorem.
(3) Prove that the following language is nonregular via the Myhill-Nerode Theorem.
L3 = { ai bj ck : i = j or i 6= k }.
(18 pts., for 66 total on the set)
Answer: To make the OR statement critical we should falsify the i = j part since it comes
first in the relevant strings. We can handle it either entirely with our choice of S or split
between S and z. In the latter case, it helps to make i = j false by fixing j = 0 which lets i
be any arbitrary positive number. So take S = a+ = {an : n ≥ 1}. Clearly S is infinite. Let
any x, y ∈ S, x 6= y, be given. Then there are positive natural numbers m, n ∈ N+ such that
x = am and y = an . Take z = cm . Then xz = am cm = am b0 cm ∈
/ L3 since it fails both the
“i = j” and “i 6= k” tests when you substitute i = m, j = 0, and k = m (noting m ≥ 1 by
the definition of S). But yz = am cn ∈ L3 since m 6= n, so L3 (xz) 6= L3 (yz). It follows that S
is an infinite PD set for L3 , so L3 is nonregular.
An example of the other way is to take S = {am bm+1 : m ≥ 0}. No one says S has to be
regular—it only has to be infinite. This automatically falsifies the i = j part. Let any x.y ∈ S,
x 6= y, be given. Then there are numbers m 6= n such that x = am bm+1 and y = an bn+1 . (Note
how this is truly a general choice from this particular S, in contrast to the fallacy mentioned
in lecture notes of making n = m + 1.) Take z = cm . The rest of the reasoning is as before.
(There are other correct answers too.)