CSE 200 Spring 2010
Calibration Homework Solutions
Chris Calabro and Russell Impagliazzo
April 11, 2010
1
Regular languages
Prove that
Reg
k
+1
is a proper superset of
Reg
k
.
It is immediate from the deFnition that
Reg
k
+1
is a superset of
Reg
k
.
We
need to show that the inclusion is proper.
To do this, we need to exhibit a language
L
k
+1
in
Reg
k
+1
, i.e., that is
accepted by a
k
+ 1 state Fnite automaton, but not in
Reg
k
, that is, cannot be
accepted by any
k
state Fnite automaton.
Let
L
k
+1
=
{
x
∈ {
0
} * 
x

= 0
mod
k
+ 1
}
of strings over a onesymbol
alphabet whose length is evenly divisible by
k
+1. Let
A
be the automaton with
states 0
, ..k
, with 0 as the start state and only accepting state, and transitions
δ
(
q,
0) =
q
+ 1
mod
k
+ 1, i.e., upon reading the next symbol, we go to the next
state, unless we are in the last, in which case we go to the initial state.
We can see by induction that 0
l
ends in state
l
mod
k
+ 1. since this is true
for
l
= 0, and if 0
l
is in state
l
mod
k
+1, 0
l
+1
ends in state (
l
mod (
k
+1))+1
mod (
k
+ 1) = (
l
+ 1)
mod (
k
+ 1).
Since the only accepting state is 0,
then 0
l
is accepted if and only if
l
mod (
k
+ 1) = 0, which is by deFnition if and only if 0
l
∈
L
k
+1
.
Thus, automaton
A
is a
k
+ 1 state machine that accepts
L
k
+1
, so
L
k
+1
∈
Reg
k
+1
.
Let
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 Winter '12
 Edmonds
 Prime number, Integer factorization, Selfreference, Prime factor, Lk+1

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