CSE 200 Spring 2010
Calibration Homework Solutions
Chris Calabro and Russell Impagliazzo
April 10, 2012
1 Regular languages
Prove that Regk+1 is a proper superset of Regk.
It is immediate from the definition that Regk+1 is a superset of Regk. We
need to show that the inclusion is proper.
To do this, we need to exhibit a language Lk+1 in Regk+1, i.e., that is
accepted by a k+ 1 state finite automaton, but not in Regk, that is, cannot be
accepted by any kstate finite automaton.
Let Lk+1 ={x∈ {0} ∗ ||x|= 0 mod k+ 1}of strings over a one-symbol
alphabet whose length is evenly divisible by k+1. Let Abe 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 0lends in state lmod k+ 1. since this is true
for l= 0, and if 0lis in state lmod k+ 1, 0l+1 ends in state (lmod (k+1))+ 1
mod (k+ 1) = (l+ 1) mod (k+ 1).
Since the only accepting state is 0, then 0lis accepted if and only if l
mod (k+ 1) = 0, which is by definition if and only if 0l∈Lk+1.
Thus, automaton Ais a k+ 1 state machine that accepts Lk+1, so Lk+1 ∈
Regk+1.
Let A0be any at most kstate automaton. Consider the states A0reaches
on λ, 0,00,000, ...0k, where λ= 00is the empty string. Since there are k+ 1
such strings, and A0has fewer than k+ 1 states, two of these end at the same
state of A0. Let these two be 0i,0j, with 0 ≤i<j ≤k. Then 0i0k+1−j=
0k+1+i−jmust end at the same state as 0j0k+1−j= 0k+1. Now 0k+1 ∈Lk+1 ,
but 0k+1+i−j6∈ Lk+1, since its length is between 1 and ksince 0 ≤i<j≤k.
Thus, if the state reached on both strings is accepting A0accepts a string not
in Lk+1, and if it is rejecting, A0rejects a string in Lk+1 . So in neither case
does A0recognize Lk+1. Since A0was an arbitrary DFA with at most kstates,
Lk+1 6∈ Regkas needed, so Regk6=Regk+1 and the inclusion is proper.
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