Challenge about NFA for {a^y : y\ne 1000} answered.

Recall that in a prior post I asked

Is there an NFA for  { a^y : y ≠ 1000 } with substantially less than 1000 states.

I will now show that any NFA for this set requires 999 states, so essntially 1000. The proof uses Ramsey Theory. I will tell you the little bit of Ramsey Theory that you need.

NO- the above is false.

There is an NFA with 60 states.  I have a complete exposition here and I have a paper (with coauthors) on cofinite unary languages and NFA's here). Generally if you want to only avoid
aⁿ the NFA an be done in sqrt(n) + O(\log n) states, but requires sqrt(n) states.

I sketch the ideas here for the 1000.

Convention- `reject 988' would mean rejecting a⁹⁸⁸.

It is easy show the following:
For all n \ge 992 there exists x,y\in N such that n = 32x +33y

For all x,y  32x+ 33y is NOT = 991.

If you have  an NFA with a 33-loop and a shortcut so you can also to 32 and back to the start
state, this NFA

accepts all y ≥ 992

rejects 991

we have no comment on anything else.

So if you prepend 9 states to that NFA you will have an NFA that

accepts all y ≥ 1001

rejects 1000

How to get all the numbers < 1000?

Use mod 3, mod 5, mod 7, mod 11 loops that only accept if the number is NOT equiv to 1000
mod 3, 5, 7, 11, Since 3*5*7*11 > 1000 we have

if y is rejected then y ≤ 1000, but y \equiv 1000 mod 3*5*7*11, so must have y=1000.

so 1000 is the only reject.

Number of states: 1 + 33 + 3 + 5 + 7 + 11 = 60 states.

I think you can do this in 58, but what is the BEST you can do?
My paper has lower bounds of sqrt(n) so in this case 32.

This is a GREAT theorem to teach to students in a course that covers regular langs and also P vs NP since the students THINK that an NFA cannot do better - AH BUT IT CAN! - so it gives a concrete example of

lower bounds are hard since someone may come along with a very clever trick you didn't think of

This semester when I had the class VOTE on whether or not there was a small NFA

48 thought that ANY NFA would require around 1000 states

One of the best students proved this! or perhaps ``proved'' this

Only 2 students knew that it could be done with MUCH LESS than 1000 states- and those two are exceptions- one is co-author on the paper (and he tells me that he originally thought it required 1000 when I first showed him the problem) and one is someone who often goes to my old course websites looking for more information and problems to work on (I like that ambition!) and came across material on this. 

I tell them `you thought this was the last lecture on reg langs. It was not. it was the first lecture on P vs NP'