Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $\Sigma$ be Radó's Busy Beaver function, and let $\Delta[\Sigma]$ denote the forward difference of $\Sigma$, such that $\Delta[\Sigma] \ (n) = \Sigma(n+1) - \Sigma(n)$ for all $n \in \mathbb{N}$. Define $\Delta^0[\Sigma] = \Sigma$, and $\Delta^{k+1}[\Sigma] = \Delta[\Delta^{k}[\Sigma]]$, for all $k \in \mathbb{N}$.

Question: Is it the case that for all $k \in \mathbb{N}$, the function $\Delta^k [\Sigma]$ eventually dominates every computable function $f: \mathbb{N} \rightarrow \mathbb{N}$?

(I believe this is the case, but do not know how to prove it. Any pointers or sources on proving this -- or disproving it, if I'm wrong?)

Here's a difference table showing all the known values for these sequences:

     n:  0      1      2      3      4      5     ...
     ------------------------------------------------
     Σ:  0      1      4      6      13    ≥4098  ...
    ΔΣ:  1      3      2      7     ≥4085  ...
   ΔΔΣ:  2     -1      5     ≥4078  ...
  ΔΔΔΣ: -3      6     ≥4073  ...
 ΔΔΔΔΣ:  9     ≥4067  ...
ΔΔΔΔΔΣ: ≥4058  ...

Known sources: "A note on Busy Beavers and other creatures" (1996), by Ben-Amram, Julstrom, and Zwick, contains the conjecture that for any computable function $f$, there exists a constant $N_f$ such that $\forall n \gt N_f, \Sigma(n+1) > f(\Sigma(n))$. In support of this they prove the weaker result that for any computable function $f$, $\Sigma(n+1) > f(\Sigma(n))$ for infinitely many values of $n$. (Considering $f(x) = 2x$, for example, the conjecture implies that $\Delta[\Sigma]$ eventually dominates every computable function.)

Motivation: In the "Busy Beaver game" class of Turing machines, which eventually halt after starting with a blank tape, the "score" of a machine is the number of $1$s remaining on the tape after halting. A variant of Kolmogorov complexity $C: \mathbb{N} \rightarrow \mathbb{N}$ defines $C(n)$ to be the least $k$ such that $n$ is the score of some $k$-state machine. (cf "Computability and Logic" by Boolos, Burgess & Jeffrey, 2007, p229.) I believe I can show that if the first difference $\Delta[\Sigma]$ eventually dominates every computable function, then this $C$ is not monotonically nondecreasing, i.e., there exist $m \lt n$ such that $C(m) \gt C(n)$.

NB: Due to receiving no satisfactory answer here, I've posted the simplest case of this question on cstheory.SE.

share|improve this question
    
Have you tried posting this question on MathOverflow? –  Quinn Culver Feb 21 '12 at 15:41
add comment

2 Answers

For the eventual conclusion "there exist $m<n$ such that $C(m)>C(n)$" it seems to be easier just to prove that for some $k$ there exists a $k$-state machine with a score higher than the number of possible $k$-state machines. Then by the pigeonhole principle there has to be a number less than this that is reached only by a machine with $>k$ states.

Depending on the details, there are at most about $(6k)^{2k}$ different $k$-state machines, and this is easily a computable function. Therefore if add some initial states that preload a large enough $k$ (which can be done in $O(\log k)$ states) we will eventually reach a machine with a score that is high enough relative to its size.

share|improve this answer
    
Good points. Which approach is easier depends on how hard it is to prove the domination result for the first difference of $\Sigma$ (and maybe that's harder than I thought, assuming it's true!). But given that result, and the fact that it's easy to show $C(\Sigma(k-1) + 1) = C(\Sigma(k)) = k$, the non-monotonicity conclusion follows readily by considering the number of $k$-state machines as a computable function of $k$ that's eventually dominated by $\Sigma(k) - \Sigma(k-1)$. –  r.e.s. Nov 14 '11 at 16:45
2  
Hmmm ... I upvoted your reply although it's not an answer to my question -- not realizing that this removes the question from the "unanswered" list. I really would like to have a good answer re the $k$th difference of $\Sigma$. I think it would be a noteworthy result even to hold for only $k = 1$ (but I suspect it either holds for no positive $k$, or else it holds for all positive $k$). –  r.e.s. Nov 15 '11 at 19:03
add comment

Use http://mathworld.wolfram.com/NewtonsForwardDifferenceFormula.html to get $D^k [z] (n)>f(k,n)z(n+k)$ for some computable $f(k,n)$ and every strictly increasing positive function $z$. Thus if $g(n)>D^k [z](n)$ for some computable $g$, then there is a computable bound on any $z(n+k)$. Contradiction.

share|improve this answer
1  
I am still puzzled. Where do you get the first sentence from? It might also help if you specify exactly what $f(k,n)$ is. –  Srivatsan Nov 14 '11 at 13:36
    
(FYI: I've switched from $D$ to the standard $\Delta$ notation for the forward difference.) –  r.e.s. Nov 16 '11 at 4:04
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.