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0 votes
2 answers
101 views

Busy-Beaver-like question for WHILE-Programs (Theoretical CS)

So this is exam-task is called "Busy WHILE-Programs" In our lecture it was proven that WHILE-Programs are turing-complete. In short a WHILE-Program only allows the following: ...
Edmundo Del Gusto's user avatar
1 vote
0 answers
89 views

What do we know about the reciprocal busy beaver series? [closed]

[Cross-posted from Math SE - apologies if this is not appropriate.] I just read these excellent lecture notes by Scott Aaronson, and I found the second homework problem at the end to be incredibly ...
tparker's user avatar
  • 1,146
8 votes
1 answer
264 views

Wanted: Concrete Example of Busy Beaver Holdout

I understand from the Wikipedia page on the Busy Beaver problem that the Busy Beaver values for 5-state 2-symbol (quintuple) Turing-machines have not been determined, because there are 'holdout' ...
Bram28's user avatar
  • 243
0 votes
0 answers
174 views

Busy Beaver Instruction Generator and Computablity

Question 1 From busy beaver uncomputablity it follows that we cannot design algorithm that for every input N, generate the appropriate set of TM instructions (the bit value of every bit in every card)...
Mendi Barel's user avatar
4 votes
4 answers
11k views

Understanding proof for Busy Beaver being uncomputable

I found this proof on http://jeremykun.com/2012/02/08/busy-beavers-and-the-quest-for-big-numbers/ and have highlighted the part I don't understand in bold. (BB(n) is defined as the number of steps ...
x squared's user avatar
  • 185
4 votes
1 answer
1k views

Busy Beaver problem - Proof by contradiction

I am trying to understand a proof regarding the Busy Beaver problem that uses a proof by contradiction approach to show $\sum(n)$ is Turing-uncomputable: Find $\sum(n) = max \{\sum(M) | M \in M(n) \...
Nubcake's user avatar
  • 249
6 votes
1 answer
1k views

Is a secondary TM sufficient to detect all loops?

Procedure: Start a secondary TM in parallel with the first, but have the second perform exactly 1 step each 2 steps the first TM performs (i.e. it runs at half speed). If the second machine ever ...
mafu's user avatar
  • 339
13 votes
3 answers
1k views

Computation of busy beaver function

The busy beaver max shifts function, $S(n)$, has known values for $n\leq4$. Is there some basic, structural reason why it's inconceivable that we will ever find $S(n)$ for $n>4$? What is so ...
PeteyPabPro's user avatar
3 votes
1 answer
173 views

Given an n-state TM, can we construct an m-state TM (m>n) to determine if it halts?

BB(n) is roughly the maximum number of new states an n-state TM can run into without halting. So for a particular n, if we know BB(n), then we can find out if an arbitrary n-state TM halts by running ...
hollow7's user avatar
  • 527