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A Busy Beaver champion derived from scratch

99 pointsby nickdrozdover 3 years ago

6 comments

throwaway81523over 3 years ago
Summary: brute force search of the space of 4-state TM programs yielded a BB champion with an initially mysterious structure. Careful examination and reverse engineering showed that the program turns out to compute a variant of the Collatz function, where instead of C(n:odd)=3n+1, we have L(3k)=0 and L(3k+r)=L(5k+r+3). There is discussion of a &quot;spaghetti conjecture&quot; that BB champions are likely to be &quot;messes&quot; (i.e. have complete control flow graphs), but this Collatz example doesn&#x27;t satisfy that criterion. Thus the author suggests that the spaghetti conjecture is false.<p>Discussion: the whole thing is very encoding dependent, but if you figure the BB machine on N states is randomly distributed on all the N-state machines, then for the types of encodings being discussed, for large N, the flow graph has (I think) a high chance of being complete but that is not guaranteed. So that&#x27;s a reason to think the conjecture is strictly false but approximately true.<p>The BB champ being a Collatz variant is iirc a well known approach to manually constructed BB champs. That is inspired by an old result (Conway&#x27;s?) found in Jeff Lagarias&#x27;s book on the 3n+1 problem. It says the generalized Collatz conjecture is Pi-0-2 complete so specific instances sound like a good place to look for functions that are on the edge of undecidability.<p>I wonder if anything is known about the arithmetic (not computational) complexity of random programs like that. I&#x27;m imaginging e.g. Shannon&#x27;s theorem that most boolean functions on N inputs require exponentially many gates to compute. The analogy would be with the likelihood of functions on some parameter, expressed as N-state programs, being Pi-0-2 complete.
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jakearover 3 years ago
Great writeup! I ended up following a few of the links and came across perhaps my favorite hand waving of all time for the validity of &quot;The Spaghetti Code Conjecture&quot;[1]. Basically:<p>For N=0,1: the problem is trivaial&#x2F;degenerate.<p>For N=2: the Turing machine must go back and forth between states or else it&#x27;d just be stuck in one state forever, which wouldn&#x27;t be very busy beaverish.<p>For N=n+1: well we just proved it up till n, so it seems pretty unlikely that n+1 would be different, things don&#x27;t really just suddenly change like that y&#x27;know?<p>QED<p>So I&#x27;m not too surprised that this conjecture is being called into question.<p>[1]: <a href="https:&#x2F;&#x2F;nickdrozd.github.io&#x2F;2021&#x2F;01&#x2F;26&#x2F;spaghetti-code-conjecture.html" rel="nofollow">https:&#x2F;&#x2F;nickdrozd.github.io&#x2F;2021&#x2F;01&#x2F;26&#x2F;spaghetti-code-conjec...</a>
Y_Yover 3 years ago
Cool post. Moving the goalposts a little bit, since it&#x27;s not the same Busy Beaver problem that&#x27;s normally considered. I think that its an interesting problem in its own right, but deserves a different label, since the beaver never halts but rather gets bored and wanders off.
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FartyMcFarterover 3 years ago
Can something really be called a Turing machine if it&#x27;s not able to halt (or continue going) by itself?<p>In this case it appears that the author considers halting to be decided externally to the program, which is not the mechanism Turing machines use.
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brileeover 3 years ago
a typo in the collatz description:<p>C (2k + 1) = C (3k + 2)<p>should be<p>C (2k + 1) = C (6k + 4)
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codeulikeover 3 years ago
So what does &quot;colour&quot; means with regards to Turing Machines?
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