Hardness of busy beaver value BB(15)

TL;DR

It is proved that knowing BB(15) is at least as hard as solving the following Collatz-related conjecture by Erd\H{o}s, open since 1979 [9]: for all n>8 there is at least one digit 2 in the base 3 representation of $2^n$.

Abstract

The busy beaver value BB(n) is the maximum number of steps made by any n-state, 2-symbol deterministic halting Turing machine starting on blank tape. The busy beaver function $n \mapsto \text{BB}(n)$ is uncomputable and, from below, only 4 of its values, BB(1) ... BB(4), are known to date. This leads one to ask: from above, what is the smallest BB value that encodes a major mathematical challenge? Knowing BB(4,888) has been shown by Yedidia and Aaronson [28] to be at least as hard as solving Goldbach's conjecture, with a subsequent improvement, as yet unpublished, to BB(27) [4,1]. We prove that knowing BB(15) is at least as hard as solving the following Collatz-related conjecture by Erdős, open since 1979 [9]: for all n > 8 there is at least one digit 2 in the base 3 representation of $2^n$. We do so by constructing an explicit 15-state, 2-symbol Turing machine that halts if and only if the conjecture is false. This 2-symbol Turing machine simulates a conceptually simpler 5-state, 4-symbol machine which we construct first. This makes, to date, BB(15) the smallest busy beaver value that is related to a natural open problem in mathematics, bringing to light one of the many challenges underlying the quest of knowing busy beaver values.

Figures (5)

• Figure 1: The first 75 powers of two assembled in base 3 by the size-6 Wang tile set introduced in thesis_sixtilesmawatam. Reading left-to-right, each column of glues (colours) corresponds to a power of two: beige glues represent ternary digit $0$, green glues $1$ and red glues $2$. For instance, the rightmost column encodes (from top to bottom) $110210021020202011202012000020112001021021222022_3 = 2^{75}$. The complexity of the patterns illustrates the complexity of answering Erdős conjecture which amounts to asking if each glue-column (except for the few first) has at least one red glue.
• Figure 2: The mul2 Finite State Transducer that multiplies a reverse-ternary represented number (base-3 written in reverse digit order) by $2$. For example, the base 10 number $64_{10}$ is $2101_3$ in base 3, which we represent in reverse-ternary with a leading zero to give the input $10120$, which in turn yields the FST output $20211$; the reverse-ternary of $128_{10}$. Transition arrows are labelled $r:w$ where $r$ is the read symbol and $w$ is the write symbol.
• Figure 3: 5-state 4-symbol Turing machine $M_{5,4}\xspace$ that halts if and only if Erdős' conjecture is false. The initial state of the machine is mul2_G, denoted '(init)'. The blank symbol is $\#$ and, since this is a busy-beaver candidate, the initial tape is empty: $\ldots \# \# \underline{\#} \# \# \ldots$ (tape head underlined). Example \ref{['ex:four']} shows the initial 333 steps of $M_{5,4}$. States mul2_F and mul2_G implement states $F$ and $G$ of the "${\bf\textrm{mul2}}\xspace$" FST in Figure \ref{['fig:fst']}, that multiplies a reverse-ternary number by 2. The other states check whether the result is a counterexample to, or one of the three special cases of, Erdős' conjecture.
• Figure 4: 15-state 2-symbol Turing machine $M_{15,2}\xspace$ that halts if and only if Erdős' conjecture is false. The initial state is mul2_G_ sim, denoted '(init)'. The blank symbol is $b$, and, since this is a busy-beaver candidate, the initial tape is empty: $\ldots b b {\pmb{\underline{b}}} b b \ldots$ (tape head position bold and underlined). States are organised into 5 columns, one for each state of $M_{15,2}\xspace$ in Figure \ref{['fig:four']}, and inherit name prefixes from $M_{15,2}\xspace$. In Lemma \ref{['lem:sim']} we prove that $M_{15,2}\xspace$ simulates $M_{5,4}\xspace$.
• Figure 5: The partial function $g$ of Lemma \ref{['lem:sim']} that defines how many steps are needed by $M_{15,2}\xspace$ to simulate one step of $M_{5,4}$, for each (state, move-direction-on-previous-step), and symbol, of an $M_{5,4}$ step. The function is partial, defined on only one entry of the final column, as the proof of Theorem \ref{['thm:four']} shows that only symbol $\#$ can be read if $M_{5,4}$ reaches state rewind coming from the left (hence by moving tape head in direction R).

Theorems & Definitions (21)

• Conjecture 1: Erdős ErdosPowers2
• Theorem 1
• Theorem 2
• Corollary 2
• Corollary 2
• Lemma 3
• proof
• Example 4
• Theorem 4
• proof