On the Salient Limits of Strings in Assembly Theory

1. Introduction

Assembly theory (AT), formulated in 2017, introduced the concept of an initial pool [1].

Definition 1. 

We call a set $P_0\left(b\right)\{0,1,\dots ,b-1\}$ that contains different basic symbols c, where b is a finite natural radix, the initial assembly pool.

The reader will find numerous results on AT published since 2017 in refs. [1,2,3,4,5,6,7,8,9].

In this short note, we extend the results of our previous study [9] to strings of any natural radix b. We consider the formation of strings $C_k^{(N,b)}$ of length N containing symbols from the initial assembly pool within the AT framework in consecutive assembly steps from basic symbols c and strings assembled in previous steps.

Definition 2. 

We call a set $P_s\left(b\right)$ that contains basic symbols and strings assembled in previous steps $\{1,2,\dots ,s\}$ the working assembly pool.

Using the Definitions 1 and 2, the assembly index of a string is the minimal achievable value of a difference between the cardinalities of the working and initial assembly pools leading to this string, since at each assembly step the cardinality of the working assembly pool increases by one. Therefore, in contrast to the working assembly pool 2, the initial assembly pool 1 must not contain strings of basic symbols. To illustrate this, consider the following mapping between such a faulty initial assembly pool containing five basic symbols and three strings of these symbols and the initial assembly pool of radix $b=8$

$$\begin{array}{cc}\hfill \cancel{P_0\left(5\right)}& \leftrightarrow P_0\left(8\right)\hfill \\ \hfill 0& \leftrightarrow 0\hfill \\ \hfill 1& \leftrightarrow 1\hfill \\ \hfill 2& \leftrightarrow 2\hfill \\ \hfill 3& \leftrightarrow 3\hfill \\ \hfill 4& \leftrightarrow 4\hfill \\ \hfill 20& \leftrightarrow 5\hfill \\ \hfill 201& \leftrightarrow 6\hfill \\ \hfill 2012& \leftrightarrow 7\hfill \end{array}$$

(1)

Now consider the string

$$C_k^{(11,5)}=\left[20123242012\right]$$

(2)

assembled beginning with the initial assembly pool $P_0\left(5\right)$ and having the assembly index $a^{(11,5)}\left(C_k\right)=7$ only two steps above $a_{\min }^{\left(11\right)}=5$. We can assemble the string

$$C_l^{(8,8)}=\left[20123247\right]$$

(3)

of length $N=8$ in 7 steps with the initial assembly pool $P_0\left(8\right)$ and then, using the mapping (1), it will correspond to the string (2). However, as we shall show in the following section, $N_{\max }\left(8\right)=73\ne 7$. In fact the latter string (3) should be assembled as

$$C_m^{(5,8)}=\left[73247\right]$$

(4)

with the assembly index $a^{(5,8)}\left(C_k\right)=5-1=4$ and with the initial assembly pool $P_0\left(8\right)$, as $2012\leftrightarrow 7$ according to (1).

The following two theorems were already stated in our previous study [9] for $b=2$. We restate them here $\forall b$ for clarity.

Theorem 1. 

A string of length $N=4$ is the shortest string that allows for more than one assembly index for all b.

Proof. 

$N=2$ provides $b^2$ available strings with unit assembly indices. $N=3$ provides $b^3$ available strings with assembly indices equal to two. Only $N=4$ provides $b^4$ strings that include b strings $C_k^{(4,b)}=[\ast \ast \ast \ast ]$ and $b(b-1)$ strings $C_k^{(4,b)}=[\ast ★\ast ★]$ with assembly indices equal to two, while the assembly index of the remaining strings is three. For example, to assemble the string $C_k^{(4,4)}=\left[0202\right]$, we need to assemble the string $\left[02\right]$ and reuse it from $P_1$, while there is nothing available to reuse, in the case of the string $C_l^{(4,4)}=\left[0123\right]$. □

Where the symbol value can be arbitrary, we write * assuming that it is the same within the string. If we allow for the $2^{\mathrm{nd}}$ possibility different from *, we write ★. Thus, $C_k^{(2,b)}=[\ast \ast ]$, for example, is a placeholder for all b strings, while $C_l^{(2,b)}=[\ast ★]$ a placeholder for all $b(b-1)$ strings.

Theorem 2. 

The smallest string assembly index $a^{\left(N\right)}\left(C_{\min }\right)$ as a function of N corresponds to the shortest addition chain for N (OEIS A003313) for all b.

Proof. 

Strings $C_{\min }$ for which $a^{\left(N\right)}\left(C_{\min }\right)=\underset{k}{min}\left(\left\{a^{(N,b)}\left(C_k\right)\right\}\right)$, $\forall k\in \{1,2,\cdots ,b^N\}$ can be formed in subsequent steps s by joining the longest string assembled so far with itself until $N=2^s$ is reached. Therefore, if $N=2^s$, then $\underset{k}{min}\left(\left\{a^{\left(2^s\right)}\left(C_k\right)\right\}\right)=s={log}_2\left(N\right)$. Only $b^2$ strings have such an assembly index in this case, including b strings

$$C_k^{(2^s,b)}=[\ast \ast \cdots ],$$

(5)

and $b(b-1)$ strings

$$C_k^{(2^s,b)}=[\ast ★\ast ★\cdots ],$$

(6)

and the assembly pathway of each of the strings (5) and (6) is unique. At each assembly step, its length doubles.

An addition chain for $N\in \mathbb{N}$ having the shortest length $s\in \mathbb{N}$ (commonly denoted as $l\left(N\right)$) is defined as a sequence $1=a_0<a_1<\cdots <a_s=N$ of integers such that $\forall j\ge 1$, $a_j=a_k+a_l$ for $l\le k<j$. The first step in creating an addition chain for N is always $a_1=1+1=2$ and this corresponds to assembling a doublet $[\ast \ast ]$ or $[\ast ★]$ from the initial assembly pool P. Thus, the lower bound for s of the addition chain for N, $s\ge {log}_2\left(N\right)$ is achieved for $N=2^s$ by strings (5) and (6) .

The second step in creating an addition chain can be $a_2=1+1=2$ or $a_2=1+2=3$. Thus, finding the shortest addition chain for N corresponds to finding an assembly index of a string containing basic symbols and/or doublets and/or triplets containing these doublets for $N\ne 2^s$ since due to Theorem 1 only they provide the same assembly indices $\{0,1,2\}$. □

2. Results

The seven-bit string is the longest string that can have the maximum assembly index $a_{\max }^{(7,2)}=7-1=6$. There are four such bitstrings containing two clear triplets and the starting bit at the end or the ending bit at the start, that is

$$[\ast \ast \ast ★★★\ast ]\mathrm{and}[★\ast \ast \ast ★★★],$$

(7)

and their lengths cannot be increased without a repetition of a doublet, which inevitably reduces the assembly index to $a_{\max }^{(8,2)}=8-2=6$.

This observation and Theorem 2 motivated us to develop a general procedure to construct the longest possible string that has the assembly index $a_{\max }^{(N,b)}=N-1$, as a function of the radix $b\ge 3$. We denote the length of this string by $N_{\max }\left(b\right)$.

After a few groping try-outs (cf. Appendices Appendix A and Appendix B) we eventually reached a stable procedure. We start with an initial balanced string of length $3b$ containing b clear triplets ordered as

$$[0001112\cdots (b-2\left)\right(b-1\left)\right(b-1\left)\right(b-1\left)\right].$$

(8)

The doublets that can be inserted into the initial string (8) can be arranged in a $b\times b$ matrix

$$\left[\begin{array}{ccccc}\cancel{\bcancel{00}}& 01& 02& \cdots & 0(b-1)\\ 10& \cancel{\bcancel{11}}& 12& \cdots & 1(b-1)\\ 20& 21& \cancel{\bcancel{22}}& \cdots & 2(b-1)\\ \cdots & \cdots & \cdots & \cdots & \cdots \\ (b-2)0& (b-2)1& (b-2)2& \cdots & (b-2)(b-1)\\ (b-1)0& (b-1)1& (b-1)2& \cdots & \cancel{\bcancel{(b-1)(b-1)}}\end{array}\right],$$

(9)

where the crossed out entries on diagonal cannot be reused, as they would create repetitions in this string. If we assume that we shall not insert doublets between the clear triplets of the string (8) and hence we can also cross out the entries on the first superdiagonal in the matrix (9).

In the $1^{\mathrm{st}}$ step, we create a string containing doublets on the first subdiagonal of the matrix (9) starting with 10

$$[102132\cdots (b-2\left)\right(b-3\left)\right(b-1\left)\right(b-2\left)\right],$$

(10)

and we append it to the string (8). With this step, we also eliminate the doublets on the second superdiagonal starting with the doublet 02, as well as the doublet $(b-1)1$. In the $2^{\mathrm{nd}}$ step, we create a string containing doublets on the third superdiagonal beginning with the doublet 03

$$[0314\cdots (b-5\left)\right(b-2\left)\right(b-4\left)\right(b-1\left)\right],$$

(11)

and append it to the string created so far. With this step, we also remove the doublet $(b-2)0$ and the middle part of the second subdiagonal containing $\{31,42,\cdots ,(b-2\left)\right(b-4\left)\right\}$. And so on.

We shall illustrate this process for $b=8$. The matrix

$$\left[\begin{array}{cccccccc}\cancel{\bcancel{00}}& \cancel{\bcancel{01}}& \cancel{02}& \begin{array}{c}\hfill 03\end{array}& \begin{array}{c}\hfill 04\end{array}& \begin{array}{c}\hfill 05\end{array}& \cancel{06}& \cancel{07}\\ \begin{array}{c}\hfill 10\end{array}& \cancel{\bcancel{11}}& \cancel{\bcancel{12}}& \cancel{13}& \begin{array}{c}\hfill 14\end{array}& \begin{array}{c}\hfill 15\end{array}& \begin{array}{c}\hfill 16\end{array}& \cancel{17}\\ \begin{array}{c}\hfill 20\end{array}& \begin{array}{c}\hfill 21\end{array}& \cancel{\bcancel{22}}& \cancel{\bcancel{23}}& \cancel{24}& \begin{array}{c}\hfill 25\end{array}& \begin{array}{c}\hfill 26\end{array}& \begin{array}{c}\hfill 27\end{array}\\ \begin{array}{c}\hfill 30\end{array}& \cancel{31}& \begin{array}{c}\hfill 32\end{array}& \cancel{\bcancel{33}}& \cancel{\bcancel{34}}& \cancel{35}& \begin{array}{c}\hfill 36\end{array}& \begin{array}{c}\hfill 37\end{array}\\ \cancel{40}& \cancel{41}& \cancel{42}& \begin{array}{c}\hfill 43\end{array}& \cancel{\bcancel{44}}& \cancel{\bcancel{45}}& \cancel{46}& \begin{array}{c}\hfill 47\end{array}\\ \cancel{50}& \cancel{51}& \cancel{52}& \cancel{53}& \begin{array}{c}\hfill 54\end{array}& \cancel{\bcancel{55}}& \cancel{\bcancel{56}}& \cancel{57}\\ \cancel{60}& \begin{array}{c}\hfill 61\end{array}& \cancel{62}& \cancel{63}& \cancel{64}& \begin{array}{c}\hfill 65\end{array}& \cancel{\bcancel{66}}& \cancel{\bcancel{67}}\\ \cancel{70}& \cancel{71}& \cancel{72}& \cancel{73}& \begin{array}{c}\hfill 74\end{array}& \begin{array}{c}\hfill 75\end{array}& \begin{array}{c}\hfill 76\end{array}& \cancel{\bcancel{77}}\end{array}\right],$$

(12)

contains all the doublets that were used to create the string of length $N_{\max }\left(8\right)=73$

$$\begin{array}{cc}\hfill & \left[000111222333444555666777\right|\begin{array}{c}\hfill 10213243546576\end{array}|\hfill \\ \hfill & \begin{array}{c}\hfill 0314253647\end{array}|\begin{array}{c}\hfill 04152637\end{array}|\begin{array}{c}\hfill 2075\end{array}|\begin{array}{c}\hfill 051627\end{array}|\begin{array}{c}\hfill 306174\end{array}\left|0\right],\hfill \end{array}$$

(13)

For $b=7$ we would obtain the string of length $N_{\max }\left(7\right)=57$

$$\begin{array}{cc}\hfill & \left[000111222333444555666\right|\begin{array}{c}\hfill 102132435465\end{array}|\hfill \\ \hfill & \begin{array}{c}\hfill 03142536\end{array}|\begin{array}{c}\hfill 041526\end{array}|\begin{array}{c}\hfill 2064\end{array}|\begin{array}{c}\hfill 0516\end{array}\left|\begin{array}{c}\hfill 30\end{array}\right].\hfill \end{array}$$

(14)

for $b=6$ we would obtain the string of length $N_{\max }\left(6\right)=43$

$$\left[000111222333444555\right|\begin{array}{c}\hfill 1021324354\end{array}\left|\begin{array}{c}\hfill 031425\end{array}\right|\begin{array}{c}\hfill 0415\end{array}\left|\begin{array}{c}\hfill 2053\end{array}\right|0],$$

(15)

for $b=5$ we would obtain the string of length $N_{\max }\left(5\right)=31$

$$\left[000111222333444\right|\begin{array}{c}\hfill 10213243\end{array}\left|\begin{array}{c}\hfill 0314\end{array}\right|\begin{array}{c}\hfill 04\end{array}\left|\begin{array}{c}\hfill 20\end{array}\right],$$

(16)

for $b=4$ we would obtain the string of length $N_{\max }\left(4\right)=21$

$$\left[000111222333\right|\begin{array}{c}\hfill 102132\end{array}\left|\begin{array}{c}\hfill 03\end{array}\right|0],$$

(17)

and $b=3$ leads to the following string of length $N_{\max }\left(3\right)=13$

$$\left[000111222\right|\begin{array}{c}\hfill 10\end{array}\left|\begin{array}{c}\hfill 20\end{array}\right].$$

(18)

The final string is always terminated by 0.

The strings of odd lengths generated by the general procedure outlined above are not only the longest, but also the most balanced. This leads to the following theorem.

Theorem 3. 

The longest length $N_{\max }\left(b\right)$ of a string composed of b different basic symbols that has the assembly index of $N-1$ is given by

$$N_{\max }\left(b\right)=3b+{(b-1)}^2=b^2+b+1$$

(19)

(OEIS A353887) and this string is nearly balanced, that is

$$N_{\max }\left(b\right)=b{N}_c\left(b\right)+1,$$

(20)

where $N_c=b+1$ is the number of occurrences of all but one symbol within the string.

Proof. 

The $N_{\max }\left(b\right)$ given by formula (19) is an odd number for all b. As shown in Table 1, the first element $3b$ is the length of the initial string (8) containing b clear triplets and ${(b-1)}^2$ is the number of entries in the doublet matrix (9) of the previous b. By definition, a string of length $N_{\max }\left(b\right)$ cannot have any repetitions; it can only contain doublets and clear triplets that do not contain these doublets. Therefore, to be the most patternless, this string must maximize Shannon entropy; must be the most balanced. For the string of the form (20) the fractions in the Shannon entropy are

$$p_0={\displaystyle \frac{N_c\left(b\right)+1}{N_{\max }\left(b\right)}},p_{1,2,\cdots ,b-1}={\displaystyle \frac{N_c\left(b\right)}{N_{\max }\left(b\right)}},$$

(21)

(where without loss of generality we assume that the symbol occuring $N_c\left(b\right)+1$ times within the string is $c=0$) and the Shannon entropy is

$$\begin{array}{cc}\hfill H_{\max }\left(b\right)=-\sum _{c=0}^{b-1}{p}_c{log}_2\left(p_c\right)& =-(b-1){\displaystyle \frac{N_{\max }\left(b\right)-1}{b{N}_{\max }\left(b\right)}}{log}_2\left({\displaystyle \frac{N_{\max }\left(b\right)-1}{b{N}_{\max }\left(b\right)}}\right)-{\displaystyle \frac{N_{\max }\left(b\right)-1+b}{b{N}_{\max }\left(b\right)}}{log}_2\left({\displaystyle \frac{N_{\max }\left(b\right)-1+b}{b{N}_{\max }\left(b\right)}}\right)=\hfill \\ \hfill & ={\displaystyle \frac{1-b^2}{b^2+b+1}}{log}_2\left({\displaystyle \frac{b+1}{b^2+b+1}}\right)-{\displaystyle \frac{b+2}{b^2+b+1}}{log}_2\left({\displaystyle \frac{b+2}{b^2+b+1}}\right)\approx {log}_2\left(b\right).\hfill \end{array}$$

(22)

The strings given by the equation (19) are not the shortest possible ones. Strings satisfying the equation (20) and satisfying $min(b{N}_c\left(b\right)+1)>N_{\max }(b-1)$ are given by $b^2+1$ (OEIS A002522). However, they do not contain all the possible doublets and furthermore their entropies are smaller than the entropies of the strings given by the equation (19).

Now, assume a contrario that a string longer than $N_{\max }\left(b\right)$ can be constructed, say of length $N_{\max }{\left(b\right)}^{\prime }=N_{\max }\left(b\right)+1$. But in this case, the corresponding $H_{\max }{\left(b\right)}^{\prime }<H_{\max }\left(b\right)$. The string of the length given by the formula (19) maximizes the Shannon entropy if it must additionally satisfy the relation (20). □

Although the case for $b=1$ (only one symbol) is degenerate, the formula (19) yields correct result; the string $\left[000\right]$ is the longest string with $a_{\max }^{(N,1)}=N-1$, as for $b=1$ we simply have $a_{\max }^{(N,1)}=a_{\min }^{\left(N\right)}$ (OEIS A003313).

3. Conclusions

There is one string of length $N_{\max }\left(1\right)=3$, four strings of length $N_{\max }\left(2\right)=7$, seventy-two strings of length $N_{\max }\left(3\right)=13$ (cf. Appendix A). Their number for $b\ge 4$ requires further research.