On the Salient Regularities of Strings of Assembly Theory

1. Introduction

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

Definition 1.

We call a set $P_0^{\left(b\right)}\{0,1,\cdots ,b-1\}$ that contains $b\in \mathbb{N}$ different basic symbols c, the initial assembly pool.

The reader will find numerous results on AT in refs. [1,2,3,4,5,6,7,8,9,10], for example. Here, we extend the results of our previous study [9] concerning bitstrings 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 $P_0^{\left(b\right)}$ within the AT framework in consecutive assembly steps from basic symbols c and strings (doublets, triplets, n-plets) assembled in previous steps. The ancient Greek verb symbállein means putting only two things (“symbols”) together [11].

In fact, any embodiment of AT, with basic symbols representing LEGO® blocks, chemical bonds, graphs, monomers, etc. assembled in any n-dimensional space ($n\in \mathbb{C}$) [12] corresponds to the string AT version. This is because in AT an assembly step always consists in joining two parts only, which can be thought of as the left and right fragments of the newly formed string. Put simply, AT explains and quantifies selection and evolution [7] but it is through the word (aka string or message), in particular a nucleotide sequence in the case of $b=4$, all AT things come into existence [13].

Definition 2.

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

An assembly step s may consist of

$$c_1\circ c_2=C_k^{(2,b)},C_l^{(N_l,b)}\circ c_2=C_k^{(N_l+1,b)},c_1\circ C_m^{(N_m,b)}=C_k^{(1+N_m,b)},C_l^{(N_l,b)}\circ C_m^{(N_m,b)}=C_k^{(N_l+N_m,b)},$$

(1)

where $c_1,c_2\in P_0^{\left(b\right)}$, $C_l^{(N_l,b)},C_m^{(N_m,b)}\in P_{s-1}^{\left(b\right)}$, and $C_k\in P_s^{\left(b\right)}$. Using Definitions 1 and 2, the assembly index (ASI) of a string is the minimal achievable value of a difference between the cardinalities of the working and initial assembly pools (ASPs) leading to this string, since at each assembly step the cardinality of the working ASP increases by one. Therefore, the working ASP 2 cannot be identified with the initial ASP 1; the initial ASP 1 must not contain strings of basic symbols (see Appendix G).

2. Results

Theorems 1 and 2 were already stated in our previous study [9] for $b=2$. We restate them here $\forall b$ for clarity.

Theorem 1.

A quadruplet is the shortest string that allows for more than one ASI for all b.

Proof. 

$N=2$ provides $b^2$ available doublets with unit ASI. $N=3$ provides $b^3$ available triplets with ASI equal to two. Only $N=4$ provides $b^4$ quadruplets that include $b^2$ quadruplets with ASI equal to two, that is b quadruplets $C_{k,\min }^{(4,b)}=[****]$ and $b(b-1)$ quadruplets $C_{l,\min }^{(4,b)}=[*★*★]$, while the ASI of the remaining $b^4-b^2$ quadruplets is three. □

For example, to assemble the quadruplet $C_{k,\min }^{(4,4)}=\left[0202\right]$, we need to assemble the doublet $\left[02\right]$ and reuse it from the first step ASP $P_1$, while there is nothing available to reuse, in the case of the quadruplet $C_l^{(4,4)}=\left[0123\right]$.

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

Theorem 2.

The minimum ASI $a^{\left(N\right)}\left(C_{\min }\right)$ 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 ASI if $N=2^s$, including respectively b and $b(b-1)$ strings

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

(2)

and the assembly pathway of each of the strings (2) 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 $k\le l<j$. Thus, an addition chain starts with one, not zero, as zero is the neutral element of addition. For the same reason, two is considered the smallest prime, as one is the neutral element of multiplication. Hence, $j=1\Rightarrow k=l=0$ and the first step in creating an addition chain for N is always $a_1=1+1=2$; the ASI of any doublet is one. The second step in creating an addition chain can be $a_2=1+1=2$, $a_2=1+2=3$, or $a_2=2+2=4$. The 1^st case does not represent the shortest addition chain, the 2^nd one corresponds to assembling a triplet based on the previously assembled doublet, and the 3^rd one corresponds to assembling a quadruplet from this doublet. Therefore, four is the smallest number achievable in two ways since $a_2=2+2=4$ and $a_3=3+1=4$, where the latter case corresponds to assembling a quadruplet by joining a basic symbol to a triplet, which is not the shortest way for assembling a quadruplet having a minimum ASI.

Thus, finding the shortest addition chain for N corresponds to finding the ASI 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 $\{0,1,2\}$. □

The assembly pathways of strings $a_{\min }^{\left(N\right)}$ of length $N\ne 2^s$ are not unique. For example, a string $C_{\min }^{(5,b)}=\left[01010\right]$ can be assembled in three steps from three working ASPs $P_3^{\left(2\right)}=\{0,1,01,0101\}$, $P_3^{\left(2\right)}=\{0,1,10,1010\}$, and $P_3^{\left(2\right)}=\{0,1,01,010\}$.

Theorem 3.

The strings $C_{\min }^{(2^s,b)}$ can contain at most two symbols if $b>1$. Other minimum ASI strings of length $N\ne 2^s$ can contain at most three symbols if $b>2$.

Proof. 

Minimum ASI strings of length $N=2^s$ are formed by joining the newly assembled string to itself, where a clear or mixed doublet is created in the first step. Minimum ASI strings of other lengths admit a doublet and a triplet containing this doublet and an additional basic symbol.

To formally prove the first part, we can also use mathematical induction on the assembly step s. If $s=1$, then the minimum ASI strings $C_{\min }^{(2,b)}$ are doublets of the form $\left[c_1{c}_2\right]$, where $c_1,c_2\in P_0^{\left(b\right)}$. If $c_1=c_2$, the string contains one distinct symbol, and if $c_1\ne c_2$, the string contains two distinct symbols. In both cases, the number of distinct symbols does not exceed two. Now assume that for some $k\in \mathbb{N}$, all minimum ASI strings $C_{\min }^{(2^k,b)}$ contain at most two distinct symbols. We must show that $C_{\min }^{(2^{k+1},b)}$ also contains at most two distinct symbols. Consider constructing $C_{\min }^{(2^{k+1},b)}$ by joining two identical minimum ASI strings $C_{\min }^{(2^k,b)}$

$$C_{\min }^{(2^k,b)}\circ C_{\min }^{(2^k,b)}=C_{\min }^{(2^{k+1},b)},$$

(3)

with each other. By the inductive hypothesis, each $C_{\min }^{(2^k,b)}$ contains at most two distinct symbols. Therefore, their concatenation also contains at most two distinct symbols. By induction, for all $s\in \mathbb{N}$, the minimum ASI string $C_{\min }^{(2^s,b)}$ contains at most two distinct symbols.

We will now show that other minimum ASI strings of length $N\ne 2^s$ can contain at most three distinct symbols if $b>2$. We provide the construction of minimum ASI strings with three symbols. In the first step $s=1$, we create a doublet $\left[c_1{c}_2\right]$ where $c_1,c_2\in P_0^{\left(b\right)}$ and $c_1\ne c_2$. Next, we combine the existing doublet $\left[c_1{c}_2\right]$ with a new symbol $c_3\in P_0^{\left(b\right)}$ where $c_3\notin \{c_1,c_2\}$. This forms a triplet $\left[c_1{c}_2{c}_3\right]$, introducing a third distinct symbol and further increasing the ASI by 1. We continue assembling by joining the longest string formed so far with itself or with previously formed strings, maintaining the minimal increase in ASI.

Assume a contrario that there exists a minimum ASI string $C_{\min }^{(N,b)}$ of length $N\ne 2^s$ that contains four or more distinct symbols. To incorporate a fourth symbol, at least one additional assembly step is required beyond what is needed for the three symbols. This additional step implies an increase in ASI, which contradicts the minimality of $C_{\min }^{(N,b)}$. Thus, Theorem 3 is proven. □

The strings having non-minimum ASI can contain all symbols. For example, the string [14]

$$C_k=\left[01234012340123401234\right],$$

(4)

has ASI $a^{(20,5)}\left(C_k\right)=6=a_{\min }^{\left(20\right)}+1$ and contains all five basic symbols $P_0^{\left(5\right)}\{0,1,2,3,4\}$.

Theorem 4.

A string containing the same three doublets has the same ASI as a string containing two pairs of the same doublets, provided that both strings have the same distributions of other repetitions and have the same lengths.

Proof. 

Without loss of generality (w.l.o.g.), consider the following two strings of the same length $N+8$ with $*★\ne 01$ and the same distributions of other repetitions (if there are any other repetitions)

$$C_k=[\cdots 01\cdots 01\cdots 01\cdots *★\cdots ],C_l=[\cdots 01\cdots 01\cdots 22\cdots 22\cdots ],$$

(5)

where $*★\ne 01$. Creating a doublet takes one assembly step. Each appending of a doublet to an assembled string counts as another assembly step. Hence, in a general case (i.e., for strings $C_k$, $C_l$ containing also other symbols), the string $C_k$ requires six additional assembly steps, the same as the string $C_l$, which completes the proof. □

Theorem 5.

A string containing the same three doublets has the same ASI as a string containing the same two triplets, provided that both strings have the same distributions of other repetitions.

Proof. 

W.l.o.g. consider the following two strings of the same length $N+6$ with the same distributions of other repetitions

$$C_k=[\cdots 01\cdots 01\cdots 01\cdots ],C_l=[\cdots 010\cdots 010\cdots ].$$

(6)

Creating a triplet takes two assembly steps. Hence, in the general case, the string $C_k$ requires four additional assembly steps, the same as the string $C_l$, which completes the proof. □

Theorem 6.

A string containing the same two triplets has the same ASI as a string containing two pairs of the same doublets, provided that both strings have the same distributions of other repetitions and have the same lengths.

Proof. 

The proof stems from Theorems 4 and 5. □

Theorem 7.

A string containing the same two quadruplets of the minimum ASI has the same ASI as a string containing the same three triplets, provided that both strings have the same distributions of other repetitions and have the same lengths.

Proof. 

W.l.o.g. consider the following two strings of the same length $N+9$ with the same distributions of other repetitions

$$C_k=[\cdots 0101\cdots 0101\cdots ★\cdots ],C_l=[\cdots 010\cdots 010\cdots 010\cdots ].$$

(7)

Creating such a quadruplet takes two assembly steps. Hence, in a general case, the string $C_k$ requires five additional assembly steps, the same as the string $C_l$, which completes the proof. □

Theorem 8.

A string containing the same two quadruplets of the maximum ASI has the same ASI as a string containing a doublet and the same two triplets based on this doublet, provided that both strings have the same distributions of other repetitions.

Proof. 

W.l.o.g. consider the following two strings of the same length $N+8$ with the same distributions of other repetitions

$$C_k=[\cdots 0001\cdots 0001\cdots ],C_l=[\cdots 110\cdots 10\cdots 110\cdots ].$$

(8)

Creating such a quadruplet takes three assembly steps. Hence, in a general case, the string $C_k$ requires five additional assembly steps, the same as the string $C_l$, which completes the proof. □

Theorem 9.

A string containing the same two doublets and the same two triplets not based on this doublet has the same ASI as a string containing a doublet and the same two triplets based on this doublet, provided that both strings have the same distributions of other repetitions and have the same lengths.

Proof. 

W.l.o.g. consider the following two strings of the same length $N+10$ with the same distributions of other repetitions

$$C_k=[\cdots 110\cdots 00\cdots 110\cdots 00\cdots ],C_l=[\cdots 110\cdots 10\cdots 110\cdots *★\cdots ],$$

(9)

where $*★\notin \{11,10\}$. In a general case, the string $C_k$ requires seven additional assembly steps, the same as the string $C_l$, which completes the proof. □

In general, Theorems 1-9 show that

• k copies of a doublet in a string decrease the ASI of this string at least by $k-1$;
• k copies of a triplet in a string decrease the ASI of this string at least by $2k-2$;
• k copies of a minimum ASI quadruplet in a string decrease the ASI of this string at least by $3k-2$;
• k copies of a maximum ASI quadruplet in a string decrease the ASI of this string at least by $3k-3$;

where, the phrase "at least" is meant to indicate that other repetitions, such as e.g. doublets forming multiple quadruplets, etc. can further decrease the ASI of the string. This observation allows us to state the following theorem.

Theorem 10.

Each $k_r$ copies of an $n_r$-plet $C_r^{(n_r,b)}$ contained in a string $C_m^{(N,b)}$ decrease its ASI at least by $\left[k_r(n_r-1)-a^{(n_r,b)}\left(C_r\right)\right]$. That is

$$a^{(N,b)}\left(C_m\right)\le N-1-\sum _{r=1}^R\left[k_r(n_r-1)-a^{(n_r,b)}\left(C_r\right)\right],$$

(10)

where R is the total number of repeated $n_r$-plets.

Proof. 

W.l.o.g. consider the following string

$$C_m^{(N,b)}=[\cdots [c_1{c}_2\cdots c_n]\cdots [c_1{c}_2\cdots c_n]\cdots ],$$

(11)

containing two copies of an n-plet $C_l^{(n,b)}=[c_1{c}_2\cdots c_n]$. The n-plet $C_l^{(n,b)}$ can be assembled in $a^{(n,b)}\left(C_l\right)$ steps and appended to the assembled string $C_m$ in one step. Consider that the ASI of the n-plet $C_l^{(n,b)}$ is $a^{(n,b)}\left(C_l\right)=n-1$, i.e. the n-plet does not have any repetitions that can be reused. Then one copy of this n-plet - as expected - does not decrease the ASI of the string $C_m^{(N,b)}$, as $1(n-1)-(n-1)=0$, while more copies k decrease it by $(n-1)(k-1)$. On the other hand, if $a^{(n,b)}\left(C_l\right)<n-1$ then even a single copy of this n-plet will decrease the ASI of $C_m$. □

For example, due to the presence of three copies of a 5-plet $\left[01001\right]$, each with $a^{(5,6)}\left(\left[01001\right]\right)=3$, in a string

$$C_k^{(24,6)}=\left[12\right|01001\left|21\right|01001\left|235\right|01001\left|52\right],$$

(12)

its ASI amounts to $a^{(24,6)}\left(C_k\right)=24-1-(3·(5-1)-3)=14$. The relation (10) provides the upper bound on ASI as it does not describe a situation in which n-plet for $n>2$ is assembled on a doublet also present in one copy in the string. For example, the string $a^{(14,9)}\left(\left[56101781014301\right]\right)=10$, while $14-1-\left(2\right(3-1)-2)=11$. We note that the maximum ASI decrease is provided by $2^s$-plets of the minimum ASI and amounts to $k(n-1)-{log}_2\left(n\right)=k(2^s-1)-s$.

Another quantity quantifying the complexity of a string is the assembly depth (ASD) defined [15] as

$$d_s^{(N_k,b)}\left(C_k\right)\max \left(d^{(N_l,b)}\left(C_l\right),d^{(N_m,b)}\left(C_m\right)\right)+1,$$

(13)

where $d_0^{(1,b)}\left(c\right)0$, and $d^{(N_l,b)}\left(C_l\right)$ and $d^{(N_m,b)}\left(C_m\right)$ are the ASDs of two substrings $C_l$, $C_m$ of the string $C_k$ that were joined in step s, where for $N\ge 4$, and if there are more assembly pathways with different depths $w_j$ leading to a string, which happens if at least two independent assembly steps are possible, the minimum pathway depth is the ASD of this string. Hence, the ASD captures the notion of an independent assembly step.

Theorem 11.

If a working ASP contains strings having the same ASD they were assembled in independent assembly steps.

Proof. 

W.l.o.g. assume a contrario that two strings $C_l$, $C_m$ in the working ASP have the same ASD, i.e., $d^{(N_l,b)}\left(C_l\right)=d^{(N_m,b)}\left(C_m\right)$, but $C_m$ was used in the assembly of $C_l$ along with a basic symbol c. Then

$$d_s^{(N_l,b)}\left(C_l\right)=\max \left(d^{(N_m,b)}\left(C_m\right),d^{(1,b)}\left(c\right)\right)+1=d^{(N_m,b)}\left(C_m\right)+1\ne d^{(N_m,b)}\left(C_m\right),$$

(14)

which contradicts our assumption and completes the proof. □

In other words, if two strings $C_l$, $C_m$ in the working ASP have the same ASD, their assembly pathways are unrelated to each other; by the defining equation (13) neither of them could have been used in the assembly pathway of the other.

Theorem 12.

The ASD of any minimum ASI string $C_{\min }^{(N,b)}$ is equal to the ASI of this string, $d_{a_{\min }}^{(N,b)}=a_{\min }^{\left(N\right)}$.

Proof. 

We need to show that $d_{a_{\min }}^{(N,b)}=a_{\min }^{\left(N\right)}$. While constructing the minimum ASI string, we start with a doublet and follow the shortest addition chain for N, joining this doublet with itself or with a basic symbol to form a triplet. At each assembly step, the ASD increases by one, as we join the assembled string with a string or a basic symbol from the working ASP and we cannot perform independent assembly steps. Since, by Theorem 2, the minimum ASI corresponds to the length of the shortest addition chain $l\left(N\right)$, we have

$${d_s}^{(N,b)}\left(C_{\min }^{(N,b)}\right)=l\left(N\right)=a_{\min }^{\left(N\right)}.$$

(15)

This completes the proof (see Appendix F for additional comments). □

Theorems 11 and 12 show that

• the working ASP of a minimum ASI string cannot contain strings assembled in independent assembly steps,
• the working ASP of a non-minimum ASI string must contain at least two such strings, and
• the assembly pathway of a maximum ASI string will tend to maximize their number in the working ASP, and hence to minimize the possible ASD, taking into account the saturation of the working ASP, as the number of distinct n-plets in the working ASP cannot exceed $b^n$.

Theorem 13.

The ASD of any maximum ASI string $C_{\max }^{(N,b)}$ satisfies

$$d_{a_{\max }}^{(N,b)}=⌈{log}_2\left(N\right)⌉.$$

(16)

Proof. 

Let $d^{\left(N\right)}{d}_{a_{\max }}^{(N,b)}$. For $N=2$ we have $d^{\left(2\right)}=1$, as we are joining basic symbols from the initial ASP. This is the base case. In an assembly tree of ASD $d^{\left(N\right)}$, the maximum number of leaves that can be combined is $2^{d^{\left(N\right)}}$, because at each assembly step, we join two substrings. Therefore, the maximum length $N_{\max }$ of a $C_{\max }$ string that can be assembled with ASD $d^{\left(N\right)}$ satisfies:

$$N_{\max }\le 2^{d^{\left(N\right)}}.$$

(17)

This implies that

$$d^{\left(N\right)}\ge {log}_2\left(N_{\max }\right),$$

(18)

and leads to the relation (16), since both $d^{\left(N\right)}$ and $N_{\max }$ are natural numbers and the latter does not have to be a power of two. We can also use mathematical induction. For $N\ge 2$ and for $N+1$ we have respectively

$$\begin{array}{cc}\hfill d^{\left(N\right)}=⌈{log}_2\left(N\right)⌉& \Rightarrow 2^{d^{\left(N\right)}-1}<N\le 2^{d^{\left(N\right)}},\hfill \\ \hfill d^{(N+1)}=⌈{log}_2(N+1)⌉& \Rightarrow 2^{d^{(N+1)}-1}-1<N\le 2^{d^{(N+1)}}-1,\hfill \\ \hfill \max \left(2^{d^{\left(N\right)}-1},2^{d^{(N+1)}-1}-1\right)<& N\le \min \left(2^{d^{\left(N\right)}},2^{d^{(N+1)}}-1\right),\hfill \end{array}$$

(19)

where $d^{\left(N\right)}\in \mathbb{N}$ implies that either $d^{(N+1)}=d^{\left(N\right)}$ or $d^{(N+1)}=d^{\left(N\right)}+1$. Hence,

$$\begin{array}{cc}\hfill d^{(N+1)}=d^{\left(N\right)}& \Rightarrow 2^{d^{\left(N\right)}-1}<N\le 2^{d^{\left(N\right)}}-1,\hfill \\ \hfill d^{(N+1)}=d^{\left(N\right)}+1& \Rightarrow 2^{d^{\left(N\right)}}-1<N\le 2^{d^{\left(N\right)}},\hfill \end{array}$$

(20)

which completes the proof. □

Theorems 12 and 13 are somehow counterintuitive. For example, the string $C_{\max }^{(11,2)}=\left[10100001110\right]$ has the ASI $a_{\max }^{(11,2)}=8$ and the ASD $d_{a_{\max }}^{(11,2)}=4$, while the string $C_{\min }^{(11,2)}=\left[10101010101\right]$ has a smaller ASI $a_{\min }^{\left(11\right)}=5$ but a larger ASD $d_{a_{\min }}^{(11,2)}=5$.

For example, the ASD of a string $C_{\max }^{(7,2)}=\left[0001110\right]$ is $d_{a_{\max }}^{(7,2)}=⌈{log}_2\left(7\right)⌉=3$ as

$$\begin{array}{cccc}00d_1=1,\hfill & 00w_1=1,\hfill & 00w_1=1,\hfill & 00w_1=1,\hfill \\ 01d_2=1,\hfill & 01w_2=1,\hfill & 01w_2=1,\hfill & 000w_2=2,\hfill \\ 11d_3=1,\hfill & 11w_3=1,\hfill & 0001w_3=2,\hfill & 0001w_3=3,\hfill \\ 110d_4=2,\hfill & 0001w_4=2,\hfill & 00011w_4=3,\hfill & 00011w_4=4,\hfill \\ 0001d_5=2,\hfill & 000111w_5=3,\hfill & 000111w_5=4,\hfill & 000111w_5=5,\hfill \\ 0001110d_6=3,\hfill & 0001110w_6=4,\hfill & 0001110w_6=5,\hfill & 0001110w_6=6,\hfill \end{array}$$

(21)

even though this string can be assembled with three larger pathway depths $w_6=\{4,5,6\}$ and the ASD of a minimum ASI string $C_{\min }^{(7,2)}=\left[0101010\right]$ is

$$01d_1=1,0101d_2=2,010101d_3=3,0101010d_4=4.$$

(22)

Similarly, the ASD of a string $C_{\max }^{(8,2)}=\left[00011101\right]$ is $d_{a_{\max }}^{(8,2)}=⌈{log}_2\left(8\right)⌉=3$ as

$$\begin{array}{cccc}00d_1=1,\hfill & 00w_1=1,\hfill & 00w_1=1,\hfill & 01w_1=1,\hfill \\ 01d_2=1,\hfill & 01w_2=1,\hfill & 01w_2=1,\hfill & 001w_2=2,\hfill \\ 11d_3=1,\hfill & 11w_3=1,\hfill & 0001w_3=2,\hfill & 0001w_3=3,\hfill \\ 0001d_4=2,\hfill & 0001w_4=2,\hfill & 00011w_4=3,\hfill & 00011w_4=4,\hfill \\ 1101d_5=2,\hfill & 000111w_5=3,\hfill & 000111w_5=4,\hfill & 000111w_5=5,\hfill \\ 00011101d_6=3,\hfill & 00011101w_6=4,\hfill & 00011101w_6=5,\hfill & 00011101w_6=6.\hfill \end{array}$$

(23)

However, the non-maximum ASI string $C_k^{(8,2)}=\left[01001011\right]$ has only two doublets that can be assembled in independent steps. Hence, its ASD cannot be decreased to $⌈{log}_2\left(8\right)⌉$

$$\begin{array}{cc}01d_1=1,\hfill & 01w_1=1,\hfill \\ 11d_2=1,\hfill & 010w_2=2,\hfill \\ 010d_3=2,\hfill & 010010w_3=3,\hfill \\ 010010d_4=3,\hfill & 0100101w_4=4,\hfill \\ 01001011d_5=4,\hfill & 01001011w_5=5.\hfill \end{array}$$

(24)

The seven-bit string is the longest string that can have the maximum ASI $a_{\max }^{(7,2)}=7-1=6$. 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

$$[***★★★*]\mathrm{and}[★***★★★],$$

(25)

and their lengths cannot be increased without a repetition of a doublet, which keeps the ASI at the same level $a_{\max }^{(8,2)}=8-2=6$.

This observation and Theorem 2 motivated us to develop a general method to construct the longest possible string having the ASI $a_{\max }^{(N,b)}\left(C_{(N-1)}\right)=N-1$, as a function of the radix b. We denote the length of this string by $N_{(N-1)}$ or $N_{(N-1)}\left(b\right)$, and we call this string a $C_{(N-1)}$ string.

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

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

(26)

The doublets that can be inserted into the initial string (26) can be arranged in a $b\times b$ matrix

(27)

where the crossed out entries on a 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 (26), we can also cross out the entries in the first superdiagonal of the matrix (27). The strings of odd lengths generated by these general methods are not only the longest but also the most balanced. This can be stated in the following theorem.

Theorem 14

($N_{(N-1)}$). The longest length of a string that has the ASI of $N-1$ is given by

$$N_{(N-1)}=3b+{(b-1)}^2=b^2+b+1$$

(28)

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

$$N_{(N-1)}=b{N}_c+1,$$

(29)

where $N_c=b+1$ is the number of occurrences of all but one symbol within the string, and its Shannon entropy is

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

(30)

The proof of Theorem 14 is given in Appendix D. A $C_{(N-1)}$ string must contain all clear triplets and all doublets and if it is generated by Method Appendix A or Appendix B it is terminated with 0 and has a form

$$C_{(N-1)}=[000111222\cdots 0].$$

(31)

Although the case for $b=1$ is degenerate, as no information can be conveyed using only one symbol ($H\left(C_{(N-1)}\right)=0$ in this case), nothing precludes the assembly of such defunct strings and the formula (28) yields the correct result; the string $\left[000\right]$ is the longest string with $a_{\max }^{(N,1)}=N-1$ by Theorem 1, as for $b=1$ the upper and the lower bound on the ASI are the same, $a_{\max }^{(N,1)}=a_{\min }^{\left(N\right)}$ (OEIS A003313). This is the only case where the maximum ASI is not a monotonically nondecreasing function of N.

For $b=3$, only two doublets can be introduced without repetitions into the initial string (26), leading to twelve unique strings of length $N_{(N-1)}=13$

$$\begin{array}{cc}\hfill & \left[000111222\right|0210],[000111222\left|1020\right],\left[20\right|21\left|000111222\right],\left[21\right|02\left|000111222\right],\left[0001112\right|02\left|22\right|10],[0001112\left|10\right|22\left|20\right],\hfill \\ \hfill & \left[21\right|000\left|20\right|111222,\left[000\right|20\left|111222\right|10],[02\left|000111222\right|10],[20\left|00\right|21\left|0111222\right],\left[21\right|0001112\left|02\right|22],[21\left|000111222\right|02].\hfill \end{array}$$

(32)

Finally, we have to multiply the cardinality of this set by $3!=6$ to account for permutations. For example, the first string $\left[0001112220210\right]$, is equivalent to five strings $\left[0002221110120\right]$, $\left[1110002221201\right]$, $\left[1112220001021\right]$, $\left[2220001112102\right]$, and $\left[2221110002012\right]$. Hence, there are seventy-two different strings of length $N_{(N-1)}\left(3\right)=13$.

Subsequently, we considered other $C_{(N-k)}$ strings of length $N_{(N-k)}$ with the maximum ASI $a_{\max }\left(C_{(N-k)}\right)=N-k$ for $k>1$.

Theorem 15

($N_{(N-k)}$). For all $b>1$ and $2\le k\le 9$ the longest length of a string that has the ASI of $N-k$ is given by

$$N_{(N-k)}=b^2+b+2k.$$

(33)

The proof of Theorem 15 is given in Appendix E. This result disproves our upper bound Conjecture 1 for $b=2$ stated in our previous study [9]. If the strings of Theorem 15 are based on strings generated by Method Appendix A or Appendix B, for $b>2$ they owe their properties to the following distributions of symbols

$$\begin{array}{cc}\hfill C_{(N-2)}& =[010000111222\cdots 10\cdots 0],\hfill \\ \hfill C_{(N-3)}& =[01010000111222\cdots 10\cdots 0],\hfill \\ \hfill C_{(N-4)}& =[0101010000111222\cdots 10\cdots 0],\hfill \\ \hfill C_{(N-5)}& =[010101000000111222\cdots 10\cdots 0],\hfill \\ \hfill C_{(N-6)}& =[01010100000011111222\cdots 10\cdots 0],\hfill \\ \hfill C_{(N-7)}& =[0101010000000111111222\cdots 10\cdots 0],\hfill \\ \hfill C_{(N-8)}& =[010101000000011011111222\cdots 10\cdots 0],\hfill \\ \hfill C_{(N-9)}& =[01010100100000011011111222\cdots 10\cdots 0].\hfill \end{array}$$

(34)

For the strings of the form (34) the fractions in the Shannon entropy are

$$p_0={\displaystyle \frac{b+k+f_0}{b^2+b+2k}},p_1={\displaystyle \frac{b+k+f_1}{b^2+b+2k}},p_{2,\cdots ,b-1}={\displaystyle \frac{b+1}{b^2+b+2k}},$$

(35)

where $f_0=3$, $f_1=-1$ if $k=5$ and $f_0=2$, $f_1=0$ otherwise, as $\left[00\right]$ is inserted into $C_{(N-5)}$, $\left[11\right]$ into $C_{(N-6)}$ and $\left[01\right]$ or $\left[10\right]$ otherwise. This leads to Shannon entropy

$$\begin{array}{cc}\hfill H\left(C_{(N-k)}\right)& =-{\displaystyle \frac{b^2-b-2}{b^2+b+2k}}{log}_2\left({\displaystyle \frac{b+1}{b^2+b+2k}}\right)-{\displaystyle \frac{b+k+f_1}{b^2+b+2k}}{log}_2\left({\displaystyle \frac{b+k+f_1}{b^2+b+2k}}\right)-{\displaystyle \frac{b+k+f_0}{b^2+b+2k}}{log}_2\left({\displaystyle \frac{b+k+f_0}{b^2+b+2k}}\right).\hfill \end{array}$$

(36)

The entropies (30) and (36) are shown in Figure 1. Radix $b=4$ is the smallest one at which the entropy (36) is a monotonically decreasing function. For $b\in \{2,3\}$ there is a local entropy minimum for $k=5$ and for $b=2$ an additional local entropy minimum for $k=2$.

Conjecture 16

($N_{\max }>N_{(N-k)}$). If $b>1$ and $N_{(N-2)}\le N\le N_{\max }$ then

$$a_{\max }^{(N,b)}=\left\{\begin{array}{cc}{a}_{\max }^{(N-1,b)}+1\hfill & \mathrm{iff}N=2l,\hfill \\ a_{\max }^{(N-1,b)}\hfill & \mathrm{iff}N=2l+1,\hfill \end{array}\right..$$

(37)

or equivalently

$$a_{\max }^{(N,b)}=⌊{\displaystyle \frac{N}{2}}⌋+{\displaystyle \frac{b(b+1)}{2}},$$

(38)

where

$$N_{\max }=\left\{\begin{array}{cc}4b^4\hfill & \mathrm{iff}b=2l,\hfill \\ 4(b^4+1)\hfill & \mathrm{iff}b=2l+1,\hfill \end{array}\right..$$

(39)

In other words, if $N\ge N_{(N-2)}$, then ASI increases by one, where N increases by two ($b(b+1)/2$ are triangular numbers, OEIS A000217).

First, we note that maximum ASI must rise. If it were constant for $N>{\widehat{N}}_{max}$, then at some even larger N it would inevitably become lower than the minimum ASI bound 2 which also rises, and this would be a contradiction. W.l.o.g. we aim to prove this conjecture for $b=2$. We note that inserting any doublet into a $C_{(N-3)}^{(12,2)}$ string (A19) at any position creates a triplet. Using the equation (10) of Theorem 10 we have

$$\begin{array}{cc}\hfill a_s& =a_{s-2}+1,N_s=N_{s-2}+2,\hfill \\ \hfill a_s& =N_s-1-\sum _{r=1}^{R_r}\left[k_r(n_r-1)-a\left(C_r^{(n_r,b)}\right)\right],\hfill \\ \hfill a_{s-2}& =N_{s-2}-1-\sum _{p=1}^{R_{s-2}}\left[k_p(n_p-1)-a\left(C_p^{(n_p,b)}\right)\right],\hfill \\ \hfill a_s-a_{s-2}& =(N_{s-2}+2)-1-\sum _{r=1}^{R_r}\left[k_r(n_r-1)-a\left(C_r^{(n_r,b)}\right)\right]-\left(N_{s-2}-1-\sum _{p=1}^{R_p}\left[k_p(n_p-1)-a\left(C_p^{(n_p,b)}\right)\right]\right)=\hfill \\ \hfill & =2-\sum _{r=1}^{R_r}\left[k_r(n_r-1)-a\left(C_r^{(n_r,b)}\right)\right]+\sum _{p=1}^{R_p}\left[k_p(n_p-1)-a\left(C_p^{(n_p,b)}\right)\right]=1,\hfill \\ \hfill & \sum _{r=1}^{R_r}\left[k_r(n_r-1)-a\left(C_r^{(n_r,b)}\right)\right]=\sum _{p=1}^{R_p}\left[k_p(n_p-1)-a\left(C_p^{(n_p,b)}\right)\right]+1,\hfill \end{array}$$

(40)

for any step s if only $N_{(N-2)}\le N_s\le N_{\max }$. Now, assume that $\forall r$, $a\left(C_r^{(n_r,b)}\right)=n_r-1$ and $\forall p$, $a\left(C_p^{(n_p,b)}\right)=n_p-1$. Then

$$\begin{array}{cc}\hfill \sum _{r=1}^{R_r}\left[(k_r-1)(n_r-1)\right]& =\sum _{p=1}^{R_p}\left[(k_p-1)(n_p-1)\right]+1,\hfill \\ \hfill \sum _{r=1}^{R_r}{n}_r{k}_r-\sum _{r=1}^{R_r}{n}_r-\sum _{r=1}^{R_r}{k}_r+R_r& =\sum _{p=1}^{R_p}{n}_p{k}_p-\sum _{p=1}^{R_p}{n}_p-\sum _{p=1}^{R_p}{k}_p+R_p+1.\hfill \end{array}$$

(41)

The proof of the Conjecture 16 must show the conditions for the equations (40) and (41) to hold. We note that the assumption used in the equation (41) is valid only for $n_r\le N_{(N-1)}$ and $n_p\le N_{(N-1)}$. The bounds of Theorems 14 and 15 and Conjecture 16 are illustrated in Figure 2.

The results thus far led us to a simple method of determining the ASI of a maximum ASI and a minimum ASD string and strengthened our Conjectures 3 and 4 stated in the previous study [9]. The method is based on unique $2^s$-plets and powers of two, as shown in Table 1. First, a maximum ASI string is sequenced, every two symbols to find the number $n_{UAD}$ of unique adjoining doublets $\times 2_{\left(b\right)}$. In particular, a $C_{(N-1)}$ string (A3) or (A4) contain the maximum of $⌊N_{(N-1)}/2⌋$ unique adjoining doublets, a $C_{(N-2)}$ string (A13) contains the maximum of $N_{(N-2)}/2-1$ unique adjoining doublets, and so on. In general, a $C_{(N-k)}$ string contains the maximum of

$$n_{UAD}=⌊{\displaystyle \frac{N_{(N-k)}}{2}}⌋-k+1=\left\{\begin{array}{cc}b(b+1)/2=\sum _{l=1}^{b}l\hfill & \mathrm{iff}k=1,\hfill \\ b(b+1)/2+1=\sum _{l=1}^{b}l+1\hfill & \mathrm{iff}k\ne 1,\hfill \end{array}\right..$$

(42)

unique adjoining doublets, where $N_{(N-k)}$ is given by the relations (28) or (33), which is independent of k.

Subsequently, these doublets form $\times 4_{\left(b\right)}$ unique adjoining quadruplets, quadruplets form $\times 8_{\left(b\right)}$ unique adjoining octuples, and so on depending on the length of the string N and the radix b, as there can be at most $b^{2^s}$ unique $2^s$-plets. The columns "last $2^s$" indicate if the assembled string should be terminated with a single substring of length $2^s$ in descending order. The empty fields in the respective columns for $N>1$ indicate that a given $\times 2^s$ substring can be interpreted as either a "regular" single $\times 2^s$ substring or a last $\times 2^s$ substring if $\times 2^s=1$.

For example, the $N_{(N-3)}$ string (A20) of length $N_{(N-3)}=18$ for $b=3$ can be assembled as

(43)

Similarly, the $N_{(N-1)}$ string (A3) of length $N_{(N-1)}=21$ for $b=4$ can be assembled, as shown in Table 1 as

(44)

For $N<15$ and for other small N this combinatorics is valid also for $b=1$, where obviously $max\left(\times 2^s\right)=1$. For example, the string of length $N=15$ can be assembled in six steps as

(45)

However, this is the 1^st exception for $b=1$ as the ASI of this string is five if it is assembled using doublet $\left[00\right]$ and triplet $\left[000\right]$. For $b=1$ the method produces OEIS A014701 sequence corresponding to the number of steps to reach 1 starting from $N_0$ and assigning $N_{s+1}=N_s-1$ if $N_s$ is odd and $N_{s+1}=N_s/2$ otherwise.

We further note that the method illustrated in Table 1 cannot be used to construct the maximum ASI string. For example, both the following two distributions of doublets for $N=6$ satisfy the distributions of Table 1. However, only the left one correctly reflects the maximum ASI of the assembled string.

(46)

as the right one can be assembled in four steps with $P_4^{\left(2\right)}=\{0,1,01,\cdots \}$. Similarly, only the top distribution of doublets below correctly reflects the maximum ASI of the assembled string for $N=10$

(47)

as the bottom one can be assembled in six steps with $P_6^{\left(2\right)}=\{0,1,11,011,\cdots \}$. Furthermore, this method tends to exaggerate the estimated maximum ASI value, that is,

$$a_{\max }^{(N,b)}\le a_{\mathrm{method}}^{(N,b)}\left(C_k\right),$$

(48)

where $a_{\mathrm{method}}^{(N,b)}$ is the ASI of a string $C_k$ determined by the method illustrated in Table 1. For example, the first six strings below contain four unique doublets instead of the required three. Therefore

$$\begin{array}{cc}\hfill & C_1=\left[00\right|10\left|01\right|11],a^{(8,2)}\left(C_1\right)=5,a_{\mathrm{method}}^{(8,2)}\left(C_1\right)=7,\hfill \\ \hfill & C_2=\left[00\right|10\left|11\right|01],a^{(8,2)}\left(C_2\right)=5,a_{\mathrm{method}}^{(8,2)}\left(C_2\right)=7,\hfill \\ \hfill & C_3=\left[00\right|01\left|10\right|11],a^{(8,2)}\left(C_3\right)=5,a_{\mathrm{method}}^{(8,2)}\left(C_3\right)=7,\hfill \\ \hfill & C_4=\left[00\right|01\left|11\right|10],a_{\max }^{(8,2)}\left(C_4\right)=6,a_{\mathrm{method}}^{(8,2)}\left(C_4\right)=7,\hfill \\ \hfill & C_5=\left[00\right|11\left|10\right|01],a^{(8,2)}\left(C_5\right)=5,a_{\mathrm{method}}^{(8,2)}\left(C_5\right)=7,\hfill \\ \hfill & C_6=\left[00\right|11\left|01\right|10],a^{(8,2)}\left(C_6\right)=5,a_{\mathrm{method}}^{(8,2)}\left(C_6\right)=7,\hfill \\ \hfill & \\ \hfill & C_7=\left[00\right|01\left|11\right|00],a_{\max }^{(8,2)}\left(C_7\right)=6=a_{\mathrm{method}}^{(8,2)}\left(C_7\right)=6.\hfill \end{array}$$

(49)

Further research should consider researching the formula equivalent to (28) that captures a quadruplet repetition, similarly as $b^2+b^1+b^0$ captures a doublet repetition.

3. Discussion

Applications of AT seem to be promising. It offers a new lens for studying the construction of biological molecules like DNA and proteins. By analyzing the steps needed to assemble these molecules from basic building blocks, researchers can gain deeper insights into the evolutionary constraints and optimizations that shape biological pathways. This perspective also sheds light on the efficient construction of cellular structures and helps to identify the minimum number of assembly steps that define biological complexity, reinforcing the idea that life is characterized by highly organized pathways. Furthermore, AT provides an essential tool for understanding the growth of complexity in biological systems over evolutionary time. By quantifying the assembly steps required to form increasingly complex organisms, scientists can map the trajectory of evolutionary development and identify key transitions that lead to higher levels of structural and functional complexity. It can guide the design and optimization of synthetic biological systems by minimizing the number of steps required to build new biological pathways, making bioengineering more efficient and scalable. The ability to model and simplify complex biological processes using AT could lead to the development of more robust and adaptable synthetic organisms.

Strings having lengths $N_{(N-1)}$ (e.g. (A3) or (A4)) are necessarily the most balanced: all but one symbol occur $b+1$ times and one symbol occurs $b+2$ times within a string $C_{(N-1)}$. However, if the length of a string is constant, it will tend to evolve to decrease the Shannon entropy [16,17] and, hence, to become less balanced. As the energy of a black hole that can be thought of as a balanced bitstring [18] can be two times the energy of the entropy variation sphere that it generates [19], this tendency to imbalance seems to be associated with the minimum energy condition. For example, the Shannon entropy of the SARS-CoV genome containing $N=29903$ nucleobases decreased from $H=1.3565$ to $1.3562$ within two years after the Wuhan outbreak [9,16]. The minimum ASI for this length of the string, given by the OEIS A003313, is $a_{\min }^{\left(29903\right)}=19$. Perhaps, entropy (36) has other local entropy minima for $b<4$ and for $k>9$ and is a monotonically decreasing function only for $b\ge 4$. This could be the reason nature has chosen the non-binary radix $b=4$ and four nucleobases to encode genetic information.