On the Certain Salient Regularities of Strings of Assembly Theory

I. 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 of unit length, the initial assembly pool (ASP).

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, quadruplets,…,n-plets) assembled in previous steps.

In fact, any embodiment of AT, with basic symbols representing LEGO® blocks, chemical bonds, graphs, monomers, etc. assembled in any n-dimensional space ($n\in \mathbb{C}$) [11] 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. The ancient Greek verb symbállein means putting only two things (“symbols”) together [12]. 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 ASP.

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.

II. 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 $\forall b$.

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 $C_{l,\min }^{(4,b)}=[*{\displaystyle ★}*{\displaystyle ★}]$, 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 working 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 $2^{\mathrm{nd}}$ possibility different from *, we write ★. Thus, $C_k^{(2,b)}=[**]$, for example, is a placeholder for all b strings, while $C_l^{(2,b)}=[*{\displaystyle ★}]$ 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 $a^{\left(N\right)}\left(C_{\mathit{min}}\right)$ as a function of N corresponds to the shortest addition chain for N (OEIS A003313) $\forall b$.

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

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

(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 $k\le l<j$. Hence, $j=1\Rightarrow k=l=0$ and the first step in creating an addition chain for N is always $a_1=1+1=2$, which is an equivalent of saying that 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 $1^{\mathrm{st}}$ case does not represent the shortest addition chain but the first step, the $2^{\mathrm{nd}}$ one corresponds to assembling a triplet based on the previously assembled doublet, and the $3^{\mathrm{rd}}$ one corresponds to assembling a minimum ASI quadruplet (2) from this doublet. Maximum ASI quadruplet can be assembled in a third step $a_3=3+1=4$, which corresponds to joining a basic symbol to a triplet. Therefore, four is the smallest number achievable in two ways according to Theorem 1.

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 $\{0,1,2\}$ with no internal repetitions. □

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 four working ASPs $P_3^{\left(2\right)}=\{0,1,01,0101\}$, $P_3^{\left(2\right)}=\{0,1,10,1010\}$, and $P_3^{\left(2\right)}=\{0,1,01,010\}$.

We note in passing that any shortest addition chain for n 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, the fundamental theorem of arithmetic can be thought of as the shortest multiplication chain for n.

Theorem 3. 

The strings $C_{\mathit{min}}^{(2^s,b)}$ can contain at most two distinct symbols if $b>1$. Other minimum ASI strings of length $N\ne 2^s$ can contain at most three distinct 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 string has a form (2) and 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. We construct $C_{\min }^{(2^{k+1},b)}$ 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 join 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 ASI increase.

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. But, 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 $*{\displaystyle ★}\ne 01$ and the same distributions of other repetitions (if there are any other repetitions)

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

(5)

where $*{\displaystyle ★}\ne 01$. 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

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

(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

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

(9)

where $*{\displaystyle ★}\notin \{11,10\}$. 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 at least $a^{(n,b)}\left(C_l\right)$ 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 based 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. For $N>3$, 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 the working ASP contains two strings $C_l$, $C_m$ having the same ASD, i.e., $d^{(N_l,b)}\left(C_l\right)=d^{(N_m,b)}\left(C_m\right)$, that were not assembled in independent assembly steps, i.e., that $C_m$ was used in the assembly of $C_l$ along with a basic symbol c in some previous step s. 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. Here, we introduce the following definition, which is also related to the independent assembly step.

Definition 3. 

We call the number of steps ${\hat{a}}_{\mathit{min}}^{\left(N\right)}$ to reach 1 starting from $N{N}_0$ and assigning $N_{s+1}=N_s-1$ if $N_s$ is odd and $N_{s+1}=N_s/2$ otherwise (OEIS A014701 sequence) the depth index (DPI).

Theorem 12. 

The maximum length N of any string that can be assembled with the ASD $d_s^{\left(N\right)}$ (13), $\forall b$, satisfies

$$N\le 2^{d_s^{\left(N\right)}}.$$

(15)

Proof. 

Assume a contrario that $N>2^{d_s^{\left(N\right)}}$. Then for $d_s^{\left(N\right)}=0$, we have $N>2^0=1$ which is a contradiction as basic symbols c have unit length $N=1$. □

Theorem 13. 

The minimum ASI-independent ASD as a function of N, $\forall b$, satisfies

$$d_{\min }^{\left(N\right)}=⌈{log}_2\left(N\right)⌉,$$

(16)

where $⌈x⌉$ denotes the ceiling function.

Proof. 

$d_s^{\left(N\right)}\ge {log}_2\left(N\right)$ follows from the relation (15). $d_{\min }^{\left(2\right)}=⌈{log}_2\left(2\right)⌉=1$ satisfies both the definition (13) and our hypothesis (16). Similarly $N=3$. Using induction on length N, assume that for some $N>3$, we can assemble a minimum ASD string with ASD (16). We need to show that for $N+1$, we can assemble a string with the ASD satisfying

$$d_{\min }^{(N+1)}=⌈{log}_2(N+1)⌉.$$

(17)

Since, by definition (13), the ASD as a function of N is monotonously nondecreasing and can increase at most by one between N and $N+1$, we have

$$d_{\min }^{(N+1)}=\left\{\begin{array}{c}{d}_{\min }^{\left(N\right)}=⌈{log}_2\left(N\right)⌉\hfill \\ d_{\min }^{\left(N\right)}+1=⌈{log}_2\left(N\right)⌉+1\hfill \end{array}\right.=⌈{log}_2(N+1)⌉,$$

(18)

where we used relations (16) and (17). Solving the relation (18) for N yields

$$d_{\min }^{(N+1)}=\left\{\begin{array}{cc}{d}_{\min }^{\left(N\right)}=s\hfill & \mathrm{iff}{2}^{s-1}<N<2^s,\hfill \\ d_{\min }^{\left(N\right)}+1=s+1\hfill & \mathrm{iff}N=2^s.\hfill \end{array}\right.$$

(19)

and completes the proof. □

The ASD does not have to be a monotonously nondecreasing function of the assembly step. For example

$$\begin{array}{cc}\hfill & \left[11\right]d_1=1;\left[110\right]d_2=2;\left[01\right]d_3=1;\left[00\right]d_4=1;\left[0001\right]d_5=2;\left[0001110\right]d_6=3.\hfill \end{array}$$

(20)

Theorem 14. 

The ASI and the ASD of a minimum ASI string $C_{\min }^{\left(2^s\right)}$ are equal to each other.

Proof. 

We use mathematical induction on the length N of the string. For the base case ($N=2^0=1$), the string consists of a single basic symbol $c\in P_0^{\left(b\right)}$. Hence, its ASI is $a_{\min }^{\left(1\right)}0$ and its ASD ${d_0}^{(1,b)}0$. Therefore, ${d_0}^{(1,b)}=a_{\min }^{\left(1\right)}=0$. Assume now that for all strings of length $2^s$ less than N, the ASD equals the minimum ASI, that is

$$d_{a_{\min }}^{(2^s,b)}=a_{\min }^{\left(2^s\right)}\forall 2^s<N.$$

(21)

For some integer s, we construct the minimum ASI string as follows. First, we assemble a doublet from two basic symbols:

$$c_1\circ c_2=C^{(2,b)},c_1,c_2\in P_0^{\left(b\right)}.$$

(22)

Its ASI is $a_{\min }^{\left(2\right)}=1$ and its ASD is $d_s^{(2,b)}=1$. Then for each $s\ge 2$ we have $C^{(2^{s-1},b)}$ with the ASI $a_{\min }^{\left(2^{s-1}\right)}=s-1$ and the ASD ${d_s}^{(2^{s-1},b)}=s-1$ and we construct $C^{(2^s,b)}$ by joining two copies of $C^{(2^{s-1},b)}$

$$C^{(2^{s-1},b)}\circ C^{(2^{s-1},b)}=C^{(2^s,b)}.$$

(23)

The ASI of the string $C^{(2^s,b)}$ is equal to

$$a_{\min }^{\left(2^s\right)}=a_{\min }^{\left(2^{s-1}\right)}+1=(s-1)+1=s,$$

(24)

and, similarly, its ASD is equal to

$$d_s^{(2^s,b)}max\left(d_s^{(2^{s-1},b)},d_s^{(2^{s-1},b)}\right)+1=(s-1)+1=s.$$

(25)

Therefore, ${d_s}^{(2^s,b)}=a_{\min }^{\left(2^s\right)}=s$ in this case. At any step, we assemble strings (2). The proof also follows from Theorems 2 and 13. □

Theorem 15. 

The assembly pathways of strings having the same ASI and ASD cannot contain independent assembly steps.

Proof. 

For the minimum ASI strings of length $N=2^s$, the proof follows from the proof of Theorem 14, as no two assembly steps can be independent. Furthermore, the set of such strings contains the strings

$$\hat{N}{2}^s+2^l,l=0,1,\cdots ,s-1\iff a_{\min }^{\left(\hat{N}\right)}=s+1,$$

(26)

of our previous study [9], which are constructed by adding $2^l$-plet to a $2^s$-plet assembled in a previous assembly step. The base case for $s=1,l=0$ describes the assembly of a triplet by joining a symbol to a doublet made in a previous assembly step, so that both the ASI and the ASD of this triplet increase by one. The case for $s=2,l=0,1$ describes the assembly of a 5-plet or 6-plet, again with unit increase of the ASI and the ASD, and so on. □

The assembly pathways of other minimum ASI strings can contain independent assembly steps. The first such case occurs for $N=7$, where the pathway

$$\begin{array}{cc}\hfill 01& d_1=1,\hfill \\ \hfill 001,0101& d_2=d_3=2,\hfill \\ \hfill 0010101& d_4=3\hfill \end{array}$$

(27)

results in a string having ma $a_{\min }^{\left(7\right)}=4$ and $d_{a_{\min }^{\left(7\right)}}=⌈{log}_2\left(7\right)⌉=3$, since both $\left[001\right]$ and $\left[0101\right]$ were assembled from the doublet $\left[01\right]$ in two independent assembly steps at the same depth $d_2=d_3=2$, which is congruent with Theorem 11.

Theorem 13 considers ASI-independent ASD, neglecting the working ASP, and hence also the AT principles. Similarly, the maximum achievable ASD would be $d_{\max }^{\left(N\right)}=N-1$, as a result of increasing the length of the string by one unit length symbol in $N-1$ steps. Hence, in general, we cannot consider the ASD apart from the ASI. 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

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

(28)

even though this string can be assembled with three larger pathway depths $w_6=\{4,5,6\}$. 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

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

(29)

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 ASI-dependent ASD cannot be decreased to $⌈{log}_2\left(8\right)⌉=3$

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

(30)

In general, the working ASP that contains a $2^d$-plet at the ASD d can also contain $\{2^{d-1}+1,2^{d-1}+2,\cdots ,2^d-1\}$-plets based on the substrings contained in the working ASP at the ASD $d-1$.

Conjecture 16. 

Strings [9] of lengths

$$\begin{array}{cc}\hfill {\tilde{N}}_3& 2^s+3·2^l,l=0,1,\cdots ,s-2\iff a_{\mathit{min}}^{\left({\tilde{N}}_3\right)}=s+2,\mathrm{and}\hfill \\ \hfill {\tilde{N}}_{7,s-3}& 2^s+7·2^{s-3}=\{15,30,60,\cdots \},s\ge 3\iff a_{\mathit{min}}^{\left({\tilde{N}}_{7,s-3}\right)}=s+2,\hfill \end{array}$$

(31)

are minimum ASI strings with assembly pathways devoid of independent assembly steps and the ASI higher than the minimum ASI-independent ASD (16).

Conjecture 17. 

If the DPI equals the minimum ASI of a string $C_k$, then the ASD of this string equals the minimum ASI-independent ASD (16) as a function of N, $\forall b$, that is

$${\hat{a}}_{\mathit{min}}^{\left(N\right)}=a_{\mathit{min}}^{\left(N\right)}\left(C_k\right)\iff d_{a_{\mathit{min}}^{\left(N\right)}}\left(C_k\right)=d_{\mathit{min}}^{\left(N\right)}=⌈{log}_2\left(N\right)⌉.$$

(32)

Conjecture 18. 

If the DPI is larger than the minimum ASI of a string $C_k$, then the ASD of this string equals its minimum ASI, that is

$${\hat{a}}_{\mathit{min}}^{\left(N\right)}>a_{\mathit{min}}^{\left(N\right)}\left(C_k\right)\iff d_{a_{\mathit{min}}^{\left(N\right)}}\left(C_k\right)=a_{\mathit{min}}^{\left(N\right)}\left(C_k\right),$$

(33)

and the ASD $d_{\mathit{min}}^{\left(N\right)}$ can be achieved in ${\hat{a}}_{\mathit{min}}^{\left(N\right)}$ assembly steps.

For example, the DPI ${\hat{a}}_{\min }^{\left(15\right)}=6$ is larger than the ASI $a_{\min }^{\left(15\right)}\left(C_k\right)=5$. Therefore, the ASD of any string $C_k$ having this ASI is $d_{a_{\min }^{\left(15\right)}}\left(C_k\right)=a_{\min }^{\left(15\right)}\left(C_k\right)=5$, whereas the minimum ASI-independent ASD $d_{\min }^{\left(15\right)}=4$ can be achieved with a string assembled in six assembly steps, which corresponds to the DPI of ${\hat{a}}_{\min }^{\left(15\right)}=6$. Certain tuples satisfying the Conjecture 18 are listed in Table 1 and illustrated in Figure 1.

In general, Theorems 11-15 show that

• the working ASP of minimum ASI strings having ASI equal to DPI cannot contain strings assembled in independent assembly steps,
• the working ASPs of other minimum ASI strings can contain at least two such strings, and therefore
• the assembly pathway of a maximum ASI string will tend to maximize the number of strings assembled in independent assembly steps in the working ASP, taking into account the saturation of the working ASP as it cannot contain more than $b^n$ distinct n-plets, and hence to minimize the possible ASD.

As shown in Figure 2(c,d), the string $C_{\max }^{(15,2)}=\left[010101000011100\right]$ has the ASI $a_{\max }^{(15,2)}=10$ and the ASD $d_{a_{\max }}^{(15,2)}=4$, while the string $C_{\min }^{(15,2)}=\left[010010100101001\right]$ has smaller ASI $a_{\min }^{\left(15\right)}=5$ but larger ASD $d_{a_{\min }}^{(15,2)}=5$. On the other hand, the ASD of the strings $C_{(N-5)}^{(16,2)}$ (A21) and (2), shown in Figure 2(a,b), is the same.

Theorem 19. 

The ASD of any maximum ASI string $C_{\mathit{max}}^{(N,b)}$, for all b corresponds to the minimum ASI-independent ASD (16) of Theorem 13, that is

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

(34)

Proof. 

Using the property of the ceiling function $\lceil x\rceil =n\iff n-1<x\le n$ valid for $n\in \mathbb{N},x\in \mathbb{R}$, we have

$$d_{a_{\max }}^{(N,b)}=⌈{log}_2\left(N\right)⌉\iff d_{a_{\max }}^{(N,b)}-1<{log}_2\left(N\right)\le d_{a_{\max }}^{(N,b)},$$

(35)

The non-strict inequality (35) corresponds to the non-strict inequality (15) valid for any N and any ASD. Therefore, we need to prove that the strict inequality $d_{a_{\max }}^{(N,b)}<{log}_2\left(N\right)+1$ holds for all $C_{\max }$ strings. Assume, for contradiction, that there exists a maximum ASI string $C_{\max }^{(N,b)}$ such that

$$d_{a_{\max }}^{(N,b)}\ge {log}_2\left(N\right)+1\Rightarrow 2^{d_{a_{\max }}^{(N,b)}}\ge 2N.$$

(36)

But this relation does not hold for the maximum ASI string $C_{\max }^{(N,b)}$. □

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

$$[***{\displaystyle ★}{\displaystyle ★}{\displaystyle ★}*]\mathrm{and}[{\displaystyle ★}***{\displaystyle ★}{\displaystyle ★}{\displaystyle ★}],$$

(37)

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 maximum 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 A and B). In both methods, we start with an initial balanced string of length $3b$ containing b clear triplets ordered as

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

(38)

The doublets that can be inserted into the initial string (38) can be arranged in a $b\times b$ matrix

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

(39)

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 (38), we can also cross out the entries in the first superdiagonal of the matrix (39). 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 20. 

The longest length of a string that has the ASI of $N-1$ is given by

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

(40)

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

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

(41)

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

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

(42)

The proof of Theorem 20 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 A or B it is terminated with 0 and has a form

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

(43)

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 (40) 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 (38), leading to twelve unique strings of length $N_{(N-1)}=13$

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

(44)

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 21. 

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

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

(45)

The proof of Theorem 21 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 21 are based on strings generated by Method A or B, for $b>2$ they owe their properties to the following distributions of symbols

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

(46)

For the strings of the form (46) the fractions in the Shannon entropy are

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

(47)

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

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

(48)

of any $C_{\max }$ string having length $N-k$, for $2\le k\le 9$.

The entropies (42) and (48) are shown in Figure 3. Radix $b=4$ is the smallest one at which the entropy (48) is a monotonically decreasing function of k. For $b\in \{2,3\}$ there is a local entropy minimum for $k=5$ and for $b=2$ an additional local entropy minimum for $k=2$. Perhaps, the entropy (48) has other local entropy minima for $b<4$ and for $k>9$.

Conjecture 22. 

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

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

(49)

or equivalently

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

(50)

where

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

(51)

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 $N>{\hat{N}}_{max}$, 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

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

(52)

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

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

(53)

The proof of the Conjecture 22 must show the conditions for the equations (52) and (53) to hold. We note that the assumption used in the equation (53) is valid only for $n_r\le N_{(N-1)}$ and $n_p\le N_{(N-1)}$. The bounds of Theorems 20 and 21 and Conjecture 22 are illustrated in Figure 1.

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 2. 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

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

(54)

unique adjoining doublets, where $N_{(N-k)}$ is given by the relations (40) or (45), 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 $\times 2^s=1$. Furthermore, each $\times 2^s$ step and all last $2^s$ steps are tantamount to one ASD level.

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

$$\begin{array}{cc}\hfill 0\circ 1=\left[01\right],0\circ 0=\left[00\right],1\circ 1=\left[11\right],1\circ 2=\left[12\right],& \\ \hfill 2\circ 2=\left[22\right],1\circ 0=\left[10\right],2\circ 0=\left[20\right]& (\times 2_{(b=3)}=7),\hfill \\ \hfill \circ \left[01\right]=\left[0101\right],\left[00\right]\circ \left[00\right]=\left[0000\right],\left[11\right]\circ \left[12\right]=\left[1112\right],\left[22\right]\circ \left[10\right]=\left[2210\right]& (\times 4=4),\hfill \\ \hfill \circ \left[0000\right]=\left[01010000\right],\left[1112\right]\circ \left[2210\right]=\left[11122210\right]& (\times 8=2),\hfill \\ \hfill \circ \left[11122210\right]=\left[0101000011122210\right]& (\times 16=1),\hfill \\ \hfill \circ \left[20\right]=\left[010100001112221020\right]& (\mathrm{last}\times 2),\hfill \\ \hfill 2-27+4+2+1+1=& 15\mathrm{steps},d_{15}=5.\hfill \end{array}$$

(55)

Similarly, the $N_{(N-1)}$ string (A3) of length $N_{(N-1)}=21$ for $b=4$ can be assembled, as shown in Table 2 as

$$\begin{array}{cc}\hfill 0\circ 0=\left[00\right],0\circ 1=\left[01\right],1\circ 1=\left[11\right],2\circ 2=\left[22\right],2\circ 3=\left[23\right],& \\ \hfill 3\circ 3=\left[33\right],1\circ 0=\left[10\right],2\circ 1=\left[21\right],3\circ 2=\left[32\right],0\circ 3=\left[03\right]& (\times 2_{(b=4)}=10),\hfill \\ \hfill \circ \left[01\right]=\left[0001\right],\left[11\right]\circ \left[22\right]=\left[1122\right],\left[23\right]\circ \left[33\right]=\left[2333\right],& \\ \hfill \circ \left[21\right]=\left[1021\right],\left[32\right]\circ \left[03\right]=\left[3203\right]& (\times 4=5),\hfill \\ \hfill \circ \left[1122\right]=\left[00011122\right],\left[2333\right]\circ \left[1021\right],\left[23331021\right]& (\times 8=2),\hfill \\ \hfill \circ \left[23331021\right]=\left[0001112223331021\right]& (\times 16=1),\hfill \\ \hfill \circ \left[3203\right]=\left[00011122233310213203\right]& (\mathrm{last}\times 4),\hfill \\ \hfill \circ 0=\left[000111222333102132030\right]& (\mathrm{last}\times 1),\hfill \\ \hfill 2-210+5+2+1+1+1=& 20\mathrm{steps},d_{20}=5.\hfill \end{array}$$

(56)

Furthermore, for $b=1$ the method produces the DPI (OEIS A014701). For example, the string of length $N=15$ can be assembled in six steps as

$$\begin{array}{cc}\hfill 0\circ 0=\left[00\right],& (\times 2_{(b=1)}=1),\hfill \\ \hfill \circ \left[00\right]=\left[0000\right]& (\times 4_{(b=1)}=1),\hfill \\ \hfill \circ \left[0000\right]=\left[00000000\right]& (\times 8_{(b=1)}=1),\hfill \\ \hfill \circ \left[0000\right]=\left[000000000000\right]& (\mathrm{last}\times 4),\hfill \\ \hfill \circ \left[00\right]=\left[00000000000000\right]& (\mathrm{last}\times 2),\hfill \\ \hfill \circ \left[0\right]=\left[000000000000000\right]& (\mathrm{last}\times 1),\hfill \\ \hfill 2-21+1+1+1+1+1=& 6\mathrm{steps},d_6=4,\hfill \end{array}$$

(57)

where obviously $max\left(\times 2^s\right)=1$. However, this is the $1^{\mathrm{st}}$ exception for $b=1$ as the ASI of this string is five if it is assembled using doublet $\left[00\right]$ and triplet $\left[000\right]$.

We further note that the method illustrated in Table 2 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 2. However, only the left one correctly reflects the maximum ASI of the assembled string.

$$\begin{array}{cccc}\hfill 0\circ 0=\left[00\right],0\circ 1=\left[01\right],1\circ 1=\left[11\right]& \hfill (\times 2_{(b=2)}=3),& \hfill 0\circ 0=\left[00\right],1\circ 0=\left[10\right],1\circ 1=\left[11\right]& \hfill (\times 2_{(b=2)}=3),\\ \hfill \left[00\right]\circ \left[01\right]=\left[0001\right]& \hfill (\times 4=1),& \hfill \left[00\right]\circ \left[10\right]=\left[0010\right]& \hfill (\times 4=1),\\ \hfill \left[0001\right]\circ \left[11\right]=\left[000111\right]& \hfill (\mathrm{last}\times 2),& \hfill \left[0010\right]\circ \left[11\right]=\left[001011\right]& \hfill (\mathrm{last}\times 2),\\ \hfill 2-24-43+1+1=& \hfill 5\mathrm{steps},d_5=3,& \hfill 3+1+1=& \hfill 5\ne 4\mathrm{steps},d_5=3,\end{array}$$

(58)

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 $N=10$

$$\begin{array}{cc}\hfill 0\circ 1=\left[01\right],0\circ 0=\left[00\right],1\circ 1=\left[11\right],1\circ 0=\left[10\right]& \hfill (\times 2_{(b=2)}=4),\\ \hfill \left[01\right]\circ \left[00\right]=\left[0100\right],\left[00\right]\circ \left[11\right]=\left[0011\right]& \hfill (\times 4=2),\\ \hfill \left[0100\right]\circ \left[0011\right]=\left[01000011\right]& \hfill (\times 8=1),\\ \hfill \left[01000011\right]\circ \left[10\right]=\left[0100001110\right]& \hfill (\mathrm{last}\times 2),\\ \hfill 2-24+2+1+1=& \hfill 8\mathrm{steps},d_8=4\\ \hfill & \\ \hfill 0\circ 0=\left[00\right],0\circ 1=\left[01\right],1\circ 0=\left[10\right],1\circ 1=\left[11\right]& \hfill (\times 2_{(b=2)}=4),\\ \hfill \left[00\right]\circ \left[01\right]=\left[0001\right],\left[10\right]\circ \left[11\right]=\left[1011\right]& \hfill (\times 4=2),\\ \hfill \left[0001\right]\circ \left[1011\right]=\left[00011011\right]& \hfill (\times 8=1),\\ \hfill \left[0001011\right]\circ \left[11\right]=\left[0001101111\right]& \hfill (\mathrm{last}\times 2),\\ \hfill 2-24+2+1+1& \hfill 8\ne 6\mathrm{steps},d_8=4,\end{array}$$

(59)

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,

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

(60)

where $a_{\mathrm{method}}^{(N,b)}$ is the ASI of a string $C_k$ determined by the method illustrated in Table 2. For example, the first six strings below contain four unique doublets instead of the required three. Therefore

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

(61)

Further research should consider researching the formula equivalent to (40) that captures a quadruplet repetition, similarly as $b^2+b^1+b^0$ captures a doublet repetition.

III. Discussion

Our study’s mathematical findings, especially the theorems concerning the ASI, DPI, and ASD, provide a framework for understanding the principles underlying the assembly of biological macromolecules such as DNA and proteins. For example, we show that DNA strand containing four nucleobases cannot represent a mininimum ASI string as it would violate Chargaff’s rules and Theorem 3 asserting that a mininimum ASI string can contain only at most two distinct symbols if $N=2^2$ and at most three, otherwise. We also show that maximum length strings without any substring repetitions, i.e., the maximum ASI strings having $a_{\max }^{(N,b)}=N-1$ are necessarily the most balanced: all but one symbol occur $b+1$ times and one symbol occurs $b+2$ times within any string $C_{(N-1)}$. However, longer maximum ASI strings $C_{(N-k)}$ become less balanced and their entropies (48) decrease. We have shown that the radix $b=4$ is the smallest one at which the entropy (48) is a monotonically decreasing function of k. Together with Theorem 3 this could be the reason nature has chosen four nucleobases to encode genetic information.

Despite the mathematical ideal of maximum entropy in balanced sequences, biological systems often tend to deviate from this balance. This deviation is evident in natural sequences, where certain nucleotides or amino acids occur more frequently than others, resulting in lower entropy. 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] ($a_{\min }^{\left(29903\right)}=19$). 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.

The observed tendency of biological sequences to evolve toward lower entropy may be associated with energy minimization. According to thermodynamic principles, systems tend to move toward states of lower free energy. In the context of AT, the energy associated with assembling a sequence can be related to the number of assembly steps. Sequences that require fewer assembly steps (lower ASI) may be energetically favorable. Thus, more complex sequences (with higher ASIs) require more assembly steps and, consequently, more energy. 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.

In essence, our theorems provide a mathematical underpinning for the observed patterns in biological sequences, offering explanations for phenomena such as the preference for radix $b=4$ in genetic encoding and the evolutionary trend toward lower entropy. The integration of AT into biological contexts opens avenues for a fundamental mathematical understanding of evolutionary processes, responding to the call for a precise and abstract mathematical theory of evolution [20].