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.

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. 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 [12].

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 $C_k\in P_s\left(b\right)$. We note that the joining operator "∘", in general, does not commute. 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 leading to this string, since at each assembly step the cardinality of the working assembly pool increases by one. Therefore, the working assembly pool 2 cannot be identified with the initial assembly pool 1; the initial assembly pool 1 must not contain strings of basic symbols (see Section H).

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 quadruplets $C_{k,\min }^{(4,b)}=[****]$ and $b(b-1)$ quadruplets $C_{l,\min }^{(4,b)}=[*★*★]$ with ASI equal to two, while the ASI of the remaining 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 pool $P_1$, 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 ★. Furthermore, we consider the degenerate case of just one basic symbol ($b=1$).

Theorem 2.

The smallest 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 $l\le k<j$. The first step in creating an addition chain for N is always $a_1=1+1=2$ and this corresponds to assembling a doublet $[**]$ or $[*★]$ from the initial assembly pool $P_0\left(b\right)$. 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 $b^2$ strings (2).

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 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 $\{0,1,2\}$. □

At least some of the following seven simple theorems are useful for further consideration.

Theorem 3.

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

Proof. 

Minimal 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. Minimal 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 minimal 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 minimal 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 minimal 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 minimal string $C_{\min }^{(2^s,b)}$ contains at most two distinct symbols.

We will now show that other minimal ASI strings of length $N\ne 2^s$ can contain at most three distinct symbols if $b>2$. We provide the construction of minimal 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 minimal 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. □

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)

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

(4)

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

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

(5)

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

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

(6)

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

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

(7)

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

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

(8)

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.

A string $C_m^{(N,b)}$ containing $k_1$ copies of an $n_1$-plet $C_1^{(n_1,b)}$, $k_2$ copies of an $n_2$-plet $C_2^{(n_2,b)}$, etc. has the ASI of at most.

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

(9)

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

Proof. 

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

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

(10)

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 $a\left(C_l^{(n,b)}\right)$ 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 $a\left(C_l^{(n,b)}\right)=n-1$, 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 $(n-1)1-(n-1)=0$, while more copies k decrease it by $(n-1)(k-1)$. On the other hand, if $a\left(C_l^{(n,b)}\right)<n-1$ 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 5-plet $\left[01001\right]$, each with $a^{(5,6)}\left(\left[01001\right]\right)=3$, in a string

$$C_k^{(24,6)}=\left[120100121010012350100152\right],$$

(11)

its ASI amounts to $a^{(24,6)}\left(C_k\right)=24-1-(4·3-3)=14$. We note that the maximum ASI decrease is provided by $2^s$-plets of the minimum ASI and amounts to $(n-1)k-{log}_2\left(n\right)=(2^s-1)k-s$.

Another quantity related to the string assembly is the assembly depth (AD) defined [13] as

$$d_s^{(N,b)}\max (d_{(s-1)L}^{(N,b)},d_{(s-1)R}^{(N,b)})+1,$$

(12)

where $d_{(s-1)L}^{(N,b)}$ and $d_{(s-1)R}^{(N,b)}$ are the ADs of two parts of this string that were joined in step s, where $d_0^{(N,b)}0$. If there are more assembly paths leading to a string with different ADs, which happens if at least two assembly steps can occur independently, in ref. [13] the minimum AD is assigned to the string. Here, we relax this assumption. Any string has a unique assembly index but can have different ADs if its ASI is not minimal.

Theorem 11.

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

Proof. 

We will prove this by mathematical induction on the length N of the string. For the base case ($N=1$) 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 AD ${d_s}^{(1,b)}0$. Therefore, ${d_s}^{(1,b)}=a_{\min }^{\left(1\right)}=0$.

Assume now that for all strings of length k less than N, the AD equals the minimal ASI, that is

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

(13)

We need to show that $d_{a_{\min }}^{(N,b)}=a_{\min }^{\left(N\right)}$. Consider two cases based on whether N is a power of two or not.

If N is a power of two, i.e., $N=2^k$ for some integer k, we construct the minimal ASI string as follows. First, we assemble a doublet from two basic symbols:

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

(14)

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

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

(15)

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

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

(16)

and the AD is equal to

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

(17)

Therefore, ${d_s}^{(2^k,b)}=a_{\min }^{\left(2^k\right)}=k$ in this case.

If N is not a power of two, we construct the minimal ASI string as follows. We take the shortest addition chain for N. Next, we build the string by combining previously assembled strings according to the addition chain. At each assembly step, the AD increases by one, as we are combining previously assembled strings (i.e. we do not perform independent assembly steps) Since, by Theorem 2, the minimal ASI corresponds to the length of the shortest addition chain $l\left(N\right)$, we have

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

(18)

also in this case.

In both cases, whether N is a power of two or not, we have:

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

(19)

This completes the proof. □

Hence, for all b, a string with $a_{\min }^{\left(N\right)}$ can be constructed only with the AD $d_{a_{\min }}^{\left(N\right)}=a_{\min }^{\left(N\right)}$.

Theorem 12.

The AD of a maximal ASI string $C_{\max }^{\left(N\right)}$ satisfies

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

(20)

where $⌈x⌉$ denotes the ceiling function.

Proof. 

Since in each assembly step, we combine two previously assembled substrings or basic symbols, the AD increases by at most one. Therefore, the AD cannot exceed the total number of assembly steps and we obtain

$$d_{a_{\max }}^{(N,b)}\le a_{\max }^{(N,b)}.$$

(21)

Now we will prove that $d_{a_{\max }}^{(N,b)}\ge ⌈{log}_2\left(N\right)⌉$. Assume, for contradiction, that there exists a maximal ASI string $C_{\max }^{(N,b)}$ such that:

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

(22)

Let $d{d}_{a_{\max }}^{(N,b)}$, so $d<⌈{log}_2\left(N\right)⌉$. In an assembly tree of AD d, the maximum number of leaves that can be combined is $2^d$, because at each assembly step, we can combine at most two subassemblies. Therefore, the maximum length $N_{\max }$ of a string that can be assembled with AD d satisfies:

$$N_{\max }\le 2^d.$$

(23)

But since $d<⌈{log}_2\left(N\right)⌉$, we have:

$$2^d<2^{⌈{log}_2\left(N\right)⌉}\le 2^{{log}_2\left(N\right)+1-ϵ}=N·2^{1-ϵ},$$

(24)

for some $ϵ>0$. This implies that

$$N_{\max }<N·2^{1-ϵ}.$$

(25)

However, since $ϵ>0$, $2^{1-ϵ}<2$, so:

$$N_{\max }<N·2^{1-ϵ}<2N.$$

(26)

But this contradicts the fact that $N_{\max }\ge N$ (since we are assembling a string of length N).More precisely, since $d<⌈{log}_2\left(N\right)⌉$, we have $d\le ⌈{log}_2\left(N\right)⌉-1$, which implies:

$$2^d\le 2^{⌈{log}_2\left(N\right)⌉-1}=2^{{log}_2\left(N\right)}·2^{-1}={\displaystyle \frac{N}{2}}.$$

(27)

Thus, the maximum length of a string that can be assembled with AD d is $N_{\max }\le N/2$, which is less than N, contradicting the assumption that we are assembling a string of length N. This completes the proof. □

For example, the string $C_{\max }^{(8,2)}=\left[00011101\right]$ can be assembled with AD $⌈{log}_2\left(8\right)⌉\le d_{a_{\max }}^{(8,2)}\le a_{\max }^{(8,2)}$ as

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

(28)

Conjecture 1.

For N such that $a_{\min }^{\left(N\right)}>a_{\min }^{(N-1)}$ the AD $d_s^{(N,b)}$ of a string $C_k^{(N,b)}$ has a range of values between the minimum ASI for N and the ASI of this string, that is

$$⌈{log}_2\left(N\right)⌉\le d_s^{(N,b)}\left(C_k\right)\le a^{(N,b)}\left(C_k\right).$$

(29)

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

$$[***★★★*]\mathrm{and}[★***★★★],$$

(30)

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

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

(31)

The doublets that can be inserted into the initial string (31) can be arranged in a $b\times b$ matrix

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

(32)

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 (31), we can also cross out the entries in the first superdiagonal of the matrix (32). 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 13

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

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

(33)

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

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

(34)

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

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

(35)

The proof of Theorem 13 is given in Appendix D. A $C_{(N-1)}$ string must contain all clear triplets and all doublets. 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 (33) 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.

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

Theorem 14

($N_{(N-2)}$). For all $b>1$ the longest length of a string that has the ASI of $N-2$ is given by $N_{(N-2)}=N_{(N-1)}+3$ or equivalently by

$$N_{(N-2)}=b^2+b+4,$$

(36)

and

$$N_{(N-2)}=(b-2)N_c+(N_c+1)+(N_c+3)=b{N}_c+4,$$

(37)

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

$$\begin{array}{cc}\hfill H\left(C_{(N-2)}\right)& =-{\displaystyle \frac{b^2-b-2}{b^2+b+4}}{log}_2\left({\displaystyle \frac{b+1}{b^2+b+4}}\right)-{\displaystyle \frac{b+2}{b^2+b+4}}{log}_2\left({\displaystyle \frac{b+2}{b^2+b+4}}\right)-{\displaystyle \frac{b+4}{b^2+b+4}}{log}_2\left({\displaystyle \frac{b+4}{b^2+b+4}}\right).\hfill \end{array}$$

(38)

The entropy $H\left(C_{(N-2)}\right)\lesssim {log}_2\left(b\right)$ for $b⪆1.6398$.

The proof of Theorem 14 is given in Appendix F. $C_{(N-2)}$ string must contain only two copies of a doublet. Hence, a clear quadruplet ($bbbb$) and a pattern binding different symbols adjoining this quadruplet, such as $[\cdots abbbbc\cdots abc\cdots ]$, $[\cdots abbbbaba\cdots ]$, etc. must be present, so that any $C_{(N-2)}$ string contains only one pair of repeated doublets $ab$, $bb$, or $\{bc,ba\}$ (See also Appendix C). For example, for $N=10$, sixteen bitstrings

$$\begin{array}{cc}\hfill & \left[0100011110\right],\left[0111100010\right],\left[0111101000\right],\left[\underline{0100001110}\right],\hfill \\ \hfill & \left[0001011110\right],\left[0001111010\right],\left[0101111000\right],\left[0111000010\right]\hfill \end{array}$$

(39)

(an additional eight are given by swapping 0 with 1) have the ASI $a=N-2=8$, where the underlined string (39) is the one that is created for $b=2$ in Appendix F.

Theorem 15

($N_{(N-3)}$). The longest length of a string that has the ASI of $N-3$ is given by $N_{(N-3)}=N_{(N-1)}+5$. or equivalently by

$$N_{(N-3)}=b^2+b+6,$$

(40)

and

$$N_{(N-3)}=(b-2)N_c+(N_c+2)+(N_c+4)=b{N}_c+6,$$

(41)

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

$$\begin{array}{cc}\hfill H\left(C_{(N-3)}\right)& =-{\displaystyle \frac{b^2-b-2}{b^2+b+6}}{log}_2\left({\displaystyle \frac{b+1}{b^2+b+6}}\right)-{\displaystyle \frac{b+3}{b^2+b+6}}{log}_2\left({\displaystyle \frac{b+3}{b^2+b+6}}\right)-{\displaystyle \frac{b+5}{b^2+b+6}}{log}_2\left({\displaystyle \frac{b+5}{b^2+b+6}}\right).\hfill \end{array}$$

(42)

The entropy $H\left(C_{(N-3)}\right)\lesssim {log}_2\left(b\right)$ for $b⪆1.8317$.

The proof of Theorem 15 is given in Appendix G. $C_{(N-3)}$ string must contain only three copies of a doublet, or two pairs of different doublets.

In general, the strings of Theorem 14, generated by Method A or B, are terminated with 0 and owe their properties to the following distributions of symbols

$$\begin{array}{cc}\hfill C_{(N-2)}& =[0100001\cdots 10\cdots 0],\hfill \\ \hfill C_{(N-3)}& =[010100001\cdots 10\cdots 0],\hfill \end{array}$$

(43)

$C_{(N-2)}$ contains three pairs of doublets $\left[00\right]$, $\left[01\right]$, and $\left[10\right]$ which are connected in such a way that only one pair can be reused from the assembly pool to decrease the maximum $N-1$ ASI by one.

Conjecture 2.

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

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

(44)

Equivalently, if $b>1$ and $N>N_{(N-2)}$ then

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

(45)

($b(b+1)/2$ are triangular numbers, A000217). In other words, if $N>N_{(N-2)}$, then ASI increments by one every time the string length N increments by two.

First, we note that maximum ASI must rise. Otherwise, if it was constant for N larger than some $N_{max}$ then at some even larger N it would become inevitably lower than minimum ASI bound which also rises, and this would be a contradiction.

W.l.o.g. we aim to prove this theorem for $b=2$. We note that inserting any doublet into a $C_{(N-3)}^{(12,2)}$ string (A20) creates a triplet. Using the equation (9) of Theorem 10 we have

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

(46)

for any step s if only $N_s\ge N_{(N-2)}$.

Assume that $\forall s$, $a\left(C_s^{(n_s,b)}\right)=n_s-1$ and $\forall p$, $a\left(C_p^{(n_p,b)}\right)=n_p-1$. Then

$$\begin{array}{cc}\hfill \sum _{s=1}^{R_s}\left[(n_s-1)(k_s-1)\right]& =\sum _{p=1}^{R_p}\left[(n_p-1)(k_p-1)\right]+1,\hfill \\ \hfill \sum _{s=1}^{R_s}{n}_s{k}_s-\sum _{s=1}^{R_s}{n}_s-\sum _{s=1}^{R_s}{k}_s+R_s& =\sum _{p=1}^{R_p}{n}_p{k}_p-\sum _{p=1}^{R_p}{n}_p-\sum _{p=1}^{R_p}{k}_p+R_p+1,\hfill \end{array}$$

(47)

The proof of the Conjecture 2 must show the conditions for the equations (46) and (47) to hold.

The bounds of Theorems 14-15 and Conjecture 2 are illustrated in Figure 1 and Figure 2.

These results disprove our upper bound Conjecture 1 for $b=2$ stated in ref. [9]. However, along with Theorem 12 they led us to the method of determining the ASI of a maximum ASI string, based on independent doublets and powers of two, as shown in Table 1. First, a string is sequenced, every two symbols to find the number of unique doublets $\times 2_{\left(b\right)}$. Subsequently, these doublets form $\times 4$ quadruplets, quadruplets form $\times 8$ octuples, and so on depending on the length of the string N. Only the first quantity, $\times 2_{\left(b\right)}$, depends on the radix b. 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 $\times 2^s=1$. It turns out that for $b>1$, the ASI depends only on the distribution of doublets ($\times 2_{\left(b\right)}$). In particular, the $N_{(N-1)}$ strings (A3) or (A4) contain the maximum of $⌊N_{(N-1)}/2⌋$ independent doublets, and the $N_{(N-2)}$ string (A14) contains the maximum of $N_{(N-2)}/2-1$ independent doublets.

For example, the $N_{(N-3)}$ string (A21) of length $N_{(N-3)}=18$ for $b=3$ can be assembled as

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

(48)

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

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

(49)

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

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

(50)

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 $\left[000\right]$.

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 satisfy the distributions of Table 1. However, only the left one correctly reflects the ASI of the assembled string.

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

(51)

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 minimal 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 [14,15] and, hence, will become less balanced. This tendency to imbalance seems to be associated with some minimum energy condition. The energy of a black hole that can be thought of as a balanced bitstring [16] can be two times the energy of the entropy variation sphere that it generates [17]. 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 [9,14]. The maximum ASI of a $C_{\max }^{(29903,4)}$ string given by the equation (45) is $a_{\max }^{(29903,4)}=14961$.

Authors’ contributions— WB: first concept of a general method for constructing the string of length $N_{(N-1)}$ leading to Theorem 13; the concept of the doublet matrix (32); outline of the general Method A; proposition of Theorem 9; a string with exactly two copies of all doublets idea and the formula for its length; numerous clarity corrections and improvements; PM: outline of the general Method B; the hint for ASI combinatorics; creation of a software supporting Conjecture 2; numerous clarity corrections and improvements; AT: formal proof of Theorem 3; proof that the Shannon entropy (35) can be approximated by ${log}_2\left(b\right)$ for large b; proof of the Theorem 11; proof of the Theorem 12; the 1^st paragraph of the discussion Section 3; numerous clarity corrections and improvements; SŁ: The remaining part of the study.