A new summary of the conceptual approach of the nearest chain would be appreciated. Especially to clear up the confusion, I will provide a link to a written computer script with the linear approach in mind.

Wikipedia Nearest chain

Example of a basic script

NOTE: The concept is to visualize the chains as a number line.

Script stripped

L = length of the string in number of characters

X = how many ropes in total.

D = hamming distance.

```
Z = X * X * 1 * L + X * D * D * D + 1
S = Z / D
B = S / D
Y = S / B
P = B * Y
```

The algorithm takes input for the length of a string to **L**, and the number of strings in total for **X**. The hamming distance is for the entrance. **re**. Entry **Z** calculates the exact amount of all possible characters in the string plus the permutations for the input **re**.

We get a set of **Z** number of characters. We divide **Z** by **re** which **S**. We obtain **second** Possible permutations of characters that can be generated in a list of **X** string instruments. In other words, **Z** divided by **Y** groupings of the same **X** must have only B possible permutations within the **Z** characters. (For the numerical chain of the center based on Hamming?)

If the algorithm is, correct (or I've cheated). The center of **Z** it's in the **S** chain that should be the **second** permutation.

The verification of mathematics.

S = S * D / D

D = Z / S

Continue if I want to find the second closest, the third closest and so on within the amount S that was swapped within the distance of Hamming.

```
NQ = S / RT
P = RT * NQ
```

*My understanding of the nearest chain is wrong when it comes to the linear approach, why is the approach not promising? In what way would you approach it?*

**TO UPDATE-
EXAMPLE
The number line starts from 485.5 and ends in 971. To find our nearest number 15 string, we go to**

NQ = S / RT

P = RT * NQ

**The result of NQ is added to 485.5 and we obtain our nearest 15th chain.**

Let's suppose, we have a habit. **Fifth** chain closest to our primary **S** what's it gonna be **NQ**. We divide **NQ** continues adding the sum of them **5 D** until an approximate **Z**. If it is not equal to Z then divide **NQ for 2** Will this check the algorithm?