I’m probably wrong, but I believed I may have solved a NP-complete problem quickly. I’m having a hard time trying to organize my thoughts for others to understand. So if anyone has a constructive suggestion please let me know with the edit suggestions or the commentary.

*Anyway, It is known that if any NP-complete problem can be solved quickly, then every problem in NP can because the definition of an NP-complete problem states that every problem in NP must be quickly reducible to every NP-complete problem (that is, it can be reduced in polynomial time).Because of this, it is often said that NP-complete problems are harder or more difficult than NP problems in general. At least according to Wikipedia.*

Anyone, who wants an idea of what I was trying to solve please take a look here and take a look at the algorithm down below. I will begin with the explanation of the algorithm.

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

The algorithm takes input for a string’s length for **L**, and the number of strings altogether for **X**. The hamming distance is for input **D**. Input **Z** calculates the exact amount of all possible characters in the string besides the permutations for input **D**.

We get a set of **Z** amount of characters. We divide **Z** by D which is **S**. We get **B** possible permutations of characters that can be generated in a list of **X** strings. In other words, **Z** divided by **Y** groupings of the same **X** should uniquely have **B** possible permutations within the **Z** characters. (For the center numerical listed string based on hamming?)

If the algorithm is, correct (or I’ve misled). The center of **Z** is at the **S** string which should be the **B** permutation.

`Z = X * X * 1 * L + X * D * D * D + 1 S = Z / D B = S / D Y = S / B CL = D * A ≤ X ≤ B * Y--just for checking P=B*Y `

More checks

S = S * D / D

D = Z / S

NQ = S / RT P = RT * NQ

Suppose, we have a custom 5th closest string to our primary **S** which is going to be **NQ**. We divide **NQ** by 2 keep adding the sum thereof **5*D** until one reaches approximate **Z**. If it doesn’t equal **Z** then divide **NQ** by 2 This will check the algorithm?

Put differently, we have 971 characters altogether in a “superstring.” With these 971 we try to find the 5th closest string. (from 323) We take **NQ** which is 64.7 and we divide it by 2. Now, 323.6 is our approximated closest string. We take our **D** which is a 3 hamming distance and multiply it by **RT** which is 5. We get 15.17. We take 15.17 and multiply it by 64 and we get about 971. 971/5=194.2 We divide it by **D** and get approximate 3. SO there are 3 other strings that are equivalent to our main solution. To check, we keep subtracting 15 from 971 until we arrive at the closest to 323.