Step by Step solution about How to resolve the algorithm Bitcoin/public point to address step by step in the Ruby programming language

The provided Ruby code is a function that translates a public point on the Bitcoin elliptic curve to a Bitcoin address. Here's a step-by-step explanation of the code:

1. Constants and Input:

   • The code defines two constants, X and Y, which represent the coordinates of a public point on the Bitcoin elliptic curve.
   • n is calculated by concatenating '00' with the hexadecimal digest of the double SHA-256 hash of the converted public point.

2. Base58 Encoding:

   • The code uses the convert function to convert the public point coordinates into a string of hexadecimal digits. This string is then packed into a binary format according to the specification provided by the i string.
   • The resulting binary string is encoded using Base58, which is a modified version of Base64 that doesn't use the ambiguous characters '0' (zero), 'O' (uppercase o), 'I' (uppercase i), and '+' (plus). This encoded string is stored in the res variable.

3. Loop and Reverse:

   • The code enters a loop that repeatedly divides the n value by 58 and takes the remainder.
   • The character corresponding to the remainder is appended to the res string.
   • This loop continues until n becomes 0.
   • Finally, the res string is reversed, and the Bitcoin address is printed.

In summary, this code converts a public point on the Bitcoin elliptic curve to a Bitcoin address using double SHA-256 hashing and Base58 encoding. The resulting address is a string that represents the cryptocurrency wallet where bitcoins can be sent or received.

#  Translate public point to Bitcoin address
#
#  Nigel_Galloway
#  October 12th., 2014
require 'digest/sha2'
def convert g
  i,e = '',[]
  (0...g.length/2).each{|n| e[n] = g[n+=n]+g[n+1]; i+='H2'}
  e.pack(i)
end
X = '50863AD64A87AE8A2FE83C1AF1A8403CB53F53E486D8511DAD8A04887E5B2352'
Y = '2CD470243453A299FA9E77237716103ABC11A1DF38855ED6F2EE187E9C582BA6'
n = '00'+Digest::RMD160.hexdigest(Digest::SHA256.digest(convert('04'+X+Y)))
n+= Digest::SHA256.hexdigest(Digest::SHA256.digest(convert(n)))[0,8]
G = "123456789ABCDEFGHJKLMNPQRSTUVWXYZabcdefghijkmnopqrstuvwxyz"
n,res = n.hex,''
while n > 0 do
  n,ng = n.divmod(58)
  res << G[ng]
end
puts res.reverse