How to resolve the algorithm RSA code step by step in the F# programming language
How to resolve the algorithm RSA code step by step in the F# programming language
Table of Contents
Problem Statement
Given an RSA key (n,e,d), construct a program to encrypt and decrypt plaintext messages strings. Background RSA code is used to encode secret messages. It is named after Ron Rivest, Adi Shamir, and Leonard Adleman who published it at MIT in 1977. The advantage of this type of encryption is that you can distribute the number “
n
{\displaystyle n}
” and “
e
{\displaystyle e}
” (which makes up the Public Key used for encryption) to everyone. The Private Key used for decryption “
d
{\displaystyle d}
” is kept secret, so that only the recipient can read the encrypted plaintext. The process by which this is done is that a message, for example “Hello World” is encoded as numbers (This could be encoding as ASCII or as a subset of characters
a
01 , b
02 , . . . , z
26
{\displaystyle a=01,b=02,...,z=26}
). This yields a string of numbers, generally referred to as "numerical plaintext", “
P
{\displaystyle P}
”. For example, “Hello World” encoded with a=1,...,z=26 by hundreds would yield
08051212152315181204
{\displaystyle 08051212152315181204}
. The plaintext must also be split into blocks so that the numerical plaintext is smaller than
n
{\displaystyle n}
otherwise the decryption will fail. The ciphertext,
C
{\displaystyle C}
, is then computed by taking each block of
P
{\displaystyle P}
, and computing Similarly, to decode, one computes To generate a key, one finds 2 (ideally large) primes
p
{\displaystyle p}
and
q
{\displaystyle q}
. the value “
n
{\displaystyle n}
” is simply:
n
p × q
{\displaystyle n=p\times q}
. One must then choose an “
e
{\displaystyle e}
” such that
gcd ( e , ( p − 1 ) × ( q − 1 ) )
1
{\displaystyle \gcd(e,(p-1)\times (q-1))=1}
. That is to say,
e
{\displaystyle e}
and
( p − 1 ) × ( q − 1 )
{\displaystyle (p-1)\times (q-1)}
are relatively prime to each other. The decryption value
d
{\displaystyle d}
is then found by solving The security of the code is based on the secrecy of the Private Key (decryption exponent) “
d
{\displaystyle d}
” and the difficulty in factoring “
n
{\displaystyle n}
”. Research into RSA facilitated advances in factoring and a number of factoring challenges. Keys of 829 bits have been successfully factored, and NIST now recommends 2048 bit keys going forward (see Asymmetric algorithm key lengths). Summary of the task requirements:
Let's start with the solution:
Step by Step solution about How to resolve the algorithm RSA code step by step in the F# programming language
Source code in the fsharp programming language
//Nigel Galloway February 12th., 2018
let RSA n g l = bigint.ModPow(l,n,g)
let encrypt = RSA 65537I 9516311845790656153499716760847001433441357I
let m_in = System.Text.Encoding.ASCII.GetBytes "The magic words are SQUEAMISH OSSIFRAGE"|>Array.chunkBySize 16|>Array.map(Array.fold(fun n g ->(n*256I)+(bigint(int g))) 0I)
let n = Array.map encrypt m_in
let decrypt = RSA 5617843187844953170308463622230283376298685I 9516311845790656153499716760847001433441357I
let g = Array.map decrypt n
let m_out = Array.collect(fun n->Array.unfold(fun n->if n>0I then Some(byte(int (n%256I)),n/256I) else None) n|>Array.rev) g|>System.Text.Encoding.ASCII.GetString
printfn "'The magic words are SQUEAMISH OSSIFRAGE' as numbers -> %A\nEncrypted -> %A\nDecrypted -> %A\nAs text -> %A" m_in n g m_out
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