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:

import "/big" for BigInt

var pt = "Rosetta Code"
System.print("Plain text:            : %(pt)")
var n = BigInt.new("9516311845790656153499716760847001433441357")
var e = BigInt.new("65537")
var d = BigInt.new("5617843187844953170308463622230283376298685")
var ptn = BigInt.zero
// convert plain text to a number
for (b in pt.bytes) {
    ptn = (ptn << 8) | BigInt.new(b)
}
if (ptn >= n) {
    System.print("Plain text message too long")
    return
}
System.print("Plain text as a number : %(ptn)")

// encode a single number
var etn = ptn.modPow(e, n)
System.print("Encoded                : %(etn)")

// decode a single number
var dtn = etn.modPow(d, n)
System.print("Decoded                : %(dtn)")

// convert number to text
var db = List.filled(16, 0)
var dx = 16
var bff = BigInt.new(255)
while (dtn.bitLength > 0) {
    dx = dx - 1
    db[dx] = (dtn & bff).toSmall
    dtn = dtn >> 8
}
var s = ""
for (i in dx..15) s = s + String.fromByte(db[i])
System.print("Decoded number as text : %(s)")