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3 years ago
{
"cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
"# GCD / Euclidean algorithm\n",
"Explaination: https://brilliant.org/wiki/extended-euclidean-algorithm/"
]
},
{
"cell_type": "code",
"execution_count": 1,
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"metadata": {},
"outputs": [
{
"data": {
"text/plain": [
"5"
]
},
"execution_count": 1,
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"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"# Returns the greatest common divisor of a and b.\n",
"def gcd(a, b):\n",
" if b > a:\n",
" return gcd(b, a)\n",
" if a % b == 0:\n",
" return b\n",
" return gcd(b, a % b)\n",
"\n",
"gcd(35,30)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"# Back substitution / Extended euclidean algorithm\n",
"Explaination: https://brilliant.org/wiki/extended-euclidean-algorithm/"
]
},
{
"cell_type": "code",
"execution_count": 2,
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"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"ax + by = gcd(a,b) ==> 35*1 + 10*-3 = gcd(35,10)\n",
"\n",
"gcd(35,10) = 5\n",
"x = 1, y = -3\n"
]
}
],
"source": [
"def gcdExtended(a, b): \n",
" # Base Case \n",
" if a == 0 : \n",
" return b,0,1\n",
"\n",
" gcd,x1,y1 = gcdExtended(b%a, a)\n",
"\n",
" # Update x and y using results of recursive call \n",
" x = y1 - (b//a) * x1 \n",
" y = x1 \n",
"\n",
" return gcd, x, y \n",
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"\n",
"a, b = 35,10\n",
"g, x, y = gcdExtended(a, b)\n",
"print(f\"ax + by = gcd(a,b) ==> {a}*{x} + {b}*{y} = gcd({a},{b})\\n\")\n",
"\n",
"print(f\"gcd({a},{b}) = {g}\")\n",
"print(f\"x = {x}, y = {y}\")"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"# Modular inverse, A^-1\n",
"Explanation: https://www.khanacademy.org/computing/computer-science/cryptography/modarithmetic/a/modular-inverses"
]
},
{
"cell_type": "code",
"execution_count": 3,
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"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"3 * 5 ≡ 1 (mod 7)\n",
"Modular inverse for 3 (mod 7) is 5\n",
"a^-1 = 5\n"
]
}
],
"source": [
"# Naive method of finding modular inverse for A (mod C) \n",
"def inverse(a, c):\n",
" for b in range(c):\n",
" if (a*b)%c == 1: return b\n",
" print(\"NO INVERSE FOUND\")\n",
"\n",
"a,c = 3,7\n",
"b = inverse(a,c)\n",
"\n",
"print(f\"{a} * {b} ≡ 1 (mod {c})\")\n",
"print(f\"Modular inverse for {a} (mod {c}) is {b}\")\n",
"print(f\"a^-1 = {b}\")"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"# Chineese remainder theorem\n",
"Source: https://rosettacode.org/wiki/Chinese_remainder_theorem#Python"
]
},
{
"cell_type": "code",
"execution_count": 4,
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"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"23\n"
]
}
],
"source": [
"from functools import reduce\n",
"def crt(n, a):\n",
" sum = 0\n",
" prod = reduce(lambda a, b: a*b, n)\n",
" for n_i, a_i in zip(n, a):\n",
" p = prod // n_i\n",
" sum += a_i * mul_inv(p, n_i) * p\n",
" return sum % prod\n",
"\n",
"def mul_inv(a, b):\n",
" b0 = b\n",
" x0, x1 = 0, 1\n",
" if b == 1: return 1\n",
" while a > 1:\n",
" q = a // b\n",
" a, b = b, a%b\n",
" x0, x1 = x1 - q * x0, x0\n",
" if x1 < 0: x1 += b0\n",
" return x1\n",
"\n",
"n = [3, 5, 7]\n",
"a = [2, 3, 2]\n",
"print(crt(n, a))"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"# Integer factorisation\n",
"Given an integer, find its prime factors"
]
},
{
"cell_type": "code",
"execution_count": 5,
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The factors of 641361 are [3, 7, 7, 4363], verified\n",
"Manual test: True\n"
]
}
],
"source": [
"# Brute force solution\n",
"def prime_factors(n):\n",
" i = 2\n",
" factors = []\n",
" while i * i <= n:\n",
" if n % i:\n",
" i += 1\n",
" else:\n",
" n //= i\n",
" factors.append(i)\n",
" if n > 1:\n",
" factors.append(n)\n",
" return factors\n",
"\n",
"# Check if the numbers in array a is the factors of n\n",
"def factor_test(a, n):\n",
" k = 1\n",
" for i in factors:\n",
" k=k*i\n",
" return k==n\n",
"\n",
"n = 641361\n",
"factors = prime_factors(n)\n",
"print(f\"The factors of {n} are {factors}, {'verified' if factor_test(factors, n) else 'ERROR'}\")\n",
"\n",
"print(\"Manual test:\", factor_test([3, 7, 7, 4363], 641361))"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
3 years ago
"# Euler function φ and Z*n\n",
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"From slides 2020-4135-l07 - Number Theory for Public Key Cryptography\n"
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]
},
{
"cell_type": "code",
"execution_count": 6,
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"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"φ(21) = 12 since the 12 items in Z*21 are each relatively prime to 21.\n",
"Z*21 = [1, 2, 4, 5, 8, 10, 11, 13, 16, 17, 19, 20]\n"
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]
}
],
"source": [
"# Relatively prime means that gcd equals 1\n",
"def is_relative_prime(a,b):\n",
" if b == 0 or a == 0: return False\n",
" return gcd(a,b) == 1\n",
"\n",
"# The set of positive integers less than n and relatively prime \n",
"# to n form the reduced residue class Z∗n.\n",
"def z_star(n):\n",
" r = []\n",
" for i in range(1,n):\n",
" if is_relative_prime(i,n):\n",
" r.append(i)\n",
" return r\n",
"\n",
"n = 21\n",
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"ar = z_star(n)\n",
"e = len(ar)\n",
"print(f\"φ({n}) = {e} since the {e} items in Z*{n} are each relatively prime to {n}.\")\n",
"print(f\"Z*{n} = {ar}\")"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Euler’s Totient\n",
"- https://www.dcode.fr/euler-totient\n",
"- https://www.ibnus.io/eulers-totient-function-using-python/\n",
"\n",
"In general:\n",
"\n",
"> $\\varphi(n) = n \\prod_{p \\mid n} \\left( 1 - \\frac{1}{p} \\right)$\n",
"\n",
"Examples:\n",
"> $\\varphi(16) = \\varphi(2^{4}) = 16*\\left(1-\\frac{1}{2}\\right) = 8$\n",
"\n",
"> $\\varphi(1125) = \\varphi(33555) = 1125\\left(1-\\frac{1}{3}\\right)\\left(1-\\frac{1}{5}\\right) = 600$\n"
]
},
{
"cell_type": "code",
"execution_count": 7,
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"𝜑(1125) = 600\n"
]
}
],
"source": [
"# Euler's Totient Function\n",
"# using Euler's product formula\n",
"def totient(n) :\n",
" result = n # Initialize result as n\n",
"\n",
" # Consider all prime factors\n",
" # of n and for every prime\n",
" # factor p, multiply result with (1 – 1/p)\n",
" p = 2\n",
" while(p * p <= n) :\n",
"\n",
" # Check if p is a prime factor.\n",
" if (n % p == 0) :\n",
"\n",
" # If yes, then update n and result\n",
" while (n % p == 0) :\n",
" n = n // p\n",
" result = result * (1.0 - (1.0 / (float) (p)))\n",
" p = p + 1\n",
"\n",
" # If n has a prime factor\n",
" # greater than sqrt(n)\n",
" # (There can be at-most one\n",
" # such prime factor)\n",
" if (n > 1) :\n",
" result = result * (1.0 - (1.0 / (float)(n)))\n",
"\n",
" return (int)(result)\n",
"\n",
"n = 1125\n",
"print(f\"𝜑({n}) = {totient(n)}\")"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Finding a generator of Z*p\n",
"- https://crypto.stanford.edu/pbc/notes/numbertheory/gen.html\n",
"- Lecture 2, page 19\n",
"\n",
"A generator of $Z_p^∗$ is an element of order $p − 1$\n",
"\n",
"To find a generator of $Z_p^∗$ we can choose a value g and test it as follows:\n",
"1. Compute all the distinct prime factors of $p − 1$ and call them $f_1, f_2, ..., f_r$\n",
"2. Then $g$ is a generator as long as $g^{\\frac{p−1}{f_i}} \\neq 1 \\mod(p)$ for $i = 1,2,,...,r$"
]
},
{
"cell_type": "code",
"execution_count": 8,
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Is 3 a generator for Z*4? True\n"
]
}
],
"source": [
"# Finding a generator g of Z∗p\n",
"def is_generator(p, g):\n",
" pf = prime_factors(p-1)\n",
" for f in pf:\n",
" if (g**((p-1)/f))%p == 1:\n",
" return False\n",
" return True\n",
"\n",
"g = 3 # Generator\n",
"p = 4 # Z*p\n",
"\n",
"print(f\"Is {g} a generator for Z*{p}? {is_generator(p, g)}\")"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
3 years ago
"# Testing for primality\n",
"From slides 2020-4135-l07 - Number Theory for Public Key Cryptography"
]
},
{
"cell_type": "code",
"execution_count": 9,
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"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"137 True\n"
]
}
],
"source": [
"from math import sqrt\n",
"# Inefficient, but accurate method\n",
"def is_prime(n):\n",
" if n % 2 == 0 and n > 2: \n",
" return False\n",
" return all(n % i for i in range(3, int(sqrt(n)) + 1, 2))\n",
"\n",
"print(137, is_prime(137))"
]
},
{
"cell_type": "code",
"execution_count": 10,
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"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"1105 is probable prime\n"
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]
}
],
"source": [
"# Fermat primality test\n",
"\n",
"# Inputs: \n",
"# n: a value to test for primality;\n",
"# k: a parameter that determines the number of \n",
"# times to test for primality\n",
"from random import randint\n",
"def fermat_prime(n, k):\n",
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" for _ in range(k):\n",
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" a = randint(2,n-2)\n",
" # Fermat’s theorem says that if a number p is prime then \n",
" # a^p−1 mod p = 1 for all a with gcd(a,p) = 1.\n",
" if not (a**(n-1))%n == 1:\n",
" return \"composite\"\n",
" return \"probable prime\"\n",
"\n",
"n = 1105 # value to test for primality;\n",
"k = 1 # number of times to test for primality\n",
"\n",
"print(f\"{n} is {fermat_prime(n,k)}\")\n",
"\n",
"# First few Carmichael numbers are: 561, 1105, 1729, 2465 \n",
"# Will output probable prime for every a with gcd(a, n) = 1"
]
},
{
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"cell_type": "markdown",
"metadata": {},
"source": [
"## Miller-Rabin\n",
"From https://en.wikipedia.org/wiki/Miller%E2%80%93Rabin_primality_test \n",
"This test is not fooled by Charmichael numbers.\n",
"\n",
"To test `n`, you choose different `a` values, and test if \n",
"\n",
"> $a^{d*2^i} \\equiv 1 \\pmod{n}$ for $i = (0, 1, 2, ... r)$\n",
"\n",
"If this happens, look at the congruence on the calculation just before you got `1`. If that value is not -1 (aka n-1), the value is composite aka not prime."
]
},
{
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"cell_type": "code",
"execution_count": 11,
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"metadata": {},
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"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"168 = 21*2^3 = 168\n",
"65536 = 1*2^16 = 65536\n",
"\n",
"16127 is probably prime.\n",
"561 is composite.\n",
"2465 is composite.\n",
"65537 is probably prime.\n"
]
}
],
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"source": [
3 years ago
"# Miller-Rabin primality test\n",
"# This test is not fooled by Charmichael numbers.\n",
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"\n",
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"from math import log2, floor\n",
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"from random import randint\n",
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"\n",
"def decompose_even_number(n):\n",
" '''Find the values r and d such that n == d * 2**r.'''\n",
" r = 0\n",
" d = n\n",
" for i in range(1, floor(log2(n)) + 1):\n",
" if (d/2).is_integer():\n",
" d = d/2\n",
" r += 1\n",
" else:\n",
" break\n",
" return (r, int(d))\n",
"\n",
"verbose = False # Print during execution\n",
"\n",
"def miller_rabin_prime(n):\n",
" ''':param n: should be odd, and n>3. No primes can be even, for obvious reasons.'''\n",
" accuracy = 3 # How many different `a` to test\n",
" r, d = decompose_even_number(n-1) # n is odd, thus n-1 is even\n",
" verbose and print(f\"Testing {n} a maximum of {accuracy} times...\")\n",
" \n",
" for _ in range(accuracy):\n",
" try:\n",
" a = randint(2, n-2) # The \"witness\". TODO: don't reuse a witness.\n",
" verbose and print(f\"\\tTesting {n} with a={a}\")\n",
"\n",
" x = (a**d) % n\n",
" if x == 1 or x == n-1:\n",
" verbose and print(f\"\\t({a}**{d})%{n} was 1 or -1 already!\")\n",
" continue # This witness was immediately 1 or -1 (aka n-1), so it can't help us any more.\n",
" for squaring in range(r-1):\n",
" x = (x**2) % n\n",
" verbose and print(f\"\\tx={x}\")\n",
" if x == n-1:\n",
" raise Exception(\"Try next witness\") # This witness can't help any more.\n",
" return \"composite\" # The witness didnt uphold the criteria for a prime, so n is composite!\n",
" except Exception as tryNext:\n",
" continue\n",
" return \"probably prime\"\n",
" \n",
"s, d = decompose_even_number(168)\n",
"print(f\"168 = {d}*2^{s} = {d*(2**s)}\")\n",
"s, d = decompose_even_number(65536)\n",
"print(f\"65536 = {d}*2^{s} = {d*(2**s)}\")\n",
"print()\n",
"\n",
"for value in [16127, 561, 2465, 65537]:\n",
" print(f\"{value} is {miller_rabin_prime(value)}.\")\n"
3 years ago
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"# Discrete logarithm problem\n",
"Given a prime `p` and an integer `y` with `0 < y < p`, find `x` such that \n",
3 years ago
"\n",
"> $y = g^{x}\\mod{p}$\n",
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"\n",
"No polynomial algorithims exists, but \"Baby-step giant-step\" is a reasonable one. \n",
"\n",
"Sources:\n",
"- https://stackoverflow.com/a/1832648/5976426\n",
"- https://en.wikipedia.org/wiki/Baby-step_giant-step\n",
"- https://gist.github.com/0xTowel/b4e7233fc86d8bb49698e4f1318a5a73\n",
"\n",
"### Square and multiply\n",
"\n",
"Used to simplify doing large exponentiations. Instead of solving $3^5$ as $3*3*3*3*3$, \n",
"\n",
"1. Convert the exponent to Binary.\n",
"2. For the first $1$, simply list the number\n",
"3. For each ensuing $0$, do Square operation\n",
"4. For each ensuing $1$, do Square and Multiply operations\n",
"\n",
"Example:\n",
"```\n",
"5 = 101 in Binary\n",
"1 First One lists Number 3\n",
"0 Zero calls for Square (3)²\n",
"1 One calls for Square + Multiply ((3)²)²*3\n",
"```\n",
"\n",
"https://www.practicalnetworking.net/stand-alone/square-and-multiply/\n"
3 years ago
]
},
{
"cell_type": "code",
"execution_count": 12,
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"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"7894352216^(1) * 7894352216^(2) * 7894352216^(16) * 7894352216^(64) * 7894352216^(1024) * 7894352216^(8192) * 7894352216^(131072) * 7894352216^(2097152) * 7894352216^(33554432) * 7894352216^(67108864)\n",
"squarings: 26\n",
"mults: 9\n",
"\n",
"z = 7894352216^102900819 mod(604604729)\n",
"z = 355407489\n",
"\n",
"7894352216^102900819 mod(604604729) = 355407489\n"
]
}
],
"source": [
"# Square and multiply\n",
"from math import log, floor\n",
"\n",
3 years ago
"def square_and_multiply(y,e,n, verbose=False):\n",
" # prep\n",
" e_bin = bin(e)[:1:-1] # e0, e1, e2, e3, ... , ek\n",
" k = len(e_bin)\n",
" z = 1\n",
" yi = y\n",
" indices = []\n",
"\n",
" for i in range(k):\n",
" ei = int(e_bin[i])\n",
" if ei == 1:\n",
" z = z*yi % n\n",
" if ei < k:\n",
" yi = yi*yi % n\n",
" if ei:\n",
" indices.append(i)\n",
"\n",
" if verbose:\n",
3 years ago
" mults = str(e_bin).count(\"1\") - 1\n",
" squarings = floor( log(e, 2) )\n",
"\n",
" print(f\"{' * '.join([f'{y}^({2**i})' for i in indices])}\")\n",
" print(f\"squarings: {squarings}\")\n",
" print(f\"mults: {mults}\")\n",
" print()\n",
" print(f\"z = {y}^{e} mod({n})\")\n",
" print(f\"z = {z}\\n\")\n",
" return z\n",
"\n",
"# y^e mod(n)\n",
"y = 7894352216\n",
"e = 102900819\n",
"n = 604604729\n",
"\n",
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"z = square_and_multiply(y,e,n, verbose=True)\n",
"\n",
"print(f\"{y}^{e} mod({n}) = {z}\")"
]
},
{
"cell_type": "code",
"execution_count": 13,
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"x = 102900819\n",
"y = g^x mod p ==> 355407489 = 7894352216^102900819 mod 604604729\n",
3 years ago
"Test: True\n"
]
}
],
"source": [
"from math import ceil, sqrt\n",
"\n",
"def bsgs(g, y, p):\n",
" '''\n",
" Solve for x in y = g^x mod p given a prime p.\n",
" If p is not prime, you shouldn't use BSGS anyway.\n",
" '''\n",
" N = ceil(sqrt(p - 1)) # phi(p) is p-1 if p is prime\n",
"\n",
" # Store hashmap of g^{1...m} (mod p). Baby step.\n",
" tbl = {pow(g, i, p): i for i in range(N)}\n",
"\n",
" # Precompute via Fermat's Little Theorem\n",
" c = pow(g, N * (p - 2), p)\n",
"\n",
" # Search for an equivalence in the table. Giant step.\n",
" for j in range(N):\n",
" h = (y * pow(c, j, p)) % p\n",
" if h in tbl:\n",
" return j * N + tbl[h]\n",
"\n",
" # Solution not found\n",
" return None\n",
"\n",
"def disc_log_test(y, g, x, p):\n",
" #return y == (g**x) % p\n",
" return y == square_and_multiply(g,x,p,False)\n",
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"\n",
"\n",
"# y = g^x mod p\n",
"y = 355407489\n",
"g = 7894352216\n",
"p = 604604729 # Must be prime\n",
"\n",
"#y = 34\n",
"#g = 7324\n",
"#p = 4363 # Must be prime\n",
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"\n",
"x = bsgs(g, y, p)\n",
"\n",
"print(f\"x = {x}\")\n",
"\n",
"print(f\"y = g^x mod p ==> {y} = {g}^{x} mod {p}\")\n",
"\n",
"print(f\"Test: {disc_log_test(y,g,x,p)}\")"
3 years ago
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
3 years ago
"# RSA\n",
"\n",
"### Key generation\n",
"\n",
"1. Let $p$ and $q$ be distinct prime numbers, randomly chosen from the set of all prime numbers of a certain size.\n",
"2. Compute $n = pq$.\n",
"3. Select $e$ randomly with $gcd(e, \\varphi(n)) = 1$.\n",
"4. Compute $d = e−1 \\mod \\varphi(n)$.\n",
"5. The public key is the pair $n$ and $e$.\n",
3 years ago
"6. The private key consists of the values $p$, $q$ (or $n = pq$) and $d$.\n",
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"\n",
"### Encryption\n",
"\n",
"The public key for encryption is $KE = (n, e)$\n",
"1. Input is any value $M$ where $0 < M < n$. \n",
"2. Compute $C = E(M,KE) = Me mod n$.\n",
"\n",
"### Decryption\n",
"\n",
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"The private key for decryption is $KD = d$ (values $p$ and $q$ are not used here).\n",
"1. Compute $D(C,KD) = Cd \\mod n = M$."
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]
},
{
"cell_type": "code",
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"execution_count": 14,
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"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"p=43, q=59, n=2537, 𝜑(n)=2436, e=5, d=1949\n",
"\n",
"Public key: n=2537, e=5\n",
"Private key: p=43, q=59, d=1949\n",
"\n",
"{'n': 2537, 'e': 5, 'p': 43, 'q': 59, 'd': 1949}\n"
]
}
],
"source": [
"import random\n",
"\n",
"def rsa_keygen(p=None, q=None, e=None, prime_range=1000, verbose=False):\n",
" # Generate a set of primes. In practice, this should be\n",
" # a large number of large primes.\n",
" primes = [i for i in range(0,prime_range) if is_prime(i)]\n",
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"\n",
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" if not p: p = random.choice(primes)\n",
" if not q: q = random.choice(primes)\n",
" n = p*q\n",
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"\n",
3 years ago
" # Euler function 𝜑(n) \n",
" phi_n = totient(n)\n",
"\n",
" # Select e randomly with gcd(e, φ(n)) = 1.\n",
" if not e: e = random.randint(1, phi_n)\n",
" while not gcd(e, phi_n) == 1:\n",
" e = random.randint(1,phi_n)\n",
"\n",
" # Compute d = e−1 mod φ(n).\n",
" d = inverse(e, phi_n)\n",
" \n",
" if verbose:\n",
" print(f\"p={p}, q={q}, n={n}, 𝜑(n)={phi_n}, e={e}, d={d}\\n\")\n",
" print(f\"Public key: n={n}, e={e}\")\n",
" print(f\"Private key: p={p}, q={q}, d={d}\\n\")\n",
"\n",
" return {\"n\":n, \"e\":e, \"p\":p, \"q\":q, \"d\":d}\n",
"\n",
"# Example from slides\n",
"rsa_keys = rsa_keygen(p=43, q=59, e=5, verbose=True)\n",
"\n",
"# Random keys\n",
"#rsa_keys = rsa_keygen(verbose=True)\n",
"print(rsa_keys)"
]
},
{
"cell_type": "code",
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"execution_count": 15,
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"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Ciphertext: 2488\n"
]
}
],
"source": [
"def rsa_encrypt(M, rsa_keys):\n",
" # e and n are the public key\n",
" e = rsa_keys['e']\n",
" n = rsa_keys['n']\n",
"\n",
" # C = M^e (mod n)\n",
" return square_and_multiply(M,e,n)\n",
"\n",
"M = 50\n",
"C = rsa_encrypt(M, rsa_keys)\n",
"print(f\"Ciphertext: {C}\")"
]
},
{
"cell_type": "code",
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"execution_count": 16,
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"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Message: 50\n"
]
}
],
"source": [
"def rsa_decrypt(C, rsa_keys):\n",
3 years ago
" # Private key\n",
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" p = rsa_keys['p']\n",
" q = rsa_keys['q']\n",
" d = rsa_keys['d']\n",
" n = p*q\n",
"\n",
" # M = C^d (mod n)\n",
" return square_and_multiply(C,d,n)\n",
"\n",
"M = rsa_decrypt(C, rsa_keys)\n",
"print(f\"Message: {M}\")"
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]
}
],
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"display_name": "Python 3",
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"file_extension": ".py",
"mimetype": "text/x-python",
"name": "python",
"nbconvert_exporter": "python",
"pygments_lexer": "ipython3",
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},
"nbformat": 4,
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"nbformat_minor": 4
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}