GordonOus

import math
from random import randint, seed
from time import time, process_time
import string

strchr = lambda x: chr(x)
strbyt = lambda x, y=0: ord(x[y])
bitlst = lambda x, y: x << y
bitrst = lambda x, y: x >> y
bitext = lambda x, y, z=1: bitrst(x, y) & int(math.pow(2, z) - 1)

GordonOus

def pohligHellmanPGH(p,g,h):
  #p is the modulus
  #g is the argument (a)
  #h is the equivalence class i.e the result of pow(g,e,p). e is what we are looking for
	F=IntegerModRing(p)
	g=F(g)
	h=F(h)
	G=[]
	H=[]
	X=[]

GordonOus

def shift_rows(s):
    s[0][1], s[1][1], s[2][1], s[3][1] = s[1][1], s[2][1], s[3][1], s[0][1]
    s[0][2], s[1][2], s[2][2], s[3][2] = s[2][2], s[3][2], s[0][2], s[1][2]
    s[0][3], s[1][3], s[2][3], s[3][3] = s[3][3], s[0][3], s[1][3], s[2][3]


def inv_shift_rows(s):
    tmp = s[13]
    s[13] = s[9]
    s[9] = s[5]

GordonOus

ssh-keygen -e -f 'bruce_rsa_6e7ecd53b443a97013397b1a1ea30e14.pub' -m 'PKCS8' > bruce.pub
openssl rsa -pubin -in bruce.pub -noout -modulus | cut -d '=' -f2 | xargs echo "ibase=16; $1"| bc | tr -d '\' | tr -d '\n'

GordonOus

ssh-keygen -e -f 'bruce_rsa_6e7ecd53b443a97013397b1a1ea30e14.pub' -m 'PKCS8' > bruce.pub
openssl rsa -pubin -in foo.pub -noout -modulus | cut -d '=' -f2 | xargs echo "ibase=16; $1"| bc | tr -d '\' | tr -d '\n'

GordonOus

import numpy as np
from PIL import Image

img1 = Image.open('lemur.png')
img2 = Image.open('flag.png')

n1 = np.array(img1)*255
n2 = np.array(img2)*255

#our images have a mode of RGB which is assumed to be an 8-bit int

GordonOus

import gmpy2
from Crypto.Util.number import long_to_bytes, bytes_to_long
from binascii import unhexlify

flag = open('output.txt','r').read().strip().split(' ')[1]
ciphertext = bytes_to_long(flag)
e = 3

#Fist method of getting cube root of ciphertext
gmpy2.get_context().precision = 4096

GordonOus

import time
import random
from pwn import *

def gen_seeds(s):
  return list(range((s - 20),(s + 20)))

possible_seeds = gen_seeds(int(time.time()))
conn = remote('IP ADDRESS',PORT)

GordonOus

from binascii import unhexlify

enc_flag = 'Put the encrypted flag extracted from the pcap hear'
ct = 'the First Ciphertext we find'
ct2 = 'the second Ciphertext we find'
flag = ''
#Note the length of all the above texts are 36 bytes after decoding from ascii
for i,j in enumerate(ct):
  flag += chr(j ^ ct2[i] ^ enc_flag[i])

GordonOus

import string
from binascii import unhexlify

encrypted = unhexlify(open('msg.enc','r').read())
plaintext = []
#brute force the encryption logic
for b in encrypted:
  for ch in string.printable:
    if ((123 * ord(ch) + 18) % 256 == b):
      plaintext.append(ch)