hacker.org

Code: Select all

possibles = __import__('string',('printable',)).printable
KEYLEN = 6
keys = [possibles[:] for i in range(KEYLEN)]

key_mapping = '''
	a dictionary for all the possible keys which returns a
	dictionary of possible outputs, i.e.
		if ctext[i] not in key_mapping['a'].values():
			keys[i % KEYLEN] != 'a'
	this can be used in the end to lookup what cyphertext value
	for the given key to return the plaintext value.

	this dictionary is made by taking every running the cypher
	with a key of all the same characters and a plaintext
	consisting of the printable characters.
'''

ctext = [int(v) for v in "2632 2718 ... 2683 647".split(' ')]

# now reduce the keys
for k in range(KEYLEN):
	key = keys[k] # list of possible keys for every KEYLENth value
	for value in ctext[k::KEYLEN]: # every key + KEYLENth value of the cyphertext
		i = 0
		while i < len(key):
			# remove the value from the list of possible keys if it can't produce a character in the output
			if value not in key_mapping[key[i]].values():
				del key[i]
			else:
				i += 1
	print(key) # will hopefully be a set of length 1 which is the correct key

Code: Select all

local memory = {126,664,235,156,145,645,95,96,97,98,99,100,101,102,103,104,0}

function C(x, y, z)
	local v1 = 1
	local v2 = 1
	local v3 = 0

	while v1 < z do
		v1 = v1*x
		v2 = v2*y
	end

	repeat
		z, v1, v2, v3 = z%v1, math.floor(v1/x), math.floor(v2/y), math.floor(z/v1)*v2 + v3
	until v2 == 0

	return v3
end

local i = 6
while true do
	if memory[i + 1] == 0 then
		break
	end
	io.write(C(10, 7, C(3, 10, memory[i + 1]) + C(5, 10, memory[i%6 + 1])), " ")
	i = i + 1
end
print()

Code: Select all

...
		# main_1
60^<0^2?		# push 6, duplicate, get mem[6], duplicate, jump 2 (to !) if mem[6] was 0 (end of string)
1g!				# end of program or jump over it
355+2v6c		# push 3 & 10; swap current char to top; call stack push; jump back to method_1_INIT
				# that means, calculate the base-3 bit string for the current char
		# main_2
555+			# push 5 & 10 (bases for bit string calculation)
3^0^6/6*-<		# get char index; duplicate; calculate MOD 6 of char index; then get respective key
6c				# jump to method_1_INIT
		# main_3
+				# add base-3 and base-5 bit string together
55+7			# push 10 & 7 (bases for bit string calculation)
2v6c			# switch bit string sum to top; jump to method_1_INIT
...