Develop Reference

SHA256 doesn't yield same result

I'm following on this tutorial and in part 2 (picture below) it shows that the SHA256 yields a result different than what I get when I ran my python code:
the string is: 0450863AD64A87AE8A2FE83C1AF1A8403CB53F53E486D8511DAD8A04887E5B23522CD470243453A299FA9E77237716103ABC11A1DF38855ED6F2EE187E9C582BA6
While the tutorial SHA256 comes to: 600FFE422B4E00731A59557A5CCA46CC183944191006324A447BDB2D98D4B408
My short python shows:
sha_result = sha256(bitconin_addresss).hexdigest().upper()
print sha_result
32511E82D56DCEA68EB774094E25BAB0F8BDD9BC1ECA1CEEDA38C7A43ACEDDCE
in fact, any online sha256 shows the same python result; so am I missing here something?

You're hashing the string when you're supposed to be hashing the bytes represented by that string.
>>> hashlib.sha256('0450863AD64A87AE8A2FE83C1AF1A8403CB53F53E486D8511DAD8A04887E5B23522CD470243453A299FA9E77237716103ABC11A1DF38855ED6F2EE187E9C582BA6'.decode('hex')).hexdigest().upper()
'600FFE422B4E00731A59557A5CCA46CC183944191006324A447BDB2D98D4B408'

You could use Gavin's "base58.py", which I believe he no longer shares it on his github page. However you probably could easily google and find different versions of it from github.
Here is one version edited a little by me:
#!/usr/bin/env python
"""encode/decode base58 in the same way that Bitcoin does"""
import math
import sys
__b58chars = '123456789ABCDEFGHJKLMNPQRSTUVWXYZabcdefghijkmnopqrstuvwxyz'
__b58base = len(__b58chars)
def b58encode(v):
""" encode v, which is a string of bytes, to base58.
"""
long_value = 0L
for (i, c) in enumerate(v[::-1]):
long_value += ord(c) << (8*i) # 2x speedup vs. exponentiation
result = ''
while long_value >= __b58base:
div, mod = divmod(long_value, __b58base)
result = __b58chars[mod] + result
long_value = div
result = __b58chars[long_value] + result
# Bitcoin does a little leading-zero-compression:
# leading 0-bytes in the input become leading-1s
nPad = 0
for c in v:
if c == '\0': nPad += 1
else: break
return (__b58chars[0]*nPad) + result
def b58decode(v):
""" decode v into a string of len bytes
"""
long_value = 0L
for (i, c) in enumerate(v[::-1]):
long_value += __b58chars.find(c) * (__b58base**i)
result = ''
while long_value >= 256:
div, mod = divmod(long_value, 256)
result = chr(mod) + result
long_value = div
result = chr(long_value) + result
nPad = 0
for c in v:
if c == __b58chars[0]: nPad += 1
else: break
result = chr(0)*nPad + result
return result
try:
import hashlib
hashlib.new('ripemd160')
have_crypto = True
except ImportError:
have_crypto = False
def hash_160(public_key):
if not have_crypto:
return ''
h1 = hashlib.sha256(public_key).digest()
r160 = hashlib.new('ripemd160')
r160.update(h1)
h2 = r160.digest()
return h2
def hash_160_to_bc_address(h160, version="\x00"):
if not have_crypto:
return ''
vh160 = version+h160
h3=hashlib.sha256(hashlib.sha256(vh160).digest()).digest()
addr=vh160+h3[0:4]
return b58encode(addr)
def public_key_to_bc_address(public_key, version="\x00"):
if not have_crypto or public_key is None:
return ''
h160 = hash_160(public_key)
return hash_160_to_bc_address(h160, version=version)
def sec_to_bc_key(sec, version="\x80"):
if not have_crypto or sec is None:
return ''
vsec = version+sec +"\x01"
hvsec=hashlib.sha256(hashlib.sha256(vsec).digest()).digest()
return b58encode(vsec+hvsec[0:4])
def bc_key_to_sec(prv):
return b58decode(prv)[1:33]
def bc_address_to_hash_160(addr):
bytes = b58decode(addr)
return bytes[1:21]
if __name__ == '__main__':
if len(sys.argv) > 1:
if sys.argv[1] == '-en':
print b58encode(sys.argv[2].decode('hex_codec'))
if sys.argv[1] == '-de':
print b58decode(sys.argv[2]).encode('hex_codec')
if sys.argv[1] == '-pub':
print public_key_to_bc_address(sys.argv[2].decode('hex_codec'))
if sys.argv[1] == '-adr':
print bc_address_to_hash_160(sys.argv[2]).encode('hex_codec')
if sys.argv[1] == '-sec':
print sec_to_bc_key(sys.argv[2].decode('hex_codec'))
if sys.argv[1] == '-prv':
print bc_key_to_sec(sys.argv[2]).encode('hex_codec')
else:
print ''
print 'Usage: ./base58.py [options]'
print ''
print ' -en converts hex to base58'
print ' -de converts base58 to hex'
print
print ' -pub public_key_to_bc_address'
print ' -adr bc_address_to_hash_160'
print
print ' -sec sec_to_bc_key'
print ' -prv bc_key_to_sec'
print
To answer your specific question, based on above code you could use this command:
hashlib.sha256('0450863AD64A87AE8A2FE83C1AF1A8403CB53F53E486D8511DAD8A04887E5B23522CD470243453A299FA9E77237716103ABC11A1DF38855ED6F2EE187E9C582BA6'.decode('hex_codec')).digest().encode('hex_codec').upper()

Python invalid literal for int() with base 2

I'm trying a simple steganography python program. Encoding works, but I got the error ValueError: invalid literal for int() with base 2: '0b#99c1f#......
The value should be binary and I can't find the error. I'm using python 2.7 an this is the code:
from PIL import Image
import binascii
import optparse
def rgb2hex(r,g,b):
return '#{:02x}{:02x}{:02x}'.format(r,g,b)
def hex2rgb(hexcode):
return tuple(map(ord,hexcode[1:].decode('hex')))
def str2bin(message):
binary = bin(int(binascii.hexlify(message), 16))
return binary[2:]
def bin2str(binary):
message = binascii.unhexlify('%x' %(int('0b' + binary, 2)))
return message
def encode(hexcode, digit):
if hexcode[-1] in ('0', '1', '2', '3', '4', '5'):
hexcode = hexcode[:-1]+digit
return hexcode
else:
return None
def decode(hexcode):
if(hexcode[-1] in ('0','1')):
return hexcode[:-1]
else:
return None
def hide(filename, message):
img = Image.open(filename)
binary = str2bin(message) + '1111111111111110'
if img.mode in ('RGBA'):
img = img.convert('RGBA')
datas = img.getdata()
newData = []
digit = 0
temp = ''
for item in datas:
if (digit < len(binary)):
newpix = encode(rgb2hex(item[0], item[1], item[2]), binary[digit])
if newpix == None:
newData.append(item)
else:
r,g,b = hex2rgb(newpix)
newData.append((r,g,b,255))
digit += 1
else:
newData.append(item)
img.putdata(newData)
img.save(filename, "bmp")
return "Completed!"
return "Incorrect Image mode"
def retr(filename):
img = Image.open(filename)
binary = ''
if img.mode in ('RGBA'):
img = img.convert('RGBA')
datas = img.getdata()
for item in datas:
digit = decode(rgb2hex(item[0], item[1], item[2]))
if digit == None:
pass
else:
binary = binary + digit
if (binary[-16:] == '1111111111111110'):
print "Success"
return bin2str(binary[:-16])
return bin2str(binary)
return "incorrect image mode"
def Main():
parser = optparse.OptionParser('usage %prog -e/-d <target file>')
parser.add_option('-e', dest='hide', type='string', help='target picture')
parser.add_option('-d', dest='retr', type='string', help='target picture')
(options, args) = parser.parse_args()
if options.hide != None:
text = raw_input("Enter a message: ")
print hide(options.hide, text)
elif options.retr != None:
print retr(options.retr)
else:
print parser.usage
exit(0)
Main()

This line causes the error:
binary = binary + digit
In there, digit is a string with hex value (e.g. #003f7), which you are concatenating with what is supposed to be a binary string (but in fact is something like: #003f7#003f7...). You then pass that value to bin2str().

Python 3: ValueError: invalid literal for int() with base 10: '0001.0110010110010102e+22'

I'm a beginner, so sorry if this is obvious.
I'm at a loss here. I've been trying to make an encryption/decryption program, but I keep getting this error. I'm aware that there are other questions on this issue, but I still can't resolve it.
Encryptor:
import binascii
def text_to_bits(text, encoding='utf-8', errors='surrogatepass'):
bits = bin(int(binascii.hexlify(text.encode(encoding, errors)), 16))[2:]
return bits.zfill(8 * ((len(bits) + 7) // 8))
def text_from_bits(bits, encoding='utf-8', errors='surrogatepass'):
n = int(bits, 2)
return int2bytes(n).decode(encoding, errors)
def int2bytes(i):
hex_string = '%x' % i
n = len(hex_string)
return binascii.unhexlify(hex_string.zfill(n + (n & 1)))
#ENCRYPTION ALGORITHM
algorithm = 61913299
#ASCII ----> NUMBERS
raw = input("Enter text to encrypt:")
binary = text_to_bits(raw)
binary = int(binary)
algorithm = int(algorithm)
encrypted = binary * algorithm
encrypted = str(encrypted)
print(encrypted)
print("Done")
Decryptor:
import sys
import time
def to_bin(string):
res = ''
for char in string:
tmp = bin(ord(char))[2:]
tmp = '%08d' %int(tmp)
res += tmp
return res
def to_str(string):
res = ''
for idx in range(len(string)/8):
tmp = chr(int(string[idx*8:(idx+1)*8], 2))
res += tmp
return res
incorrectpasswords = 0
password=("password")
originpassword = password
x = 1
algorithm = 61913299
while x==1:
passwordattempt =input("Enter Password:")
if passwordattempt == password:
print("Correct")
x = 2
if passwordattempt!= password:
print("Incorrect")
incorrectpasswords = incorrectpasswords + 1
if incorrectpasswords > 2:
if x == 1:
print("Too many wrong attempts, please try again in one minute.")
time.sleep(60)
encrypted = input("Enter numbers to unencrypt:")
encrypted = int(encrypted)
one = encrypted / algorithm
size = sys.getsizeof(one)
one = str(one).zfill(size + 1)
one = int(one)
unencrypted = to_str(one)
x = unencrypted
For the conversion between binary and text, and text and binary, I used some code I found online.

I believe your code is not working because:
one = encrypted / algorithm
generates a float
to turn your string back into a number you should apply
eval(one)
or
float(one)
instead of
int(one)
(You can also turn it into an int after applying float or eval)
alternatively you might be able to get it by using integer division // as opposed to / , which will make one the type int by flooring the decimal result of the divison, but I'm not sure if that is the behavior you are looking for
Example in python 3 shell:
>>> import sys
>>> one = 15/25
>>> size = sys.getsizeof(one)
>>> one = str(one).zfill(size+1)
>>> one
'00000000000000000000000.6'
>>> type(one)
<class 'str'>
>>> one = eval(one)
>>> one
0.6
>>> type(one)
<class 'float'>

Converting exponential to float

This is my code, trying to convert the second field of the line from exponential into float.
outputrrd = processrrd.communicate()
(output, error) = outputrrd
output_lines = output.split('\n')
for line in output_lines:
m = re.search(r"(.*): ", line)
if m != None:
felder = line.split(': ')
epoch = felder[0].strip(':')
utc = epoch2normal(epoch).strip("\n")
#print felder[1]
data = float(felder[1])
float_data = data * 10000000
print float_data
resultslist.append( utc + ' ' + hostname + ' ' + float_data)
But, the program stops with this error:
File "/opt/omd/scripts/python/livestatus/rrdfetch-convert.py", line 156, in <module>
data = float(felder[1])
ValueError: invalid literal for float(): 6,0865000000e-01
Does anyone know the reason?

The easy way is replace! One simple example:
value=str('6,0865000000e-01')
value2=value.replace(',', '.')
float(value2)
0.60865000000000002

The reason is the use of comma in 6,0865000000e-01. This won't work because float() is not locale-aware. See PEP 331 for details.
Try locale.atof(), or replace the comma with a dot.

The float is correct, just use format to display it as you want, i.e.:
print(format(the_float, '.8f'))

I think it is useful to you:
def remove_exponent(value):
"""
>>>(Decimal('5E+3'))
Decimal('5000.00000000')
"""
decimal_places = 8
max_digits = 16
if isinstance(value, decimal.Decimal):
context = decimal.getcontext().copy()
context.prec = max_digits
return "{0:f}".format(value.quantize(decimal.Decimal(".1") ** decimal_places, context=context))
else:
return "%.*f" % (decimal_places, value)

Simply by casting string into float:
new_val = float('9.81E7')

This work for me, try it out.
def remove_exponent(value):
decial = value.split('e')
ret_val = format(((float(decial[0]))*(10**int(decial[1]))), '.8f')
return ret_val

I had a similar issue trying to convert from string in scientific/exponential notation to a float number (that can result also in exponential notation if too long)
num = '-8e-05'
def scientific_to_float(exponential):
split_word = 'e'
e_index = exponential.index('e')
base = float(exponential[:e_index])
exponent = float(exponential[e_index + 1:])
float_number = base * (10 ** exponent)
return float_number
scientific_to_float(num) # return -8e-05 float number

Custom Python Encryption algorithm

Hey, I have been working on this for a while, and I can remebr my brother stepped me through this very same alogorithm.
Basicly, it just adds the ascii values of both the characters from the key, and the phrase.
I can encrypt it with this:
def encrypt(key, string):
encoded = ''
for i in range(len(string)):
key_c = ord(key[i % len(key)])
string_c = ord(string[i % len(string)])
encoded += chr((key_c + string_c) % 127)
return encoded
But I can't seem to remember what we did as far as decrypting. Its difficult to revers a mod :P
Any ideas?

That's simple, let's see how it works. First of all, the encrypted message is obtained by subtracting the key.
enc = msg + key (mod 127)
How can we obtain the original message? That's easy, subtract the key in both sides
enc - key = msg + key - key (mod 127)
And here we get:
enc - key = msg (mod 127)
For more details, please refer to Modular arithmetic, I think it should belong one of group/field/ring. I'm not an expert in math, for further reading, you should check out Number theory. Here is the refined code:
def encrypt(key, msg):
encryped = []
for i, c in enumerate(msg):
key_c = ord(key[i % len(key)])
msg_c = ord(c)
encryped.append(chr((msg_c + key_c) % 127))
return ''.join(encryped)
def decrypt(key, encryped):
msg = []
for i, c in enumerate(encryped):
key_c = ord(key[i % len(key)])
enc_c = ord(c)
msg.append(chr((enc_c - key_c) % 127))
return ''.join(msg)
if __name__ == '__main__':
key = 'This_is_my_awsome_secret_key'
msg = 'Hello world'
encrypted = encrypt(key, msg)
decrypted = decrypt(key, encrypted)
print 'Message:', repr(msg)
print 'Key:', repr(key)
print 'Encrypted:', repr(encrypted)
print 'Decrypted:', repr(decrypted)
Output
Message: 'Hello world'
Key: 'This_is_my_awsome_secret_key'
Encrypted: '\x1dNV`O\nkO`fD'
Decrypted: 'Hello world'

Decryption is the same, except with minus instead of plus.

But you do not need to inverse the mod, just the + key_c, right? So just add 128, subtract key_c, and do modulo 127 again to keep in range. (instead of the last line, all the other lines are the same as with encrypting.