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Perfroming AES CBC 128 Cannot get the same result same the ONline tools

I using getting the encrypted message from a third party sever.
I am using the tools in enter link description here
to test and its correct
The Encrypted Text =
"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"
Key = "1234567812345678"
And the result
After AES Decrypted Output (Base64):
And after Decode Plain Text
THe result is :
{"upPacketSN":-1,"upDataSN":-1,"topic":"v1/up/ad","timestamp":1619672797621,"tenantId":"2000034792","serviceId":"","protocol":"lwm2m","productId":"15044315","payload":{"APPdata":"MTIzNDU2NzgsNTM0OCwwNywwOCwyNTUsMTc5LDEyOCwwMDAwLDQyNTksMSw2LDAsMCwwLDAsMCwwLDI0LDMuNzAsMy43MCwxLDEsNiwsMCwwLDAsMCwwLDAsMCwwLDAsMCwtMjU1LC0yNTUsLTI1NSwxLjY5LDEuOCw0OTk4LDYwLDE0NDAsNjA="},"messageType":"dataReport","deviceType":"","deviceId":"523ede8bb7e34dd4a1bd74028d63749e","assocAssetId":"","IMSI":"undefined","IMEI":"864162041961023"}
One of the thing is that IV must be NULL to get the correct answer.
And so if i implement in Python AES CBC using pycryptodome library in the following like:
class AES_CBC:
def add_to_16(self, value):
while len(value) % 16 != 0:
value += '\0'
return str.encode(value) # 返回bytes
#解密方法
def decrypt_oralce(self, key, text):
# 初始化加密器
# 偏移量 16个0
iv = "0000000000000000"
aes = AES.new(self.add_to_16(key), AES.MODE_CBC, self.add_to_16(iv))
#优先逆向解密base64成bytes
base64_decrypted = base64.decodebytes(text.encode(encoding='utf-8'))
#
decrypted_text = str(aes.decrypt(base64_decrypted), encoding='utf-8') # 执行解密密并转码返回str
unpad = lambda s : s[0:-ord(s[-1])]
#PADDING = '\0'
#print decrypted_text.rstrip(PADDING) #zeropadding只见诶去掉结尾\0
# print(unpad(decrypted_text))
return unpad(decrypted_text)
if __name__ == '__main__':
aes = AES_CBC()
#加密
key = "1234567812345678"
enc_msg = "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"
#解密
dec_text = aes.decrypt_oralce(key, enc_msg)
print(key)
print(dec_text)
EDITED:
i got a different result using same IV 16'0'
and
the result from the web is
{"upPacketSN":-1,"upDataSN":-1,"topic":"v1/up/ad","timestamp":1619687373640,"tenantId":"2000034792","serviceId":"","protocol":"lwm2m","productId":"15044315","payload":{"APPdata":"MTIzNDU2NzgsNTM0OCwwNywwOCwyNTUsMTc5LDEyOCwwMDAwLDQ1MDQsMSw2LDAsMCwwLDAsMCwwLDI0LDMuNzAsMy43MCwxLDEsNiwsMCwwLDAsMCwwLDAsMCwwLDAsMCwtMjU1LC0yNTUsLTI1NSwxLjY5LDEuOCw0OTk4LDYwLDE0NDAsNjA="},"messageType":"dataReport","deviceType":"","deviceId":"523ede8bb7e34dd4a1bd74028d63749e","assocAssetId":"","IMSI":"undefined","IMEI":"864162041961023"}
while the result of my code is :
KE#`QS[UDc~
1,"upDataSN":-1,"topic":"v1/up/ad","timestamp":1619687373640,"tenantId":"2000034792","serviceId":"","protocol":"lwm2m","productId":"15044315","payload":{"APPdata":"MTIzNDU2NzgsNTM0OCwwNywwOCwyNTUsMTc5LDEyOCwwMDAwLDQ1MDQsMSw2LDAsMCwwLDAsMCwwLDI0LDMuNzAsMy43MCwxLDEsNiwsMCwwLDAsMCwwLDAsMCwwLDAsMCwtMjU1LC0yNTUsLTI1NSwxLjY5LDEuOCw0OTk4LDYwLDE0NDAsNjA="},"messageType":"dataReport","deviceType":"","deviceId":"523ede8bb7e34dd4a1bd74028d63749e","assocAssetId":"","IMSI":"undefined","IMEI":"864162041961023"}
can anyone help me which part i get wrong?
Thanks

Currently the code is not using a "null-IV", which means an array filled with bytes set to zero. Instead it is using an array filled with '0' characters, which have value 0x30 in hexadecimals or 48 in decimals; distinctly not zero.
To create a null-IV please have a look here on how to indicate byte values within byte arrays in Python. The same trick is used within the add_to_16 loop, where the \0 escape is used to indicate a zero byte.
Note that padding a key or IV is very bad practice. Those need to consist of randomized bytes. Beware that working crypto code is not the same thing as secure crypto code, which should probably be your goal.

How to obfuscate files without using any external libraries?

I am currently trying to obscure the contents of files or simply txt files without using any libraries. I know that this won't be very secure at all but what I basically want is a program that asks you what the password to "encrypt" it with is, then it asks what the files name is and then it finds that file and "encrypts" it. Then another program is used to "decrypt" it so it asks for the password and filename and then "decrypts" it. I don't care about actual security so if it can be easily opened it's fine I just need it so it doesn't just open if you click the file.
On top of that I don't want it to use ANY libraries so no pycrypto or anything like that.
I am on 64 bit windows.
I also am a complete beginner in tthe world of code and only know basic things such as how to get user input, print stuff, if loops and while loops.
Thanks in advance!

I don't know if this qualifies as an "external library" in your mind, but if you're on a linux machine you probably have the gpg command available to you. This is a reasonably* secure encryption protocol, which you could access from python - or directly from the command line, if you just want the files protected and you don't care about having it done through python.
Alternatively, you could bang together a trivial mechanism for obscuring a file's contents based on a known password. For example, you could "stretch" the password to the length of the file text (multiply the string by (1 + (text length / password length)) and then zip the two together. This gives you a bunch of tuples, which can by converted to their ordinal value (ord('f')=>102, for example) and xored together (ord('f')^ord('b')=>4) and converted back to chars (chr(4) => the unprintable '\x04'). The resulting chars are your cyphertext.
All of this is trivial to break, of course, but it's easy to implement, and decryption is trivial.
*intentional understatement :)

You can try using the password as a key to encrypt it. May be a logical operation on the file on a binary level such as or, and, or others will be able to encrypt it very simply -- but it won't be secure like you mentioned.

You can use XTEA (xTended Tiny Encryption Algorithm) by copying the python code (xTended Tiny Encryption Algorithm) code into your project, it is only 28 lines of python. It has been subjected to cryptanalysis and shown to be reasonably secure.
import struct
def crypt(key,data,iv='\00\00\00\00\00\00\00\00',n=32):
def keygen(key,iv,n):
while True:
iv = xtea_encrypt(key,iv,n)
for k in iv:
yield ord(k)
xor = [ chr(x^y) for (x,y) in zip(map(ord,data),keygen(key,iv,n)) ]
return "".join(xor)
def xtea_encrypt(key,block,n=32,endian="!"):
v0,v1 = struct.unpack(endian+"2L",block)
k = struct.unpack(endian+"4L",key)
sum,delta,mask = 0L,0x9e3779b9L,0xffffffffL
for round in range(n):
v0 = (v0 + (((v1<<4 ^ v1>>5) + v1) ^ (sum + k[sum & 3]))) & mask
sum = (sum + delta) & mask
v1 = (v1 + (((v0<<4 ^ v0>>5) + v0) ^ (sum + k[sum>>11 & 3]))) & mask
return struct.pack(endian+"2L",v0,v1)
def xtea_decrypt(key,block,n=32,endian="!"):
v0,v1 = struct.unpack(endian+"2L",block)
k = struct.unpack(endian+"4L",key)
delta,mask = 0x9e3779b9L,0xffffffffL
sum = (delta * n) & mask
for round in range(n):
v1 = (v1 - (((v0<<4 ^ v0>>5) + v0) ^ (sum + k[sum>>11 & 3]))) & mask
sum = (sum - delta) & mask
v0 = (v0 - (((v1<<4 ^ v1>>5) + v1) ^ (sum + k[sum & 3]))) & mask
return struct.pack(endian+"2L",v0,v1)
Attribution, code from: ActiveState Code » Recipes

Transpostion decryption is not working with some keys?

Why is this transposition decryption code not working with some keys?
def transencrypt(word,key):
'''Traspositon encryption function. This function is used to encrypt a line
using the transposition encryption method. To know how transpositon encryption
works you can visit here https://en.wikipedia.org/wiki/Transposition_cipher.'''
count1=0
count2=0
encrypted=''
encryptbox=['']*key
while count1<key:
count2=count1
while count2<len(word):
encryptbox[count1]+=word[count2]
count2+=key
encrypted+=encryptbox[count1]
count1+=1
return encrypted
def transdecrypt(word,key):
'''This Function is for the decrypting the encrypted strings encrypted by
transencrypt().This function only requires the encrypted string and the key
with which it has been decrypted.'''
import math
count1=0
count2=0
decrypted=''
col=int(math.ceil(len(word)/key))
decryptbox=['']*col
while count1<col:
count2=count1
while count2<len(word):
decryptbox[count1]+=word[count2]
count2+=col
decrypted+=decryptbox[count1]
count1+=1
return decrypted
print(transencrypt('hello world',5))
print(transdecrypt('h dewlolrol',5))
OP's original code source
I have tried encrypting "hello world" with key 5, but at the time of decrypting I am getting the wrong result. Using other keys works fine.

The problem is that the string length (11) doesn't divide evenly into the key (5), so the string "hello world" encodes into the groups h d-ew-lo-lr-ol i.e. "h dewlolrol". Which is fine, but the decrypt routine chops "h dewlolrol" into h d-ewl-olr-ol and generates the wrong result, "heoo wlldlr".
A couple of possible ways to fix this:
1) Replace the encryptbox string with an array and pad the encryption units into even width segments: h d-ew -lo -lr -ol i.e. "h dew lo lr ol "
This will allow your decrypt routine to work but you'll end up with spaces at the end of the decryption and the encrypted string will be a different size than the original.
OR
2) Dynamically adjust your decryption logic to figure out, based on the length of the remaining string to decode, and remaining number of expected segments, how much the segments must shrink. This means your decrypt routine can't be as similar to the encrypt routine as it is now. But it will allow you to handle the output of the current encrypt routine and the encrypted string can remain the same length as the original.
Below is a rough rework along the lines of approach #2 above -- you can see it allows the encrypt routine to remain simple but the decrypt routine has to be more complex to make up for it:
import math
def transencrypt(string, key):
'''
Transpositon encryption function. This function is used to encrypt a line
using the transposition encryption method. To learn how transpositon encryption
works you can visit here https://en.wikipedia.org/wiki/Transposition_cipher.
'''
encrypted = ''
length = len(string)
for start in range(key):
for offset in range(start, length, key):
encrypted += string[offset]
return encrypted
def transdecrypt(string, key):
'''
This function is for the decrypting the strings encrypted by
transencrypt(). This function only requires the encrypted
string and the key with which it was decrypted.
'''
decrypted = ''
length = len(string)
width = int(math.ceil(length / key))
for start in range(width):
offset = start
remaining_key = key
remaining_length = length
remaining_width = width
while offset < length:
decrypted += string[offset]
offset += remaining_width
remaining_key -= 1
if remaining_key > 0:
remaining_length -= remaining_width
remaining_width = int(math.ceil(remaining_length / remaining_key))
return decrypted[:length]
if __name__ == '__main__':
import sys
string = sys.argv[1]
key = int(sys.argv[2])
print(transencrypt(string, key))
print(transdecrypt(transencrypt(string, key), key))
**OUTPUT*
> python3 test.py "hello world" 5
h dewlolrol
hello world
>

RSA Encryption using Python

I'm trying to RSA encrypt a word 2 characters at a time padding with a space using Python but not sure how I go about it.
For example if the encryption exponent was 8 and a modulus of 37329 and the word was 'Pound' how would I go about it? I know I need to start with pow(ord('P') and need to take into consideration that the word is 5 characters and I need to do it 2 characters at a time padding with a space. I'm not sure but do I also need to use <<8 somewhere?
Thank you

Here's a basic example:
>>> msg = 2495247524
>>> code = pow(msg, 65537, 5551201688147) # encrypt
>>> code
4548920924688L
>>> plaintext = pow(code, 109182490673, 5551201688147) # decrypt
>>> plaintext
2495247524
See the ASPN cookbook recipe for more tools for working with mathematical part of RSA style public key encryption.
The details of how characters get packed and unpacked into blocks and how the numbers get encoded is a bit arcane. Here is a complete, working RSA module in pure Python.
For your particular packing pattern (2 characters at a time, padded with spaces), this should work:
>>> plaintext = 'Pound'
>>> plaintext += ' '      # this will get thrown away for even lengths
>>> for i in range(0, len(plaintext), 2):
group = plaintext[i: i+2]
plain_number = ord(group[0]) * 256 + ord(group[1])
encrypted = pow(plain_number, 8, 37329)
print group, '-->', plain_number, '-->', encrypted
Po --> 20591 --> 12139
un --> 30062 --> 2899
d --> 25632 --> 23784

If you want to efficiently code the RSA encryption using python, my github repository would definitely to understand and interpret the mathematical definitions of RSA in python
Cryptogrphic Algoritms Implementation Using Python
RSA Key Generation
def keyGen():
''' Generate Keypair '''
i_p=randint(0,20)
i_q=randint(0,20)
# Instead of Asking the user for the prime Number which in case is not feasible,
# generate two numbers which is much highly secure as it chooses higher primes
while i_p==i_q:
continue
primes=PrimeGen(100)
p=primes[i_p]
q=primes[i_q]
#computing n=p*q as a part of the RSA Algorithm
n=p*q
#Computing lamda(n), the Carmichael's totient Function.
# In this case, the totient function is the LCM(lamda(p),lamda(q))=lamda(p-1,q-1)
# On the Contrary We can also apply the Euler's totient's Function phi(n)
# which sometimes may result larger than expected
lamda_n=int(lcm(p-1,q-1))
e=randint(1,lamda_n)
#checking the Following : whether e and lamda(n) are co-prime
while math.gcd(e,lamda_n)!=1:
e=randint(1,lamda_n)
#Determine the modular Multiplicative Inverse
d=modinv(e,lamda_n)
#return the Key Pairs
# Public Key pair : (e,n), private key pair:(d,n)
return ((e,n),(d,n))

Base64 and non standard

I try to create a python client for bacula, but I have some problem with the authentication.
The algorithm is :
import hmac
import base64
import re
...
challenge = re.search("auth cram-md5 ()", data)
#exemple ''
passwd = 'b489c90f3ee5b3ca86365e1bae27186e'
hm = hmac.new(passwd, challenge).digest()
rep = base64.b64encode(hm).strp().rstrip('=')
#result with python : 9zKE3VzYQ1oIDTpBuMMowQ
#result with bacula client : 9z+E3V/YQ1oIDTpBu8MowB'
There's a way more simple than port the bacula's implemenation of base 64?
int
bin_to_base64(char *buf, int buflen, char *bin, int binlen, int compatible)
{
uint32_t reg, save, mask;
int rem, i;
int j = 0;
reg = 0;
rem = 0;
buflen--; /* allow for storing EOS */
for (i=0; i >= (rem - 6);
if (j

To verify your CRAM-MD5 implementation, it is best to use some simple test vectors and check combinations of (challenge, password, username) inputs against the expected output.
Here's one example (from http://blog.susam.in/2009/02/auth-cram-md5.html):
import hmac
username = 'foo#susam.in'
passwd = 'drowssap'
encoded_challenge = 'PDc0NTYuMTIzMzU5ODUzM0BzZGNsaW51eDIucmRzaW5kaWEuY29tPg=='
challenge = encoded_challenge.decode('base64')
digest = hmac.new(passwd, challenge).hexdigest()
response = username + ' ' + digest
encoded_response = response.encode('base64')
print encoded_response
# Zm9vQHN1c2FtLmluIDY2N2U5ZmE0NDcwZGZmM2RhOWQ2MjFmZTQwNjc2NzIy
That said, I've certainly found examples on the net where the response generated by the above code differs from the expected response stated on the relevant site, so I'm still not entirely clear as to what is happening in those cases.

I HAVE CRACKED THIS.
I ran into exactly the same problem you did, and have just spent about 4 hours identifying the problem, and reimplementing it.
The problem is the Bacula's base64 is BROKEN, AND WRONG!
There are two problems with it:
The first is that the incoming bytes are treated as signed, not unsigned. The effect of this is that, if a byte has the highest bit set (>127), then it is treated as a negative number; when it is combined with the "left over" bits from previous bytes are all set to (binary 1).
The second is that, after b64 has processed all the full 6-bit output blocks, there may be 0, 2 or 4 bits left over (depending on input block modulus 3). The standard Base64 way to handle this is to multiply the remaining bits, so they are the HIGHEST bits in the last 6-bit block, and process them - Bacula leaves them as the LOWEST bits.
Note that some versions of Bacula may accept both the "Bacula broken base64 encoding" and the standard ones, for incoming authentication; they seem to use the broken one for their authentication.
def bacula_broken_base64(binarystring):
b64_chars="ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/"
remaining_bit_count=0
remaining_bits=0
output=""
for inputbyte in binarystring:
inputbyte=ord(inputbyte)
if inputbyte>127:
# REPRODUCING A BUG! set all the "remaining bits" to 1.
remaining_bits=(1 << remaining_bit_count) - 1
remaining_bits=(remaining_bits<<8)+inputbyte
remaining_bit_count+=8
while remaining_bit_count>=6:
# clean up:
remaining_bit_count-=6
new64=(remaining_bits>>remaining_bit_count) & 63 # 6 highest bits
output+=b64_chars[new64]
remaining_bits&=(1 << remaining_bit_count) - 1
if remaining_bit_count>0:
output+=b64_chars[remaining_bits]
return output
I realize it's been 6 years since you asked, but perhaps someone else will find this useful.