rsa padding done

src-py/plexcryptool/scripts/authur1.py

@@ -6,7 +6,7 @@ Since this (auth) hash did not have a name before, I gave it the name 'authur1'
 
 @author: Christoph J. Scherr <software@cscherr.de>
 @license: MIT
-@source: https://git.cscherr.de/PlexSheep/plexcryptool/src/branch/master/plexcryptool/authur1.py
+@source: https://git.cscherr.de/PlexSheep/plexcryptool/
 """
 import argparse
 import random

src-py/plexcryptool/scripts/basic-decrypt.py

@@ -5,7 +5,7 @@ you also need the chiffrat.txt file in the same directory.
 
 @license: MIT
 @author: Christoph J. Scherr <software@cscherr.de>
-@source: https://git.cscherr.de/PlexSheep/plexcryptool/src/branch/master/plexcryptool/basic-decrypt.py
+@source: https://git.cscherr.de/PlexSheep/plexcryptool/
 """
 MAX_IN_LINE = 16
 DE_MOST_COMMON = ['E', 'N', 'I', 'R', 'A']

src-py/plexcryptool/scripts/common.py

@@ -7,6 +7,7 @@ Common tools that may be used by many
 """
 
 from math import floor
+from typing import Literal
 
 def byte_xor(ba0: bytearray, ba1: bytearray) -> bytearray:
     """
@@ -27,3 +28,19 @@ def calc_byte_len(n: int) -> int:
         return 1 + floor(len)
     else: 
         return floor(len)
+
+def int_to_bytearray(n: int, size: int|None = None, endianness = 'big') -> bytearray:
+    """
+    convert an integer to a bytearray
+
+    you can specify a size if you want, if the number is too big an Exception will be raised.
+    size is in bytes
+    """
+    brepr: bytes = n.to_bytes(calc_byte_len(n), endianness)
+    ba: bytearray = bytearray(brepr)
+    if size is not None and len(ba) > size:
+        raise Exception("Number is larger than requested")
+    while size is not None and size > len(ba):
+        ba: bytearray = bytearray(1) + ba
+
+    return ba

src-py/plexcryptool/scripts/md5-analyzer.py

@@ -4,8 +4,7 @@ A small script to help analyze the md5 hash function.
 
 @author: Christoph J. Scherr <software@cscherr.de>
 @license: MIT
-@source: https://git.cscherr.de/PlexSheep/plexcryptool/src/branch/master/plexcryptool/md5-analyzer.py
+@source: https://git.cscherr.de/PlexSheep/plexcryptool/
-TODO do I need to implement md5 by myself for the assignment?
 """
 import argparse
 import hashlib

src-py/plexcryptool/scripts/oaep-ciphertext.txt

@@ -1 +0,0 @@
-0976f89569d94414fed4f7714cd85c840dbf338781926f5d1f6284688c97eaeef7f42b3d844c70468cfb960092235855da5d5c66107dfee4204c1a78512ee00306da957a856a085affe40b2ab32fc90a8c8e059feea9a83e753618aeae04884aa1658afe59a69c30ec5c014a5da06ff484143ca1ea884b44a3f92d063db8503f

src-py/plexcryptool/scripts/oaep_rsa.py

@@ -4,7 +4,7 @@ Perform RSA-OAEP
 
 @author: Christoph J. Scherr <software@cscherr.de>
 @license: MIT
-@source: https://git.cscherr.de/PlexSheep/plexcryptool/src/branch/master/plexcryptool/trash-hash.py
+@source: https://git.cscherr.de/PlexSheep/plexcryptool/
 """
 
 import argparse

src-py/plexcryptool/scripts/rsa_padding.py

@@ -0,0 +1,104 @@
+"""
+Rsa padding script for croptology lectures
+
+@author: Christoph J. Scherr <software@cscherr.de>
+@license: MIT
+@source: https://git.cscherr.de/PlexSheep/plexcryptool/
+"""
+
+import argparse
+
+import plexcryptool
+import common
+
+# predefined constants
+P = 2**206 - 5
+Q = 2**226 -5
+N = P*Q
+D = 0xaffe0815
+CIPHERTEXTS: list[int] = [
+    0x78766a52455329b486aaa414c3a029834a7e4b6ed87019dce4056f4d8999b137404d9ec4df28da201c9b0bc142deb1d86ff94d83becc,
+    0x670b865216dfd0aacd5f7fa8802e704fa82f3fb9c7dbe3eb5a9ec308a1a2288648b15d5cc8ba2f54b245a972aea977932c9c84cf6422,
+    0x61d5f2a4298bff3d6ebcd78830fb9181d97235623819eb7c60b92dcdf836a6cf731c60187e72f471c05d1c6eab216c3f6032af3c5370,
+    0x3651009d02a0c72b9bc206c57d12277594d9eaad28bb3de5d661670b42f1cfafe688b9674e34d4ad79db898205417086e7e1877b9ef1,
+    0x96e51d4675c6be5b14ec0cf2a9e9a9610a99d632723b3f1fcfc6b36806f5d74045f47622817cc35f6ffe9afe29f0aa236cbe12371651,
+    ]
+
+def padding_check(cihpertext: int, private_key: tuple[int, int], verbose: bool = False) -> bool:
+    """
+    Check the padding of a given cihpertext
+    """
+    decrypted: int = pow(cihpertext, private_key[0], private_key[1])
+    # the ciphertexts seem to all be 54B long, so I have just hardcoded that
+    cleartext: bytearray = common.int_to_bytearray(decrypted, 54)
+    if verbose:
+        print(f"original:\t{cihpertext:054x}")
+        print(f"decrypted:\t{cleartext.hex()}")
+
+    index: int = 0
+    padding_active: bool = False
+    for b in cleartext:
+
+        if index == 0 and b != 0:
+            # the first byte has to be 0x00
+            if verbose:
+                print(f"error:\t\tThe first byte is not 00: {b:02x}")
+            return False
+        elif index == 1:
+            # the second byte is the padding type, we only accept 01 and 02 type padding
+            if b == 1:
+                padding_active = True
+                pad_type = b
+                if verbose:
+                    print(f"type:\t\t01 deterministic padding")
+            elif b == 2:
+                padding_active = True
+                pad_type = b
+                if verbose:
+                    print(f"type:\t\t02 nondeterministic padding")
+            else:
+                if verbose:
+                    print(f"error:\t\tThe second byte is not 01 or 02 (block type): {b:02x}")
+                return False
+        elif padding_active:
+            if b == 0:
+                # padding has ended
+                padding_active = False
+                continue
+            if pad_type == 1 and b != 0xff:
+                if verbose:
+                    print(f"error:\t\tPadding of type 01 contains non ff byte: {b:02x}")
+                return False
+            elif pad_type == 2:
+                if b == 0:
+                    # must be random non 00
+                    if verbose:
+                        print(f"error:\t\tPadding of type 02 contains 00 byte: {b:02x}")
+                    return False
+
+
+
+        index += 1
+    
+    if padding_active:
+        if verbose:
+            print(f"error:\t\tPadding has not been ended with 00 byte")
+
+
+    return True
+
+def main():
+    parser = argparse.ArgumentParser(prog="rsa-padding", description='validates a rsa padding')
+    parser.add_argument('-v', '--verbose', action="store_true",
+                    help='append some auth hashed stuff')
+    args = parser.parse_args()
+
+    results: list[bool] = []
+    for c in CIPHERTEXTS:
+        b = padding_check(c, (D, N), args.verbose)
+        print(f"Passed:\t\t{b}")
+        plexcryptool.cplex.printing.seperator()
+
+
+if __name__ == "__main__":
+    main()

src-py/plexcryptool/scripts/trash-hash.py

@@ -4,7 +4,7 @@ Dirty hash we covered in an excercise for Week 16 of 2023 in cryptography
 
 @author: Christoph J. Scherr <software@cscherr.de>
 @license: MIT
-@source: https://git.cscherr.de/PlexSheep/plexcryptool/src/branch/master/plexcryptool/trash-hash.py
+@source: https://git.cscherr.de/PlexSheep/plexcryptool/
 """
 import math
 import argparse