aaron/htb-santa-ctf
1
0
Fork 0
Code Issues Pull Requests Projects Releases Wiki Activity

fix wording and typo

Browse Source
This commit is contained in:
aaron
2021-12-04 05:27:04 +01:00
parent 97dd63e956
commit dc76ab7403
1 changed files with 4 additions and 3 deletions
Download Patch File Download Diff File Expand all files Collapse all files

7
crypto/XMASSpirit/README.md
View File

@@ -15,13 +15,14 @@ HTB{4ff1n3_c1ph3r_15_51mpl3_m47h5}
## How to solve ## How to solve
- It's a known plaintext attack. - The vulnerability is a known plaintext attack.
- The encryption works by applying `(a*byte+b)%n` for each byte of the pdf. - The encryption works by applying `(a*byte+b)%n` for each byte of the pdf.
- Where `a` and `b` remain static during the entire encryption process. - Where `a` and `b` remain static during the entire encryption process.
- The PDFv5 Header is known (`00000000: 2550 4446 2d31 2e35 0a25 e2e3 cfd3 0a37 %PDF-1.5.%.....7`) - The PDFv5 Header is known (`00000000: 2550 4446 2d31 2e35 0a25 e2e3 cfd3 0a37 %PDF-1.5.%.....7`)
- From the encryption algorithm the followin constraints can be derived:
- Constraint 1: Factor `a` has to be a number between 1..256 while not sharing a common divisor with 256 - Constraint 1: Factor `a` has to be a number between 1..256 while not sharing a common divisor with 256
- Constraitn 2: Factor `b` has to be a number between 1..256 - Constraint 2: Factor `b` has to be a number between 1..256
- So it's possible to attack the encryption by iterating and trying to create a pair (a, b) that matches the entire encrypted pdf header. - So it's possible to attack the ciphertext by iterating and trying to create a pair (a, b) that encrypts the pdf5 header bites in such a way that it matches the ciphertext.
```python ```python
def get_factors(ct:bytes, n:int=256) -> (int, int): def get_factors(ct:bytes, n:int=256) -> (int, int):