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https://github.com/flyq/puzzle-chaos-theory
https://github.com/flyq/puzzle-chaos-theory
Last synced: 20 days ago
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- Host: GitHub
- URL: https://github.com/flyq/puzzle-chaos-theory
- Owner: flyq
- Created: 2024-01-30T23:33:50.000Z (11 months ago)
- Default Branch: main
- Last Pushed: 2024-02-07T00:11:46.000Z (11 months ago)
- Last Synced: 2024-12-08T20:43:30.837Z (26 days ago)
- Language: Rust
- Size: 4.88 KB
- Stars: 0
- Watchers: 2
- Forks: 0
- Open Issues: 0
-
Metadata Files:
- Readme: README.md
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README
# puzzle-chaos-theory
**DO NOT FORK THE REPOSITORY, AS IT WILL MAKE YOUR SOLUTION PUBLIC. INSTEAD, CLONE IT AND ADD A NEW REMOTE TO A PRIVATE REPOSITORY, OR SUBMIT A GIST**
Trying it out
=============
Use `cargo run --release` to see it in action
Submitting a solution
=====================
[Submit a solution](https://xng1lsio92y.typeform.com/to/UYMwUsgG)
[Submit a write-up](https://xng1lsio92y.typeform.com/to/NGwTHlVz)
Puzzle description
==================
|___ /| | / / | | | | | |
/ / | |/ / | |_| | __ _ ___| | __
/ / | \ | _ |/ _` |/ __| |/ /
./ /___| |\ \ | | | | (_| | (__| <
\_____/\_| \_/ \_| |_/\__,_|\___|_|\_\
Bob designed a new one time scheme, that's based on the tried and true method of encrypt + sign. He combined ElGamal encryption with BLS signatures in a clever way, such that you use pairings to verify the encrypted message was not tampered with. Alice, then, figured out a way to reveal the plaintexts...
# Writeup
https://hackmd.io/@liquan/H1srq-D5T
## ElGamal encryption in ECC
### Key generation
Alice:
- secret key: $x$
- public key: $H = x\cdot G$
### Encryption
Bob:
- $M = F(m)$, the $m$ is the message, and the $M$ is the element in $G$, $F$ is an invertible function
- ephemeral key: $y$
- shared secret: $S = H \cdot y = x \cdot y \cdot G$
- $C_1 = y \cdot G$
- $C_2 =S + M$
- public $C_1, C_2$
### Decryption
Alice:
- $S = C_1 \cdot x = x \cdot y \cdot G$
- $M = C_2 - S$
- $m = F^{-1}(M)$
## Solution
The Struct and method:
- Sender(Bob): $y$ and $C_1 = y \cdot G_1$
- Receiver(Alice): $H$
- Message: $M$
- ElGamal: $(C_1, C_2)$ in $G_1$, $Hash((C_1, C_2))$ in $G_2$ group
- Sender's `send()`: new $C_1$, $C_2 = H \cdot y + M = S + M$
- Sender's `authenticate()`: $y \cdot H((C_1, C_2))$
- Auditor's `check_auth()`: $e(G_1, y\cdot Hash((C_1, C_2))) = e(C_1, Hash((C_1, C_2)))$
Now, we have blob, which is $C_1, (C_1, C_2), y\cdot Hash((C_1, C_2)), H$, and some messages, how to know which message is the pre-image of (C_1, C_2)?
$e(x \cdot y \cdot G_1, Hash((C_1, C_2))) = e(x \cdot G_1, y \cdot Hash((C_1, C_2)))$
$S = x \cdot y \cdot G_1 = C_2 - M$, so we just need to try different messages to get different S, and check if the pairs equal.