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# ASCON Marchombre
**Catégorie:** Cryptanalyse - **Difficulté:** Facile
**Description:**
* Clef : *`00456c6c616e61206427416c2d466172`*
* Nonce : *`0`*
* Message chiffré : *`ac6679386ffcc3f82d6fec9556202a1be26b8af8eecab98783d08235bfca263793b61997244e785f5cf96e419a23f9b29137d820aab766ce986092180f1f5a690dc7767ef1df76e13315a5c8b04fb782`*
* Données associées : *`80400c0600000000`*
**Solution:**
Pour ce challenge, quelques données nous sont fournies ainsi que le nom de la méthode de chiffrement : ASCON.
En cherchant rapidement sur internet, nous trouvons un [package pypi](https://pypi.org/project/ascon/) à installer pour pouvoir l'utiliser dans un programme python.
Il n'y a pas grand chose à détailler ici à part donner le script utilisé.\
Voici donc le code que j'ai utilisé pour résoudre ce challenge en suivant des exemples trouvés sur internet :
```python
import ascon
key = bytes.fromhex("00456c6c616e61206427416c2d466172")
nonce = bytes.fromhex("00000000000000000000000000000000")
data = bytes.fromhex("80400c0600000000")
cipher = bytes.fromhex("ac6679386ffcc3f82d6fec9556202a1be26b8af8eecab98783d08235bfca263793b61997244e785f5cf96e419a23f9b29137d820aab766ce986092180f1f5a690dc7767ef1df76e13315a5c8b04fb782")
print(f"Key: {key}")
print(f"Nonce: {nonce}")
print(f"Ciphertext: {cipher}")
print(f"Data: {data}")
# Decrypt
plaintext = ascon.decrypt(key, nonce, data, cipher, variant="Ascon-128")
print(f"Plaintext: {plaintext}")
```
Lorsque nous exécutons ce script, voici le résultat :
Nous pouvons donc récupérer le flag en fin d'exécution en remplaçant le *`\xe8`* par son caractère unicode : *`è`*
🚩 FLAG
```
404CTF{V3r5_l4_lum1èr3.}
```
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