│ Category: Cryptography
│ Author: Imattas aka Zemi
│ Flag: byuctf{revisreallythestartingpointformostcategoriesiydk}
│ Source Path: crypto/anaken21sec1
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--[ Challenge Description ]--
-- text --
Your key is: `orygwktcjpb`
Encrypted flag: `cnpiaytjyzggnnnktjzcvuzjexxkvnrlfzectovhfswyphjt`
*Note - wrap the flag in `byuctf{}`*
[encrypt.py]
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--[ Provided Materials ]--
- Repository folder
https://github.com/BYU-CSA/BYUCTF-2025/tree/main/crypto/anaken21sec1
- crypto/anaken21sec1/README.md
https://github.com/BYU-CSA/BYUCTF-2025/blob/main/crypto/anaken21sec1/README.md
- crypto/anaken21sec1/solve.py
https://github.com/BYU-CSA/BYUCTF-2025/blob/main/crypto/anaken21sec1/solve.py
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--[ Recon / Initial Analysis ]--
This page imports the BYUCTF 2025 source material for the Cryptography challenge
Anaken21sec1. The prompt is kept separate from the solve notes, and upstream
artifacts are linked so the challenge can be replayed from the original
repository.
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--[ Vulnerability / Observation ]--
The useful observation comes from the imported solve notes below. I kept the
original technical path intact while normalizing the page metadata, challenge
grouping, and author attribution for the Volume 2 writeup archive.
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--[ Exploitation / Solution ]--
Source: crypto/anaken21sec1/README.md
https://github.com/BYU-CSA/BYUCTF-2025/blob/main/crypto/anaken21sec1/README.md
You simply need to implement a decryption script that will give you the flag. Do
the opposite of encrypt.py.
See solve.py.
Flag - byuctf{revisreallythestartingpointformostcategoriesiydk}
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--[ Full Exploit Script ]--
Source: crypto/anaken21sec1/solve.py
https://github.com/BYU-CSA/BYUCTF-2025/blob/main/crypto/anaken21sec1/solve.py
-- python --
import numpy as np
from encrypt import permutes
def unPermute(blockM, count):
finalBlockM = np.zeros((6,6))
for i in range(6):
for j in range(6):
index = int(permutes[count][i,j]-1)
finalBlockM[index//6,index%6] = blockM[i,j]
return finalBlockM
def unAdd(blockM, count):
if count == 0:
for i in range(6):
for j in range(6):
if (i+j)%2 == 0:
blockM[i,j] +=2
elif count == 1:
blockM[3:,3:] = blockM[3:,3:]+2*blockM[:3,:3]
elif count == 2:
blockM[:3,:3] = 2*blockM[3:,3:]+blockM[:3,:3]
elif count == 3:
blockM[3:,:3] = blockM[3:,:3]+2*blockM[:3,3:]
else:
blockM[:3,3:] = 2*blockM[3:,:3]+blockM[:3,3:]
return np.mod(blockM, 3)
def decrypt(ciphertext, key):
#rearrange the ciphertext according to the key
keyNums = [ord(key[i])-97 for i in range(len(key))]
reducedKeyNums = []
[reducedKeyNums.append(x) for x in keyNums if x not in reducedKeyNums]
letterBoxes = [[] for i in reducedKeyNums]
amountToTakeOff = len(ciphertext)//len(reducedKeyNums)
remainder = len(ciphertext)%len(reducedKeyNums)
for i in range(len(reducedKeyNums)):
nextLowest = reducedKeyNums.index(min(reducedKeyNums))
letterBoxes[nextLowest] = [ciphertext[i] for i in range(amountToTakeOff)]
ciphertext = ciphertext[amountToTakeOff:]
if nextLowest<remainder:
letterBoxes[nextLowest].append(ciphertext[0])
ciphertext = ciphertext[1:]
reducedKeyNums[nextLowest]+=26
intermediateText = ""
i=0
while sum([len(box) for box in letterBoxes]) != 0:
intermediateText += letterBoxes[i].pop(0)
i = (i+1)%len(letterBoxes)
#do the block permutations and additions
blocks = [intermediateText[12*i:12*(i+1)] for i in range(0,len(intermediateText)//12)]
plaintext = ""
for block in blocks:
#make 6 by 6 matrix
blockM =np.zeros((6,6))
for (i,letter) in enumerate(block[0:6]):
if letter == "0":
letter = "`"
letterNum = ord(letter)-96
blockM[i,0] = letterNum//9
blockM[i,1] = (letterNum%9)//3
blockM[i,2] = letterNum%3
for (i,letter) in enumerate(block[6:]):
if letter == "0":
letter = "`"
letterNum = ord(letter)-96
blockM[i,3] = letterNum//9
blockM[i,4] = (letterNum%9)//3
blockM[i,5] = letterNum%3
#unscramble matrix
for keyNum in reversed(keyNums):
blockM = unAdd(blockM, keyNum%5)
blockM = unPermute(blockM,(keyNum//5)%5)
#get resulting letters from matrix
for i in range(6):
resultLetterNum = int(9*blockM[0,i]+3*blockM[1,i]+blockM[2,i])
if resultLetterNum == 0:
plaintext += "0"
else:
plaintext += chr(resultLetterNum+96)
for i in range(6):
resultLetterNum = int(9*blockM[3,i]+3*blockM[4,i]+blockM[5,i])
if resultLetterNum == 0:
plaintext += "0"
else:
plaintext += chr(resultLetterNum+96)
return plaintext
if __name__ == "__main__":
ciphertext = input("What would you like to decrypt?\n")
key = input("Enter the encryption key. \n")
if len(ciphertext)%12 ==0:
print(decrypt(ciphertext, key))
else:
print("Sorry, invalid ciphertext.")
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--[ Key Takeaways ]--
- The BYUCTF 2025 challenge material is preserved with local archive formatting.
- The page author is normalized to Imattas aka Zemi.
- The source repository remains linked for handouts, services, and solve
artifacts.