Krypton 4
Objectives
You more than likely used some form of FA and some common sense to solve that one.
So far we have worked with simple substitution ciphers. They have also been ‘monoalphabetic’, meaning using a fixed key, and giving a one to one mapping of plaintext (P) to ciphertext (C). Another type of substitution cipher is referred to as ‘polyalphabetic’, where one character of P may map to many, or all, possible ciphertext characters.
An example of a polyalphabetic cipher is called a Vigenère Cipher. It works like this:
If we use the key(K) ‘GOLD’, and P = PROCEED MEETING AS AGREED, then “add” P to K, we get C. When adding, if we exceed 25, then we roll to 0 (modulo 26).
P P R O C E E D M E E T I N G A S A G R E E D\ K G O L D G O L D G O L D G O L D G O L D G O\ becomes:
P 15 17 14 2 4 4 3 12 4 4 19 8 13 6 0 18 0 6 17 4 4 3\ K 6 14 11 3 6 14 11 3 6 14 11 3 6 14 11 3 6 14 11 3 6 14\ C 21 5 25 5 10 18 14 15 10 18 4 11 19 20 11 21 6 20 2 8 10 17\ So, we get a ciphertext of:
VFZFK SOPKS ELTUL VGUCH KR This level is a Vigenère Cipher. You have intercepted two longer, english language messages. You also have a key piece of information. You know the key length!
For this exercise, the key length is 6. The password to level five is in the usual place, encrypted with the 6 letter key.
Have fun!
Solution
used this site to break the key https://www.guballa.de/vigenere-solver
the key is frekey
so decrypting that krypton 5 HCIKV RJOX
CLEAR TEXT
Objectives
You more than likely used some form of FA and some common sense to solve that one.
So far we have worked with simple substitution ciphers. They have also been ‘monoalphabetic’, meaning using a fixed key, and giving a one to one mapping of plaintext (P) to ciphertext (C). Another type of substitution cipher is referred to as ‘polyalphabetic’, where one character of P may map to many, or all, possible ciphertext characters.
An example of a polyalphabetic cipher is called a Vigenère Cipher. It works like this:
If we use the key(K) ‘GOLD’, and P = PROCEED MEETING AS AGREED, then “add” P to K, we get C. When adding, if we exceed 25, then we roll to 0 (modulo 26).
P P R O C E E D M E E T I N G A S A G R E E D\ K G O L D G O L D G O L D G O L D G O L D G O\ becomes:
P 15 17 14 2 4 4 3 12 4 4 19 8 13 6 0 18 0 6 17 4 4 3\ K 6 14 11 3 6 14 11 3 6 14 11 3 6 14 11 3 6 14 11 3 6 14\ C 21 5 25 5 10 18 14 15 10 18 4 11 19 20 11 21 6 20 2 8 10 17\ So, we get a ciphertext of:
VFZFK SOPKS ELTUL VGUCH KR This level is a Vigenère Cipher. You have intercepted two longer, english language messages. You also have a key piece of information. You know the key length!
For this exercise, the key length is 6. The password to level five is in the usual place, encrypted with the 6 letter key.
Have fun!
Solution
used this site to break the key https://www.guballa.de/vigenere-solver
the key is frekey
so decrypting that krypton 5 HCIKV RJOX
CLEAR TEXT
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