How Do You Solve an Affine Cipher with Given Mappings?

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SUMMARY

The discussion focuses on solving an affine cipher using given mappings where ciphertext 'a' maps to plaintext 'E' and ciphertext 'v' maps to plaintext 'T'. The encryption function derived is E(x) = (147x + 10) mod 26, based on the equations formed from the mappings. The user successfully applies the modular multiplicative inverse to isolate the variable 'a' and subsequently solves for 'b'. The final encryption function is verified against the original mappings, confirming its correctness.

PREREQUISITES
  • Understanding of affine ciphers and their encryption functions
  • Familiarity with modular arithmetic, specifically mod 26
  • Knowledge of the modular multiplicative inverse
  • Basic algebraic manipulation skills
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  • Learn about modular arithmetic and its applications in encryption
  • Explore the concept of the modular multiplicative inverse in detail
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Students studying cryptography, mathematicians interested in modular arithmetic, and anyone looking to understand or implement affine ciphers.

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Homework Statement


7) You are attempting to break an affine cipher. You believe that the ciphertext ‘a’ maps to the plaintext letter ‘E’ and that the ciphertext ‘v’ maps to the plaintext ‘T’. Determine the encryption function used based on these two pieces of information.

Homework Equations



Ciphertext “a” = 0, Plaintext “E” = 4
Ciphertext “v” = 21, Plaintext “T” = 19
E(x) = (ax + b) mod m; Given: m = 26.

The Attempt at a Solution



0 = (a(4) + b) mod 26 => Eq 1. Plug in 4 for x, 0 for E(x), 26 for m.
21 = (a(19) + b) mod 26 => Eq 2. Plug in 19 for x, 21 for E(x), 26 for m.
21 = (a(15)) mod 26 => Subtract Eq. 1 from Eq. 2.
7(21) = (7a(15)) mod 26 => Multiply both sides by modular multiplicative inverse of 15 mod 26, which is 7. Given from Reference Sheet.
147 = (a(105)) mod 26 => Simplified.
147 = (a) => Eq. 3. Identity:1 = aa-1 mod m.
0 = (147(4) + b) mod 26 => Plug answer (a) in from Eq. 3 to Eq. 1 to solve for b.
0 = (588 + b) mod 26 => Simplified.
b = 10. STUCK HERE, I USED WOLFRAM ALPHA TO CALCULATE THIS
0 = (147(4) + 10) mod 26 => Check.
21 = (147(19) + 10) mod 26 => Check.
Encryption Function: E(x) = (147x + 10) mod 26

Please help, I might be completely off base on my methods after Eq. 3 to solve for b.
 
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I don't really know anything about ciphers. But I'm guessing you would want to use 52 instead of 26 since you are using both upper and lower case letters.

Disclaimer: I may not know what I'm talking about:shy:
 

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