Hill Cipher for the Holidays.

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Given the matrix A = ∣ 3 5 2 7 ∣ m o d 26 A= \begin{vmatrix}{3} && {5} \\ {2} && {7}\end{vmatrix} \bmod{26} , use a Hill cipher to decipher the message V O Q F D T T B N L Q M VOQFDTTBNLQM using modulo 26, where A → 0 , B → 1 , … , Z → 25 , A \rightarrow 0, B \rightarrow 1, \ldots , Z \rightarrow 25, and enter the result as a string of integers.

What does the deciphered message state?

Hill cipher


The answer is 701515247141183024.

Rocco Dalto by Rocco Dalto , 67, USA

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1 solution

Rocco Dalto
Dec 23, 2017

Note: det ⁡ ( A ) = 11 \: \det(A) = 11 and gcd ⁡ ( 11 , 26 ) = 1 \gcd(11,26) = 1 , so for each block X j X_{j} the system A X j = B j m o d 26 A X_{j} = B_{j} \bmod{26} has a unique solution, where ( 1 < = j < = 6 ) (1 <= j <= 6) .

Using six blocks for V V I X H U I E Y C R I VVIXHUIEYCRI we obtain:

21 14 21\: 14

16 5 16 \: 5

3 19 3 \: 19

19 1 19 \: 1

13 11 13 \: 11

16 12 16 \: 12

A ∗ X j = B j ⟹ X j = A − 1 ∗ B j A * X_{j} = B_{j} \implies X_{j} = A^{-1} * B_{j}

Using the A − 1 : A^{-1} \textbf{:}

C = A ∣ I = [ 3 5 1 0 2 7 0 1 ] m o d 26 C = A|I = \left [\begin{array}{cc|cc} 3 & 5 & 1 & 0 \\ 2 & 7 & 0 & 1 \\ \ \end{array} \right] \bmod{26}

Using the row operations: 9 ∗ R o w 1 9 * Row_{1} and 24 ∗ R o w 1 + R o w 2 ⟹ 24 * Row_{1} + Row_{2} \implies

C = [ 1 19 9 0 0 21 8 1 ] m o d 26 C = \left [\begin{array}{cc|cc} 1 & 19 & 9 & 0 \\ 0 & 21 & 8 & 1 \\ \ \end{array} \right] \bmod{26}

Using the row operation: 5 ∗ R o w 2 5 * Row_{2} and 7 ∗ R o w 2 + R 1 ⟹ 7 * Row_{2} + R_{1} \implies

C = [ 1 0 3 9 0 1 14 5 ] m o d 26 C = \left [\begin{array}{cc|cc} 1 & 0 & 3 & 9\\ 0 & 1 & 14 & 5 \\ \ \end{array} \right] \bmod{26}

⟹ \implies

A − 1 = [ 3 9 14 5 ] m o d 26 A^{-1} = \left [\begin{array}{cc|c} 3 & 9\\ 14 & 5 \\ \ \end{array} \right] \bmod{26}

Note: A ∗ A − 1 = I A * A^{-1} = I

Deciphering the first block [ 21 14 ] : \left [\begin{array}{cc|c} 21 \\ 14 \\ \ \end{array} \right] \textbf{:}

X 1 = A − 1 ∗ B 1 m o d 26 = [ 3 9 14 5 ] ∗ [ 21 14 ] m o d 26 = [ 7 0 ] m o d 26 X_{1} = A^{-1} * B_{1} \bmod{26} = \left [\begin{array}{cc|c} 3 & 9\\ 14 & 5 \\ \ \end{array} \right] * \left [\begin{array}{cc|c} 21 \\ 14 \\ \ \end{array} \right] \bmod{26} = \left [\begin{array}{cc|c} 7 \\ 0 \\ \ \end{array} \right] \bmod{26} :

Deciphering the second block [ 16 5 ] : \left [\begin{array}{cc|c} 16 \\ 5 \\ \ \end{array} \right] \textbf{:}

X 2 = A − 1 ∗ B 2 m o d 26 = [ 3 9 14 5 ] ∗ [ 16 5 ] m o d 26 = [ 15 15 ] m o d 26 X_{2} = A^{-1} * B_{2} \bmod{26} = \left [\begin{array}{cc|c} 3 & 9\\ 14 & 5 \\ \ \end{array} \right] * \left [\begin{array}{cc|c} 16 \\ 5 \\ \ \end{array} \right] \bmod{26} = \left [\begin{array}{cc|c} 15 \\ 15 \\ \ \end{array} \right] \bmod{26}

Repeat the same procedure for the remaining four blocks.

Our deciphered blocks are:

7 0 7 \: 0

15 15 15 \: 15

24 7 24 \: 7

14 11 14 \: 11

8 3 8 \: 3

0 24 0 \: 24

Our plain text message is:

H A P P Y H O L I D A Y HAPPYHOLIDAY

As a string of integers we have: 701515247141183024 \boxed{701515247141183024} .

H A P P Y H O L I D A Y ! HAPPY \: HOLIDAY \: !

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