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Hill Cipher

Tool to decrypt/encrypt with Hill cipher, a ciphering system similar to affine cipher but using a matrix for the gradient.

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Hill Cipher -

Tag(s) : Cryptography,Substitution Cipher

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# Hill Cipher

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## Hill Decoder

Also on dCode: Affine Cipher

## Hill Encoder

Also on dCode: Affine Cipher

## Matrix Inversion

Also on dCode: Inverse of a Matrix

Tool to decrypt/encrypt with Hill cipher, a ciphering system similar to affine cipher but using a matrix for the gradient.

### How to encrypt using Hill cipher?

Encryption with Hill cipher necessitates a matrix M and an alphabet.

Let the plain text be DCODE and the matrix M (size 2): $$M = \begin{pmatrix} 2 & 3 \\ 5 & 7 \end{pmatrix}$$ and the alphabet ABCDEFGHIJKLMNOPQRSTUVWXYZ

One splits the text into n-grams with n the matrix size, if needed, one completes with letters.

The matrix M is a 2x2 matrix, DCODE becomes DC,OD,EZ (one adds a Z to complete the bigram)

One substitutes the letters of the plain message with their rank in the alphabet starting from 0.

With ABCDEFGHIJKLMNOPQRSTUVWXYZ one gets A=0, B=1, ..., Z=25. One can also use ZABCDEFGHIJKLMNOPQRSTUVWXY to get A=1, B=2, ... Y=25, Z=0.
Groups of letters DC, OD, EZ become the groups of values (3,2), (14,3), (4,25)

For each group of values P of the plain text (equivalent to a vector of size n) one compute the matrix calculation: $$M.P \equiv C \mod 26$$ where C is the group of ciphered values and 26 the alphabet length.

$$\begin{pmatrix} 2 & 3 \\ 5 & 7 \end{pmatrix} \begin{pmatrix} 3 \\ 2 \end{pmatrix} \equiv \begin{pmatrix} 12 \\ 3 \end{pmatrix} \mod 26$$

From cipher values C, one can find the cipher letters with their rank in the alphabet.

12 is equal to M and 3 is equal to D.
So, DCODEZ is encrypted MDLNFN.

### How to decrypt Hill cipher?

Decryption needs to know the matrix and the alphabet used. Decryption involves matrix computations such as matrix inversion, of arithmetic such as modular inverse.

To decipher, one has to compute the matrix inverse modulo 26 (where 26 is the alphabet length), this needs the matrix to be invertible.

Using the example matrix, one computes the inverse matrix (modulo 26) : $$\begin{pmatrix} 2 & 3 \\ 5 & 7 \end{pmatrix}^{-1} \equiv \begin{pmatrix} -7 & 3 \\ 5 & -2 \end{pmatrix} \equiv \begin{pmatrix} 19 & 3 \\ 5 & 24 \end{pmatrix} \mod 26$$

Decryption consists in encrypting the cipher text with the inverse matrix.

Note that not all matrices can be adapted to hill cipher. The determinant of the matrix has to be coprime with 26. For a 2x2 matrix, the 4 numbers (a,b,c,d) must satisfy the conditions ad-bc is coprime with 26.

### How to recognize Hill ciphered text?

The ciphered message has a small index of coincidence and similar ngrams can be coded using the same letters.

### How to decipher Hill without matrix?

dCode proposes to test around 8000 combinations of 2x2 matrices and alphabets.

### What are the variants of the Hill cipher?

Hill is already a variant of Affine cipher. Few variants, except the use of large size matrices.

### When Hill cipher have been invented?

In 1929 by Lester S. Hill