The Secure Hill Cipher
Jeff Overbey
MA464-01 • Dr. Jerzy Wojdyło • April 29, 2003
Based on
S. Saeednia. How to Make the Hill Cipher Secure.
Cryptologia 24 (2000) 353–360.
The Scenario
The Scenario
Alice
The Scenario
Bob
The Scenario
Oscar
The Scenario
The Scenario
Insecure Channel
Sender
Recipient
Nosy Neighbor
Hill Cipher
Beforehand…
 Alice and Bob privately share
matrix K, invertible over Zm
To encrypt a matrix X over Zm…
 Compute Y = KX
 Send Y
To decrypt Y…
 Compute X = K–1Y
Why the Hill Cipher Isn’t Secure
Insecure Channel
Sender
Recipient
Nosy Neighbor
Why the Hill Cipher Isn’t Secure
“Hey Bob,
wassup?”
Insecure Channel
Sender
Recipient
Nosy Neighbor
Why the Hill Cipher Isn’t Secure
 7 1 22 18 
Insecure Channel

X   4 14 0 20
 24 1 18 15 
Sender
Recipient
Nosy Neighbor
Why the Hill Cipher Isn’t Secure
11 23 10 
Insecure
Channel 

K  16 19 24
 20 4 23
Sender
Recipient
Nosy Neighbor
Why the Hill Cipher Isn’t Secure
Insecure Channel
Y = KX
Sender
Recipient
Nosy Neighbor
Why the Hill Cipher Isn’t Secure
6 2
19 5
Channel
Y  Insecure
10 20
4 14 
 6 21 22 5 
Sender
Recipient
Nosy Neighbor
Why the Hill Cipher Isn’t Secure
19 5 6 2 
10

Insecure
Channel
20
4
14


 6 21 22 5 
Sender
Recipient
HEYBOBWAS
Nosy Neighbor
Why the Hill Cipher Isn’t Secure
19 5 6 2 
10

Insecure
Channel
20
4
14


 6 21 22 5 
23 10 
11 Recipient
K  16 19 24
 20 4 23
Sender
HEYBOBWAS
Nosy Neighbor
Why the Hill Cipher Isn’t Secure
19 5 6 2 
10

Insecure
Channel
20
4
14


 6 BOB,
21 22 5 
HEY
Sender
WASSUP?
Recipient
hahahaHAHAHAHA…
Nosy Neighbor
Hill Cipher
Secure Hill Cipher
Beforehand…
 Alice and Bob privately share
matrix K, invertible over Zm
To encrypt a matrix X over Zm…
 Compute Y = KX
 Send Y
To decrypt Y…
 Compute X = K–1Y
Beforehand…
 Alice and Bob privately share
matrix K, invertible over Zm
To encrypt a matrix X over Zm…
 Choose a vector t over Zm
 Form permutation matrix Pt
 Compute Kt = Pt–1KPt
 Compute Y = KtX and u = Kt
 Send (Y,u)
To decrypt (Y,u)…
 Compute t = K–1u
 Compute Pt
 Compute Kt–1 = Pt–1K–1Pt
 Compute X = Kt–1Y
Note that (Kt)–1 = (K–1)t
Permutation Matrices
I=
1
0
0
0
0
1
0
0
0
0
1
0
0
0
0
1
Permutation Matrices
I=
1
0
0
0
0
1
0
0
0
0
1
0
0
0
0
1
P23 =
1
0
0
0
0
1
0
0
0
1
0
0
0
0
0
1
Permutation Matrices
I=
1
0
0
0
0
1
0
0
0
0
1
0
0
0
0
1
A permutation matrix is a
matrix with exactly one 1 in
each row and column and
zeros elsewhere.
P23 =
P12,23 =
1
0
0
0
0
1
0
0
0
1
0
0
0
0
0
1
0
0
1
0
0
1
0
0
1
0
0
0
0
0
0
1
Permutation Matrices
1
0
P23 M  
0

0
0 0 0  1 2 3 4   1 2 3 4 
0 1 0   5 6 7 8   9 10 11 12 

1 0 0   9 10 11 12   5 6 7 8 

 

0 0 1  13 14 15 16  13 14 15 16 
0
0
P12,23 M  
1

0
1 0 0  1 2 3 4   5 6 7 8 
0 1 0   5 6 7 8   9 10 11 12 

0 0 0   9 10 11 12   1 2 3 4 

 

0 0 1  13 14 15 16  13 14 15 16 
Vector Representation
P23 =
P12,23 =
1
0
0
0
0
1
0
0
1
3
0
1
0
0
2
0
0
0
1
4
0
0
1
0
0
1
0
0
2
3
1
0
0
0
1
0
0
0
1
4
Hill Cipher
Secure Hill Cipher
Beforehand…
 Alice and Bob privately share
matrix K, invertible over Zm
To encrypt a matrix X over Zm…
 Compute Y = KX
 Send Y
To decrypt Y…
 Compute X = K–1Y
Beforehand…
 Alice and Bob privately share
matrix K, invertible over Zm
To encrypt a matrix X over Zm…
 Choose a vector t over Zm
 Form permutation matrix Pt
 Compute Kt = Pt–1KPt
 Compute Y = KtX and u = Kt
 Send (Y,u)
To decrypt (Y,u)…
 Compute t = K–1u
 Compute Pt
 Compute Kt–1 = Pt–1K–1Pt
 Compute X = Kt–1Y
The Easy Way to Encrypt

Require t to be the vector representation of a
permutation matrix

N.B.: This is for example only—it is not practical “in the field.”
An Example
“Hey Bob, wassup?”
 7 1 22 18 
X   4 14 0 20
 24 1 18 15 
11 23 10 
K  16 19 24
 20 4 23
An Example
Choose a permutation vector:
3
t  1 
 2 
3 rows, since K is 3×3
Form permutation matrix Pt:
0 0 1 
Pt  1 0 0
0 1 0
 7 1 22 18 
X   4 14 0 20
 24 1 18 15 
11 23 10 
K  16 19 24
 20 4 23
An Example
Find
Pt–1:
0 1 0
1
Pt  0 0 1 
1 0 0 
 7 1 22 18 
X   4 14 0 20
 24 1 18 15 
11 23 10 
K  16 19 24
 20 4 23
3
0 0 1 


t  1  Pt  1 0 0


 2 
0 1 0
Compute Kt = Pt–1KPt :
0 1 0 11 23 10  0 0 1 
Kt  0 0 1  16 19 24 1 0 0
1 0 0  20 4 23 0 1 0
An Example
Find
Pt–1:
0 1 0
1
Pt  0 0 1 
1 0 0 
Compute Kt = Pt–1KPt :
19 24 16 
K t   4 23 20 
 23 10 11 
Permutation of K
 7 1 22 18 
X   4 14 0 20
 24 1 18 15 
11 23 10 
K  16 19 24
 20 4 23
3
0 0 1 


t  1  Pt  1 0 0


 2 
0 1 0
An Example
Compute Y = KtX:
19 24 16   7 1 22 18 
Y   4 23 20  4 14 0 20
 23 10 11   24 1 18 15 
Compute u = Kt:
11 23 10   3
u  16 19 24  1 
 20 4 23  2
 7 1 22 18 
X   4 14 0 20
 24 1 18 15 
11 23 10 
K  16 19 24
 20 4 23
3
0 0 1 


t  1  Pt  1 0 0


 2 
0 1 0
19 24 16 
K t   4 23 20 
 23 10 11 
An Example
Compute Y = KtX:
15 7 4 22 
Y   2 8 6 0 
 23 18 2 25
Compute u = Kt:
 24 
u   11 
 6 
 7 1 22 18 
X   4 14 0 20
 24 1 18 15 
11 23 10 
K  16 19 24
 20 4 23
3
0 0 1 


t  1  Pt  1 0 0


 2 
0 1 0
19 24 16 
K t   4 23 20 
 23 10 11 
An Example
Send (Y,u) to Bob
 7 1 22 18 
X   4 14 0 20
 24 1 18 15 
11 23 10 
K  16 19 24
 20 4 23
3
0 0 1 


t  1  Pt  1 0 0


 2 
0 1 0
19 24 16 
K t   4 23 20 
 23 10 11 
15 7 4 22 
Y   2 8 6 0 
 23 18 2 25
 24 
u   11 
 6 
The Scenario
 24 
15 7 4 22 


Y   2 8 6 0  u   11 
 6 
 23 18 2 25
Sender
Recipient
Nosy Neighbor
The Scenario
 24 
15 7 4 22 


Y   2 8 6 0  u   11 
 6 
 23 18 2 25
Sender
Recipient
Nosy Neighbor
The Scenario
 24 
15 7 4 22 


Y   2 8 6 0  u   11 
 6 
 23 18 2 25
Sender
Recipient
Nosy Neighbor
The Scenario
 24 
15 7 4 22 


Y   2 8 6 0  u   11 
 6 
 23 18 2 25
Sender
Recipient
Nosy Neighbor
An Example
11 23 10 
K  16 19 24
 20 4 23
Receive (Y,u):
15 7 4 22 
Y   2 8 6 0 
 23 18 2 25 
 24 
u  11 
 6 
An Example
Compute
K–1:
15 25 16 
K 1  14 5 0 
 6 0 11
15 7 4 22 
Y   2 8 6 0 
 23 18 2 25
11 23 10 
K  16 19 24
 20 4 23
 24 
u   11 
 6 
Compute t = K–1u and derive Pt:
15 25 16  24 3
0 0 1 
t  14 5 0  11   1   Pt  1 0 0
 6 0 11  6   2
0 1 0
An Example
Compute
Kt–1
=
Pt–1K–1Pt:
 5 0 14 
1
K t   0 11 6 
 25 16 15
Compute X = Kt–1Y:
 5 0 14 15 7 4 22 
X   0 11 6   2 8 6 0 
 25 16 15  23 18 2 25
15 7 4 22 
Y   2 8 6 0 
 23 18 2 25
11 23 10 
K  16 19 24
 20 4 23
 24 
u   11 
 6 
3
0 0 1 


t  1  Pt  1 0 0


 2 
0 1 0
An Example
Compute
Kt–1
=
Pt–1K–1Pt:
 5 0 14 
1
K t   0 11 6 
 25 16 15
Compute X = Kt–1Y:
 7 1 22 18 
X   4 14 0 20
 24 1 18 15 
“Hey Bob, wassup?”
15 7 4 22 
Y   2 8 6 0 
 23 18 2 25
11 23 10 
K  16 19 24
 20 4 23
 24 
u   11 
 6 
3
0 0 1 


t  1  Pt  1 0 0


 2 
0 1 0
An Example
But that was just a
regular Hill cipher with
a fancy key…
The Actual Really Secure Hill Cipher
Jeff Overbey
MA464-01 • Dr. Jerzy Wojdyło • April 29, 2003
Based on
S. Saeednia. How to Make the Hill Cipher Secure.
Cryptologia 24 (2000) 353–360.
The Actual Really Secure Hill Cipher
or
How to Secure the Secure Hill Cipher
Ensuring Security

X should have two columns

Why? Suppose K is n × n and X is n × s.




If s  n…
If 2 < s < n…
If s = 2…
Pt should be different for each encryption

Theoretically, this can be ensured by choosing a
different t for each encryption

We did this by requiring t to be a vector representation of
a permutation matrix, but this is not the best solution.
The  Function
Let n denote the set of n × n permutation matrices.
: Zmn  n
Ideally,
  is onto
  is 1-1
The  Function
Let n denote the set of n × n permutation matrices.
: Zmn  n
Is it possible?
 Zmn = mn
 n = n!

Zmn n
An “ideal” function DNE.
y = 26n
y = n!
The  Function
Ideas?
Awkward silence…
Apologies for the horrible background image pun.
Apologies for the horrible background image pun.