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P. 81
1 Random substitution as a model for the cyclic keys formation
Let us consider the definition and basic properties of a random substitution (permutation) [13, 14],
that will be used further to assess the probability properties of round key sequences.
In combinatorics, a permutation is an ordered set of numbers 1,2, ,n , that is a bijection on the
set {1,2, , } n , which puts the i-th elements of the set in correspondence to the i number. The number
n in this case is called the order (degree) of permutation [13, 14].
The substitution s of arbitrary set Y = { ,yy 2 ,..., }y is a rule that each element y of set Y puts
1
n
i
in correspondence some other element ()sy [13, 14]:
i
y y ... y
s = 1 2 n .
( ) sy
( )
sy 1 ( ) ... sy n
2
In group theory the substitution is a bijection of this set into itself, i.e. substitution s degrees n is
y
considered as a permutation of the elements of the set Y = { , yy 2 ,..., } and for all i = 1,2,...,n
1
n
corresponding ()sy ∈ .
Y
i
The function ()sy value for a specific element y ∈ Y will be called the implementation of
i
i
substitution s in i-th point.
The composition of substitutions s and s degree n is defined as the consistent fulfillment of
w
u
the set Y elements permutation [13, 14]:
y y ... y
ss y
s s = 1 2 n , ( ( )) Y∈ .
u
w
( ( )) ss y
( ( ))
w
i
u
ss y 1 u ( ( )) ... ss y n
2
u
w
w
w
u
Concerning operations of sequential substitutions execution the set of all !n permutations degree
n forms a group, called the symmetric group and denoted as S = { , ,..., }ss 2 s ! n .
1
n
By definition [13, 14], random substitution (permutation) s is the random vector
x
{ ( ), ( ),..., ( )}s y s y 2 s y n , where all elements take values from the Y set and the probability of a
x
1
x
x
match of any two elements is equal to 0. In other words, a random substitution is randomly chosen
permutation from the set S
n
y y ... y
s = 1 2 n , s ∈ S , x∈ {1,2,..., !}n ,
x s x ( )y 1 s x ( ) ...y 2 s x ( )y n x n
defined by a set (vector) of random values { ( ), ( ),..., ( )}s y s y 2 s y n , that match probabilities satisfy
x
x
1
x
the following condition:
j
∀ i ≠ ∈ {1,2,..., }: (n P s x ( )y = s x (y j )) = .
0
i
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