=Paper= {{Paper |id=Vol-2145/p15 |storemode=property |title=Post-quantum Key Exchange Protocol Using High Dimensional Matrix |pdfUrl=https://ceur-ws.org/Vol-2145/p15.pdf |volume=Vol-2145 |authors=Richard Megrelishvili,Melkisadeg Jinjikhadze,Maksim Iavich,Avtandil Gagnidze,Giorgi Iashvili }} ==Post-quantum Key Exchange Protocol Using High Dimensional Matrix == https://ceur-ws.org/Vol-2145/p15.pdf
     Post-quantum Key Exchange Protocol Using High
                  Dimensional Matrix
           Richard Megrelishvili                          Melkisadeg Jinjikhadze                                   Maksim Iavich
       I. J. Tbilisi State University                  Akaki Tsereteli State University                          Caucasus University
              Tbilisi, Georgia                                Kutaisi, Georgia                                     Tbilisi, Georgia
      richard.megrelishvili@tsu.ge                           mjinji@yahoo.com                                     m.iavich@scsa.ge


          Avtandil Gagnidze                                     Giorgi Iashvili
           Georia University                                  Georia University
            Tbilisi, Georgia                                   Tbilisi, Georgia
       gagnidzeavto@yahoo.com                                 g.iashvili@scsa.ge


    Abstract— Active work is being done to create and develop                But this problem can easily be solved by quantum computers
quantum computers. Google Corporation, NASA and the                          using Shor algorithm [1,2].
Universities Space Research Association (USRA) have teamed up                The security of RSA algorithm relies on factorization problem,
with DWAFE, the manufacturer of quantum processors. D-Wave                   but this problem can be easily solved using quantum computers
2X is a quantum processor that contains 2,048 physical qubits.
1152 qubits from the whole number of qubits are used to perform
                                                                             [3].
the calculations. As we see, quantum computers can easily solve              Active work is being done to create and develop quantum
the problem of calculating the discrete logarithm used in Diffie-            computers. Google Corporation, NASA and the Universities
Hellman algorithm. So it can break Diffie-Hellman algorithm.                 Space Research Association (USRA) have teamed up with
    When quantum computers are released all existing crypto                  DWAFE, the manufacturer of quantum processors. D-Wave 2X
systems will be useless, because there will be no way to transfer the        is a quantum processor that contains 2,048 physical qubits. 1152
key securely.                                                                qubits from the whole number of qubits are used to perform the
    In the article is proposed the new key exchange method using             calculations. As we see, quantum computers can easily solve the
high dimensional matrix, this method is safe against attacks                 problem of calculating the discrete logarithm used in Diffie-
implemented using quantum computers. The case concerns the
matrix function and algorithm for cryptographic keys exchange
                                                                             Hellman algorithm. So it can break Diffie-Hellman algorithm.
with open channel. For the algorithm is offered the method of                When quantum computers are released all existing crypto
building a high dimensional matrix multiplicative group.                     systems will be useless, because there will be no way to transfer
    The arising of this goal is that traditional key exchange                the key securely [4,5].
methods are vulnerable to quantum computer attacks.                          In the article is proposed the new key exchange algorithm using
    Keywords— post-quantum cryptography, attacks, a matrix one-              high dimensional matrix. This algorithm is safe against
way function, Abelian multiplicative group, asymmetric                       quantum computer attacks.
cryptography, high dimensional matrix finite field                           The case concerns the matrix function and algorithm for
                       I. INTRODUCTION                                       cryptographic keys exchange with open channel.
One of the fundamental problems of cryptography is the safe                        For this is offered the method building a high dimensional
communication over the listening channel. Messages need to be                matrix multiplicative group.
encrypted and decrypted, but for this, both parties need to have                  The arising of this goal is that traditional key exchange
a common key. If this key is transmitted via the same channel,               methods are vulnerable to quantum computer attacks.
then the listening side will also receive it, and the meaning of                  One-way function (OWF) is a function whose value is easy
the encryption will disappear.                                               to calculate for any argument, but it is “difficult” to find an
Diffie-Hellman algorithm allows the two parties to obtain a                  argument for the given value of the function. The word
common secret key using an unprotected, but spoofed,                         "difficult" is to understand the complexity of the computation.
communication channel. The received key can be used to                       In other words, finding the relevant argument of the given
exchange messages using symmetric encryption.                                function in real time is difficult even with the modern
The security of forming a common key in the Diffie-Hellman                   computing techniques. The irreversibility of function does not
algorithm follows from the fact that, although it is relatively              mean that the function is one-way [6,7].
easy to calculate exponents modulo a prime number, it is very                     The existence of one-way functions is the basis for the idea
difficult to calculate discrete logarithms. For large prime                  of asymmetric cryptography. It (one-way function) is the
numbers of hundreds and thousands of bits, the task is                       foundation       of    asymmetric     cryptography,     personal
considered unsolvable, since it requires a tremendous amount                 identification, authentication, and other fields of information
of computational resources.                                                  protection. Although there is no theoretical proof of the
                                                                             existence of one-way functions in general, there are several
  Copyright held by the author(s).




                                                                        83
“possible pretendents” (eg, multiplication and factorization,                             𝑣1 𝑎11 + 𝑣2 𝑎21 + ⋯ + 𝑣𝑛 𝑎𝑛1            𝑢1
squaring and module rooting, discreet exponent and                                        𝑣1 𝑎12 + 𝑣2 𝑎22 + ⋯ + 𝑣𝑛 𝑎𝑛2            𝑢2
                                                                            𝑣𝐴1 = (                              ⋮            )=( ⋮ )      (10)
logarithmization), whose one-wayity (ie the difficulty of
finding the argument for the value of function) at this time real                        𝑣1 𝑎𝑛1 + 𝑣2 𝑎𝑛2 + ⋯ + 𝑣3 𝑎𝑛3             𝑢𝑛
and is actively used in information exchange protocols.                     The number of unknowns in the system of linear equations is
    As we have mentioned, one-way functions are actively used               the square of number of equations. Obviously, the system can
in the algorithms for developing a cryptographic open key. The              not be solved in limited time, if the size of the matrix is large
initial idea (1976) belongs to Whitfield Duffie and Martin                  enough. Size of the matrix must be chosen considering Grover’s
Helman. Based of their idea was established the first practical             algorithm. Classically, searching requires a linear search, which
wel-known Diffie-Helman-Merkel method, which enabled the                    is O(N) in time. Grover's algorithm needs O(N1/2) time, it is
development of a common cryptographic key using the open                    considered as fastest quantum algorithm for searching an
(unprotected) channel. A year later, the first RSA algorithm of             unsorted information. This algorithm provides a quadratic
asymmetric encryption was formed. The RSA in fact, resolved                 speedup [10,11].
the problem of exchange information with open channel. Both                  One fact must be taken into consideration if the 𝐴1 matrix
algorithms are not safe against quantum computers attacks. Are              contains the internal recurrence, or if each of its rows are in a
proposed quantum key exchange protocols, but quantum                        certain recurrence with the previous row, then the task of
computers are needed to implement them [8,9].                               solving the system will be replaced by a simpler task that is easy
                                                                            to solve. It is so important that it puts itself in doubt the one-
               II. ONE-WAY MATRIX FUNCTION                                  way character of our function and requires the existence of
                                                                            Abelian multiplicative matrix group with a high order, that is
     The new one-way function for the development of common                 free from the recurrence of the inside.
cryptographic keys is based on high order cyclic matrix groups,                          III. FINITE MATRIX GROUPS CONSTRUCTION
with the power 𝑒 = 2𝑛 − 1 , where the n is row dimension of                 Let's consider (1 + 𝛼)𝑗 , where j = 0,1,2, ⋯, and α represents
the square matrix. Let's assume that "A" is the above matrix                the root of primitive polynomial in the 𝐺𝐹(2𝑛 ) field odule with
group, while A is the initial 𝑛 × 𝑛 matrix, then "A" = A={𝐴, 𝐴2 ,           the module p (x).
            𝑛
𝐴3 , … , 𝐴2 −1 = 𝐼}                             (1)                              (1 + 𝛼)0 = 1                               1
where I represents an identity matrix.                                           (1 + 𝛼)1 = 1 + 𝛼                           11
One-way function and algorithm for common key development                        (1 + 𝛼)2 = 1 + 𝛼 2                         101
are as follows:                                                                  (1 + 𝛼)3 = 1 + 𝛼 + 𝛼 2 + 𝛼 3 1111
     The sender chooses 𝐴1 ∈ A secret matrix to send to the                     (1 + 𝛼)4 = 1 + 𝛼 4                         10001
          receiving party via open channel the                                   (1 + 𝛼)5 = 1 + 𝛼 + 𝛼 4 + 𝛼 5 110011
                            u𝑢1 = 𝑣𝐴1                            (2)        The polynomial coefficients generated the structure known as
          vector where 𝑣 ∈ 𝑉𝑛 vector is known (𝑉𝑛 – is a vector             Serpinsky Triangle. The derived structure contains a number of
          space on GF field);                                               sub-structures that can be used as a generator (generating
       The receiving party shall, on the other hand, choose                matrix) for multiplicative groups, ie primitive elements. Such
           𝐴2 ∈ A secret matrix and send to the sender                      is, for example,
                             𝑢2 = 𝑣𝐴2                           (3)                                                        1 1 1 1 1
           vector;                                                                     1 1 1                               1 0 0 0 0
                                                                            𝑃3 = (1 0 0) (9), 𝑃5 = 1 1 0 0 0 (11)
       Sender calculates 𝑘1 = 𝑢2 𝐴1                            (4)
           vector;                                                                     1 1 0                               1 0 1 0 0
       Receivier calculates 𝑘2 = 𝑢1 𝐴2                         (5)                                                       (1 1 1 1 0)
                                                                            And many more. Their natural powers create Abelian
           where 𝑘1 and 𝑘2 – are secret keys;                               multiplicative cyclic group.
       Obviously, 𝑘1 = 𝑘2 = 𝑘, because                                     For example:
                   𝑘 = 𝑣𝐴1 𝐴2 = 𝑣𝐴2 𝐴1                          (6)                          1 1 1                       1 0 1
           because of the commutativeness of the "A" group. The                  𝑃31 = (1 0 0) , 𝑃32 = (1 1 1),
           𝑣𝐴𝑖 = 𝑢                                       (7)                                 1 1 0                       0 1 1
           is one-way fast function.                                                          0 0 1                      1 1 0
     Let 𝑣 = (𝑣1 , 𝑣2 , 𝑣3 , ⋯ , 𝑣𝑛 ) ∈ 𝑉𝑛                      (8)               𝑃33 = (1 0 1) , 𝑃34 = (0 0 1),
     and                                                                                      0 1 0                      1 0 0
     𝑢 = (𝑢1 , 𝑢2 , 𝑢3 , ⋯ , 𝑢𝑛 ) ∈ 𝑉𝑛 are non-secret vectors from                           0 1 1                       0 1 0
the                 above                algorithm             and               𝑃35 = (1 1 0) , 𝑃36 = (0 1 1),
         𝑎11 ⋯ 𝑎1𝑛                                                                           1 1 1                       1 0 1
𝐴1 = ( ⋮        ⋱        ⋮ )∈A                                   (9)                                  1 0 0
                                                                                             0
         𝑎𝑛1 … 𝑎𝑛𝑛                                                               𝑃37 = 𝑃3 = (0 1 0)                                   (12)
is a secret matrix. Then, according to algorithm the following                                        0 0 1
                                                                                 It’s easy to confirm that
system is formed:
                                                                                 𝑃30 , 𝑃31 , 𝑃32 , 𝑃33 , 𝑃34 , 𝑃35 , 𝑃36




                                                                       84
    is an Abelian multiplicative group.                                                 𝑃3,2 = 𝑃36 × 𝑃36 + 𝑃36 × 0 + 0 × 𝑃36 = 𝑃35 ,
Lets keep the structure of 𝑃3 matrix and extend it by elements                               𝑃3,3 = 𝑃36 × 𝑃36 + 𝑃36 × 0 + 0 × 0 = 𝑃35 .
of set (2) as follows:                                                                                 𝑃33 𝑃32 𝑃34
                         𝑗    𝑗
                   𝑃3𝑖 𝑃3 𝑃3                                                                 or 𝑃 = (𝑃34 𝑃35 𝑃35 ) (15) ( see pic. 2).
    𝑃32 (𝑖, 𝑗) = (𝑃3𝑗       0      0 ), where i,j=0..6. (13)                                           𝑃32 𝑃35 𝑃35
                       𝑗     𝑗
                     𝑃3   0𝑃3
Fore example, when 𝑖 = 5 and 𝑗 = 6, we have
                𝑃35 𝑃36 𝑃36
   𝑃32 (5,6) = (𝑃36 0     0)=
                𝑃36 𝑃36 0                                                              pic.2: 𝑃 = [𝑃32 (5,6)]2
   0 1 1        0 1 0          0 1 0
 (1 1 0) (0 1 1) (0 1 1)                                                               Using the software package we have developed it has been
   1 1 1        1 0 1          1 0 1                                                   confirmed that the 𝑷𝟑𝟐 (𝟓, 𝟔) matrix is a primitive element. Its
   0 1 0                                                                               natural powers generate Abelian multiliplicative group, whose
 (0 1 1)            0             0         ( (14)                                                𝟐
                                                                                       power is 𝟐𝟑 −𝟏 .
   1 0 1
   0 1 0        0 1 0                                                                       The elements of [𝑃32 (5,6)]𝑘 when k = 73, 146, 219, 292,
 (0 1 1) (0 1 1)                  0                                                    365, 438, 511 are diagonal matrices (see pic. 3):
( 1 0 1         1 0 1                    )                                                    𝑃34 0       0    𝑃31 0       0      𝑃35 0       0
                                                                                                     4                1
   When 𝑖 = 0 and 𝑗 = 1, we have (pic. 1):                                                  ( 0 𝑃3 0 ) , ( 0 𝑃3 0 ) , ( 0 𝑃35 0 ),
                                                                                               0    0 𝑃34       0    0 𝑃31         0     0 𝑃35
                                                                                                          2                6
                                                                                                        𝑃3 0      0      𝑃3 0        0
                                                                                                      ( 0 𝑃32 0 ) , ( 0 𝑃36 0 ),
                                                                                                         0   0 𝑃32        0    0 𝑃36
                                                                                                          3                0
Pic.1: 𝑃32 (5,6) 𝑎𝑛𝑑 𝑃32 (0,1)                                                                          𝑃3 0      0      𝑃3 0        0
                                                                                                      ( 0 𝑃33 0 ) , ( 0 𝑃30 0 )
                𝑃30 𝑃31 𝑃31                                                                              0   0 𝑃33        0    0 𝑃30
  𝑃32 (0,1) = (𝑃31 0      0)=
                𝑃31 𝑃31 0
  1 0 0         1 1 1        1 1 1
 (0 1 0) (1 0 0) (1 0 0)
  0 0 1         1 1 0        1 1 0                                                     Pic.3 : [𝑃32 (5,6)]𝑘 , 𝑘 = 73, 146, 219, 292, 365, 438, 511
  1 1 1
 (1 0 0)            0            0         (15)                                             Also 𝑃32 (0,1) matrix is a primitive element, and elements
  1 1 0                                                                                of the set [𝑃32 (0,1)]𝑘 , when k=73, 146, 219, 292, 365, 438,
  1 1 1         1 1 1                                                                  511, are diagonal matrices(pic 4):
 (1 0 0) (1 0 0)                 0
( 1 1 0         1 1 0                   )                                                   𝑃33    0    0      𝑃36 0     0   𝑃32 0           0
                  Consider 𝑃 = [𝑃32 (5,6)]2 =                                              (0     𝑃33                6
                                                                                                        0 ) , ( 0 𝑃3 0 ) , ( 0 𝑃32           0 ),
               𝑃35 𝑃36 𝑃36       𝑃35 𝑃36 𝑃36
                                                                                             0     0   𝑃33      0   0 𝑃36     0   0         𝑃32
           = (𝑃36 0      0 ) × (𝑃36 0       0 ) (16)                                                    5
                                                                                                      𝑃3 0        0    𝑃31 0    0
                 6    6            6    6
               𝑃3 𝑃3 0           𝑃3 𝑃3 0                                                             ( 0 𝑃35 0 ) , ( 0 𝑃31 0 )
    If we take into consideration that the set                                                         0    0 𝑃35       0  0 𝑃31
                                                                                                        4                0
0,𝑃30 , 𝑃31 , 𝑃32 , 𝑃33 , 𝑃34 , 𝑃35 , 𝑃36 is a field, it is easy to assure that                       𝑃3 0        0    𝑃3 0     0
each sub-matrix of the 𝑃 matrix is in the the same set:                                              ( 0 𝑃34 0 ) , ( 0 𝑃30 0 )
 𝑃1,1 = 𝑃35 × 𝑃35 + 𝑃36 × 𝑃36 + 𝑃36 × 𝑃36 = 𝑃33 ,                                                      0    0 𝑃34       0  0 𝑃30
𝑃1,2 = 𝑃35 × 𝑃36 + 𝑃36 × 0 + 𝑃36 × 𝑃36 = 𝑃32 ,
 𝑃1,3 = 𝑃35 × 𝑃36 + 𝑃36 × 0 + 𝑃36 × 0 = 𝑃34 ,
 𝑃2,1 = 𝑃36 × 𝑃35 + 0 × 𝑃36 + 0 × 𝑃36 = 𝑃34 ,
𝑃2,2 = 𝑃36 × 𝑃36 + 0 × 0 + 0 × 𝑃36 = 𝑃35 ,
             6        6                              5
𝑃2,3 = 𝑃3 × 𝑃3 + 0 × 0 + 0 × 0 = 𝑃3 ,
𝑃3,1 = 𝑃36 × 𝑃35 + 𝑃36 × 𝑃36 + 0 × 𝑃36 = 𝑃32 ,                                           pic.4: [𝑃32 (5,6)]𝑘 , 𝑘 = 73, 146, 219, 292, 365, 438, 511




                                                                                  85
Set of non-zero elements of the diagonal matrices represents the                                      𝑃302
                                                                                                         𝑃312 𝑃312
perturbation of the group 𝑃30 , 𝑃31 , 𝑃32 , 𝑃33 , 𝑃34 , 𝑃35 , 𝑃36 (called as                        1
                                                                                    𝑃33 (0,1) = (𝑃32 0          0 )=
primary group) and one of the elements is 𝑃30 .                                                     1      1
Finally, we can conclude that empirically we proved the                                            𝑃32 𝑃32 0
                                                                                            0
following fact:                                                                           𝑃3 0        0       𝑃30 𝑃31                𝑃31  𝑃30                𝑃31     𝑃31
                                                                                                 0
The second order (𝑖, 𝑖 + 1) expansion 𝑃32 (𝑖, 𝑖 + 1) , 𝑖 = 0. .5,                        ( 0 𝑃3 0 ) (𝑃31 0                           0 ) (𝑃31                0        0)
of the matrix 𝑃3 is a primitive element and creates the Abelian                            0    0 𝑃30         𝑃31 𝑃31                0    𝑃31                𝑃31      0
multiplicative finite group 𝐹(𝑃32 (𝑖, 𝑖 + 1)) , with the power                              0    1
                                                                                          𝑃3 𝑃3 𝑃3     1
  2
23 – 1.                                                                             = (𝑃31 0          0)           0                                         0                  (18)
Below we can see other primitive elements that are results of                               1    1
                                                                                          𝑃3 𝑃3 0
expansion of 𝑃3 matrix:
                                𝑃30 𝑃31 𝑃31                                               𝑃30 𝑃31 𝑃31         𝑃30 𝑃31                𝑃31
                 𝑃32 (0,1) = (𝑃31 0              0 ),                                    (𝑃31 0       0 ) (𝑃31 0                     0)                      0
                                                                                            1    1
                                𝑃31 𝑃31 0                                             ( 𝑃3 𝑃3 0               𝑃31 𝑃31                0                                      )
                                𝑃31 𝑃32 𝑃32
                                                                                         Let's consider [𝑃33 (0,1)]𝑘 set. It has the same basic
                 𝑃32 (1,2) = (𝑃32 0               0 ),
                                  2       2
                                                                                    structure 𝑃3 as the primary group, as well as the first and second
                                𝑃3 𝑃3 0                                             expansion matrices taken from the primary group. It is expected
                                𝑃32 𝑃33 𝑃33                                         that this set is characterized by the same properties as the
                 𝑃32 (2,3) = (𝑃33 0               0)                                primary group has. Indeed, experimentally, it also has diagonal
                                𝑃33 𝑃33 0                                           matrices, whose diagonal elements represent one of the
                                                                                    perturbations of the primary group.
                                 𝑃33       𝑃34 𝑃34
                    𝑃32 (3,4) = (𝑃34        0   0)                                       For       the       set    [𝑃33 (0,1)]𝑘        diagonal         matrices            are
                                                                                                          2   2
                                                                                                   𝑗∙(22∙3 +23 +1)                          32
                                 𝑃34       𝑃34 0                                    [𝑃33 (0,1)]         , 𝑗 = 1,2,3, ⋯ , 2 − 1.
                                 𝑃34       𝑃35 𝑃35                                                 32
                                                                                         When 𝑗 = 2 − 1, we get the final element of the set
                    𝑃32 (4,5) = (𝑃35        0   0)                                  [𝑃33 (0,1)]𝑘 :
                                 𝑃35       𝑃35 0
                                 𝑃35       𝑃36 𝑃36                                                             32       2∙32    32
                                                                                           [𝑃33 (0,1)](2 −1)∙(2 +2 +1) = [𝑃33 (0,1)](2 −1) =
                                                                                                                                                                   33

                    𝑃32 (5,6) = (𝑃36        0   0)                                             [𝑃30 (0,1)]0         0             0
                                 𝑃36         6
                                           𝑃3 0                                            =(        0        [𝑃30 (0,1)] 0
                                                                                                                                  0      ) (19)
                                 𝑃36       𝑃30 𝑃30
                                                                                                     0              0       [𝑃30 (0,1)]0
                    𝑃32 (6,0) = (𝑃30        0   0 ) (17)
                                 𝑃30         0
                                           𝑃3 0
                                                                                          We see that this is an Identity matrix. Therefore 𝑃33 (0,1)
      In order to get higher order primitive elements, we still                     is a primitive element and creates the Abelian multiplicative
retain the structure of 𝑃3 matrix and put into the elements of the                                              3
                                                                                    finite group with power 23 − 1.
group 𝐹(𝑃32 (𝑖, 𝑖 + 1)) . We get 33 order matrix (call it a third
                                                                                          Definition: We call the following matrix
order expansion).
      For example, if we use the elements of group 𝐹(𝑃32 (0,1))
                                                                                                                    𝑃3𝑖𝑘−1     𝑃3𝑖+1
                                                                                                                                  𝑘−1    𝑃3𝑖+1
                                                                                                                                            𝑘−1
for the first and second expansions of the matrix of 𝑃3 ,
respectively [𝑃32 (0,1)]0 [𝑃32 (0,1)]1 matrices, we get the                              𝑃3𝑘 (𝑖, 𝑖 + 1) = (𝑃3𝑖+1
                                                                                                              𝑘−1                0          0 )                  (20)
following matrix (pic. 5):                                                                                          𝑃3𝑖+1
                                                                                                                       𝑘−1     𝑃3𝑖+1
                                                                                                                                  𝑘−1       0

                                                                                       where 𝑃3𝑖𝑘−1 ∈ 𝐹(𝑃3𝑘−1 (𝑖, 𝑖 + 1)), as the kth order (𝑖, 𝑖 + 1)
                                                                                    expansion of the 𝑃3 matrix.
                                                                                     Theorem: 𝑃3𝑘 (𝑖, 𝑖 + 1) is a primitive element and creates the
                                                                                    abelian multiplicative finite group 𝐹(𝑃3𝑘 (𝑖, 𝑖 + 1)) with
                                                                                               𝑘
                                                                                    power 23 – 1.
pic.5. 𝑃33 (0,1)                                                                                                                         𝑗(22∙3
                                                                                                                                                 𝑘−1     𝑘−1
                                                                                                                                                       +23         +1)
                                                                                       In general matrices [𝑃3𝑘 (𝑖, 𝑖 + 1)]                                              , where
                                                                                                         𝑘−1
                                                                                    𝑗 = 1,2,3, ⋯ , 23 − 1 are diagonal matrices and diagonal
                                                                                    elements are one of the permutations of the elements of the
                                                                                    primary group.




                                                                               86
                     𝑘−1
    When 𝑗 = 23            − 1, we get                                                    one way function, that we offer. So the key exchange method is
                   𝑗(22∙3
                         𝑘−1
                             +23
                                𝑘−1
                                    +1)=
                                                                                          got and it is secure against quantum computers attacks.
[𝑃3𝑘 (𝑖, 𝑖 + 1)]
                           𝑘−1           𝑘−1     𝑘−1
                     (23         −1)∙(22∙3     +23     +1)=                                                   ACKNOWLEDGEMENT
= [𝑃3𝑘 (𝑖, 𝑖 + 1)]
                           𝑘−1
                     (23         −1)                                                      The Work Was Conducted as a Part of Research Grant of Joint
= [𝑃3𝑘 (𝑖, 𝑖 + 1)]
                           0                                                              Project of Shota Rustaveli National Science Foundation and
     [𝑃3𝑘–1 (𝑖, 𝑖 + 1)]                      0                        0                   Science & Technology Center in Ukraine [№ STCU-2016-08]
                                                       0
=            0                    [𝑃3𝑘–1 (𝑖, 𝑖 + 1)]                  0
                                                                              0
    (        0                               0                [𝑃3𝑘–1 (𝑖, 𝑖 + 1)] )                                       REFERENCES
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matrices with the same characteristic polynom. It is also
important the use of other baseline matrices, which enlarges a
new type of primitive structures. Elements of abelian
multiplicative matrix groups can be used in implementation of




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