A new algorithm for multiplying two Dirac numbers - arXiv

A new algorithm for multiplying two Dirac numbers Aleksandr Cariow1, Galina Cariowa1 1

Faculty of Computer Sciences and Information Technologies, Żołnierska 52 71-210 Szczecin, Poland {atariov, gtariova}@wi.zut.edu.pl tel. +48 91 4495573

Abstract. In this work a rationalized algorithm for Dirac numbers multiplication is presented. This algorithm has a low computational complexity feature and is well suited to FPGA implementation. The computation of two Dirac numbers product using the naïve method takes 256 real multiplications and 240 real additions, while the proposed algorithm can compute the same result in only 88 real multiplications and 256 real additions. During synthesis of the discussed algorithm we use the fact that Dirac numbers product may be represented as vector-matrix product. The matrix participating in the product has unique structural properties that allow performing its advantageous decomposition. Namely this decomposition leads to significant reducing of the computational complexity. Keywords: Dirac numbers, multiplication of hypercomplex numbers, fast algorithms.

1.

Introduction

Recently hypercomplex numbers [1] are used in various fields of data processing including digital signal and image processing, machine graphics, telecommunications and especially in public key cryptography [2-10]. The most popular are quaternions, octonions and sedenions [1]. Perhaps the less popular are the Pauli, Kaluza and Dirac numbers [11]. This numbers are mostly used in solving different physical problems in electrodynamics, field theory, etc. Anyway, hypercomplex arithmetic is a very important issue in modern data processing applications. Among other operations in hypercomplex arithmetic, multiplication is the most time consuming one. The reason for this is, because the addition of N -dimensional hypercomplex numbers only requires N real additions; the multiplication of these numbers already requires N ( N − 1) real additions and N 2 real multiplication. It is easy to see that the increasing of dimensions of hypercomplex number increases the computational complexity of the multiplication. Therefore, reducing the computational complexity of the multiplication of hypercomplex numbers is an important scientific and engineering problem. Several efficient algorithms for the multiplication of hypercomplex numbers have been reported in the literature [12-17]. Our previous work [12] proposed an algorithm for computing product of two Dirac numbers which has lower computational complexity compared with the direct (schoolbook) method of computations. In this paper we propose another algorithm for this purpose. Compared with our previous algorithm, the proposed algorithm has lower multiplicative complexity.

2.

Preliminary Remarks A Dirac number is defined as follows [12]: 15

d = d0 +

∑d i

n n

(1)

n =1

where d 0 and {d n }, n = 1,...,15 - are real numbers, and {in }, n = 1,...,15 - are the imaginary units that commute with real numbers during multiplication. At that i1 , i2 , i3 , i4 - the main imaginary units, and the remaining imaginary units are composite ones and are expressed in terms of the main imaginary units by the formula:

1

is = i p iq Lir , where 1 ≤ p < q L < r ≤ n . All products of imaginary units on each other are entirely determined by the predetermined rules for multiplication main imaginary units on each other:

i 2p = ε p , iq i p = α pq i p iq , p < q , p, q = 1, 2, ..., n , where ε p , α pq are equal to −1 , 1 , or 0 . The results of all possible products of the Dirac numbers imaginary units can be summarized in the following table [12]: × 1 i1 i2 i3 i4

1 1 i1 i2 i3 i4

i1 i1 1 − i5 − i6 − i7

i2 i2 i5 −1 − i8 − i9

i3 i3 i6 i8 −1 − i10

i4 i4 i7 i9 i10 −1

i5 i5 i2 i1 i11 i12

i6 i6 i3 − i11 i1 i13

i7 i7 i4 − i12 − i13 i1

i8 i8 i11 − i3 i2 i14

i9 i9 i12 − i4 − i14 i2

i10 i10 i13 i14 − i4 i3

i11 i11 i8 i6 − i5 − i15

i12 i12 i9 i7 i15 − i5

i13 i13 i10 − i15 i7 − i6

i14 i14 i15 − i10 i9 − i8

i15 i15 i14 i13 − i12 i11

i5 i6 i7 i8 i9 i10 i11 i12 i13

i5 i6 i7 i8 i9 i10 i11 i12 i13

− i2 − i3 − i4 i11 i12 i13 i8 i9 i10

− i1 − i11 − i12 i3 i4 i14 i6 i7 i15

i11 − i1 − i13 − i2 − i14 i4 − i5 − i15 i7

i12 i13 − i1 i14 − i2 − i3 i15 − i5 − i6

1 i8 i9 i6 i7 i15 i3 i4 i14

− i8 1 i10 − i5 − i15 i7 − i2 − i14 i4

− i9 − i10 1 i15 − i5 − i6 i14 − i2 − i3

− i6 i5 i15 −1 − i10 i9 − i1 − i13 i12

− i7 − i15 i5 i10 −1 − i8 i13 − i1 − i11

i15 − i7 i6 − i9 i8 −1 − i12 i11 − i1

i3 − i2 − i14 − i1 − i13 i12 −1 − i10 i9

i4 i14 − i2 i13 − i1 − i11 i10 −1 − i8

− i14 i4 − i3 − i12 i11 − i1 − i9 i8 −1

− i13 i12 − i11 − i4 i3 − i2 − i7 i6 − i5

i10 − i9 i8 − i7 i6 − i5 − i4 i3 − i2

i14 i15

i14 i15

− i15 − i14

− i10 − i13

i9 i12

− i8 − i11

i13 i10

− i12 − i9

i11 i8

− i4 −i 7

i3 i6

− i2 − i5

i7 i4

− i6 − i3

i5 i2

1 i1

− i1 −1

Consider the problem of multiplying two Dirac numbers: 15

d = d (1) ⋅ d ( 2) = d 0 +

∑

15

in d n , d (1) = a0 +

n =1

∑

15

in an , d ( 2) = b0 +

n =1

∑i b

n n

.

n =1

The operation of multiplication of Dirac numbers can be represented more compactly in the form of vector-matrix product: Y16×1 = B16 X16×1 (2) where

X16×1 = [a0 , a1 ,..., a15 ]Τ , Y16×1 = [d 0 , d1 ,..., d15 ]Τ ,

B16

 b0 b  1  b2   b3  b4   b5 b  6 b = 7  b8  b9  b10 b  11 b12 b  13 b14  b15

b1 b0

− b2 b5

− b3 − b2

− b4 b6

b5 − b3

b6 b7

b7 − b4

− b8 − b11

− b9 − b12

− b10 − b13

− b11 − b8

− b12 − b9

− b13 − b10

b14 − b15

b5

b0

− b1

b8

− b11

b9

− b12

− b3

− b4

− b14

− b6

− b7

− b15

− b10

b6 b7

− b8 − b9

b11 b12

b0 − b10

− b1 b13

b10 b0

− b13 − b1

b2 − b14

b14 b2

− b4 b3

b5 − b15

b15 b5

− b7 b6

b9 − b8

b2

− b1

b0

− b11

b8

− b12

b9

− b6

− b7

− b15

− b3

− b4

− b14

b13

b3 b4

b11 b12

− b8 − b9

− b1 b13

b0 − b10

− b13 − b1

b10 b0

b5 − b15

b15 b5

− b7 b6

b2 − b14

b14 b2

− b4 b3

− b12 b11

b11

b3

− b6

− b2

b5

− b14

b15

b0

b10

− b9

b1

b13

− b12

− b4

b12 b13

b4 − b14

− b7 b15

b14 b4

− b15 − b7

− b2 − b3

b5 b6

− b10 b9

b0 − b8

b8 b0

− b13 b12

b1 − b11

b11 b1

b3 − b2

b8 b9

− b6 − b7

b3 b4

b5 − b15

− b2 b14

b15 b5

− b14 − b2

b1 − b13

b13 b1

− b12 b11

b0 − b10

b10 b0

− b9 b8

b7 − b6

b10

b15

− b14

− b7

b4

b6

− b3

b12

− b11

b1

b9

− b8

b0

b5

b15 b14

b10 − b13

− b13 b10

− b9 b12

b12 − b9

b8 − b11

− b11 b8

b4 b7

− b3 − b6

b2 b5

b7 b4

− b6 − b3

b5 b2

b0 − b1

− b15  b14  b13   − b12  b11   − b10  b9  , − b8   b7  − b6   b5  − b4   b3  − b2   − b1   b0 

2

The direct multiplication of two Dirac numbers requires 256 real multiplications and 240 real additions. Our previous paper [12] reported an algorithm for multiplication of two Dirac numbers with 128 real multiplications and 160 real additions. In this paper we introduce the new algorithm, which reduce multiplicative complexity to 88 real multiplications at the cost of 96 extra real additions compared with our previous algorithm.

3.

Synthesis of a rationalized algorithm for computing Dirac numbers product

First and foremost, we rearrange the columns of the matrix in the following order {1, 2, 3, 4, 6, 7, 9, 12, 11, 14, 15, 5, 16, 8, 10, and 13}. Next, we rearrange the rows of obtained matrix in the same manner. Next, let us multiply by (−1) every element in the lower half of the rows in the obtained matrix. As a result, we obtain the following matrix:

~ B16

 b0  b  1  b2   b3  b5   b6  b  8 b =  11  − b10  − b13   − b14 −b  4  − b15 −b  7  − b9   − b12

b1

− b2

− b3

b5

b6

− b8

− b11

− b10

− b13

b14

− b4

− b15

b7

− b9

b0 b5 b6 b2 b3

b5 b0 − b8 − b1 b11

b6 b8 b0 − b11 − b1

− b2 − b1 b11 b0 − b8

− b3 − b11 − b1 b8 b0

− b11 − b3 b2 − b6 b5

− b8 − b6 b5 − b3 b2

− b13 − b14 − b4 − b15 − b7

− b10 − b15 − b7 − b14 − b4

− b15 − b10 b9 b13 − b12

b7 b9 b10 − b12 − b13

b14 b13 − b12 − b10 b9

− b4 − b12 − b13 b9 b10

− b12 − b4 b14 − b7 b15

b11 b8 − b13 − b10

b3 − b6 b14 − b15

− b2 b5 − b4 b7

− b6 b3 − b15 b14

b5 − b2 b7 − b4

b0 b1 − b9 − b12

b1 b0 − b12 − b9

− b9 − b12 − b0 − b1

− b12 − b9 − b1 − b0

− b4 b7 b2 − b5

− b14 b15 b3 − b6

b7 − b4 − b5 b2

b15 − b14 − b6 b3

b10 b13 b8 b11

− b15 − b7 − b14 − b4 − b12

− b10 b9 b13 − b12 − b4

b9 b10 − b12 − b13 − b14

b13 − b12 − b10 b9 b7

− b12 − b13 b9 b10 b15

− b4 b14 − b7 b15 b10

− b7 b15 − b4 b14 b13

− b2 − b3 − b5 − b6 − b8

− b5 − b6 − b2 − b3 − b11

− b0 b8 b1 − b11 − b3

− b8 − b0 b11 b1 b2

b1 − b11 − b0 b8 b6

b11 b1 − b8 − b0 − b5

b3 − b2 b6 − b5 − b1

− b9

b7

b15

− b4

− b14

b13

b10

− b11

− b8

b6

− b5

− b3

b2

− b1

− b12  − b9  − b7   b15  − b4   b14  b13   b10   b11  b8   b6  − b5   b3  − b2   − b0   −b 0 

Then we can rewrite expression (2) in following form: ( 2) ~ (1) Y16×1 = E16 P16 B16 P16 X16×1

(3)

where

E16 = diag (1,1,1,1,1,1,1,1,−1,−1,−1,−1,−1,−1,−1,−1) ,

(1) P16

1   1      1   1     1   1     1     1 = , 1     1   1     1   1    1     1   1  

3

( 2) P16

 1   1     1   1     1   1     1     1 = . −1     −1   −1     −1   − 1    −1     −1   −1  

~ It is easy to see that B16 has the following structure: A ~ B16 =  8  B8

B8  , − A 8 

where

 b0 b1 b  1 b0  b2 b5  b b6 A8 =  3  b5 b2   b6 b3 b b 11  8 b11 b8 − b10 − b  13 − b14  −b B8 =  4 − b15   − b7 −b  9 − b12

− b2

− b3

b5

b6

− b8

b5

b6

− b2

− b3

− b11

b0

b8

− b1

− b11

− b3

− b8

b0

b11

− b1

b2

− b1

− b11

b0

b8

− b6

b11

− b1

− b8

b0

b5

b3

− b2

− b6

b5

b0

− b6

b5

b3

− b2

b1

− b13

b14

− b4

− b15

b7

− b9

− b10

− b15

b7

b14

− b4

− b12

− b15

− b10

b9

b13

− b12

− b4

− b7

b9

b10

− b12

− b13

b14

− b14

b13

− b12

− b10

b9

− b7

− b4

− b12

− b13

b9

b10

b15

− b12

− b4

− b14

b7

b15

b10

− b9

b7

b15

− b4

− b14

b13

− b11  − b8  − b6   b5  , − b3   b2  b1   b0  − b12  − b9  − b7   b15  . − b4   b14  b13   b10 

As was shown in [18, 19], the matrix having such a structure can be efficiently factorized. This factorization reduces the number of multiplications by 25%. Than a computational procedure for the multiplication of Dirac numbers can be represented as follows:

Y16×1 =

( 2) E16 P16 (T2×3

 A8 − B8  (1) ⊗ I 8 )diag − ( A 8 + B 8 ) (T3×2 ⊗ I 8 )P16 X16×1   B8 

(4)

where I N - is an identity N × N matrix, sign „ ⊗ ”– denotes tensor product of two matrices [20],

4

T2×3

 b0 + b10 b +b  1 13  b2 + b14  b +b A8 − B8 =  3 4  b5 + b15   b6 + b7  b +b  8 9 b11 + b12

1 0 1 0 1   =  , T3×2 = 0 1 0 1 1 1 1

b1 + b13

− b2 − b14

− b3 + b4

b5 + b15

b6 − b7

− b8 + b9

b0 + b10

b5 + b15

b6 − b7

− b2 − b14

− b3 + b4

− b11 + b12

b5 + b15

b0 + b10

b8 − b9

− b1 − b13

− b11 + b12

− b3 + b4

b6 + b7

− b8 − b9

b0 − b10

b11 + b12

− b1 + b13

b2 − b14

b2 + b14

− b1 − b13

− b11 + b12

b0 + b10

b8 − b9

− b6 + b7

b3 + b4

b11 + b12

− b1 + b13

− b8 − b9

b0 − b10

b5 − b15

b11 + b12

b3 + b4

− b2 + b14

− b6 − b7

b5 − b15

b0 − b10

b8 + b9

− b6 − b7

b5 − b15

b3 + b4

− b2 + b14

b1 − b13

 − b0 + b10  −b +b  1 13  − b2 + b14  −b +b − ( A8 + B8 ) =  3 4  − b5 + b15   − b6 + b7  −b +b  8 9 − b11 + b12

− b11 + b12  − b8 + b9  − b6 + b7   b5 − b15  , − b3 + b4   b2 − b14  b1 − b13   b0 − b10 

− b1 + b13

b2 − b14

b3 + b4

− b5 + b15

− b6 − b7

b8 + b9

− b0 + b10

− b5 + b15

− b6 − b7

b2 − b14

b3 + b4

b11 + b12

− b5 + b15

− b0 + b10

− b8 − b9

b1 − b13

b11 + b12

b3 + b4

− b6 + b7

b8 − b9

− b0 − b10

− b11 + b12

b1 + b13

− b2 − b14

− b2 + b14

b1 − b13

b11 + b12

− b0 + b10

− b8 − b9

b6 + b7

− b3 + b4

− b11 + b12

b1 + b13

b8 − b9

− b0 − b10

− b5 − b15

− b11 + b12

− b3 + b4

b2 + b14

b6 − b7

− b5 − b15

− b0 − b10

− b8 + b9

b6 − b7

− b5 − b15

− b3 + b4

b2 + b14

− b1 − b13

b11 + b12  b8 + b9  b6 + b7   − b5 − b15  . b3 + b4   − b2 − b14  − b1 − b13   − b0 − b10 

Let we rearrange the columns of the matrix A 8 − B 8 in the following order: {1, 2, 4, 7, 5, 3, 8, 6}. After such permutation we multiply by (-1) every element of the sixth and eighth columns of the resulting matrix. The rows of the obtained matrix we rearrange in following order: {1, 7, 3, 4, 5, 6, 2, 8}. Then we multiply by (-1) every element of the sixth and seventh rows of the resulting matrix too. As a result, we obtain the following matrix: b1 + b13 − b3 + b4 − b8 + b9 b5 + b15 b2 + b14 − b11 + b12 − b6 + b7   b0 + b10  b +b b11 + b12 − b2 + b14 b0 − b10 − b6 − b7 − b3 − b4 b1 − b13 − b5 + b15   8 9  b2 + b14 b5 + b15 b8 − b9 − b3 + b4 − b1 − b13 − b0 − b10 − b6 + b7 b11 − b12    b +b b6 + b7 b0 − b10 b2 − b14 b11 + b12 b8 + b9 b5 − b15 b1 − b13  B′8 =  3 4 .  b5 + b15 b2 + b14 − b11 + b12 − b6 + b7 b0 + b10 b1 + b13 − b3 + b4 − b8 + b9    b1 − b13 − b5 + b15 b8 + b9 b11 + b12 − b2 + b14 b0 − b10   − b6 − b7 − b3 − b4 − b − b − b0 − b10 − b6 + b7 b11 − b12 b2 + b14 b5 + b15 b8 − b9 − b3 + b4   1 13  b8 + b9 b5 − b15 b1 − b13 b3 + b4 b6 + b7 − b10 + b0 b2 − b14   b11 + b12 It is easy to see that the matrix B′8 possesses a structure that provides “good” factorization too.

A B′8 =  4 B4

B4  , A4 

where

b0 + b10 b +b A4 =  8 9 b2 + b14   b3 + b4

b1 + b13

− b3 + b4

b11 + b12

− b2 + b14

b5 + b15

b8 − b9

b6 + b7

b0 − b10

− b8 + b9   b5 + b15  − b − b b0 − b10  , B4 =  6 7 − b1 − b13 − b3 + b4    b2 − b14   b11 + b12

b2 + b14

− b11 + b12

− b3 − b4

b1 − b13

− b0 − b10

− b6 + b7

b8 + b9

b5 − b15

− b6 + b7  − b5 + b15  , b11 − b12   b1 − b13 

As follows from [18, 19], such block-structural form reduces the number of multiplications by 50%. This brings to the following factorization:

5

1 B′8 = (H 2 ⊗ I 4 ) (S (40) ⊕ S (41) )(H 2 ⊗ I 4 ) (5) 2 1 1  where sign “ ⊕ ”– denotes direct sum of two matrices and H 2 =   is the 2 × 2 Hadamard matrix [20]. 1 − 1

S (40) = A 4 + B 4 , S (40)

S (41) = A 4 − B 4 , S (41)

 b0 + b5 + b10 + b15  −b −b +b +b = 6 7 8 9 − b1 + b2 − b13 + b14   b3 + b4 + b11 + b12 b0 − b5 + b10 − b15  b +b +b +b = 6 7 8 9 b1 + b2 + b13 + b14  b3 + b4 − b11 − b12

b1 + b2 + b13 + b14

− b3 − b4 + b11 + b12 − b0 + b5 − b10 + b15 b6 + b7 + b8 + b9 b1 − b2 + b13 − b14

− b3 + b4 − b11 + b12 b1 − b2 − b13 + b14 − b6 + b7 + b8 − b9 b0 + b5 − b10 − b15

− b3 + b4 + b11 − b12 − b1 − b2 + b13 + b14 b6 − b7 + b8 − b9 b0 − b5 − b10 + b15

b3 + b4 + b11 + b12 b0 + b5 + b10 + b15 b6 + b7 − b8 − b9

− b6 + b7 − b8 + b9  b0 − b5 − b10 + b15  , − b3 + b4 + b11 − b12   b1 + b2 − b13 − b14 

b6 − b7 − b8 + b9  b0 + b5 − b10 − b15  . − b3 + b4 − b11 + b12   − b1 + b2 + b13 − b14 

Next we rearrange the columns of the matrix −( A 8 + B8 ) in the following order: {1, 2, 4, 7, 5, 3, 8, 6}. Then we multiply by (-1) every element of the sixth and eighth columns of the resulting matrix. We rearrange the rows of the obtained matrix in the following order: {1, 7, 3, 4, 5, 6, 2, 8}. Then we multiply by (-1) every element of the sixth and eighth rows of the resulting matrix. As a result, we obtain the following matrix:

 − b0 + b10  −b +b  1 13  − b3 + b4  −b +b B′8′ =  8 9  − b5 + b15   b2 − b14 − b + b  11 12  b6 − b7

− b1 + b13 − b0 + b10 − b6 + b7 − b11 + b12 − b2 + b14 b5 − b15 − b8 + b9 b3 − b4

b3 + b4

b8 + b9

− b6 − b7 − b0 − b10 b2 + b14 b11 + b12 b8 + b9 − b5 − b15 − b1 − b13

b11 + b12 − b2 − b14 − b0 − b10 b6 + b7 − b3 − b4 − b1 − b13 b5 + b15

− b5 + b15 b2 − b14 − b11 + b12 b6 − b7 − b0 + b10 − b1 + b13 − b3 + b4 − b8 + b9

− b2 + b14 b5 − b15 − b8 + b9 b3 − b4 − b1 + b13 − b0 + b10 − b6 + b7 − b11 + b12

b11 + b12 b8 + b9 − b5 − b15 − b1 − b13 b3 + b4 − b6 − b7 − b0 − b10 b2 + b14

b6 + b7  − b3 − b4  − b1 − b13   b5 + b15  b8 + b9   b11 + b12  − b2 − b14   − b0 − b10 

It is easy to see that the matrix B′8′ has the same type of structure as in the previous case:

C B′8′ =  4 D 4

D4  , C4 

where

− b0 + b10 − b + b C 4 =  1 13  − b3 + b4   − b8 + b9

− b1 + b13 − b0 + b10 − b6 + b7 − b11 + b12

b3 + b4 − b6 − b7 − b0 − b10 b2 + b14

b8 + b9   − b5 + b15  b −b b11 + b12  , D 4 =  2 14 − b11 + b12 − b2 − b14    − b0 − b10   b6 − b7

− b2 + b14 b5 − b15 − b8 + b9 b3 − b4

b11 + b12 b8 + b9

− b5 − b15 − b1 − b13

b6 + b7  − b3 − b4  . − b1 − b13   b5 + b15 

Then we can write:

1 B′8′ = (H 2 ⊗ I 4 ) (S (42) ⊕ S (43) )(H 2 ⊗ I 4 ) 2

(6)

where

S (42) = C 4 + D 4 , S (42)

− b0 − b5 + b10 + b15 − b + b + b − b =  1 2 13 14  − b3 + b4 − b11 + b12   b6 − b7 − b8 + b9

− b1 − b2 + b13 + b14 − b0 + b5 + b10 − b15 − b6 + b7 − b8 + b9 b3 − b4 − b11 + b12

b3 + b4 + b11 + b12

− b6 − b7 + b8 + b9 − b0 − b5 − b10 − b15 − b1 + b2 − b13 + b14

b6 + b7 + b8 + b9  − b3 − b4 + b11 + b12  , − b1 − b2 − b13 − b14   − b0 + b5 − b10 + b15 

6

S (43) = C 4 − D 4 , S (43)

− b0 + b5 + b10 − b15 − b − b + b + b =  1 2 13 14  − b3 + b4 + b11 − b12   − b6 + b7 − b8 + b9

− b1 + b2 + b13 − b14

b3 + b4 − b11 − b12

− b0 − b5 + b10 + b15

− b6 − b7 − b8 − b9

− b6 + b7 + b8 − b9

− b0 + b5 − b10 + b15

− b3 + b4 − b11 + b12

b1 + b2 + b13 + b14

− b6 − b7 + b8 + b9  b3 + b4 + b11 + b12  , b1 − b2 + b13 − b14   − b0 − b5 − b10 − b15 

Lastly we rearrange the columns of the matrix B8 in the following order: {4, 2, 3, 8, 1, 6, 7, 5}. We rearrange the rows of the obtained matrix in the following order: {1, 6, 7, 5, 4, 2, 3, 8}. Then we multiply by (-1) every element of the fifth, sixth, seventh and eighth rows of the resulting matrix too. As a result, we obtain the following matrix:

 − b4 − b  13 − b14  −b B′8′′ =  12 − b10   − b7 −b  9 − b15

− b13

b14

− b12

− b10

b7

− b9

− b4

− b12

b14

− b7

b10

b15

− b12

− b4

b13

− b9

b15

b10

− b14

b13

− b4

− b15

b9

− b7

b7

− b9

− b15

b4

b13

− b14

b10

b15

b9

b13

b4

b12

b15

b10

b7

b14

b12

b4

b9

− b7

− b10

b12

b14

− b13

− b15  b9  b7   − b10  , b12   − b14  − b13   b4 

The matrix thus obtained has the following structure:

 − b4 − b F4   13 E = , 4  − b14 − E4   − b12

E B′8′′ =  4  F4

− b13

b14

− b4

− b12

− b12

− b4

− b14

b13

− b12  − b10  −b b14  , F4 =  7  − b9 b13    − b4  − b15

b7

− b9

b10

b15

b15

b10

b9

− b7

− b15  b9  . b7   − b10 

Then we can write:

B′8′′ = (T2×3 ⊗ I 4 )(S (44) ⊕ S (45) ⊕ S (46) )(T3×2 ⊗ I 4 )

(7)

S (44) = E 4 − F4 , S (45) = −(E 4 + F4 ) , S (46) = F4 .

S (44)

 − b4 + b10  b −b =  7 13  b9 − b14  − b12 + b15

− b7 − b13

b9 + b14

− b4 − b10

− b12 − b15

− b12 − b15

− b4 − b10

− b9 − b14

b7 + b13

− b12 + b15   b4 + b10  − b9 + b14  (5)  b7 + b13 , S =  b9 + b14 − b7 + b13  4   − b4 + b10  b12 + b15

S (46)

− b10 −b = 7  − b9   − b15

b7

− b9

b10

b15

b15

b10

b9

− b7

− b7 + b13

b9 − b14

b4 − b10

b12 − b15

b12 − b15

b4 − b10

− b9 + b14

b7 − b13

b12 + b15  − b9 − b14  , − b7 − b13   b4 + b10 

− b15  b9  . b7   −b10 

Combining (5), (6), and (7) and taking into account all manipulations with rows and columns in each matrix, we obtain a following vector-matrix procedure: (1) ( 2 ) (1) ( 4) ( 2) ( 2) ~ ( 2) (3) ~ (1) (1) Y16×1 = E16 P16 W16 ×24 P24 W24×28 D 28 W28×24 P24 W24×16 P16 X16×1

(8)

where

~ (1) (1) ( 2) W16 ×24 = T2×3 ⊗ I 8 , W24×16 = T3×2 ⊗ I 8 , W24×28 = (I 2 ⊗ ( H 2 ⊗ I 4 )) ⊕ (T2×3 ⊗ I 4 ) ,

7

~ ( 2) 1 (0) 1 (1) 1 ( 2) 1 (3) ( 4) (5) (6) ( 2) W28 ×24 = (I 2 ⊗ ( H 2 ⊗ I 4 )) ⊕ (T3×2 ⊗ I 4 ) , D 28 = diag ( S 4 , S 4 , S 4 , S 4 , S 4 , S 4 , S 4 ) . 2

(3) P24

( 4) P24

2

2

2

1    1     1   1     1   −1     1     −1   1     1   1     1  =   1   − 1     1   −1     1     1   1    1   1     1    1    1  

1    1     1   1     1   −1    −1      1   1     1   1     1  =   1   −1     1   −1     1     1   1     1   − 1     −1     −1   − 1 

8

Unfortunately matrices S (40) , S (41) , S (42) and S (43) can no longer be effectively factorized. Their block composition is not conducive to a reduction of computational complexity. As for the matrix S (44) , its block structure after some modifications can be reduced to a convenient form. If we rearrange the columns and rows of the matrix S (44) in the following order: {1, 2, 4, 3} and then multiply by (-1) every element of the last column and every element of the last row, we obtain the matrix

 − b4 + b10  b −b =  7 13 − b12 + b15   − b9 + b14

~ S 4( 4)

− b7 − b13

− b912 + b15

− b4 − b10

− b9 + b14

− b9 − b14

− b4 + b10

b12 + b15

b7 − b13

− b + b A 2 =  4 10  b7 − b13

− b9 − b14  b12 + b15   A 2 = − b7 − b13   B 2  − b4 − b10 

− b7 − b13  − b + b , B 2 =  12 15  − b4 − b10   − b9 + b14

B2  , A2 

− b9 − b14  , b12 + b15 

whose structure has a “good” factorization property. Then we can write:

~ 1 S 4( 4) = (H 2 ⊗ I 2 ) (S (20) ⊕ S (21) )(H 2 ⊗ I 2 ) 2

(9)

S (20) = A 2 + B 2 , S (21) = A 2 − B 2 . In turn, if we rearrange the columns and rows of the matrix S (45) as {1, 2, 4, 3} and then multiply by (-1) every element of the last column and the every element of the last row, we obtain the matrix

~ S 4(5)

 b4 + b10  b +b =  7 13  b12 + b15  − b9 − b14

− b7 + b13 b4 − b10 − b9 + b14 − b12 + b15

b + b C 2 =  4 10 b7 + b13

b12 + b15

− b9 − b14 b4 + b10 b7 + b13

− b9 + b14  − b12 + b15  C 2 = − b7 + b13  D 2  b4 − b10 

− b7 + b13   b +b , D 2 =  12 15  b4 − b10  − b9 − b14

D2  , C2 

− b9 + b14  , − b12 + b15 

which has a “good” block structure too. We can write:

~ 1 S 4(5) = (H 2 ⊗ I 2 ) (S (22) ⊕ S (23) )(H 2 ⊗ I 2 ) 2

(10)

S (22) = C 2 + D 2 , S (23) = C 2 − D 2 . Let we rearrange columns of the matrix S (46) in following way {1, 4, 3, 2}. Then multiply by (-1) every element of the last row, we obtain the matrix

~ S 4( 6 )

− b10 −b = 7  − b9   b15

− b15 b9 b7 b10

− b9 b15 b10 b7

b7  b10  E 2 = b15   F2  − b9 

F2  − b10  , E2 =  − b − E2   7

− b15  − b , F2 =  9  b9   b15

b7  . b10 

Then we can write [18, 19]:

~ S 4( 6) = (T2×3 ⊗ I 2 )(S (24) ⊕ S (25) ⊕ F2 )(T3×2 ⊗ I 4 )

(11)

9

S (24) = E 2 − F2 , S (25) = −(E 2 + F2 ) . Combining (9), (10), and (11) and taking into account all manipulations with rows and columns in each matrix at this stage of synthesis of algorithm, we obtain a new updated matrix-vector procedure: (1) ( 2 ) (1) ( 4) ( 2) ( 6) (3) (3) ~ (3) ( 5) ~ ( 2 ) (3) ~ (1) (1) Y16×1 = E16 P16 W16 ×24 P24 W24×28 P28 W28×30 D30 W30×28 P28 W28×24 P24 W24×16 P16 X16×1

(12)

where ( 5) P28

( 6) P28

1  1  1   1   1   1 ⊕ ⊕ , = I16 ⊕   1  1  1        −1   −1   1   1  1  1   1   1   1  ⊕ ⊕ , = I16 ⊕    1  1  1       −1   −1   − 1 

~ (3) (3) W28 × 30 = I16 ⊕ (I 2 ⊗ (H 2 ⊗ I 2 )) ⊕ (T2× 3 ⊗ I 2 ) , W30× 28 = I16 ⊕ (I 2 ⊗ (H 2 ⊗ I 2 )) ⊕ (T3× 2 ⊗ I 2 ) , 1 1 1 1 1 1 1 1 ( 3) ( 3) ( 3) ( 3) ( 3) , D16 D 30 = D16 ⊕ D14 = diag{ S (40) , S (41) , S (42) , S 4(3) } , D14 = diag{ S (20) , S (21) , S 2( 2) , S (23) , S (24 ) , S (25) , F2 } , 2 2 2 2 2 2 2 2

− b + b − b + b S (20) =  4 10 12 15  b7 − b9 − b13 + b14 b + b + b + b S (22) =  4 10 12 15  b7 − b9 + b13 − b14

− b7 − b9 − b13 − b14  (1) − b4 + b10 + b12 − b15 ,S =  − b4 − b10 + b12 + b15  2  b7 + b9 − b13 − b14 − b7 − b9 + b13 + b14  b + b − b − b , S (23) =  4 10 12 15  b4 − b10 − b12 + b15   b7 + b9 + b13 + b14

 b −b S (24) =  9 10 − b7 − b15

− b7 − b15  (5) b9 + b10 , S = b9 − b10  2 b7 − b15

− b7 + b9 − b13 + b14  , − b4 − b10 − b12 − b15  − b7 + b9 + b13 − b14  , b4 − b10 + b12 − b15 

− b7 + b15  . − (b9 + b10 )

Let us now consider the second order matrices, which were formed as a result of the last decompositions. Matrices S (20) , S (21) , S (22) , S (23) , and F2 can not be factorized, but the matrix S (24) possesses a structure that provides “good” factorization:

 b −b S (24) =  9 10 − b7 − b15

− b7 − b15  a b  = . b9 − b10  b a 

Then we can write:

1 ( s0 ⊕ s1 )H 2 (13) 2 Consider now the matrix S (25) . If we multiply by (-1) every element of the last row of this matrix, we obtain a new matrix which can be successfully factorized. − b7 + b15   b +b ~ S 2(5) =  9 10 , − b + b  7 15 b9 + b10  S (24) = H 2

Then we can write:

~ 1 S 2(5) = H 2 ( s 2 ⊕ s3 )H 2 (14) 2 Combining all partial factorizations into a single whole and taking into account all manipulations with rows and columns in each matrix at fourth stage of synthesis of algorithm, we obtain a final matrix-vector procedure:

10

(1) ( 2 ) (1) ( 4) ( 2) ( 6) (3) ( 4 ) ( 4 ) ~ ( 4 ) ~ (3) ( 5) ~ ( 2 ) (3) ~ (1) (1) Y16×1 = E16 P16 W16 ×24 P24 W24×28 P28 W28×30 P30 W30 D30 W30 W30×28 P28 W28×24 P24 W24×16 P16 X16×1 (15)

~ ( 4) ( 4) W30 = I 24 ⊕ (I 2 ⊗ H 2 ) ⊕ I 2 , W30 = I 24 ⊕ (I 2 ⊗ H 2 ) ⊕ I 2 , P30 = I 27 ⊕ (−1) ⊕ I 2 , 1 1 1 1 1 1 1 1 1 1 1 1 ( 3) D 30 = diag ( S (40) , S (41) , S (42 ) , S (43) , S (20) , S (21) , S (22 ) , S (23) , s 0 , s1 , s 2 , s3 , F2 ) , 2 2 2 2 2 2 2 2 2 2 2 2

s0 = −b7 + b9 − b10 − b15 , s1 = b7 + b9 − b10 + b15 , s2 = −b7 + b9 + b10 + b15 , s3 = b7 + b9 + b10 − b15 . Fig. 1 shows a data flow diagram representation of the rationalized algorithm for computation of the Dirac numbers product. In this paper, data flow diagrams are oriented from left to right. Straight lines in the figures denote the operations of data transfer. Points where lines converge denote summation. The dashed lines indicate the sign change operation. We deliberately use the usual lines without arrows on purpose, so as not to clutter the picture. The circles in these figures show the operation of multiplication by a variable (or constant) inscribed inside a circle. In turn, the rectangles indicate the matrix–vector multiplications with the matrix inscribed inside a rectangle.

4.

Estimation of computational complexity

We calculate how many real multiplications (excluding multiplications by power of two) and real additions are required for realization of the proposed algorithm, and compare it with the number of operations required for a direct evaluation of matrix-vector product in Eq. (2). Let us look to the data flow diagram in Figure 1. It is easy to verify that all the real multiplications which to be performed to computing the product of two Dirac numbers are realized only during multiplying a vector of data by the quasi-diagonal matrix D30 . It can be argued that the multiplication of a vector by the matrix D30 requires 88 real multiplications and some trivial multiplications by the power of two. Multiplication by power of two may be implemented using convention arithmetic shift operations, which have simple realization and hence may be neglected during computational complexity estimation. Now we calculate the number of additions required in the implementation of the algorithm. To count the number of additions required to perform matrix-vector multiplications with matrices S (40) , S (41) S (42) , and S (43) we introduce the following notation:

c1,1 = b0 + b5 , c1,2 = b10 + b15 , c1,3 = b1 + b2 , c1, 4 = b13 + b14 , c1,5 = −b3 + b4 , c1,6 = −b11 + b12 , c1,7 = −b6 + b7 , c1,8 = −b8 + b9 , c2,1 = b6 + b7 , c2,2 = b8 + b9 , c2,3 = b3 + b4 , c2, 4 = b11 + b12 , c2,5 = b1 − b2 , c2,6 = b13 − b14 , c2,7 = b0 − b5 , c2,8 = b10 − b15 . Then the matrix S (40) can be represented as follows:

S (40)

 c1,1 + c1, 2  − c2,1 + c2, 2 =  − c 2, 5 − c 2, 6   c2,3 + c2,4

c1,3 + c1,4

c1,5 + c1,6

− c 2, 3 + c 2, 4

c 2, 5 − c 2, 6

− c2,7 − c2,8

c1,7 − c1,8

c2,1 + c2, 2

c1,1 − c1, 2

c1,7 + c1,8   c 2 , 7 − c 2 ,8  c1,5 − c1,6   c1,3 − c1, 4 

To carry out the multiplication of the matrix S (40) by the corresponding vector we must perform 44 additions, namely: – 16 additions which are necessary to calculate all the elements ci , j , i, j = 1, 2, ... ,8 , – 16 additions which are necessary to calculate all the sums ci , j + ci , k , i, j , k = 1, 2, ... ,8 , – 12 additions arising from the direct matrix-vector multiplication by applying the general rule for matrixvector multiplication. Next we estimate the number of real additions that needed for the matrix-vector multiplication with the matrices S (41) , S (42) , S (43) :

11

− y0

x0

x2

y1 y2 y3

1 ( 0) S 2 4

x1 x3

x4 x5

H2

x6

y4 y5 y6 y7 y8

H2

1 (1) S4 2

x7 x8

x9

y9

1 ( 2) S 2 4

x10

y10

y11 y12 y13 y14 y15

x11

x12 x13

H2

x 14

H2

1 ( 3) S 2 4

x15 1 ( 0) S 2 2 H2

1 (1) S2 2

H2

1 ( 2) S 2 2 H2

H2

H2

1 ( 3) S 2 2 1 s 2 0 1 s 2 1 1 s 2 2 1 s 2 3

H2

H2 H2

F2

Fig. 1. Data flow diagram for rationalized Dirac numbers multiplication algorithm in accordance with the procedure (15).

12

S (41)

S (42)

S (43)

c2,7 + c2,8  c2,1 + c2, 2 =  c1,3 + c1, 4  c2,3 − c2, 4  − c1,1 + c1, 2  − c2, 5 + c2, 6 =  c1,5 + c1,6   − c1,7 + c1,8 − c2,7 + c2,8  − c1,3 + c1,4 =  c1,5 − c1,6   c1,7 + c1,8

c2, 5 + c2, 6

c1,5 − c1,6

c2, 3 + c2, 4

− c1,3 + c1, 4

c1,1 + c1, 2

− c1,7 − c1,8

c2,1 − c2, 2

c2,7 − c2,8

− c1,7 − c1,8   c1,1 + c1, 2  , c1,5 + c1,6   − c2,5 + c2,6 

− c1,3 + c1, 4

c 2 ,3 + c 2 , 4

− c 2 , 7 + c 2 ,8

− c2,1 + c2,2

c1,7 + c1,8

− c1,1 − c1, 2

− c1,5 + c1,6

− c2, 5 − c2, 6

− c2, 5 + c2, 6

c2, 3 − c2 , 4

− c1,1 + c1, 2

− c2,1 − c2, 2

c1,7 − c1,8

− c2,7 − c2,8

c1,5 + c1,6

c1,3 + c1,4

c2,1 + c2, 2   − c2, 3 + c2, 4  , − c1,3 − c1, 4   − c2,7 − c2,8  − c2,1 + c2,2   c 2 ,3 + c 2 , 4  . c 2 ,5 + c 2 , 6   − c1,1 − c1, 2 

In order to implement the multiplication of the matrices S (41) , S (42) , S (43) by the appropriate vectors we need to perform only 28 additions for each of these matrix-vector products because the elements ci , j have already been calculated. To calculate the number of additions required when performing matrix-vector multiplication with the matrices S (20) , S (21) , S (22) , and S (23) we introduce the following notation:

p1,1 = −b4 + b10 , p2,1 = b7 − b9 , p1,2 = −b12 + b15 , p2,2 = −b13 + b14 , p1,3 = −b7 − b9 , p2,3 = −b4 − b10 , p1,4 = −b13 − b14 , p2,4 = b12 + b15 . Then the matrix

S (20)

can be represented as follows:

 p1,1 + p1, 2 S (20) =   p2,1 + p2, 2

p1,3 + p1, 4   p 2 ,3 + p 2 , 4 

In order to carry out the multiplication of the matrix S (20) by an appropriate vector we need to perform only 14 additions, namely: – 8 additions which are necessary to calculate all the elements pi , j , i, j = 1, 2, 3, 4 , – 4 additions which are necessary to calculate all the sums pi , j + pi , j +1 , i, j = 1, 2, 3, 4 . – 2 additions arising from the direct matrix-vector multiplication by applying the general rule for matrixvector multiplication. In order to carry out the multiplication of the matrices S (21) , S (22) , S (23) by appropriate vectors we need to perform only 6 additions for each of these matrix-vector products because the elements pi , j have already been calculated:

 p1,1 − p1,2 S (21) =  − p1,3 + p1, 4

− p2,1 + p2,2  ( 2) − p2,3 + p2,4 , S =  p 2, 3 − p 2, 4  2  p2,1 − p2, 2

p1,3 − p1,4   − p 2, 3 − p 2 , 4 (3)  , S2 =  − p1,1 + p1, 2   − p1,3 − p1,4

− p2,1 − p2,2  . − p1,1 − p1, 2 

To calculate elements s0 , s1 , s2 , s3 we need to perform only 4 additions because

s 0 = −b7 + b9 − b10 − b15 = −(b7 − b9 ) − (b10 + b15 ) = − p 21 − c12 , s1 = b7 + b9 − b10 + b15 = (b7 + b9 ) − (b10 − b15 ) = − p13 − c 28 , s 2 = −b7 + b9 + b10 + b15 = −(b7 − b9 ) + (b10 + b15 ) = − p 21 + c12 , s3 = (b7 + b9 ) + (b10 − b15 ) = − p13 + c28 .

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In order to carry out the multiplication of the matrix F2 by an appropriate subvector we need to perform 2 additions. Thus for multiplying the data vector by the quasi-diagonal matrix D30 , we need to perform 88 multiplications and 166 additions. The Fig. 1 shows that the implementation of the remaining part of the algorithm requires only 90 additions. Thus using the proposed algorithm the number of real multiplications to calculate the Dirac number product is reduced threefold compared to schoolbook method of calculation. The number of real additions required using our algorithm is 256. Therefore, the total number of arithmetic operations for proposed algorithm is approximately 30% less than that of the direct evaluation.

5.

Conclusion

In this paper, we have presented an original algorithm that allows us to compute the product of two Dirac numbers with reduced multiplicative complexity. The proposed algorithm saves 40 real multiplications compared to the algorithm [12] and 168 real multiplications compared to the schoolbook algorithm. Unfortunately, the number of real additions in the proposed algorithm is somewhat greater than in the algorithm [12], but the total number of arithmetical operations is still less than in the schoolbook algorithm. For applications where the “cost” of a real multiplication is greater than that of a real addition, the new algorithm is always more computationally efficient than our previously published algorithm, and it is generally more efficient than direct method.

References 1. 2.

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