Network operator matrix

Example

Consider the following mathematical expression

The graph for Expression (1), is presented in Fig. 1.

On the edges we place unary operations

ρ 1 ( z ) = z ;
ρ 3 ( z ) = − z ;
ρ 6 ( n ) = { ε − 1 , if  z > ln ⁡ ε e z ,  otherwise ;
ρ 12 ( z ) = sin ⁡ z ;

In the inner and sink nodes we place binary operations
χ 0 ( z ′ , z ″ ) = z ′ + z ″ ;
χ 1 ( z ′ , z ″ ) = z ′ × z ″ ;

Expression 1 can be presented in the PC memory as a NOM

Any mathematical expression can be presented as a network operator matrix.
To calculate the mathematical expression by the network operator matrix the node vector is used.
Each component of node vector corresponds to the nodes of the network operator graph. Initially each component is equal to the argument for the given node or the unit element of binary operation.
For addition and multiplication the number of operation equals its unit element. For addition it is 0, for multiplication it is 1. The node vector for the given example

z = [ x 1   q 1   x 2   0   1   0   1   0 ] T

The calculation by matrix Ψ is performed for nonzero nondaigonal elements ψ ( i , j ) ≠ 0 by

z j = χ ψ j , j ( z j , ρ ψ i , j ( z i ) )

Follow the rows of the matrix.
For the Row 1 we have ψ 1 , 4 = 1 , that is i = 1 , j = 4 , ψ j , j = ψ 4 , 4 = 0 .

Take arguments from z = [ x 1   q 1   x 2   0   1   0   1   0 ] T

z 1 = x 1 , z 4 = 0 then

z 4 = χ ψ 4 , 4 ( z 4 , ρ ψ 1 , 4 ( z 1 ) ) = χ 0 ( z 4 , ρ 1 ( z 1 ) ) = 0 + x 1 = x 1

Further in Row 1 we have ψ 1 , 5 = 1

z 5 = χ ψ 5 , 5 ( z 5 , ρ ψ 1 , 5 ( z 1 ) ) = χ 1 ( z 5 , ρ 1 ( z 1 ) ) = 1 × x 1 = x 1

ψ 1 , 8 = 12

z 8 = χ ψ 8 , 8 ( z 8 , ρ ψ 1 , 8 ( z 1 ) ) = χ 0 ( z 8 , ρ 12 ( z 1 ) ) = 0 + sin ⁡ x 1 = sin ⁡ x 1 .

As a result after Row 1 we have z = [ x 1   q 1   x 2   x 1   x 1   0   1   sin ⁡ x 1 ] T .

For the Row 2 we have ψ 2 , 5 = 1
z 5 = χ ψ 5 , 5 ( z 5 , ρ ψ 2 , 5 ( z 2 ) ) = χ 1 ( z 5 , ρ 1 ( z 1 ) ) = x 1 q 1

then follow the rows

ψ 3 , 6 = 3
z 6 = χ ψ 6 , 6 ( z 6 , ρ ψ 3 , 6 ( z 3 ) ) = χ 0 ( z 6 , ρ 3 ( z 3 ) ) = 0 + ( − x 2 ) = − x 2

ψ 4 , 8 = 1

z 8 = χ ψ 8 , 8 ( z 8 , ρ ψ 4 , 8 ( z 4 ) ) = χ 0 ( z 8 , ρ 1 ( z 4 ) ) = sin ⁡ x 1 + x 1

ψ 5 , 7 = 1

z 7 = χ ψ 7 , 7 ( z 7 , ρ ψ 5 , 7 ( z 5 ) ) = χ 1 ( z 7 , ρ 1 ( z 5 ) ) = 1 x 1 q 1 = x 1 q 1

ψ 6 , 7 = 6

z 7 = χ ψ 7 , 7 ( z 7 , ρ ψ 6 , 7 ( z 6 ) ) = χ 1 ( z 7 , ρ 6 ( z 6 ) ) = x 1 q 1 e − x 2

ψ 7 , 8 = 1

z 8 = χ ψ 8 , 8 ( z 8 , ρ ψ 7 , 8 ( z 7 ) ) = χ 0 ( z 8 , ρ 1 ( z 7 ) ) = sin ⁡ x 1 + x 1 + x 1 q 1 e − x 2 .