Mathematical Justification of the LPS Algorithm

The above algorithm processes all $G_n\times G_n$ image pixels for the following reason. Let

$$\left[ \!\! \begin{array}{c} i \\ j \end{array}\!\! \right]... ... \left[ \!\! \begin{array}{c} x \\ y \end{array}\!\! \right]$$ (8)

We will show that

T_ij = x (9)

Multiplying out the above matrix formula, we have

  

T_ij = iG_n-2 + jG_n-1 (10)

= (G_-n+1x + G_n-3y)G_n-2

+ (G_-n x+ G_n-2y)G_n-1 (11)

= (G_-n+1G_n-2 + G_-nG_n-1 )x

+ (G_n-3G_n-2 + G_n-2G_n-1)y (12)

= (G_-n+1G_n-2 + G_-nG_n-1 )x (13)

+ (G_n-3 + G_n-1)G_n-2y (14)

The values in Eq. (8-14) are all modulo G_n. One of the factors in (14) satisfies

$$G_{n-3} + G_{n-1} = G_n \equiv 0 \pmod {G_n}$$ (15)

by (3); so T_ij is independent of y. The coefficient of x in (13) satisfies

$$G_{n-2} G_{-n+1} + G_{n-1} G_{-n} \equiv 1 \pmod{G_n}$$ (16)

for all n, because

G_n-2 G_-n+1 + G_n-1 G_-n + G_n G_-n+2 = 1 (17)

for all n; this is easily established by mathematical induction. Finally, because three consecutive terms of the G sequence are relatively prime (5), $M \left[ \!\! \begin{array}{c} 0 \\ y \end{array}\!\! \right]$ takes G_n distinct values for $0 \leq y<G_n$. Linearity allows us to conclude that $\left[ \!\! \begin{array}{c} i \\ j \end{array}\!\! \right]$ ranges over all $G_n\times G_n$ pixel positions.