Reduction

Figures with
small
values for N and D are quite simple, comparable to situation a). On the
other
hand, figures with big values may become fairly complex. But one can
observe
that such complex figures have roughly simpler forms with some small
pattern on
top of it. In the sequel we will show how complex figures relate
mathematically
to simpler figures.

Consider
situation
b) at the Analysis section. The chunk R^{K}
L^{K} can be
considered as a small pattern inside the figure. The position
before and after drawing this pattern might be changed, but the
orientation
remains the same. However, the other 2 types of chunks do change the
orientation to the right (for R^{K+1}
L^{K}) or to the
left (for R^{K}
L^{K+1}). A new,
reduced figure can be constructed by ignoring the chunks R^{K} L^{K}, and by
replacing chunk R^{K+1}
L^{K} with a
single direction R,
and similarly replacing R^{K} L^{K+1} with L.
This new
figure has a recipe with N’ = E directions.

We will
prove,
with some geometrical support, that the new denominator D' equals D mod
(2*N’),
and that the replaced chunks correspond to the recipe based on pair
(N’, D’).

Consider a 2-dimensional
coordinate
system with a horizontal x-axis and a vertical y-axis. Define the points
(t*D, t*D) where t
runs from 0 to
N
. The
first N points
correspond to the points for constructing the recipe. All these points
lie on a
straight line at equidistant pieces. Along
the x-axis we focus on multiples of D.
Therefore, we can scale the points to (t, t*D). Along the y-axis we
focus on
multiples of N. Also this axis is scaled, but now by a factor N,
yielding
points (t, t*D/N) .
The points still lie on a straight line at equidistant pieces running
from (0,0) to (N,D).
(Note the
analogy of drawing a line on a graphics display.) Consider the D rows
between
coordinates y and y+1 where y runs
from 0 to D-1. Each row covers
either K points
or K+1 points from
the line. A row with
even y-coordinate
contains
points that yield an R direction, and a row
with odd y-coordinate contains points
yielding an L direction. An even row
and an odd row together relate to a chunk;
so, it
is clear that a recipe has D/2 chunks.

The rows cut
the line in
equal parts. Each part has a length along the x-axis of N/D = K + E/D . The right hand
side has
2 terms: K is the minimum number of points that are always present in a
row; the term E/D is
responsible for the (K+1)^{th}
points after accumulation of some rows.
Considering N = N(K) as a function of K, the row location of the
additional
point does not depend on K, because E/D and K are independent. So, we
can take
K=0, and consider the problem of the E additional points (t, t*D/E)
that are
located equidistant on a line from (0,0) to (E, D). Now, the line is
transformed, such that from each point the y-value t*2 is subtracted.
This is
possible since E<D/2 . All the transformed points remain
equidistant at a
straight line. Furthermore, a point at an even row lands again
in an even row, and the
same holds for the odd rows. So, the directions of the points are
preserved.
The end coordinate of the line becomes (E, D-2*E) . The transformation
is
repeated until the end coordinate becomes (E, D mod (2*E)), being the
end
coordinate (N’, D’) of the reduced line. The
configuration of the line from
(0,0) to (N’, D’) is analoguous to the
configuration of the original line from
(0, 0) to (N, D) . Therefore, we conclude that the reduced line
corresponds to
a figure based on the numbers N’ and D’.

Now, consider situation c) being a bit trickier because the reduced number of directions N’ equals (F+1)/2 + (F-1)/2 = F . We will show that the reduced value D’ equals D mod (2*F). For this proof the line configuration is transformed extensively. As in situation b) the line can be reduced to: (0, 0) -> (E, D) . Subtract the line (0, 0) -> (E, 2*E), so that all directions are preserved. This gives (0, 0) -> (E, D-2*E) with an end point below the x-axis. So, add the constant 2*E-D to all y-coordinates yielding (0, 2*E-D) -> (E, 0) . Applying a mirror transformation with respect to x = E/2 results again in a line through the origin: (0, 0) -> (E, 2*E-D). Since E> 2*E-D we can reduce this line, just as we reduced the original line by subtracting D from N . This gives the line (0, 0) -> (D-E, 2*E-D). From the endpoint's y-coordinate we can subtract multiples of 2*(D-E) . Since we can also add once 2*(D-E), the y-coordinate can be reduced to D mod (2*(D-E)) = D mod (2*F).

The conclusion is that we can reduce a figure with N and D to a reduced figure with N’ and D’ as follows. Define E = N mod D, so that:

if E < D/2, then N’ = E and D’ = D mod (2*N’) if E > D/2, then N’ = D-E and D’ = D mod (2*N’) otherwise no reduction is possible.