Sierpinksi-like triangle
mates with
pachinko-like machine
Following selected funnel-shaped dependencies,
a large size-10 triangle recursively breaks down into
a size-4-triangle-like arrangement of medium size-4 triangles.
This amounts to subdivision by a linear factor of 3.
Turns out that most top-row letters have no
effect on the corners that matter here.
o - o - o - o - o - o - o - o - o - o
\ / \ / \ / \ / \ / \ / \ / \ / \ /
o o o o o o o o o
\ / \ / \ / \ / \ / \ / o - o - o - o
o o o o o o \ / \ / \ /
\ / \ / \ / o o o o - o
o - o - o - o - o - o - o is equivalent to \ / \ / is equivalent to \ /
\ / \ / \ / \ / \ / \ / o o o
o o o o o o \ /
\ / \ / \ / \ / o
o o o o
\ / \ /
o - o - o - o o - o - o - o o - - - - - o
\ / \ / \ / \ / \ / \ / \ /
o o o if o - o - o is equivalent to \ /
\ / \ / \ / \ / \ /
o o o - o \ /
\ / \ / \ /
o o o
As a check, the relevant sub-triangles in the examples
do follow the same rules as size-2 triangles.
X Y Z Z X Z X Z X Y X - - Z - - X - - Y X - - - - - - - - Y
Z X Z Y Y Y Y Y Z \ / \ / \ / \ /
Y Y X Y Y Y Y X \ / \ / \ / \ /
Y Z Z Y Y Y Z Y - - Y - - Z \ /
X Z X Y Y X ---> \ / \ / ---> \ /
Y Y Z Y Z \ / \ / \ /
Y X X X Y - - X \ /
Z X X \ / \ /
Y X \ / \ /
Z Z Z
Y Z X Z X Z Y Z Y Y Y - - Z - - Y - - Y Y - - - - - - - - Y
X Y Y Y Y X X X Y \ / \ / \ / \ /
Z Y Y Y Z X X Z \ / \ / \ / \ /
X Y Y X Y X Y X - - X - - Y \ /
Z Y Z Z Z Z ---> \ / \ / ---> \ /
X X Z Z Z \ / \ / \ /
X Y Z Z X - - Z \ /
Z X Z \ / \ /
Y Y \ / \ /
Y Y Y
Any top-row combination of x,y,z can be obtained by beginning with all x
and converting one position at a time.
Thinking of x,y,z as 0,1,2 with modulo-3 wraparound addition,
here is a summary of how converting any position
maintains the 3-corner rule just the same
in a medium size-4 triangle as in the equivalent small size-2 triangle.
(0 = no change, 1 = change to the next letter,
2 = change to the twice-next letter = change to the previous letter)
drop a 1 on a corner, as in a pachinko machine
|
V
0 0 0 0 1 0 0 0
\ /
0 0 0 2 0 0 1 0
---> \ / which is equivalent to \ /
0 0 1 0 2
\ /
0 2
drop a 1 adjacent to a corner, like a pachinko
|
V
0 0 0 0 0 1 0 0
\ / \ /
0 0 0 2 2 0 0 0
---> \ / \ / which is equivalent to
0 0 2 1 0
\ /
0 0
Thanks to symmetry, only the above two cases and
the following 6 representations need to be considered.
(Actually, there are only 2 truly unique cases below,
one where the original top corner letters are the same
and one where they differ.)
.- - - - - - - - - -. .- - - - - - - - - -.
. x x y x . x x z x
. \ / ---> \ / . \ / ---> \ /
. x z . x y
. .
. .- - - -. .- - - - -. . .- - - -. .- - - - -.
0 0 1 0 1 0 0 0 2 0 2 0
\ / + \ / = \ / \ / + \ / = \ /
0 2 2 0 1 1
.- - - - - - - - - -. .- - - - - - - - - -.
. y x x x . y x z x
. \ / ---> \ / . \ / ---> \ /
. z x . z y
. .
. .- - - -. .- - - - -. . .- - - -. .- - - - -.
1 0 2 0 0 0 1 0 1 0 2 0
\ / + \ / = \ / \ / + \ / = \ /
2 1 0 2 2 1
.- - - - - - - - - -. .- - - - - - - - - -.
. z x x x . z x y x
. \ / ---> \ / . \ / ---> \ /
. y x . y z
. .
. .- - - -. .- - - - -. . .- - - -. .- - - - -.
2 0 1 0 0 0 2 0 2 0 1 0
\ / + \ / = \ / \ / + \ / = \ /
1 2 0 1 1 2
The effects of each addition ripple down for either no effect
or for a combined effect of changing two corners
in opposite directions.
This preserves the required condition where all three corners
are either the same or all different,
Bonus answer, incompletely justified:
The recursive effect clearly scales up by factors of 3
to include all top rows of size $n = 1{+}3^k$.
Without explanation, the following diagram is a tilted
representation of changing a single letter.
It convinced me that this is a 3-fold Sierpinski triangle
and that no further sizes will work.
Each marked diagonal shows that changing a top letter (by 1)
will result in a change at the diagonal's level
only if the top change is at a corner.
The resultant change (by 2, same as -1)
will be opposite in value.
n = 2 4 10 28
. / . / . . . . . / . . . . . . . . . . . . . . . . . / . . . . . . . . . . . . . . ...
1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 ..
. 2 2 / 1 1 . 2 2 / 1 1 . 2 2 . 1 1 . 2 2 . 1 1 . 2 2 / 1 1 . 2 2 . 1 1 . 2 2 . 1 1 . ...
/ 1 / . 2 . . 1 / . 2 . . 1 . . 2 . . 1 . . 2 . . 1 / . 2 . . 1 . . 2 . . 1 . . 2 . . ..
. 2 1 2 2 1 2 / . . 1 2 1 1 2 1 . . . 2 1 2 2 1 2 / . . 1 2 1 1 2 1 . . . 2 1 2 2 1 ...
/ 1 1 . 1 1 / . . . 2 2 . 2 2 . . . . 1 1 . 1 1 / . . . 2 2 . 2 2 . . . . 1 1 . 1 1 ..
. 2 . . 2 / . . . . 1 . . 1 . . . . . 2 . . 2 / . . . . 1 . . 1 . . . . . 2 . . 2 ...
. 1 2 1 / . . . . . 2 1 2 . . . . . . 1 2 1 / . . . . . 2 1 2 . . . . . . 1 2 1 . ..
. 2 2 / . . . . . . 1 1 . . . . . . . 2 2 / . . . . . . 1 1 . . . . . . . 2 2 . ...
. 1 / . . . . . . . 2 . . . . . . . . 1 / . . . . . . . 2 . . . . . . . . 1 . . ..
. 2 1 2 1 2 1 2 1 2 2 1 2 1 2 1 2 1 2 / . . . . . . . . 1 2 1 2 1 2 1 2 1 1 2 ...
/ 1 1 . 2 2 . 1 1 . 1 1 . 2 2 . 1 1 / . . . . . . . . . 2 2 . 1 1 . 2 2 . 2 2 ..
. 2 . . 1 . . 2 . . 2 . . 1 . . 2 / . . . . . . . . . . 1 . . 2 . . 1 . . 2 ...
. / . . . . . . . . . . .
. / . . . . . . . . . . . .
Further musing:
Seems like some hay could be made from the fact that these triangles
obey the same rules regardless of which side is considered top
and which direction is considered down.