Let's first define some terminology. We'll say that
- two rectangles are at the same elevation with respect to any given axis iff there is a dimensional hyperplane perpendicular to the given axis that crosses through both (note that this is not a transitive property),
- a -dimensional rectangle visibility graph forms a grid iff, when drawing all visibility lines parallel to any one axis, the connected components of the resulting graph satisfy the following properties:
- for any two components and
- for every component, there is a line parallel to the axis that passes through all of its hypercubes,
- for any two components and the 1:1 mapping that preserves this ordering satisfies the following two properties:
- the graph formed by under this mapping goes to the graph formed by ,
- iff and are at the same elevation,
- a hypercube grid is a grid where all components mentioned above are hybercubes.
I believe following holds true, but I do not have a proof yet:
If
is representable as a rectangle visibility graph in
dimensions, then it is also representable as a
hypercube grid in
dimensions.
If this statement holds, and if
were representable with
then
would be representable as a bar visibility graph with a line passing through all bars.
For the sake of contradiction, assume this to be the case. Pretend that the following graph is a representation of
Let
be the "protrusion" of each line out from the left side of the shaded region, for example given the pretend graph above,
Likewise let
be the protrusion of each line on the right side of the shaded region, i.e. in this example
Now pick two large numbers
and
and for every pair
draw a rectangle with two opposite corners at
This would then be a valid representation of
However, by
post #87 above,
is not representable in 2 dimensions, a contradiction.
Therfore, either my statement above is false, or
is not representable with
or both.
I believe my statement is true however, and that
is therefore not representable with