# How do you draw 4D, 5D, 6D hypercubes not topologically correct?

Asked by inunsure (423) April 25th, 2012

So I dont care of its topology but I want to make sure the right vertexes are connected to the right vertexes and how many to have and how many lines.

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Going from one dimension to the next is an iterative process. Start in zero dimensions with a point. To go from zero dimensions to one dimension, take two points and join them with a segment. In two dimensions, draw two segments and join corresponding points. You get a cube by joining corresponding points of two squares. A tesseract is obtained by connecting corresponding points of two cubes. You can continue this way for 5 and 6 dimensions, though it would be difficult to visualize. The number of points doubles each time. Wikipedia article.

1) I don’t think you mean topological here. Are you sure that’s the right word?
2) What @LostInParadise said is correct. Another way to approach that is like this:
For an n-cube you have 2^n points.
Each point is represented by a tuple: (x1, ..., xn) where each xi = 0 or 1.
Two points are connected if they are the same except for one of the x’s. More precisely: if a = (a1, ..., an) and b = (b1, ..., bn) then there is an edge between a and b if there is only one i for which ai != bi.
E.g. using a 4-cube
a = (0, 1, 1, 0)
b = (0, 1, 0, 1)
c = (1, 1, 1, 0)
There is an edge between a and b (since only a3 != b3) and between a and c (since only a1 != c1) but there is no edge between b and c (since b1 != c1, b3 != c3 and b4 != c4).

roundsquare (5512)

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