## Hypercube offsets

In my previous posts, each drawing consisted of two offset copies of the previous drawing. For example, here are the drawings for $n=3$ and $n=4$:

You can see how the $n=4$ drawing contains an exact copy of the $n=3$ drawing, plus another copy with the fourth red element added to every set. The second copy is obviously offset by one unit in the vertical direction, because every set gained one element and hence moved up one row. But how far is the second copy offset horizontally? Notice that it is placed in such a way that its second row from the bottom fits snugly alongside the third row from the bottom of the original copy.

In this case we can see from counting that the second copy is offset five units to the right of the original copy. But how do we compute this number in general?

The particular pattern I used is that for even $n$, the second copy of the $(n-1)$-drawing goes to the right of the first copy; for odd $n$ it goes to the left. This leads to offsets like the following:

Can you see the pattern? Can you explain why we get this pattern?

Assistant Professor of Computer Science at Hendrix College. Functional programmer, mathematician, teacher, pianist, follower of Jesus.
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### 2 Responses to Hypercube offsets

1. Theo says:

Double plus one, double minus one?

• Brent says:

Yes — though I think it becomes a bit more elegant if you think of the rightward offsets as positive and the leftward ones as negative.