Yesterday was my 63rd birthday. I bought a strawberry dream cake at Loblaw's and we ordered felafels from the place around the corner. (Prices are going up these days!)
Through social media I heard about readcomiconline.li . It has a lot of comic books I remember from my childhood, stuff like Donald Duck.
I was thinking about dimensions. With zero dimensions, you have a point. With one, you have one segment with two points on the end. With two, you have one square with four side segments and four corners (except that if you take the sides individually, you have a total of eight end points). With three dimensions, you have one cube with six face squares, twelve edges (or 24 if you take the faces individually), and eight vertices (or 24 points). So I made a table:
Dimensions
0 1 2 3 4 5
Points 1 2 4 8 16 32
(ends, corners, vertices)
(1) (2) (8) (24) (64) (160)
Segments 1 4 12 32 80
(sides, edges)
(1) (4) (24) (96) (320)
Squares 1 6 24 80
(faces)
(1) (6) (48) (240)
Cubes 1 8 40
(sub-cubes)
(1) (8) (80)
Tesseracts 1 10
Super-tesseracts 1
How did I determine what the numbers must be for four and five dimensions? By seeing the common patterns! For n dimensions you have 2^n points. A segment has 2 ends, a square has four corners, a cube has 8 corners, suggesting a linear increase (2, 4, 6, 8, 10...). To figure out the bracketed numbers for each dimension, take the unbracketed numbers from the previous dimension and multiply by 4, then 6, then 8... Viewing successive diagonal lines, the bracketed numbers will surpass their unbracketed counterpart by 2, then 3, then 4...
In addition if you look at the unbracketed numbers at dimension n, their total sum will be 3^n!
That's the sort of thing I sometimes think about when my mind is wandering...
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