New video(s): Binary, Hanoi and Sierpinski

Hum... The fractals? Cool! ???

That is indeed a beautiful relation. I feel like there is already ample internet coverage of this fact though :)

3blue1brown

The moment I saw "binary" and "sierpinski" I thought you were going to share another related relationship. That is, a binary cellular automata in a triangular grid using XOR generate sierpinksi's triangle. That is, imagine an infinite triangular grid filled with zeros. Then if you place a 1 and propagate it out following an XOR rule in one direction you get the triangle. Each triangular cell gets its value from the XOR of the two cells above it. So the first row is "1" and then the next row is "11" and then the next row is "101" and then the next row is "1111" and the next row is "10001" and so forth. I discovered this my freshman year of college and at first I thought it was a major novel discovery. I later learned that cellular automata like this have been explored a lot so surely many others have also noticed this.

It get's easier, a lot easier. You now have not just two places to place the first ring, but three. So you could for example just put away the first ring on the 4th peg, and solve the rest as you would with 3 pegs, just without that first ring. Trivially, with each additional peg, you would have one ring less to solve for. As there really isn't any room for inefficiency, there isn't really also any room for improvement - unless constrained. Anything else would add the make way for inefficiencies, and trivialize the whole thing.

In the second video, when you say "it's pretty fun to sit back and watch this play out" it seems like it would be even more impactful to maintain a color-coding of the tower blocks throughout the movements that correspond to the same colorization of the corresponding ternary place in the counter. Same thing would apply to the animations for the binary version in video 1.

Impressive relationships... I was wondering what happens if you have more than 3 pegs. Does the problem gets easier to solve or more difficult?

MedNait

It's really neat that in the unconstrained version of the problem, you can take the bottom side of the triangle and solve the problem in 2^n - 1 steps. But the constrained version forces you to tour the entire triangle, and take 3^n - 1 steps.

Max Goldstein

You know Keith?! He's a wonderful lecturer; I take all his classes that I can fit into my requirements. You two would make an incredible duo teaching a class.

I've never seen that visual explanation of base numbers before, and it's almost upsetting that I haven't, because it makes the whole thing so obvious and intuitive! As a tangent, the discussion of binary reminded me of John Gustafson's Unums; specifically Unums type Two and it's reversible division trick. I wonder what you think of it? <a href="http://ubiquity.acm.org/article.cfm?id=3001758" rel="nofollow noopener" target="_blank">http://ubiquity.acm.org/article.cfm?id=3001758</a>

Job van der Zwan