The Tale of Three Triangles 

It's not clear at all what the heck you mean at 8:08 how does m plus n choose m equals waterfal which isnjust a coordinate point..notnthe same thing..unless you meant something else..

@@leif1075 It's applying the coordinate system to the entries in the triangle with the top as [0,0], the second row as [1,0] and [0,1], the third row as [2,0], [1,1], and [0,2], and so on. Notice that in row 0, m+n = 0. In row 1, m+n = 1, etc. To go down m to the left and n to the right, you reach row m+n, entry n. In the triangle, the value of the entry at row r, entry e is r choose e, hence m+n choose n. All it's doing is describing the triangle in terms of the created coordinate system by mapping each entry to a point on the grid of said coordinate system. It's exactly the same thing as taking a 5x5 square of items and assigning each item x and y coordinates in the Cartesian system.

@@Qazqi Sorry but what you said doesn't sound at all like wjwt he said in the video..and what rows?? A triangle has 3 vertices and the emlpty space within the triangle and that's it.

Its all NAND!!! Jokes aside this was an awesome video. (And the music selection was on point as well :) )

Was this video made for SOME1? You are awesome aswell. The quality of content in SOME is great, it's almost as if every good teacher was just waiting to be encouraged to teach.

What competition? Miss Universe?

@@leif1075don't you know about SOME1. It's sort of a competition organised by 3b1b to encourage online maths related content. The "winners" are shown in a 3b1b video and get prizes I think.

Are you saying that I can't become General Grevious and flip it off? Lame

"Base eight is just like base ten really, if you're missing two fingers." - Tom Lehrer

1000-7

@@plasmaballin every base is base 10 if you think about it

@@1224chrisng what about bases 11 to ∞?

Actually these explanations are not really new. The Sierpinski's triangle can be found in many other places. For example using a Wolfram's automaton or by drawing truth tables of logical operators. These appearances were studied by many mathematicians. The truth is that it represents a property of a special family of groups. The more geometric examples (the classic definition and the chaos game) are related to symetries in a group of geometric transformations. And the more numerical examples (Pascal's triangle, automaton and logic operators) are linked to a group that comes from modular arithmetic.

the work is indeed illuminating, but coordinate system is just regular Cartesian system with one axis a bit rotated over origin. In 3blue1brown linear algebra series this coordinate transformation is described very nicely. The only novelty here is to put origin on top and orient axes downwards. But yeah, this is only because I know about coordinate system transformations, for those who don't this indeed looks like an invention.

heh, illuminating. Watch those puns, clank

@@uprola I think it's really just functioning as the first quadrant of R^2. Angles aren't involved, so it doesn't matter what angle the axes form.

What this video shows is how you can totally master a math problem if you have the right ideas. It's really valuable, even if it isn't a novel result.

Huge respects :)

Also, I’d like to note that the “leading decimal is always 0 for points inside the triangle” is still true for (1, 0) and (0, 1) if you instead write them as (0.11111111..., 0) and (0, 0.11111111...). In general, when working with functions defined by decimal expansions (such as the “is_in_triangle” function defined in the video), you usually have to specify one of two rules: “terminate the expansion if possible” or “always use an infinite expansion”. Although I haven’t thought it through completely, I suspect that the latter is what is needed for your triangle functions to work.

Okay, having thought it through a bit more, I don’t think _either_ method of defining decimal expansions works. Take the example of (½, ½). It is rotationally symmetric to (½, 0), which is definitely a member of the fractal, so (½, ½) “should” be in the fractal. However, both of the expressions (0.1, 0.1) and (0.01111111..., 0.01111111...) have some binary place values with ones in common. Thus, I propose that a point is in the fractal if there exists ANY binary representation of its coordinates which share no place values; (0.1, 0.01111111...) being one such representation for our example (½, ½).

I think this arises because of the way we compute the quantity "1-x". In decimal, it's pretty easy : Assuming x is less than 1, just take the decimal expansion as a number, take its 9's complement, add 1 to the last digit (if it exists), and put the decimal point back in. 0.276 becomes 276, then 723, then 724, then 0.724. If you neglect to add 1, it's kinda like doing an approximation : You know 1-0.276 is around 0.723. If a decimal expansion has n digits before terminating, that approximation has error exactly 10^-n. You can apply this to any base, including binary. In base b, you do the same thing, except you take the (b-1) complement, and the error is exactly b^-n. The problem is that every number with a terminating decimal expansion can be thought of as having an infinite number of zeros after said expansion. If you add a finite number of zeros the method does work, for example 0.27600 ==> 27600 ==> 72399 ==> 72400 ==> 0.724. But if you add an infinite number of zeros, then the last digit doesn't exist so we can't add 1 to it, and we end up with trailing zeros. 0.276000... ==> 276000... ==> 723999... ==> ? ==> 0.723999... In this case the "approximation" of not adding 1 goes to 0 so it's "infinitely good", if not completely equal.

In the case of binary, you take the 1's complement, so for values with infinitely many digits you're completely sure that no digits are in common, but when the values terminate there's the problem of the last 1. Considering that values only terminate if they are a multiple of a power of 2 (in our case a negative power of 2), I think this corresponds to the boundary of the completely black parts of the triangle.

I think the shape produced by that is subtly different, like Sierpinski's triangle, but with every other cell a zero. Since xor is the same thing as mod-2 addition, we should see mod-2 Pascal's triangle if we only consider following cells: _ _ _ x _ _ _ _ _ x _ x _ _ _ x _ x _ x _ x _ x _ x _ x A cell receives no information from the cell directly above it, meaning that the "hole" in the first row is a zero. You can then by induction extend this for all further "holes" in the structure, since they can only receive information from outside of the triangle (initialized to zeroes) or from the holes one row above (proven to be zeroes earlier)

​@@ZeroViruzz The value at a place p on row r is the xor of the values at places p-1 and p+1 on row r-1. It actually does give you the Sierpinski's triangle, slightly stretched horizontally: 1. From the wonderful video above we know that the even/odd Pascal triangle gives you the Sierpinski triangle. 2. You can also define Pascal's triangle by making each entry the sum of its neighbours in the previous row. 3. The "xor of neighbours" rule is the same rules that generates the even/odd Pascal triangle.

​@@FineDesignVideos Yes, if you only look at every other cell, alternating each line, that would be the Sierpinski's triangle. However I disagree that you can call the result a Sierpinski's triangle when it has additional holes Sierpinski's does not

@@ZeroViruzz I think it would be a bit unfair to not call this a Sierpinski triangle. It's the same in structure as the Pascal triangle. The only difference is that with the Pascal triangle you can put entries at half positions (first row's 1 at position 0, the second row's 1s are at -1/2,1/2, third row's 101 at -1,0,1). A discrete grid can't do this, the best it can do is to multiply each position by 2 so that every position is integral and then put the same entries there, and that's what the cellular automaton manages.

This is just the result of pascal's triangle mod 2, since an XOR gate is identical to single bit binary addition without a carry. An infinite lattice of XOR gates, with a single logical 1 as input, will send logical 1 through the lattice in the shape of a Sierpinski triangle, because it performs the exact same operation as coloring all the odd numbers in an infinite Pascal's triangle.

Something that helps me are the subtitles, even the automatic ones. idk why some videos don't have them

@@Lokonarua It used to be the case that viewers could just create subtitles for youtube vids. This allowed a lot of those vids to be subtitled in all sorts of different languages and they were commonly known as "community captions". Unfortunately youtube got rid of them I think a year or two ago which I think was a really shitty decision esp for speakers of different languages and the deaf/HoH community. I really have no clue why yt made such a dumb decision.

Chaos usually finds a way of being astoundingly elegent.

I was thinking when the bitwise AND of the two components is 0, but they're just the same thing. Now to see what happens if we try to plot (x, y) where x & y == 0 in different coordinate systems.

This is awesome! It really lets me 'feel' the connection/reason at an intuitive level.

It also appears in the Nim Sum table of two integers, as it's another way of defining XOR.

Hey they don't need to understand it. I've made plenty of math shirts. The trick is to not make them look like math shirts; then people come up and ask what your shirt is. They thought they were gonna talk about art and then -- boom -- math ambush.

This is very true. I could've fixed it; unfortunately I rushed the ending since the deadline was coming up and did not perfect those details. Thanks for the feedback!

Hello, fellow nerd

I think it's more likely to be related to NAND, since the combinations that work are 00, 01, and 10.

Yes, XOR is basically the sum modulo 2. So it retains even/odd properties

Look up Rule 90 cellular automaton. The rule is essentially "new cell = left XOR right". If you start it with a lone live cell, you get the Sierpinski triangle.

It did. Popped up in my feed, no doubt because i have watched several videos in 3B1B's selection. So I did sort of self select it. All hail our Algorithm™️ Overlord! PS Thanks to 3B1B for the SOME. If only these sorts of videos had been around when I was young. We are lucky to live in a time where such riches are to be found.

14 Ganondorfs disliked

I programmed the chaos game version in my uni days. Also tried modifying the program to use four and five points. Four just filled in the square. Five got a fractal pentagon.

I used ManimCE to produce the raw animations programmatically and DaVinci Resolve to chain them together. Both are free but have a nontrivial learning curve.

@@robintruax612 thanks! great work

Technically the chaos game (or its deterministic version) isn't (necessarily) the same as the gasket either. If you start at a point that's not already in the gasket, you will *never* actually land inside the gasket - you'll just get closer and closer as you take more steps. Also, the first deterministic version you stated depends on every sequence of base 3 digits eventually appearing; while it is strongly believed that this is true for pi and e, it's not actually known to be true!

@@japanada11 Yeah, that's why I had it start on the 3 corners.

The Rule 90 1D cellular automaton also has XOR in its update function and produces the same pattern if started with a single live cell.

Play the chaos game with any triangle (doesn't need to be equilateral) and it will produce a Sierpinksi triangle with the same 'shape'. The critical point is that the triangle has three vertices corresponding to 00, 01, and 10.With more vertices, something else has to happen.

@@alex.mojaki Of course, I know that. I just don't get how that proof proves it. Every point has a non-0 probability of happening and so if you keep generating points for long enough then you'll get them. Wouldn't this proof have to say why points within the Sierpinksi triangle are more likely to happen than points outside?