+PBS Infinite Series I'm sure Trixelian actually meant the last minute before the ad and the questions from last week... you know the part where you basically told us the entire video before was a lie?
I really liked the fact that throughout I'm sitting thinking, "yeah, but parallel lines fail this", then suddenly you pause to say, really it's complex projective. So fair enough, all my quibbles are silenced, then you pulled a face as if to say "shhh, I know what I'm doing". It really made me smile.
1. The contraction step was brilliant! Even coming up with it is difficult. 2. The fact that "multiplying" the indexes from that map [0 1 0 1], through that triangle produces anything useful at all is mindblowing. How would anyone even come up with that? And its distributive? Seriously,... that's gold.
It's not hard to get the number of blue and red triangles as well as parallelograms from the fact that the triangle can be contracted as shown in 4:30. But I'm missing the proof that the triangle can always be contracted like that without ever creating holes. I believe that it is true, but somehow I feel that you jumped over a part of the proof that is not completely obvious.
There are only three possibilities for what can lie directly next to a red triangle on a particular side: Either it's another red triangle, a parallelogram or nothing / "the outside". The last possibility can never create holes, so we'll skip that. 1) If it's a red triangle, then the two triangles we are looking at are congruent and both shrink at the same speed, meaning they will never "shear" and produce a hole. 2) Since the parallelograms here always have exactly two red sides that are opposite each other, every parallelogram forms a sort of "tunnel" to another red triangle or the edge of the whole big triangle (notably, they cannot change direction: If the parallelogram has Left-Right orientation (meaning the red sides point left/right), every other parallelogram touching it will have that orientation as well. Therefore, every parallelogram can shrink at the same speed as the two triangles it connects to, or the one it connects to in case it connects to the outside. There will be no shearing between those two edges either. And blue edges do not shrink at all, they're just moved together with the shapes they belong to. So no shearing there either. I hope I covered all possible cases that could cause holes in that construction.
Challenge: We know that the sum of the areas of the parallelograms must be equal to the area of the whole minus the areas of the triangles: N^2 - (K^2 + (N-K)^2) = N^2 - (K^2 + N^2 - 2*N*K + K^2) = 2*N*K - 2*K^2 = 2*K*(N-K). Every palallelogram has area 2, so the number of palallelograms must be K*(N-K), meaning the number of red edges multiplied by the number of blue edges.
I just signed up for curiositystream because of this video, I kept finding the adds on youtube interesting. I can't believe someone actually created such an excellent source of science videos. The first one I watched (The Ultimate Formula) was the best I have ever scene on the subject of the standard model. Thanks for convincing me to sign up.
why does the triangle puzzle work? i realize the answer might be complicated, but is there some way of relating the rules of the puzzle to the algebra it describes? are there other puzzles that correspond to other algebras? what about algebras that arise in space with a different topology? do those have their own puzzles or conjectures? what if the space is curved?
Interestingly, the Grassmanian Gr(2,4) of planes in 4 space (the topic of this video) actually *is* curved, which is precisely why four lines (linear constraints) determine 2, not 1, element of the Grassmanian. When we embed Gr(2,4) into a higher dimensional projective space via the Plucker embedding, we obtain a variety of degree 2 (in the same way that a conic section has degree 2 in the plane: if you intersect it with enough hyperplanes, you end up with 2 points). The general algebra corresponding to an algebraic variety is called its Chow ring, and is closely tied into the cohomology of the space.
That is absolutely mind blowing! I'm baffled at how those dimension indices form a set with a distributive operation! And the product is calculated through some sort of puzzle?! Mind blowing! (Oh, and about the bonus question: Take N as the number of blue edges and K as the number of red edges. Since the total area in triangle units is (N+K)^2. The blue triangles fill an area of N^2 and the red triangles fill up K^2 triangle units, leaving (N+K)^2-N^2-K^2 to be filled in with parallelograms. However, the parallelogram has an area of 2, so the number of parallelograms is ((N+K)^2-N^2-K^2)/2. )
I remember how 2 years ago I used to be an ardent follower of PBS spacetime when I came to know about this channel, and that it was no longer making videos. Eventually through channels such as 3b1b etc. Now I have lost my interest in physics and I am teaching myself mathematics. And now I feel disheartened that this channel doesn't make any more videos .
To take a stab at the challenge (6:48): Imagine trying to make an ever larger equilateral triangle using solely smaller, congruent equilateral triangles. If you want to add another "layer" (make it larger) you would have to put however many triangles existed in the previous layer plus 2. (If it helps, look up a triangular Rubik's puzzle-- because that's honestly where I got the idea). Because the top layer is a single triangle, this has the effect of making each layer's number of triangles an odd number. If N is the total number of red edges (R) and blue edges (B) per side, then the area of the full triangle (in triangle units) will be the sum of the first N odd natural numbers. Let's call this Z and reintroduce the parallelograms. The total number of triangles are given by R^2 + B^2. So you would subtract this from Z. The figure you have left represents the equivalent number of triangle units "trapped" in parallelograms. Since you know that the parallelograms are the equivalent of two triangles, you would divide this answer by two and have the number of parallelograms in the puzzle.
Naviron Ghost I do. The Subscriptions tab gets cluttered with videos that I already watched and if you don't watch them right away they can be a pain to go through. I can keep notifications for the videos in my inbox for days and at times months until I get around to watching them.
Based on your template, "the collection of all lines that touch a fixed ___ and are contained in a ___", it looks like all the relevant puzzles will have two 1's, e.g. 00100010. What would triangular puzzles whose left and right sides had more than two 1's, e.g. 01100010110 correspond to in terms of Schubert's counting project?
Higher dimensional subspaces. "Line" matches "2" because we're working in projective geometry, which always eats exactly one dimension. So puzzles of size N with five 1s on each side are computing the number of 4-dimensional projective subspaces of (N-1)-dimensional projective space satisfying some intersection conditions, and so on.
So transcendental numbers can't be algebraically constructed over a finite sequence of operations. However some transcendental numbers such as e can be contructed over an infinite sequence of operations, but this sequence is countable (as e can be defined by a sum 1/n!). Are there any transcendental numbers that cannot be constructed over a countably infinite set of operations? ie. Its impossible to write an infinite formula for them. And would they form a further category?
All numbers have countable number of decimal digits. If the whole number would require uncountably infinite operations to compute, so would at least 1 of its digits. If only finite number of them need uncountable computation, then the number can be written as the number made of all those uncountably computable digits as a rational constant + number made of all the countably computable digits. Therefore the number would have infinite number of those uncountably infinite-computable digits. Wow... math is fun :-D I've already figured out some of their properties even without knowing whether they exist or even make sense :-D
In addition to the reply above there are only a countable number of numbers precisely described by a finite combination of letters and numbers. In other words, the set of numbers you can exactly define in, say, the English language with digits 0 to 9 is a countable set. Therefore the set of numbers that can never be finitely described in English is uncountable. Note that the number e is transcendental but still describable so is not in the set of indescribable numbers.
That's why it's instructive to also know the construction of reals by Cauchy sequences; you can always pinpoint some specific number by a sequence of rationals converging to that number. Which is just a different way of saying that every real number can be described by (possibly infinite) decimal expansion. As Doug has noted, you can quickly determine if a set is countable or not by answering the question whether you can uniquely describe any of its elements with only a finite string.
Will you talk about projective geometry and grassmanians in the next one? If a viewer understands this video they're extremely close to making these (very) abstract ideas in algebraic geometry tangible, and I personally would love to see that.
10:00 Where strings like [1 0 0 1] were mapped to sides of the triangle gadget: For the inputs, an orientation was given, but nothing was said about the orientation of the output string. Both answers in the example were symmetrical. Is there a consistent orientation, like always clockwise around the triangle? Are the answers always symmetrical?
At 7:50 : "The second type of line that intersects both L1 and L2 are the lines conttained within this plain." Yes, but only if you exclude those that are parrallel but not equal to L1 or L2.
The number of parallelograms is K(blue edges)*N-K(red edges). So, if you had 4 blue edges and 2 red edges (3:31), you would have 8 parallelograms. 2 red and blue edges each would give you 4 parallelograms.
For the challenge problem (how many parallelograms are in the puzzle), I think the answer is (N^2 - K^2 - (N-K)^2) / 2. Since N^2 is the total number of triangle units, K^2 the number of blue triangles, and (N-K)^2 the number of red triangles, N^2 - K^2 - (N-K)^2 is the number of triangle units covered by parallelograms. But each parallelogram covers 2 units, so one must divide by 2 to get the number of parallelograms.
If the vector is [1 0 1 0], then divide the edge into 4 equal parts. From the bottom, color the quarter blue (corresponding to 1), the next quarter as red (corresponding to zero), the next as blue and finally as red
Interesting video. I was totally confused by it - but that's MY problem! ;) In response to Goldfish's question... It's my understanding [however wrong it may be, of course!] that one cannot define a given irrational number by its place between 2 rational numbers. The reason is that between ANY 2 rational numbers, there are more irrational numbers than there are counting numbers in the universe! So there can be no such mapping. I very much enjoy this series - even if I don't understand a lot of it. Since this IS PBS, I'd also like to say that I've gotten a world of enjoyment & wonder from a certain NOVA program which aired way back in 1984 (I know... I'm an old man at 54!) Anyway, it's called "A Mathematical Mystery Tour" & it's awesome! While it naturally doesn't go into great detail, it does cover the history of math, with a lot of mind-bending ideas being talked about. I don't even know if it's still available to buy, but it's a super video for anyone to enjoy. It's ideal to get kids into the "wow" of math without bogging them down in equations & details. Another really good one is "The Great Math Mystery." Don't worry - it's from 2015! Anyway, I found the 2015 one one pbs.org, & you CAN find the older one - if you look around! AND I don't work for PBS!!! I'm just sayin' they're great videos. I love programs that both inform me & kinda blow my mind, & especially "A Mathematical Mystery Tour" certainly does that! If I were a teacher, I'd show it in class (with permission, of course!) In my 10th grade biology class, the teacher showed Cosmos (with the irreplaceable Carl Sagan) every week for its full run. I'll just end by pointing out that, even though there ARE cracks in the foundations of math (something the 1984 NOVA show talks about), it is STILL "the most certain body of knowledge we posses." Even more certain than any branch of the hard sciences (ALL of which MUST use math!) Anyway, thanks for a great series & keep up the good work! tavi.
My question is from 7:09 Let us say I have 'n 'number of blue lines on a side of an equilateral triangle and 'k' number of red ones, for any given values of 'n' and 'k', is the arrangement of red and blue triangles inside the triangle and parallelograms always definite or do other combinations exist? I see this to be the case for arbitrarily small values, but don't know where to start with for values greater than those. Thanks!
What I mean by definite arrangement here is- is the arrangement of red and blue triangles and parallelograms for any given values of 'n' and 'k' unique?
In the example she went through, even specifying the locations of the red and blue lines for 2 sides of the triangle left enough wiggle room for 2 distinct arrangements of the triangles and parallelograms. See 10:00 et seq.
Also the last statements about stand in of line in 3 space for two dimensional subspace of 4 dimension complex projective space. You can do a video on projective space for people that don't understand the relation of P^n = A^n+1 considering lines thru origin A= affine /euclidean space , and P = projective space. Its weird because if you work over non algebraicilly closed field the problem becomes much harder. There is so much connection between grassmannian , complex projective spaces and subspaces, and schubert calculus . That i am still a beginner at figuring out the connections completely. I hope you guys continue with this its a vast subject that incorporates so much different math in algebraic geometry
If there are K^2 blue triangles, (N-K)^2 red triangles and the total area is equal to N^2, then the remaining area that is unaccounted for is N^2-K^2-(N-K)^2=2K(N-K). Since the area of one parallelogram is the same as two triangles, so two units of area. Therefore there must be K(N-K) parallelograms inside the large triangle. This is the same as the square root of the product of the number of blue and red triangles, or the product red and blue segments along one edge of the big triangle.
Come back Kelsey and Infinite Series. Y'all were so good that I would wait until I had enough time to actually think about the content...and then you were gone :(
I feel like the last couple of minutes of this video could have done with a bit more explanation. Something I particularly felt was missing was how you quantised lines; in a continuous space, how are do end up with a finite number of lines between two other lines? Are you only counting how many you need to form an orthogonal basis set?
I wonder, is there some equivalent of that puzzle in higher dimensions - e.g. a tetrahedron, made up of smaller tetrahedra and [some other shape to take the parallelogram's place, maybe an octahedron?), with faces coloured in a similar way to the 2D version (with the octahedron maybe having opposite faces the same and adjacent faces the opposite colour? No wait, that wouldn't work...), and that has similar properties?
Thanks for another fascinating video! I'm wondering if the process of specializing lines mentioned at 1:20 is in any way related to the process of contracting Knuston's & Tao's puzzle mentioned at 4:27, since both - at least in the visual analogy presented here - involve bringing together elements that previously were not touching while maintaining certain properties of the overall picture. Disclaimer: I've only taken one college math class so far, so I am very much an amateur and apologize in advance for any glaring mistakes born of that.
best math video definitely never really understood this. Is it possible you can continue this to another few videos. One for how they compute the twisted cubics problem. Not only that i am having a hard time see the relation between the grassmannian , and schubert calculus.
I think of it like this. There are two similar triangles in each puzzle. A red triangle and a blue one. The two triangles are partitioned into equal-sized triangles similar to the original triangles and spread apart in such a way that the two triangles can then be fit together. The parallelograms are inserted as spacers that also indicate how the triangles were spread apart. There is a lot of math in this simple puzzle. I wish I had a physical representation of it. I want to explore it.
I have a question. I would like to know how the puzzle can talk about Schubert Calculus. As in, how do the rules of the puzzle and finally solving it inform finding the solutions to Schubert's problem?
Sometimes i have delusions of grandeur and think about going for a PhD. Then I watch these videos and I'm immediately brought down to earth. hahaha. great stuff and keep it up.
I'm not certain I understand how you can go from 'two lines in 3D space' to 'two lines in 3D space that intersect at a single point' - It seems like [0,1,0,1]x[0,1,0,1] specifically refers to finding all lines in 3D space that intersect both given lines in 3D space; but if the two given lines are skew, then how does the result of 'All lines in 3D space through a point' + 'all lines in a plane'. If there is no point of intersection, how does either of those groups make sense?
HeavyMetalMouse: I was also confused by this, but I think the answer is to remember that the resulting summands from the puzzle are descriptions of sets, but it's not a given that there are any elements in those sets. In the case of skew lines like you mentioned, the set of lines through the intersection point just has size zero. However, since there's no plane that goes through the two skew lines either doesn't that mean the other set in the sum is empty as well? But there are lines that intersect the two original skew lines you started with... They just don't lie in a plane. Really, it's not necessary to have two lines to uniquely define a plane, it only takes a line and a point. Two intersecting lines lie on a plane since you can define the plane by the first line and any point on the second line (except the intersection point), and then the second line is necessarily in the plane because it goes through both that point (by definition) and the first line (because they are intersecting lines). But any line going through that point that does not intersect the first line is not going to be on that plane (not necessarily anyway, unless the two lines are parallel). So for skew lines that don't intersect there is no plane on which they both lie. This would seem to indicate that both sets in the sum given by the triangle puzzle are empty, and yet there definitely are lines that intersect two skew lines in 3D space... And it doesn't seem like there's anything inherent in the triangle puzzle that requires the two lines intersect either, as demonstrated by the example with four lines in which there are (mysteriously) two pairs of intersecting lines which don't seem to co-intersect. Can someone help me understand what I'm missing? Maybe this is just an artifact of the simplification from the true domain of the triangle puzzles, which is the complex projective space mentioned by Kelsey at 11:45.
*"I'm not certain I understand how you can go from 'two lines in 3D space' to 'two lines in 3D space that intersect at a single point' "* Neither did David Hilbert, which is why he made a rigorous explanation of that his Problem #15. See 1:39 et seq. *" If there is no point of intersection, how does either of those groups make sense?"* In the case she was considering, it was assumed that Schubert's method had already reduced the problem to one involving intersecting lines.
So if you colored the puzzle with four colors, and may need two additional shapes, you'd be solving for projective quaternion space? (But that this is probably as far as it generalizes, as octonions and higher are increasingly weird and not as flexible.)
The addition and multiplication notation is a bit confusing, I assume this is Boolean algebra and your solving for X so to speak where X is the number of conditions that satisfy the equation? I'm thinking about this in regards of projections, for example if you asked me to mentally find a line that intersected lines in 3d space I would just project it onto 2d space and just rotate until I see a point where all the lines cross. Sooo essentially every single element has a projection in lower space. This projection is represented by all the possible states that logic graph (triangles) can occupy. Given 2 inputs that graph can have multiple solutions, so you plug in your entities and then reduce it down to the identity of a line and the coefficient remaining is the solution?
I don't understand the proof for Conjecture 1. It's saying if you contract a triangle which you know has x number of blue edges a certain way, you'll end up with a triangle with x number of blue edges. How does this prove that all triangles using rules mentioned in the video give a triangle with the same number of blue edges?
Interesting video but I still can’t figure out why the link between puzzles and the descriptions of intersections being mapped to the edges of the puzzles are actually connected. That is the video does a good job proving how two edges of a triangle puzzle describe the possible interiors and third edge, and it does a good job describing what the mapping is from, say, [0,1,1,0] to “all lines that intersect a given lines on a given plane”, and mentions (but does show how to prove) that these puzzles form a ring algebra with distributive properties, etc. That’s all cool. :) But what the video doesn’t mention at all from what I can tell is explaining why the triangle puzzles and the intersecting line categories are connected in the first place. Sure you can label a given intersecting line scenario with the binary code and then use the code in a triangle puzzle, but how do you know that the results of the triangle puzzle have anything to do with the results you are looking for in the intersecting line question? That’s where I’m still lost, it doesn’t look like they even talked about that at all here. What is the underlying connection between the puzzles and the intersecting line scenarios that allows solving the puzzles to produce results that match solving the intersecting line question?
Let P be the number of Parallelograms, B be that of Blue triangles and R the Red ones. Since Area(triangle) = 1 and Area(parallelogram) = 2, we have (B+R)^2 = B^2 + R^2 + 2P. As a result P = BR which is product of (number of blue triangles) and (number of red triangles).
OK. I see what she said, and I see why it works. But how did we get there from here? I'd be very interested in a distillation of the Knutson-Tau paper.
I am so egar to figure this connection out. Also this video only explained intersections of points, lines , planes, spaces,...etc. Do you know how this video relates to just contact /tangencies instead of interesections problem. (like the twisted cubic in the begining is more just a touching/contact/tangent number problem which is a special case of intersection.
I attempted a bijection by constructing paths going from 1 side to another crossing only lines of the same colour to prove corollary 1. Still the shrinking result is brilliant.
The total number of triangle units is N^2 The total number of red triangles is (N-K)^2 The total number of blue triangles is K^2 The number of triangle units not accounted for by the red and blue triangles, x, is given by: x=N^2-(N-K)^2-K^2=2NK-2K^2 There are two triangle units per parallelogram, so there are x/2 parallelograms, or NK-K^2=K(N-K) parallelograms, which is the product of the number of red edges per side and the number of blue edges per side
Also i am a little confused on how one is lead to this triangle configuration working. I see how to do it algorithmically from this video to some extent. However is there any other shapes within shapes that one could uses to compute faster in certain case... and the justification of why it works its shaky but so great of a video
The challange: the area of the paralelegrams is n^2-k^2-(n-k)^2=n^2-k^2-n^2+2nk-k^2=2nk-2k^2. Since each paralelegram has twice the area oof a triangle, there are nk-k^2,or k(n-k) paralelegrams in each puzzle.
I'm sure this says something profound about our perception of complexity, and hints at strategies for dimensionality reduction, but I just don't know the context to put it in.
I enjoyed these videos a lot more before I had 2 separate math classes in university. I spend 9 hours/week listening to math and that does not include homework...
I like your videos Kelsey. I've tried to work out your sign language, perhaps you could do a video on your code. Watching your hands makes me think you're from Italian stock :-)
The formula for the number of parallelograms I found was (N^2-(K^2+(N-K)^2))/2 so for a triangle with a side length of 5 and 2 blue edges, it would be: (5^2-(2^2+(5-2)^2))/2 parallelograms (5^2-(2^2+3^2))/2 (5^2-(4+9))/2 (5^2-13)/2 (25-13)/2 12/2 6 parallelograms
I hope you understand what i am talking about. Just being tangent or contact " i think this is called contact geometry" and seems to be a specialized case of intersection theory of this video.
This video reminded me that there is beautiful, but relatively simple to get into, mathematics out there. From all my years of schooling, it often feels like the only math I haven't yet grasped must be really complicated, but this video shows that that's just not true.
This series really needs to be a thing again.
I thought I understood. Then I reached the last minute of the video.
Then you understood! It's tricky stuff! The last minute was just some extra details.
+PBS Infinite Series I'm sure Trixelian actually meant the last minute before the ad and the questions from last week... you know the part where you basically told us the entire video before was a lie?
Kelsey, your intelligence is sick.
Amazingly, I understood the last minute. I was totally shocked that I did. She explained it very well.
طططضظل❤❤❤❤❤@@pbsinfiniteseries
I really liked the fact that throughout I'm sitting thinking, "yeah, but parallel lines fail this", then suddenly you pause to say, really it's complex projective. So fair enough, all my quibbles are silenced, then you pulled a face as if to say "shhh, I know what I'm doing". It really made me smile.
This was my favorite episode: when algebra and geometry intertwine so beautifully it's mesmerizing. I sincerely thank you Kelsey!
1. The contraction step was brilliant! Even coming up with it is difficult.
2. The fact that "multiplying" the indexes from that map [0 1 0 1], through that triangle produces anything useful at all is mindblowing. How would anyone even come up with that? And its distributive? Seriously,... that's gold.
If this video were a symphony, it'd be unfinished.
this is the symphony that Schubert wrote but never finished.
Fan Club alright, how can I join ? o.O
It's not hard to get the number of blue and red triangles as well as parallelograms from the fact that the triangle can be contracted as shown in 4:30.
But I'm missing the proof that the triangle can always be contracted like that without ever creating holes. I believe that it is true, but somehow I feel that you jumped over a part of the proof that is not completely obvious.
There are only three possibilities for what can lie directly next to a red triangle on a particular side: Either it's another red triangle, a parallelogram or nothing / "the outside". The last possibility can never create holes, so we'll skip that.
1) If it's a red triangle, then the two triangles we are looking at are congruent and both shrink at the same speed, meaning they will never "shear" and produce a hole.
2) Since the parallelograms here always have exactly two red sides that are opposite each other, every parallelogram forms a sort of "tunnel" to another red triangle or the edge of the whole big triangle (notably, they cannot change direction: If the parallelogram has Left-Right orientation (meaning the red sides point left/right), every other parallelogram touching it will have that orientation as well. Therefore, every parallelogram can shrink at the same speed as the two triangles it connects to, or the one it connects to in case it connects to the outside. There will be no shearing between those two edges either.
And blue edges do not shrink at all, they're just moved together with the shapes they belong to. So no shearing there either.
I hope I covered all possible cases that could cause holes in that construction.
Challenge:
We know that the sum of the areas of the parallelograms must be equal to the area of the whole minus the areas of the triangles:
N^2 - (K^2 + (N-K)^2) = N^2 - (K^2 + N^2 - 2*N*K + K^2) = 2*N*K - 2*K^2 = 2*K*(N-K). Every palallelogram has area 2, so the number of palallelograms must be K*(N-K), meaning the number of red edges multiplied by the number of blue edges.
Yes!!!! Infinite series uploaded!!
I just signed up for curiositystream because of this video, I kept finding the adds on youtube interesting. I can't believe someone actually created such an excellent source of science videos. The first one I watched (The Ultimate Formula) was the best I have ever scene on the subject of the standard model. Thanks for convincing me to sign up.
why does the triangle puzzle work? i realize the answer might be complicated, but is there some way of relating the rules of the puzzle to the algebra it describes? are there other puzzles that correspond to other algebras? what about algebras that arise in space with a different topology? do those have their own puzzles or conjectures? what if the space is curved?
Interestingly, the Grassmanian Gr(2,4) of planes in 4 space (the topic of this video) actually *is* curved, which is precisely why four lines (linear constraints) determine 2, not 1, element of the Grassmanian. When we embed Gr(2,4) into a higher dimensional projective space via the Plucker embedding, we obtain a variety of degree 2 (in the same way that a conic section has degree 2 in the plane: if you intersect it with enough hyperplanes, you end up with 2 points).
The general algebra corresponding to an algebraic variety is called its Chow ring, and is closely tied into the cohomology of the space.
Thanks for the technical commentary at the end of the episode (before the question round).
That is absolutely mind blowing! I'm baffled at how those dimension indices form a set with a distributive operation! And the product is calculated through some sort of puzzle?! Mind blowing!
(Oh, and about the bonus question: Take N as the number of blue edges and K as the number of red edges. Since the total area in triangle units is (N+K)^2. The blue triangles fill an area of N^2 and the red triangles fill up K^2 triangle units, leaving (N+K)^2-N^2-K^2 to be filled in with parallelograms. However, the parallelogram has an area of 2, so the number of parallelograms is ((N+K)^2-N^2-K^2)/2. )
K × N, to be more concise. 😉
When you perform the shrinking step, the rhombuses disappear, and leave behind only triangles in the color of your choosing.
I remember how 2 years ago I used to be an ardent follower of PBS spacetime when I came to know about this channel, and that it was no longer making videos. Eventually through channels such as 3b1b etc. Now I have lost my interest in physics and I am teaching myself mathematics. And now I feel disheartened that this channel doesn't make any more videos .
12:00 ahhh, thank you. I was on brink of posting a comment asking about the fact that sometimes they lines don't intersect each other.
Start at 11:50 , but the face she makes right at 12:00 is priceless xD
Back off, get your own sandwich.
:[
To take a stab at the challenge (6:48):
Imagine trying to make an ever larger equilateral triangle using solely smaller, congruent equilateral triangles. If you want to add another "layer" (make it larger) you would have to put however many triangles existed in the previous layer plus 2. (If it helps, look up a triangular Rubik's puzzle-- because that's honestly where I got the idea). Because the top layer is a single triangle, this has the effect of making each layer's number of triangles an odd number. If N is the total number of red edges (R) and blue edges (B) per side, then the area of the full triangle (in triangle units) will be the sum of the first N odd natural numbers. Let's call this Z and reintroduce the parallelograms. The total number of triangles are given by R^2 + B^2. So you would subtract this from Z. The figure you have left represents the equivalent number of triangle units "trapped" in parallelograms. Since you know that the parallelograms are the equivalent of two triangles, you would divide this answer by two and have the number of parallelograms in the puzzle.
Who needs notification when you're always on TH-cam..
Couch King th-cam.com/video/A8MO7fkZc5o/w-d-xo.html (appropriate)
Who needs notifications when there's a Subscriptions tab? :P
Naviron Ghost I do. The Subscriptions tab gets cluttered with videos that I already watched and if you don't watch them right away they can be a pain to go through. I can keep notifications for the videos in my inbox for days and at times months until I get around to watching them.
Couch King you forgot the question mark
Based on your template, "the collection of all lines that touch a fixed ___ and are contained in a ___", it looks like all the relevant puzzles will have two 1's, e.g. 00100010. What would triangular puzzles whose left and right sides had more than two 1's, e.g. 01100010110 correspond to in terms of Schubert's counting project?
Higher dimensional subspaces. "Line" matches "2" because we're working in projective geometry, which always eats exactly one dimension. So puzzles of size N with five 1s on each side are computing the number of 4-dimensional projective subspaces of (N-1)-dimensional projective space satisfying some intersection conditions, and so on.
...I'm not sure where you lost me there.
So transcendental numbers can't be algebraically constructed over a finite sequence of operations.
However some transcendental numbers such as e can be contructed over an infinite sequence of operations, but this sequence is countable (as e can be defined by a sum 1/n!).
Are there any transcendental numbers that cannot be constructed over a countably infinite set of operations?
ie. Its impossible to write an infinite formula for them.
And would they form a further category?
All numbers have countable number of decimal digits. If the whole number would require uncountably infinite operations to compute, so would at least 1 of its digits. If only finite number of them need uncountable computation, then the number can be written as the number made of all those uncountably computable digits as a rational constant + number made of all the countably computable digits. Therefore the number would have infinite number of those uncountably infinite-computable digits.
Wow... math is fun :-D I've already figured out some of their properties even without knowing whether they exist or even make sense :-D
In addition to the reply above there are only a countable number of numbers precisely described by a finite combination of letters and numbers. In other words, the set of numbers you can exactly define in, say, the English language with digits 0 to 9 is a countable set. Therefore the set of numbers that can never be finitely described in English is uncountable. Note that the number e is transcendental but still describable so is not in the set of indescribable numbers.
That's why it's instructive to also know the construction of reals by Cauchy sequences; you can always pinpoint some specific number by a sequence of rationals converging to that number. Which is just a different way of saying that every real number can be described by (possibly infinite) decimal expansion. As Doug has noted, you can quickly determine if a set is countable or not by answering the question whether you can uniquely describe any of its elements with only a finite string.
As numbers are defined by their definition, wouldn't the existence of indescribable numbers be, by definition, impossible?
+Dustin Rodriguez not at all. It is possible to prove that a thing exists even if you can't define that thing in a enumerable way.
11:45 was like "WHOOPS dropped it haha know you know *wink wink*.....SPEAKING OF HIGHER DIMENSIONS"
Will you talk about projective geometry and grassmanians in the next one? If a viewer understands this video they're extremely close to making these (very) abstract ideas in algebraic geometry tangible, and I personally would love to see that.
10:00 Where strings like [1 0 0 1] were mapped to sides of the triangle gadget: For the inputs, an orientation was given, but nothing was said about the orientation of the output string. Both answers in the example were symmetrical. Is there a consistent orientation, like always clockwise around the triangle? Are the answers always symmetrical?
At 7:50 : "The second type of line that intersects both L1 and L2 are the lines conttained within this plain." Yes, but only if you exclude those that are parrallel but not equal to L1 or L2.
Johannes Lippmann for this problem's purpose, parallel lines intersect at infinity
Which is a sneaky way of saying "We didn't want to deal with it." I see.
The number of parallelograms is K(blue edges)*N-K(red edges). So, if you had 4 blue edges and 2 red edges (3:31), you would have 8 parallelograms. 2 red and blue edges each would give you 4 parallelograms.
For the challenge problem (how many parallelograms are in the puzzle), I think the answer is (N^2 - K^2 - (N-K)^2) / 2. Since N^2 is the total number of triangle units, K^2 the number of blue triangles, and (N-K)^2 the number of red triangles, N^2 - K^2 - (N-K)^2 is the number of triangle units covered by parallelograms. But each parallelogram covers 2 units, so one must divide by 2 to get the number of parallelograms.
People really learn shit you’ll never need
I didn't get the connection between puzzles and 0/1 vectors, can somebody explain where it comes from?
If the vector is [1 0 1 0], then divide the edge into 4 equal parts. From the bottom, color the quarter blue (corresponding to 1), the next quarter as red (corresponding to zero), the next as blue and finally as red
the intelligence involved in developing these ideas is just mindblowing for me
Interesting video. I was totally confused by it - but that's MY problem! ;) In response to Goldfish's question... It's my understanding [however wrong it may be, of course!] that one cannot define a given irrational number by its place between 2 rational numbers. The reason is that between ANY 2 rational numbers, there are more irrational numbers than there are counting numbers in the universe! So there can be no such mapping. I very much enjoy this series - even if I don't understand a lot of it. Since this IS PBS, I'd also like to say that I've gotten a world of enjoyment & wonder from a certain NOVA program which aired way back in 1984 (I know... I'm an old man at 54!) Anyway, it's called "A Mathematical Mystery Tour" & it's awesome! While it naturally doesn't go into great detail, it does cover the history of math, with a lot of mind-bending ideas being talked about. I don't even know if it's still available to buy, but it's a super video for anyone to enjoy. It's ideal to get kids into the "wow" of math without bogging them down in equations & details. Another really good one is "The Great Math Mystery." Don't worry - it's from 2015! Anyway, I found the 2015 one one pbs.org, & you CAN find the older one - if you look around! AND I don't work for PBS!!! I'm just sayin' they're great videos. I love programs that both inform me & kinda blow my mind, & especially "A Mathematical Mystery Tour" certainly does that! If I were a teacher, I'd show it in class (with permission, of course!) In my 10th grade biology class, the teacher showed Cosmos (with the irreplaceable Carl Sagan) every week for its full run. I'll just end by pointing out that, even though there ARE cracks in the foundations of math (something the 1984 NOVA show talks about), it is STILL "the most certain body of knowledge we posses." Even more certain than any branch of the hard sciences (ALL of which MUST use math!) Anyway, thanks for a great series & keep up the good work! tavi.
My question is from 7:09
Let us say I have 'n 'number of blue lines on a side of an equilateral triangle and 'k' number of red ones, for any given values of 'n' and 'k', is the arrangement of red and blue triangles inside the triangle and parallelograms always definite or do other combinations exist? I see this to be the case for arbitrarily small values, but don't know where to start with for values greater than those. Thanks!
What I mean by definite arrangement here is- is the arrangement of red and blue triangles and parallelograms for any given values of 'n' and 'k' unique?
In the example she went through, even specifying the locations of the red and blue lines for 2 sides of the triangle left enough wiggle room for 2 distinct arrangements of the triangles and parallelograms. See 10:00 et seq.
Also the last statements about stand in of line in 3 space for two dimensional subspace of 4 dimension complex projective space.
You can do a video on projective space for people that don't understand the relation of P^n = A^n+1 considering lines thru origin A= affine /euclidean space , and P = projective space. Its weird because if you work over non algebraicilly closed field the problem becomes much harder. There is so much connection between grassmannian , complex projective spaces and subspaces, and schubert calculus . That i am still a beginner at figuring out the connections completely. I hope you guys continue with this its a vast subject that incorporates so much different math in algebraic geometry
Just found this channel and i am addicted.
At 7:56, the line can't be parallel to L1 or L2, right?
Yes. Throughout the video, it's best to ignore parallel lines/planes. It simplifies matters quite a bit.
Thanks, I was wondering the same thing!
I think this was the most complex video I watched and still kind of got what you were talking about. So i guess.. good job by both of us. 🤔
If there are K^2 blue triangles, (N-K)^2 red triangles and the total area is equal to N^2, then the remaining area that is unaccounted for is N^2-K^2-(N-K)^2=2K(N-K). Since the area of one parallelogram is the same as two triangles, so two units of area. Therefore there must be K(N-K) parallelograms inside the large triangle. This is the same as the square root of the product of the number of blue and red triangles, or the product red and blue segments along one edge of the big triangle.
Awesome video and presentation
Come back Kelsey and Infinite Series. Y'all were so good that I would wait until I had enough time to actually think about the content...and then you were gone :(
I feel like the last couple of minutes of this video could have done with a bit more explanation. Something I particularly felt was missing was how you quantised lines; in a continuous space, how are do end up with a finite number of lines between two other lines? Are you only counting how many you need to form an orthogonal basis set?
I wonder, is there some equivalent of that puzzle in higher dimensions - e.g. a tetrahedron, made up of smaller tetrahedra and [some other shape to take the parallelogram's place, maybe an octahedron?), with faces coloured in a similar way to the 2D version (with the octahedron maybe having opposite faces the same and adjacent faces the opposite colour? No wait, that wouldn't work...), and that has similar properties?
There is, yes it involves octahedra, and one can exploit it to give a direct proof that the multiplication defined by the puzzles is associative.
Thanks for another fascinating video! I'm wondering if the process of specializing lines mentioned at 1:20 is in any way related to the process of contracting Knuston's & Tao's puzzle mentioned at 4:27, since both - at least in the visual analogy presented here - involve bringing together elements that previously were not touching while maintaining certain properties of the overall picture. Disclaimer: I've only taken one college math class so far, so I am very much an amateur and apologize in advance for any glaring mistakes born of that.
Wow, that was a such a confusing set of problems/puzzles- but explained really well.
best math video definitely never really understood this. Is it possible you can continue this to another few videos. One for how they compute the twisted cubics problem. Not only that i am having a hard time see the relation between the grassmannian , and schubert calculus.
It's amazing the triangle puzzle is able to compute the correct answer, but why does it work?
I think of it like this. There are two similar triangles in each puzzle. A red triangle and a blue one. The two triangles are partitioned into equal-sized triangles similar to the original triangles and spread apart in such a way that the two triangles can then be fit together. The parallelograms are inserted as spacers that also indicate how the triangles were spread apart. There is a lot of math in this simple puzzle. I wish I had a physical representation of it. I want to explore it.
This leaves me one question...
Where was this series when I was going to school?
where was kelsey*
Please do an episode about Hilbert's problems! Thank you.
I have a question. I would like to know how the puzzle can talk about Schubert Calculus. As in, how do the rules of the puzzle and finally solving it inform finding the solutions to Schubert's problem?
Sometimes i have delusions of grandeur and think about going for a PhD. Then I watch these videos and I'm immediately brought down to earth. hahaha. great stuff and keep it up.
Saw this thing high AF and surprisingly came easy to me
I'm not certain I understand how you can go from 'two lines in 3D space' to 'two lines in 3D space that intersect at a single point' - It seems like [0,1,0,1]x[0,1,0,1] specifically refers to finding all lines in 3D space that intersect both given lines in 3D space; but if the two given lines are skew, then how does the result of 'All lines in 3D space through a point' + 'all lines in a plane'. If there is no point of intersection, how does either of those groups make sense?
HeavyMetalMouse how do the ones and zeroes correspond to the different spaces? That was confusing to me.
HeavyMetalMouse: I was also confused by this, but I think the answer is to remember that the resulting summands from the puzzle are descriptions of sets, but it's not a given that there are any elements in those sets. In the case of skew lines like you mentioned, the set of lines through the intersection point just has size zero. However, since there's no plane that goes through the two skew lines either doesn't that mean the other set in the sum is empty as well? But there are lines that intersect the two original skew lines you started with... They just don't lie in a plane.
Really, it's not necessary to have two lines to uniquely define a plane, it only takes a line and a point. Two intersecting lines lie on a plane since you can define the plane by the first line and any point on the second line (except the intersection point), and then the second line is necessarily in the plane because it goes through both that point (by definition) and the first line (because they are intersecting lines). But any line going through that point that does not intersect the first line is not going to be on that plane (not necessarily anyway, unless the two lines are parallel). So for skew lines that don't intersect there is no plane on which they both lie.
This would seem to indicate that both sets in the sum given by the triangle puzzle are empty, and yet there definitely are lines that intersect two skew lines in 3D space... And it doesn't seem like there's anything inherent in the triangle puzzle that requires the two lines intersect either, as demonstrated by the example with four lines in which there are (mysteriously) two pairs of intersecting lines which don't seem to co-intersect. Can someone help me understand what I'm missing? Maybe this is just an artifact of the simplification from the true domain of the triangle puzzles, which is the complex projective space mentioned by Kelsey at 11:45.
*"I'm not certain I understand how you can go from 'two lines in 3D space' to 'two lines in 3D space that intersect at a single point' "*
Neither did David Hilbert, which is why he made a rigorous explanation of that his Problem #15. See 1:39 et seq.
*" If there is no point of intersection, how does either of those groups make sense?"*
In the case she was considering, it was assumed that Schubert's method had already reduced the problem to one involving intersecting lines.
Ah... Well that's... unsatisfying. :)
So if you colored the puzzle with four colors, and may need two additional shapes, you'd be solving for projective quaternion space? (But that this is probably as far as it generalizes, as octonions and higher are increasingly weird and not as flexible.)
4:20 Is this really a general proof?
The addition and multiplication notation is a bit confusing, I assume this is Boolean algebra and your solving for X so to speak where X is the number of conditions that satisfy the equation?
I'm thinking about this in regards of projections, for example if you asked me to mentally find a line that intersected lines in 3d space I would just project it onto 2d space and just rotate until I see a point where all the lines cross.
Sooo essentially every single element has a projection in lower space. This projection is represented by all the possible states that logic graph (triangles) can occupy. Given 2 inputs that graph can have multiple solutions, so you plug in your entities and then reduce it down to the identity of a line and the coefficient remaining is the solution?
Isn't it Kd subspace in N-1d space? Like 2d in 3d? Because we have K ones, but N-1 total dimensions, since we start counting at 0(a point)..
Soooo when's the torus-equivariant cohomology version :D
I don't understand the proof for Conjecture 1. It's saying if you contract a triangle which you know has x number of blue edges a certain way, you'll end up with a triangle with x number of blue edges. How does this prove that all triangles using rules mentioned in the video give a triangle with the same number of blue edges?
10:48 Why did you say either?
What if the lines are skewed lines. They don't have any intersection point. How would we find a single line that intersects the four lines?
Good question; depending on the configuration, it is possible that no such line exists!
Interesting video but I still can’t figure out why the link between puzzles and the descriptions of intersections being mapped to the edges of the puzzles are actually connected. That is the video does a good job proving how two edges of a triangle puzzle describe the possible interiors and third edge, and it does a good job describing what the mapping is from, say, [0,1,1,0] to “all lines that intersect a given lines on a given plane”, and mentions (but does show how to prove) that these puzzles form a ring algebra with distributive properties, etc. That’s all cool. :)
But what the video doesn’t mention at all from what I can tell is explaining why the triangle puzzles and the intersecting line categories are connected in the first place. Sure you can label a given intersecting line scenario with the binary code and then use the code in a triangle puzzle, but how do you know that the results of the triangle puzzle have anything to do with the results you are looking for in the intersecting line question? That’s where I’m still lost, it doesn’t look like they even talked about that at all here. What is the underlying connection between the puzzles and the intersecting line scenarios that allows solving the puzzles to produce results that match solving the intersecting line question?
making all those hand waved arguments rigorous would be nightmares...
WOW. Mind blown - this is amazing!
This is the first episode I’ve watched that makes me just scratch my head and go “I don’t get it”
Let P be the number of Parallelograms, B be that of Blue triangles and R the Red ones. Since Area(triangle) = 1 and Area(parallelogram) = 2, we have (B+R)^2 = B^2 + R^2 + 2P. As a result P = BR which is product of (number of blue triangles) and (number of red triangles).
OK. I see what she said, and I see why it works. But how did we get there from here? I'd be very interested in a distillation of the Knutson-Tau paper.
Gorgeous video!
I am so egar to figure this connection out. Also this video only explained intersections of points, lines , planes, spaces,...etc. Do you know how this video relates to just contact /tangencies instead of interesections problem. (like the twisted cubic in the begining is more just a touching/contact/tangent number problem which is a special case of intersection.
I attempted a bijection by constructing paths going from 1 side to another crossing only lines of the same colour to prove corollary 1. Still the shrinking result is brilliant.
Calculus and the Maiden.
The Unfinished Calculus.
The Trout Calculus.
Ave Calculus.
I could go on, you know.
PlayTheMind even though you wrote a big comment, still you are first!
Calculinati
PlayTheMind Die Freunde von calculus.
PlayTheMind please don't
Hi! Please make a video related predicting next number in random number series use in rollete software. Is it breakable. If yes then please explain
The total number of triangle units is N^2
The total number of red triangles is (N-K)^2
The total number of blue triangles is K^2
The number of triangle units not accounted for by the red and blue triangles, x, is given by: x=N^2-(N-K)^2-K^2=2NK-2K^2
There are two triangle units per parallelogram, so there are x/2 parallelograms, or NK-K^2=K(N-K) parallelograms, which is the product of the number of red edges per side and the number of blue edges per side
Also i am a little confused on how one is lead to this triangle configuration working. I see how to do it algorithmically from this video to some extent. However is there any other shapes within shapes that one could uses to compute faster in certain case... and the justification of why it works its shaky but so great of a video
The challange: the area of the paralelegrams is n^2-k^2-(n-k)^2=n^2-k^2-n^2+2nk-k^2=2nk-2k^2. Since each paralelegram has twice the area oof a triangle, there are nk-k^2,or k(n-k) paralelegrams in each puzzle.
I see 14 intersecting lines at 0:30
I will have to watch this again some time.
I'm sure this says something profound about our perception of complexity, and hints at strategies for dimensionality reduction, but I just don't know the context to put it in.
4:53 would have been the perfect time to insert a discrete Zelda chest opening theme, just saying.
just imagine the first 20 seconds of v=0jWRf8x1DTY playing over the video
Guillaume Lagueyte probably not worth getting their channel flagged by Nintendo
Oh snap.
:(
Sweet video! I love seeing how to solve crazy problems!
what is 4D complex space? how is it different to just 4D space (R^4)?
Jino Tzino both spaces you mentioned are isomorphs, meaning they're technically the same
Relationships between seemingly unrelated areas of math are always surprising. But some (like this) make you wonder how the heck it can be true!
I enjoyed these videos a lot more before I had 2 separate math classes in university. I spend 9 hours/week listening to math and that does not include homework...
I like your videos Kelsey. I've tried to work out your sign language, perhaps you could do a video on your code. Watching your hands makes me think you're from Italian stock :-)
It was very hard to follow what happened in 3D space with lines. Rotating the view could have helped
11:08 Are you sure that THAT multiplication is associative?
伊藤那由多 yea becuase it just represents the intersection of sets witch is assosiative : p
You got the point!
The formula for the number of parallelograms I found was
(N^2-(K^2+(N-K)^2))/2
so for a triangle with a side length of 5 and 2 blue edges, it would be:
(5^2-(2^2+(5-2)^2))/2 parallelograms
(5^2-(2^2+3^2))/2
(5^2-(4+9))/2
(5^2-13)/2
(25-13)/2
12/2
6 parallelograms
I enjoy this channel because it shows me how much I don't know. :D
I think there are 2NK-2K^2 parallelograms, where N>K
I miss this show.
The video blows my mind!
That was crazy hahaha. Since we're on the subject of Hilbert and vectors... A video on abstract vector spaces??
I read Logicomix years ago and it's definitely worth a read, I thought it was amazing to read.
Good evening from Singapore! Wishing you'd have a nice evening =)
Cool, love the videos!
I hope you understand what i am talking about. Just being tangent or contact " i think this is called contact geometry" and seems to be a specialized case of intersection theory of this video.
I love math, even though it hates me
This video reminded me that there is beautiful, but relatively simple to get into, mathematics out there. From all my years of schooling, it often feels like the only math I haven't yet grasped must be really complicated, but this video shows that that's just not true.
K*(2N-K)/2 parallelograms in each puzzle
6:05 Implicit geometric proof that the sums of triangular numbers are squares