I felt that I had to take a break from all the heavy algebra that dominated recent videos. Today’s video is about a fairly recent chance discovery of some very beautiful mathematics by the mathematician Steve Morton. After I read about this discovery in an article by Steve Humble and his colleague Erhard Behrens in the Mathematical Intelligencer it occurred to me that this beautiful mathematics can be used as a very nifty visual key to some hidden self-similar patterns in Pascal’s triangle. Today is the first time that I tell anybody about this insight. Enjoy and please let me know how this video worked for you :)
@Mathologer: I tried with 9 colors. Each color has a number from 1 to 9. The result of the addition of 2 colors is determined by the addition of the digit until the result is a 1 digit number. For example color 3 + color 8 = color 2. Then I tried the shortcut you proved by applying the rule 9^n+1 to find the row of the triangle but it did not work. It works for length 2 for sure but not for length 10 and 82. Do I make a mistake or the shortcut just work with 2 or 3 items ?
8:40 Up to left-right symmetry and color permutations, there are only 10 top rows to check: 1 way to fill with 1 color (aaaa) 5 ways to fill with 2 colors (abbb, babb, aabb, abab, abba) 4 ways to fill with 3 colors (aabc, abac, abca, baac)
Wow that blew my mind that Pascal's triangle mod 2 gives the Sierpinski triangle, and mod 3 makes a similar pattern. I wrote a script to play with this more and it seems that mod any prime number it gives a nice Sierpinski-like pattern, and mod composite numbers give more messy overlapping patterns that relate to the factors. It's always amazing when seemingly disparate areas of math connect together in unexpected ways!
I noticed the same thing! This means that “special” numbers only exist when M = p is a prime number. Otherwise, zero rows (mod M) fail to exist, due to M being composite and generating extra patterns where the zeroes would be
The sliding magenta triangles at 26:04 made my jaw drop. The concept, while amazing as are most automata, wasn't hyper-noteworthy, but the idea of these little protein-like slider triangles infixing colors upon corner matching was just a revolutionary conceptualization for me. Now I want to reconsider all sorts of automata and the Wolfram/Cook Turing completeness in light of little slider machines.
Mistake @ 8:58: bottom row has 2 copies of: Blue, Blue, Red, Yellow and Blue, Blue, Red, Blue The ones missing are: Blue, Blue, Yellow, Yellow, and Blue, Blue, Yellow, Blue The trick still works for these 2, of course. Great video :)
Mathologer: explains rules about two colours added -> the third one me: ok, addition mod 3, easy Mathologer: two same colours -> that colour me: hmm me: something something group theory??? Mathologer: -(a+b) mod 3 me: takes maths degree certificate off wall
@@PurpleCoin7777 yes, my thoughts exactly, immediately at 0:45, but a moment later also "what? that doesn't work! let's continue watching" and finally exactly 19 minutes of being sidetracked(?) in interesting ways, i could continue my first thought at 19:45 "oh, that simple!" :-) (but i have an excuse, i like mathematics, but have not studied it)
Such puzzles and mathematical "mysteries" are like a concentrated food for always hungry young mathematical intellects - they're teaching them to mathematical beauty, elegance and aesthetics, and hone their logic, heuristic ability, pattern recognition, intuition (I saw the shortcut too, yesss!), making seemingly impossible connections between semantically distant objects or facts. For not such young minds it is pure pleasure and recreational activity to tackle such problems. So, thank you, sir, for making my day.
@VibratorDefibulator I agree on the old side, speaking for myself. I also use these brain limbering exercises (as I absorb them) as posers to the young'uns i get the chance to interact with.
Amazing video and I noticed a neat trick as an extension of this. If you have a number n, and want to know all mod(k), where k is all the numbers up to n, then take pascals triangle and extend out the top row by (n-1) zeros on the LHS and add an extra zero to the RHS after the 1 giving a row length of n+1 for the top row and that puts our value n, at the bottom position of a special triangle for mod(n) ... in fact you are at the bottom of many special triangles and can walk down each mod(k) by just cutting the top row off and our value in question will remain at the bottom of the new special triangle mod(k-1). Plus since the LHS is zero, the RHS is our value mod(k). Hope you find this trick neat and you really are inspiring and I hope to watch many more of your videos.
8:40 This is probably the more straightforward Mathologer homework question ever. Let's start by defining an essentially different triangle. Let's say two triangles are essentially the same if we can get from one to the other by relabeling colors or by flipping across the vertical axis. Some have mentioned rotational symmetry but I'm not including that. Now let's canonize the colorings: We'll relabel the colors so that the top left hex is always red and the next color as we go rightwards is always yellow. If there is no next color, we're OK because we know there's only one solid red triangle. Let's also canonize the orientation: We'll also say that if we can flip the triangle before relabeling to get more red hexes on the left half of the top row after relabeling, we do. If we had larger triangles, we would need a more complex rule using yellow hexes as a tie-breaker, but triangles of 4 are too small to ever need that. Now all we need to do is count the triangles that have red in their upper left, yellow next, and can't be flipped before relabeling to get more red hexes in the left half of the top row. There are 10.
You can actually further reduce it to 6 by realizing that the way the 'addition' is defined still works whichever side of the triangle you choose as the top, so triangles related by rotational symmetry can also go in the same group. EDIT: I see you actually mentioned rotational symmetry already but chose not to include it. I think 10 is really the correct answer anyway, since the rotational symmetry depends on the specific sum rule and wouldn't apply in general.
Always love proofs that are related to mathematical induction. So simple, but also so beautiful...:) I didn't realise at first why pascal triangles might be related (except the 2 objects add to one part), but when i saw the pattern animation of how the colours added; it struck me that we could interpret addition results through modular arithmetic. Simply brilliant.
I recently solved that problem on codewars and came up with the 10 rule myself (looking at the results of many different triangles). Helped me solve it fast enough
1 - trivial 1 1 - still trivial *1 2 1 - the middle ones have two as their greatest common factor* *1 3 3 1 - GCF = 3* 1 4 6 4 1 - GCF = 2, NOT 4 *1 5 10 10 5 1 - GCF = 5* 1 6 15 20 15 6 1 - GCF = 1 *1 7 21 35 35 21 7 1 - GCF = 7* 1 8 28 56 70 56 28 8 1 - GCF = 2 So, when P is prime, doing Pascal mod P makes the triangles of size P^k + 1 special because the GCF of the middle numbers in the P'th row is P itself
17:11 mathologer-instead of 1+powers of 3, we take 1, can you guess it? me-powers of 2(word to word exact) mathologer-powers of 2. me-OMG i am right. btw I just watched a video of mathologer relating to reuleaux triangles and then I got hitched to this channel.I love mathematics and you are just increasing my affection for it.
This is the best thing ever. On our last two COCI programming competitions we had triangulations of polygons (with 3 possible colors of edges) where each triangle had all sides differently colored.
23:56 Challenge for M = 4+ colors As far as I can tell, there are no special numbers for M = 4 colors. (In particular, the fractal for Pascal’s Triangle mod 4 is similar to Sierpinski’s Triangle, except that the huge “hole” triangle in the middle is filled in with a smaller copy of Sierpinski’s Triangle - Wolfram Alpha was a great help here. This means that we never get an “empty” row of zero coefficients, which would be required for our special trick) However, when it comes to M = 5 colors, it looks like the first “empty” row of Pascal’s Triangle mod 5 occurs on the 6th row (which looks like 1, 5, 10, 10, 5, 1 before modding). That is, 6 is the smallest non-trivial special number for 5 colors Notice that 6 = 5^1 + 1… Do we detect a pattern here? To find the next special number, we would have to jam together 5 rows of size-6 triangles, whose corners would form a super triangle of size-6 as well. The overall triangle would have a size of 5^2 + 1 = 26. Thus, the special numbers for 5 colors are of the form 5^n + 1 A similar non-existence issue occurs for M = 6 colors (as we saw for 4 colors). The rows of Pascal’s Triangle mod 6 where we might hope to find a gap, are instead filled in with smaller periodic patterns. I suspect this is caused by elements of Z/6Z that don’t have order 6… In other words, I conjecture that special numbers for M colors exist if and only if M = p is prime. For instance, the 8th row of Pascal’s Triangle 1 7 21 35 35 21 7 1 consists entirely of multiples of 7 (aside from the 1’s at the end), which means that 8 is the first special number for M = 7 colors (and will ultimately lead to a special number formula of 7^n + 1) Proof of special triangles iff M = p is prime: First, we will prove the “reverse” direction. That is, let M = p be our given prime number. Then the (p+1)-th row of Pascal’s Triangle looks like: 1 p (p choose 2) … (p choose 2) p 1 In general, the entries of this row are of the form (p choose k) = p! / [k! (p-k)!], where k = 0 to p. For simplicity, let’s call the left side A, and the denominator on the right D: A = p! / D, so that p! = A * D Notice that p divides p!, so p must divide AD. However, since D = k! (p-k)! consists of factors strictly smaller than p (for k ≠ 0 or p), it is clear that p does not divide D. Thus, since p is prime and p divides AD, it must be that p divides A In other words, A = (p choose k) = 0 (mod p), for all 0 < k < p (this excludes the ends of the row, where A = 1). This means that p+1 is a “special” number for M = p colors. In other words, if we define color addition as addition modulo p, then all the “middle” coefficients will cancel p rows down (since they will all be multiples of p), which means that triangles with p+1 rows will satisfy our trick of just adding the top 2 corners, as desired (And ultimately, all triangles of size p^n + 1 will be special as well, which can easily be shown by following the same “super triangle” method we’ve used several times before) Next, let’s prove the “forward” direction, by way of contraposition. Let our number of colors M be composite… (Might come back to this later. The basic idea is that as you move along a row of Pascal’s Triangle, you multiply by (M-k+1) / k to find the next entry. At first, these entries will be multiples of M, and multiplying by (M-k+1) will never change that. However, if M is composite, then some k < M will divide it, and eventually break our multiple of M. That is, we will find some (M choose k) ≠ 0 mod M, which means that our coefficient cancellation trick does not work. We also need to think about how to extend this to other rows, and/or why we only consider the (M+1)-th row in the first place…)
You only need to check 27 since only the difference between the colors matter. So you can ignore the first one and use relative values to that one instead of absolute ones. This leads to 3^3=27 combinations. Greetings from Germany :)
It is weird watching a video of something I work with in my tool kit and having a book beside me with a very similar example. The book is, Artificial Life: An Overview, MIT Press, 1997. My toolkit is my mind full of math and algorithms I use to make art. And that's why I absorb as much as I can from Mathologer. Thank you!
A variation I do is I would start as you did with N random integer numbers in a row of a given range. The next row I would choose a selection of cells from the previous row relative to the cell in the current row that I am evaluating. (Best to think of a row as a ring, end connects to start) The values of the selected cells would go through a function that had modulus the output which would be mapped to a value to be given to the current cell. I would do this in Microsoft Excel for fun years ago.
The colour rule also works for width 1. The cell at the foot of the side 1 triangle always is identical to both ends if the top row, and that satisfies the rule. That presumably means that zero is a power of three ;)
9:03 On the last row, the 3rd and 6th are the same pattern. Blue-blue-yellow-blue seems to be missing, but so does blue-blue-yellow-yellow, so there must be another duplicate somewhere.
Great! You go quite deep in this problem. I didn’t find the reduction argument as compelling as I would have liked and had to pause and think a fair bit.
I guessed it right! I am taking my free subscription! 😊 I was actually looking at the 2 corners but had no idea why. Just somehow felt right and consistent with the first example.
2 minutes into the video amd here's what's on my mind: the rule is the same as the average modulo 3, meaning (x+y)/2. If you assign a number to each color, and use this average, the answer will be the number for the corresponding color. For example: (1+0)/2 =1/2 = 2 (inverse of 2 modulo 3 is 2). Average of x and x will be x. This also reminds me of tricolorability of knots. (Or whatever it's called. I forget)
that's clever. I think division by two is multiplication by -1. In the field of remainder classes mod 3, 1/2=-1. Ah, I see, you got that already (2=-1)
Sorry if the answer is obvious, but can't figure it out... using the mirror and color symmetry I got down to 15 patterns, but what is the rotational Symmetry?
The pattern I found for the numbers at 9:20 was that the next term was three times the previous term minus 2. Or t(n)= (3*t(n-1)) -2. so (2*3)-2=4, 4*3-2=10, 10*3-2=28 28*3-2=82; the next term would be 82*3-2=244.
4:50 already every power of 9 plus one is special, because the triangles can be collapsed in like so: every 9th hexagon is sufficient to account for 9 generations in the future, so ignore the eight between that and the previous one. 10 is good, and a group of 82 would behave just like a group of 10 (with 8 between each one, 8*9 = 72, which is 82-10), and generation 10 of 82 hexagons would behave just like generation 1 of 10 hexagons. likewise generation 82 (assuming the first row is generation 1) is determined only by the top corners in the same way. Therefore 82 behaves like 10, and so 9^3 + 1 will behave to 82 the way 82 behaves to 10, and so on. My first expectation is that since there are 9 possible rules that this is the root behind why 9 is so special, but it may be that 4 generations also obey this property (powers of 3, plus 1 instead)
This one clicked so hard in my head that I ended up laughing out loud, and I was calling out answers before the video. My family must think I'm nuts. :)
I love this sort of thing. Excellent and even insightful video despite starting from such a simple rule, a deep dive into the "why" of things, as much of the best math starts. Am I the only one who feels an odd moment of terror when I realize that sierpinski's triangle is going to show up again though? I've seen it in so many surprising places and yet can't understand why, so my guess is that my brain fears it as some intelligent unknown, like some lovecraftian entity lmao.
I paused and, (after being stuck for quite a while staring at triangles of a different size and doing modulo math on those and not finding anything), managed to find the rule that it had to do with powers of 3 when I looked at your examples from 1:53, as that gave me a starting triangle of the correct size so there *was* a pattern to find where things worked out. I didn't think of it as working for triangles with starting rows of size 3^n+1, instead I thought of it as "going down 3^n steps", thus removing the "+1" and making the pattern feel simpler to me.
For the width 4 triangles, there are 10 "essentially" different triangles, if your symmetries are permuting colors and reflections across the vertical axis. I used Burnside's lemma twice.
I just did it. With rotations, it's 6. Here they are: 1111 1112 1121 1122 1231 1213 Each digit represents a color, and each group of 4 digits represents a starting row. There are many other starting rows you can choose, but once you build the triangle, you will see that one of the other two sides matches one of these 6 patterns, once you adjust the colors or reflect.
@@cubicardi8011 no. These are just elements from a set that we're acting on with some bigger group. They don't form a groul themselves or anything like that. The way I found them was use Burnside's lemma on the 81 starting rows for the triangles with permutations of colors. All 81 are fixed by the identity element, and then swapping two colors fixes 1 color (and we have 3 possible swaps), and swapping all 3 colors together fixes no triangle (we have 2 ways to do this). Burnside says we have (81+1+1+1+0+0)/6 = 14. Then I just wrote them down and did the reflection part, and I got (14+6)/2 = 10. Then finally I just drew the possible triangles and saw which sides appear together, and got 6 different possible triangles.
For the purposes of the question, do two width-4 triangles count as different or the same if one is the other rotated by 120 degrees? For example, A A B A gives you a triangle with one of the other sides showing A C C A. So in a sense A A B A and A B B A are the same triangle.
Reminds me of the matrix exponential. The power expansion of { e^(A) * e^(B) } for commutative matrices A & B can be rearranged to fit the binomial theorem. Fun!
I like the connection of this pattern with Sierpinski's triangle. I was expecting it to be a direct Sierpinski triangle, not a "higher order" one with many more smaller copies of itself in the central triangle though. I also saw the symmetries of rotating the triangle and running the rules that way. I now wonder if it's possible to "run the rules in reverse" and build up the triangle that way? It doesn't really look it's possible with this rule set, but is there one that can?
A cool way to state the rule for that last 3D example is this: Assign the quaternion units to the four different colours, eg. red = 1, yellow = i, blue = j and green = k. Three colours now give the fourth by the rule that the product of the four associated quaternions must be real(1 or -1). Alternatively, C4 = |C1 C2 C3| .
I would say there are fourteen (14) essentially different coloring patterns. First we remove the 3 full colors, then the remaininng 78 can all be mirror reflected down to 39. add the 3 full colors back in for 42. But then I realized it doesn't matter which color is which, since the 'addition' of colors is circular so if you label the colors A B and C then a triangl starting w/ AABC will end in B but if you cycle the numbers round to CCAB then the bottom will also cycle to A proving another threefold symmetry and cutting 42 down to 14. ^-^ thanks for the challenge Mathologer, now to the rest of the video :P -Tuonra
For the 3D version, the fourth color d we put on top of three colors a, b and c is a⊕b⊕c where ⊕ is the nim-addition en.wikipedia.org/wiki/Nimber#Addition_and_multiplication_tables . We also have a⊕b⊕c⊕d = 0, which explains why rotating any such pyramid gives another pyramid of the same type.
Great video. Want to see more videos on origami. Or about the "Golden ratio" for example with such a "Golden sword". 1 - 1(one way) 2=1+1 (2w) 3=1+1+1=1+2=2+1 (3w, prime numbers) 4=1+1+1+1=1+1+2=2+1+1=1+2+1=2+2 (5w, cquare number) 5=1+1+1+1+1=... (8w) ... 9=...=2+2+2+1+1+1 (..w, cquare number) 9-4=2+2+2+1+1+1-2-2=2+1+1+1=5 (prime number).
A real delight to watch, thank you. I suppose the triangle is also a 1D cellular automaton with the generations running down. Similarly the pyramid could be a 2D cellular automaton with each horizontal slice being a generation. I used to love playing Conway's Game of Life. With n neighbours and c colours, I think there might be c^(c^n) possible generation rules so these particularly symmetric ones are special cases in a sense.
The 3d-rule corresponds to adding in the group Z_2 x Z_2. If the bottom layer is a b c d e f then the top ball will be a+d+f+2(b+c+e) = a+d+f. Then we can do the collapsing argument giving the special sizes 2^n+1. Rotational symmetries: If the hexagons in a small triangle have numbers a, b, c, then the negative mod m game says that a + b = -c (mod m). But that is equivalent to a + c = -b and b + c = -a, which are the rotations. Since all the small triangles can be rotated, the whole figure can as well. For the pascal (positive) mod 2 game, a+b = c = -c, since 1=-1 in Z_2, so the rotations work here as well. The same applies for the 3d game: a+b+c = d = -d.
By the way, I learned when I was a teen that if you take the rows of pascal's triangle, mod 2, not only do you have a fractal, but if you turn that into a digital sound stream, say go out 1024 elements then skip down and do the same, it plays a little tune all made up from the same note in different octaves and beats that are also at multiples of that frequency.
I played with this a while ago and found that if you colour Pascal's triangle with two colour according to whether the entries are zero mod N, if N is prime you get a regular pattern of triangles of size N. If N is not prime, you get a superposition of the patterns corresponding to the prime factors of N. So for smallish N you can often look at the pattern and get an idea of what its prime factors are.
@@Mathologer by the way, I think I was sort of generating Pascal's triangle at an angle or backwards. The actual rule was a 1024 bit shift register and a exor gate, I think the result is playing the Sierpinski triangle but starting with click then a pure tone the whole width of the shift register and up, working your way toward that ever smaller tip, then it starts over... If you make the shift register something other than a power of two it sounds chaotic, things a bit like distorted guitar and down to noise. I also tried mod3 when I did it in software, but I didn't find that as interesting. I didn't make it up myself. I learned it from a binary clock that some college kid made and gave his dad. Every so often it would play this song and his dad was enough of a nerd to explain to 15 year old me how it worked. So I went home and programmed it into my TRS-80 model I in assembly language and it worked.
Knowing there is some trick to guessing quickly I looked back at the examples. I noticed that the top left and right corners added to the bottom one in each. Let's see if that hypothesis is right!
9:17 Yeah; I spotted the pattern. The gap between the special numbers gets tripled, each time. Thus, the next special number, after 82, is: 82 + (3 * 54) = 82 + 162 = 244. The general formula, thus, is: S(n) = S(1) + 2 * 3^(n-2) = 2 + 2 * 3^(n-2). *EDIT:* Or, as Mathologer put it: S(n) = 3^(n-1) + 1, which is, actually, a lot simpler form. 👍🏻😎
Fractal patterns on Australian shell is fascinating, like the dots on the shell of a Chinese turtle that triggered the first 'magic square', 2000 bc. The turtle was found walking up the shore of river Lo, and the magic square gave rise to the mystery of Lo Shu.
This also seems related to path counting in lattices. Specifically, in mod m, canceling occurs at a point B for a starting value A for which there are exactly m paths from A to B. I presume, and am currently too lazy to check, that there is a combinatorial way to compute the number of paths from one point to another in such a simple lattice (factorials will be everywhere probably), so these facts should be provable by algebraic manipulations of those path counts and, in fact, might tie into some combinatorics structures as well. Thoughts?
8:34 I think that there are 27 essentialy different 4-triangles because there are 3 axes of symmetry on the triangle, and the rules apply from all directions, so you can create two other triangles by rotation.
8:33 Number of essentially different width-4 triangles I believe there should be 10 different cases. Let’s label the colors A, B, and C for convenience, then consider the top row: (1) 4 of a kind (AAAA) (2) 3 of a kind + corner (AAAB) (3) 3 of a kind + middle (AABA) (4) 2 pair, sorted (AABB) (5) 2 pair, nested (ABBA) (6) 2 pair, alternating (ABAB) (7) 1 pair, take 1-2 (AABC) (8) 1 pair, take 1-3 (ABAC) (9) 1 pair, take 1-4 (ABCA) (10) 1 pair, take 2-3 (BAAC) And that’s it! There always has to be at least one pair in a row of 4, since there are only 3 colors (See the Hexagonal Pidgeon Hole Principle, for reference 😅) What does bother me, however, is that I can’t yet see a “formula” for getting from 81 to these 10 unique cases… The colors have 3! = 6 symmetries, and then there’s reflecting about the middle of the row, which gives us another 2 symmetries. This might suggest 6*2 = 12 symmetries overall, but clearly 12 does not divide 81 (and there are overlaps) For instance ABAB can be transformed into BABA either by swapping the colors A and B, or by reflection of the row, so calling it 12 “full” symmetries would clearly be over counting It’s been a while since I took group theory, but I’ve got a sneaky suspicion it applies here somewhere… 🤔 Edit: I found an answer to my “formula” question. Each case has a number of symmetries that divides 12, as follows: 3 symmetries: case (1) 6 symmetries: cases (4), (5), (6), (9), and (10) 12 symmetries: cases (2), (3), (7), and (8) (The 3 symmetry case is monotonic and only cares about “color rotations” rather than pairwise swaps, and the 6 symmetry cases are already symmetric about the middle of the row up to color permutations, so they don’t care about row reflections) Thus we get 3(1) + 6(5) + 12(4) = 3 + 30 + 48 = 81 total combinations What would you call this in group theory… Some sort of coset decomposition?
A nice fact from combinatorics or field theory is that the middle binomial coefficients can only all be divisible by n when n is prime. So for 2 and 3, we get powers of 2 and 3 +1 as special numbers, for 4 colors there should be no special numbers (except 2), but then for 5 each power of 5 +1 is special, and so on only for every prime
@8:34 I find 6 essentially different width 4 triangles. Then, with a little program I find the number of different triangles for widths 1, 2, 3, ... 1 1 2 2 3 3 4 6 5 13 6 31 7 76 8 214 9 592 10 1726 11 5056 12 15019 13 44665 14 133630 15 399715 16 1197970 17 3590629 18 10768363 It seems to get close to tripling each time. My program finds rotated triangles by calculating the entire triangle from the top row to get the left and right sides. I can see that there's an efficient way to calculate left and right sides using the embedded 3^n+1 sub-triangles. .. I don't have time to program it though.
8:50 I paused the video and figured that you only need to check 45 of the 81 listed. 9 are symmetrical at the base so 81-9 is 72. of those 72, each one is non symmetrical at the base and thus has an exact mirror image that does not need to be checked. so 72/2 = 36. then adding back the 9 symmetrical cases that must be checked gave me 45. I'm pretty dumb so please tell me if/how i am wrong.
I actually tried to think of the most intelligent thing to do when he asked the question, so I took those 2 corners and added them together because it felt right. How the hell did I get that right off the bat?
One of the first things i noticed is that the entire triangle works with the same rules if rotated under either of its other 2 sides. That is, every colour of every row is determined by the colours in the row above, no matter the rotation.
I felt that I had to take a break from all the heavy algebra that dominated recent videos. Today’s video is about a fairly recent chance discovery of some very beautiful mathematics by the mathematician Steve Morton. After I read about this discovery in an article by Steve Humble and his colleague Erhard Behrens in the Mathematical Intelligencer it occurred to me that this beautiful mathematics can be used as a very nifty visual key to some hidden self-similar patterns in Pascal’s triangle. Today is the first time that I tell anybody about this insight. Enjoy and please let me know how this video worked for you :)
Thank you!
Wow, your content is amazing, and even more so, because it is actually original. How are we worthy?
Schöne Grüße aus Deutschland :)
@Mathologer: I tried with 9 colors. Each color has a number from 1 to 9. The result of the addition of 2 colors is determined by the addition of the digit until the result is a 1 digit number. For example color 3 + color 8 = color 2.
Then I tried the shortcut you proved by applying the rule 9^n+1 to find the row of the triangle but it did not work.
It works for length 2 for sure but not for length 10 and 82.
Do I make a mistake or the shortcut just work with 2 or 3 items ?
Mathologer - love your videos. Will this work for prime powers to get the special numbers?
@@alre9766 try -(3 +8) mod 9 = 7. Does it work then?
One of Mathologer best videos.
The great graphic design,
The clear explanation,
The general idea, all are simply perfect!
@@rosiefay7283 just watch on 2x speed if you feel it's too slow
@@valeriobertoncello1809 meanwhile I'm here pausing lol
8:40 Up to left-right symmetry and color permutations, there are only 10 top rows to check:
1 way to fill with 1 color (aaaa)
5 ways to fill with 2 colors (abbb, babb, aabb, abab, abba)
4 ways to fill with 3 colors (aabc, abac, abca, baac)
and the counts
(aaaa) -3
(abbb) - 12, (babb) - 12, (aabb) - 6, (abab) - 6, (abba) - 6
(aabc) - 12, (abac) - 12, (abca) - 6, (baac) - 6.
Giving the 81 triangles.
taking into account rotation symmetry, only 6 checks needed:
aaaa
abca
aaab (checks abab for free)
aaba (checks abba for free)
aabb (checks aabc for free)
abbc (checks abac for free)
@@pocarski What do you mean by rotation, and how are your 4 free pairs equivalent?
This video is a great example of why Mathologer is the BEST Mathematics channel all over TH-cam. Thank you Burkard, for making my day!!
really love how much production quality goes into this (all the animation programming, etc)!
this guy is the best. He is everything I ever wanted in a teacher.
I just won a lifetime subscription to Mathologer! So cool!
Wow that blew my mind that Pascal's triangle mod 2 gives the Sierpinski triangle, and mod 3 makes a similar pattern. I wrote a script to play with this more and it seems that mod any prime number it gives a nice Sierpinski-like pattern, and mod composite numbers give more messy overlapping patterns that relate to the factors. It's always amazing when seemingly disparate areas of math connect together in unexpected ways!
I noticed the same thing! This means that “special” numbers only exist when M = p is a prime number. Otherwise, zero rows (mod M) fail to exist, due to M being composite and generating extra patterns where the zeroes would be
Wow very cool
The sliding magenta triangles at 26:04 made my jaw drop. The concept, while amazing as are most automata, wasn't hyper-noteworthy, but the idea of these little protein-like slider triangles infixing colors upon corner matching was just a revolutionary conceptualization for me. Now I want to reconsider all sorts of automata and the Wolfram/Cook Turing completeness in light of little slider machines.
This is one of the best mathologer videos so far, incredible. I inspires to play with math.
Saturday morning fun with Mathologer! What a pleasant surprise! 😎
Mistake @ 8:58: bottom row has 2 copies of:
Blue, Blue, Red, Yellow
and
Blue, Blue, Red, Blue
The ones missing are: Blue, Blue, Yellow, Yellow, and Blue, Blue, Yellow, Blue
The trick still works for these 2, of course.
Great video :)
Mathologer: explains rules about two colours added -> the third one
me: ok, addition mod 3, easy
Mathologer: two same colours -> that colour
me: hmm
me: something something group theory???
Mathologer: -(a+b) mod 3
me: takes maths degree certificate off wall
I thought addition mod 3 at first too haha
I thought the same at first glance, but there is no identity element :s
@@PurpleCoin7777 yes, my thoughts exactly, immediately at 0:45, but a moment later also "what? that doesn't work! let's continue watching"
and finally exactly 19 minutes of being sidetracked(?) in interesting ways, i could continue my first thought at 19:45 "oh, that simple!" :-)
(but i have an excuse, i like mathematics, but have not studied it)
At first I was thinking of addition mod3 and then taking the result times 2. It works in mod3 but I guess it doesn't generalize to other modulos.
@@valeriobertoncello1809 Yeah, that's because in mod 3, 2 and -1 are the same.
Such puzzles and mathematical "mysteries" are like a concentrated food for always hungry young mathematical intellects - they're teaching them to mathematical beauty, elegance and aesthetics, and hone their logic, heuristic ability, pattern recognition, intuition (I saw the shortcut too, yesss!), making seemingly impossible connections between semantically distant objects or facts.
For not such young minds it is pure pleasure and recreational activity to tackle such problems.
So, thank you, sir, for making my day.
:)
@VibratorDefibulator I agree on the old side, speaking for myself. I also use these brain limbering exercises (as I absorb them) as posers to the young'uns i get the chance to interact with.
It's beautiful how complex behavior emerges from simple rules. Great video!
I love it when pure math connects to natural phenomena like these Pascalian triangles and Fibonacci numbers, both connected to the growth of shells.
I loved it as always. Didn't want it to end.
Amazing video and I noticed a neat trick as an extension of this. If you have a number n, and want to know all mod(k), where k is all the numbers up to n, then take pascals triangle and extend out the top row by (n-1) zeros on the LHS and add an extra zero to the RHS after the 1 giving a row length of n+1 for the top row and that puts our value n, at the bottom position of a special triangle for mod(n) ... in fact you are at the bottom of many special triangles and can walk down each mod(k) by just cutting the top row off and our value in question will remain at the bottom of the new special triangle mod(k-1). Plus since the LHS is zero, the RHS is our value mod(k). Hope you find this trick neat and you really are inspiring and I hope to watch many more of your videos.
At last a Mathologer video that makes me feel special.
Are you a triangle with 3^n + 1 rows?
@@simo4875 Terve, kaima 🙋🏼♂️!
Absolutely amazing!
blown away!
what a great presentation, Mathologer!!
You made my day!
"Are there any other special numbers?" Well obviously 2 works :p
And 1 :D
And4
1 does not work
That’s exactly what I thought
@@dhu2056 Why not? If the same hexagon is all three corners, they're all the same color, which follows the rule
This is a gem of a video, congratulations. This is the definition of quality
You miss so much to talk about this more!!
Do part 2 !!!!
That proof for width 10 triangles was absolutely beautiful!
8:40 This is probably the more straightforward Mathologer homework question ever. Let's start by defining an essentially different triangle. Let's say two triangles are essentially the same if we can get from one to the other by relabeling colors or by flipping across the vertical axis. Some have mentioned rotational symmetry but I'm not including that.
Now let's canonize the colorings: We'll relabel the colors so that the top left hex is always red and the next color as we go rightwards is always yellow. If there is no next color, we're OK because we know there's only one solid red triangle.
Let's also canonize the orientation: We'll also say that if we can flip the triangle before relabeling to get more red hexes on the left half of the top row after relabeling, we do. If we had larger triangles, we would need a more complex rule using yellow hexes as a tie-breaker, but triangles of 4 are too small to ever need that.
Now all we need to do is count the triangles that have red in their upper left, yellow next, and can't be flipped before relabeling to get more red hexes in the left half of the top row.
There are 10.
You can actually further reduce it to 6 by realizing that the way the 'addition' is defined still works whichever side of the triangle you choose as the top, so triangles related by rotational symmetry can also go in the same group.
EDIT: I see you actually mentioned rotational symmetry already but chose not to include it. I think 10 is really the correct answer anyway, since the rotational symmetry depends on the specific sum rule and wouldn't apply in general.
@@Reddles37 Thanks, but I said "some have mentioned rotational symmetry but I'm not including that".
Pascal's triangle is so cool! There are all kinds of interesting patterns in it. This color thing is about the best I've seen.
11:00 - a proof by induction, my favorite. 🙂
Always love proofs that are related to mathematical induction. So simple, but also so beautiful...:)
I didn't realise at first why pascal triangles might be related (except the 2 objects add to one part), but when i saw the pattern animation of how the colours added; it struck me that we could interpret addition results through modular arithmetic. Simply brilliant.
Yes. Induction is a beautiful tool, indeed. 🙂
"Did you get it? Without pausing the video?"
Yes!
"Or just guessing?"
Oh...
Wow! I actually got it! In the time given, and for the right reason! I couldn't *justify* the reason, but, still kind of impressed with myself.
Great video! From the beginning of the video I waited for cellular automaton to be mentioned. Still waiting :)
What better way to begin a lazy Sunday morning than with some simple and beautiful mathematics presented so eloquently?
I recently solved that problem on codewars and came up with the 10 rule myself (looking at the results of many different triangles). Helped me solve it fast enough
Have you got a reference ? :)
@@Mathologer www.codewars.com/kata/coloured-triangles
and this
www.codewars.com/kata/5a331ea7ee1aae8f24000175
@@nowifi4u thanks
@@nowifi4u That's great, thank you very much :)
Fantastic video thank you for increasing my love of maths with every video you make
I don't know if this video is beautifully awesome or awesomly beautiful...
I'm here to brag about my bragging rights for choosing yellow arbitrarly
@oH well,lord! same
@oH well,lord! exactly
I guessed it by reading this comment
@@jogiff Big brain time
Me too! :-)
1 - trivial
1 1 - still trivial
*1 2 1 - the middle ones have two as their greatest common factor*
*1 3 3 1 - GCF = 3*
1 4 6 4 1 - GCF = 2, NOT 4
*1 5 10 10 5 1 - GCF = 5*
1 6 15 20 15 6 1 - GCF = 1
*1 7 21 35 35 21 7 1 - GCF = 7*
1 8 28 56 70 56 28 8 1 - GCF = 2
So, when P is prime, doing Pascal mod P makes the triangles of size P^k + 1 special because the GCF of the middle numbers in the P'th row is P itself
Yep only works for primes. For non primes you can't fine special numbers
Found the same!
Can you use this as a way to find prime numbers?
beautiful!
I did not know that
17:11
mathologer-instead of 1+powers of 3, we take 1, can you guess it?
me-powers of 2(word to word exact)
mathologer-powers of 2.
me-OMG i am right.
btw I just watched a video of mathologer relating to reuleaux triangles and then I got hitched to this channel.I love mathematics and you are just increasing my affection for it.
The snail shell thing is amazing.
I Finally caught up with mathogoger!
This is the best thing ever. On our last two COCI programming competitions we had triangulations of polygons (with 3 possible colors of edges) where each triangle had all sides differently colored.
Did they mention the art gallery theorem ? Its proof features a nice application of 3-colouring triangulations :)
We haven't heard of that, but I shall come up with a cool problem about it someday :)
11:30 ach, das ist beinahe das Schönste, was ich an Erklärungen gesehen habe. Außerdem auch ist es so ein typischer "waaaas?" Moment
His shirt doesn't have a 27
How odd...
@@scudlee :)
@@Mathologer This was such an amazing video! This is really satisfying maths
But the number of numbers on his shirt is even!
It's a hint to the homework. 81/3 = 27
23:56 Challenge for M = 4+ colors
As far as I can tell, there are no special numbers for M = 4 colors. (In particular, the fractal for Pascal’s Triangle mod 4 is similar to Sierpinski’s Triangle, except that the huge “hole” triangle in the middle is filled in with a smaller copy of Sierpinski’s Triangle - Wolfram Alpha was a great help here. This means that we never get an “empty” row of zero coefficients, which would be required for our special trick)
However, when it comes to M = 5 colors, it looks like the first “empty” row of Pascal’s Triangle mod 5 occurs on the 6th row (which looks like 1, 5, 10, 10, 5, 1 before modding). That is, 6 is the smallest non-trivial special number for 5 colors
Notice that 6 = 5^1 + 1… Do we detect a pattern here? To find the next special number, we would have to jam together 5 rows of size-6 triangles, whose corners would form a super triangle of size-6 as well. The overall triangle would have a size of 5^2 + 1 = 26. Thus, the special numbers for 5 colors are of the form 5^n + 1
A similar non-existence issue occurs for M = 6 colors (as we saw for 4 colors). The rows of Pascal’s Triangle mod 6 where we might hope to find a gap, are instead filled in with smaller periodic patterns. I suspect this is caused by elements of Z/6Z that don’t have order 6…
In other words, I conjecture that special numbers for M colors exist if and only if M = p is prime. For instance, the 8th row of Pascal’s Triangle
1 7 21 35 35 21 7 1
consists entirely of multiples of 7 (aside from the 1’s at the end), which means that 8 is the first special number for M = 7 colors (and will ultimately lead to a special number formula of 7^n + 1)
Proof of special triangles iff M = p is prime:
First, we will prove the “reverse” direction. That is, let M = p be our given prime number. Then the (p+1)-th row of Pascal’s Triangle looks like:
1 p (p choose 2) … (p choose 2) p 1
In general, the entries of this row are of the form
(p choose k) = p! / [k! (p-k)!],
where k = 0 to p. For simplicity, let’s call the left side A, and the denominator on the right D:
A = p! / D, so that p! = A * D
Notice that p divides p!, so p must divide AD. However, since D = k! (p-k)! consists of factors strictly smaller than p (for k ≠ 0 or p), it is clear that p does not divide D. Thus, since p is prime and p divides AD, it must be that p divides A
In other words, A = (p choose k) = 0 (mod p), for all 0 < k < p (this excludes the ends of the row, where A = 1). This means that p+1 is a “special” number for M = p colors. In other words, if we define color addition as addition modulo p, then all the “middle” coefficients will cancel p rows down (since they will all be multiples of p), which means that triangles with p+1 rows will satisfy our trick of just adding the top 2 corners, as desired
(And ultimately, all triangles of size p^n + 1 will be special as well, which can easily be shown by following the same “super triangle” method we’ve used several times before)
Next, let’s prove the “forward” direction, by way of contraposition. Let our number of colors M be composite…
(Might come back to this later. The basic idea is that as you move along a row of Pascal’s Triangle, you multiply by (M-k+1) / k to find the next entry. At first, these entries will be multiples of M, and multiplying by (M-k+1) will never change that. However, if M is composite, then some k < M will divide it, and eventually break our multiple of M. That is, we will find some (M choose k) ≠ 0 mod M, which means that our coefficient cancellation trick does not work. We also need to think about how to extend this to other rows, and/or why we only consider the (M+1)-th row in the first place…)
You only need to check 27 since only the difference between the colors matter. So you can ignore the first one and use relative values to that one instead of absolute ones. This leads to 3^3=27 combinations.
Greetings from Germany :)
27 definitely suffice, but when you take symmetry into consideration you can get away with even less :)
It is weird watching a video of something I work with in my tool kit and having a book beside me with a very similar example. The book is, Artificial Life: An Overview, MIT Press, 1997.
My toolkit is my mind full of math and algorithms I use to make art. And that's why I absorb as much as I can from Mathologer. Thank you!
A variation I do is I would start as you did with N random integer numbers in a row of a given range.
The next row I would choose a selection of cells from the previous row relative to the cell in the current row that I am evaluating. (Best to think of a row as a ring, end connects to start) The values of the selected cells would go through a function that had modulus the output which would be mapped to a value to be given to the current cell. I would do this in Microsoft Excel for fun years ago.
The colour rule also works for width 1. The cell at the foot of the side 1 triangle always is identical to both ends if the top row, and that satisfies the rule.
That presumably means that zero is a power of three ;)
9:03 On the last row, the 3rd and 6th are the same pattern. Blue-blue-yellow-blue seems to be missing, but so does blue-blue-yellow-yellow, so there must be another duplicate somewhere.
2nd and 5th on the same bottom row (one step to the left)
Great! You go quite deep in this problem. I didn’t find the reduction argument as compelling as I would have liked and had to pause and think a fair bit.
I guessed it right! I am taking my free subscription! 😊
I was actually looking at the 2 corners but had no idea why. Just somehow felt right and consistent with the first example.
I actually almost used my initial example also as the challenge example (to see how many people would pick up on it :)
2 minutes into the video amd here's what's on my mind: the rule is the same as the average modulo 3, meaning (x+y)/2. If you assign a number to each color, and use this average, the answer will be the number for the corresponding color. For example: (1+0)/2 =1/2 = 2 (inverse of 2 modulo 3 is 2). Average of x and x will be x.
This also reminds me of tricolorability of knots. (Or whatever it's called. I forget)
that's clever. I think division by two is multiplication by -1. In the field of remainder classes mod 3, 1/2=-1. Ah, I see, you got that already (2=-1)
I think it was of maps (aka graphs) not knots but I'm not sure
@@sinom no, it's for knots. See here: en.m.wikipedia.org/wiki/Tricolorability
See the second rule in the rules section.
@8:50. There are 6 different patterns. They remaining ones are redundant due to rotational and mirror symmetry.
and color symmetry? One of the four can be chosen, so there are only 3^3 ways
@@johannesh7610 well yes. That symmetry is the most apparent though, so i overlooked mentioning it
@@TazPessle yes. The rotational one is not so obvious :)
Sorry if the answer is obvious, but can't figure it out...
using the mirror and color symmetry I got down to 15 patterns, but what is the rotational Symmetry?
@@simo4875 the process works all three ways. Rotate a triangle and check
The pattern I found for the numbers at 9:20 was that the next term was three times the previous term minus 2. Or t(n)= (3*t(n-1)) -2. so (2*3)-2=4, 4*3-2=10, 10*3-2=28 28*3-2=82; the next term would be 82*3-2=244.
It leads to the same result:
t(n)= 3.t(n-1) - 2 t(n) - 1 = 3.(t(n-1) - 1), so t(n) - 1 = 3^n (t(0) - 1).
As t(0) = 2, t(n) = 3^n + 1.
Pefect, amazing, bravo!!! 👏👏👏
4:50 already every power of 9 plus one is special, because the triangles can be collapsed in like so: every 9th hexagon is sufficient to account for 9 generations in the future, so ignore the eight between that and the previous one. 10 is good, and a group of 82 would behave just like a group of 10 (with 8 between each one, 8*9 = 72, which is 82-10), and generation 10 of 82 hexagons would behave just like generation 1 of 10 hexagons. likewise generation 82 (assuming the first row is generation 1) is determined only by the top corners in the same way. Therefore 82 behaves like 10, and so 9^3 + 1 will behave to 82 the way 82 behaves to 10, and so on. My first expectation is that since there are 9 possible rules that this is the root behind why 9 is so special, but it may be that 4 generations also obey this property (powers of 3, plus 1 instead)
This one rocks! Very fun.
This one clicked so hard in my head that I ended up laughing out loud, and I was calling out answers before the video. My family must think I'm nuts. :)
I love this sort of thing. Excellent and even insightful video despite starting from such a simple rule, a deep dive into the "why" of things, as much of the best math starts.
Am I the only one who feels an odd moment of terror when I realize that sierpinski's triangle is going to show up again though? I've seen it in so many surprising places and yet can't understand why, so my guess is that my brain fears it as some intelligent unknown, like some lovecraftian entity lmao.
I paused and, (after being stuck for quite a while staring at triangles of a different size and doing modulo math on those and not finding anything), managed to find the rule that it had to do with powers of 3 when I looked at your examples from 1:53, as that gave me a starting triangle of the correct size so there *was* a pattern to find where things worked out.
I didn't think of it as working for triangles with starting rows of size 3^n+1, instead I thought of it as "going down 3^n steps", thus removing the "+1" and making the pattern feel simpler to me.
This video was really awesome and accessible. Really fun!
For the width 4 triangles, there are 10 "essentially" different triangles, if your symmetries are permuting colors and reflections across the vertical axis. I used Burnside's lemma twice.
How many with rotations counting only once?
I just did it. With rotations, it's 6. Here they are:
1111
1112
1121
1122
1231
1213
Each digit represents a color, and each group of 4 digits represents a starting row. There are many other starting rows you can choose, but once you build the triangle, you will see that one of the other two sides matches one of these 6 patterns, once you adjust the colors or reflect.
@@f5673-t1h do those 6 elements have a special Name in Group theory? How do calculate them?
@@cubicardi8011 no. These are just elements from a set that we're acting on with some bigger group. They don't form a groul themselves or anything like that.
The way I found them was use Burnside's lemma on the 81 starting rows for the triangles with permutations of colors. All 81 are fixed by the identity element, and then swapping two colors fixes 1 color (and we have 3 possible swaps), and swapping all 3 colors together fixes no triangle (we have 2 ways to do this). Burnside says we have (81+1+1+1+0+0)/6 = 14.
Then I just wrote them down and did the reflection part, and I got (14+6)/2 = 10. Then finally I just drew the possible triangles and saw which sides appear together, and got 6 different possible triangles.
@@f5673-t1h Can you give the details for the Burnside calculations? I'm not familiar with that and I'm not sure how to count the permutations.
For the purposes of the question, do two width-4 triangles count as different or the same if one is the other rotated by 120 degrees?
For example, A A B A gives you a triangle with one of the other sides showing A C C A.
So in a sense A A B A and A B B A are the same triangle.
You deserve over a million subscribers!
Reminds me of the matrix exponential. The power expansion of { e^(A) * e^(B) } for commutative matrices A & B can be rearranged to fit the binomial theorem. Fun!
I'm 34 and I just learned about Pascal's triangle yesterday.. Very interesting indeed.
I like the connection of this pattern with Sierpinski's triangle. I was expecting it to be a direct Sierpinski triangle, not a "higher order" one with many more smaller copies of itself in the central triangle though.
I also saw the symmetries of rotating the triangle and running the rules that way. I now wonder if it's possible to "run the rules in reverse" and build up the triangle that way? It doesn't really look it's possible with this rule set, but is there one that can?
Why is math so beautiful?
A cool way to state the rule for that last 3D example is this: Assign the quaternion units to the four different colours, eg. red = 1, yellow = i, blue = j and green = k. Three colours now give the fourth by the rule that the product of the four associated quaternions must be real(1 or -1). Alternatively, C4 = |C1 C2 C3| .
Yet another amazing video:)
Glad you like it :)
Great video - I really learned a lot from this one!
I would say there are fourteen (14) essentially different coloring patterns. First we remove the 3 full colors, then the remaininng 78 can all be mirror reflected down to 39. add the 3 full colors back in for 42. But then I realized it doesn't matter which color is which, since the 'addition' of colors is circular so if you label the colors A B and C then a triangl starting w/ AABC will end in B but if you cycle the numbers round to CCAB then the bottom will also cycle to A proving another threefold symmetry and cutting 42 down to 14. ^-^ thanks for the challenge Mathologer, now to the rest of the video :P
-Tuonra
Getting close but you can still do a bit better :)
Very clean
Excellent as well as beautiful
For the 3D version, the fourth color d we put on top of three colors a, b and c is a⊕b⊕c where ⊕ is the nim-addition en.wikipedia.org/wiki/Nimber#Addition_and_multiplication_tables . We also have a⊕b⊕c⊕d = 0, which explains why rotating any such pyramid gives another pyramid of the same type.
Yes, basically, the rotation property works because it works at the level of the rules :)
Great video. Want to see more videos on origami. Or about the "Golden ratio" for example with such a "Golden sword".
1 - 1(one way)
2=1+1 (2w)
3=1+1+1=1+2=2+1 (3w, prime numbers)
4=1+1+1+1=1+1+2=2+1+1=1+2+1=2+2 (5w, cquare number)
5=1+1+1+1+1=... (8w)
...
9=...=2+2+2+1+1+1 (..w, cquare number)
9-4=2+2+2+1+1+1-2-2=2+1+1+1=5 (prime number).
A real delight to watch, thank you.
I suppose the triangle is also a 1D cellular automaton with the generations running down. Similarly the pyramid could be a 2D cellular automaton with each horizontal slice being a generation. I used to love playing Conway's Game of Life.
With n neighbours and c colours, I think there might be c^(c^n) possible generation rules so these particularly symmetric ones are special cases in a sense.
I just noticed that Jimmy Gervais
had already made a related comment below.
The 3d-rule corresponds to adding in the group Z_2 x Z_2. If the bottom layer is
a
b c
d e f
then the top ball will be a+d+f+2(b+c+e) = a+d+f. Then we can do the collapsing argument giving the special sizes 2^n+1.
Rotational symmetries: If the hexagons in a small triangle have numbers a, b, c, then the negative mod m game says that a + b = -c (mod m). But that is equivalent to a + c = -b and b + c = -a, which are the rotations. Since all the small triangles can be rotated, the whole figure can as well.
For the pascal (positive) mod 2 game, a+b = c = -c, since 1=-1 in Z_2, so the rotations work here as well. The same applies for the 3d game: a+b+c = d = -d.
how good is this argument? 10/10
By the way, I learned when I was a teen that if you take the rows of pascal's triangle, mod 2, not only do you have a fractal, but if you turn that into a digital sound stream, say go out 1024 elements then skip down and do the same, it plays a little tune all made up from the same note in different octaves and beats that are also at multiples of that frequency.
Yes, lots of fun games you can play with this idea :)
I played with this a while ago and found that if you colour Pascal's triangle with two colour according to whether the entries are zero mod N, if N is prime you get a regular pattern of triangles of size N. If N is not prime, you get a superposition of the patterns corresponding to the prime factors of N. So for smallish N you can often look at the pattern and get an idea of what its prime factors are.
@@Mathologer by the way, I think I was sort of generating Pascal's triangle at an angle or backwards. The actual rule was a 1024 bit shift register and a exor gate, I think the result is playing the Sierpinski triangle but starting with click then a pure tone the whole width of the shift register and up, working your way toward that ever smaller tip, then it starts over...
If you make the shift register something other than a power of two it sounds chaotic, things a bit like distorted guitar and down to noise.
I also tried mod3 when I did it in software, but I didn't find that as interesting.
I didn't make it up myself. I learned it from a binary clock that some college kid made and gave his dad. Every so often it would play this song and his dad was enough of a nerd to explain to 15 year old me how it worked. So I went home and programmed it into my TRS-80 model I in assembly language and it worked.
Knowing there is some trick to guessing quickly I looked back at the examples. I noticed that the top left and right corners added to the bottom one in each. Let's see if that hypothesis is right!
20:04 My first thought was 2(a+b). My next thought just seconds later was 3-(a+b) which is just the same as -(a+b) mod 3
Yes, but also 2 = -1 mod 3 so your first thought was right
@@trueriver1950 Right! But 2 =/= -1 mod m in general
9:17 Yeah; I spotted the pattern. The gap between the special numbers gets tripled, each time. Thus, the next special number, after 82, is: 82 + (3 * 54) = 82 + 162 = 244. The general formula, thus, is:
S(n) = S(1) + 2 * 3^(n-2) = 2 + 2 * 3^(n-2).
*EDIT:* Or, as Mathologer put it:
S(n) = 3^(n-1) + 1, which is, actually, a lot simpler form.
👍🏻😎
I was *_INSTANTLY_* reminded of the Sierpiński triangle 😎.
🔺
🔺🔺
Yay I have lifetime subscription just by rewinding the video once and finding out the pattern and the answer in half a minute lol
Fractal patterns on Australian shell is fascinating, like the dots on the shell of a Chinese turtle that triggered the first 'magic square', 2000 bc. The turtle was found walking up the shore of river Lo, and the magic square gave rise to the mystery of Lo Shu.
This also seems related to path counting in lattices. Specifically, in mod m, canceling occurs at a point B for a starting value A for which there are exactly m paths from A to B. I presume, and am currently too lazy to check, that there is a combinatorial way to compute the number of paths from one point to another in such a simple lattice (factorials will be everywhere probably), so these facts should be provable by algebraic manipulations of those path counts and, in fact, might tie into some combinatorics structures as well. Thoughts?
On a scale of 1-10 for how beautiful that recursive argument is, because 10 is just a special case of 4, the rating I give is 4.
??? what a strange comment :)
Army of Won -- Your comment deserves at least a '7,' too.
The first thing I noticed is if you rotate the triangle so a different edge is on top, it still follows the rules.
That's because every group of three that are all touching are either all different colours, or all the same colour. Rotating doesn't change that.
Got the yellow right..! Love this episode, very interesting and something that i actually do understand..!:)
These triangleceptions made my day
8:34 I think that there are 27 essentialy different 4-triangles because there are 3 axes of symmetry on the triangle, and the rules apply from all directions, so you can create two other triangles by rotation.
Woah, I was able to guess the shortcut rule! Do I get bragging rights? 🙃
Sure :)
I guessed it too!
same, checked the previous examples afterwards to confirm this is always the case
I guessed the rule also! I don't care if I get the right to brag, I will anyway. I think I'll use it as a party trick.
Great job, you win a life subscription too. Just hit the button to claim your prize
8:33 Number of essentially different width-4 triangles
I believe there should be 10 different cases. Let’s label the colors A, B, and C for convenience, then consider the top row:
(1) 4 of a kind (AAAA)
(2) 3 of a kind + corner (AAAB)
(3) 3 of a kind + middle (AABA)
(4) 2 pair, sorted (AABB)
(5) 2 pair, nested (ABBA)
(6) 2 pair, alternating (ABAB)
(7) 1 pair, take 1-2 (AABC)
(8) 1 pair, take 1-3 (ABAC)
(9) 1 pair, take 1-4 (ABCA)
(10) 1 pair, take 2-3 (BAAC)
And that’s it! There always has to be at least one pair in a row of 4, since there are only 3 colors
(See the Hexagonal Pidgeon Hole Principle, for reference 😅)
What does bother me, however, is that I can’t yet see a “formula” for getting from 81 to these 10 unique cases…
The colors have 3! = 6 symmetries, and then there’s reflecting about the middle of the row, which gives us another 2 symmetries. This might suggest 6*2 = 12 symmetries overall, but clearly 12 does not divide 81 (and there are overlaps)
For instance ABAB can be transformed into BABA either by swapping the colors A and B, or by reflection of the row, so calling it 12 “full” symmetries would clearly be over counting
It’s been a while since I took group theory, but I’ve got a sneaky suspicion it applies here somewhere… 🤔
Edit: I found an answer to my “formula” question. Each case has a number of symmetries that divides 12, as follows:
3 symmetries: case (1)
6 symmetries: cases (4), (5), (6), (9), and (10)
12 symmetries: cases (2), (3), (7), and (8)
(The 3 symmetry case is monotonic and only cares about “color rotations” rather than pairwise swaps, and the 6 symmetry cases are already symmetric about the middle of the row up to color permutations, so they don’t care about row reflections)
Thus we get 3(1) + 6(5) + 12(4) = 3 + 30 + 48 = 81 total combinations
What would you call this in group theory… Some sort of coset decomposition?
A nice fact from combinatorics or field theory is that the middle binomial coefficients can only all be divisible by n when n is prime. So for 2 and 3, we get powers of 2 and 3 +1 as special numbers, for 4 colors there should be no special numbers (except 2), but then for 5 each power of 5 +1 is special, and so on only for every prime
Great video! A masterpiece.
3:20 ...wait so my gut instinct was RIGHT??? When I saw all the examples I thought "maybe the corners are related!" and it worked somehow???
Yeah me too
@8:34 I find 6 essentially different width 4 triangles.
Then, with a little program I find the number of different triangles for widths 1, 2, 3, ...
1 1
2 2
3 3
4 6
5 13
6 31
7 76
8 214
9 592
10 1726
11 5056
12 15019
13 44665
14 133630
15 399715
16 1197970
17 3590629
18 10768363
It seems to get close to tripling each time.
My program finds rotated triangles by calculating the entire triangle from the top row to get the left and right sides. I can see that there's an efficient way to calculate left and right sides using the embedded 3^n+1 sub-triangles. .. I don't have time to program it though.
8:50 I paused the video and figured that you only need to check 45 of the 81 listed. 9 are symmetrical at the base so 81-9 is 72. of those 72, each one is non symmetrical at the base and thus has an exact mirror image that does not need to be checked. so 72/2 = 36. then adding back the 9 symmetrical cases that must be checked gave me 45. I'm pretty dumb so please tell me if/how i am wrong.
I actually tried to think of the most intelligent thing to do when he asked the question, so I took those 2 corners and added them together because it felt right. How the hell did I get that right off the bat?
One of the first things i noticed is that the entire triangle works with the same rules if rotated under either of its other 2 sides. That is, every colour of every row is determined by the colours in the row above, no matter the rotation.