The Mosaic Problem - How and Why to do Math for Fun

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  • เผยแพร่เมื่อ 8 ก.ย. 2024

ความคิดเห็น • 177

  • @JohnDoe-ti2np
    @JohnDoe-ti2np ปีที่แล้ว +256

    According to combinatorialist Gian-Carlo Rota, Richard Feynman was fond of giving the following advice on how to be a genius. You have to keep a dozen of your favorite problems constantly present in your mind, although by and large they will lay in a dormant state. Every time you hear or read a new trick or a new result, test it against each of your 12 problems to see whether it helps. Every once in a while there will be a hit, and people will say: "How did he do it? He must be a genius!"

  • @yaitz3313
    @yaitz3313 11 หลายเดือนก่อน +71

    You should submit the (n,2) and (n,3) cases to the Online Encyclopedia of Integer Sequences. They're two interesting sequences of integers, which is exactly what the OEIS tries to collect.

    • @SirCutRy
      @SirCutRy 10 หลายเดือนก่อน +2

      Do it!

    • @TymexComputing
      @TymexComputing หลายเดือนก่อน +1

      My name is Sloane - i am a retired IT professor - i advise not to send any more sequences to my encyclopeadia as its already too slow..

  • @darkkupo5162
    @darkkupo5162 ปีที่แล้ว +59

    I've had a back burner problem throughout most of my post secondary education.
    The problem started because one day my dad made a joke about how the square root of 99 should be 33. This made me wonder if there were any numbers consisting of just a single repeating digit (greater than 10) that is the square of an integer. It's pretty easy to just check and see no such number exists; however, it gets interesting if you ask for which numerical bases can you find a repeating number that is the square of an integer in that base. for example 11 (4) in base 3 is a square of the integer 2
    more formally you're asking for what integer b does there exist n such that: sqrt((b^n - 1)/(b-1)) is an integer.

    • @NolanGrayson151
      @NolanGrayson151 10 หลายเดือนก่อน +2

      Hey, saw this comment and hope you don't mind me interjecting some of my own thoughts! That last part is if you restrict yourself to only 1 as the repeating digit. If you want other digits to be allowed, you're asking: for what integers b do there exist n,d with 1 = 2 such that sqrt(d * (b^n-1)/(b-1)) is an integer (d is the repeating digit here). Or in other words, for what b do there exist n,d,k with k>=1, n>=2, 1

  • @matthew6666
    @matthew6666 ปีที่แล้ว +82

    Nice video! I think I've always had some of these "back burner problems", which might explain a lot. I hope some students see this and get inspired with their own back burner problems.

  • @PaulMurrayCanberra
    @PaulMurrayCanberra ปีที่แล้ว +59

    My back-burner problem for a while was the question "What does it mean to double a probability?" It led me eventually to find the logistical function.

    • @invincible9240
      @invincible9240 7 หลายเดือนก่อน

      could u elaborate pls

  • @jackhanke343
    @jackhanke343  ปีที่แล้ว +27

    Thank you all for watching! There is a small correction at 11:22, as the notation t_{i, 2} should be t_{5-i, 2}.

  • @hashtags_YT
    @hashtags_YT ปีที่แล้ว +14

    Taking things a step further, imagine if teachers or professors employed this strategy of starting with a difficult question and slowly developing it, would certainly help make some concepts much easier to grasp.

  • @jackdinsmore4326
    @jackdinsmore4326 11 หลายเดือนก่อน +8

    Great video! Problems like the mosaic problem are actually very useful in physics. Counting all the mosaics with loops allows you to find the temperature at which a liquid evaporates into a gas in some simple models (we would call the generating function a cluster expansion, and the liquid gas model I was thinking of is called the Ising model). Just goes to show that sometimes your back burner problem might not be as esoteric as it seems at first!

  • @yellowthegreat691
    @yellowthegreat691 ปีที่แล้ว +35

    Very inspirational video, top tier content

  • @78Mathius
    @78Mathius ปีที่แล้ว +6

    Your voice is good for TH-cam and the animation was as visually stimulating as your voice over. On top of that, the content was interesting and the script well written. I hope you make regular math videos.

  • @AllenKnutson
    @AllenKnutson ปีที่แล้ว +2

    Incidentally, there is a much-better studied nearby problem, which concerns the probability that *all* squares are in a perimeter. Check out the vast literature on the "six-vertex model". Now there are additional questions to ask: how many loops do you expect? How is the boundary likely connected to itself (when one follows the paths)?

  • @spacelem
    @spacelem ปีที่แล้ว +5

    I already have a degree in pure maths at undergraduate, and an applied maths MSc, so I've been exposed to a lot of maths. However, I did not enjoy probability & statistics at uni (some poor lecturers maybe), and avoided it. Now however, it's really important to my job; also my major hobby is tabletop roleplaying, where there are many different dice rolling mechanisms. I therefore get a lot of fun out of investigating interesting dice mechanisms, and trying to calculate through various means (simulation, exact calculation, convolution and Fourier transforms...) what the various probabilities are. I also get to have fun programming it in different ways. This has been both fun and useful!
    Incidentally, for this problem my initial naive approach would be build a function that can determine if any given mosaic does not contain a loop, then simulate a bunch of random mosaics and apply that function to them and count the results. After a while I'd have a good estimate of P(0 loops), then P(>= 1 loop) is just its complement.

  • @hobbified
    @hobbified ปีที่แล้ว +1

    As a computer programmer I very often run into people online asking questions like "I'm learning , what should I write?" Or "I want to contribute to open-source, what project should I contribute to?" As a participant in various nerdy hobbies I run into people asking "okay, I'm new to this hobby and I just bought all the equipment they say I need to get started [ouch!]... now what do I do?"
    My answer to all of those people is very much related to this video: explore what interests *you*. Have something *you* want to accomplish. Wanting to learn a language, or partake in a hobby, or absorb a mathematical concept, aren't really the kind of goals that you can get invested in for their own sake. They're means to an end. You need to go into the game with an end in mind, something that drives you personally, and then you'll have a reason to knock down everything that stands in your way - and to discover something new that you didn't know you cared about in the beginning.
    Ordinary people can probably sustain one or two such interests. Really *interesting* people cultivate several and find ways to synthesize them.

  • @purplenanite
    @purplenanite ปีที่แล้ว +7

    I have my own "backburner" problem, which is kinda similar - I am going to try using some of the techniques in this video - thanks!
    My problem is this:
    there is an infinite grid of land and water. each tile has a p% chance to be water, and a (1-p)% chance to be land. Define a lake to be a contiguous stretch of water (corners don't count).
    1) When p=0.5, what is the average size of lake?
    2) What is the average size of lake as a function of p?

    • @dogeteam2235
      @dogeteam2235 ปีที่แล้ว +1

      wow comment if you succed I'm going to try too... but y'know

    • @purplenanite
      @purplenanite ปีที่แล้ว

      @@dogeteam2235 Yeah, I get it.
      I feel it's possible, but i do not know how.
      I call it "the riddle of the lake" because it sounds like something the sphinx would ask you and i find it funny.

    • @FreeAsInFreeBeer
      @FreeAsInFreeBeer ปีที่แล้ว +1

      Warning: spoilers about your backburner problem ahead!
      This problem is usually called "percolation". You don't need too much background to show that the lakes are always infinite for some threshold probability p, and that p must be larger than 1/4. With some more thinking you can show that the critical value is p = 0.5.

    • @purplenanite
      @purplenanite ปีที่แล้ว

      @@FreeAsInFreeBeer
      I am aware, but the *average* size of lake still stays finite even if there is an infinite lake - as in p=0.9 still has a valid avg lake size.
      For example, the probability of a lake of size 1 occurring is (1-p)^4 * p^1, (the probability of four land tiles and 1 water).
      If there is an infinite lake (this proof is trivial and is left as an exercise to the reader), then the average size of lake inside a bounded area A is approximately
      1*(A) (the one lake of size A) +
      (1-p)^4 p^1 A * 1 (many small lakes of size 1)
      divided by /
      1+((1-p)^4 p^1 A) (total number of lakes)
      if p=0.9, this is (A +0.00009A)/(1+0.00009A)
      in the limit as A->infinity, this limits to 11112.111...
      Of course, that one small lake is not the only lake, so this isn't 100% accurate
      So there is still a value, even if there exists an infinite lake.

    • @gabitheancient7664
      @gabitheancient7664 ปีที่แล้ว

      this reminds me of percolation

  • @danielyuan9862
    @danielyuan9862 ปีที่แล้ว +1

    One of my friends had a backburner problem himself. To state it in the most concise way: for a given k, what's the largest number n such that the numbers 1 to n can be partitioned into k sets, such that within each set, the greatest common divisor of every pair of numbers in the set is the same? For example, when k=2, the largest n is 8, since the numbers 1 to 8 can be split into two sets {1, 3, 4, 5, 7} and {2, 6, 8}.
    I had a point of obssession on this and generated code that has solved values all the way to k=15. For those of you who want to try to doe this yourself, brute forcing will get you to k=6. I had to use advanced techniques to grind my way to k=15.
    k f(k)
    1 3
    2 8
    3 15
    4 24
    5 29
    6 47
    7 49
    8 62
    9 79
    10 95
    11 120
    12 127
    13 142
    14 161
    15 168
    I might adopt the problem in the video as a backburner problem myself. Who knows? It might be my next obssession 😉

    • @danielyuan9862
      @danielyuan9862 ปีที่แล้ว

      Another backburner problem that I ended up abandoning is this: suppose you have an infinite minesweeper board, with a certain mine density p (meaning every square has a probability p of being a mine). I'm using the fact that if a clicked square does not have any surrounding mines, it automatically clicks the squares around it. If the density is low enough, there is a non-zero chance that the board will blow up infinitely, never to be seen again. So my question to you is this: what's the minimum probability where that doesn't happen, that the board never blows up infinitely?
      For those who want to try this, I've simulated this and hypothesize that the answer is somewhere between 0.1 and 0.11, give or take 0.01.

  • @smergthedargon8974
    @smergthedargon8974 ปีที่แล้ว +3

    Followup suggestion: Equivalent problems for hexagonal and triangular tiles - perhaps even a third for non-Euclidean tiling mosaics, if you'd feeling crazy!

  • @user-th5wd4fu9s
    @user-th5wd4fu9s 9 หลายเดือนก่อน

    I was doing this without realising the benefits, now when i think, they really helped me

  • @mikechad27
    @mikechad27 9 หลายเดือนก่อน

    Bro that intro was so good. I'm not in high school yet, but that made me remind to find something that interest me in math and that would also motivate me to study. One example is those mob farm Minecraft videos. I could not understand a single thing about the spawn rate analysis.

  • @mxxz1247
    @mxxz1247 ปีที่แล้ว +10

    Great video! The artwork was lovely and the message great. I am kind of biased as I mainly specialize in combinatorics, but that's a really neat problem. :) I had some of these "backburner problems" on and off again, but I didn't really do that on purpose or all-to frequent. Instead, my motivation in a course was sometimes focused on one single topic (e.g. Galois Theory in my Algebra course), so much so that it was hard to stay as focused for the rest of the class. I guess that's where more open-ended "backburner problems" would have come in handy.

  • @WhiterockFTP
    @WhiterockFTP ปีที่แล้ว +3

    wow. what a perfectly produced video, every second of it was just flawless.

  • @sdjhgfkshfswdfhskljh3360
    @sdjhgfkshfswdfhskljh3360 ปีที่แล้ว

    2:50 "what is the probability": first idea is to launch tons of randomized simulations.
    It will give rough value, which (hopefully) can help to find precise value.

  • @FineAndAndy
    @FineAndAndy 9 หลายเดือนก่อน

    Very nice problem! Two potential generalizations that could be interesting:
    1) What about different sets of tiles? Either by defining different entry-exit points for the square and then taking all ways of connecting them (e.g., dividing the sides into thirds rather than halves), or just by restricting which shapes of tiles to use (e.g., only the diagonals).
    2) Instead of "probability of a closed shape", what about "average area enclosed"? This would need some reasonable way to handle one enclosed shape contained inside another.

  • @moocowpong1
    @moocowpong1 ปีที่แล้ว +2

    I have a weird hunch that the full problem ties into TQFTs. You really want to be paying attention not just to fully closed loops of tiles, but open loops which cleanly connect one point on the boundary of your rectangle to another. As you add tiles, these have a chance of either continuing to make a larger open loop, or being closed off to form a closed loop, i.e. a connected shape. That feels like the same kind of thing TQFTs work with to me - although, the field has some pretty major different assumptions behind it, so who knows. At the very least it seems like it might be an interesting place to look for inspiration.

    • @jackhanke343
      @jackhanke343  ปีที่แล้ว +2

      Your hunch is correct, the authors of the paper I attached in the sources section also have papers that use various tile games and apply them to ideas in TQFT.

    • @T3sl4
      @T3sl4 10 หลายเดือนก่อน

      I was just thinking, on the infinite grid, you could have a line that comes in "from infinity", say from the left "edge", crosses the entire plane, and exits "to infinity" on the right "edge", therefore dividing the plane in half, without a defined curvature, and therefore the infinite enclosure problem is ambiguous or undefined. Compare to the analogous geometry circumstance: a straight line is also the limb of a circle of infinite curvature, where the enclosure is a degenerate case, or a matter of trivial definition (either the upper or lower plane is inside or outside; both possibilities are equally valid).
      What I don't know, is if there is a reasonable definition to get around this wrinkle; or, if indeed the probability of such a connected line is approximately zero even across infinite rows...

  • @squidwithcheese
    @squidwithcheese 4 หลายเดือนก่อน

    I finally got around to watching this video, and I can say that I’m glad I did!

  • @adityakhanna113
    @adityakhanna113 ปีที่แล้ว

    Haven't watched the whole thing yet but your point of recontextualization is really important! One of the ways is how you suggest: by having a problem to which that context can be repeatedly applied. Even if you don't have a pet problem, just thinking about new information and connecting it to what you know can help recontextualize things!

    • @adityakhanna113
      @adityakhanna113 ปีที่แล้ว

      Also your foley work is incredible

    • @adityakhanna113
      @adityakhanna113 ปีที่แล้ว

      I don't even have to 2x this video and am fully engaged! Great work! Extremely impressive

  • @billionai4871
    @billionai4871 11 หลายเดือนก่อน

    This is a really interesting concept for many math students. It absolutely would not have worked for me, cause ADHD doesn't understand the expression "backburner", but you did leave with itching with a different mosaic problem: given an MxN grid of randomly distribute tiles, what is the average length of the longest continuous line in the grid.
    I have some python to code now, lmao.

  • @BrianSpurrier
    @BrianSpurrier 9 หลายเดือนก่อน

    I think my first back burner problem was back in high school. I saw some post about a bakery that sold square donuts, and wondered how much more volume you would get with them in the same box out of it.
    I was just taking calculus around then, so I didn’t know the volume of a torus but figured it out, and was surprised that it was just a plain ratio between them, rather than being some messy function based on the 2 radii of the torus.
    I wondered if this was true for any polygon, and then had to figure out what exactly a polygonal torus was. I was stuck for a bit until one of my teacher’s mentioned the fact that a torus has the same volume as the cylinder you would get by cutting it open and unbending it. And then I realized that the shapes I was looking at could be unfolded into prisms in the same way, and the volume was just the area of the tube section times the perimeter of the larger polygon. The same way that a torus is the area of a smaller circle times the circumference a larger one. And that ratio was just (tan(π/n)/π)^2
    It was a very simple problem, but it was the first time I actually wanted to solve something for no reason other than curiosity and I used whatever knowledge I had at the time to do it. And it lead me to thinking about a lot more than the initial random post that started this, and made me think more like a mathematician. Even know, at least 6-7 years later, online I still use a square Toruhedron (they didn’t have a name so I made one) to represent myself a lot because it means something to me

  • @FunkyTurtle
    @FunkyTurtle ปีที่แล้ว +6

    Loved this journey through your intriguing problem! The animations were incredible and the explanations were very clear despite the tricky subjects.
    Regarding the second "further question": if the grid is infinite, wouldn't every shape be enclosed by and infinite amount of other shapes? this type of math is new to me but it seems like more and more shapes must pop up across the entire infinite grid, unless maybe the chances of a shape get too small as the shape increases... Interesting indeed.

    • @Stirdix
      @Stirdix ปีที่แล้ว

      Yeah, my first instinct was that it wouldn't converge. At the very least, there are definitely *some* configurations for which the answer is ill-defined, so unless that set of configurations has probability zero, then the question is ill-posed.

    • @jackhanke343
      @jackhanke343  ปีที่แล้ว

      @@Stirdix @FunkyTurtle I agree, and I had the exact question of is this question ill-defined in my script. I decided against it in the video so that the interested viewers would have to figure out which configurations they cared about. Thanks for the comment!

  • @mikaackermann4072
    @mikaackermann4072 ปีที่แล้ว +4

    I really like your approach and hope to see more in the near future!

  • @liambohl
    @liambohl ปีที่แล้ว +1

    I only know basic combinatorics, so when you showed the exact formula for t_n,2, i was shocked to see square roots! Anyway, i love the message and presentation of this video.

  • @joselor7614
    @joselor7614 10 หลายเดือนก่อน

    Maybe linking your problem to graph theory could open the way for a general solution in the n*m case : if you consider the vertices to be the sides of the squares and the edges to be the mosaic tiles, it may be possible to consider the probability of a certain non oriented graph having a cycle ( a closed shape in the mosaic case ). Thinking about it in this way may be useful because there are many useful results and definitions in graph theory that could make the problem easier to solve

  • @AgentM124
    @AgentM124 10 หลายเดือนก่อน

    You would think that on the one hand, it's probability 1 that a tile is enclosed in a shape. Because you can think of it that eventually there will be some huge shape that encloses all of it in between. But then again, the chance such a huge enclosing shape exists will tend to probability 0. Because it takes only 1 tile to mess up that enclosing shape.
    Definitely an interesting one.

  • @jeffhanke1662
    @jeffhanke1662 ปีที่แล้ว +2

    Amazing work Jack! Great artwork Kim!

  • @Illum1nato
    @Illum1nato ปีที่แล้ว

    Great video. I never really thought about the concept of a "backburner problem", but after watching your video I noticed that the subjects I enjoyed most were the subjects where I already had such a problem. I am definitely going to actively implement this in my further studies. Thanks for sharing this with the world. This video should definitely go viral!

  • @RubikxsMan
    @RubikxsMan 10 หลายเดือนก่อน

    I’ve never realised that the back burner problems I’ve had were probably helping me.
    In relation to the question, I can see some way to expand into the m,n cases but it would require classifying a whole lot of other properties of the rectangles. No where near as practically as you have done it here.

  • @ImaginaryMdA
    @ImaginaryMdA ปีที่แล้ว +7

    For me this was the Collatz conjecture. Particularly useful to get through p-adic integers.

  • @circuitcraft2399
    @circuitcraft2399 11 หลายเดือนก่อน

    My own thoughts on the problem:
    First, the parity requirement for interior-ness is not well-defined on an infinite grid, since there could be infinitely many closed loops around a given square.
    Regarding the original problem, perhaps there is some kind of inductive construction based on mosaics with a path from a certain point on the boundary to another. These seem easier to characterize, and might yield interesting insights.

  • @felixcourbois6624
    @felixcourbois6624 ปีที่แล้ว +1

    Commenting for the algorithm. Amazing content ! Deserve more views

  • @PretzelBS
    @PretzelBS ปีที่แล้ว

    My current back burner is minesweeper probability calculations. They’re A LOT harder then I initially thought

  • @screamingfungus_
    @screamingfungus_ ปีที่แล้ว

    Yo, you have the expanding-spray-can-in-Paint-trick as banner. Dope!

  • @sebastianmestre8971
    @sebastianmestre8971 26 วันที่ผ่านมา

    You can solve the n by k cases for small k using matrix exponentiation. To get an exact formula you can use the eigenvectors of the matrix. I did it for the 2 by n case and got the same formula that you got.

    • @jackhanke343
      @jackhanke343  20 วันที่ผ่านมา

      Interesting, what matrix was it?

  • @altairmislata5068
    @altairmislata5068 ปีที่แล้ว +1

    Nice video and problem! But I think you went wrong calculating t_5,2 because overcounting. Think that in the t_4,2 * 6^2 you are arleady counting some mosaics that end in a little diamnod shape and then counting that again in the t_3,2 * 6^(2*3) case. Nevertheless, using the Inclusion-Exclusion Principle there has to be a way of getting t_5,2 correctly.
    (Btw I'm learning english so if I just make a mistake I'll apriciate if you can teach me what did I wrote wrong and I hope you get my point)

  • @AndreasWilfer
    @AndreasWilfer 10 หลายเดือนก่อน

    This is my favorite of the #some3 submissions of 2023, as it is both useful and motivating to me personally.

  • @Aequorin628
    @Aequorin628 ปีที่แล้ว

    I was going to make a video just like this about my own backburner problem, but the hook of talking about how having a backburner problem is helpful never occurred to me! This video really makes me wish I committed to doing something for SoME.
    Also, I like the term backburner problem I'm stealing it

  • @UCXEO5L8xnaMJhtUsuNXhlmQ
    @UCXEO5L8xnaMJhtUsuNXhlmQ ปีที่แล้ว

    11:32 is an incredible equation that arose so naturally from the video so far. Bravo on the great communication

    • @ant961Handle
      @ant961Handle ปีที่แล้ว

      It's also wrong, although the general formula is correct.

  • @mauithedog10
    @mauithedog10 ปีที่แล้ว +1

    Amazing video! I’ve always had these back burner problems from time to time, but never thought to think of it this way! It always leads me to new areas of math and keeps me engaged, even though most of the time I don’t solve my original question.
    For anyone out there willing to give it a try, here’s a vague one I’ve been going at for a while: is there a formula that gives the dimensions that minimizes surface area but maximizes volume for any solid? For example, the solid that does this the best is obviously a sphere, which is always optimized. But what if I gave you a cylinder? Well then it’s only optimized when height =2*radius. You can solve this similarly (using sorta basic calculus) for other solids like cones. But what about any solid? Frustrums? Ice Cream Cone Shapes? The closest I got was that for all the solids I tested, at the “optimized point” the surface area^3 is always proportional to the volume^2 (again related to the sphere). I’ve been tugging at this one for a while and not sure if I can make it any more general than that. (For example, could you easily solve for the constant of proportionality just by knowing the shape). But anyone is welcome to give it a try!

    • @smugless191
      @smugless191 ปีที่แล้ว

      I might try to find the minimum surface area, maximum volume torus dimensions.

    • @smugless191
      @smugless191 ปีที่แล้ว

      Actually, I think that might be just a sphere lol, and I'm guessing the optimized non self intersecting torus has a radius such that it's almost intersecting with itself, R = r.

  • @gasparliboreiro4572
    @gasparliboreiro4572 ปีที่แล้ว

    I've got some backburner problems, but I feel like life itself has become my backburner problem, I get into every detail about every topic in mathematics, science and everything actually, just because EVERYTHING lies in the "is just logical" reality, even a person's illogical behaviour can be understood as a logical behaviour under human circumstances

  • @sdjhgfkshfswdfhskljh3360
    @sdjhgfkshfswdfhskljh3360 ปีที่แล้ว

    One fun category of problems is optimization problems.
    In some cases, it is not practically possible to find perfect solution.
    But it is clearly possible to find better and better solutions - with increasing difficulty.
    I formulated such problem for my training with programming.
    Amount of candidate solutions for my case is huge - approximately 2^500000.
    No brute force or semi-brute force (AI) can help to navigate such space.
    However, easiest solution still can be found in several hours.

  • @oinkoink3669
    @oinkoink3669 11 หลายเดือนก่อน

    A great explanation and I loved how you took us on the journey of your thinking and learning process 😊

  • @dominiquelaurain6427
    @dominiquelaurain6427 ปีที่แล้ว +8

    Hi Jack. I am at @10:00 watching and I comment before watching more. Good luck for the very friendly some3 "competition". ;-) you are good and your "backburner" idea is good...and old so the competition is not the motive. I had one in my childhood (how to pack circles in one open rectangle) but never succeed to solve it 555. I have more backrunners now but I prefer to use my imagination to work out new maths ideas (in, mathematical billiards), than learning "tips and tricks" (reference to "fish and chips") . Not anymore for me. You can of course generalize your backrunner by changing the tile shape (see "The Hat" discovery for 2023) or the drawings (see Wang tiles, Hankin polygons, random mazes, random path....). CU

  • @MrJaffjunior
    @MrJaffjunior ปีที่แล้ว

    At time 10:10 of the video, the number of connected shapes of t(5,2) is equal to 2 times t(4,2), minus symetric cases in t(5,2), plus one case that uses all tiles.
    t(5,2) = 2 x t(4,2) - Symetric cases in t(5,2) + 1
    The multiplication by 2 is to account for all the horizontal mirror reflexions.

  • @nosuchthing8
    @nosuchthing8 ปีที่แล้ว

    I picked a heck of a time to start watching a wicked hard math video...late at night..😳

  • @Alex-02
    @Alex-02 10 หลายเดือนก่อน +2

    7:18 I got a bigger answer: 32070. I could be wrong but I think you may have forgotten the diagonal rectangle shape that adds 6^2 * 2

    • @kuiperbolic
      @kuiperbolic 9 หลายเดือนก่อน +1

      Seconded!

    • @kuiperbolic
      @kuiperbolic 9 หลายเดือนก่อน

      Next day realization from glancing at my sketch: these rectangles are actually not valid. They require a tile in the middle which has 2 diagonal lines on opposite corners!

    • @Alex-02
      @Alex-02 9 หลายเดือนก่อน +1

      @@kuiperbolicOh you’re right! Thanks for pointing it out this had been stuck in my head for a while!

  • @nicemathproblems
    @nicemathproblems ปีที่แล้ว

    Love the idea of this.... Do math you are interested in and all math will become more interesting! Thanks for the video and introducing this fun problem.

  • @julesfouchy9444
    @julesfouchy9444 ปีที่แล้ว

    Thank you for the video ! It is so good it makes me want to pick up math studies again !

  • @yoavshati
    @yoavshati ปีที่แล้ว

    Haven't watched the whole video yet, but the intro reminded me of Vihart's videos

  • @ifroad33
    @ifroad33 ปีที่แล้ว +6

    I enjoy maths until I feel like I can’t apply it to any programming problems that I may encounter. Linear Algebra is super useful for a lot of optimizations in programming, but when it comes to abstract things, such as topology for example, I have a hard time finding motivation to learn it.

    • @chaotickreg7024
      @chaotickreg7024 ปีที่แล้ว +2

      You can't render that stuff?

    • @ifroad33
      @ifroad33 ปีที่แล้ว +2

      @@chaotickreg7024 That's true, if you're into rendering and visualization of data then I suppose it can definitely be a fun task, but I was thinking more generally about programming. Using algorithms for searching lists of items, or finding intersections between different geometric shapes, etc. These problems can be solved in many different ways, but to find the best algorithms there is a lot of mathematics involved which is what I think is interesting. But there is a limit to where I feel like I can apply different mathematical concepts to these problems, and that's where I have a hard time motivating myself to learn any further.

    • @alexismiller2349
      @alexismiller2349 ปีที่แล้ว

      That's a shame, hope you get can get over it

  • @kaydenlimpert2779
    @kaydenlimpert2779 หลายเดือนก่อน

    here's a definition for an open which doesn't over count, the definition will be found when clicking "read more"
    the new definition for an open can be anything if a loop isn't possible, else pick anything that isn't the mosaic which makes a loop

  • @rubiks6
    @rubiks6 ปีที่แล้ว

    I wanted to solve for 2π/7 because I thought cutting circles was interesting. I spent 2 years in correspondence with a mathematics professor from the University of Maryland (and others) addressing this number from different angles and I went into many fields and learned just so much cool stuff. I never became a mathematician but I have so many good concepts in my head and I use them in my daily thinking. I am a theologian.

    • @georgen9755
      @georgen9755 ปีที่แล้ว

      Clock = clock.tick(60)

    • @rubiks6
      @rubiks6 ปีที่แล้ว

      @@georgen9755- I don't know what you're trying to convey. Looks like a line of computer code.

  • @xenathcytrin202
    @xenathcytrin202 ปีที่แล้ว +2

    For the last problem, if youre on an infinite grid, isnt the probability that there is another shape that encloses everything you have examined so far 1? and the probability that that shape is also enclosed is 1, and so on.
    How then could you ever conclude how many shapes a tile is enclosed within?

  • @AOSABS
    @AOSABS ปีที่แล้ว +1

    I definitely obsessed over one of these types of problems. Mine was to take a regular polygon with “a” equally spaced points and “b” sides, and connect all the dots excluding those of the edge. Then find an equation to describe how many lines there were. It took me a few years and getting to algebra 2 before I got it!

  • @haukur1
    @haukur1 ปีที่แล้ว +3

    Be sure to pick a problem that is at least mildly tractable, and not something people have been struggling with for decades or centuries like integer factorization.

  • @Lukeisun7
    @Lukeisun7 10 หลายเดือนก่อน

    Amazing video! I took a combinatorics class in my undergrad and loved it. I'm a bit rusty now though lol. We used the Brualdi textbook and it was not my favorite.

  • @PaulMurrayCanberra
    @PaulMurrayCanberra ปีที่แล้ว +1

    It is very much the case that it is much easier to go from the specific to the general, and this is why attempting to learn math from wikipedia is so incredibly frustrating. The math articles generally have any specific pictures and examples removed, and just present the general findings as algebra and as references to other similarly context-free pages. I still don't get what a wreath product is.

  • @leonsoderberg6094
    @leonsoderberg6094 ปีที่แล้ว

    Amazing video🔥🔥🔥

  • @adarshkishore6666
    @adarshkishore6666 ปีที่แล้ว +1

    This was awesome!

  • @JTGROT
    @JTGROT ปีที่แล้ว

    congrats on 1K subs! 🎉🎉🎉

  • @brighamhellewell6479
    @brighamhellewell6479 ปีที่แล้ว

    your content is so good you should have more fans and a patreon.

  • @HaphazardDisastard
    @HaphazardDisastard ปีที่แล้ว

    I occasionally think up random problems like this, and I generally solve them with a bit of programming (often a Monte Carlo simulation).
    While I enjoy probabilities, I find them difficult to process when I only ever think about them every few months. It just evanesces from my mind over time, and my understanding lessens. It's rather annoying.

  • @amitlanis3104
    @amitlanis3104 ปีที่แล้ว

    Great video👏👏 Thank you for creating this quality content for us.

  • @johnchristian5027
    @johnchristian5027 ปีที่แล้ว

    Nice video!

  • @ant961Handle
    @ant961Handle ปีที่แล้ว

    the formula given for the case of t(5,2) is inconsistent with the general formula given later (t(5,2) is even defined in terms of itself), the general formula is correct I think there was just a mix up with the index notation when considering the 5x2 example. I think what happened is that the index i was mixed up with the length of the connected shape that included the 2 extra squares on the end but i was actually introduced as the length of the region that does not contain a connected shape. Still an excellent video though, I keep coming across combinatorics problems I can't make significant progress because of the large degree of overlap and having to consider the inclusion exclusion principle very carefully so it was interesting to get a taste of the standard ways to solve these kinds of problems with linear recurrence relations and generating functions.

  • @micayahritchie7158
    @micayahritchie7158 ปีที่แล้ว

    Once i saw you write down a recurrence relation my first thought was eould generating functions give me a bice closef form here

  • @kkski4817
    @kkski4817 ปีที่แล้ว

    Fun video! I want to get into the problem myself now 🎉

  • @visolate
    @visolate ปีที่แล้ว

    Fantastic video mate.

  • @Number_Cruncher
    @Number_Cruncher ปีที่แล้ว

    Great.

  • @coopikoop
    @coopikoop ปีที่แล้ว

    This is so weird. I have my first real analysis class tomorrow, and I purchased the textbook you showed a picture of not even 6 hours ago

  • @dominiquelaurain6427
    @dominiquelaurain6427 ปีที่แล้ว +1

    I forgot : I use "cocalc" for computer help. Good for symbolic computation in my case : I use it to verify (and cheat a little). For example : you can code in Python "f1 = (x + y)^2 - (x^2 + 2*x*y + y^2) ; print "f1 = ",f1.full_simplify()," vs 0" to check an easy identity ;-) WARNING : using a CAS is NOT a good idea for backrunner users......it is YOU human, who need how to guess and use tools you need, not a chatbot.

  • @zerauj_kcire
    @zerauj_kcire ปีที่แล้ว

    You're describing the significance learning by the psychologists :) it's pretty nice and it will be great if you check it out all of the ideas of Vigotsky and its contemporaries :)

  • @thatotherdavidguy
    @thatotherdavidguy 7 หลายเดือนก่อน

    I'm curious how the numbers chance if you add in two additional tile types: one where you have both corner cuts and ones where you have the two cross-cuts where one passes over the other (a bridge). For the n,2 case it wouldn't allow for any new shapes but it would change the denominator [now there are 8 tiles] and numerator [anything that uses the cross-cut or the corner cut can also use the dual] but doesn't allow for any new shapes. Definitely for 4 and up it adds new shape possibilities.

  • @danielyuan9862
    @danielyuan9862 ปีที่แล้ว

    14:35 saying that a, b, and c are the recipricals of roots of 1-228x+2699x^2-7758x^3 is unnecessary, as those are just the roots of x^3-228x^2+2699x-7758

  • @jonathanjam1158
    @jonathanjam1158 ปีที่แล้ว

    I have a question related to your further question related to the infinite tiling. Is it possible that for any set of tiles, it's probability of being inclosed fully inside at least one closed shape is one?
    If so, your question regarding the parity inclosing shapes in the infinite plane would be undefined, since we could proove by induction that every tile would have probability one of being inclosed by infinitely many boundaries.

  • @RobinFiveWords
    @RobinFiveWords 6 หลายเดือนก่อน

    If you start sketching ideas for a problem, go to save your file, and realize you already have a file of the same name from when you did the same thing two years ago ... you may have a backburner problem.

  • @samylahlou
    @samylahlou ปีที่แล้ว

    Excellent video !

  • @allanwrobel6607
    @allanwrobel6607 ปีที่แล้ว +2

    Interesting: I would like to ask , on the infinite plain, at what distance between connected shapes would the probability exceed 0.5?

  • @alyssadoubrovsky1465
    @alyssadoubrovsky1465 ปีที่แล้ว

    I believe your answer for the possibilities of connectedness on a 3x3 grid is off by 24, there’s four types of loop that can form on a 3x3 grid and you failed to count the reflections of one of them

    • @ant961Handle
      @ant961Handle ปีที่แล้ว

      I just calculated it and I get 31998 as well. I find 4 types of loop and the largest loop has mirror and rotational symmetry so no need to multiple by 4. The other cases are multiplied by 4 to account for reflections/rotations.

  • @SirTasie
    @SirTasie ปีที่แล้ว +1

    On the infinite grid, if enclosed means enclosed by an odd number of shapes, wouldn’t the probability that any cell is enclosed be undefined? If you you zoom out far enough you would eventually find, by pure random chance, a shape that encloses all previously seen cells, and if you then zoom even farther out won’t you find a shape that encloses the previous biggest shape, and so on forever. If enclosed instead meant within the boundaries of any number of closed shapes, then the probability of a cell being enclosed would be 1 or 100%, or at least approach 1 as the size of the grid approaches infinity, right?
    In the odd number of shapes means enclosed case, I think the probability is undefined because you would never reach an all enclosing shape so you could never determine if a cell is enclosed by an even or odd number of shapes. Infinity doesn’t have a parity, right?
    Explain why I’m wrong because I currently don’t know how to check.

    • @jackhanke343
      @jackhanke343  ปีที่แล้ว

      I think you are right! I originally included that this question may be ill-defined in my script, but decided against it to encourage viewers to think about it. Regardless, there may be a way to formulate this question to get some kind of answer, I just have no idea what that might be.

  • @rounitkamal2832
    @rounitkamal2832 ปีที่แล้ว

    Really enjoyed it

  • @dolos9250
    @dolos9250 9 หลายเดือนก่อน

    this is a good video

  • @ajaykiran6
    @ajaykiran6 ปีที่แล้ว

    This will be one of your viral videos

  • @omargaber3122
    @omargaber3122 6 หลายเดือนก่อน

    genius!

  • @SuperCityscan
    @SuperCityscan ปีที่แล้ว

    I had a similar problem to this, but it was given as a programming problem.
    First the mosaic had 3 lines with different primary colour, The three lines would overlap each other creating new colours or don't. There weren't random tiles but a set of tiles and I (we as a gorup actually) had to build a n*m grid of tiles with the best scoring according to what colour and how long circles it had.
    The solution wasn't that good in my opinion and ever since then I wanted to make it right.

  • @lexibyday9504
    @lexibyday9504 11 หลายเดือนก่อน

    I've still never personally solved my fun math problem. I want a non repeating arangement of coloured tiles where no to tiles the same colour ever touch. I have realised that this is only possible if the shape of the tiles is a non repeating pattern but which shape is the best shape for this puzzle?

  • @1.4142
    @1.4142 ปีที่แล้ว

    Conway also said something similar

  • @claytonharting9899
    @claytonharting9899 ปีที่แล้ว

    Please correct me if I’m wrong, but wouldn’t the probability that any given tile is enclosed by a connected perimeter on the infinite grid be 1? Because no matter where you are on the grid, somewhere out in infinity, there is a ring of connected tiles enclosing your local area?

  • @xl000
    @xl000 ปีที่แล้ว

    Also See Knuth's work
    Especially Vol 4A and 4B

  • @David-gj6dc
    @David-gj6dc 11 หลายเดือนก่อน

    My back burner question for part of my undergrad was about finding numbers that are palindromes in multiple bases, in particular base 2 and 10, like 585 (1001001001). I wanted to know if there are infinite of them, and if I could prove it. I never made much rigorous mathematical progress though but did write code to search for them

    • @btd6vids
      @btd6vids 3 หลายเดือนก่อน

      Trivial example but:
      5 (101) in base 2 is 11 in base 4
      17 (10001) in base 2 is 101 in base 4
      in general 4^n +1 is a palindrome in both base 2 and base 4, and you can extend that to the case where the larger base is any perfect power of the smaller one.

  • @JavierSalcedoC
    @JavierSalcedoC ปีที่แล้ว

    I would have guessed around a 2x50 grid to get 50% chance of getting a closed shape. it's just a 2x2 diamond after all

  • @Witcheridoo
    @Witcheridoo ปีที่แล้ว

    This is a great some3 video! Did you use Manim?

  • @truefiasco2637
    @truefiasco2637 ปีที่แล้ว

    doesn't the area enclosed by an odd amount of perimeters have to be 50% because there has to be the same amount of shapes?