11:19 it is usually written as omega*2 rather than 2*omega because of its multiplication definition.
@mostly_mental12 күн бұрын
I'm following the convention of the book I learned this from (A Course in Game Theory by Thomas Ferguson). And it seems to be pretty common in my other game theory books.
@yinq538413 күн бұрын
From Mathemaniac as well. Great video! If we alter the rule from "twice" to "three times", will we be dealing with another sequence?
@mostly_mental13 күн бұрын
Good question. If we modify the code at it 5:39 to use 3 instead of 2, we'll get a related sequence ( oeis.org/A003411 ), where each term a_n is given by a_(n - 1) + a_(n - 4), and all the winning moves seem to belong to that sequence too. And using 4 instead, we get a sequence with a_n = a_(n - 1) + a_(n - 6) ( oeis.org/A003413 ). That suggests there's a nice generalization where a factor of k gives us a sequence with a_n = a_(n - 1) + a_(n - (2k - 2)). There's probably a clever analog of the Zeckendorf representation that gives us a pretty proof, but I don't know it off-hand.
@yinq538412 күн бұрын
@@mostly_mental Thank you so much! I tried something like a_n = a_(n-1) + 2 a_(n-2) before I posted the comment but it didn't work of course. I never thought of a sequence of high order. Then is it possible to generalize to non-integer case, for example 5/2 or e?
@mostly_mental12 күн бұрын
@@yinq5384 You certainly can. For every ratio I've tried, it looks like you get a sequence with the same sort of behavior (P positions form a sequence, and winning moves from N positions in that same sequence). There almost certainly won't be a nice recurrence relation for every non-integer value. But these sorts of linear recurrence relations are really exponential, so I wouldn't be too surprised if there's a relationship between the chosen ratio and the base of that exponential. If you're interested, this might make for a good research problem.
@Roberth911able14 күн бұрын
That's very good explanation. Thank you:)
@mostly_mental14 күн бұрын
Glad you liked it. Thanks for watching!
@pawebielinski490315 күн бұрын
Hello, I'm from Mathemaniac, gonna recommend the channel to my students :)
@mostly_mental15 күн бұрын
I'm glad you like it. Thanks for watching!
@appybane848119 күн бұрын
13:44 can't you just remove 13?
@mostly_mental19 күн бұрын
If the opponent takes 13, you can immediately remove the remaining 21 and win. The winning play isn't just to take a Fibonacci number, it's to take the smallest Fibonacci number in the Zeckendorf representation.
@johnchessant301220 күн бұрын
here from Mathemaniac! this is really awesome
@mostly_mental20 күн бұрын
Glad you like it. Thanks for watching!
@Djake3tooth21 күн бұрын
The other (nim) cube-roots of 2 are ω⊗2 and ω⊗3, because exponentiation distributes from the right on multiplication, just as we'd expect . And you gave the hint that 1, 2 and 3 are all cube-roots of 1 :)
@mostly_mental20 күн бұрын
That's exactly right. Well done!
@leventelikhanecz22582 ай бұрын
are there any assisting arcs to help manual tiling of spectre tiles like those on penrose kite and darts? in this video at 10:06 (anyhow, the video begins with the same decorated tiles.) kzbin.info/www/bejne/amnWdKtjrKemaZY
@mostly_mental2 ай бұрын
Not that I know of, but I wouldn't be surprised if there's a way to do it. When we're tiling, arcs as on the Penrose tiles, and curved sides as on a spectre, serve the same purpose. They're both there to ensure sides meet up in the correct direction. The wiki page for Penrose tilings ( en.wikipedia.org/wiki/Penrose_tiling ) shows how you can force the rhombic Penrose tiles to be aperiodic with either method.
@leventelikhanecz22582 ай бұрын
@@mostly_mental thanx your prompt answer. i guess for the weakly chiral, strictly chiral arcing thing comes with physical tiling. where i can flip the stones by accident. on the basic straight lines spectre stone would be sufficient to mark the sides. on screen i can avoid flipping.
@unrealunreal74952 ай бұрын
E' spiegato benissimo!
@mostly_mental2 ай бұрын
Glad you liked it. Thanks for watching!
@ArslanRozyjumayev2 ай бұрын
Pure Gold!
@mostly_mental2 ай бұрын
Glad you liked it. Thanks for watching!
@pma14702 ай бұрын
beautiful ❤
@mostly_mental2 ай бұрын
Glad you liked it. Thanks for watching!
@mr.theking24843 ай бұрын
If monotiles cannot exist in 3d (which we do not know yet), I don't think the same would apply to 4d or any other nD where n=x^2, but I have no reason to believe this other than intuition
@mostly_mental3 ай бұрын
It's certainly possible, but I'd be a bit surprised. My guess is that either there are monotiles in every dimension (with possible exceptions of 4D, 8D, or 24D, since those are always weird), or there's some cutoff where it's possible for nD or lower but not (n+1)D or higher.
@Mumpie-JI3 ай бұрын
Great video
@mostly_mental3 ай бұрын
Glad you liked it. Thanks for watching!
@sumdumbmick4 ай бұрын
if zero is the empty set, {}, and the constructor for Surreals starts off by dividing two empty sets, {|}, then the construction of the Surreals in fact only ever generates different names for zero: {} = 0 {|} = {{}|{}} = 0 {0|} = 1 = {{}|{}} = 0 {1|} = 2 = {{}|{}} = 0 {0|1} = 1/2 = {{}|{}} = 0 and no matter what you do, this is how every 'surreal' is constructed. which means you're an idiot.
@sumdumbmick4 ай бұрын
add to this the fact that using the binomial theorem to expand the definition e reveals that e is the value of a non-Real Levi-Cevita series, despite the fact that e is supposed to be Real, and... oops. you're kinda wrong about everything, aren't you? I mean, if the Reals are continuous, as per Dedekind Completeness, then not only should you ask yourself where the hell the Surreals are supposed to exist, but now we also need to ask how e can be non-Real despite existing on the Real axis. but you're too dumb to have even noticed these problems in the first place, so... why are you attempting to educate others about things you do not remotely understand? watching your video is like trying to learn how to read from a cockroach running across a newspaper.
@jeffdodson4 ай бұрын
Who is the guy on the Mostly Mental videos on KZbin? I learned great gobs of stuff from his video on the Spectre tile. I would love to be able to email him, and bounce some ideas around with him. He introduces himself on his videos as Foxy or Doxy or something like that.
@mostly_mental4 ай бұрын
Hello, Foxy here. I'm just some guy on the internet who likes talking about math. My contact info is on my channel page, though I can't promise I'm very responsive.
@Syklhun-e1x4 ай бұрын
Ces formes je les ai decouvert en 2013
@mostly_mental4 ай бұрын
It's certainly possible. The authors weren't the first people ever to look at these shapes. I went to a talk with Craig Kaplan, and he mentioned that in the research process he stumbled across a list of shapes for a completely unrelated problem which happened to also include the hat. This group just happened to be the first to prove and publish the aperiodic property.
@Atticusrobins4 ай бұрын
You should write a book. I’d buy so many copies so I could have one at all times
@mostly_mental4 ай бұрын
I'm glad you like it. Thanks for watching!
@vedantsingh57864 ай бұрын
can you provide the proof that the sprague grundy value of a position in the game of twins is the nim sum of the sizes of piles
@mostly_mental4 ай бұрын
Let's (for example) say we have a game of nim with piles 1, 3, and 4. Those have nim sum 6, which means there's a move which reaches a position with each value from 0 to 5. Now look at the corresponding game of twins, with red coins in spots 1, 3, and 4. Take the move corresponding to each of those nim moves, and you'll reach positions where the nim sums of the red coins are each value from 0 to 5. Also note there's no way to reach a position with nim sum 6, since the corresponding move in nim would also have Sprague-Grundy value 6, which isn't allowed. So we can always reach positions with each smaller nim value, but not the same value. And the end position (with all blue coins) has nim sum 0. That's exactly the relation we used to define our Sprague-Grundy values. So the Sprague-Grundy value of the twins game is the same as nim sum as the piles.
@vedantsingh57864 ай бұрын
@@mostly_mental Thanks a lot! Can you provide me with some resources that can help me understand how to formally prove stuff like these (game theory, number theory stuff)? Love your content btw
@mostly_mental4 ай бұрын
@@vedantsingh5786 Glad you like it. If you're looking for a proper textbook, I first learned about the nimbers from "A Course in Game Theory" by Thomas Fergesen. I also highly recommend "Winning Ways for Your Mathematical Plays" by Berlekamp, Conway, and Guy. There are also plenty of good free resources online (I usually just search for "Combinatorial Game Theory PDF"). I don't know much number theory, so I can't help you with resources there.
@AfroSnackey4 ай бұрын
how did you define the length rad3?
@mostly_mental4 ай бұрын
The hat and turtle were both discovered on a grid of kites, which are really just two 30-60-90 right triangles glued together along the hypotenuse. So the ratio of sides is sin(60), or sqrt(3).
@amitganguli80114 ай бұрын
Great explanation. In our remodeling we were considering using Penrose tiles. Then the hat and soon after spectre were discovered, now we are looking for a tile maker who'd make this for us.
@bongo50_5 ай бұрын
I found this very interesting and you explained it very clearly. Thanks for making this video.
@mostly_mental5 ай бұрын
Glad you like it. Thanks for watching!
@lexinwonderland57415 ай бұрын
Yay! You're back!! I'm always excited to see your videos, friend :) Nim and game-theoretic extensions of number systems (like what Conway and Knuth developed) are truly an underappreciated love of mine. Thank goodness you have Aurora as your assistant!!
@mostly_mental5 ай бұрын
It's good to be back. Game theory is always fun to talk about, and I'm glad there are people who want to hear about it. And yes, Aurora is the best, and I couldn't do it without her.
@ShalevWen5 ай бұрын
I would like to see a video about Chomp
@mostly_mental5 ай бұрын
Excellent suggestion. I'm not sure I know much more than you do (and I'm pretty sure it's still an open problem, so no one else does either), but I'll do some research and see if anyone's published some partial results that could make for an interesting video.
@lexinwonderland57415 ай бұрын
@@mostly_mental hey, even if you aren't an expert on Chomp, you'd still have the most thorough KZbin education video about it. I encourage you to do it!! (I dont know what Chomp is, either, I just know that I love game theory and especially your videos)
@JulianEpsilon5 ай бұрын
9:15 proof If f_n is the highest fib which goes into some number, then [f_(n-1)]+ [f_(n-3)] + [f_(n-5)] ... + [1] is the largest number we can construct without using f_n and following the rules. we can expand each term[f_(n-2) + f_(n-3)]+ [f_(n-4) + f_(n-5)] + [f_(n-6) + f_(n-7)]... + [1 + 0 ] . This sequence is the sum of the first n-2 terms and it sums to f_(n) - 1 . (easy to show via induction.. f_1 + f_2 + f_3 = f_5 - 1, therefore f_1 + f_2 + f_3 + f_4= f_5 + f_4 - 1 = f_6 - 1) So the biggest number we can create without f_n is less than f_n and therefore less than the number we are trying to sum to. This means f_n must be used in our representation. then we can use the same process to deduce that in order to reach (our target - f_n) we must add the largest fib which fits into (our target - f_n). So in order to get a valid representation, we're going to get a single list of numbers we must use, therefore unique.
@mostly_mental5 ай бұрын
Looks right to me. Nicely done.
@Vierkantswortel25 ай бұрын
Great explanation! Glad you're back!
@mostly_mental5 ай бұрын
Glad you like it. Thanks for watching!
@purplenanite5 ай бұрын
Aurora seems like a good sport!
@mostly_mental5 ай бұрын
She really is the best. She's taught me so much cool math. I'm lucky to have her as my assistant.
@AdriandeSilva-rl3lg5 ай бұрын
You know another thing that's also mental, only completely so, it's the topological inversion of a sphere inside out...
@mostly_mental5 ай бұрын
Yeah, that one's absolutely wild. I watched that video ~15 years ago, and it was a big part of my falling in love with topology.
@MichaelDarrow-tr1mn5 ай бұрын
cool facts about hyperbolic geometry: almost every path is close to straight equidistants (paths that stay the same distance from lines) converge to a horocycle (circle of infinite radius) as the guiding line goes away
@eamonnmccallum22826 ай бұрын
is there a link to that game ? what great visual add. thanks for the vid
@mostly_mental6 ай бұрын
I'm glad you liked it. HyperRogue can be downloaded at zenorogue.itch.io/hyperrogue (or it's available on Steam).
@nihilisticnirvana6 ай бұрын
i love your channel!!!!!!!! please don't stop making videos, they're great and i love learning math without the pressure of exams!
@mostly_mental6 ай бұрын
Glad you like it. Thanks for watching!
@mrflibble57177 ай бұрын
Good explanation, simple approach. Nice!
@mostly_mental7 ай бұрын
Glad you liked it. Thanks for watching!
@RickyMud7 ай бұрын
Just turned in. Find it strange to be ending on hackenbush so I’m curious where you started
@mostly_mental7 ай бұрын
I started with impartial games and the nimbers (which I find more interesting than the surreals). If you're curious, you can check out the rest of the series here: kzbin.info/aero/PLH5zdqQODdBiGrWszPScMO2Dvp0Ix_vpV
@azzy6377 ай бұрын
Does one orbit refer to one shell/one layer of the sphere? This video is the perfect blend of maintaining the technicality of the concept along with making it easy for the viewers. Thank you so much for such a wonderful video.
@mostly_mental7 ай бұрын
I'm glad you liked it. The orbit of a point is the set of points you can reach from that point using any sequence of a's and b's and their inverses. Everything in this video is referring to the hollow outer shell of the sphere, but it's not too hard to extend it all to the solid sphere (excluding the center, which has to be handled separately).
@azzy6376 ай бұрын
Understood. Thanks for the quick reply 😊
@guyvan10007 ай бұрын
I highlighted the need to see this video on the Wikiversity article "Surreal number" and will attempt to include a link to this video on the Wikipedia article with the same name. Nothing on any of these wikis come close to explaining surreal numbers this well.
@mostly_mental7 ай бұрын
I'm glad you liked it, and I appreciate the enthusiasm, but I'd rather not be directly linked in those articles. Maybe just add a section summarizing the connection to Hackenbush?
@guyvan10007 ай бұрын
@@mostly_mental I will be happy to oblige. Just to be clear, you want me to remove the links from Wikipedia or Wikiversity out to this KZbin video? I will wait until you verify that this is you wish (because I have no idea why you don't want the link). I control the Wikiversity page, so deleting that link there will be no problem. I will also delete the link I made at Wikipedia, but I can't guarantee that a Wikipedia editor won't reinsert it. I think I will just quietly revert my Wikipedia edit without comment. If I say anything it will draw attention to what they may decide is a good link from Wikipedia to this KZbin video.
@mostly_mental7 ай бұрын
@@guyvan1000 Thank you. I'm not a primary source on the topic, so I'm a bit uncomfortable being listed as a reference. Besides, everything I know comes from the books I mentioned at the end, which are both already listed on the Wikipedia page.
@guyvan10007 ай бұрын
@@mostly_mental There is a difference between an external link and a primary source on Wikipedia, but I am not well enough versed in Wikipedia rules to know if a link out of the Wikipedia page is proper. Wikiversity has entirely different standards (one might say no standards...) I would like to include a link from Wikiversity to your video, but won't do it without your permission.
@mostly_mental7 ай бұрын
@@guyvan1000 I think I'm okay with a link on Wikiversity, so long as the math is the primary focus and not the video.
@yingxu16948 ай бұрын
Amazing video!
@mostly_mental8 ай бұрын
Glad you like it. Thanks for watching!
@Dance_with_nandini.8 ай бұрын
How to measure the curve of the road ?
@mostly_mental8 ай бұрын
If you want to measure the length, the easiest way is to notice that the length of each arc of the curve must be the same as the length of one side of the square (which is 2). If you want to know how curved it is, the best measure is the radius of curvature, which essentially finds the radius of circle that best approximates the curve at each point. There's a formula here en.wikipedia.org/wiki/Radius_of_curvature (which comes out to cosh(x + c)^2).
@atamir83398 ай бұрын
sick bikies
@okbutuwilldie73909 ай бұрын
Nice
@pranjaljain124210 ай бұрын
thanks for the explanation of crossing paths I was struggling with it.
@Marek_Chojnacki11 ай бұрын
Good series! Too bad it seems to have been discontinued(?).
@mostly_mental11 ай бұрын
I'm still here. There's more to come when life gets less complicated.
@Marek_Chojnacki11 ай бұрын
@@mostly_mental I understand. Farewell
@suhana.a.a794911 ай бұрын
Sir, will you provide the reference book please?
@mostly_mental11 ай бұрын
I'm largely working from "Knot Knotes" by Justin Roberts.
@suhana.a.a794911 ай бұрын
@@mostly_mental Thank you so much sir for your reply. Your presentation and explanations are really amazing. Actually i thought of taking this topic as my master's project.🙏😊
@mostly_mental11 ай бұрын
@@suhana.a.a7949 I'm glad you like it. Thanks for watching!
@Trismhmm11 ай бұрын
By the way, keep Going! You are incredibly gifted at explaining complex ideas 🫂🛐⚡
@carlkuss Жыл бұрын
One thing that strikes me here if I am understanding it correctly is that you have this kind of spectrum in which the hat side consists of a continuum of hats between chevron and spectre and the turtle side consists of a continuum of turtles between spectre and comet with all these intermediary tiles having the property of tiling the plane only in an aperiodic fashion but involving a certain number of reflected versions (proportion phi to the fourth power) with the spectre in the middle which can be tiled periodically or periodically but which heads a family of spectre mutants having chiral figures taking the place of its non-chiral line segments. This infinite family of spectre mutants has the property of tiling the plane only periodically. Cool!
@carlkuss Жыл бұрын
Also: that the hat in the strict sense, like the turtle in the strict sense, is constituted by an assemblage of parts that you get from the (periodic!) tiling of the plane into hexagons with their corresponding dual (tiling in triangles). So that the whole aperiodic tiling is there in your face amidst this simple (periodic) tiling into hexagons and triangles. If I am right about that. Weird!
@mostly_mental Жыл бұрын
@@carlkuss Yeah, you've summed it all up pretty well. It's wild that stitching triangles and hexagons together is all you need to get so much complexity.
@brandonmassaro3222 Жыл бұрын
I’m having to pick some of this up for research, great video!
@mostly_mental Жыл бұрын
Glad you like it. Thanks for watching!
@pokemonjourneysfan5925 Жыл бұрын
hello
@mateussteffler Жыл бұрын
another question, do you know where can i find something more specific about the "ein stein" problem that the hat broke? an article talking about this problem or something? thanks :)
@mostly_mental Жыл бұрын
"Einstein" is just another name for "aperiodic monotile". "Ein stein" literally translates to "one stone", so it's bit of a play on words.
@mateussteffler Жыл бұрын
hey, i would really appreciate if you could share your references for the aperiodic tiling with trapezoide tile demonstration, starting at 1:34. thank you!
@mostly_mental Жыл бұрын
strauss.hosted.uark.edu/papers/AHT.pdf describes the whole process in a bit more depth.
@mateussteffler Жыл бұрын
thank you so much man! :)
@shaunokane9600 Жыл бұрын
I appreciated the series. Thank you kindly.
@mostly_mental Жыл бұрын
Glad you liked it. Thanks for watching!
@calarquist3617 Жыл бұрын
Please keep doing videos about this topic, i was looking for so long!! Thanks🎉🎉🎉
@mostly_mental Жыл бұрын
I'm glad you liked it. I'm currently in the middle of a series on knot theory, but I plan to come back to game theory when I have the time.