A Colorful Unsolved Problem - Numberphile
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- Опубликовано: 26 фев 2019
- James Grime on the Hadwiger-Nelson problem.
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Extra footage from this interview: • A Colorful Problem (ex...
The Four Color Map Theorem: • The Four Color Map The...
More on James Grime (you can book him for talks): singingbanana.com
More James Grime videos: bit.ly/grimevideos
The de Grey paper: arxiv.org/abs/1804.02385
A good write up on this topic via Quanta Magazine: www.quantamagazine.org/decade...
The Marijn Heule counterexample: www.cs.utexas.edu/users/marijn...
More links & stuff in full description below ↓↓↓
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What if you connect the two center points together?
Hey Numberphile, if we possibly figured out the mathematical equation for the theory of everything, where do you recommend we try getting it tested?
How do we know that the grids presented generate walks that are posible in the 2d plane?
Can't wait until some mathematician finds an indescribably monstrous number that has something to do with this infinite plane.
can you make one with a Voronoi diagram of the pattern at 8:00 ? :D
James Grime means automatic click
he's so cute lol
while James_Grime:
click
True
i’ve been wanting to draw him for so long, he matches my style perfectly
Ayo?
With a name like "de-grey", what other field could he be working in but anti-aging?
nice
Anti women
idk, coloring graphs?
Bleaching
1. A conspiracy theorist who hates CGP Grey
2. A commenter who points out the mistakes, flaws, and misinformation in CGP Grey's videos
3. A person allergic to the color grey
This episode of Numberphile brought to you by the Ministry of Silly Walks
Darn. Beat me to it.
o man...
Every step I take
Every move I make
Every hexagon
Every time I walk
I'll be counting shades
All in all you're just another hexagon in the tile
Puff Daddy
The Police
After he said "Every step you take", I immediately scrolled down to find a comment about it :)
first thought after his explanation
looked in comment section
sure found it
This week on "EXTREME COLOURING with Dr. James Grime"...
There seems to be some vague hope in these videos, that, somehow in the comment section, some unknown mega talented kid just posts a solution to all of this.
It has happened. Some viewer did solve one thing
Many of these problems are not super difficult to require mega talented people, specially with the computing power we have today, they just don't attract the attention needed for it to be solved.
@@eriks1765 when? I need to know
I think I could take a fair crack at a theory of everything, but it's nothing new. There's nothing new under the sun.
The fun is in the trying.
[ THIS WALK IMPOSSIBLE ]
YEET!
This sentence no verb
RIP English
Who neeed grammarz?
[ REDACTED ]
Wait, there's a scientist called Grey who's field is anti-aging? That's a sign.
Maybe its our grey?
And he's colouring paths with great success
He claims we will have the first anti-aging therapies by 2037.
I feel that there should be more introductory courses to graph theory in education. There are so many problems that benefit from thinking about them in graph theory, it is doing students a disservice to leave them in the dark on it for so long.
Graph theory should be everywhere in math because of its insane versatility
0.53
James: its not something that has been solved
Me: MUST. SOLVE.
I love the framed brown paper by Ron Graham in the background
I was looking for that comment xD
@@dennismuller1141 D:
I love your profile picture.
It would be nice if they had drawn the graph at 5:10 to scale. It's impossible to draw without crossing edges.
Is it possible to draw this impossible walk with unit lenght?
@@RouteNRide
The points represent area on continuous map. It doesn't matter how you lay out the graph as map.
@@RouteNRide In the 4 colour theorem, what we can do is to represent each area by a single point and draw vertices between points that represent adjacent areas, in order to make a graph that will be finite and planar (vertices won't cross each other). Coloring the graph is finding a function that assigns a color to a single point such that no _adjacent points_ (i.e. points that are linked by one vertex) are the same color.
Now, take the infinite amount of points that are in the plane and draw a infinite number of vertices between points of distance 1. Here, many vertices will cross each other; thus, if you try to think of it as adjacent countries it would make no sense and that's what confuses you : _there are no areas to think about._
Now, in order to prove that this infinite graph can't be colored by 4 colors, it's sufficient to take some _finite part_ of that graph that can't be colored by 4 colors. That finite part may not be planar as well - _otherwise by the 4 color theorem it would possible to find a 4-coloring_ - and there would still be no areas to think about.
@@RouteNRide The problem is more than just painting a maps area, it is also trying to determine where the actual areas are. There are infinitely many possibilities. It could be a map of Europe, a checkerboard, or many regions at planck length size. Since there are infinite maps, they are using points in order to generalize the possible areas there could be. You can think of the points as spots on the map which we can deduce the color, and every other point not shown on the graph to be 'any color'. This is done so that you will solve the problem for ALL possible 2-color maps. Trying to solve every single possible 2-color map is impossible.
You could also think about the counterexample like this: Suppose I have a map with only two colors in which from any starting point, every unit step (say 1-meter) I make will be in a different colored region. If I make a walk with an EVEN number of 1-meter steps, I will end up on a spot with the same color I started on. If I make a walk with an ODD number of 1-meter steps, I will end up on a spot with a different color from which I started on. All I have to do to show that this is impossible is to find a walk of EVEN steps and a walk of ODD steps that both end up on the EXACT same spot. The most simplest example was shown in the video, as an equilateral triangle. From one corner, you can reach another corner in both one step and two steps.
I agree. Maybe not initially, but morph the not-to-scale one into the real one. It would give better context to the next section too
Dr. James Grimes picks the most authentic and beautiful subjects. I enjoy listening to him, every time. Sometimes I watch the SingingBanana as well! :)
I’m still waiting for another big breakthrough
to happen with the Riemann Hypothesis
There you are again. You're like my own personal Justin Y.
Evariste Galois I’d love to see it Monsieur Galois but it seems utterly intractable short of Wiles-esque solo run at the problem, ergo it’s not exactly a few papers away from a solution. Though with that noted, twin primes got an enormous shot in the arm in 2013 out of no where that brought it the hope of a general solution from a near zero probability to something we could expect in the ball park, so even when there’s no hope, there is still hope.
I'm currently using the RH as a side-problem so I don't get burnt out on what I really want to make happen (though it is fun to daydream about solving it every once and a while). Wrestling with it part-time for the last 2 years or so, I think we're going to need to see a fundamental change in how we think about objects like the zeta function and their definitions in order to prove the RH.
I've decided it's undecidable.
@@timh.6872 You're actually working on that kind of stuff? Wow nice
Oh yeah my boy Aubrey! I was looking for news on his aging work one day about a year ago and found that he had solved this in his free time. Cool dude. :)
It's usually way over my head, but James is so easy to listen to and learn.
Way to go Aubrey de Grey! I like his work against aging through SENS. It's super cool to see him making mathematical breakthroughs and how it inspired them to find an even better walk
Correction - 32 degrees offset for the Moser Spindle. I also suggest checking out the Golomb graph as another comprehensible proof that the answer is at least four. It has ten points rather than seven, but it has a lot more symmetry and an easier mental proof: the offset triangle must have all three nodes of different colors. In order to complete the hexagon in three colors, all three nodes which connect to the triangle must be the same. Ergo, four colors required.
I used to fail almost all of my math classes since year 9 (which was alright though 'cause I originally wanted to study chemistry or foreign languages anyway)
BUT especially Dr. Grime somehow managed to make me a numberphile aswell, his motivation and his enthusiasm are thrilling! Last maths exam for example, I got a score of 90 and I unironically started reading my maths books and doing fun algebra stuff in my spare time.
Odd, but I appreciate it.
thing I love about puzzles like this is how important parsimony is. finding how few is needed sets a wonderful base line for understanding the situation. even better, it makes it simple to teach others about! gods, i love number puzzles :^)
I'm a simple person. I see James Grime, I click.
how did you find the video
i still cannot wrap my head around the depths of insipidity one would have to fall to to find themselves unabashedly compelled to leave such trite, cookie-cutter comments such as this. what a dreary existence it must be...
@@noncanadian I believe you too are using pre existing things
TiredSounds never. everything i've ever thought and said is 100% original. they call me genesis on the streets because no man alive has caught me being derivative xD but seriously, what exactly are you insinuating by saying i too am using pre existing things? do you mean that i am using the basic axioms of logic and language? are you seriously attempting to undermine my original point with such absurd pedantics? i suppose if you wanted to be that disingenuous then yes you are right. everything anyone does is based on pre exising things. however, by taking a more charitable interpretation of my original response, i think it would not be difficult to parse that i was simply railing against the mad lib style of commenting and how unoriginal and boring it is
That one student in class who doesn't want himself to be understood by others in order to keep himself in talks because otherwise he knows he has nothing else to be proud of.
I was literally watching the 4 color theorem video luke 3 hours ago and got this surprise! Thanks numberphile!
Yes! Give me more of that James Grimes
I'd love to see a video on the Nurse Staffing Problem at some point. It's one of those issues that seems so mundane and easy that turns out to require super complicated math.
0:23, you're the world's best map maker
nice profile pic
As a native Michigander, I gotta call you out on your 4-color US map - you colored Michigan's Lower Peninsula blue and the Upper Peninsula purple. They're the same state!
This is such a wonderful problem, and video!
Seems so weird you would need such an insanely long walk to suddenly need the 5th colour. The hex tiling is genius though, did not occur to me that it works with hexagons slightly smaller than 1.
Yea, it is really weird to me too. If 2 colors required 3 points, 3 colors required 7 points, and 4 colors required 1000+ points...it just seems super arbitrary. Maybe 5 colors also requires around 1000 points, or maybe it will just grow another 3 magnitudes like before.
Thats your guess. If you want to study the research, you are allowed.
@@c0mmment ONE proof required 1000+ points with 4 colors, but no one said this is the "optimal" proof with the smallest amount of points. The "optimal" might as well be around 50 points or less.
It's like TREE(3)
Singing banana and numberphile ! Perfect combination
Very interesting topic. Great video, thanks.
Brown paper from the Graham's number video posted on the wall behind James, that's awesome lol!!
5:10 "not to scale" lol
James Grime talking about Aubrey De Grey? Two of my favorite human beings!
So very glad that this man is back in new numberphiles. His video's are always the most sympatic to watch
There is a problem with the counter example to three: The steps have different lengths. You would have to show that you can squeeze the graph in a way that makes all steps equally long.
yes, that is why it says "not to scale"
not sure why they did that though, it's not that hard to show what it looks like to scale
Why is it described as the steps between nodes are unit length but then the three color counterexample has differently lengthed edges connecting them?
"Not to Scale" ... Watch the video.
because its not to scale, as the real version looks a bit confusing (not really though). Searching Moser Spindle in google will show you the real prove
Dark there’s no scaling that can make different lengthed edges all unit length though
Pieter van Nes thx I think I will look into this one a bit more
well I mean scaling as in a linear transformation
Imagine being famously successful in one field, entering another as a free-time hobby, and just shitting on the scholars of that field.
So Gauss?
That's the dream
Math is not about being better, it's about finding solutions.
Marlon Oswald yea but humans aren’t computers and competition inspires.
Einstein was once a famous patent officer, until he dabbled in physics.
Nice! I've been waiting for an expanded skill tree for FFX
0:51 There is the sheet used for explaining the Graham's Number at the background. Well preserved😍😍.
That's what so fascinating about math. Simple problem but complex solution. 533 points!
So far. Maybe we'll find a smaller one which might lead to better general understanding of the problem or make it easier to make a 6 colour impossible walk
Considering the problem requires you to walk the same distance each step, I don't see how these graphs are useful unless you can also prove the graph can be represented on the plane with every edge having a length of 1. It's quite easy to come up with graphs that do not satisfy this property. Fortunately, the Moser Spindel does have this property, though you have to draw the graph in a very different way than shown in this video, thus the "proof" in this video is incomplete.
Agreed! He should've mentioned that the four colour theorem of maps and restricts the number of connections each vertex can have and to which other vertices. You can't just connect 10 dots all to each other and claim that 10 colours are required because you cannot make a map which has 10 areas of land all connected to one another, 4 is the max.
I don't understand your point, care to elaborate?
Can you give an example of a graph with a finite number of vertices and edges with weights of 1 unit that can't be drawn in plane?
edit: nevermind, I found one: (0,1) (1,2) (2,3) (3, 0) (0,2)
@@eugenajechiloae8262 Your example actually can be drawn in the plane. It would look like two triangles back-to-back with 1 and 3 on opposite ends. However, an example would be (0,1) (0,2) (0,3) (1,4) (2,4) (3,4). The first three connections can be used to triangulate any common connections, so 0 and 4 would have to be in exactly the same location.
Search moser spindle. It is possible to place the vertices at unit lengths along the walks. But it is easier to see the colours etc when spread out this way. Hence “not to scale”.
I admit that I misunderstood it at first with the last proof, but just imagine the insane beauty it would be if there was a way of colouring the plane in n colours - except for a single point. The whole plane can be coloured in n colours and satisfy the walk but a single point on the infinite plane would screw it up. Granted, that's not what was said but what I understood at first, but wouldn't that be one of the most amazing things ever?
My intuition tells me that's impossible due to the density of R^2: whatever finite argument you have for requiring that point to have a different shade, reapply the same to another point infinitesimally ever so slightly adjacent to the first. That second point will also have to be colored the different shade. I suppose this may not hold if the argument involves a continuum number of lines of reasoning within it, but I don't know about metaproofs stuff enough to say if that is likely. I'm guessing it's not?
Really astonishing
1930s movies: _oh I don’t think so_
Well, technically color film releases existed in the early 1920s, it was just do expensive and complicated that it was only used for short sequences until 1929.
The four colour map you refer to actually only works if every country is made of one part. If you have a lot of countries that are not continuous (like the US with Alaska ) you can make up maps that do not adhere to this rule.
Not some criticism, just a footnote
@TheDerpy Kitty And in fact is equivalent to assuming a planar graph.
Yup those do appear to look right, I would reckon there probably is a way to count connection lines between colors and plug them into a formula however it likely would be a solution where you have to compare 2 sets that see each other and see if they equalize out you would need and upper and lower bounds and will need like an "n" like n-1 or n+1 in it, this nearly reminds me of the match stick thing where you add one to every open end and grow it out! Taking any step does make the process seem infinite, why not do geometric patterns with like straight lines and use some slants, that will surely mess the walk up enough to eventually create a set of infinite slivers kinda like how knights move, the idea would be with that try to create situations in the fewest steps possible that makes you have to use 1 extra color than you wanted to use, so the idea is if you wanted to use 3 colors but example 5 was proven but 4 was not yet then you can try doing with 3 but if had to use 4 then if you were not forced to do 5 then you at least proved 4 finally, kinda like trying to force it to behave a certain way and just let it edge you on into equilibrium is my take on this, however would be nice to make a formula here! I just have certain feels about things when they look a certain way! I know myself I don't make formulas but I definitely can use them when presented to me xD.
Oh! We also need a pocast with James Grime.
For a few minutes I thought I had found a very simple solution for 5 colors, using just the classic coordinate grid to color tiles. Upon looking at it again, I actually just proved the opposite. No solutions for 5 or 6 colors exist on a uniform square grid. You would need some kind of color scheme that uses other shapes, or tiles of different sizes.
I have to be honest I don't understand how the proof for 3 colours and up work. The patterns don't comply with the 1 unit rule of the puzzle. As far as i can see if you were to create that diagram with only unit length rods between nodes the diagram itself would be impossible and therefore it cannot make any valid paths.
Am i missing something?
I think that it's more about graph theory than it is about maps (they used maps to make it more intuitive i guess) but they didn't clarify so maybe a quick google search could confirm that.
Yeah it's a bit deceptive. The Moser Spindle he is talking about actually looks a bit different (which is what they meant with not to scale). What they've shown is a simpler step before it's contracted in a way that all connected points are at unit distance.
Try to think of the map as choices of colours to step on, as apposed to actually coloured tiles.
They describe the perfect situation and show that it can't be done, and you're complaining that it can't be done so it can't be a proof?
For three colors, your road is in shape of an equidistant triangle. You start at any point on plane you want and mark it with color, lets say green. You move to any point that is 1 unit length away and color it with blue. You have a "base" for your triangle and you move to any of the two points that make it an equidistant traingle. You have to choose color green because you have only two colors available. Now you want to go back to the start point, drawing the equidistant triangle. You will land on green.
Since so many people are getting confused by this, please note:
1) The paths can intersect in this problem (unlike in the graph representation of the 4-Color Theorem).
2) The graphs are not drawn to scale (i.e., with the same distance between every vertex) to make them easier to interpret for the viewer. My intuition tells me that finite graphs, unless pathologically constructed, are likely to be embeddable with edges of uniform length into the 2D plane. In any case I'm sure the published solutions were checked for this requirement.
[edited with addenda]
only just noticed the red and blue square color editing in the outro, very sneaky easter egg
The way you drew the Moser spindel it is absolutely unclear that it can be realized with edges of unit length.
I tried to draw this to scale starting with the base line of three points and I ended with two pairs of back to back equilateral triangles, one on the left and one on the right whose furthest points (the top central line on the not-to-scale diagram) were three units apart. No way can they be one unit apart. (I thought. Wrongly.) Finding a diagram of the Moser Spindle I realised that you can treat these two pairs of triangles as hinged at the joining point and rotate them both upwards until the far points have narrowed to a distance of one unit.
Impossible with edges of unit length!
@@OrenLikes Yes it is possible. That is the point of the Moser spindle.
@@OrenLikes why did you comment this after somebody already explained how to do it?
0:50 Michigan counted as two colours, top kek
Michigan and West Michigan
I live in Michigan, and I don't doubt you, but just out of curiosity, how are the people different in the upper peninsula?? in general terms
Technically the 4 color theorem doesn’t apply if you have non-contiguous regions. Fortunately, 4 colors can still fill in the US despite Michigan’s *ahem* “issue”
@@ericsbuds in general the UP is more rural. the largest cities are Marquette, St. Ignace, Sault Ste. Marie (the soo), and Houghton.
Honestly when you get north of Lansing (the capital), you quickly see country looking an awful lot like the UP. In-between the largest cities in the northern lower peninsula (Traverse City, Gaylord, Mackinac City, Alpena), you're pretty much getting the flavor of the UP - long distances with nothing but trees, maybe a gas station or a bar, usually a lot of snow.
edit: this contrasts to the larger cities in the south-most quarter of the state: Lansing, Detroit, Grand Rapids, Jackson, Battle Creek.
Thomas Urech
You could join the two parts of Michigan and then separate them again after you colored them.
James Grime is back!
I loved watching your shadow in this episode, i don't know why.
5:02
The lines on the diagram don't have the same length (1 unit) though?
That's what "Not to scale" means. If the lines are actually unit length they have to cross each other but that makes it harder to tell what's connected to what.
How do we know that the graph used in to prove that 3 colors is not possible can be drawn on a flat plane, with the distance of each two connected vertices being exactly 1 unit? I'm not saying it's not true, but it would be nice to have a proof in the video.
Try to draw it. You'll see why.
Moser spindle. Do you even know what google is? FFS! Use your brain.
The real Moser spindle works. The map shown in the video is an approximation of it which is a bit easier to follow
Interestingly the wikipedia article on this seems to have an extension to 3 dimensions with a result that is out of date. It lists a lower bound of 6 colours for the 3d equivalent problem, but clearly if 5 colours are now known to be needed for the plane then 7 colours are an obvious lower bound for three dimensions as the plane map needing 5 colours could be given thickness 1/2 and the space above and below would both need to be different colours from the 5 already used and from each other.
I think the easy answer to the four colour map theorem is a central triangle with a triangle on each of the sides (so 4 triangles). However ways you try to fit more triangles into it (to make it interact with a 5th triangle), the same shape starts to happen in some way or another.
he says the guy who solved it does math "for fun" as if there is a practical application for this lol
6:42 And he is just 12 years old.
I knew you guys would do this!!!!
It's called graph colouring problem in computer science. It's also one of NPA Problem if I recall correct. Minimum no of colors used to color graph is called chromatic number of graph.😊
0:49 you made Michigan two colors instead of one
From a map coloring perspective, those are actually two different regions.
Michigan should be two different states.
Yeah, but that's a restriction to the theorem. It only works with continuous areas.
@@timh.6872 From a map coloring perspective the state is the scope, as such Michigan should be one color.
Grookey Gang
Sobble gang
Would have been good to see a unit distance version of the Moser spindle. It's obvious after you see one (on Wikipedia) - "ah, the lines can cross, of course"
With the 3 color example, those not being to scale is really important... it turns it into equilateral triangles... If each step is a unit length...I'd like to see another example...
Never clicked faster
Maps with exclaves/enclaves can break the 4-colour rule.
James sarcastically asking, “what’s the problem?” is everything
Nice video content!!
How about simulating the walk at the size of the earth with each walk take about 1 meter
Even the 5 colour map isn't very big, it just involves a LOT of doubling back on yourself
5:10 you can't just show me a graph and label it "not to scale" when the scale is important. For all I know, this graph might not be possible with constant edge-lengths.
CaesarsSalad so google it. They even gave you the name of it on the screen.
(Repeating a reply to another comment making the same point). Yes; to be relevant to the proof, the steps have to have length = 1. However, it doesn't matter if edges cross -- edges represent steps between nodes. With that in mind, it's not difficult to draw this map to scale; and it becomes clear that any map can be drawn to scale, even if it is difficult to do so.
ANOTHER JAMES VIDEO
HYYYPPPEEEEE
i ve got a question, how would a image for the 3 color example look like to scale, cause i tried imagining and there should be 4 triangles which have all the same sidelength, but that way the top connection cant have the length 1, or am i missing sth?
Edit: found my mistake: they cant be seen as areas, cause the paths are able to intersect
"Not to scale"
What? Isn't that kind of important since we're taking unit steps on the plane? Why would it not be to scale?!
It doesn't matter whether everything looks to be one unit so long as mathematically it is 1 unit in length
@@theproz997 That's like making a 30-60-90 triangle and saying "Each side is one unit in length. Take my word for it."
I want to see it. And I have seen a much better diagram since writing this.
@@RipleySawzen you have a fundamental misunderstanding of what they're trying to prove and how they're going about it. Your triangle is a false equivalency because it's objectively wrong. As long as their diagram is consistent, regardless of how what they decree as 1 unit in length it works. It has to be mathematically 1 unit in length, it doesn't have to APPEAR 1 unit in length
@@theproz997 It is not false equivalency, and you're strawmanning what I said. The fact that you blindly believe their diagram is accurate is what's hilarious here. It's like if I drew a right triangle with three sides of length 1. I never stated that their diagram was incorrect. In fact, I did not state a single damn thing. Therefore, nothing I said can be wrong. Therefore, you are wrong!~
@@RipleySawzen you said "that's like making a 30-60-90 triangle and saying ..." That is a claim that you made. because you seem to think that it's similar to what they did, you clearly think that their diagram is wrong since ur example is as well. There's no need to lash out at others for your poor understanding of math.
Colorful? More like colourful amirite?
Meh, who cares? Besides the u is really redundant.
Brady's really living the Fitotron 5000 lifestyle! Looking good!
I'm not sure I understand the difficulty of this question because it could defined in terms of the variables presented but are not explicitly talked about. The answer of how many colors to use depends on a pretty simple issue of, find the largest value of from a given point how many neighbor points share your original point's neighbors. Since you define finite regions of a plane to a point, and an arbitrary distance, how many colors are possible depends on your defined plane. So.. all you need to do is define your plane such that at least one instance where you 6 neighbors of the original point are also neighbors of each-other. Now you have a plane that requires 6 colors. The question is, what is your plane? Is it a 2 dimensional or 3 dimensional plane? Surely a 1 dimensional plane as a much more limited colorset of 2 required. So a 2 dimensional plane has... 5? Does a 3 dimensional plane have a different minimum?
The most confusing thing is that you have 'examples' which have variable distances between points, which don't mean anything unless it was maybe 3 dimensional.
In fact, is there a question? If your examples are all 100% spot on, then I am correct in my assumption that you are increasing the amount of dimensions the plane has to accommodate an arbitrary unit of distance. and my defined plane where a point has 6 neighbors who are themselves all neighbors of eachother, then you are just modifying the plane's properties to justify the 5 or 6.
The graph at 5:10 should have been drawn so that the lines actually looked as if they were the same length, as it currently looks it's not convincing at all.
I feel like the example for 3 can't be constructed on a plane... At least not with unit distances
The example consists only of the 7 points in the plane. The line segments that people are treating as edges in a graph are just there to show the distances between the points.
Google "Moser Spindle"
James, never leave us.
James Grime Podcast!!!
Mate, colourful is spelled wrong.
Colourful is spelt correctly and colorful is spelled correctly. Now don’t force choke me or anything, but in your comment the space is on the wrong side of the comma. It should be on the colourful side. Come back to the colourful side, Vader!
What Luke said, Lord Vader. No, not /that/ Luke. Also, thou shouldst have writ "Colourful is spelled wrongly". Yrs truly, The Society for the Conservation of the Adverb.
So both colourful and colourful are correct??
My auto correct always make it into first one. It was difficult to write *colorful*
“Colourful” and “spelt” are British spelling, while “colorful” and “spelled” are US spelling. But it looks like spelled is becoming more universally accepted. Anyway, I couldn’t resist the Star Wars gag!
“Colourful” is spelled wrong
depends on if you are using British or American spelling. I guess in this case it would be wrong though
It's american english. They threw out some letters to make newspapers cheaper to produce.
You spelt “spelt” wrong too.
rixterz
An Aussie and an Englishman would not spell it with only one “u”.
@@ragnkja colorful: American
Colourful: Canadian/British
And?
You guys keep me sane
what do you call that fancy framed type of map in the background? i like those circles
I solved this like 70 years ago
I'm wondering, when you say "[...] found a smaller example of structure" (around 7:30), is this kind of research purely empirical/intuitive/random? Or are there some theories guiding toward a structure that work?
Numberphile video - Good.
Numberphile video with James Grime - Best Video On The Internet.
Saw the thumbnail and title. Gotta assume its about the graph coloring problem / graph theory in general.
8:30 this is the mosaic on the floor of the Dance Hall at the Ministry of Silly Walks
I want a poster of James Grime in my room, now!
I totally agree that seeing the actual Moser spindle would have made more sense; I didn't buy any of it until I looked it up. Basically, two 60 degree rhombi with one common acute point, and the other acute points one unit apart (about 16 degrees offset).
Has anyone attacked this problem with a more regular set of points that will be highly connected? As a first try, I'm thinking of a rectangular mesh of spacing 1/5 (0.2) of the step unit. Then, each point (not near the edge) will have unit connections to twelve other points, at (0, +/-5), (+/=5,0), (+/-3,+/-4), and (+/-4, +/-/3) relative to itself. Next would be 1/25, multiplying the previous offsets by five and adding in (+/-7, */-12), and (+/-12, +/-7). Each point would then be connected to twenty other points, located an average of 18 degrees apart, with the largest step being less than 21 degrees - pretty darn equally spaced. I would expect this sort of grid to be more easily programmable/expandable, and (should a 5 or 6 color answer be possible, as I strongly suspect based on the wide acceptable range of units-to-province-diameters for the hexagon solution), it may actually provide some insight into the shape of the provinces. Further refinements would be to include more pythagorean triples, and setting the spacing to 1/LCM(hypotenuses). (hyponeni?)
He's alive!
How is the moser spindle a unit distance graph??
Okay I totally didn't expect Aubrey de Grey to come up :D
Can you talk about E8 and the math that there’s behind it in the next video?
Bring memories of 3-COL and 3-SAT from my theoretical computer science