A Colorful Unsolved Problem - Numberphile
video description
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.
Date: 2022-04-09
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Comments and reviews: 9
Sarainia
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.
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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.
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Slidell
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)
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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)
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gedstrom
Two things about this presentation confused me at first until I could wrap my head around what was being said. First, talking about the four-color theorem in the beginning threw in a confusion factor because this is NOT about the four-color theorem. Second, not drawing the Moser Spindle to scale threw in more confusion. Of course, drawing the spindle to scale means that the connecting lines of some of the nodes need to cross, making the drawing a little more complicated. Once I got myself straightened out on those two points, things made a LOT more sense. Of course, I still haven't been able to find a smaller network than the one in the video that requires 5 colors, but at least I understand better what needs to be done. The question for me: -What is the basic quality that the 533 point network has that forces 5 colors? -
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Two things about this presentation confused me at first until I could wrap my head around what was being said. First, talking about the four-color theorem in the beginning threw in a confusion factor because this is NOT about the four-color theorem. Second, not drawing the Moser Spindle to scale threw in more confusion. Of course, drawing the spindle to scale means that the connecting lines of some of the nodes need to cross, making the drawing a little more complicated. Once I got myself straightened out on those two points, things made a LOT more sense. Of course, I still haven't been able to find a smaller network than the one in the video that requires 5 colors, but at least I understand better what needs to be done. The question for me: -What is the basic quality that the 533 point network has that forces 5 colors? -
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Starkswood
The original problem specifies 'no two points AT UNIT DISTANCE' on a plane, and yet the solutions are all topological: the steps of the walk are allowed to be variable lengths. That is the same as saying the plane is a self-adaptive 3D surface. In other words, these 'proofs' work because one of the parameters was ignored. The original problem seems to be practical-real-world, yet the proposed solutions (apart from the hexagonal tessellation) are purely theoretical. Mathematicians sometimes do this, they forget the original purpose of the problem and go off into the realms of what can be imagined but not physically demonstrated.
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The original problem specifies 'no two points AT UNIT DISTANCE' on a plane, and yet the solutions are all topological: the steps of the walk are allowed to be variable lengths. That is the same as saying the plane is a self-adaptive 3D surface. In other words, these 'proofs' work because one of the parameters was ignored. The original problem seems to be practical-real-world, yet the proposed solutions (apart from the hexagonal tessellation) are purely theoretical. Mathematicians sometimes do this, they forget the original purpose of the problem and go off into the realms of what can be imagined but not physically demonstrated.
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Mat
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.
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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.
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Darkener
I don't know if I was the first person to think of this so I won't say a made this but this is a similar but easier question then that
[Sorry for inconsistent grammar]
can you arrange n points on a grid so that the lines between 2 points don't touch another point (assuming that the point isn't infinitely small and the line isint infinitely thin)
Now ofcourse the obvious answer is to put the points evenly on the outside of a circle but I'm excluding that solution it's to simple
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I don't know if I was the first person to think of this so I won't say a made this but this is a similar but easier question then that
[Sorry for inconsistent grammar]
can you arrange n points on a grid so that the lines between 2 points don't touch another point (assuming that the point isn't infinitely small and the line isint infinitely thin)
Now ofcourse the obvious answer is to put the points evenly on the outside of a circle but I'm excluding that solution it's to simple
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Slidell
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.
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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.
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PrivateSi
Hi, this comment is actually about a possible proof of The Four Colour Theorem. The standard graph solution is much more complicated than mine. In 2D, the maximum number of nodes that can be connected to each other (each to each) without connectors crossing is 4. I know it's not an actual mathematical proof but it is a graphical abstraction of the same problem. I'd like to know if I'm correct, perhaps you could tell me, I don't know any proper mathematicians.
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Hi, this comment is actually about a possible proof of The Four Colour Theorem. The standard graph solution is much more complicated than mine. In 2D, the maximum number of nodes that can be connected to each other (each to each) without connectors crossing is 4. I know it's not an actual mathematical proof but it is a graphical abstraction of the same problem. I'd like to know if I'm correct, perhaps you could tell me, I don't know any proper mathematicians.
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Philip
James. don't 2 rules need to be established before attempting a solution? it appears you are doing the problem of '2' colors [America] without incorporating any rules. for either the size of the 'step' OR the colored area. Without any rules, how do you know if the 3rd step you took BACK to green is too long to make? Or where the color boundary is when making the 2nd step? Taking arbitrary step lengths in a boundry-less area doesn't seem Mathematical. Just sayin.
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James. don't 2 rules need to be established before attempting a solution? it appears you are doing the problem of '2' colors [America] without incorporating any rules. for either the size of the 'step' OR the colored area. Without any rules, how do you know if the 3rd step you took BACK to green is too long to make? Or where the color boundary is when making the 2nd step? Taking arbitrary step lengths in a boundry-less area doesn't seem Mathematical. Just sayin.
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