[HN Gopher] A non-constructive proof of the Four Colour Theorem ___________________________________________________________________ A non-constructive proof of the Four Colour Theorem Author : ColinWright Score : 122 points Date : 2022-12-21 15:36 UTC (7 hours ago) (HTM) web link (arxiv.org) (TXT) w3m dump (arxiv.org) | scythe wrote: | Most of this proof is very dense. But Theorem 7 seems like an | interesting tactic that allows the authors to infer nonexistence | from rarity: | | > _Theorem 7. If there is a map L which cannot be 4-coloured then | only an exponentially small fraction of the maps with n edges can | be 4-coloured._ | | It's one of those things that seems obvious in retrospect -- of | course large random graphs should be likely to contain any | particular graph! But it hadn't occurred to me in my own (very | amateur) considerations of the problem. | | Will be interesting to see if it's correct. We've gone from an | argument too long for a human to read to one in just seven pages. | bodhiandphysics wrote: | This is in fact a very common proof strategy in combinatorics, | often known as the probabalistic method. Naturally it was | invented by Paul Erdos | sweezyjeezy wrote: | I think that's a bit different - in the probabilistic method | you show that if you randomly sample a graph (say) then the | probability of it having some property is > 0 - therefore | there must exist some graph having the property. | | Theorem 7 in their paper is trying to count the number of | graphs that contain a hypothetical 4-color counterexample if | it exists. It's done by hitting a suitable generating | function with a big analysis hammer to bound its | coefficients, not by probabilistic means. | zeroonetwothree wrote: | I haven't read this carefully yet but I'm initially skeptical | that it's that easy. | naasking wrote: | From the abstract: | | > The approach uses a singularity analysis of generating | functions for particular sets of maps, and Tutte's enumerative | and asymptotic work on planar maps and their chromatic | polynomials | | Err yeah, so easy... | burnished wrote: | That quote is so dense with words that individually are old | friends, but paired up that way are as strangers to me. | mkaic wrote: | That was a surprisingly prosaic way to communicate this | feeling, are you by chance a writer? If not, you should | write more! | ectopod wrote: | Of writing, prosaic means unpoetic, i.e. dull. | thombat wrote: | He might just be Buster Scruggs' less melodious & more | bookish brother. | lairv wrote: | So until now every proofs of this theorem were still computer- | assisted right ? | gentle wrote: | Yes, that's right. | NelsonMinar wrote: | My memory is that the Four Colour Theorem was one of the first | computer search-assisted proofs which caused all sorts of | epistemological questions. Specifically how to know whether the | software had no bugs and ran correctly, and how re-running to get | the same result only lowered the probability and did not yield | absolute certainty. Seems sort of a quaint concern now. | | (Apologies for no reference. I hope I'm not misremembering.) | Dove wrote: | You're remembering entirely correctly! I gave a talk on the | Four Color Theorem as part of my master's degree in 2001, and | it was a somewhat active epistemic debate among mathematicians | at the time. I took the position that examining code was no | different than examining a proof, that running it through a | computer was no different than running it through your brain, | and that mathematicians are not shy at all about "automating" | processes over uncountable infinities in proofs, so why should | we be shy about automating them over thousands mechanically? | | There is a fun faith-shaking element to the story, though. The | Four Color Theorem, as a proof, actually contains a couple | thousand smaller proofs. These were generated and checked by a | computer, but it is humanly _possible_ to check them with a | great deal of effort. There was a guy -- I forget his name, and | Google isn 't helping me! -- who, as his Ph. D. thesis, | actually went through them and checked them. And IIRC, he found | about a dozen errors! They weren't fatal -- he was able to | repair them -- but that whole process didn't really reassure | people. ;) | Dove wrote: | AMS has a good summary: | https://blogs.ams.org/mathgradblog/2014/06/29/color-theorem | anonymousDan wrote: | How did the errors arise, some bug in the theorem prover? | Dove wrote: | That's a really good question. I may have known the answer | to it twenty years ago, but I don't now. XD | | What I can tell you is that it wouldn't have been so | straightforward as that. We're not talking about an axiom- | to-result symbolic proof piece of software like you might | attempt to write now. It wasn't a computer proof so much as | a computer-assisted proof. Lots of human analysis to | describe specific properties, and then a computer program | to check that a long list of geometric configurations had | those properties. | | I don't know exactly what the program output as its proof, | but I can only imagine the problems would have been small | assumptions built into its design failing in corner cases. | The theorem is notoriously difficult to reason about | rigorously, and there have been a number of accepted proofs | over the years that turned out to be false. It really | doesn't help the whole situation that not only is it the | first major computer-assisted theorem, it just happens to | deal with a result that people are already wary of. | puzzledobserver wrote: | The original computer assisted proof of the four colour | theorem was written in IBM 370 assembler [0]. This was | naturally susceptible to programming errors such as those | discussed above. Gonthier's subsequent certified proof | requires trust in a much smaller body of code and hardware | [1]. | | [0] https://projecteuclid.org/journals/illinois-journal-of- | mathe... | | [1] https://www.ams.org/notices/200811/tx081101382p.pdf | SilasX wrote: | The same issues come up with interactive zero-knowledge | proofs[A], where someone proves that they possess certain | private information without leaking said information. | | The proofs have the form: | | 1) Give the prover a challenge that they only have a 50% | chance[B] of passing if they don't possess the private info. | | 2) Repeat as necessary with new, independent challenges. | | 3) If they keep passing, hooray! The probability that it's a | faker decays exponentially. You can get really high confidence, | very quickly! | | Doing more rounds is just like re-running the computer proof. | | [A] I first learned about the issue from reading the Scott | Aaronson lectures that became _Quantum Computing Since | Democritus_ | | [B] Doesn't have to be 50%, any figure will give the | exponential decay and only require a linear increase in the | number of rounds, so long as the probability stays the same | between all rounds. | OJFord wrote: | I'm not a mathematician, but when I was in school I read a lot of | 'pop math', and fancied myself a lot better at it than I was (or | am). I was fascinated by things like the FCT, and recall writing | a document ...I don't think I kidded myself _proving_ it, but at | least 'understanding' it. | | The rough sketch was to show that it was not possible to | construct a map that required _five_ colours: that if you take a | blob of one colour, wrap it in three others, all four are | touching the others, but you can 't possibly add a fifth since it | either only touches the outer three, or it has to prevent two of | them touching to get to the inner one (then the non-touching ones | could be recoloured the same). | | To this day, although I of course realise I was having no | groundbreaking thoughts and it did not constitute a proof, I do | not understand why the actual proofs are so complex, or require | so much computation? It eluded far greater minds for over a | century - what am I missing that makes it so problematic? | nicky0 wrote: | Your proof sketch actually suggested that it was not possible | to construct a map _with five regions_ that require five | colours. | | Now you need to generalise that to maps with infinitely many | regions. | OJFord wrote: | But as a result there could never be an area of an infinitely | large map that had five regions each needing a different | colour. I understand people are saying but you take two such | sub-maps, put them next to each other, and maybe the | 'outside' colours are the same - but in that case you could | always just recolour them individually before putting them | together? | ColinWright wrote: | What you've said here might be a reasonable reply in your | head, but for me, reading its parent, and this reply, I | can't work out what you're trying to say. | | You need to be _really_ careful about things like | "infinitely large map" ... if you want to talk about really | large things then they are really large. But if you want to | talk about infinite sized things then you need to be | careful. There are traps in infinities, and this theorem is | about finite maps (or equivalently, graphs). | | Proving that a small, local sub-section doesn't require | five colours is helpful, but no one has been able to extend | that, because there is a kind of "spooky action at a | distance" that can happen. Local decisions about colours | can lead to implications across the other side of the | structure. Proving that your local decisions can always be | made in a way that doesn't cause problems elsewhere is the | problem. | | You are saying the sorts of things I've seen many times | before, but you need to try to make complete, precise | statements about _exactly_ what you propose. So far all you | 've said is that you can't arrange five regions that all | touch each other, and that's true. Now, exactly how do you | extend that to a map with billions of regions? | sebzim4500 wrote: | I have no idea what that guy is talking about, but your | post reminded me of a fun trick. If you have a colouring | of every finite subgraph then you can show the existence | of a colouring of the whole graph by using Tychonoff's | theorem. | ColinWright wrote: | That's cool. | OJFord wrote: | It was the comment I replied to that introduced | infinitely large maps - but regardless I wasn't trying to | be as clever as you might be trying to assume (and of | course not finding it :)) - I would have said the same | about 'really large' or indeed 'just a bit larger than | the 4 or 5 regions I described initially'. | | > there is a kind of "spooky action at a distance" that | can happen. Local decisions about colours can lead to | implications across the other side of the structure. | Proving that your local decisions can always be made in a | way that doesn't cause problems elsewhere is the problem. | | This is the bit I can't grasp - I honestly appreciate the | responses and attempts to explain it, but it just | (mistakenly, I can accept even if I don't _understand_ ) | seems counter-intuitive to me. As in, intuitive that it | extends, or that there's no such problem. | | Maybe I just need to play with it on paper a bit, stumble | into such a spooky action to get it. | | > Now, exactly how do you extend that to a map with | billions of regions? | | I don't have the background for the precision you'd like | I'm afraid (if I did, I'm sure I'd understand anyway!) | but loosely - divide and conquer; a similar thing happens | at the interface of all the 'sub-maps'? | ColinWright wrote: | >> _there is a kind of "spooky action at a distance" that | can happen_ | | > _This is the bit I can 't grasp_ | | OK. Let me give you a collection of statements, each of | which I can expand on, but which might give you a sense | of the problem. | | * Instead of colouring regions in a map we can colour | vertices of a graph; | | (put a vertex in each region and connect vertices if | their regions share a border) | | * If a graph can be three-coloured then it can be four- | coloured; | | (Any three colouring is a four colouring, but with no | vertices of the fourth colour) | | * An instance of colouring an arbitrary graph can be | converted into an instance of colouring a planar graph; | | (There is a "widget" that uncrosses edges) | | * Given a number to factor, we can construct a graph such | that three-colouring the graph will produce a | factorisation[0]; | | (Graph three-colouring is NP-Complete, so this is even | easier than that) | | Broadly, we can construct a graph that has exactly one | (up to permutation of colours) valid three-colouring. As | you start to colour it you find that you have to make | choices, on average about $4/3$ choices per vertex[1]. | But the wrong choice results in not being able to | complete the colouring, and you don't know which choice | or choices were wrong. | | So you get this "spooky action at a distance" thing going | on. | | Sometimes a planar graph can't be coloured with three | colours, and sometimes it can but a colouring is hard to | find. When we allow the extra colour it turns out that | it's always possible, but it's not obvious that that | should be the case. | | That's an #ISS-level view of the ideas ... hope it helps. | | [0] https://www.solipsys.co.uk/new/FactoringViaGraphThree | Colouri... | | [1] Beigel, R.; Eppstein, D. (2005), "3-coloring in time | O(1.3289^n)"[2] | | [2] | https://en.wikipedia.org/wiki/Graph_coloring#cite_note-15 | | PS: I've upvoted you because I'm sure other people have | the same questions, they're good questions, and they | deserve answers. | ColinWright wrote: | It's (relatively) easy to prove that you can't have five | regions that all touch each other. If you could then that would | mean that a map requires (at least) five colours, but that's | not the only way a map could require five colours. Even so, | it's certainly the first thing to try. | | But colours-at-distance can be forced, so proximity isn't | needed to create the need for another colour. | | The theorem is difficult for two meta-reasons. One is that | there are arbitrarily many planar configurations, and you need | to prove that they are _all_ four-colourable. The other is that | whether or not a planar graph is _three_ colourable is NP- | Complete, and showing a planar graph is five-colourable is | fairly easy, so it 's on the boundary between fairly easy and | really hard. | | The proof that every planar graph (equivalently every map on a | plane) is five colourable proceeds by showing that there is a | small set of sub-graphs such that every graph must contain one | (an unavoidable set), and then showing for each of those that | it cannot be in a minimal counter example. This is easily done | by hand. | | Appel and Haken's proof of the four-colour theorem proceeded | the same way, but the unavoidable set was large. The | computation was to confirm that the set was unavoidable, and | that each element was reducable. This involved a huge amount of | tedious bookkeeping, best done by machine, but it was really | just "doing the sums". The real work was creating a finite | amount of bookkeeping that would then constitute a proof. | siftrics wrote: | Thanks for the insightful comment. | | >The proof that every planar graph (equivalently every map on | a plane) is five colourable proceeds by showing that there is | a small set of sub-graphs such that every graph must contain | one (an unavoidable set), and then showing for each of those | that it cannot be in a minimal counter example. This is | easily done by hand. | | Isn't this missing a crucial point? You seem to be saying: | | 1. There is some (small) set S of graphs, all of which are | 5-colorable | | 2. Every planar graph has at least 1 sub-graph in S | | 3. Therefore, all planar graphs are 5-colorable | | But, from statements 1 and 2, one can only conclude that | every planar graph _has a sub-graph_ that is 5-colorable, not | that the entire graph is 5-colorable. | | Don't you also need to prove that the complement of the sub- | graphs is 5-colorable? | lmm wrote: | You don't just show that the graphs in S are 5-colourable. | You show that they couldn't possibly be part of a minimal | non-colourable graph (e.g. if a graph containing this | subgraph is not 5-colourable, you can replace this subgraph | with this simpler subgraph, and the larger graph will still | be non-5-colourable). | ColinWright wrote: | My comment was: | | > _The proof that every planar graph (equivalently every | map on a plane) is five colourable proceeds by showing that | there is a small set of sub-graphs such that every graph | must contain one (an unavoidable set), and then showing for | each of those that it cannot be in a minimal counter | example._ | | You said: | | > Isn't this missing a crucial point? You seem to be | saying: | | > 1. There is some (small) set S of graphs, all of which | are 5-colorable | | This is not what it's saying. It's saying that these graphs | cannot be in a minimal counter-example. That's different | from simply saying that they _are_ 5-colourable. | | > 2. Every planar graph has at least 1 sub-graph in S | | > 3. Therefore, all planar graphs are 5-colorable | | No, you've missed a bit of my comment. I said: | | > _showing for each of those that it cannot be in a minimal | counter example._ | | So we have an unavoidable set U. So for every graph G there | is an element S of U that is a sub-graph of G. | | But we have also shown for every element S of G that any | graph containing S cannot be a minimal counter-example. | We're not showing that these elements S of U are themselves | 5-colourable, we are showing that any G containing S is not | a counter-example to the theorem. | | So now suppose the theorem is false. That means there is a | graph that requires 6 colours. From among those choose a | minimal one, so M is a minimal counter-example. But the set | U is unavoidable, so there is a S in U that is a subgraph | of M. But any graph containing S cannot be a minimal | counter-example, and we have our contradiction. | | Hence there are no counter-examples, and the theorem is | true. | | For the Five Colour Theorem the unavoidable set U has only | four graphs, so it's easy to check by hand. For the Four | Colour Theorem in the original proof there were around 2000 | elements of U, so it was infeasible to check by hand. That | has been reduced to 633 graphs in U, but it's still tough | to check by hand. | | The paper in the original submission here takes a different | approach. | siftrics wrote: | Ah, thank you. The key to my misreading was this part: | | > showing for each of those that it cannot be in a | minimal counter example | | I read "those" as "those subgraphs" rather than "those | graphs". | | Thanks. | Someone wrote: | That shows (loosely) that there is no map with five countries | that cannot be four-colored. It says nothing about maps with | six, seven, etc. countries. An actual map may have billions of | countries and need not even contain "a blob of one colour, | wrapped in three others". | | Loosely because | | a) it only shows it for such maps that contain such a blob. Who | says other maps with five countries are easier to four-color? | | b) it doesn't rigorously show it. That you (and me, and | everybody who looked at it) can't find a way to draw that fifth | area that touches all others doesn't say much. what are your | arguments for showing you looked hard enough? | | For a similar case, read about the Jordan curve theorem. In | layman's terms, it says closed curves on a plane have an inside | and an outside part | (https://en.wikipedia.org/wiki/Jordan_curve_theorem). Naive | readers think it's so trivial to not require a proof, and it | took mathematicians a while to figure out that it needed one. I | don't know the details, but mathematicians may have observed | that that theorem also is true on a sphere but isn't true on a | torus (a closed loop 'around the hole' in the torus doesn't | have an interior and an exterior). So, they needed an | explanation as to why it's true on a plane or a sphere but not | on a torus (the answer essentially boils down to "because a | torus has a hole in it", but that requires a rigorous | definition of what a hole is) | OJFord wrote: | If you don't have such a blob then you don't even need four? | ColinWright wrote: | This is in the grandparent comment: | | "An actual map may have billions of countries and need not | even contain "a blob of one colour, wrapped in three | others"." | | In fact every map of interest _will_ contain a blob wrapped | in at least three others. | | But they are right that the problem is showing that the | theorem holds for maps with billions of regions, or more. | Decisions you make early in the process of colouring the | map can turn out to be wrong, but only much later as you | are trying to colour the last few. | | In short, you are saying "I can't see how it can go wrong", | but that's not enough ... there are things that might | happen that you weren't able to imagine, and that's part of | why proofs like this are hard. | | Also, the theorem is true, and that makes it hard to | explain why it might go wrong. Because we now know that it | doesn't. | | The grand-parent comment also says: | | "... the Jordan curve theorem ... is true on a sphere ..." | | The stronger form is not true on a sphere. The stronger | form says that every topological sphere in 3-space has an | interior that's retractable to a point and an exterior | that's retractable to a shell. But the Alexander Horned | Sphere[0] is a counter example. | | In short, weird things you didn't suspect can happen, so | proofs are sometimes harder than you think, even for | "obvious" things. | | 0: https://en.wikipedia.org/wiki/Alexander_horned_sphere | bodhiandphysics wrote: | in some sense because there might be strongly non-local | effects... an assignment of color in one part of the graph can | have an effect on what colors are allowed in a very different | part of the graph. Planarity assures some degree of "locality", | you can't arbitrarily connect one point to another point. | Phemist wrote: | https://imgur.com/a/WJKb9fT | | I have always wondered about this sample. Consider the possible | 4-colours to be red, blue, green and orange, as in the image. We | need to determine the colour of the black square. All 4 colours | are joined in this inverted cross like formation and the center | of the cross (a gray pixel atm) kind of indicates an infinity | point. If the space in which these colours are drawn is not | discrete, then they can all be said to join at the center of the | cross. | | Is this sample just not allowed by the rules of the "game"? | | Consider it an extreme case of gerry-mandering... | bodhiandphysics wrote: | The graph has to be planar, which means no to edges intersect | (or you can move the edges so that no two edges intersect). The | complete graph with 5 or more vertices isn't planar. | blevin wrote: | Your example adds more constraints than the original problem, | by pre-assigning colors to regions and then asking how to solve | for the black region. | | In the original problem statement, you could solve by re- | assigning your green region to be orange, your red region to be | blue, your black region to green, and the white regions to red. | Also note that regions that meet at a single point (as in your | example) are not considered to be adjacent. | Phemist wrote: | aha, so infinities are not allowed. Fair enough, in that case | the gray pixel needs to be filled with one of the colours and | the two colours connected to that specific one become | interchangeable. | Phemist wrote: | so in essence there is an additional constraint, that each | touching edge has to be of a given minimal size. Each | contiguous "blob" with a given circumference of N * this | minimal size, can only connect to at most N other blobs. | NotYourLawyer wrote: | Meeting at a point doesn't count, so for example, blue could be | recolored to be red. | Phemist wrote: | This is btw equivalent to a graph with 5 nodes, where all nodes | are connected to every other node, which would quite obviously | be impossible to colour with just 4 colours. | klyrs wrote: | It's not, though. If the four corners are sharp, blue & red | are not adjacent, nor are green & orange. You can re-paint | blue to red and orange to green, and you're left with a | proper 3-coloring. Alternately, if the grey pixel takes | another color (say, red) so that blue, green and orange are | all adjacent to it, then green & orange are not adjacent -- | re-paint orange to green, and you get a 4-coloring. Finally, | if the grey pixel is its own region, you're effectively back | in the first case and you can paint it black to obtain a | 3-coloring. | drbacon wrote: | The "countries" are supposed to share an edge. An edge, no | matter how small, breaks the cross. | Phemist wrote: | https://blog.artsper.com/wp- | content/uploads/2022/08/escher-2... | | Consider the gray pixel in the center a placeholder, much | like Escher's place holder in the above image (because he | couldn't think of how to realistically depict what turns out | to be the infinite fractal nature of the center) | | > An edge, no matter how small breaks the cross. | | My very weak intuition says that an edge of infinite | smallness would not. | bell-cot wrote: | An "edge of infinite smallness" is a point. | | (I don't recall details - but the preconditions of the Four | Colour Theorem rule out all the fuzzy sets, infinitely | crooked lines, "this region is all points with one rational | and one irrational coordinate", and other trickery that | folks who've had a bit of math might otherwise be tempted | by.) | ColinWright wrote: | "... an edge of infinite smallness ..." is not a well- | defined concept, and is not allowed in the original | formation of the problem. Otherwise take a circle and | divide it into N segments "like a pizza". They all meet in | the middle in an "edge of infinite smallness", so that | would require N colours. | | Now make N as large as you like. | | So allowing this makes the problem uninteresting, and | precluding it makes the problem interesting and hard. | pfortuny wrote: | The thing is: those four regions can be recoloured (change | green by orange). | qsort wrote: | If it's correct, then it's a wonderful approach. | | If it's not correct, then HN is the place to find where's the | error. | | By elimination of disjunction, upvote ;) | onos wrote: | Agree on the first part. If correct, it is a wonderful | achievement, apparently built upon the work of at least two | others. | EGreg wrote: | Hardly I doubt HN is that place | | Can this also prove the more general Hadwiger's conjecture? | feet wrote: | With the amount of incorrect comments I see on HN, definitely | not. I'm not saying that I'm not included in that group, but | I think we all are | notafraudster wrote: | Waterloo has a Department of Combinatorics and Optimization? | Amazing. | mwnorman2 wrote: | Not only does Waterloo have a world-class C-&-O department, | they also had a hero from Bletchley Park - Dr William Tutte, a | kind gentle genius that never bragged about his accomplishments | (he didn't talk about it, even to his family!) He broke the | 'Lorenz' cipher (generally regarded as much more difficult than | Enigma) Almost all modern Graph-theory and Matroid analysis is | built on his work. --- Mike Norman, B.Math UofW 1986 | bodhiandphysics wrote: | Tutte's are the central results in this work | Syzygies wrote: | I'm a math professor who has lost years to this question. I've | met various authors involved in successful proofs. My 1976 grad | school application from the University of Illinois came | postmarked "Four Colors Suffice" (Appel, Haken). I've also been | inundated with well-intended crank proofs of every tar pit like | this. One can tell the difference without focusing one's eyes. | | These authors have long, established careers, with 2,000 | MathSciNet citations between them. Waterloo is a hotbed of | combinatorics; I doubt that they posted this before passing the | paper around a bit. | | Graph theory as a branch of mathematics is like photography as an | art form: Anyone can point a camera, and one sees a lot of | incredibly weak graph theory if one digs deep enough into the | regional conference circuit. | | This work is deep, relying on an understanding of generating | functions that many casual practitioners don't possess. If the | proof has a flaw, the issue will be technical and difficult to | uncover. I love how people take the obvious bet that this is | likely wrong because so many proof attempts have failed. In a | prediction market, I'd bet on this proof being right. | ruuda wrote: | > In a prediction market, I'd bet on this proof being right. | | Let's create a question on Metaculus! | spekcular wrote: | Your post feels like a dispatch from bizarro-world. | | There is a good amount of second-order evidence that the proof | is wrong. Further, I skimmed the introduction, and it seems the | indicated approach cannot possibly work. My guess is that both | authors are senile. (They're quite old.) | | If you give me decent odds, I'd be happy to bet against you | regarding the proof's correctness. Would you take 1:1? | [deleted] | ogogmad wrote: | The tone of your comment is pretty harsh. What kind of 2nd- | order evidence do you see? And why did you learn from the | introduction that makes you so sure? It's only an | introduction, which might only give you an oversimplified | summary of their ideas. | | Are both of them senile? | spekcular wrote: | Yeah, my guess is that both are senile if they're putting | this on the arxiv. | | Second-order evidence: Old mathematicians (~80). Famous | problem. Weird, imprecise writing style. No mention of the | proof by mainstream mathematical news sources | (breakthroughs are usually accompanied by excited | blogging/tweeting). No acknowledgements directed at other | mathematicians who have checked or commented on the proof. | | [deleted argument here; replaced by more precise comment | below] | | My tone is harsh because I think the best thing to do is to | quietly ignore it, similar to how the community treated | Atiyah's claims of a RH proof at the end of his life. | ogogmad wrote: | > The argument looks like it's based on large n | asymptotics, so even assuming everything works correctly | the strongest statement they can hope to show is that the | theorem is true for all n > n_0, where n_0 is some large | constant. But there is no mention of this fact. The | theorem is claimed for all n. | | Have you seen this comment? | https://news.ycombinator.com/item?id=34083099 | spekcular wrote: | Yes. How does it bear on what I wrote? | eganist wrote: | You're claiming: | | > the argument looks like it's based on large n | asymptotics, so even assuming everything works correctly | the strongest statement they can hope to show is that the | theorem is true for all n > n_0, where n_0 is some large | constant. but there is no mention of this fact. the | theorem is claimed for all n. | | They're claiming: | | > Theorem 7. If there is a map L which cannot be | 4-coloured then only an exponentially small fraction of | the maps with n edges can be 4-coloured. | | These claims appear mutually exclusive. | spekcular wrote: | See above clarification about the bad writing. | adgjlsfhk1 wrote: | The point is that if there were a small graph that was a | counter-example, then there would be large graph counter- | examples (and the percentage of them would increase with | graph size), so proving that the 4 color theorem is true | for large graphs implies that it is true for all graph | sizes. | adgjlsfhk1 wrote: | Some counterpoints: 2 mathematicians greatly increases | chances of senility. Blogs/twitter also correlate much | stronger with author age than truth of statement. | | >The argument looks like it's based on large n | asymptotics, so even assuming everything works correctly | the strongest statement they can hope to show is that the | theorem is true for all n > n_0, where n_0 is some large | constant. But there is no mention of this fact. The | theorem is claimed for all n. | | This is completely wrong. A proof for all n>n_0 is a | proof for all n, since any counter-examples have to exist | as subgraphs of arbitrarily large graphs. | spekcular wrote: | I don't think asymptotic estimates of that form suffice | to treat this problem. (Where else in combinatorics has | an argument of this form succeeded? What intuitive reason | is there to expect it to succeed here?) | | Specifically I think section 4 is basically nonsense. (I | see Sniffnoy has already pointed this out below.) | | (Re: your comment, Theorem 7 is going to fail below the | smallest counterexample, right? This is bad, imprecise | writing - a red flag.) | ColinWright wrote: | I caught a rumour of this paper about 6 months ago, so I'm | pretty sure that in the meantime it's been evaluated and | assessed by colleagues. So I'm pretty sure it's going to be | right, with only small fixable things that need tidying up. | ngvrnd wrote: | well put. | Sniffnoy wrote: | Reposting my comment from /r/math: | | OK, section 4.2 makes no sense to me. So they have this set Q, | right? Which is the set of plane graphs that can be written | G=A*B*C*D, where * denotes joining at a cut vertex. Their goal is | to show that every G[?]Q can be 4-colored, and then use that to | argue that a nonzero fraction of plane graphs can be 4-colored, | and then argue (via the generating functions and asymptotics) | that this means that they all can be. | | Except... if they can show that every G[?]Q can be 4-colored, | then all of this argument with generating functions and | asymptotics is unnecessary, right? Because surely G can be | 4-colored iff A, B, C, and D all can; and since A, B, C, and D | can be any plane graph, that would be the end of it -- if you | have a plane graph A and you want to 4-color it, all you'd have | to do is stick a 3-path hanging off the end of it to put it in Q | (indeed, in the subset they call \overline{Q}), and then 4-color | that. The rest of the argument would appear to be unnecessary! | | Meanwhile, their argument for why graphs in Q can be 4-colored | doesn't seem to make any sense to me. It's inductive, but they | haven't written it in a clear manner that makes it exactly | apparent how they inductive hypothesis is used. I honestly do not | see how they are justifying the statement "It follows from the | Induction Hypothesis that S_1 is 4-colorable" -- it doesn't look | like any work is being done, it all looks circular to me! | | Am I missing something here? | sebzim4500 wrote: | From my reading, it looks like can't make up their mind about | whether Q = {X . Y: X, Y graphs} or Q = {X . Y : X, Y in Q}. | Neither definitions seems to work everywhere. | spekcular wrote: | You're not missing anything. | Strilanc wrote: | Given the history of false proofs of the four color theorem | surviving scrutiny for decades, I think I'll wait for this to be | verified by a computer proof language such as Lean before | trusting it. | | (This is intended seriously, but I do love the ironic twist | compared to historical hesitancy to accept the computer assisted | proof.) | civilized wrote: | A short paper like this will almost certainly be either | verified or refuted by the traditional method of mathematicians | reading and analyzing the proof's logic. Computer-aided proof | is pretty niche in mathematics. | krsrhe wrote: ___________________________________________________________________ (page generated 2022-12-21 23:00 UTC)