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The bunkbed conjecture is false
- solidsnack9000 2y agoIt is helpful that the post links to the Wikipedia page: https://en.wikipedia.org/wiki/Bunkbed_conjecture https://en.wikipedia.org/wiki/Bunkbed_conjecture Reading that and then rereading the post, it all made a whole more sense: why the conjecture is intuitively appealing and why the computational approach doesn't readily result in a proof.
- knodi123 2y agoI still don't get it. So you make this graph, with a random number on each edge. Then you duplicate that graph, stack them like a bunkbed, and connect every lower vertex to the one above it. Then you delete all edges with a number below some threshold. Wouldn't the upper and lower graphs still be identical? Since you had the same edge weights in both? So the route from any upper node X to any upper node Y should be 1 less than the route length between X and the point beneath Y. What did I miss?
- cochne 2y agoThere is no thresholding, you delete edges randomly based on the probability assigned.
- knodi123 2y agoThanks, that's an important distinction.
- TheRealPomax 2y agoYou missed the part where the conjecture is about subgraphs: after forming the "bunk bed", we start randomly deleting edges based on edge probability, so there may no longer be a direct path from x to y in what was originally the upper graph.
- torstenvl 2y agoThat seems irrelevant since then there would be no path in the lower graph either. This must be true because deletions are based on probability and the corresponding edges between upper graph and lower graph are required to have the same probability.
- TheRealPomax 2y agoEach edge is independent from each other edge: you don't roll a die and then go "4: remove every edge that's 4 or less", you make a separate roll for each edge.
- arjvik 2y agoSame probability, but still independently sampled.
- calfuris 2y agoHaving the same probability of deletion doesn't mean that they will necessarily share the same fate.
- torstenvl 2y agoAh. The Wikipedia description of the conjecture does not say that each edge is independently randomly deleted (or not) based on the probability assigned to that edge. I just assumed the probabilities were the weights in a weighted graph and had some other effect (e.g., likelihood of success traversing that edge, comparative likelihood of choosing that edge vs another when traversing the graph, etc.).
- yread 2y agoI don't know. To me it sounds like it's obvious the conjecture doesn't hold. if you have a path in the upper bunk that gets broken you are screwed but if that path is broken you have the option to switch to path in the lower bunk at ANY moment. So you have n+1/n higher chance of a break but n-1 ways how to avoid it
- supernewton 2y agoThe conjecture is true for all small graphs that they tried, so if it's "obviously" false to you then something went wrong with your intuition somewhere.
- zmgsabst 2y agoOr everyone else’s, since his intuition arrived at the right asymptotic answer.
- tech_ken 2y agoI'm pretty sure you're sampling the "bedposts" as well, the upper and lower subgraphs aren't just connected at every node; I understood the rough (and incorrect) intuition to be like: P[nodes in upper bunk connected] > P[nodes in lower bunk connected] * P[sufficient "bedpost" edges survive connecting upper and lower] Since P[nodes in upper bunk connected] = P[nodes in lower bunk connected] And by definition P[sufficient "bedpost" edges survive connecting upper and lower]<1
- winwang 2y agoIntuitively (subjectively) why this argument doesn't work: if you switch to the path in the lower bunk, you can switch to the upper bunk "at any moment", regardless of if there is a path which exists on the lower bunk. In this example, a1 is connected to both c1 and c2: a1 b1 -- c1 | | | a2 -- b2 -- c2 If, instead of deleting edges, we assigned a distance d = 1/p, then you'd expect (and probably easily prove) that the distance to the target vertex is at most the distance to its partner in the other bunk. The fact that this intuitive "problem translation" doesn't actually translate to probabilities is quite surprising (to me)!
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- auggierose 2y agoYeah, the introductory sentence in the post doesn't really do a good job explaining what the bunkbed conjecture is. If I was interested in it, I would have to look it up somewhere else.
- keepamovin 2y agoWow, funny how our intuitions fail on objects with 7523 vertices and 15654 edges
- fbn79 2y agoand funny 7523 is a prime number.
- f1shy 2y agoBut 16564 not!
- keepamovin 2y agoNice! 15654 = 2*3*2609 and (7253/15654)**(phi*pi)*100 ~ 2 haha :)
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- y-curious 2y agoIt was always obvious for me. But they did call me gifted in the third grade.
- keepamovin 2y agoHahaha it was obvious that it was false for you???
- dataflow 2y ago> This is completely obvious, of course! Could someone explain why the conjecture seemed obviously true? I have no background with it, but just reading the description here, I was caught off guard by the sentence. What made it seem obvious?
- keepamovin 2y agoWhy is it "obviously true" (that lower probabilities of connection between than on levels)? Because if it wasn't true then that would imply that V on different levels (as reached presumably by a canonical DFS tree // spanning tree) were closer (in terms of paths) than V on the same level, which would mean that those V should have "more likely" been on the same level (as however measured). TL;DR - 'obviously true' because there's a level between them so (on average) of course!
- jprete 2y agoThe upper and lower bunk graphs are symmetrical in their probabilities, so it would intuitively seem like connecting to the same bunk's point would be easier/more likely than connecting to the other bunk's point. The first would require zero (or an even number of) crossings, the latter would require one (or an odd number of) crossings. Every crossing would seem to only decrease the odds of the path existing, or at best leave it the same. To me it smells like a trap, though. This feels like exactly the kind of problem where some algorithmically chosen infinite graph would break the conjecture, possibly through some kind of NP-hardness-proof-style transformation of the problem into another form, and who knows, maybe the Axiom of Choice gets involved too (but probably not based on the headline!).
- sweezyjeezy 2y agoThe (disproven) conjecture is for finite graphs.
- joelignaatius 2y agoI may be missing something. If theres edges u1 to v1 and symmetrical graph u2 to v2 then any post has to be between u1 to u2 for any u. Which means traversing a post would essentially be adding an edge to the graph (no matter the probability). It seems intuitively obvious (which is stated here as incorrect) that going through a post would be more expensive.
- geminger 2y agoSo it's been debunked.
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- andrewflnr 2y agoWhile I wouldn't accept a probabilistic "proof" of a theorem like this, it does seem clear to me that those results are important for directing the focus of research, especially in cases where they go against intuition. Given that most of the math community was barking up the wrong tree, even if these guys only had the probabilistic result, surely that would eventually help someone find the right proof? That's at least worth publishing as a letter or something, right? Ed: reflecting on my first sentence, I guess I tend to think of proofs as being at least as important a product of math as the raw truth of a statement. A fact is a fact, but a proof represents (some level of) understanding, and that's the good stuff. Experiments are potentially good science, but usually not math.
- bubblyworld 2y agoCan you elaborate? It's possible to prove probabilistic results about discrete objects rigourously - in fact this is a reasonably common technique (going back to work by Paul Erdos, and probably further). It can give you concrete information about the existence or non-existence of these objects (if something exists with non-zero probability in a finite set there has to be at least one example). It's not a "less valid" way to prove things, just a particular choice of approach. Sometimes it makes things easier, sometimes it doesn't.
- ilya_m 2y agoWe are talking about different kinds of "probabilistic" proofs. Igor Pak's blog post and the root message in this thread refer to statements that are validated only probabilistically. The probabilistic method, by Erdos and others, is a proof as rigorous as any. "Probabilities" in the this method are just a convenient apparatus for counting objects. (As an aside, there's important distinction with statements that are believed to be true based on empirical evidence, such as hardness of factoring or the Riemann conjecture. If the bunkbed conjecture was refuted based on Monte Carlo simulations, it'd be possible to reduce the level of uncertainty arbitrarily low by throwing more resources at running more simulations.)
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- IshKebab 2y agoThis post could really do with a couple of diagrams of what these graphs are. I think I got it from the description but I'm not sure...
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- willcipriano 2y agoIt does not in fact give you extra space in your room to do activities.
- tsimionescu 2y agoThis is an interesting case of a conjecture holding true for small objects but breaking down for huge ones, without specifically adding that size in somehow. Sometimes we tend to have this intuition that if a rule applies to all low numbers, than it must apply to all numbers, that there can't be some huge number after which it breaks down (unless of course it explicitly includes that huge number, such as the rule "all numbers are smaller than a billion billion billion"). This is such a powerful intuition, even though it's obviously wrong, that rules that break it are even seen as a problem. For example there is the so-called "hierarchy problem" in physics, which states something like "there is no reason why the weak force is so much stronger than gravity". As if 2 times as strong is somehow fundamentally different than it being 10^24 times as strong from a mathematical perspective. And this has ended up being called a major problem with the standard model, even though it's completely normal from a mathematical perspective.
- throwawayiionqz 2y agoMany examples of conjectures true for all n, until a very large one is eventually found : https://mathoverflow.net/questions/15444/examples-of-eventual-counterexamples https://mathoverflow.net/questions/15444/examples-of-eventua...
- snarkconjecture 2y agoThere are two kinds of naturalness principle in physics, sometimes called "technical naturalness" and "Dirac naturalness" respectively. Dirac naturalness is as you describe: skepticism towards extremely large numbers, end of story. It has the flaw you (and every other person who's ever heard it) point out. Technical (or t'Hooft) naturalness is different, and specific to quantum field theory. To cut a long story short, the "effective", observable parameters of the Standard Model, such as the mass of the electron, are really the sums of enormous numbers of contributions from different processes happening in quantum superposition. (Keywords: Feynman diagram, renormalization, effective field theory.) The underlying, "bare" parameters each end up affecting many different observables. You can think of this as a big machine with N knobs and N dials, but where each dial is sensitive to each knob in a complicated way. Technical naturalness states: the sum of the contributions to e.g. the Higgs boson mass should not be many orders of magnitude smaller than each individual contribution, without good reason. The Higgs mass that we observe is not technically natural. As far as we can tell, thousands of different effects due to unrelated processes are all cancelling out to dozens of decimal places, for no reason anyone can discern. There's a dial at 0.000000.., and turning any knob by a tiny bit would put it at 3 or -2 or something. There are still critiques to be made here. Maybe the "underlying" parameters aren't really the only fundamental ones, and somehow the effective ones are also fundamental? Maybe there's some reason things cancel out, which we just haven't done the right math to discover? Maybe there's new physics beyond the SM (as we know there eventually has to be)? But overall it's a situation that, imo, demands an explanation beyond "eh, sometimes numbers are big". If you want to say that physical calculations "explain" anything -- if, for example, you think electromagnetism and thermodynamics can "explain" the approximate light output of a 100W bulb -- then you should care about this.
- dudeinjapan 2y agoI already learned this from Step Brothers: https://www.youtube.com/watch?v=ulwUkaKjgY0 https://www.youtube.com/watch?v=ulwUkaKjgY0
- ykonstant 2y agoThat does seem extremely counter-intuitive; can the counterexample be adapted to provide a sequence of counterexamples with unbounded number of edges?
- r0uv3n 2y agoI guess you can just add some edges extra vertices and edges and give all the extra edges probability 1 of being deleted.
- User3456335 2y agoThe paper seems to have a definition where bed posts are never deleted, i.e. they are all assigned probability 1 in which case the conjecture is obviously true. The counterexample seems to rely on correlations between edge deletions which makes no sense because the deletions should be independent (in the definition I'm seeing on Wikipedia). I could be wrong here because I haven't read it in detail but on first sight, it looks like there are some serious issues with mathematical rigour here.
- JohnKemeny 2y agoYou can probably replace any transversal vertex with a large clique (depending only on p) such that the probability that the clique remains connected and at least one post remains. (I.e., you can "model" undeletability by adding sufficiently many twin posts.) Ps., do you mean "in which case the conjecture is obviously false"?
- User3456335 2y agoSurely, if the poles can't be deleted then which node is chosen from a bunkbed will not affect the connectedness probability, which would make it impossible to find a counterexample, right? What am I missing here? PS Depends on what conjecture we refer to here.