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How to avoid being hit by a laser in a room of mirrors [video]
- xwdv 5y agoIncredible, but I don’t see a way to quickly figure out what the safe spot would be given any configuration of a room.
- deleted 5y ago[deleted]
- jekub 5y agoIt works the other way, the shooter position is fixed as well as your position. Your are allowed to place a finite amount of blockers to make your position safe. The video comment say that it is always possible to make your place safe with at most 16 blockers. There is a link to the actual code used.
- mitko 5y agoYou could try projecting rays from start through the obstacles to see where they intersect after a limited number of reflections. This should narrow it to a few candidate spots
- mehwoot 5y agoThere isn't always a safe spot, but you can generate a room that makes a safe spot- https://www.youtube.com/watch?v=jJ6FD59U0_E https://www.youtube.com/watch?v=jJ6FD59U0_E
- laurent92 5y agoThe area covered by the lasers should remain painted, so we could see all the spots which haven’t been hit ever.
- tdeck 5y agoI'm curious why you couldn't just surround yourself with the 16 objects. Perhaps they're meant to be much smaller than the person?
- ordu 5y agoThis way allow you to hide. A tight group of obstacles will give away your position and it will be demolished with a bazooka. Though an attacker might calculate the possible location of your body and use a grenade. But it might be unsafe for himself.
- neitherboosh 5y agoThe posted solution seems to work even when the objects have zero area, and are just points in space. If you just surrounded the target with 16 points, there would either be gaps or the points would be infinitesimally close to the target and would occupy the same space.
- ric2b 5y agoSo that means that any possible laser coming out of the shooter's position eventually passes through one of these 16 specific infinitesimally small points _before_ ever hitting the target? And that is always possible for any position of shooter and target? That's bonkers, wow.
- Pyramus 5y agoBoth person and obstacles are meant to be infinitesimally small, i.e. points with mass zero, which makes one really appreciate the beauty of the solution.
- bunabhucan 5y agoThis video explains it. His pinned comment explains the mod arithmetic: https://www.youtube.com/watch?v=jJ6FD59U0_E https://www.youtube.com/watch?v=jJ6FD59U0_E
- parhamn 5y agoThanks for the share! I took some quick (read, not proof-read) snippets on this if anyone wants to see: https://synth.app/s/uiElJvnnk https://synth.app/s/uiElJvnnk
- Pyramus 5y agoI wish he had explained the mod arithmetic more, the periodicity of 4 is really at the heart of the solution.
- nullc 5y agoIf the room was an octagon would you need 64 objects?
- seaish 5y agoThey did do it with a hexagon with joined sides. Not sure if the octagon is solvable. https://youtu.be/qhqdAiypldc https://youtu.be/qhqdAiypldc
- schappim 5y agoGreen Circle: You expect me to talk? Red Circle: No, Mr Green Circle, I expect you to d̵i̵e̵ find the spot in the mirror room where you won't get hit!
- Lio 5y agoSurely it's more like: [Red Circle]: I suggest a Duel of Titans, "mano a mano" the only true test for gentlemen. [Valet leading Green Circle to room of mirrors]: If you kill him, all this be mine. Monsieur, Good shooting. Although it does also rather make me think of Buster Keaton standing still as half a house falls over him and the cameraman closes his eyes.
- Yeri 5y agoThis is stressful to watch.
- deleted 5y ago[deleted]
- irjustin 5y agoFeels like the difference between playing the video game and watching TAS bot do the run =/
- rcfaj7obqrkayhn 5y agofor other people that also didn't get it... there's a green circle on the top right side corner
- blu_ 5y agoWhat about scatter from the objects?
- infogulch 5y agoThe design assumes absorbing obstacles. He did make a separate video with reflecting obstacles, but the green circle is not safe. https://youtube.com/shorts/DyRPASsynhI https://youtube.com/shorts/DyRPASsynhI
- wnkrshm 5y agoAbsorbing obstacles will technically not protect you either from receiving any light at all. Diffraction will 'push' some light into the geometric optical shadows.
- mike_d 5y agoHere is the explanation: https://www.youtube.com/watch?v=jJ6FD59U0_E https://www.youtube.com/watch?v=jJ6FD59U0_E
- parhamn 5y agoI took some quick (read, not proof-read) snippets on this if anyone wants a quick run down: https://synth.app/s/uiElJvnnk https://synth.app/s/uiElJvnnk
- techdragon 5y agoThank you. So much more useful to grasp this relatively simple geometry problem with a few step by step pictures. Wish the link was to this instead of the YouTube clip.
- fxtentacle 5y agoThat page is great, but the synth.app homepage seems weird to me. "Browser from thefuture. Augment your work and your mind with the internet, don't just browse it." What exactly is synth.app you offering? I couldn't figure it out from scrolling around their homepage a bit.
- parhamn 5y agoThanks for this feedback! Def need to go back to the landing page and build it out at some point.
- codetrotter 5y agoThis is very neat. I feel strongly however, that each of the buttons that you have for opening the video ought to directly link to the time stamp in question. So for the button you have for time stamp 02:46 you’d link to https://www.youtube.com/watch?v=jJ6FD59U0_E&t=2m46s https://www.youtube.com/watch?v=jJ6FD59U0_E&t=2m46s and for the button for time stamp 03:30 you’d link to https://www.youtube.com/watch?v=jJ6FD59U0_E&t=3m30s https://www.youtube.com/watch?v=jJ6FD59U0_E&t=3m30s and so on. For the very first button however, linking the very start of the video is nice though.
- infogulch 5y agoHere's one with a close up of 'the circle with plot armor', as one comment describes: https://youtu.be/jgpjnrFEUFI https://youtu.be/jgpjnrFEUFI
- bayesian_horse 5y agoDon't go into the room.
- shapefrog 5y agoBreak the mirrors.
- taneq 5y agoCalm down, Alexander! ;)
- effingwewt 5y agoDuck? Does this always assume a 2-D room? I wonder what the simulation would look like for a 3-D? Makes me think of Catherine Zeta Jones in 'Entrapment', or Vincent Cassel in 'Ocean's 12'.
- fjfaase 5y agoI guess, you would need (at least) four times sixteen, that is 64, blockers.
- fouronnes3 5y agoAhah, your comment has the same vibe as this great fiction of how would Richard Feynmann pass a "lateral-thinking puzzle" interview question. https://ericlippert.com/2011/02/14/what-would-feynman-do/ https://ericlippert.com/2011/02/14/what-would-feynman-do/
- pvaldes 5y agoSwitch one receives a green cable, light C in the room has a green cable, Switch two has a yellow cable, same as light A in the room...
- starfallg 5y agoTurn off the power.
- zzt123 5y agoThis was beautiful. I never imagined the existence of a solution was always guaranteed, given that this includes the case where you are using point-particle blockers.
- uuidgen 5y agoI think the solution can be reached following from a set of simple observations about a single line of sight: - in a square room all angles are right - with wall at right angles the laser light will always create an inscribed square and return to the source for any given angle - there is exactly one angle in a range <0,90> at which the light bounced from a single wall will go through an arbitrary point in the room - there are 4 walls which gives at most 4 squares to be blocked - the blocking point can be set anywhere on the inscribed square before the target - this gives at most 2 points per square (we can consider them left- and right- -hand directed) hence 8 points necessary - at start we selected only angles from the <0,90> degree range for the ease of calculation, so there are also symmetric versions in the (90,180> range - which gives us at most 16 points to block all the possible squares in both directions
- zzt123 5y agoThis line of reasoning reminds me of that one IMO geometry problem… I’ll unwind this once I’m less brain foggy.
- gampleman 5y agoI wonder if there is a way to solve it such that both the shooter and the target are safe?
- akdor1154 5y agoIt should be more like "how to find a shadow in a room full of mirrors", shouldn't it? Which is even more amazing.
- lmilcin 5y agoWell... no. These are two separate problems. This example doesn't show all places where the shadow is. It only shows, when you place yourself and a light, an example arrangement of occluders that will prevent you from being hit.
- akdor1154 5y agoWhy are they separate problems? Maybe I should have phrased "how to create a complete shadow in a room full of mirrors".
- SamBam 5y ago> This example doesn't show all places where the shadow is. I'm curious. Does it not? Because, from the descriptions of how to place all the blockers in order to create a shadow on that one spot, it did seem to me as if that one spot would be the only shadow in the room. With 16 blockers can you have more that one discrete shadow?
- lmilcin 5y agoHave you proven there is going to be only one shadow? Is it possible there are symmetries? I tell you that these are two different problems, just read it again: 1, given light source and an object, find position of occluders so that the object is never illuminated and 2, given light source and occluders, find all places that are not getting illuminated,
- NKosmatos 5y agoThere was an interactive web page where you could place single point light sources and obstacles freely in a 2D room and then see in real time the shadows and where are the safe points. Too bad I can't remember the URL and can't find this old page. Anyone has a link to share?
- azalemeth 5y agoFun problem. Can't help but ask what happens if the room is circular, or shaped in such a way that it doesn't easily tessellate, however...
- mnmmn123456 5y agoThe solution is great, but it shows a fundamental flaw in mathematics, that I am thinking about a lot: You have to be smart, no doubt, but you also have to know the answer to get to the solution - It requires so many steps first until "the solution / the proof is clear". If you go down a different route (as I would have done), then you'd likely not succeed and certainly be in "rough" unknow territory, making it difficult for others to help or assess your approach. And this is expected: The hard problems will take the smartest people to solve eventually. I wish we could be more honest about that, that Math is a lot about "memorizing problems". And that was always the case (over history): At some point in history all this was a frontier. And there will be people willing to fight on this frontier and I am not disputing that intelligence/logical thinking is also required. I am just wondering if all this couldn't be put to more practical use. But then maybe it is and as fascinating as it is, most of us move on from Math and work on real world engineering tasks, eventually and from time to time still look back to these problems, as we do today. And absorb them, adding them to our repertoire and looking/assessing for real world application, which likely is limited / none.
- antonfire 5y agoI think a bit of disillusionment with mathematics is healthy, but it's misguided to call this kind of thing "a fundamental flaw in mathematics". For one thing, in my experience, working on real life problems involves a whole lot more "memorizing problems" than working on mathematics. And thank goodness for that! When I drive over a bridge or install an app, I really don't want to hear that whoever made it has an aversion to memorizing problems and solutions. Solving a real world problem is usually boring. Making a real life thing well is usually about correctly putting together many pieces that some other people have put together, in a way that's roughly similar to how somebody else has already put a lot of those pieces together before. Most of the work is well-trodden and uncreative. I think part of the reason mathematics feels like it's about "memorizing problems" is that getting something deeper than that out of it is a habit/skill. A very important aspect of the vague notion of "mathematical maturity" is a habit of looking at a solution to a problem with the mindset of "how could I have come up with this?". That is, unpacking a problem and solution into some deeper understanding or way of thinking that led to it. As opposed to filing the problem and solution away "as is" into some toolbox to be referenced later. A lot of "gotcha!" solutions to mathematical problems are unsatisfying precisely for this reason. In a lot of cases once you've read a problem and read a solution to that problem, and understood the solution, you've done about 10% of the work. The remaining 90% is this difficult work of unpacking and repacking lessons from that solution into something that actually deepens your understanding of what the problem is about. Sometimes reading the solution is actually doing negative work. In other words, if you look at this problem and look at this solution and think "neat, but I don't get anything out of it beyond 'neat'", that's normal and fine, and probably correct. But that doesn't mean there isn't anything in it beyond 'neat', and a big part of learning mathematics well is to dig deeper than that even when the problem and solution don't force you to. Whether that digging is worth it is up to you. (It's often kind of a crapshoot in terms of payoff.) But that's a more complicated situation than "this shows a fundamental flaw in mathematics".
- deleted 5y ago[deleted]
- dkersten 5y agoEncircle yourself with the circles? That would be finite and if you're thin enough, might even be minimal? Certainly easier to find the configuration than the actual solution. (ok ok, it looks like the blockers are point-size, so this wouldn't work of couse)
- tantalor 5y agoNot finite. Not even countably finite.
- odd_perfect_num 5y agoOr even countably infinite :) One could argue that all finite sets are countable since it's not too hard to find a surjection from the naturals to a given finite set. (Intended to be mildly humorous.)
- tantalor 5y agoOops, that's what I meant, not even countably in-finite
- dkersten 5y agoI wrote the top of my comment before realising that the points were points and bot circles with a radius.
- soapdog 5y agoquestion: "How to avoid being hit by laser in a room of mirrors?" me, a connoisseur: "This requires further reflections..."
- GistNoesis 5y agoSo in an icosahedral room, let me guess : one would need 696729600 blockers ?
- LightG 5y agoThanks, this is exactly what I needed at the right time.
- anamexis 5y agoDoes anyone know if this is solvable in 3 dimensions (inside a mirrored cube)?
- mabbo 5y agoBased on the full explanation video, I suspect it probably is. https://www.youtube.com/watch?v=jJ6FD59U0_E https://www.youtube.com/watch?v=jJ6FD59U0_E But perhaps instead of 4^2 blockers, you'd need 4^3. But, I'm making guesses based on intuition, which makes for bad math.
- raldi 5y agoI’d like to see a version of this where the barriers are colored and the lasers are monochrome, but when a laser hits the barrier, instead of disappearing, it takes on that color. That way, you could see which barrier saves the target at any point.
- ghostbrainalpha 5y agoThat's a really cool idea.
- raldi 5y agoThe lasers could even leave a dim colored trail behind them, so that all the deadly space in the room gets shaded over time.
- vmception 5y agoI think this just helped me solve bullet hell games where it is tempting to evade so much when you can just stay in one part of the screen danmaku!
- j2kun 5y agoMade a demo here: https://j2kun.github.io/assassin-puzzle/index.html https://j2kun.github.io/assassin-puzzle/index.html Blog post: https://jeremykun.com/2018/07/24/visualizing-an-assassin-puzzle/ https://jeremykun.com/2018/07/24/visualizing-an-assassin-puz... Excellent video that I learned of this problem from: https://www.youtube.com/watch?v=a7gp9c2p0UQ https://www.youtube.com/watch?v=a7gp9c2p0UQ
- deleted 5y ago[deleted]
- kazinator 5y agoThe chosen visualization complicates it. In a perfectly square room, a beam in any direction, originating from some interior point p0, will hit the four mirrors and form a rectangle, passing through the original p0 at the same angle. When tracing a beam, we don't have to consider more than five mirror strikes. Moreover, this rectangle can only pass through another distinct point p1 exactly once. If we know that if p1 lies on a beam rectangle that is interrupted by an obstacle, there are only two possibilities: p1 lies on the unobstructed part of the rectangle between p0 and the obstruction. Or else it lies on the shaded part. We can determine which rectangles that pass through p0 also pass through p1, and then for each one determine whether p1 is in the shaded part. I suspect this can be done with some computational geometry (with a fair number of cases in it) without resorting to a brute-force ray-casting technique.
- EricRiese 5y agoI thought this was going to be a new album by The Flaming Lips.
- 1_over_n 5y agoHow about just wrapping yourself in mirrors :)