A visibility problem, how many guards are enough?
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- Опубліковано 19 чер 2024
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The video explains the proof of 'The Art Gallery Problem'. There are different versions of this problem (such as where guards can be placed) but we will focus on the most well known one where guards are forced to be in corners (and can see in all directions at once).
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Simple solution: stop using weirdly shaped museums
Ailis Catach then the museum would be empty
Alucard how exactly?
Nye Simpson would be a really boring museum without any large exhibits that don’t block the view
Sounds about right
It's atr
“Assuming your guards have 360 degree vision and can only be place in corners” why? Why would I assume any of that?
be silent comrade, we do not discuss such unimportant matters here.
well he did mention it'd be better to use 360 degree cameras but that he was sticking with guards. and hes saying only at corners so that the proof is a lot easier and to just have a solution
@@somerandomguyontheinternet9100 He said "at once" as in: they have to be able to see everything at any moment.
welcome to math
@@kuhmuh2357 ya bro
one operating the cameras
I'm not sure a single guard would be able to pay attention to, say, 50 cameras simultaneously, though
@@fernando47180 give him cocaine
@@fernando47180 who said 50? just =>n/3
One operating cameras and one on the ground to respond quickly, using radios
@@fernando47180 One guard operating 50 cameras with shape detection software
"My job gets me really dizzy."
"What? What do you do?"
"Oh, I'm just a 360° security guard."
"How is that possible?"
"I spin."
"How does spinning... Oh.. That must suck."
"No not really I love my job, it spins in the family."
This comment not having a reply after months is completely understandable.
You really threw a spin on that
Gyro be all
*Spins rapidly*
@@x_sphinz2982 look at what you've created.
No matter how many guards you put in a room. Nothing can stop a four man ECM rush.
Just watch the lazers
Except for that one guy that gets meleed by a guard, instantly killing them and launching them halfway across the gallery
@@Lopeped-Cring Just have one guy with Inspire for that case,lol
You dont exactly need low concealment build to do ECM rush XD
E
@@skell6134 Inspire doesn't work during stealth btw. Me and my friend learnt that the hard way trying to defy gravity.
Really thought you were going to triangulate that triangle into a Triforce. Missed opportunity
That would be cute but that goes against the principles shown in the video
@@selectivepontification8766 YoU mUsT bE fUn At PaRtIeS
@@selectivepontification8766Not sure. The induction would still work. It would just be needlessly complicated.
E
I go too far calculating the perfect preplanning of a heist in PAYDAY 2.
This makes me wonder how different the problem would be if the guards could be placed in places other than the vertices, *but* they have a limited range of vision of a given radius.
Im guessing you could draw imaginary intersections points by extending all vertices to a huge length, and look at where each and every one of these lines intersect. These would be the new points of interest since they tell us where we can get more information.
For the radius thing I have no idea where to start lmao
@@ifroad33 Good point.
E
Very - even just the question of how many you need to cover a circle is quite complex for small vision ranges.
This reminds me of the illumination problem discussed in Numberphile
Maybe because these two problems are basically isomorphic (same)? lol
Me too, I thought it was the same problem
They are kind of the same
@@aidarosullivan5269 they aren't "isomorphic" they are equivalent.
@@aidarosullivan5269 it is similar, but not the same. Illumination problem defines walls (sides of polygon) as mirrors; it reflects light, whereas art gallery problem define its walls to be non-reflective.
I watched this video twice 3 years apart, before taking a college-level discrete maths course and right after. Now, being able to understand the terminology and logic just made me appreciate this video that much more.
So, I guess there is something much deeper in the video than what I saw in it. Because I did not take a course in mathematics
E
Yea, im watching this after taking discrete maths aswell and it is nice to see all the proofs which I once did used in a problem. The induction explanation in the video feels backwards because formaly you first make a conjecture and then use the trivial case to proove the conjecture.
The thing with Ocarina's stealth section, is that it could have been easily been done with *one* single guard, who just guards any single choke point that you HAVE to go. Like, with that first square with the two fountains, just have someone lean against the wall in the left corner. Boom, no way to get through without being seen. If you want to assume that they're preparing for an intruder who could actually attack them, just add a second guard in the same spot. The only reason you would need the entire area in LoS, would be if you wanted to... I don't know, stop some weird kid from vandalizing the fountains?
E
You can have 10, 15 or even 20 guards in a room, cameras, sensors, tripwires and mines.
But Snake will always find a way.
😂
SketchyTh0ughts da box
Just don't put a cardboard box in the museum, and you'll be fine.
Because he can reload saves
Snake sneaking in and seeing 4 well dressed men with clown masks
Zero.
Because it's a closed room.
ua-cam.com/video/k-oVuQpjG3s/v-deo.html
ua-cam.com/video/ar0xLps7WSY/v-deo.html
If there's no light source it is indeed very difficult to observe the entire room when it's closed off from the outside. That aside, the question was how many guards was required to observe the entire room at once, so whether or not a break-in is plausible is besides the point.
when did it ever mention that?
No, the entire room has to be observed since they forgot to install a roof, so thieves can drop in and land at any location in the gallery from their helicopters.
@@dumb214 Wouldnt be able to do that without breaking their legs tho ? Unless someone conviniently place stuff for them to land on tho
That's how you TEACH 😍 I can re-explain the whole thing without having to watch it ever again. Wish my math teachers were like you sir. Sad part is this isn't there for my syllabus 😣
approved.
Your math teacher didn't make a lesson to be watched by thousands, they probably didn't even have the powerful video editing to assist in understanding
E
Something I didn't mention but wanted to at least say here is that for galleries in the shape of an 'orthogonal polygon', where every corner makes a 90 degree angle (or 270 if we're talking internal angles) then the upper limit is n/4 (rounded down) instead of n/3. The proof is very similar so if you want a challenge you can give that a try.
I'd say that's pretty simple.
Instead of making triangles, you make rectangles. Instead of 3 colors, you use 4 colors. You apply the same proofs used here - and you get n/4 rounded down.
you smart
E
Can you design a room made of mirrors that has a space that doesn’t get light?
@@MsTwissy Yeah, a room full of mirrors with the lights off
10:33 Minimum 2 guards are required here
right-most blue and top-most red or green
@@aasyjepale5210 We don't know for sure if the blue is able to see all of the bottom triangle (there may be a slither missing) as we aren't told exact lengths of the sides of the bottom right corner.
Top green and the red below is a solution we know can be correct, also they're close enough to chat to each other.
Prove it.
@@precumming hmmm
No shit sherlock
my first thought of triangulating a triangle was to use the bisector of one of its angles (splitting it into two, smaller triangles) until you mentioned that it's already a triangle and doesn't need to be split up.
thank you brain for wanting everything to be complicated.
This being called The Art Gallery problem made me think this was about the hiest in Payday 2
Guard: “Why am I spinning in circles in the corner of the room???”
Me: “Don’t worry about it, you are still getting paid”
4:06 can definitely be done with 2 guards...just dont only place them at verticies. The figure can be adjusted slightly to make your point, but as it currently is shown you can put a guard on the bottom line between 2 of the upper point’s angles and the last guard where they can see the other point.
4:00 you can observe everything with 2 guards if they weren't in the corners
Some of the middle triangle would be out of view (draw straight lines from corners)
actually no and i have proof
*i got none but pretty sure im right my brain say so*
true
This can be done with 2 guards in this case. Proof by example:
Lets divide the room into a large rectangular area at the bottom of the picture, and the three triangular bumps at the top.
Imagine three friends, each one standing at the top of a bump and looking down at the far wall. They see most of the gallery, but not all of it, and cannot see each other.
Just to help with the explanation, imagine that the room is dim and the center friend turns on a flashlight, illuminating the center cone of view.
The left friend, standing at the top of the left bump, can see part of the flashlight's beam. If the left friend turns on a flashlight, too, there will be a bright patch illuminated by both flashlights. That is the area that those two friends share and can both see.
The same thing happens on the right. There is an area that the center and right friend share.
If you put a guard in a shared area, the guard can see both of the friends that share the area.
If you put one guard on the left and one guard on the right, those guards will be able to see all three friends and also the whole rest of the room.
QED
Even if they were in the corners u could still use 2
I am very jealous about you. You have potential and excellent explaining skills.
Heartly congratulations. Your videos are awesome.
Replying just to stress this more. Since the very first time I watched you I felt something different, more clear and objective explanations. Congratulations.
Thank you!
Look at triangles that are covered by multiple guards in a color and in different colors. I think you could use patterns with shared triangles to optimize guard placement (by minimizing triangle sharage).
4:07: In this case two guard are enough because of simple mathematical solutions. Since it is true you only need one guard for a normal surface you can start looking for them. So instead of seeing this as some kind of rectangle with three triangles, you can extend the legs of the triangles and if the legs overlap you can use that specifik region to place a guard. This guard will be capable of seeing the left triangle as well as the right.
4:07 You only need to, the line of the right side of the left triangle and the line of the left side of the middle triangle touch eachother inside of the polygone. Just put 1 guard at the intersection to watch over both the triangles.
Came here to say that, good spot
The only issue is that they have to be placed on the corners :/
1:16
"The art gallery problem is both easy, by easy I mean understandable, and also hard, I'll explain what that means later."
We know what hard means it means hard
*lenny face*
Shows shape: How many guards?
Me: Two!
Explains a lot of theory: We can prove at least 10/3 rounded down, thus 3.
Me: ...Two!
Informs: There's no algorithm that can prove the exact amount yet.
Me: TWO! ... Two, 2... TWO! Z! [in Roman] II... TWO, dangit!
Me disappointed...
There is no efficient algorithm. There are algorithms that can solve it, but they start to take a very long time as the shapes get complex. Complex meaning you have no chance of solving them just by looking at them.
@@Satheo05 Probably (long time since I watched the video), but that isn't the point. As the shapes become more complicated, they quickly are impossibly difficult to solve.
Another example of a problem that seems simple but isn't is factoring primes. For example, can you figure out what two prime numbers multiplied together give you 10? It's pretty simple - 2 and 5. But we don't have an efficient algorithm to solve those problems. We can do it easily for small numbers, but not for large numbers. If you had a way to efficiently factor large numbers, you could easily make millions of dollars by showing the NSA or a large bank how.
Always look forward to a new major prep video. Keep the good work up
This has quickly become one of my favorite channels! Continue what you are doing, its awesome content!
Payday 2 taught me everything i need to know
You know I often forget this guy used to do educational videos
So addicted to your videos ! Great job keep it up 👍
Guys, the thermal dr... Oops, wrong heist.
bain never shuts up
ah, i see you're a man of culture as well
Donacdum
Did anyone bring a medic bag?
Creating perfect camera/guard system with no blindspots:
Payday gang with loud approach:
5:50 This is handwavy, and wrong. Think of a polygon starting with a square of side length 1 and cut out a smaller square of length >0.5 from its top right corner. Let the bottom left vertex of the 1x1 square be A, its adjacent vertices be B & C, and let the bottom left vertex of the smaller cut-out square be A'. If you start with vertex A, which is 90 degree (less than 180 degree), you cannot directly connect B and C. When you "shift" BC and hit A', by connecting A and A', you don't get a 5-gon and triangle, but two 4-gons. This way, you'd need strong induction.
I thought you were gonna solve the NP hard problem, but I'm not disappointed 👍🏻
Egg
I really enjoyed this video. Thanks a lot. You explained it in a very enjoyable manner
Even once I've seen your videos, these video are so relaxing that i watch them again.
What i'm liking about these kinds of videos, is that more than just explaining the problem and going through how to solve it, they're also explaining how to use and make proofs and why they're so important.
Well done man you really do have a talent in explaining not only engineering concepts, but STEM concepts in general.
Speaking of P versus NP problem I really do wish you make an entire series about it not just one video.
Thanks a lot
Side note, you'd actually need to use strong induction to prove any polygon can be triangulated.
In regular induction, you prove the theorem for n+1 given the theorem for n, n> some a for the inductive step.
In strong induction, you prove for n+1, given a,a+1,a+2,..., for n>a, for the inductive step.
For example, the shape you showed for n=5 needs n=3 and n=4. The proof for n=4 was not sufficient.
this is so cool! i love your content
1AM UA-cam is a magical place
man keep doing the good work v good content
At least 40 and they all get EOD suits an miniguns
You + Explaining Maths = Pure Happiness ❤️
Everybody: awesome video!
Me: big ass dominoes
What
Explosive Pineapple 6:20
Going from watching your second channel to watching this is a sharp transition.
great video, pretty much all of the concepts you explained i learned this year at Ohio State University. i’m an engineering major
4:09 2 guards would be enough there, though, as long as you removed the argument that they had to be situated at the corners, so it's a bad example. If you bent any of those middle points to not be fully visible from the outside, you would have a point though. 4:36, again only 2 guards needed here. You are only connecting the triangles of points 2 distance away, when the lines for the triangles could be placed between any 2 vertices that don't force you to draw outside the shape for them to be reached.
This said, the easiest way to determine the visibility of an area is to extend all lines extending from vertices to the edge of the shape. All points found within the lines extending from a vertex can see that vertex. If there is a vertex inside these lines, extend a line from the original vertex using along which this interfering vertex would fall to the edge of the shape. This will create a number of shapes that can see certain amount of vertices. Find the point or a point that can see the greatest amount of vertices, and shade every point it can see. Then look at the remaining area and look for further overlap of vision. This does a remarkably better job than drawing those triangles.
@Stratowind I extended the edges on both the right triangular extensions and you are incorrect. They cross before the edge of the figure.
4:09 ok but that’s a condition. There would only be one guard needed if he could see through walls but that’s not the point.
4:36 is only a demonstration of the rule he expressed, not a minimal amount of guards
@@abl9643 The original question asked how many guards would be needed to see you regardless of where you were. Originally, this didn't require the guards to be standing in corners. Later on, he adds the condition that the guards would be standing in the corners. Also, we are trying to narrow the upper limit which, given the condition was not originally there but later added, should not require the guards to have to be standing in the corner.
i just realized that Zach looks super happy and excited during his vids but he also looks like he hasn't slept in 3 days
I was able to do alot what you were saying in my head. But didnt know the words for it >.< and the visuals you provided were my imagination exactly
Already liked this channel, but the fact that you love ocarina of time just made me a fan forever
Slight sidenote: finding an efficient, polynomial-time solution to an NP-hard problem wouldn't just be a better algorithm. It would be a paradigm shift, a world-changing, million-dollar prize winning, revolutionary breakthrough that would change the future of computing, challenge our perception of proofs as a concept, earn you a tenured position at the prestigious university of your choice, and grant you everlasting fame.
Thanks, Gaston!
Here's the solution to the problem given an example from ocarina of time: Go to the place at night and a guard will stop you in your tracks. Only 1 guard is needed. Now why they only have the night shift, who knows.
Nice, but was looking for guard placement ANYWHERE - not limited to vertices or edges.
Thank you!
What about smoothly curving art galleries; is there anything that can be said about those? (If the curves don’t need to be smooth you can get fractals, which can need infinite guards)
You can use smoothed analysis to come up with a fixed parameter tractable algorithm where the fixed parameter is the number of reflex vertices.
i would've done a similar aproch, but more like an algorithm, plus it needs you to be able to tell if everything is covered.
1. split the shape into non-overlapping triangles
2. place a guard in the center of every triangle (i mean why only limit them to corners?)
3. the whole shape is now guaranteed to be covered.
4. go through each guard and remove them, if the shape is still completely covered let the guard remian removed, if not, place him back where he was.
5. do this for every guard until none can be removed anymore
6. tada!
no idea if this would actually work, and i'm too lazy to program it.
Regarding the induction proof of the polygons; wouldn't it be easier to start with a triangle and connect a 4th vertex to two vertices of the already existing triangle? This easily proves any shape can be triangulated. It also proves the coloring as the 2 vertices you connect the new vertex to can only use 2 different colors, so you the new vertex will just be the 3rd color. I also feel like this resonates better with the classic m+1 idea of induction. I suppose that is kind of what you did, but backwards.
This is all very reminescent of Sperner's Lemma, which I closely studied as part of a school project to optimize fair share. I love visually appealling mathematical curiosities!
The reminds me of the Yiga Clan Hideout in BoTW except the guards move in shapes and can be lured away with bannanas.
I can’t look at this without thinking of payday. Thanks youtube recommendations!
🤔 The guards have 360° vision, but in which dimension? What if their vision is limited to a plane parallel to the ground?
Assuming that the room itself has dimension 3, this would mean that we need indefinitely many guards to guard any room.
Or is the room as well two-dimensional?
Probably, because the proof only was about two-dimensional shapes. Do we have proof that this also applies to rooms with 3 or more dimensions?
What about 42?
If we have a room in dimension 42, how many guards do we need to watch it? As many as the quantity of roads a man must walk down?
The answer very very likely depends on the dimension of the guards vision.
Yes.
I think so too.
Too underrated. Nice man
No way I've seen engineer/STEM Zach before Zach star himself? And I never noticed?!
If there is less then 90 degrees a guard must be place to see it let's call 90 degrees h h < n/3 =if h n go with h
NP-hard problem, pff...
Let's take this on!
Now I have something to do on bored afternoons ;p
4:07 you can watch whole area with just two guards, take away the middle and right guard and place one guard at the bottom in which he has angle for both middle and right spike areas of the room
Its reminded me of topics like pigeon hole and vertex coloring without same on adjacent 🤔🤔🤔🤔nice topic and nice explanation
Love this video! The math was really interesting and I learned a lot.
But, I feel like one thing that kinda frustrated me was the lack of sense. The ‘rule’ of only being able to place guards in corners held back the potential of placement a lot, and a fair few of the problems presented that required 3-4 guards only required 2 if a guard had been placed in a non-corner to observe more. That, plus the fact that the actual distance of how far you could observe a would-be intruder doesn’t make sense. Museum hallways and rooms are usually of a size that allow you to see a human figure at the end of it unless it would be a more absurd size, which most of the examples didn’t have.
Still an absolutely amazing math video but if it were to be applied in an actual scenario or be attempted to be solved with just common sense it would kinda fall off.
11:11 Two guards here minimum, too.
One in the corner of the top area, the bird head shaped bit.
And another JUST below that, in the main room.
This would make a great puzzle game
after 10 minutes I have realized this is NOT a Payday video
For every example shown the most guards needed to observe 100% of the area is 2.
You're overlooking overlap of zones of observation.
In the example of the last illustration place a guard at the point at the north most R plus the point at the east most B all areas can be seen when their overlapping areas are totaled.
The combination of the north most G plus the corner designated G just south of that would also combine to observe 100% of the area.
Even the two R placements previously alluded to, together, would also work.
I'll let you figure out what guard post would work if you have a guard at the furthest north west B point.
(Hint its an R that i previously mentioned.)
At 4:07 two guards would actually be enough instead of three like the video says. Like this: if instead of the two rightmost guards only one is placed approx. 3/4 of the way from the left near the bottom side. I presume based on 1:55 that the guards don't have to be placed at the main vertices. One would have to make the rectangular region of this gallery a bit shorter in vertical direction for 3 to be required. Not that this changes the message of the video in any way, but it just happens to be the geometry in that particular picture.
You're right, if we assume guards must be placed on vertices then three are required. I should've extended the triangular regions cause then three guards really would've been required no matter where they could be placed.
11:54 That determining the minimum number of guards is *NP-hard* means that if in the future someone were to discover an efficient way to solve it, then that same algorithm could also efficiently solve every problem in the huge class NP (problems where you can quickly check whether a given answer is correct).
It would have huge implications (P=NP), and go against most experts' intuitions on what it means for a problem to be "hard".
I think you described a NP-complete problem ?
For n=3x, n=3x+1 and n=3x+2, it is possible to create a sawtooth pattern that requires x guards
Since n/3 rounded down equals x, for every number of sides in a polygon, there is at least 1 such polygon that requires n/3 rounded down guards.
Stable video👍
It’s simple. You just need to make sure that your architect isn’t a complete idiot
0:36 . i mean you can just use one guard. If we assume the boundries are impassable like the game, the guard would just have to stand at either the entrance or exit.
This will help me in tactical FPS games
my guess is that if you can draw a point from the center of the shape and it never leaves the bounds of the shape and then goes back into it, then it’s only 1 guard. as for anything else i have no idea
Hi
uses a bunch of math and theorys: gets possible but not lowest number.
me using my eyes: "2 seems fine"
Just a guess: num_guards=1+floor(num_concave_corners/2), assuming free placement of guards within the region.
No matter how many guards you add, the Payday gang will find a way
breaking 3d objects into triangles is how the entire computer 3d rendering field began. today it continues to be that way, just really tiny triangles
The placing in corners bit is a tad disingenuous, because it's only for solving that formula, there are scenarios in which one wallless guard could cover more room than a vertecie bound guard.
Although, i believe this could be solved by making derivatives of shapes(a term i just made up) by drawing each line to infinity and the counting new shapes in-between other pre-existing shapes which have lines coinciding without re-entering the original shape, or parallel.
I'd have to spend some extra time on it, but for example the one you reffered to as 'honeycombing' would result in 3 because you need 1 guard, + the 2 intersecting external lines in the 'derivative'.
If the rules were different (guards weren't restricted to vertices), 4:08, 4:20, would each only need 2 guards as well...
Actually, because of the angles, 2 is sufficient in one case even assuming all the rules are kept.
In 4:20, place guard 1 at the topmost vertex, and place guard 2 one vertex down from guard 1.
However, 4:08 doesn't work with the rules. If you could place 2 guards anywhere however, it would work.
Place guard 1 on the bottom line at the point where the right triangle hypotenuse is extended to, and place guard 2 anywhere to the left where the entire far left triangle can be seen
I still can't believe it I got the right answer before you revealed it with explanation.
At 4:09, why are 3 guards needed?
Keep the guy on the top left where he is, but instead of having the remaining 2 guards have one guard on the bottom line, directly below the mid-point of the top right horizontal line.
That new guard would be able to see into both the middle and right triangular peaks (as well as the entire rectangular body) and that top left guard could see the remaining top left triangle.
is it still NP if you remove the vertex requirement?
Payday guards would need this
Well, that and pray that their mission is silent only
Crooks won't stop to restart if they could just blow through the rest of it
for the second shape, only 2 guards are needed. keep the one on the far left where he is and put another guard along the bottom wall and he can see the other 2 triangle areas
I found out about him from his other channel, and I keep forgetting he’s like actually fucking smart
I know a rule is to only place guards at corners, but for the gallery at 4:00 two guards would suffice if you remove the two on the left and replace it with one at the bottom where the extended lines of the inner walls of the two triangle extrusions meet.
Actually in the 2 vertices.example (looks like a castle) 2 would be enough
If theycwere both.placed on the bottom wall around 1/3 and 2/3 across
Thanks, now I can calculate how many security camera my home need.
4:03 No, one guard could be standing right at the bottom wall to observe both triangles, because their angle meet infront of the wall. One other would need to observe the left triangle.