A visibility problem, how many guards are enough? 

Simple solution: stop using weirdly shaped museums

“Assuming your guards have 360 degree vision and can only be place in corners” why? Why would I assume any of that?

one operating the cameras

No matter how many guards you put in a room. Nothing can stop a four man ECM rush.

"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."

Really thought you were going to triangulate that triangle into a Triforce. Missed opportunity

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.

This reminds me of the illumination problem discussed in Numberphile

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?

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.

You can have 10, 15 or even 20 guards in a room, cameras, sensors, tripwires and mines. But Snake will always find a way.

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.

Zero. Because it's a closed room.

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 😣

10:33 Minimum 2 guards are required here

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.

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.

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.

I go too far calculating the perfect preplanning of a heist in PAYDAY 2.

This being called The Art Gallery problem made me think this was about the hiest in Payday 2

I am very jealous about you. You have potential and excellent explaining skills. Heartly congratulations. Your videos are awesome.

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!

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

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.

4:00 you can observe everything with 2 guards if they weren't in the corners

Guard: “Why am I spinning in circles in the corner of the room???” Me: “Don’t worry about it, you are still getting paid”

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

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.

Payday 2 taught me everything i need to know

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.

So addicted to your videos ! Great job keep it up 👍

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.

You know I often forget this guy used to do educational videos

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.

Creating perfect camera/guard system with no blindspots: Payday gang with loud approach:

I thought you were gonna solve the NP hard problem, but I'm not disappointed 👍🏻

Guys, the thermal dr... Oops, wrong heist.

this is so cool! i love your content

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...

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.

1AM RU-vid is a magical place

At least 40 and they all get EOD suits an miniguns

You + Explaining Maths = Pure Happiness ❤️

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.)

Thanks, Gaston!

man keep doing the good work v good content

Going from watching your second channel to watching this is a sharp transition.

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

Gta taught me that you need more guards

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.

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.

great video, pretty much all of the concepts you explained i learned this year at Ohio State University. i’m an engineering major

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.

Everybody: awesome video! Me: big ass dominoes

Nice, but was looking for guard placement ANYWHERE - not limited to vertices or edges. Thank you!

3:00 Upper Limit is very easy, just need 2 guards. One at the corner of the L shape on the right, then the other guard is at the corner of the big L shape on the left. So both guards can see end to end and overlapping. I hope I explain it right.

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.

Already liked this channel, but the fact that you love ocarina of time just made me a fan forever

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!

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.

The reminds me of the Yiga Clan Hideout in BoTW except the guards move in shapes and can be lured away with bannanas.

🤔 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.

You can use smoothed analysis to come up with a fixed parameter tractable algorithm where the fixed parameter is the number of reflex vertices.

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

Too underrated. Nice man

3 guards. One on the ground floor, one on the cameras, and one as backup moving between cams and the floor randomly.

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

Payday guards would need this

Its reminded me of topics like pigeon hole and vertex coloring without same on adjacent 🤔🤔🤔🤔nice topic and nice explanation

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

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.

Stable video👍

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)

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.

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

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

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.

after 10 minutes I have realized this is NOT a Payday video

No way I've seen engineer/STEM Zach before Zach star himself? And I never noticed?!

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.

No matter how many guards you add, the Payday gang will find a way

NP-hard problem, pff... Let's take this on! Now I have something to do on bored afternoons ;p

This will help me in tactical FPS games

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

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

This would make a great puzzle game

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.

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".

For the first one can't you put 2 guards for th whole thing in the top left part of the extra area and one the very most bottom right

The figure you showed rgb triangles actually needs just 2 guards for whole security

Why can’t you just check if two guards can see each other after computing n/3?

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.

Am I the only one who noticed how the main room layout used in this example has a stupidly easy 2 guard solution that doesn't require any maths to work out given the stipulations of corners only and 360 vision? Far Right Blue, and Top Left Green. That covers the entire gallery with two guards.

is it still NP if you remove the vertex requirement?

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.

That whole vertex thing was introduced way too late. All the examples given previously with the different shapes had the guard in the middle, and no vertices were ever mentioned. Then all of a sudden, it matters that guards are placed on vertices.

Man, I wish steel wool watched this before making security breach