Alright 2 things to add here 1) Be sure to checkout stemerch.com for the new recommended booklist as well as STEM related apparel, the floating globe, and more! Also you'll notice another tab for 'recommended textbooks' which are good for self study, this is a common question I get as well so will keep that updated. 2) Spoilers down below but this is where I want to acknowledge what I mention at the end of the video. I never explained WHY there is repetition after you reflect the midpoint back to the main room and the best explanation I got is quite a mouthful and it involves modular arithmetic. Let's say that the target is at point (x,y) (and we're just going to consider the horizontal reflections). After you reflect it to the right, the reflection lands at (2-x,y) and after another reflection it goes to (2+x,y) then (4-x,y) then (4+x,y) and so on. As you can see, reflected points all land at (2n +- x,y) (as in the x coordinates are just even numbers plus or minus x). Adding in some modular arithmetic you'll notice these are all congruent to x (mod 1), or -x (mod 1) . That's all that can happen as you reflect a point about the left or ride side of those 1x1 squares actually, the point either stays the same (mod 1), or it becomes negative (mod 1). Now again, all the x coordinates of the reflected targets can be written as 2n +- x, meaning the midpoints would be (2n+x+u)/2 = n + x/2 + u/2 and (2n-x+u)/2 = n - x/2 + u/2 (assuming the shooter has coordinates (u,v)). So we have two sets of midpoints and these midpoints go on forever with n. When those points are reflected back, as we've seen, they either stay the same mod 1, or become negative mod 1. So we have 4 different results, n+x/2+u/2, n-x/2+u/2, -n-x/2-u/2, and -n+x/2-u/2 (all mod 1). This seems like infinitely many points still because of n, but n can be dropped from all of these because it doesn't change the value mod 1. For example 1+x/2+u/2 = 2+x/2+u/2 mod 1, meaning they are the same point in the original 1x1 room. So we have 4 answers, x/2+u/2, -x/2+u/2, -x/2-u/2, and x/2-u/2 (all mod 1), these are all the x coordinates after the midpoints are reflected back, there can be nothing more. So that's 4 different x coordinates and the same thing can be done for the y coordinates, leaving us with 16 points in total. Then to finally to answer the other question of how can you be safe with less than 16 blockers, it can happen if the x (or y) coordinates of the shooter and target add to 1. Because then x/2+u/2 and -x/2 - u/2 are now congruent mod 1, and the other two are also congruent. So what was 4 different x coordinates becomes 2.
So, I'm guessing the next video is going to be about modular arithmetic then? Also, shouldn't it be mod 4, as they repeat after 4 reflections? I'm confused.
@@RussellSubedi he used the continuous version of mod not the normal number theory one. It's like how you can always reduce an angle to it's representation mod 2pi. 3pi = pi (mod 2pi).
@@thedoublehelix5661 What's the normal number theory one? I thought mod just gave you the least number greater than 0 left after repeated subtraction. Doesn't that definition rope everything in?
Man i hate when im in a perfectly square room made of mirrors and there is someone trying to shoot me with a laser which will never lose its energy as it reflects, thank you for the video.
Just guessing without thinking too hard about it, but I'm going to guess it's 16 because of this: You have 4 sides, so you have 2^4 ways of using those sides - you're using each side either an odd or even number of times. A higher number of odd/even times just ends up cancelling them down to simplify into one of the basic 16 situations. To simplify, let's just look at the left and right walls. Either you're hitting both an even number of times, the left odd and right even, the left even and right odd, or both an odd number of times. Whether the combination is left 2 and right 1 or left 10 and right 9, the same point ends up being it's center point. That's the 4 dots you put at first.
This use of reflections is also one method for calculating bank and kick shots in pool off of one or multiple rails (just factor in a little bit of physics about speed and friction).
I’ve played enough laser tag with mirrors to know that as long as you can see them in the mirror you can shoot them. So you would just need enough blocks to block off line of sight. Simplified.
Yes but if you're in a room where all the walls are mirrors, there's an infinite number of reflections. So there's no way to know how to place a finite number of blockers to break line of sight with every reflection.
That is an incredible puzzle. At first I was convinced the answer would have to be uncountably infinite, then after seeing the first step of the proof I thought it would be countably infinite, as we can represent each possibility on a NxN grid. And then the pattern starts repeating, very cool stuff!
@@proot. Thanks for pointing that out. I was going off of the idea that they were like how they were represented in the video graphics, but after watching the video again I realize that he mentions that a couple of times.
The death of the universe will happen in a finite, countable number of years, so since they're going over an infinite, uncountable amount of possible mirror-room scenarios, this video should still be playing quintillions of years after all life forms are long gone kinda suspicious that it's only 11 minutes then /j
Assuming you are an infinitely small point, you would only need one, since the laser can never naturally reflect into the corner, only get infinitely closer to it.
I am just realizing that this is the same guy who makes all the comedy sketches that I watch. I had no clue about this, and am doubting life as I know it. I never thought of him as a mathematician.
This is actually very similar to a coding interview I got from Google! The difference was that the laser did lose energy and you had to count all of the ways the laser hit the target, and the shooter was a target. I still landed on the reflection strategy, tho. First instinct: countably infinite, were gonna talk about Ulam's spiral and counting rational fractions
@@fakharyarkhan5848 Yeah, for those worried at home Google interview questions are nowhere NEAR this wild, and they make it a point to try and not rely on "aha" moments
Is there any way for you to link the problem? I'm curious to see what the correct answers are because I'm sure I must be missing something, but it seems to me that if it loses, say, 10% of its outset energy that means 10 reflections which I'm fairly sure means 4^10 possibilities.
Here is a (greatly altered) version of the problem. Let's assume that there are no blockers, but the shooter can place static mirrors before shooting the laser. If the laser's path will not be affected by hitting the target, is it possible that the shooter does not shoot themselves?
Oh damn, now I'm totally not scared to go with 16 blocks into the perfect square room, made of perfectly reflecting glass with someone else who has a laser gun.
Question! This only applies to a 2-dimensional plane, so I'm genuinely curious how it would apply if you had another axis, making it 3-dimensional; in this example, the ceiling and floor would also be made out of mirrors, and the shooter is able to aim for them to go over or under your blocker points. I hypothesize because math reasons that, instead of 16, you'd perhaps need 64 points to block it? The number 16 is a little too convenient not to be produced by some kind of exponential formula- that's my guess, at least. Curious if that's right or wrong?
i don’t know the answer but an alternative to 4^dimension or 16=4^2 could be 16=2^2^2 or d^d^d so 3d plane would be 3^27..? just my guess i don’t think it’s right because i don’t understand why it repeats after the 4th (i understand the model i just don’t get why it’s *4* in particular)
@@cnidocyte its cause when the laser has been reflected 4 times, it will have the same angle as when it started. and for each angle, there is only one point on each corresponding wall that will lead to the target. so you only have to block each angle once. I'm not 100% on this one, this is just a guess. But in a one dimensional space, you'd only need one blocker, since there is only one line the laser can follow. So I think the way to define the number of blockers is 16^(n-1) where n is the number of dimensions. 16^(1-1) = 1 blocker 16^(2-1) = 16 blockers 16^(3-1) = 256 blockers, the 16 we had earlier only repeated on the vertical dimension.
Man don't we all experience a time in life where we're in a perfect room of mirrors with an enemy that's frozen in place and shoots lasers as we get only a limited amount of blockers
So my theory is that if the floor and ceiling were also mirrors you would need 36 blocker points if you were the laser victim and a ton of windex if you were the laser-assassin
Thank you for this. I’ll remember it the next time I’m in a two-dimensional room full of mirrors with a man shooting me with non-energy-consuming lasers and I’m free to place single points to stop the laser.
Now I’m curious how many you would need if the space was 3d rather than 2d. I’m gonna guess 64, as the number needed for a single axis is four, and as there would be 3 axis, 4^3=64
@@cros108 so how it works is the 2 opposing faces will reflect the same image after its 2nd reflection So the square has 4 sides (2 opposing faces) so that us why the number is 16 (4^2) 😄 A cube has 6 sides (3 opposing faces) hence the 3 in 4^3 😄👍 Hope that clears things up
No, it would be 4^4, if in 2d you need 16 balls in one line and the third dimension line is the same lenth, you would need sixteen times sixteen, not 4^3, that means four times sixteen. Obviously suposing that this theory is correct
the main theme i see with every video he makes is to look at things in a different way. very good lesson to learn, helps you solve seemingly impossible questions
Fun Fact: If you put the mirror walls in front of the shooter all ways and connect them together the laser will be trapped and can’t get out so you only need 4 - 8
Before watching, I have 3 ideas for an amount to protect myself. 1: if blockers can be any shape of any size: 1. Just a wall through the whole room. 2: if blockers can be rectangles of limited size, 3. Triangular around me, trapping me. 2: if they have to be circles of equal size to my footprint circle, 6. Hexagonally around me. Touching me as well as themselves, leaving no gap.
@@unquickify yes but it will be endless death room imagine entering a room where a laser is flying everywhere all of sudden you are killed instantly by the laser.
@@snowjix bending doesn't lose energy. You lose velocity (change of direction), but your speed stays the same. In fact maybe you gain speed depending on the relationship between the mass, gravity and angle of approach and exit of significant gravitational field.
@@tubax926 sure, i get that. I worded myself wrong. But my main point is the one i was hoping you would focus on. Does the bending make it impossible to block?
It’d only take like 8 to make a circle of blockers around your circle. Actual solution has a much cooler answer with reflections and grids, but least points is just a circle around you.
@HatCatWC If the points are infinitely small, then you'd need an infinite amount of points for any solution. I think think the points were exactly the size in the video.
Bro, ever since a kid I imagined being able to shoot lasers out of my eyes, and I would think about where they would bounce if I were to shoot them at whatever I happened to be looking at. I would always imagine them traveling at light speed, giving me no time to react if I mispredicted their path, and I would wonder if it would be possible to survive if they bounced infinitely off any surface 😂 and idk man this vid reminded me of that
Thank You, I will always be safe when I'm in a square room with mirror walls which also has a shooter with a laser and I have the ability to place blockers and nothing within the room has any volume. Great tutorial!
that reflection demonstration actually also does a very good job at explaining bounce shots in Pool as well, instead of a lazer and mirrors you have the pool cue and the edges of the table. Any blockers would be other balls
Watching the set up, it immediately reminded me of a scene in the anime Zatch Bell, where it was this same exact scenario of being in a room of reflective surfaces, and the protagonist figured out that they can predict to stand in specific spots to avoid a laser shot by his opponent. Looked it up, it’s the fight in episode 16 of season 1.
I know its not the point of the video, but I have trouble really getting bought into the problem since the easier solution is just to encircle either yourself or the shooter with pillars and it would only take 8 based off the diagram, less if you just stand next too a wall, which kind of defeats the question since you're just overcomplicating a problem to demonstrate an unnessacary solution to the hypothetical
if you do math and go for minimal, so yeah 16 is good, but you can also build a circle of 1 meter radius around the shooter or even smaller full of blockers. and that is it
another way to see the importance of this problem: Whats the minimum ammount of bulletproof structures to make a room sniper proof (considering stationary shooters and targets) and How to become invisible in a mirror wall room
i dont know why this got me so hard, but at the start when the little i pops up saying "the globe your staring at" is so god dam funny cus i was indeed staring at the globe
@@zachstar It doesn't matter that the blocker is infinitely small. So are you and the shooter, and nowhere in the statement of the problem did it state that one cannot place a blocker on the same point as you or the shooter, thus placing a blocker on either of those points blocks every beam from every angle and thus is a valid solution.
Btw if the room is a cube u just need to cube the amount for one axis which is 4 in this case and when u cube 4 its 64 so in total u need 64 blockers in this case to block every possible combination of lasers in the 3rd dimension
Yo, quick question. Would some of the "blockers" also catch the path of some of the reflections? I almost feel that some points are doing more work than credit is given. Im not a genius by any means but to me it makes sense that you would also incorporate the blockers in the reflections
@@freshrockpapa-e7799 I did not, as I watched I got further confused/frustrated as more and more blockers appeared but wouldn't be reflected. I'm aware of the rule of 4 but I see what looks to me like some blockers interrupting a laser and instead they still just use the middle point. I can understand not wanting to over populate the screen with redundant data but when there is a translation of straight line to "bouncy" line showing the blockers in one but not the other just seems silly to me unless there is more information I don't know as I didn't finish the video
@@catbear22 The blockers you mentioned don’t actually interrupt the laser and that’s why they use the middle point. The blockers just look like they’re in the path of the laser because they’re bigger for us to visualize them. The points are actually all infinitely small so the laser wasn’t getting blocked.
You ofc don't need an infinate amount. Youd need just enough to either surround yourself, or the shooter. With your visuals youd need maybe 6 to completely surround one or the othet.
Did anyone else do something similar to this and mentally try to follow hypothetical laser beams in rooms when waiting/bored? I did this all the time as a kid and did it less often as phones came into that role of killing time rather than imagining a laser beam darting around the room lol
Oh my god finally I've found somebody able to put it into words. I still get it, especially while I'm not doing anything visual (i.e., I'm lying down to sleep, I'm unstacking dishes, or I'm having a conversation with somebody about something abstract), or just bored. I always called it my holding pattern, just a thing that the brain does when there isn't enough other stuff going on to use up all the free computing space. Let me ask you this: do you tend to get it to places that you are _not currently in,_ but that you _have been in recently?_ When I was a kid and I'd spend a week up at my grandparents' house, the pattern would be of back home at my parents' house. When I came back, suddenly the lasers would be bouncing around their place. When I got off school for summer, the corridors and various classrooms would become my places to mentally invisage. It wasn't intentional, but it just seemed like I was always experiencing the laser bouncing thing in the last significant location. The current one, I'd never do. I could be sitting in a dentist's office and imagine it in the reception area just downstairs, but not in the room I was in itself. I'm genuinely excited to know somebody else does this. I never even thought about it as a thing I do. But I never even thought to check if it was a thing other people did. I wonder if it has a name or if it's been described academically?
@@MarkusAldawn from what I've seen from talking to other people is that most people just come up with their own thing. The other guy I knew who did this would select something in a room (such as a person, box, etc.) And they would imagine how many could be stacked to fill that rooms entire volume. Mine was specifically more like trying to imagine if I had a laser beam coming out of or into my eye and the whole room was perfectly reflective, what were all the possible paths the beam could travel and what sort of obstructions\paths would have to occur in order to have a beam connect two points in a room with an infinite number of bounces. And yeah, I think it is a pattern of tasks the brain does when you aren't using enough of your active attention to feel satisfied in some way? Hard to really put it in simple terms but I'm glad you and I are able to bond a little bit and share experiences about weird mind behaviors. One thing that really clarifies and makes me more aware of those kinds of things is getting better at meditating, that really seems to bring about so many answers but it takes forever if you are anything like me and struggled to meditate for at least a decade.
It would only take as many lasers as it needs to surround the original circle... if the circles are all equally sized than it would only take 6 blockers.
Can't you just put like 8 around you in a 3x3 square with you in middle or the shooter in the middle blocking any possible path if you just didn't have a hole? or even better decide one or two which no matter where the shooter shoots it just can't reach you and have less than 8?
Good video, but you didn't actually solve the problem: "What is the minimum number of blockers required?" All you have done is proved it is 16 or less. You state that sometimes it is less, but give no reason or justification for when or why it would or would not be less than 16. Do you have any follow up links about this detail?
@@nogussy Sorry, I should I have been more clear. Even in your pinned comment, all you do is show: in certain cases the minimum is 8 or less (and then 4 or less if both the x and y axis have this special property). In the original case and the special cases of being congruent mod 1 you don’t actually prove that it is the lowest possible number. You merely give a proof that it is a sufficient number, but not a necessary number. It is hypothetically possible that it could be done with even fewer blockers. Do you have a proof that these numbers are necessary? For example maybe in a non-congruent mod 1 case it could be done with exactly 14 blockers. It probably can’t be, but a proof is still needed to show this to fully answer the question being asked.
4 would seem to be the minimum when you converge on the x and y special case, as there are no further variables to reduce. I don't know how how or if that would convert to a proof. Its inductive rather than deductive.
Alright 2 things to add here
1) Be sure to checkout stemerch.com for the new recommended booklist as well as STEM related apparel, the floating globe, and more! Also you'll notice another tab for 'recommended textbooks' which are good for self study, this is a common question I get as well so will keep that updated.
2) Spoilers down below but this is where I want to acknowledge what I mention at the end of the video. I never explained WHY there is repetition after you reflect the midpoint back to the main room and the best explanation I got is quite a mouthful and it involves modular arithmetic. Let's say that the target is at point (x,y) (and we're just going to consider the horizontal reflections). After you reflect it to the right, the reflection lands at (2-x,y) and after another reflection it goes to (2+x,y) then (4-x,y) then (4+x,y) and so on. As you can see, reflected points all land at (2n +- x,y) (as in the x coordinates are just even numbers plus or minus x).
Adding in some modular arithmetic you'll notice these are all congruent to x (mod 1), or -x (mod 1) . That's all that can happen as you reflect a point about the left or ride side of those 1x1 squares actually, the point either stays the same (mod 1), or it becomes negative (mod 1).
Now again, all the x coordinates of the reflected targets can be written as 2n +- x, meaning the midpoints would be (2n+x+u)/2 = n + x/2 + u/2 and (2n-x+u)/2 = n - x/2 + u/2 (assuming the shooter has coordinates (u,v)). So we have two sets of midpoints and these midpoints go on forever with n. When those points are reflected back, as we've seen, they either stay the same mod 1, or become negative mod 1. So we have 4 different results, n+x/2+u/2, n-x/2+u/2, -n-x/2-u/2, and -n+x/2-u/2 (all mod 1). This seems like infinitely many points still because of n, but n can be dropped from all of these because it doesn't change the value mod 1. For example 1+x/2+u/2 = 2+x/2+u/2 mod 1, meaning they are the same point in the original 1x1 room. So we have 4 answers, x/2+u/2, -x/2+u/2, -x/2-u/2, and x/2-u/2 (all mod 1), these are all the x coordinates after the midpoints are reflected back, there can be nothing more. So that's 4 different x coordinates and the same thing can be done for the y coordinates, leaving us with 16 points in total.
Then to finally to answer the other question of how can you be safe with less than 16 blockers, it can happen if the x (or y) coordinates of the shooter and target add to 1. Because then x/2+u/2 and -x/2 - u/2 are now congruent mod 1, and the other two are also congruent. So what was 4 different x coordinates becomes 2.
So, I'm guessing the next video is going to be about modular arithmetic then?
Also, shouldn't it be mod 4, as they repeat after 4 reflections? I'm confused.
@@RussellSubedi he used the continuous version of mod not the normal number theory one. It's like how you can always reduce an angle to it's representation mod 2pi. 3pi = pi (mod 2pi).
@@thedoublehelix5661 I'm not really familiar with it, so got confused. Still looking forward to a video on it though.
I'm gonna pretend I understood it
@@thedoublehelix5661 What's the normal number theory one? I thought mod just gave you the least number greater than 0 left after repeated subtraction. Doesn't that definition rope everything in?
*Me when I saw the title:* oh well you just wait until the light dissipates.
*the video immediately:* It never loses its energy.
No it says curiositystream immediately
You may have outsmarted me, but I have outsmarted your outsmarting
@@DoodleNoodle129 I'm gonna pretend that I understand how that makes sense with the context.
@@dannyboi7286 he thought he outsmarted the problem, but his outsmarting was outsmarted by the lack of energy dissipation
@@DoodleNoodle129 joseph
Man i hate when im in a perfectly square room made of mirrors and there is someone trying to shoot me with a laser which will never lose its energy as it reflects, thank you for the video.
Damn, it's so annoying, im so grateful that this video exist.
Also I hate being a point in a 2 dimmensional space
SHOOTER MOVES TO A DIFFERENT LOCATION
Oh shi
@@RoeiCohen go to a corner and make a barrier xddd
Hate it when that happens man
Mathematicians: Uses this information for intellectual purposes
Me: uses this information to beat my friends in air hockey
Except this time your the blocker and it’s not just a point but a thick thing which I don’t know the name of, but I see your point.
@@tomwanders6022 puck, the word you're looking for is puck
@@deathsheir2035 ty good sir. So its the same name in german.
@@tomwanders6022 in Spanish we call it "That thing used to play Air Hockey"
@@ARandomMinecraftVillager huh I’be usually heard it being called “el disco”
If DVD screens taught ne something: yes if you stay on corners.
Yes, just put a blocker there to be safe.
@@mouthlesshater that’s a good wisdom
Also all hail Winner
winner how do you feel joining TPOT
Tcher jhigl seghrt fhcwerd ghthicmus!!!!
The solution was surprising and surprisingly satisfying.
I didn't watch it. It depends on the shape of the room.
Very much so!
Satisfyingly surprising and surprisingly satisfying.
It's actually surprisingly simple.
@@manuell3505 well, if you’d watch in then you’d know he was using a square.
Kid, “math is so stupid why do we have to learn it, I’m never going to use it?”
“Yes you will, because lasers.”
the sun is a deadly laser
Not anymore there's a blanket~
“No”
Not yet
Why didn’t he just put the blockers in a circle around the target? Am I missing something?
"Switching to your pistol is faster than Reloading."
True but carrying ammo uses less space
@@largeavocado never enter combat without a sidearm
Am I the only confused one?
@@dannyboi7286 No
@@dannyboi7286 profile picture fits.
Just guessing without thinking too hard about it, but I'm going to guess it's 16 because of this: You have 4 sides, so you have 2^4 ways of using those sides - you're using each side either an odd or even number of times. A higher number of odd/even times just ends up cancelling them down to simplify into one of the basic 16 situations.
To simplify, let's just look at the left and right walls. Either you're hitting both an even number of times, the left odd and right even, the left even and right odd, or both an odd number of times. Whether the combination is left 2 and right 1 or left 10 and right 9, the same point ends up being it's center point. That's the 4 dots you put at first.
Ah, yes, now I just need to prepare 16 blockers to be carried with me everywhere I go in case I land in this super relatable situation again.
wait........ again?!
*prepares 1 blocker*
I’m prepared.
*hides on corner*
"Can't we just do this?"
"No we can't."
"Ok, but we could if we turned this problem into an entirely different one."
nice 69 likes
Yeah for an example you could just suround yourself with the circles
@@jonathanholtlajer2949 or surround the Lazer
@your mother it would?
@your mother then they wouldnt block anything?
This use of reflections is also one method for calculating bank and kick shots in pool off of one or multiple rails (just factor in a little bit of physics about speed and friction).
I’ve played enough laser tag with mirrors to know that as long as you can see them in the mirror you can shoot them. So you would just need enough blocks to block off line of sight. Simplified.
That’s a cool analogy actually.
Ever bounced a laser off three mirrors and hit someone?
I thought about the same thing lol
Yes but if you're in a room where all the walls are mirrors, there's an infinite number of reflections. So there's no way to know how to place a finite number of blockers to break line of sight with every reflection.
@@ZachAttack6089 just do a 360 and so long as you can’t see them in the reflection at all no matter where you turn then you aight.
The amount of peope saying "just put 4/5/6 blockers around you" is genuinely worrying
Me: just surrounds myself with blockers that slightly overlap
Blocker is a point
so you need infinite blockers
@@koshakvesely8186 but there isnt anything said about a point having a specific size
@@RiskierGoose340 A point is an idealized, primitive notion. It does not have any physical size
@@koshakvesely8186 ok then, u win this time
the first thing i thought was hiding in the corner and adding 2 blockers to block the 2 sides
That is an incredible puzzle. At first I was convinced the answer would have to be uncountably infinite, then after seeing the first step of the proof I thought it would be countably infinite, as we can represent each possibility on a NxN grid. And then the pattern starts repeating, very cool stuff!
Me: Eight to surround the target
I had exactly the same line of thought! this problem is fascinating
@@savant_fou9483 You actually only need 6 if you position them right. You arrange the circles in a hexagon instead of a square.
@@megaparsec4 Those aren't circles, but points with zero width. He could just shoot the laser between any two of the six.
@@proot. Thanks for pointing that out. I was going off of the idea that they were like how they were represented in the video graphics, but after watching the video again I realize that he mentions that a couple of times.
"We'll work up to infinity" I don't know man, I haven't got all day, or the lifetime of the universe for that matter
The death of the universe will happen in a finite, countable number of years, so since they're going over an infinite, uncountable amount of possible mirror-room scenarios, this video should still be playing quintillions of years after all life forms are long gone
kinda suspicious that it's only 11 minutes then /j
@@zypper7213 whooosh
@un ko double whooosh
bruh the universe will like die in 2021 (btw after 2020 dec 31 59:59 it will be 2020 dec 31 60:00).
That card that said “the globe you are staring at” actually caught me off guard
Late to the party, but does this apply to rooms with differing number of sides?
Would an octoganal room require 64 blockers? Triangular 9?
Can’t I just stand in a corner and put like 3 blockers together to completely block the laser
Assuming you are an infinitely small point, you would only need one, since the laser can never naturally reflect into the corner, only get infinitely closer to it.
Or just surround yourself with 8 without even moving
@@randomperson1844 you mean surround yourself with infinite points?
@@Pihsrosnec imagine you are a square
You surround yourself in 8 squares
You are protected from all sides
The blocker, you are infinetly small. And the laser is super precise
I am just realizing that this is the same guy who makes all the comedy sketches that I watch. I had no clue about this, and am doubting life as I know it. I never thought of him as a mathematician.
This is actually very similar to a coding interview I got from Google! The difference was that the laser did lose energy and you had to count all of the ways the laser hit the target, and the shooter was a target. I still landed on the reflection strategy, tho.
First instinct: countably infinite, were gonna talk about Ulam's spiral and counting rational fractions
Did they expect you to solve that one on the spot?
@@fakharyarkhan5848 No, it was an online coding challenge thing; I think I had a day to solve it?
@@AMTunLimited oh ok that's more fair. That sounds like a pretty cool variation of the problem then.
@@fakharyarkhan5848 Yeah, for those worried at home Google interview questions are nowhere NEAR this wild, and they make it a point to try and not rely on "aha" moments
Is there any way for you to link the problem? I'm curious to see what the correct answers are because I'm sure I must be missing something, but it seems to me that if it loses, say, 10% of its outset energy that means 10 reflections which I'm fairly sure means 4^10 possibilities.
Thanks! Now, I know what to do if I'm stuck in a room of mirrors and someone's trying to shoot me with a laser!
Starting from today I'm gonna always take 16 bricks with me! You never know...
"Why do you have 16 bricks with you?"
"Well let's say I'm a single point..."
Can we extend this solution for 3D? I mean, I am ready to be still considered a single point, but in a room with mirror walls and ceilings
@@rampukaradhikari4716 It's easily extended to n-dimensional by the same argument, and it's 4^n (except for n = 1 🙂)
@@rampukaradhikari4716 That would mean we'd have 64 points?
@@rampukaradhikari4716 4for the x coor x 4 for the y coor x 4 for the z coor
I’m so used to seeing sketch comedy from you, but this is really cool too I’ll definitely be keeping an eye out for more of this stuff then!
Here is a (greatly altered) version of the problem. Let's assume that there are no blockers, but the shooter can place static mirrors before shooting the laser. If the laser's path will not be affected by hitting the target, is it possible that the shooter does not shoot themselves?
Oh damn, now I'm totally not scared to go with 16 blocks into the perfect square room, made of perfectly reflecting glass with someone else who has a laser gun.
This is cool stuff.
If you had controlled force field.
You could save a bunch of energy with only 16 points instead of a sphere.
Question! This only applies to a 2-dimensional plane, so I'm genuinely curious how it would apply if you had another axis, making it 3-dimensional; in this example, the ceiling and floor would also be made out of mirrors, and the shooter is able to aim for them to go over or under your blocker points.
I hypothesize because math reasons that, instead of 16, you'd perhaps need 64 points to block it? The number 16 is a little too convenient not to be produced by some kind of exponential formula- that's my guess, at least. Curious if that's right or wrong?
i don’t know the answer but an alternative to 4^dimension or 16=4^2 could be 16=2^2^2 or d^d^d so 3d plane would be 3^27..? just my guess i don’t think it’s right because i don’t understand why it repeats after the 4th (i understand the model i just don’t get why it’s *4* in particular)
@@cnidocyte It repeats after the 4th in all three dimensions giving 64 blockers!
@@cnidocyte its cause when the laser has been reflected 4 times, it will have the same angle as when it started.
and for each angle, there is only one point on each corresponding wall that will lead to the target. so you only have to block each angle once.
I'm not 100% on this one, this is just a guess. But in a one dimensional space, you'd only need one blocker, since there is only one line the laser can follow. So I think the way to define the number of blockers is 16^(n-1) where n is the number of dimensions.
16^(1-1) = 1 blocker
16^(2-1) = 16 blockers
16^(3-1) = 256 blockers, the 16 we had earlier only repeated on the vertical dimension.
The blocker points could just expand to touch both the ceiling and floor, thereby covering those spots
you can solve it by making the blockers pillars that are touching the ceiling and floor. ye?
Here’s the solution: Tell the shooter that they might have green eyes.
is this a reference to that one logic problem with all the green-eyed people?
Run.
bahaha ik that riddle
TED ED
Since he does have green eyes, everyone is legally allowed to leave the room.
If there even is a way out.
That was a major call-out to me staring at the globe lol
Man don't we all experience a time in life where we're in a perfect room of mirrors with an enemy that's frozen in place and shoots lasers as we get only a limited amount of blockers
So my theory is that if the floor and ceiling were also mirrors you would need 36 blocker points if you were the laser victim and a ton of windex if you were the laser-assassin
What I would do is just go into the corner and set up like 4 blockers really close to me until enough is needed to block the whole way
that is infinite blockers
Thank you for this. I’ll remember it the next time I’m in a two-dimensional room full of mirrors with a man shooting me with non-energy-consuming lasers and I’m free to place single points to stop the laser.
Now I’m curious how many you would need if the space was 3d rather than 2d. I’m gonna guess 64, as the number needed for a single axis is four, and as there would be 3 axis, 4^3=64
Sounds right 😄
Something tells me it would be 6^3, because 6 faces it can reflect off instead of 4 edges
@@cros108 so how it works is the 2 opposing faces will reflect the same image after its 2nd reflection
So the square has 4 sides (2 opposing faces) so that us why the number is 16 (4^2) 😄
A cube has 6 sides (3 opposing faces) hence the 3 in 4^3 😄👍
Hope that clears things up
@@Zoltria Ohhh I'm an idiot. Thanks for that explanation.
No, it would be 4^4, if in 2d you need 16 balls in one line and the third dimension line is the same lenth, you would need sixteen times sixteen, not 4^3, that means four times sixteen. Obviously suposing that this theory is correct
Gotta say, knowing Zach only from his skits and not checking the channel name when clicking the video, seeing him pop up was an absolute shock lmao
I love how “off the wall” these videos are
the main theme i see with every video he makes is to look at things in a different way. very good lesson to learn, helps you solve seemingly impossible questions
0:06 ah yes, very relatable
I love CuriosityStream! I've been subscribed for years now.
How to avoid the laser:
Step 1. Turn off the Laser.
Step 2. avoid
Step 3. evade
Wait, but you can't move.
@@Penguinza oh ok
alternately: enclose yourself in mirrors facing outside
Instructions unclear, I am now going into my first professional laser tag tournament for 50,000 dollars as winnings.
Fun Fact: If you put the mirror walls in front of the shooter all ways and connect them together the laser will be trapped and can’t get out so you only need 4 - 8
infinite*
Before watching, I have 3 ideas for an amount to protect myself.
1: if blockers can be any shape of any size: 1. Just a wall through the whole room.
2: if blockers can be rectangles of limited size, 3. Triangular around me, trapping me.
2: if they have to be circles of equal size to my footprint circle, 6. Hexagonally around me. Touching me as well as themselves, leaving no gap.
“How many blockers do u need”
Me: 8
Person: how?
Me: shield
The blockers are points
@@Owen_loves_Butters wat is that supposed to mean
@@randomperson1844 It means they have no width, a shield would take an uncountably infinite amount of blockers.
@@Owen_loves_Butters so how will something that small block a laser?
@@randomperson1844 The laser is the same width
As a kid I often wondered what would happen in a mirror ball with a laser beam. As a teen I still wonder what would happen
Laser would run out of energy
@@vyor8837 but what if it didn’t un out of energy?
@@sco0t26Newton would come back to life
@@Meso.Botamia cool let’s ask him
@@sco0t26 the ball would get steadily hotter until melting occurred.
Counter solutions: put the blocker on top of the attackers head so that their dead and can’t shoot you anymore
Me, an intellectual: *grabs a hammer*
Not allowed!
The shards mean more mirrors.
@@unquickify yes but it will be endless death room imagine entering a room where a laser is flying everywhere all of sudden you are killed instantly by the laser.
that’s some insane bad luck
the shooter is just gonna drop the pointer and it will hit you
Me, an omnipotent lad: Gets some cloth.
It’s all fun and games until light starts bending
It should bend about those blockers, just extremely slightly. To a billionth decimal place maybe
@@tubax926 but since we are talking about light that never loses energy, that might make this task in the video impossible. Maybe, i dunno.
@@snowjix bending doesn't lose energy. You lose velocity (change of direction), but your speed stays the same. In fact maybe you gain speed depending on the relationship between the mass, gravity and angle of approach and exit of significant gravitational field.
@@tubax926 sure, i get that. I worded myself wrong. But my main point is the one i was hoping you would focus on. Does the bending make it impossible to block?
@@snowjix depends, lots of factors. The distance between the blocker and the walls, the mass of the blockers, the size of the blockers.
It’d only take like 8 to make a circle of blockers around your circle. Actual solution has a much cooler answer with reflections and grids, but least points is just a circle around you.
@HatCatWC If the points are infinitely small, then you'd need an infinite amount of points for any solution. I think think the points were exactly the size in the video.
I love how he immediately knew we were staring at the globe. Caught me off guard and gave me a laugh ngl
This would make a fun flash game
If you just put 6 blockers around you or the shooter, no matter the size of objects you will always be safe
He said the blockers are single points, he only used circles because you cant see single points
he asked for the minimum amount needed
@@Dinoboy676 yes, but it just becomes 8 +++
+x+
+++
@@jimmypatton4982 they are points with no size, it requires an infinite amount of blockers to form an enclosure like you describe
Bro, ever since a kid I imagined being able to shoot lasers out of my eyes, and I would think about where they would bounce if I were to shoot them at whatever I happened to be looking at. I would always imagine them traveling at light speed, giving me no time to react if I mispredicted their path, and I would wonder if it would be possible to survive if they bounced infinitely off any surface 😂 and idk man this vid reminded me of that
Glad I'm not the only one, lol
I’m in algebra 2 rn and we have been talking about i and this reminds me of it because after just 4 it repeats itself
Thank You, I will always be safe when I'm in a square room with mirror walls which also has a shooter with a laser and I have the ability to place blockers and nothing within the room has any volume. Great tutorial!
Theoretically, you could simply surround the shooter with the blockers and use only 8-10 depending on size of the blocker/shooter.
0:47 the reflection on the bottom doesn't have equal angles.
It’s just an illustration
cursed haha
That one laser plays by its own rules.
@@zachstar it's a heat seeking laser
-This is it! I have finally beat you!
-Hang on bro lemme put me blockers riiiiight here. Ok go on
-???
Zach makes comedy, and solves puzzles, what can’t this man do
that reflection demonstration actually also does a very good job at explaining bounce shots in Pool as well, instead of a lazer and mirrors you have the pool cue and the edges of the table. Any blockers would be other balls
Watching the set up, it immediately reminded me of a scene in the anime Zatch Bell, where it was this same exact scenario of being in a room of reflective surfaces, and the protagonist figured out that they can predict to stand in specific spots to avoid a laser shot by his opponent.
Looked it up, it’s the fight in episode 16 of season 1.
E
Neo from the Matrix: “are you challenging me?”
I like how i was staring at the globe and then it said. “Globe you’re staring at” at the top right.
I think a animation of the shooter and target ger moving, with the blockers responding would be cool
The whole time I’m just saying “just put a circle around the target or the shooter!”
The circle = infinite
@@pittyconor2489
Blocker = O
You = o
OOO
OoO
OOO
yes I stole this reply
@@KayBolt the blockers are infinetly small
@@pittyconor2489 Oh right. BUT!!! What if they are not
@@KayBolt bruh
So could we generalize that in d dimensions the number of blockers needed is 4^d, such as 4^3=64 in 3 dimensions?
@Mátyás Barczai Exceptions, I suppose.
I know its not the point of the video, but I have trouble really getting bought into the problem since the easier solution is just to encircle either yourself or the shooter with pillars and it would only take 8 based off the diagram, less if you just stand next too a wall, which kind of defeats the question since you're just overcomplicating a problem to demonstrate an unnessacary solution to the hypothetical
no, he isn't overcomplicating, you just didn't understand the rules. She shooter can just shoot between your 8 blockers
Awesome video!
if you do math and go for minimal, so yeah 16 is good, but you can also build a circle of 1 meter radius around the shooter or even smaller full of blockers. and that is it
another way to see the importance of this problem:
Whats the minimum ammount of bulletproof structures to make a room sniper proof (considering stationary shooters and targets) and
How to become invisible in a mirror wall room
or just make the walls with a material that doesn't bounce bullets 5 times lol
@@TanmaiKhanna Yeah this is big brain time.
I found him through the skits and I recognized the voice right away omg
Now I know what to do when I get stuck in that room again, thanks dude
See you all when this comes on our recommended again in 6 years
My man went from explaining cool scientific puzzel/ questions to makeing hollywood grade skits
I wonder how this holds up when you consider non-rectangular rooms; how does that impact it?
I think there was a PBS infinite series video on this.
I'm still mad about PBS stopping this channel.
@@TheZenytram aren't we all :(
R.I.P. Infinite Series, you had a great run
Replace the square with a software, the laser with a bug, the shooter is the end user, the blocker are the checks that you do
i dont know why this got me so hard, but at the start when the little i pops up saying "the globe your staring at" is so god dam funny cus i was indeed staring at the globe
May I suggest an alternative solution: one point. Placed at the location of the "shooter", immediately stopping any laser
It wouldn't work, unless the shooter is aiming at itself.
me trying to sleep:
my brain: Could you avoid being hit by a laser if you were in a room of mirrors?
1:20 that’s not what “countably infinite” means 😑
Wouldn’t 2 blockers be enough to put yourself in a corner ?
This video has significantly enhanced my billiards playing skills
Me:staring at globe
Video: globe your staring at link
Me: where are the cameras?
Just place a blocker right on top of that light source so the laser gets instantly blocked no matter its direction
EZ
The blocker is an infinitely small point, you'll have blocked one angle, but there are uncountably many more to choose from.
@@zachstar It doesn't matter that the blocker is infinitely small. So are you and the shooter, and nowhere in the statement of the problem did it state that one cannot place a blocker on the same point as you or the shooter, thus placing a blocker on either of those points blocks every beam from every angle and thus is a valid solution.
Btw if the room is a cube u just need to cube the amount for one axis which is 4 in this case and when u cube 4 its 64 so in total u need 64 blockers in this case to block every possible combination of lasers in the 3rd dimension
Yo, quick question. Would some of the "blockers" also catch the path of some of the reflections? I almost feel that some points are doing more work than credit is given. Im not a genius by any means but to me it makes sense that you would also incorporate the blockers in the reflections
Of course they do? Did you not watch the full video?
@@freshrockpapa-e7799 I did not, as I watched I got further confused/frustrated as more and more blockers appeared but wouldn't be reflected. I'm aware of the rule of 4 but I see what looks to me like some blockers interrupting a laser and instead they still just use the middle point. I can understand not wanting to over populate the screen with redundant data but when there is a translation of straight line to "bouncy" line showing the blockers in one but not the other just seems silly to me unless there is more information I don't know as I didn't finish the video
@@catbear22 The blockers you mentioned don’t actually interrupt the laser and that’s why they use the middle point. The blockers just look like they’re in the path of the laser because they’re bigger for us to visualize them. The points are actually all infinitely small so the laser wasn’t getting blocked.
You ofc don't need an infinate amount. Youd need just enough to either surround yourself, or the shooter. With your visuals youd need maybe 6 to completely surround one or the othet.
meanwhile me: * surounds myself with blockers *
Infinitely small points.
Did anyone else do something similar to this and mentally try to follow hypothetical laser beams in rooms when waiting/bored? I did this all the time as a kid and did it less often as phones came into that role of killing time rather than imagining a laser beam darting around the room lol
Oh my god finally I've found somebody able to put it into words.
I still get it, especially while I'm not doing anything visual (i.e., I'm lying down to sleep, I'm unstacking dishes, or I'm having a conversation with somebody about something abstract), or just bored. I always called it my holding pattern, just a thing that the brain does when there isn't enough other stuff going on to use up all the free computing space.
Let me ask you this: do you tend to get it to places that you are _not currently in,_ but that you _have been in recently?_ When I was a kid and I'd spend a week up at my grandparents' house, the pattern would be of back home at my parents' house. When I came back, suddenly the lasers would be bouncing around their place. When I got off school for summer, the corridors and various classrooms would become my places to mentally invisage. It wasn't intentional, but it just seemed like I was always experiencing the laser bouncing thing in the last significant location. The current one, I'd never do. I could be sitting in a dentist's office and imagine it in the reception area just downstairs, but not in the room I was in itself.
I'm genuinely excited to know somebody else does this. I never even thought about it as a thing I do. But I never even thought to check if it was a thing other people did. I wonder if it has a name or if it's been described academically?
@@MarkusAldawn from what I've seen from talking to other people is that most people just come up with their own thing. The other guy I knew who did this would select something in a room (such as a person, box, etc.) And they would imagine how many could be stacked to fill that rooms entire volume.
Mine was specifically more like trying to imagine if I had a laser beam coming out of or into my eye and the whole room was perfectly reflective, what were all the possible paths the beam could travel and what sort of obstructions\paths would have to occur in order to have a beam connect two points in a room with an infinite number of bounces.
And yeah, I think it is a pattern of tasks the brain does when you aren't using enough of your active attention to feel satisfied in some way? Hard to really put it in simple terms but I'm glad you and I are able to bond a little bit and share experiences about weird mind behaviors. One thing that really clarifies and makes me more aware of those kinds of things is getting better at meditating, that really seems to bring about so many answers but it takes forever if you are anything like me and struggled to meditate for at least a decade.
@@maverick9708i used to do this with things like popcorn. But most of the time i would close my eyes and go to the wonderland for a while
mental screensaver
It would only take as many lasers as it needs to surround the original circle... if the circles are all equally sized than it would only take 6 blockers.
I was just thinking of surrounding the shooter in like 6 blockers.
So it is possible but I have ask them to wait for me to solve a math problem then place the blockers before they start killing me.
Can't you just put like 8 around you in a 3x3 square with you in middle or the shooter in the middle blocking any possible path if you just didn't have a hole? or even better decide one or two which no matter where the shooter shoots it just can't reach you and have less than 8?
Good video, but you didn't actually solve the problem: "What is the minimum number of blockers required?" All you have done is proved it is 16 or less. You state that sometimes it is less, but give no reason or justification for when or why it would or would not be less than 16. Do you have any follow up links about this detail?
Pinned comment
@@nogussy Sorry, I should I have been more clear. Even in your pinned comment, all you do is show: in certain cases the minimum is 8 or less (and then 4 or less if both the x and y axis have this special property). In the original case and the special cases of being congruent mod 1 you don’t actually prove that it is the lowest possible number. You merely give a proof that it is a sufficient number, but not a necessary number. It is hypothetically possible that it could be done with even fewer blockers. Do you have a proof that these numbers are necessary?
For example maybe in a non-congruent mod 1 case it could be done with exactly 14 blockers. It probably can’t be, but a proof is still needed to show this to fully answer the question being asked.
@@jinxer42 Oh, that I wouldn't know
4 would seem to be the minimum when you converge on the x and y special case, as there are no further variables to reduce. I don't know how how or if that would convert to a proof. Its inductive rather than deductive.
I had the right feeling from the get-go, but forgot to square my initial idea of 4.
Only need 5 or 6 in a circle around me to block everything
nope