This is why hitting a vertex is assumed to end the line, otherwise it simply goes back the other way and hits the assassin. I guess there's other ways for the assasin to get himself though, like shooting at a precisely 90 degree angle to any wall (a 1-2 in football term's), but presumably we can assume the assassin is at least smart enough not to '1-2' himself of the first wall.
Wait... so... but... they just... how am I... why would they... do they...? I thought they... but... I wanted to... So there's just not a solution and there's never going to be one since they shut down Infinite Series? Not a single closure.
not only that, but you are on the end of "you are commenting before watching the whole video", she's mentions the solution is written up in the description. There wasn't a plan to make another video.
The play tester's answer: one guard immediately on top of the assassin, clipping into the start of the hitscan trajectory. If the guard can't be immediately on top of the assassin but is allowed to have width, you can use just eight guards by boxing the assassin with four and then wedging the gaps with the other four.
If you interpret a bounce on the wall of the billiard table as entering a flipped version of the original billiard table, it works the same as having 4 billiard tables on a Torus (because 2 reflections in either direction cancels out). This leads to 4 latices of points representing the target. Every possible shot on the target is a line from A to a copy of a target point at some square in the grid. And choosing a guard creates another set of 4 latices that happens to lie on as many shots as possible. If the shot lines are folded back upon themselves into the 1,1 square, the intersections are the best points to choose for a guard. The question of the Assassin puzzle can be reworded as a question of whether there are a finite number of intersections that cover all the paths from A to T. I don't know how to solve this for the general case, but for a simpler case where A=0,0 and T =(1,1) all the reflected versions of T are at the same location, and the problem is the same as if it were on a torus but due to the rule that paths that reach a corner (other than T) are discarded results in all the points of the lattice with an even number for x or y are discarded. In this odd grid, where all shots must travel an odd number of squares in both the x and y direction, all shots must also pass through the centre of the square which is halfway to the target. Which means that this special case only requires one guard in the centre of the room. So I didn't solve the problem but I did solve an easier version of it. I suspect that the true answer is 4 guards but don't really know how to proceed from here.
Abram Thiessen the answer is in a link in the video description, it is explained very well. You were really close to be correct, but the real answer is 16 bodyguards
Take the initial square representing the room and reflect it a few times to make a 2x2 group with matched sides such that bouncing off a wall can still be represented as moving through it to the next square. This 2x2 group then tiles to fill the plane as in the example shown in the video. Now turn the target into a lattice representing its positions in all of the tiles, and the possible shots are lines from the (single point) assassin to every point in the target lattice. The challenge then translates into finding a finite set of points such that every shot line goes through at least one of the points OR one of its equivalent lattice points in some other tile. (I had more started here, but then I realized that I neglected to treat the blocking points as lattices, which completely wrecked my solution. :( )
there is a simple solution to the problem: just use any wall of your choice (they all work equally well if A and T arent on the same perpendicular of the wall, then you choose a different wall ) now for your chosen wall just bounce it once to hit the target. now bounce twice using the chosen wall and the opposite wall. this can be continued forever getting a more and more acute angle. thus its not possible to have a finite number of blockers.
judgeomega interesting idea, but it is incorrect The answer explained is in a link in the video description. If you have any problems finding the link, I can put it in here for you. SPOILER FOR THE ANSWER: you need 16 bodyguards
Lukas Henke incorrect, The answer explained is in a link in the video description. If you have any problems finding the link, I can put it in here for you. SPOILER FOR THE ANSWER: you need 16 bodyguards
Finding that the lines should be parallel is an extra information that in not in the problem statement. You probably already did some problem in school asking if a triangle is special, but with a drawing clearly not to scale.
Oh ok, I thought he meant the boundaries of the room (sides of the square). I see what he means now. He's right. They are supposed to be parallel. The animation at 1:28 is pretty off--like the triangles in school you were talking about.
So easy. Every time it bounce it's like entering a new square but that was mirrored (ie a flipped copy using the plane it bounced on as the symetry axis). So we work with 4 squares. The original one plus x-flipped copy plus y-flipped plus x-and-y-flipped copy. Then we duplicate this big square (of 4 squares) to cover the plane. Each big square contains 4 targets (all virtual except the original one but all copies represent a path to hit the target using a certain number of bounces). Even if we consider the small subset of targets located at (x+2×k,y) if the y position of the assassin is different from the y position of original target, there's an infinite number of paths. If they are on the same y position, we consider the vertical set of square and the (x,y+2×k) virtual targets. If the assassin is not on the same position than the target that case will show an infinite number of paths. So we need an infinite number of bodyguards. It's a countable infinite (using the same arguments than when we say that fractions are)
I was thinking the same way, but I couldn't convince myself, since you are also copying bodyguards. So I thought of a different kind of proof. Just ignore two opposing sides of the squared and use the remaining two to bounce the bullet. You can make different trajectories by making different number of bounces. So 0 bounces is straight to the target, 1 bounce is against one side to the target, 2 bounces is from one side to the other side to the target, .... I would then claim that no triplet of trajectories will intersect in the same point. I don't have any proof for this (I'm not entirely sure if it's even true), but it feels more intuitive.
TheRMeerkerk That is exactly what happen in the 1st case. The only exception is when they are on the same height (y position). In that case there's only one trajectory this way. That's why you have to bounce vertically in that case (2nd case). At least one case works because they don't share the same point. BTW I made the hypothesis that you can't have 2 points in the same position. Otherwise you just put a bodyguard on the assassin and the whole problem is over.
I had some extra thinking and while the construction of the squares... are ok there's still a problem to proove that you can't block all those paths with a limited number of bodyguards. I found an example of a tricky situation: -assassin at (0.5 0.25) -target at (0.5 0.75) -bodyguards at (0 0.5) (0.5 0.5) (1 0.5) The 3 guards can block all line of fire that bounce only horizontally (or vertically). Of course you can still fire and using both vertical and horizontal bounces together but it shows that a finite number of bodyguards can block an infinite number of paths!
The answer explained is in a link in the video description. If you have any problems finding the link, I can put it in here for you. SPOILER FOR THE ANSWER: you need 16 bodyguards
The answer explained is in a link in the video description. If you have any problems finding the link, I can put it in here for you. SPOILER FOR THE ANSWER: you need 16 bodyguards
If you put 12 body gaurds around point T, of the same radius as target T: it may be an inelegant solution, but it get the job done. 6 for the kissing number of a circle, 6 more for all of the spaces between the body gaurds. Double redundancy. And that boring solution would work for any polygon (assuming euclidean space).
So I'm treating people like cylinders of equal height and girth; and assuming that they all have bullet proof umbrellas, and that there aren't any mines or otherwise floor-based threats (because we are solving a 2dimensional problem)
Ian Schimnoski in this problem you must consider everyone as points with 0 dimensions. That is why making a circle does not work, you need an infinite amount of points to make a circle The answer explained is in a link in the video description. If you have any problems finding the link, I can put it in here for you. SPOILER FOR THE ANSWER: you need 16 bodyguards
May I propose a variant of this puzzle? Say, there's a security guard G who won't stop bragging "When I die, I will do so saving someone's life!". Now imagine a killer K who is so annoyed by this that he decides to kill G. He lures him into a square room and intends him to shoot him dead, proving him wrong. The question is: Can you place a finite number of VIPs in the room such that none stands in the way of K shooting G, but however K tries to kill G, the latter will stand in the way of some VIP? (I don't know what the answer is, but given that the two problems are quite similar, I wouldn't be surprised if the answers were equally similar.)
it looks like you should start by mirroring the initial square twice (once on the x and then the two squares you get on the y axis) until you get 4 squares with 4 "targets". Then all reflections are translated into straight lines inside these 4 squares. You can then tile the whole thing as with the torus to get an infinite plane. Then the question translates to the question of: do you see an finite number of points from A in the resulting infinite plane? I remember there was an youtube video about it and i think that the answer is no, the number of points you see is the number of rational numbers, so infinite, so then you would need an infinite number of guards. A smaller infinite than the number of real numbers but still infinite.
I like your construction, and that's definitely the correct answer to the question "do you see an finite number of points from A in the resulting infinite plane?". However, that question is not quite identical to the original, since each guard can block multiple (indeed, infinite) lines of sight, as each is also copied into a lattice. Maybe it would be fruitful to look into what sets of angles could be blocked by a given guard?
But your construction doesn't take into account that some paths intersect each other. So if you place a guard at those intersecting points you can block multiple paths using only a few guards. Though i'm not sure if it can be done using only finitely many guards.
Jay-r Villanueva The shot can bounce any real number of times. This means the target can get hit from every possible angle, and you'd need infinitly many guards to cover it. Sure, every guard blocks infinitly many shots, but not every shot is blocked. This is hard to explain in a yt-comment (especially since english ain't my native tounge), so I'd advise you to take a sheet of paper and draw the square from 7:52. Then go infinite, draw the same square next to it, and next to that again... Just remember to align the red side with the red side, green with green etc Finally, draw a shot from your original assassin to every target. Since there are infinitly many targets on the x-axis alone, you'd need infinitly many guards to cover them all.
Nope, you first need to show that your construction directly corresponds to the problem. Your approach doesn't really capture the property of the square namely paths can intersect one another. Thus they are not really mathematically equivalent.
I have a solution. Solution A: T steals the gun Solution B: T dodges the bullet Solution C: A shoots himself Solution D: T drags A into the bullet Solution E: T somehow escapes Solution F: The cops move to keep T safe (though I wouldn't recommend this) Solution G: The cops arrest A
technically no there is no positioning of the guards such that they can block any shot the assassin takes. you can create a pseudo-torus square tessellation of the grid by reflecting the room across any wall. this pseudo-torus will have the property that every other reflection will be identical to the original. If we ignore the reflections and just focus on the copies, what we have is just a normal torus where the relevant points are limited to one quarter of the square. If we then tessellate the grid with this torus, we see that for any two copies the will be one exactly half way between them. For example, between the copy immeadiately above and the copy immeadiately to the left there is the copy both above and to the left. and there is one half way between that one and any other. this yields an infinite set of possible paths for the bullet to take an eventually hit the target. if you were to fill the room with a Gogol guard points, the target would likely die of starvation or natural causes before being hit by the bullet but he would eventually be hit by it.
Am I way off base but isn't it obviously "no"? If A shoots vertically the shot can bounce off the ceiling 0 times (direct shot), 1 time, 2 times (with one bounce off the bottom), 3 times (two bounces off the bottom) etc. Since there are an infinite number of options, - with an infinite number of them being prime so there wouldn't be overlap with previous paths - wouldn't that be an infinite number of blockers to block the shots?
So the answer is no. If you flip it on the x axis, then flip the pair of squares on the y-axis, you end up with a torus topology. When this set of 4 squares is tiled and embedded on the plane it's obvious to see that the assassin can shoot a straight line to the target in any of an infinite number of squares, i.e. there are an infinite number of ways to shoot the target.
The finiteness would be determined by the radius represented by each bystander. Additionally the minimum number would be dependent on the target's distance from the two closest walls. And if the assassin is closer than the width of one bystander, then the answer is no.
hacked2123 The answer explained is in a link in the video description. If you have any problems finding the link, I can put it in here for you. SPOILER FOR THE ANSWER: you need 16 bodyguards
Make a circle of bodyguards, and you're done. Is it an elegant solution using math? No. Is it a solution that works on the problem as it was presented to us? Yes.
EVEN MADSEN remember that the assassin, bodyguards and target are points with 0 dimensions. To make a circle you need an infinite amount of points The answer explained is in a link in the video description. If you have any problems finding the link, I can put it in here for you. SPOILER FOR THE ANSWER: you need 16 bodyguards
So if you reflect a billiard table across one of its borders and then flip the two new ones across one of their connected borders you get a square that you can tile the plane with that can act exactly as a billiard table. In this square you will have 4 targets, if we call the position of the origin target (a,b) then the position of the other 3 will be (2-a,b), (a,2-b) and (2-a,2-b). Now bullet shot by the assassin acts like a line on this x,y plane which means that given the position of the assassin lets say(c,d) any shot is uniquely defined by the slope of the line which we will call m. now since the plane is tiled there are infinitely many targets which all take the form of (2*q+x, 2*p+y) where q and e are integers. I am not really sure how this leads to a solution but I am sure someone in the comments can build on this to reach it :).
Yakov Reznikov The answer explained is in a link in the video description. If you have any problems finding the link, I can put it in here for you. SPOILER FOR THE ANSWER: you need 16 bodyguards
aikimark1955 The answer explained is in a link in the video description. If you have any problems finding the link, I can put it in here for you. SPOILER FOR THE ANSWER: you need 16 bodyguards
Wait did she mention whether the ball has dimension?? also how big are the guards and target ??? if the target is 2d and the ball and guards are 0d then there needs to be an infinite amount of guards right? if they're all 0d points and the rays are 1d then idk but it seems more likely this is what she meant since this is a more interesting question IMO.
tyler 44 both the assassin and the target are points in 0d. The bodyguards are pints in 0d too. The shot is a line in 1d. And the answer to the puzzle is in a link in the video description. SPOILERS FOR THE ANSWER: you need 16 bodyguards.
All I get from this is that leadership refuses to let dude kill target and dude is forced to socialize with target inside a room for a long time while leadership try to reach a pathetic consensus
I wish she had explained in the video what she explained in her blog. The solution itself is fascinating! But now the video stands as a 9-minute advertisement for her blog
My gut tells me that given a subset of this problem were the assassin is only allowed an even number (0 included) of bounces off parallel walls we would only need one guard at the center of mass.
First attempt at some kind of way to think about this problem. I think another interesting question would be to see what the probability of a randomly spawning a winnable arrangement of this game would be. It's probably easier to start this problem imagining 0 guards and moving forward but commenting so I can come back. --- Obviously, placing a closed circle around the target will prevent a hit from any assassin. In fact, if the target is an element of a closed set (but not a border point) and the guards are the other elements in the set, a hit from any assassin is impossible. Assume now, that there is exactly one point (x,y) 'missing' from our circle, then we can place an assassin on the straight line path from (x,y), thus for every missing point there exists a configuration of this game where a hit is possible. If the only straight line path from the 'missing' point is to the corner and there is no assassin along it, then a hit is impossible. Similarly, if the only straight line path from the 'missing' point is 90 degrees to a side and there is no assassin along it, then a hit is impossible. Now, consider the case where the straight-line path from the 'missing' point is not to a corner and not 90 degrees to a side. Let (tx,ty) be the coordinates of the target and (px,py) be the point of collision of the straight-line path from the missing point to the wall. The angle of the segment (tx,ty)--(px,py) does not form a 90-degree angle with the sides. Divide the plane into two sections, one for all (x,y) < (tx,ty) and one for all (x,y) >= (tx,ty). One of these sections (call it S1) will contain the target, the straight-line segment, and (px,py). The other will contain none of these (call it S2). Let theta be the angle of intersection of the straight-line segment with the side. We consider the point of collision on S1 (px,py) and locate the corresponding point on S2 (call it px',py'). Now the problem reduces to finding if there is an assassin on the straight-line path of (px',py') with angle theta on plane S2. If there is not, repeat this process, creating a new S1 and S2.
Shairoz Sohail The answer explained is in a link in the video description. If you have any problems finding the link, I can put it in here for you. SPOILER FOR THE ANSWER: you need 16 bodyguards
Reflect the square along x and y axis to end with 4 squares and then treat the quadrant as a single torus and follow the steps per that taken by the instructor. You'll arrive at the answer "No" a countably infinite number of points is needed to block all possible path between the assassin and the Target
Plo Hij The answer explained is in a link in the video description. If you have any problems finding the link, I can put it in here for you. SPOILER FOR THE ANSWER: you need 16 bodyguards
analytics! no do this numerically - project from A to the wall in all 360 degrees (you wana use 2 pi radians) in steps of like 0.00000001 degrees (more or less based on the circumference of A and T) then from the point of intersection at the walls apply the reflection rule iteratively and see how many trajectories intersect with T each time you find one you put a B0 B1 B2 at say the midpoint from wall to T .... i don't think its has non polynomial time solution (non NP problem) so it can be brute forced "reasonably quickly"
I think the key lies on choosing the middle for the assassin. Then by fast thinking, i thibk only 3 are needed depending if the assassin can hit or not himself.
No, there isn't a finite number of points because you can make an infinite number of unique shots. in a 360 degrees. more work below first given the assumptions, when a shot is fired it will hit either the target or one of the 4 vertices. By flipping and copying the square on it's one of its axis then on the other, you can now tile the 2x2 much like the way they did with the torus. Now, with that you can make a line a line with any real angle and the line will continue until it hits either the target or until it hits a vertex of the square. Another way to look at it, you can draw a line for every vertex and target to the assassin. Since the tile representation should be able to extend infinitely then there should be infinite possible shots. (it is under the assumption that the security guards do not take up more than a point of space)
One would need infinite guards. Given that the bullet can ricochet an infinite amount of times, start by shooting directly at the target. Then lower the angle to make the shot ricochet once. Then twice ... Geometrically it's pretty easy to see that there's an infinite amount of shooting angles even if you were given just two parallel walls.
Since a circle around the target (that is, the collection of all possible trajectories the "bullet" could be coming from) is made up of an uncountably infinite amount of points intuitively I feel like no finite number of "blockers" could block it all, assuming of course that the bullet could come from any possible direction. I don't know if that makes any sense - not kidding; I failed pre calculus, I'm terrible at math -, but I have this feeling.
About two minutes in. I'd say yes, you can block due to the corners' cancellation creating an asymmetry in the solution space. I'd say if you put a sphere around this, no, not short of arranging the guards in a way that surrounds the target, effectively isolating it from the puzzle. Looking forward to finding out why I'm wrong on all of this :)
Ok, this might be way to simple an answer... but to me it makes sense. I order to “stop” any incoming line one has to “draw” a circle around the target or assassin. Here we would define a circle as an infinite set of points. So the answer is no...
My initial guess would be that the answer is "no", you need infinite bodyguards as there are infinitely many points on the side you can hit. But thinking longer about it, this is very likely wrong, as i don't see why there should be infinitly many ways to hit the target. Oh well, I assume the answer is some natural number times 4 as you can probably build a torus by glueing several similar rooms together. Or not, I suck at doing puzzles.
P K You are right! The answer explained is in a link in the video description. If you have any problems finding the link, I can put it in here for you. SPOILER FOR THE ANSWER: you need 16 bodyguards
Yes. Because people aren't points and have volume (or area). If VIP is 1x1 metres then you need exactly 4 1x1 guards to shield it from all possible bullets. If you insist on making the people volumeless and arealess 1d points and the ray a 1d ray, both in a 2d plane, then the answer is obviously no, as a finite amount of points cannot block an infinite (as all possible angles are infinite) amount of rays. To do that you would need to construct a wall, which is the same as an infinite amount of points if taken in a 2d plane. You can see this in your torus example. Even if you discard all of the flipped versions of the room for mental simplicity, you can still hit the target (of which there are infinitely many) with any possible shot.
Surround the Target? Or the Assassin, for that matter. But since points do not have any size... is this strategy possible using a finite number of points?
The solution is that there are 9 points. Take the parity of the number of bounces on each pair of sides as well as the middle bounce side for odd values, and the midpoint of any equal parity-side double pair will be in the exact same spot. This can be shown by looking at the "equivalent" points in the "parallel universes" which are generated by reflections across the grid lines as explained in the video(this trick is pretty common in math). If 2 points are equivalent, that means that they are an even number of reflections both horizontally and vertically away from each other. However, reflecting twice will point a coordinate at +-2 away from its original spot. Thus, 2n reflections will put it at 2n distance away(parallel to grid lines), so the midpoint will be n away from the original midpoint, and thus in an equivalent point. The 9 "base" coordinates give us the different initial values that are added to the +-2s. Thus we just take the 9 possible bounce parities, and find their respective midpoints. Took me way longer than it should have to figure this out rip, but I couldn't find an answer posted in the comments section so
Oh I just found the solution in the description, but her solution has repeat points. In the diagram she shows at the really end, where you can see "the original plane", it's pretty clear that the midpoints of the segments connecting the white circle to every black dot is in the same position in every cell. My solution is completely correct and is the minimum number of points.
Your explanation is too convoluted for me. Do you maybe mean the same solution that was given by TU in one of the commets on www.math3ma.com/blog/is-the-square-a-secure-polygon?
@@saldownik I didn't read though their comment exactly, but it seems like they found an error in their solution farther down the thread. Basically, the only difference between my solution and the officially posted solution is that I believe that some of their midpoints are the exact same point. In particular, imagine the points in question are (0,1/3) and (1,2/3). Their direct midpoint is (1/2,1/2). However, if we bounce off the top edge, then the bottom, and then reach the assassin, this will have the exact same midpoint. Furthermore, if we bounce off the bottom edge, and then the top, this will also have the same midpoint! This should be relatively easy to see by drawing it out, since this example is very symmetric. However, you should be able to see how this would apply to other examples in the same way. My intuition for my solution was different from their solution, so it looks a lot different, but the basic idea is the same, and they explain it in a lot more detail.
You end up with at most eight lattices. Both assassin and target each have at most four lattices consisting of the nominal lattice, the lattice reflected on x, the lattice reflected on y, and the lattice reflected on x and y. When I graph this out I see an infinite number of trajectories. If we are talking about euclidean points here, then no a finite set of single points could not block an infinite number of trajectories. Am I missing something? Think of them as being in a room with mirrors on the wall. Assuming you could see infinitely far, it is conceivable that you would see an infinite number of assassins and targets.
Jordan Munroe The answer explained is in a link in the video description. If you have any problems finding the link, I can put it in here for you. SPOILER FOR THE ANSWER: you need 16 bodyguards
No, you would need an infinite number of points to block it. Mainly because you are defining the guards as single dimension points. You cannot surround the target with a point because in between 2 points is always another infinite number of other points. You would need to define the guards as at least 2d constructs to be able to block the shot with a finite number of guards.
Shayle Thorne The answer explained is in a link in the video description. If you have any problems finding the link, I can put it in here for you. SPOILER FOR THE ANSWER: you need 16 bodyguards
So if you draw a line (a set of infinite points) across the room between the assassin and target, is there a point you can remove that wouldn't allow the target to be hit? Imagining it, no matter which point you remove it adds a possible path to hit the target, so infinite - 1 guards isn't enough therefore a finite number can't be enough either.
Kieron George The answer explained is in a link in the video description. If you have any problems finding the link, I can put it in here for you. SPOILER FOR THE ANSWER: you need 16 bodyguards
are the bodyguards and target meant to have some amount of area? because i feel like this could be very hard if they behave like normal points on a coordinate plane which occupies a single point with no area, as opposed to a body that does have area.
To make a circle you need and infinite amount of points The answer explained is in a link in the video description. If you have any problems finding the link, I can put it in here for you. SPOILER FOR THE ANSWER: you need 16 bodyguards
Julio Toboso García The answer explained is in a link in the video description. If you have any problems finding the link, I can put it in here for you. SPOILER FOR THE ANSWER: you need 16 bodyguards
I know an easy solution: let target point T be T(tx, ty) then let the block points B1 - B8 be: B1 (tx - epsilon, ty + epsilon) B2 (tx , ty + epsilon) B3 (tx + epsilon, ty + epsilon) B4 (tx - epsilon, ty ) B5 (tx + epsilon, ty ) B6 (tx - epsilon, ty - epsilon) B7 (tx , ty - epsilon) B8 (tx + epsilon, ty - epsilon) with 0 < epsilon < R/reel numbers then the point T cannot be hit without passing through one of the B points epsilon is like an indevisible unit like the planck length, so point T is perfectly surrounded by the points B1 - B9 without any gaps left but if you exclude hyperreal numbers then this solution is invalid ^^
Benjamin S The answer explained is in a link in the video description. If you have any problems finding the link, I can put it in here for you. SPOILER FOR THE ANSWER: you need 16 bodyguards
well you can garuntee it with seven and the location of the assassin does matter in the corner it could be 3-4. or even 2 if in the very corner.. i think the proper way would be to require diameter distance between the subject and target to any blocker.
Fulla Beaverhausen The answer explained is in a link in the video description. If you have any problems finding the link, I can put it in here for you. SPOILER FOR THE ANSWER: you need 16 bodyguards
Where am I wrong: use only 2 opposite sides of the table. Bounce off the border as many times as there are guards in a zig zag fashion. To hit the target. Sinxe there will always be a new zigzag line on which there is no guard yet (because there will be new paths which don't go through previous interactions {prime number paths}) there will always exist a new path towards the target
Csaba Dunai interesting idea, but your conclusion is incorrect The answer explained is in a link in the video description. If you have any problems finding the link, I can put it in here for you. SPOILER FOR THE ANSWER: you need 16 bodyguards
Is she suggesting that we transform the square into a torus and then graph the path from the assassin to target on the lattice?? Then we would have to know how many lattice points from A-T on the graph which will equal the amount of Blockers right??
Oh, I see. We can combine 4 small squares to make a big one. And that big one will be replicated infinitely many times on a plane like in the example with torus. Each point will have 4 copies per big square unless it is on one of its sides or in the corner. But it isn't even important. The important thing is that if we place a (V)ictim somwhere in a plane and a (K)iller somwhere else, then we can draw infinitely many straight lines with different angle from each copy of K to that V. Since two different straight lines may share maximum one point (and that point is already taken by Victim), any Guard can protect only from one such line. That's why you need infinitely many Guards. The only finite solution is to make the Victim a Guard itself (but he still gonna die). But in real life there would be a finite solution just because a real guard isn't just a single point. A real guard is an object with some surface area, so you can make a ring out of them around the victim to protect him from any side.
Count of Darkness The answer explained is in a link in the video description. If you have any problems finding the link, I can put it in here for you. SPOILER FOR THE ANSWER: you need 16 bodyguards
I mean this seems pretty simple. Wouldn't it always be possible to place a small number of points forming a protective circle around the target point therefor blocking off 360 degrees of entry?
firemaker282 no because points have no dimensions, you would need an infinite amount of points to make a circle The answer explained is in a link in the video description. If you have any problems finding the link, I can put it in here for you. SPOILER FOR THE ANSWER: you need 16 bodyguards
vcostor remember that the bodyguards, target and assassin are points with 0 dimensions. So you can’t surround T with bodyguards, to do that you would need an infinite amount of points The answer explained is in a link in the video description. SPOILER FOR THE ANSWER: you need 16 bodyguards
After putting a decent amount of thought into it, I don't think it even makes for it to bounce off in that direction. I think there's multiple interpretations of how the light might reflect. I think you can make a pretty strong argument that at the corner, light should bounce off in all directions. You can also make a case that for any line, as you approach the corner from a particular side, the limit of all the angles stays the same. So you could also interpret the light reflecting in two separate directions, both outside the bounds of the square. I'm not 100% sure any of these solutions are more or less likely than another.
I think you could work it out using limits and that any line that hit a corner directly bounces back along the path it came from, according to law of reflection. So the assassin would shoot himself, essentially.
Joshua Khupmuanlian the guards, target and assassin are points with 0 dimension. So to make a circle you would need an infinite amount of points The answer explained is in a link in the video description. If you have any problems finding the link, I can put it in here for you. SPOILER FOR THE ANSWER: you need 16 bodyguards
Mickoes the target, bodyguards and assassin are points with 0 dimensions. So you would need an infinite amount of points to make a circle around the target. The answer explained is in a link in the video description. SPOILER FOR THE ANSWER: you need 16 bodyguards
I don't get the puzzle. Isn't a simple straight line of guards between the assassin and target enough to block the shots? Or is a straight line of guards not finite?
lbandit points have no dimension. So to make a line you would need an infinite amount of points The answer explained is in a link in the video description. If you have any problems finding the link, I can put it in here for you. SPOILER FOR THE ANSWER: you need 16 bodyguards
For any natural number n, there are finitely many shots that will hit the target after hiting the walls exactly n times. So you can list all the shots that will hit the target, you just have to list all the shots that hit the target after hitting the walls 0 times, then all the shots that hit the target with 1 wall hit, then all that do 2 wall hits, and so on. That means there are countably many shots that hit the target, while there are uncountably many angles that the assassin can shot to. So, under the conventional measure, the probability is 0.
In your solution, you gave the definition of a lattice, but you did not specify that the axes could be rotated and rescaled so that they are possibly no longer Cartesian coordinates.
They left us with a final boss at the end of the series.
xD
and im just starting to watch them
@@SrmthfgRockLee maybe u know something similar to this ?:v
"...and see you soon" :(
Daniel Hurtado yes... I guess we will never see the solution...
Technically, that was pretty soon, she just talked about something else.
Err...the solution is already up, guys. She's linked it in the description - www.math3ma.com/mathema/2018/5/17/is-the-square-a-secure-polygon
25 11 2018
@@ArawnOfAnnwn she didnt lol
*What if the assassin hits himself accidentally?*
just assume it passes through.
Friendly Fire! :p
This is why hitting a vertex is assumed to end the line, otherwise it simply goes back the other way and hits the assassin. I guess there's other ways for the assasin to get himself though, like shooting at a precisely 90 degree angle to any wall (a 1-2 in football term's), but presumably we can assume the assassin is at least smart enough not to '1-2' himself of the first wall.
That's equivalent to a guard being at the assassin's position so it can be handled.
then that means he counts as one of the body guards
"see you soon"
soon*
HEY SEAN MURRAY, GET OVER HERE!
Wait...
so...
but...
they just...
how am I...
why would they...
do they...?
I thought they...
but...
I wanted to...
So there's just not a solution and there's never going to be one since they shut down Infinite Series?
Not a single closure.
The solution is in the video description.
I never thought I'd be on the end of "you should read the description", but here I am.
Thanks!
also, did none of the videos mentions have clickable links?
not only that, but you are on the end of "you are commenting before watching the whole video", she's mentions the solution is written up in the description. There wasn't a plan to make another video.
What if the target isn't QPU aligned?
love your sm64 reference ^^
Amazing. An SM64 meme that fits for once
i just had to watch that long ass video to understand this, it was worth it tho.
Ah. The Revolver Ocelot problem.
Wow sh*t, you're right. Nice catch :)
Uejji It's pretty good, isn't it?
kept you waiting, huh?
You just made my day. Thank you!
The .45 assassin puzzle, the strongest puzzle in the world
This channel provided fantastic content. I hope it comes back
The play tester's answer: one guard immediately on top of the assassin, clipping into the start of the hitscan trajectory. If the guard can't be immediately on top of the assassin but is allowed to have width, you can use just eight guards by boxing the assassin with four and then wedging the gaps with the other four.
The target, the shooter and the bodyguards are points.
That was my solution too, though I wanted to surround the target with bodyguards. You know... like a presidential tour
just read my solutiuon in the comments. i did exactly what you suggest but mathmatically correct ^^
5:28 But first we need to talk about parallel universes.
I did it, I just watched every single Infinite Series episode there is
I smell the Orchard Problem. I'm surprised there's no shout out to the Numberphile video about it.
If you interpret a bounce on the wall of the billiard table as entering a flipped version of the original billiard table, it works the same as having 4 billiard tables on a Torus (because 2 reflections in either direction cancels out). This leads to 4 latices of points representing the target.
Every possible shot on the target is a line from A to a copy of a target point at some square in the grid. And choosing a guard creates another set of 4 latices that happens to lie on as many shots as possible. If the shot lines are folded back upon themselves into the 1,1 square, the intersections are the best points to choose for a guard. The question of the Assassin puzzle can be reworded as a question of whether there are a finite number of intersections that cover all the paths from A to T.
I don't know how to solve this for the general case, but for a simpler case where A=0,0 and T =(1,1) all the reflected versions of T are at the same location, and the problem is the same as if it were on a torus but due to the rule that paths that reach a corner (other than T) are discarded results in all the points of the lattice with an even number for x or y are discarded. In this odd grid, where all shots must travel an odd number of squares in both the x and y direction, all shots must also pass through the centre of the square which is halfway to the target. Which means that this special case only requires one guard in the centre of the room.
So I didn't solve the problem but I did solve an easier version of it. I suspect that the true answer is 4 guards but don't really know how to proceed from here.
Abram Thiessen the answer is in a link in the video description, it is explained very well. You were really close to be correct, but the real answer is 16 bodyguards
8:37 "And if it is possible, then, and this is really the key, what is that finite number?" That's a dead giveaway, lol.
What a cool video. I love this idea. I’m going to get working on it right now.
Take the initial square representing the room and reflect it a few times to make a 2x2 group with matched sides such that bouncing off a wall can still be represented as moving through it to the next square. This 2x2 group then tiles to fill the plane as in the example shown in the video. Now turn the target into a lattice representing its positions in all of the tiles, and the possible shots are lines from the (single point) assassin to every point in the target lattice. The challenge then translates into finding a finite set of points such that every shot line goes through at least one of the points OR one of its equivalent lattice points in some other tile.
(I had more started here, but then I realized that I neglected to treat the blocking points as lattices, which completely wrecked my solution. :( )
HebaruSan The answer explained is in a link in the video description. SPOILER FOR THE ANSWER: you need 16 bodyguards
there is a simple solution to the problem:
just use any wall of your choice (they all work equally well if A and T arent on the same perpendicular of the wall, then you choose a different wall )
now for your chosen wall just bounce it once to hit the target.
now bounce twice using the chosen wall and the opposite wall.
this can be continued forever getting a more and more acute angle.
thus its not possible to have a finite number of blockers.
had exactly the same idea, funny considering people are stumped by this problem and some acing this by just one good thought :/
judgeomega interesting idea, but it is incorrect
The answer explained is in a link in the video description. If you have any problems finding the link, I can put it in here for you. SPOILER FOR THE ANSWER: you need 16 bodyguards
Lukas Henke incorrect, The answer explained is in a link in the video description. If you have any problems finding the link, I can put it in here for you. SPOILER FOR THE ANSWER: you need 16 bodyguards
It seems that if the target hugged a wall or got in a corner.
The lines aren't parallel and they should be since it's a square and it's really bothering me.
Finding that the lines should be parallel is an extra information that in not in the problem statement. You probably already did some problem in school asking if a triangle is special, but with a drawing clearly not to scale.
MrPingouin1 Tai'Danae says the room is square. It also says so in the video description. Cheers.
I think they were talking about the fact that given the law of reflection at 1:18, a path in the square should form parallel lines.
Oh ok, I thought he meant the boundaries of the room (sides of the square). I see what he means now. He's right. They are supposed to be parallel. The animation at 1:28 is pretty off--like the triangles in school you were talking about.
Mathemeticians posing phisics problems leads to tennis balls filling every part of infinite hotel room spacetime.
So easy. Every time it bounce it's like entering a new square but that was mirrored (ie a flipped copy using the plane it bounced on as the symetry axis). So we work with 4 squares. The original one plus x-flipped copy plus y-flipped plus x-and-y-flipped copy. Then we duplicate this big square (of 4 squares) to cover the plane. Each big square contains 4 targets (all virtual except the original one but all copies represent a path to hit the target using a certain number of bounces). Even if we consider the small subset of targets located at (x+2×k,y) if the y position of the assassin is different from the y position of original target, there's an infinite number of paths. If they are on the same y position, we consider the vertical set of square and the (x,y+2×k) virtual targets. If the assassin is not on the same position than the target that case will show an infinite number of paths.
So we need an infinite number of bodyguards.
It's a countable infinite (using the same arguments than when we say that fractions are)
I was thinking the same way, but I couldn't convince myself, since you are also copying bodyguards. So I thought of a different kind of proof. Just ignore two opposing sides of the squared and use the remaining two to bounce the bullet. You can make different trajectories by making different number of bounces. So 0 bounces is straight to the target, 1 bounce is against one side to the target, 2 bounces is from one side to the other side to the target, ....
I would then claim that no triplet of trajectories will intersect in the same point. I don't have any proof for this (I'm not entirely sure if it's even true), but it feels more intuitive.
TheRMeerkerk That is exactly what happen in the 1st case. The only exception is when they are on the same height (y position). In that case there's only one trajectory this way. That's why you have to bounce vertically in that case (2nd case). At least one case works because they don't share the same point.
BTW I made the hypothesis that you can't have 2 points in the same position. Otherwise you just put a bodyguard on the assassin and the whole problem is over.
I had some extra thinking and while the construction of the squares... are ok there's still a problem to proove that you can't block all those paths with a limited number of bodyguards.
I found an example of a tricky situation:
-assassin at (0.5 0.25)
-target at (0.5 0.75)
-bodyguards at (0 0.5) (0.5 0.5) (1 0.5)
The 3 guards can block all line of fire that bounce only horizontally (or vertically).
Of course you can still fire and using both vertical and horizontal bounces together but it shows that a finite number of bodyguards can block an infinite number of paths!
The answer explained is in a link in the video description. If you have any problems finding the link, I can put it in here for you. SPOILER FOR THE ANSWER: you need 16 bodyguards
So unfortunate Maryam Mirzakhani passed away in 2017......
RIP Maryam Mirzakhani
Once you find the connection between the initial room and the tiled torii, it becomes a line-of-sight-through-an-orchard puzzle.
That was my immediate thought too. Hope it's right!
The answer explained is in a link in the video description. If you have any problems finding the link, I can put it in here for you. SPOILER FOR THE ANSWER: you need 16 bodyguards
If you put 12 body gaurds around point T, of the same radius as target T: it may be an inelegant solution, but it get the job done.
6 for the kissing number of a circle, 6 more for all of the spaces between the body gaurds.
Double redundancy.
And that boring solution would work for any polygon (assuming euclidean space).
So I'm treating people like cylinders of equal height and girth; and assuming that they all have bullet proof umbrellas, and that there aren't any mines or otherwise floor-based threats (because we are solving a 2dimensional problem)
Ian Schimnoski in this problem you must consider everyone as points with 0 dimensions. That is why making a circle does not work, you need an infinite amount of points to make a circle
The answer explained is in a link in the video description. If you have any problems finding the link, I can put it in here for you. SPOILER FOR THE ANSWER: you need 16 bodyguards
new video from my favourite channel!
Oh dear...
I have some bad news:
ruclips.net/video/UIwnCoqx91Q/видео.html
RIP pbs infinite series
The Surface of the Torus is (xy,8)
May I propose a variant of this puzzle?
Say, there's a security guard G who won't stop bragging "When I die, I will do so saving someone's life!". Now imagine a killer K who is so annoyed by this that he decides to kill G. He lures him into a square room and intends him to shoot him dead, proving him wrong. The question is:
Can you place a finite number of VIPs in the room such that none stands in the way of K shooting G, but however K tries to kill G, the latter will stand in the way of some VIP?
(I don't know what the answer is, but given that the two problems are quite similar, I wouldn't be surprised if the answers were equally similar.)
it looks like you should start by mirroring the initial square twice (once on the x and then the two squares you get on the y axis) until you get 4 squares with 4 "targets". Then all reflections are translated into straight lines inside these 4 squares. You can then tile the whole thing as with the torus to get an infinite plane. Then the question translates to the question of: do you see an finite number of points from A in the resulting infinite plane? I remember there was an youtube video about it and i think that the answer is no, the number of points you see is the number of rational numbers, so infinite, so then you would need an infinite number of guards. A smaller infinite than the number of real numbers but still infinite.
Tried the puzzle myself, then came to the commentsection for the answer. Yours matches mine perfectly.
I like your construction, and that's definitely the correct answer to the question "do you see an finite number of points from A in the resulting infinite plane?". However, that question is not quite identical to the original, since each guard can block multiple (indeed, infinite) lines of sight, as each is also copied into a lattice. Maybe it would be fruitful to look into what sets of angles could be blocked by a given guard?
But your construction doesn't take into account that some paths intersect each other. So if you place a guard at those intersecting points you can block multiple paths using only a few guards. Though i'm not sure if it can be done using only finitely many guards.
Jay-r Villanueva The shot can bounce any real number of times. This means the target can get hit from every possible angle, and you'd need infinitly many guards to cover it. Sure, every guard blocks infinitly many shots, but not every shot is blocked.
This is hard to explain in a yt-comment (especially since english ain't my native tounge), so I'd advise you to take a sheet of paper and draw the square from 7:52. Then go infinite, draw the same square next to it, and next to that again...
Just remember to align the red side with the red side, green with green etc
Finally, draw a shot from your original assassin to every target.
Since there are infinitly many targets on the x-axis alone, you'd need infinitly many guards to cover them all.
Nope, you first need to show that your construction directly corresponds to the problem. Your approach doesn't really capture the property of the square namely paths can intersect one another. Thus they are not really mathematically equivalent.
I have a solution.
Solution A:
T steals the gun
Solution B:
T dodges the bullet
Solution C:
A shoots himself
Solution D:
T drags A into the bullet
Solution E:
T somehow escapes
Solution F:
The cops move to keep T safe (though I wouldn't recommend this)
Solution G:
The cops arrest A
I don't have to imagine it. I live it. Every damn day.
technically no there is no positioning of the guards such that they can block any shot the assassin takes. you can create a pseudo-torus square tessellation of the grid by reflecting the room across any wall. this pseudo-torus will have the property that every other reflection will be identical to the original. If we ignore the reflections and just focus on the copies, what we have is just a normal torus where the relevant points are limited to one quarter of the square. If we then tessellate the grid with this torus, we see that for any two copies the will be one exactly half way between them. For example, between the copy immeadiately above and the copy immeadiately to the left there is the copy both above and to the left. and there is one half way between that one and any other. this yields an infinite set of possible paths for the bullet to take an eventually hit the target. if you were to fill the room with a Gogol guard points, the target would likely die of starvation or natural causes before being hit by the bullet but he would eventually be hit by it.
Am I way off base but isn't it obviously "no"? If A shoots vertically the shot can bounce off the ceiling 0 times (direct shot), 1 time, 2 times (with one bounce off the bottom), 3 times (two bounces off the bottom) etc. Since there are an infinite number of options, - with an infinite number of them being prime so there wouldn't be overlap with previous paths - wouldn't that be an infinite number of blockers to block the shots?
So the answer is no.
If you flip it on the x axis, then flip the pair of squares on the y-axis, you end up with a torus topology. When this set of 4 squares is tiled and embedded on the plane it's obvious to see that the assassin can shoot a straight line to the target in any of an infinite number of squares, i.e. there are an infinite number of ways to shoot the target.
The finiteness would be determined by the radius represented by each bystander. Additionally the minimum number would be dependent on the target's distance from the two closest walls. And if the assassin is closer than the width of one bystander, then the answer is no.
hacked2123 The answer explained is in a link in the video description. If you have any problems finding the link, I can put it in here for you. SPOILER FOR THE ANSWER: you need 16 bodyguards
Make a circle of bodyguards, and you're done. Is it an elegant solution using math? No. Is it a solution that works on the problem as it was presented to us? Yes.
EVEN MADSEN remember that the assassin, bodyguards and target are points with 0 dimensions. To make a circle you need an infinite amount of points
The answer explained is in a link in the video description. If you have any problems finding the link, I can put it in here for you. SPOILER FOR THE ANSWER: you need 16 bodyguards
Yes, someone just said *It should be named "finite series"*, just because infinite series ended, AND IM 4 YEARS LATE WTF
So if you reflect a billiard table across one of its borders and then flip the two new ones across one of their connected borders you get a square that you can tile the plane with that can act exactly as a billiard table. In this square you will have 4 targets, if we call the position of the origin target (a,b) then the position of the other 3 will be (2-a,b), (a,2-b) and (2-a,2-b). Now bullet shot by the assassin acts like a line on this x,y plane which means that given the position of the assassin lets say(c,d) any shot is uniquely defined by the slope of the line which we will call m. now since the plane is tiled there are infinitely many targets which all take the form of (2*q+x, 2*p+y) where q and e are integers. I am not really sure how this leads to a solution but I am sure someone in the comments can build on this to reach it :).
Yakov Reznikov The answer explained is in a link in the video description. If you have any problems finding the link, I can put it in here for you. SPOILER FOR THE ANSWER: you need 16 bodyguards
The target has a simple strategy to avoid being shot without the bodyguards. They just have to hide in the corner of the room.
In addition to the corner stop, you also need to add a condition that the bounced path can not go back to the assassin. The assassin is another stop.
aikimark1955 The answer explained is in a link in the video description. If you have any problems finding the link, I can put it in here for you. SPOILER FOR THE ANSWER: you need 16 bodyguards
Wait did she mention whether the ball has dimension?? also how big are the guards and target ??? if the target is 2d and the ball and guards are 0d then there needs to be an infinite amount of guards right? if they're all 0d points and the rays are 1d then idk but it seems more likely this is what she meant since this is a more interesting question IMO.
tyler 44 both the assassin and the target are points in 0d. The bodyguards are pints in 0d too. The shot is a line in 1d.
And the answer to the puzzle is in a link in the video description.
SPOILERS FOR THE ANSWER: you need 16 bodyguards.
All I get from this is that leadership refuses to let dude kill target and dude is forced to socialize with target inside a room for a long time while leadership try to reach a pathetic consensus
I wish she had explained in the video what she explained in her blog. The solution itself is fascinating! But now the video stands as a 9-minute advertisement for her blog
My gut tells me that given a subset of this problem were the assassin is only allowed an even number (0 included) of bounces off parallel walls we would only need one guard at the center of mass.
First attempt at some kind of way to think about this problem. I think another interesting question would be to see what the probability of a randomly spawning a winnable arrangement of this game would be. It's probably easier to start this problem imagining 0 guards and moving forward but commenting so I can come back.
---
Obviously, placing a closed circle around the target will prevent a hit from any assassin. In fact, if the target is an element of a closed set (but not a border point) and the guards are the other elements in the set, a hit from any assassin is impossible.
Assume now, that there is exactly one point (x,y) 'missing' from our circle, then we can place an assassin on the straight line path from (x,y), thus for every missing point there exists a configuration of this game where a hit is possible.
If the only straight line path from the 'missing' point is to the corner and there is no assassin along it, then a hit is impossible. Similarly, if the only straight line path from the 'missing' point is 90 degrees to a side and there is no assassin along it, then a hit is impossible.
Now, consider the case where the straight-line path from the 'missing' point is not to a corner and not 90 degrees to a side. Let (tx,ty) be the coordinates of the target and (px,py) be the point of collision of the straight-line path from the missing point to the wall. The angle of the segment (tx,ty)--(px,py) does not form a 90-degree angle with the sides. Divide the plane into two sections, one for all (x,y) < (tx,ty) and one for all (x,y) >= (tx,ty). One of these sections (call it S1) will contain the target, the straight-line segment, and (px,py). The other will contain none of these (call it S2).
Let theta be the angle of intersection of the straight-line segment with the side. We consider the point of collision on S1 (px,py) and locate the corresponding point on S2 (call it px',py'). Now the problem reduces to finding if there is an assassin on the straight-line path of (px',py') with angle theta on plane S2. If there is not, repeat this process, creating a new S1 and S2.
Shairoz Sohail The answer explained is in a link in the video description. If you have any problems finding the link, I can put it in here for you. SPOILER FOR THE ANSWER: you need 16 bodyguards
Seems fitting that the last episode is a challenge question.
Reflect the square along x and y axis to end with 4 squares and then treat the quadrant as a single torus and follow the steps per that taken by the instructor. You'll arrive at the answer "No" a countably infinite number of points is needed to block all possible path between the assassin and the Target
Plo Hij The answer explained is in a link in the video description. If you have any problems finding the link, I can put it in here for you. SPOILER FOR THE ANSWER: you need 16 bodyguards
Just stand in a circle around the dude
analytics! no do this numerically - project from A to the wall in all 360 degrees (you wana use 2 pi radians) in steps of like 0.00000001 degrees (more or less based on the circumference of A and T) then from the point of intersection at the walls apply the reflection rule iteratively and see how many trajectories intersect with T each time you find one you put a B0 B1 B2 at say the midpoint from wall to T .... i don't think its has non polynomial time solution (non NP problem) so it can be brute forced "reasonably quickly"
I think the key lies on choosing the middle for the assassin. Then by fast thinking, i thibk only 3 are needed depending if the assassin can hit or not himself.
wow I think this is the closest I have come to understanding a quotient set
Hitting a vertex would mean that the assassin shoots theirself also since it would reflect back along the same course.
love this video! Keep it up Infinite Series!!!
No, there isn't a finite number of points because you can make an infinite number of unique shots. in a 360 degrees. more work below
first given the assumptions, when a shot is fired it will hit either the target or one of the 4 vertices.
By flipping and copying the square on it's one of its axis then on the other, you can now tile the 2x2 much like the way they did with the torus.
Now, with that you can make a line a line with any real angle and the line will continue until it hits either the target or until it hits a vertex of the square.
Another way to look at it, you can draw a line for every vertex and target to the assassin. Since the tile representation should be able to extend infinitely then there should be infinite possible shots.
(it is under the assumption that the security guards do not take up more than a point of space)
If a trajectory enters a corner, its path will reverse, making it rather less desirable for the assassin to make that shot.
RIP infinite series
And i just startedw atching it....
One would need infinite guards. Given that the bullet can ricochet an infinite amount of times, start by shooting directly at the target. Then lower the angle to make the shot ricochet once. Then twice ... Geometrically it's pretty easy to see that there's an infinite amount of shooting angles even if you were given just two parallel walls.
Since a circle around the target (that is, the collection of all possible trajectories the "bullet" could be coming from) is made up of an uncountably infinite amount of points intuitively I feel like no finite number of "blockers" could block it all, assuming of course that the bullet could come from any possible direction.
I don't know if that makes any sense - not kidding; I failed pre calculus, I'm terrible at math -, but I have this feeling.
About two minutes in. I'd say yes, you can block due to the corners' cancellation creating an asymmetry in the solution space. I'd say if you put a sphere around this, no, not short of arranging the guards in a way that surrounds the target, effectively isolating it from the puzzle. Looking forward to finding out why I'm wrong on all of this :)
corners do not matter, they just bounce twice, and you are incorrect
Watching this while waiting for the newest pbs eons vid to come out
Ok, this might be way to simple an answer... but to me it makes sense.
I order to “stop” any incoming line one has to “draw” a circle around the target or assassin.
Here we would define a circle as an infinite set of points.
So the answer is no...
Bert Goethals The answer explained is in a link in the video description. SPOILER FOR THE ANSWER: you need 16 bodyguards
My initial guess would be that the answer is "no", you need infinite bodyguards as there are infinitely many points on the side you can hit. But thinking longer about it, this is very likely wrong, as i don't see why there should be infinitly many ways to hit the target. Oh well, I assume the answer is some natural number times 4 as you can probably build a torus by glueing several similar rooms together.
Or not, I suck at doing puzzles.
P K You are right! The answer explained is in a link in the video description. If you have any problems finding the link, I can put it in here for you. SPOILER FOR THE ANSWER: you need 16 bodyguards
Yes. Because people aren't points and have volume (or area). If VIP is 1x1 metres then you need exactly 4 1x1 guards to shield it from all possible bullets.
If you insist on making the people volumeless and arealess 1d points and the ray a 1d ray, both in a 2d plane, then the answer is obviously no, as a finite amount of points cannot block an infinite (as all possible angles are infinite) amount of rays. To do that you would need to construct a wall, which is the same as an infinite amount of points if taken in a 2d plane.
You can see this in your torus example. Even if you discard all of the flipped versions of the room for mental simplicity, you can still hit the target (of which there are infinitely many) with any possible shot.
Yeah. Make a circle around it, and block the gaps with B.
Surround the Target? Or the Assassin, for that matter. But since points do not have any size... is this strategy possible using a finite number of points?
The solution is that there are 9 points. Take the parity of the number of bounces on each pair of sides as well as the middle bounce side for odd values, and the midpoint of any equal parity-side double pair will be in the exact same spot. This can be shown by looking at the "equivalent" points in the "parallel universes" which are generated by reflections across the grid lines as explained in the video(this trick is pretty common in math). If 2 points are equivalent, that means that they are an even number of reflections both horizontally and vertically away from each other. However, reflecting twice will point a coordinate at +-2 away from its original spot. Thus, 2n reflections will put it at 2n distance away(parallel to grid lines), so the midpoint will be n away from the original midpoint, and thus in an equivalent point. The 9 "base" coordinates give us the different initial values that are added to the +-2s. Thus we just take the 9 possible bounce parities, and find their respective midpoints.
Took me way longer than it should have to figure this out rip, but I couldn't find an answer posted in the comments section so
Oh I just found the solution in the description, but her solution has repeat points. In the diagram she shows at the really end, where you can see "the original plane", it's pretty clear that the midpoints of the segments connecting the white circle to every black dot is in the same position in every cell. My solution is completely correct and is the minimum number of points.
Your explanation is too convoluted for me. Do you maybe mean the same solution that was given by TU in one of the commets on www.math3ma.com/blog/is-the-square-a-secure-polygon?
@@saldownik I didn't read though their comment exactly, but it seems like they found an error in their solution farther down the thread. Basically, the only difference between my solution and the officially posted solution is that I believe that some of their midpoints are the exact same point. In particular, imagine the points in question are (0,1/3) and (1,2/3). Their direct midpoint is (1/2,1/2). However, if we bounce off the top edge, then the bottom, and then reach the assassin, this will have the exact same midpoint. Furthermore, if we bounce off the bottom edge, and then the top, this will also have the same midpoint! This should be relatively easy to see by drawing it out, since this example is very symmetric. However, you should be able to see how this would apply to other examples in the same way. My intuition for my solution was different from their solution, so it looks a lot different, but the basic idea is the same, and they explain it in a lot more detail.
You end up with at most eight lattices. Both assassin and target each have at most four lattices consisting of the nominal lattice, the lattice reflected on x, the lattice reflected on y, and the lattice reflected on x and y. When I graph this out I see an infinite number of trajectories. If we are talking about euclidean points here, then no a finite set of single points could not block an infinite number of trajectories. Am I missing something?
Think of them as being in a room with mirrors on the wall. Assuming you could see infinitely far, it is conceivable that you would see an infinite number of assassins and targets.
Jordan Munroe The answer explained is in a link in the video description. If you have any problems finding the link, I can put it in here for you. SPOILER FOR THE ANSWER: you need 16 bodyguards
No, you would need an infinite number of points to block it. Mainly because you are defining the guards as single dimension points. You cannot surround the target with a point because in between 2 points is always another infinite number of other points. You would need to define the guards as at least 2d constructs to be able to block the shot with a finite number of guards.
Shayle Thorne The answer explained is in a link in the video description. If you have any problems finding the link, I can put it in here for you. SPOILER FOR THE ANSWER: you need 16 bodyguards
16 is the minimum number of guards
This took SOOO long, and I felt so accomplished when I finally got it XD. (The middle hints were what gave it).
JoJo Ray what answer did you get?
The answer explained is in a link in the video description. SPOILER FOR THE ANSWER: you need 16 bodyguards
So if you draw a line (a set of infinite points) across the room between the assassin and target, is there a point you can remove that wouldn't allow the target to be hit?
Imagining it, no matter which point you remove it adds a possible path to hit the target, so infinite - 1 guards isn't enough therefore a finite number can't be enough either.
Kieron George The answer explained is in a link in the video description. If you have any problems finding the link, I can put it in here for you. SPOILER FOR THE ANSWER: you need 16 bodyguards
are the bodyguards and target meant to have some amount of area? because i feel like this could be very hard if they behave like normal points on a coordinate plane which occupies a single point with no area, as opposed to a body that does have area.
just create a human shield around the the target (or the assassin)
If by "human shield" you mean a circle, then it's not a finite number of dots anymore.
SOZ
To make a circle you need and infinite amount of points
The answer explained is in a link in the video description. If you have any problems finding the link, I can put it in here for you. SPOILER FOR THE ANSWER: you need 16 bodyguards
Can you link the “This video” in the dooblido?
I can’t see the link in the video either 🤔
Julio Toboso García The answer explained is in a link in the video description. If you have any problems finding the link, I can put it in here for you. SPOILER FOR THE ANSWER: you need 16 bodyguards
I know an easy solution:
let target point T be T(tx, ty)
then let the block points B1 - B8 be:
B1 (tx - epsilon, ty + epsilon)
B2 (tx , ty + epsilon)
B3 (tx + epsilon, ty + epsilon)
B4 (tx - epsilon, ty )
B5 (tx + epsilon, ty )
B6 (tx - epsilon, ty - epsilon)
B7 (tx , ty - epsilon)
B8 (tx + epsilon, ty - epsilon)
with 0 < epsilon < R/reel numbers
then the point T cannot be hit without passing through one of the B points
epsilon is like an indevisible unit like the planck length, so point T is perfectly surrounded by the points B1 - B9 without any gaps left
but if you exclude hyperreal numbers then this solution is invalid ^^
Benjamin S The answer explained is in a link in the video description. If you have any problems finding the link, I can put it in here for you. SPOILER FOR THE ANSWER: you need 16 bodyguards
Finally a new video.
Torus Tells Time
Tantalising Topology Tries To Transcribe The Trancedental Titan Talk
Try Toppling The Total Times The Trancedental Titan Takes T's Throughout The Theme Of Allitera... damn that's hard
If you play pool you'll snooker the target in the corner and use 2 body guards to block all angles
Could have asked whether a billiard ball could be blocked by other billiard balls. But sensibly went with the violent version which is better PR.
well you can garuntee it with seven and the location of the assassin does matter in the corner it could be 3-4. or even 2 if in the very corner.. i think the proper way would be to require diameter distance between the subject and target to any blocker.
Miz Eve The answer explained is in a link in the video description. SPOILER FOR THE ANSWER: you need 16 bodyguards
take a coin surround it by seven other coins all touching.. thats what I meant. I was assuming they can touch creating a perfect wall.
Finally you reached a nice pace and now the series is over....
Fulla Beaverhausen The answer explained is in a link in the video description. If you have any problems finding the link, I can put it in here for you. SPOILER FOR THE ANSWER: you need 16 bodyguards
Illumination problem was on an episode of Numb3rs
What about placing T at a position which (At y = infinity) is the golden ratio?
Would placing more than 4 Guards around A be necessary?
J H The answer explained is in a link in the video description. SPOILER FOR THE ANSWER: you need 16 bodyguards
Where am I wrong: use only 2 opposite sides of the table. Bounce off the border as many times as there are guards in a zig zag fashion. To hit the target. Sinxe there will always be a new zigzag line on which there is no guard yet (because there will be new paths which don't go through previous interactions {prime number paths}) there will always exist a new path towards the target
Csaba Dunai interesting idea, but your conclusion is incorrect
The answer explained is in a link in the video description. If you have any problems finding the link, I can put it in here for you. SPOILER FOR THE ANSWER: you need 16 bodyguards
Is she suggesting that we transform the square into a torus and then graph the path from the assassin to target on the lattice?? Then we would have to know how many lattice points from A-T on the graph which will equal the amount of Blockers right??
oh nvm she says the sides have identity unlike torus
wow just read the proof so elegant.
Oh, I see. We can combine 4 small squares to make a big one. And that big one will be replicated infinitely many times on a plane like in the example with torus. Each point will have 4 copies per big square unless it is on one of its sides or in the corner. But it isn't even important. The important thing is that if we place a (V)ictim somwhere in a plane and a (K)iller somwhere else, then we can draw infinitely many straight lines with different angle from each copy of K to that V. Since two different straight lines may share maximum one point (and that point is already taken by Victim), any Guard can protect only from one such line. That's why you need infinitely many Guards. The only finite solution is to make the Victim a Guard itself (but he still gonna die).
But in real life there would be a finite solution just because a real guard isn't just a single point. A real guard is an object with some surface area, so you can make a ring out of them around the victim to protect him from any side.
Count of Darkness The answer explained is in a link in the video description. If you have any problems finding the link, I can put it in here for you. SPOILER FOR THE ANSWER: you need 16 bodyguards
I mean this seems pretty simple. Wouldn't it always be possible to place a small number of points forming a protective circle around the target point therefor blocking off 360 degrees of entry?
firemaker282 no because points have no dimensions, you would need an infinite amount of points to make a circle
The answer explained is in a link in the video description. If you have any problems finding the link, I can put it in here for you. SPOILER FOR THE ANSWER: you need 16 bodyguards
this is something like the film Equlibrium they'd like that
what would happen if the room is a rectangle?? will the number of bodyguards increase?
4-6 depending on size of blockers. surround T or A so no line touches.
vcostor remember that the bodyguards, target and assassin are points with 0 dimensions. So you can’t surround T with bodyguards, to do that you would need an infinite amount of points
The answer explained is in a link in the video description. SPOILER FOR THE ANSWER: you need 16 bodyguards
Why isn't the reflection line at a corner the line purpendiculair to the bisector of that corner?
After putting a decent amount of thought into it, I don't think it even makes for it to bounce off in that direction. I think there's multiple interpretations of how the light might reflect. I think you can make a pretty strong argument that at the corner, light should bounce off in all directions. You can also make a case that for any line, as you approach the corner from a particular side, the limit of all the angles stays the same. So you could also interpret the light reflecting in two separate directions, both outside the bounds of the square.
I'm not 100% sure any of these solutions are more or less likely than another.
I think you could work it out using limits and that any line that hit a corner directly bounces back along the path it came from, according to law of reflection. So the assassin would shoot himself, essentially.
pretty sure i've seen this for rooms that have areas that light cannot hit when its from a specific source... is it the same thing?
that's the illumination problem they mentioned
What if the guards stand in a circle around the target (o_o)?
Joshua Khupmuanlian the guards, target and assassin are points with 0 dimension. So to make a circle you would need an infinite amount of points
The answer explained is in a link in the video description. If you have any problems finding the link, I can put it in here for you. SPOILER FOR THE ANSWER: you need 16 bodyguards
couldnt you just, you know, place guards in circle around the Target?
Simon Pelz not if they were tightly surrounding the target being equally small
Just put more, finite is not defined here. We'd probably could build a larger circle around with all the required dots.
Mickoes the target, bodyguards and assassin are points with 0 dimensions. So you would need an infinite amount of points to make a circle around the target.
The answer explained is in a link in the video description. SPOILER FOR THE ANSWER: you need 16 bodyguards
@@AndresFirte Then don't call them "body guards". Call them "points".
@@quillaja No, U.
I don't get the puzzle. Isn't a simple straight line of guards between the assassin and target enough to block the shots? Or is a straight line of guards not finite?
lbandit points have no dimension. So to make a line you would need an infinite amount of points
The answer explained is in a link in the video description. If you have any problems finding the link, I can put it in here for you. SPOILER FOR THE ANSWER: you need 16 bodyguards
Surround the A with B/C/D etc untill the circle is closed, there all angles are closed
Ronin But then it wouldn't be a finite number of security guards.
what is probablity that a random shot will hit the target?
For any natural number n, there are finitely many shots that will hit the target after hiting the walls exactly n times. So you can list all the shots that will hit the target, you just have to list all the shots that hit the target after hitting the walls 0 times, then all the shots that hit the target with 1 wall hit, then all that do 2 wall hits, and so on. That means there are countably many shots that hit the target, while there are uncountably many angles that the assassin can shot to. So, under the conventional measure, the probability is 0.
amazing solution
Very cool! Excellent job!
In your solution, you gave the definition of a lattice, but you did not specify that the axes could be rotated and rescaled so that they are possibly no longer Cartesian coordinates.
Links to the mirzhakani and numberphile videos?