It shouldn't be that way... School conditions us to be scared of math but once you actually look at it from a fun and games angle it's actually really interesting and useful in certain applications.
@@melaniemedina8037 Must depend on the school/teachers/kids. Nobody was "scared" of math at my school. Maybe some who were less than average but that's about it
@@user-go5he1fe1g It is actually pretty obvious to those who have played Minecraft, that's the first thought that came into my mind. It doesn't surprise me if you saw comments similar to mine :)
The infected question is actually similar to multiplying water in minecraft, the best way is to place a water source across the diagonal of the cube, so every block becomes a water source
That's what I immediately thought! And i figured the most efficient way to infect students would be put the infected kids diagonally across the room, just like the way you would fill an area with water in minecraft
The police thief problem at 8:26 is similar to a concept in chess called triangulation where you lose a tempo with your peices and force your opponent to play an unfavorable move due to him having no good move(like the Red Red city of police and the robber in this example)
I found a different way to think about the infection problem: Suppose all in infected students are inside a rectangle smaller then the 5 * 5 grid. Since all healty students outside the rectangle share at most a single border with an infected person, the infection can never spread outside the rectangle. With 4 or fewer infected persons you can allways draw rectangles around them, so that there are remaining students outside the rectangles. Only with 5 infected students, that are placed along the diagonal, the surrounding rectangle is equal to the complete grid, so everyone can be infected (and also will be).
I might have interpreted this wrong but what about 4 people along the diagonal being infected (ignoring the center)? Now you cant draw a rectangle around everyone and have students outside of it.
MajorPrep For the thief problem, what prevents the thief from doing the same thing as the police? That is, go to the neutral city and switch states so once again the police can’t land on the same color city as that of the thief. Seems to me that then there is no guarantee that the thief can be caught.
@karagi101 If the thief has the opportunity to go through that city on their own before the cop then they will change the state of the game to one they'd lose. the cop could just follow them and it'd result in that same pin. But either way from the beginning the cop has the ability to go through the city first and from there the thief couldn't get to it themselves (at least in the layout from the video).
I think this solution works but nerds expanding a little: draw bounding boxes around each subset of students who don't neighbour each other, including diagonal neighbours. Each subset won't be able to infect beyond their bounding box, and with n infected students the biggest bounding box you can construct has an area of n² - so 4 students can't create a 5x5 bounding box, but 5 can.
@@zachstar I think the OP means 'rectangle' not in terms of an outer hull of all the points, but a connected sub-matrix with no 0 rows or columns. In your example of 4 along the diagonal, there will always be 1 row/column which is all 0s, either making a single 4x4 'rectangle', or two disconnected 'rectangles' either a 1x1 & a 3x3 or two 2x2 ones. Each of those separated connected-sub-matrices cannot expand beyond their own borders. The 0 row/column can never become populated with any infected students. Thus, the problem can be thought of in terms of matrices, matrix degrees, and linear independence (at least for figuring out the minimal n; the reasoning doesn't quite work when the # of initial infected students is allowed to be larger than the minimal n, since 0 rows/columns can then sometimes become infected).
2:16 Fun Fact: This puzzle can be visualized in Minecraft using Water since they share the same mechanics. Goal would be to make the hole filled with all source blocks.
1. Thou Shant Lewd Kaori Any wise person could increase the complexity of the problem while preserving that basic odd edge which is the key to solve this problem. In this way your brute force approach shall be in force forever.
Never thought this would become useful, but the infection problem is just minecraft water spread mechanics. At least 2 adjacent water sources are needed to form a new one, and the least amount you can fill a square with is a diagonal of them, which would be the same as any given side. This just multiplies it by 4 and uses the perimeter. That's hilarious
Diagonal is only one possible solution. It can be done without putting any sources on the diagonal. Here's an N=5 solution that doesn't use the diagonals. 0 1 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0
@@sbunny8 Or a more systematic way of doing it is to cover two connected sides of the square like an "L" as in: 00001 00001 00001 00001 11111 But then that would be wasting students since there are infected students that already have two neighbours at the start. So the most efficient way is to draw the "L" without infected students that already have two neighbours: 00001 00000 00001 00000 10101
The way I looked at the disease question is that when you initially select the infected desks, you are immediately choosing the parameters of how far the infection can spread. If you have a 5x5 square, and you put a student on (1,2) a student on (1,4) a student on (2,3) and a student on (2,5) then your parameters are going to be where the farthest x and y points meet. Since you're given a maximum of 4 people to infect, your parameters can never exceed 4x4, or 16 squares.
and the furthest you can place your edges are at the corners. But you can’t have students at just (1,1) (2,2) (4,4) and (5,5), because they would just create 2 x 2 squares. You have to connect them which would mean you would have to have five students at least
@@Mr48two if that happened he would still be caught because if he touches the red then the police is on either neutral or and it will be your move or he will have to turn back and get cornered
Imagine someone getting caught and not being able to understand why they caught him... "But the math checks out!" Would be another case to study the Dunning-Kruger effect
Thank you so much! I've tried with at least a dozen classrooms by know and I couldn't figure out how there were always survivors! I've used this info four times now and it has worked without fail!
This was a surprisingly insightful video, connecting the simple idea of whether something is possible, to how to identify and understand invariants, to the final revelation of Euler's Characteristic, which I've seen before but never really understood. Thanks!! 😊 You're on par with 3Blue1Brown, Mathologer, et al., but a little briefer, more collected, and more to-the-point. It makes your videos a little easier to digest, not requiring quite as much time spent. Truly great work, MajorPrep! Thanks again!
For anyone wondering how the V-E+F=2 works, here's a short explanation: When there's only 1 vertex, there is obviously no edges and the only face is the outer section, so V-E+F=1-0+1=2. If you add another vertex a new edge is also created, so the sum doesn't change. You can keep adding new vertices like this, and as long as you don't connect any old ones V-E is always going to stay the same (and since you're not creating any new faces F is not going to change either). If you connect two existing vertices, though, *only* an edge is going to be created - thus V-E is going to decrease by one. Meanwhile, a face is inevitably going to be created, and the total sum is going to remain two.
For the students infected one, I'm surprised he didn't mention that the students have to be in a continuous diagonal line for maximum infection. And that the students form the edge of the square/rectangle, so it's impossible for the infection to move outward
Labelling rows with letter and columns with numbers, place infected students at a1 c1 e1 e3 and e5. This would create an L shape and the rest of the students would be infected spreading from the corner
@@BryanLu0even other "shapes" work 00100 00000 10101 00000 00100 Works too But I think the l shape, is the on, that Takes the longes time to reach every cell
At 14:56 you could also just teleport right away and start from the outside point and weave your way around. Both ways work because you're starting at an odd numbered node.
@@zachstar Lmao. It's all good. Always interesting to see new ways to view and approach a problem. Even ones where it's not complicated, but ideas can vary. Also, I would've done the same thing too lol. Love your videos. Keep up the good work, but dont overwork yourself.
The reasoning behind why it's possible with the teleport hole, is that it just makes the two "nodes" the same (since you can travel between them without restrictions), so you can just add their degree and get 14, which is even. Then you just start in one of the odd nodes :).
Anyone who plays chess can solve this puzzle 8:56 easily, we have the same concept in our game. Sometimes enemy king (let's call it black) is in the worse position, but you can't pushem away or change the situation on a board with your king, because black repeats your moves and don't let your king walk in. This triangle in the right bottom is the key, in chess we call that triangulation, when black can repeatedly do only 2 moves to repeat the position, but you can do 3 and so make the black go away. Surprised that knowledge of chess helped me with some math problem, chess actually is a great game to sharpen your mind because quite often you've got to solve such puzzles on a board to win. Great video btw, it took me several hours to watch it and solve everything
For the first problem, you can connect all dots in 3 lines. Just connect all the columns then tilt the lines and expand. The line will then intercept at some point
As always, awesome video, man! I usually save your videos to watch them right after I wake up. Helps me wake up with the perspective I want for the day.
It has probably already been said, but something like the problem about the thief actually happens a lot in chess. When you reach the latest stages of the game with very few pieces (kings and pawns), you can find yourself in positions that would benefit one side or another depending on who has to move next. So the strategy involved is to "lose a tempo/move" by triangling with your king (using 3 moves to come back to the same spot, instead of 2) so you get the same position handing the turn to the other player and therefore forcing him to let you some space on the board. It's kind of hard to portrait really, but it is not that of an advanced tip you learn in chess, but still it made me instantly find the solution for this problem
I subbed three years ago when I decided to go into engineering and didn't know what to major in (ended up choosing electrical). Glad I stayed subbed because your videos are becoming more and more interesting. Happy to see that you have gained so many subscribers because you seriously have such high content quality.
I came up with a slightly different method for the infected students. I noted that in order to completely span the length of a side, you need at least n/2 + 1/2 students, arranged in a X O X O X straight line. (I think for Even n, it's n/2 + 1, X O X O X X). You need to do this for horizontal, and vertical, otherwise the infection won't be able to reach every row/column. While you can share an infected student between the two lines (e.x. lines are top row and leftmost column, sharing the top left corner infected student), that still leaves n/2 + 1/2 + n/2 - 1/2 = n. For evens, it works out to n+1. So, it's not possible to infect all students with fewer than n infections.
The student infection question was very intuitive to me because its how water works in minecraft, if 2 water sources flow immedietly into an adjacent square, they become a water source block, and the most common way to fill a pool is to place water on 2 edges, which automatically fills the center and the opposite sides.
1:53 This problem actually has a pretty interesting application in Minecraft: Water blocks technically have two different states: Source water blocks that can be collected, and rushing water blocks that cannot. Rushing water can be converted to source water using the same exact rules shown, and it is also proven there that the absolute minimum source water blocks needed to fill a pool of nxn size has to be n blocks.
The first one was a challenge our math teacher once gave us. Me, not knowing the point of the riddle, just connected three points downwards, and then FOLDED the paper, so that the line actually connected all nine. I just thought of controlling space to get what I want, because I didn't got it right. Big Brain or dumb failure of the rules? He said its good. Sorry for the bad sentences, I usually don't speak english very often.
You could also solve it by backtracking the pencil along existing line segments. You're not taking the pencil off of the paper and are not creating additional line segments, merely re-tracing existing segments.
Couple of alternative solutions I came up with: For the infection grid one, another way to solve it is to recognise that the infection can only be shared horizontally or vertically. Therefore, for any given student to become infected they must share a row of column with an already infected student. Some students that do share a row or column with an infected student may not become infected, but in any case where a student does not share a row nor a column with any other infected student, they can not become infected. Since there are five rows and five columns, we cannot place four infected points in such a way that this is satisfied. And for the door one, my solution was very similar to yours, I just didn't know the technical mathematics for it. I first analysed a room with 5 doors. I noticed that if we do not start in that room, we would eventually become trapped there. This is because we would, at some point, enter through one door, and exit through another. Then, however later, we will again enter through a third door, then exit through a fourth When we next enter that room, we would be trapped. However, this is not a problem if this is our final room. Nor is it a problem if it is our first room, because, we would exit the room first, rather than entering, making the pattern exit, enter, exit, enter, exit, and we can successfully leave. However, there are 3 rooms with 5 doors. We can only have one be the start and one be the finish, leaving the other the trap us forever. Then we starve to death.
Another way to prove the infected students involves looking at efficiency. The most efficient use of two tiles would be the max amount of squares two sources can infect. If you look at all possible arrangements of two sources, then the most efficient path would be the two squares connected by a corner, as this leads to two new infected. The new infections will always be generated inwards. This means that for efficiency, you would need to start at any corner, and work your way to the other. Since there will never be an infection on the sides that do not lead to a corner with another infected, there can never be spread in that direction, meaning that the most efficient amount of starting tiles (the smallest amount tiles to fill in the square), must be n.
3:20 the infection can never grow outward towards the edge without there being an initial infection, so you must have infected nodes on the edges, then work from there
For the cop and robber one, I saw that bottom corner and realized I could catch him there, but gave up when i couldn’t think of a way to force him there, I didnt quite think of that
For the infection example: the perimeter solution is a nice mathematical proof for it, but you could also word it differently. Given any n×n grid, the number of infected students must be equal to or greater than the number of diagonals in the grid. The diagonal configuration means that you'll start of with all sides of an infected square adjacent to a non-infected one, meaning no edges are shared between infected squares. It also means that at least 6 non-infected squares will be infected, then 4, then 2. Funnily enough, the only reason I know of the solution I outlined above was due to a slightly different problem for someone I was tutoring in a math class geared towards teachers teaching math for K-5.
The example at 10:00 is basically opposition in Chess. You triangulate with your King to end up 2 squares away from the opposing King to force them to either move back or move aside, similar to syncing up with the same type of city as the thief. The cornering of the thief is another Chess concept called zugzwang. The thief has no good moves to make since every move is a forced loss. You use opposition to force the opponent into zugzwang in both.
My mother once had me try to solve the 1st puzzle on this list and while it did take me like 20 minutes (I refused to give up!), the phrase "thinking outside the box" occurred to me as I was taking a bathroom break (lol), so I asked my mother if the dots were restrictive. As in, was I ONLY allowed to draw directly from dot to dot. She said no, I didn't have to... after which it pretty much solved itself xP. But yeah, that even occurring to me did take a long time...
When you add the hole in the house problem you are merging the 2 nodes, basically adding up the numbers on them. Thus you remove 2 odd vertices adding an even one and the graph is traversable.
There is actually a much simpler solution to the room problem as discussed in the video that I tried to work out while the video was paused. The rooms in the video can be put in two categories: Rooms with an odd number of doors and rooms with an even number of doors. Now, considerwhat would happen when you tried to 'complete' one of these rooms in isolation. Within each room, you MUST alternate between leaving the room and entering the room, because you cannot leave a room two times in a row. Now, consider the even-numbered room. There are two starting positions: inside and outside. Because you must always enter after you leave, or leave after you enter, you are either left inside the room if you started inside, or outside the room if you started outside. Now, however, consider the odd numbered room. If you start inside, your final move must always be to leave, and if you start outside, your final move must always be to enter, so this means that you always have to start inside the odd numbered room and outside the even numbered room if you don't want to get stuck inside each room. However, the problem contains 3 rooms with an odd number of doors, so the problem is impossible because you can only complete 2 odd-numbered rooms before getting stuck indefinitely, completing one by starting inside it and escaping and completing the other last by getting stuck inside. The hole fixes the problem because it allows you to 'transform' 2 odd-numbered rooms into even numbered rooms, while also allowing you to start outside those even-numbered rooms as to prevent getting stuck.
the problem at 5:18 (infected students) felt realy easy to me. I could just tell how it would work for some reason. Then I realized, it is the same as water spreading mechanics in minecraft.
Pls pls pls read this bcoz I took a long time to think it and literally, I am an 11yr old. 11:07 I can draw it Without taking the pencil off the paper or without drawing the same edge twice I will use a vertex twice bcoz it wasn't mentioned It is a pentagon Let the topmost point be A The down left one B The down right one C The one in the middle of the pentagon D The one between A and B will be E And the one between A and C will be F So start from F Go to E Go to A Go to D Go to B Go to C Again go to F Then go to A Ta da!!!!
The infected students are like water source blocks in Minecraft. It seems that many Minecrafters don't know how many water source blocks they need to make a pool of water. A diagonal line of source blocks will fill a square pool, but when you want to extend the square to a rectangle, you only need 1 source block for every 2 new rows. So for an X by Y pool, with X being greater than Y, you only need X+(Y-X)/2 (rounded up) water source blocks. Although you can start with only 2 buckets of water and never need to worry about running out because you can scoop up water in the process of making the pool, you still waste time if you do more water collection and placement than necessary.
as a computer science major this is giving me so much painful flashbacks to regular expressions and algorithmic math then having to write proofs for them
The way i see the box problem is this (sorry for my english) Disease can spread only in a boundaries of your X-es, like if you could draw a rectangle in the box that includes all the X-es, disease cannot spread more than that, and so with 4 X-es you can cover maximum 4x4 square, and its not enough for spreading disease to all
This puzzle was in Soviet magazine "Sciens and life" I think about 1976 - it's easy ,the lines must go a little bit further to form a triangle with a line in centre. But there was also a genius solution that if you imagine that these dots are lifting in the air and look at them above - only one line would be enough 😊
The structure of the sick students puzzle is actually the same as the water source mechanics in Minecraft. And my gamer brain immediately went "No, I obviously need at least 5 original sources, down the diagonal" :D
0:34 You can impress your friends even more by following up with the question: "now, do it with THREE straight lines". Hint: you need large-enough paper or large-enough dots to do it.
It’s funny because the infected problem is exactly like trying to create water sources in Minecraft. So I have a natural thought process on that already. You put them diagonally and that’s the most efficient way to cover a large area but it doesnt reach the end.
and I was watching the node-stuff asking myself the whole time: will he mention Euler at some point... wasn't disappointed but had to wait quite a while...
for the question with the students, an easy way to look at it is to imagine minecraft water sources, imagine a 5x5 water grid, you can't make the full grid source blocks with only 4 buckets of water.
There's a version of the thief/police problem in one game community I'm in, called "parity". In the game Crypt of the Necrodancer you're on a square grid and can move once per beat, and there are some enemies that also move every beat after you do. So if you're on the same color tile as them on a checkerboard grid, you can't approach them safely, but if you're on the opposite color you always can. The solution is generally to do some action that lets you stay on the same tile for one move, so you can get next to them and attack. If anyone is interested, look up how to kill the red bat in necrodancer, there's a good video that shows the concept and explains it better than I could.
Do the infected problem like you would fill a large area of water in minecraft 1) Put one in the corner 2) Expands from the two edges, but to save space, alternate between infected and non-infected 3) Profit However, you need 5 infecteds to do this, therefore, it is impossible.
I have discrete math course flashbacks. All of this is fun and games, until you are graded for it.
ikr
My discrete final is in 10 hours, very much so not looking forward to it :|
It shouldn't be that way... School conditions us to be scared of math but once you actually look at it from a fun and games angle it's actually really interesting and useful in certain applications.
Yea
@@melaniemedina8037 Must depend on the school/teachers/kids. Nobody was "scared" of math at my school. Maybe some who were less than average but that's about it
"Is it possible to start with 4 students and infect the whole class?"
2020: Say no more.
worse ror what happened to this guy?
worse ror
Patient Zer0 lives while thousands of other people died...
Quite ironic
666 likes, seems fitting
@@Zack_Zander whos Patient 0
@@DanksterPaws I'm pretty sure patient zero just refers to the first human recorded to have a particular disease
3:20 imagine pausing the video to try to solve it for like an hour jsut to come back, unpause it, and see its impossible
but if you're smart enough you will understand the impossibility and start trying to prove it
Who would do that
That is comendable engagement, and you likely understand why better than those who just shake their heads in agreement. Stay a skeptic in all matters!
if you couldn't figure out that it's impossible then you couldn't solve it
Mohammed jawahri that doesn’t mean u can’t solve it
If you get a pen thicc enough, you can connect all of them with 1 line
If the dots are big enough, you can use only three really long lines at slight angles.
@@ryancw714 Or if the piece of paper is large enough
Or if all dots are overlapped
Or if you can fold the paper
A line has no width.
1:53 the third problem is actually how water works in Minecraft
C o p i e d
qsaification I was going to write the exact same thing!!!
**looks at number of likes** nice
@@user-go5he1fe1g It is actually pretty obvious to those who have played Minecraft, that's the first thought that came into my mind. It doesn't surprise me if you saw comments similar to mine :)
@@qsaification oh k
The infected question is actually similar to multiplying water in minecraft, the best way is to place a water source across the diagonal of the cube, so every block becomes a water source
That's what I immediately thought! And i figured the most efficient way to infect students would be put the infected kids diagonally across the room, just like the way you would fill an area with water in minecraft
I love that that’s what I thought of as well
Just wanted to comment that!
Ikr!!
life imitates art..
there is something unique about this channel
It has 1 odd node.
Yeah. It's like a hybrid of 3b1b n socratica but something more
Everything about this channel is perfect but what makes it unique is HIM ⬆️
Yeah - confusion
Yeah it can predict the future
3:16 that moment when you've placed enough water sources in Minecraft to know the answer immediately.
I was thinking same
The classroom problem is the infinite water source problem in minecraft.
Just sayin’...
Yooo thats what I was thinking.. spooky
Wie alle Mainguaftler das selbe denken xd
I thought that too :D
_uses Minecraft to simulate this problem_
Damn all you, I thought I was original.
Theory: whenever someone says "no pun intended" in a video/presentation... The pun was intended.
3:06 and that's why you sit in the corner of the class room.
Best chance to survive a zombie apocalypse
@@mayocult2025 taking classes online is an even better strategy
@@HPD1171 true
@@HPD1171 now we all have to take online classess
HPD1171 Welp,
The police thief problem at 8:26 is similar to a concept in chess called triangulation where you lose a tempo with your peices and force your opponent to play an unfavorable move due to him having no good move(like the Red Red city of police and the robber in this example)
I was thinking about that as well haha, came to the timed comment section to see if anyone else spotted it.
Do you play chess?
th-cam.com/video/bE6ENkdnG0c/w-d-xo.html
Puzzle 3 in this video is the most insane version of this ever.
@@null3007me too haha
In germany we call this Zugzwang
Can't believe you didn't mention triangulation in chess in the thief/cop problem.
This problem reminded me exactly the same thing
Magnus carlsen has joined the chat.
Yes, I thought of the same.
I thought exacly the same too !
Bobby Fischer has entered the chat
13:31 ah yes, I didn't see the answer until you made that simple shape! Thank you
I found a different way to think about the infection problem: Suppose all in infected students are inside a rectangle smaller then the 5 * 5 grid. Since all healty students outside the rectangle share at most a single border with an infected person, the infection can never spread outside the rectangle. With 4 or fewer infected persons you can allways draw rectangles around them, so that there are remaining students outside the rectangles. Only with 5 infected students, that are placed along the diagonal, the surrounding rectangle is equal to the complete grid, so everyone can be infected (and also will be).
I might have interpreted this wrong but what about 4 people along the diagonal being infected (ignoring the center)? Now you cant draw a rectangle around everyone and have students outside of it.
MajorPrep For the thief problem, what prevents the thief from doing the same thing as the police? That is, go to the neutral city and switch states so once again the police can’t land on the same color city as that of the thief. Seems to me that then there is no guarantee that the thief can be caught.
@karagi101 If the thief has the opportunity to go through that city on their own before the cop then they will change the state of the game to one they'd lose. the cop could just follow them and it'd result in that same pin. But either way from the beginning the cop has the ability to go through the city first and from there the thief couldn't get to it themselves (at least in the layout from the video).
I think this solution works but nerds expanding a little: draw bounding boxes around each subset of students who don't neighbour each other, including diagonal neighbours. Each subset won't be able to infect beyond their bounding box, and with n infected students the biggest bounding box you can construct has an area of n² - so 4 students can't create a 5x5 bounding box, but 5 can.
@@zachstar I think the OP means 'rectangle' not in terms of an outer hull of all the points, but a connected sub-matrix with no 0 rows or columns. In your example of 4 along the diagonal, there will always be 1 row/column which is all 0s, either making a single 4x4 'rectangle', or two disconnected 'rectangles' either a 1x1 & a 3x3 or two 2x2 ones. Each of those separated connected-sub-matrices cannot expand beyond their own borders. The 0 row/column can never become populated with any infected students. Thus, the problem can be thought of in terms of matrices, matrix degrees, and linear independence (at least for figuring out the minimal n; the reasoning doesn't quite work when the # of initial infected students is allowed to be larger than the minimal n, since 0 rows/columns can then sometimes become infected).
"Is this possible to infect all of the students?"
"Answer: *NO"*
Coronavirus: Hold my Chinese..
Holy crap that was funny how did no one like
@@collindrummer8116 cuz they're chinese
@@collindrummer8116 because this video is months old and this got commented only a day ago?
Ah, math that i dont understand.
Perfect.
h
it was explained perfectly thou
U actually reminded me of me math hw for tomorrow, god bless you
Why, it's actually easy
Yes marvelous
2:16 Fun Fact: This puzzle can be visualized in Minecraft using Water since they share the same mechanics. Goal would be to make the hole filled with all source blocks.
The police-theif game at 8:26 can win you a bet
How exactly will you propose the bet?
Be this thief so I can prove to you that I can capture you as a cop for $x
@@NameisU Or I bet the cop will/will not catch the thief
Can it really? how dumb are people? you can solve it with very minimal bruteforcing/pattern recognition...
@@NameisU Yes, what elde would the bet be? lmao
1. Thou Shant Lewd Kaori Any wise person could increase the complexity of the problem while preserving that basic odd edge which is the key to solve this problem. In this way your brute force approach shall be in force forever.
The problem of the infected students could be viewed as Minecraft water sources spreading. It's really the same concept.
Actually yes bro it's same concept 👍
LMAO didn't think of that
Never thought this would become useful, but the infection problem is just minecraft water spread mechanics. At least 2 adjacent water sources are needed to form a new one, and the least amount you can fill a square with is a diagonal of them, which would be the same as any given side. This just multiplies it by 4 and uses the perimeter.
That's hilarious
@@qbwkp because Minecraft is an accurate model of the universe.
Just VACCINATE YOUR CHILDREN
Diagonal is only one possible solution. It can be done without putting any sources on the diagonal.
Here's an N=5 solution that doesn't use the diagonals.
0 1 0 0 0
1 0 0 0 1
0 0 0 1 0
0 0 0 0 0
0 1 0 0 0
@@erinelizabeth9545 Someone mentions Minecraft water physics
You:
@@sbunny8 Or a more systematic way of doing it is to cover two connected sides of the square like an "L" as in:
00001
00001
00001
00001
11111
But then that would be wasting students since there are infected students that already have two neighbours at the start. So the most efficient way is to draw the "L" without infected students that already have two neighbours:
00001
00000
00001
00000
10101
The way I looked at the disease question is that when you initially select the infected desks, you are immediately choosing the parameters of how far the infection can spread. If you have a 5x5 square, and you put a student on (1,2) a student on (1,4) a student on (2,3) and a student on (2,5) then your parameters are going to be where the farthest x and y points meet. Since you're given a maximum of 4 people to infect, your parameters can never exceed 4x4, or 16 squares.
Like water source blocks in Minecraft
and the furthest you can place your edges are at the corners. But you can’t have students at just (1,1) (2,2) (4,4) and (5,5), because they would just create 2 x 2 squares. You have to connect them which would mean you would have to have five students at least
Idk if it was assumed that the classroom was a square, but my solution was: square root of area - one
I CAN DO THE CONNECT THE 9 DOTS IN 1 LINE!!!
*Uses a crayon without the wrapper and uses the side*
A line has no width in geometry.
@@windows_sky r/ihavereddit
Y’all eat your crayons with or without the wrapper?
@@avamatthews1459 you monster
@@avamatthews1459 imagine eating the crayons 1 and 1
2:27
*MINECRAFT WATER PHYSICS WANTS TO KNOW YOUR LOCATION*
Only be a robber in states with an even number of cities!
Life hack...
What?
But what if the robber followed the police into that void?
@@Mr48two if that happened he would still be caught because if he touches the red then the police is on either neutral or and it will be your move or he will have to turn back and get cornered
Imagine someone getting caught and not being able to understand why they caught him... "But the math checks out!" Would be another case to study the Dunning-Kruger effect
gotcha i.imgur.com/RYDDp16.png
Thank you so much! I've tried with at least a dozen classrooms by know and I couldn't figure out how there were always survivors! I've used this info four times now and it has worked without fail!
This was a surprisingly insightful video, connecting the simple idea of whether something is possible, to how to identify and understand invariants, to the final revelation of Euler's Characteristic, which I've seen before but never really understood. Thanks!! 😊
You're on par with 3Blue1Brown, Mathologer, et al., but a little briefer, more collected, and more to-the-point. It makes your videos a little easier to digest, not requiring quite as much time spent. Truly great work, MajorPrep! Thanks again!
If we bring the question at 3:13 to Minecraft "Can you use only 4 water bucket to fill the 5x5 grid?" it will be a lot easier to think
For anyone wondering how the V-E+F=2 works, here's a short explanation:
When there's only 1 vertex, there is obviously no edges and the only face is the outer section, so V-E+F=1-0+1=2. If you add another vertex a new edge is also created, so the sum doesn't change. You can keep adding new vertices like this, and as long as you don't connect any old ones V-E is always going to stay the same (and since you're not creating any new faces F is not going to change either).
If you connect two existing vertices, though, *only* an edge is going to be created - thus V-E is going to decrease by one. Meanwhile, a face is inevitably going to be created, and the total sum is going to remain two.
For the students infected one, I'm surprised he didn't mention that the students have to be in a continuous diagonal line for maximum infection. And that the students form the edge of the square/rectangle, so it's impossible for the infection to move outward
Labelling rows with letter and columns with numbers, place infected students at a1 c1 e1 e3 and e5. This would create an L shape and the rest of the students would be infected spreading from the corner
@@BryanLu0 Good find!
It don't have to be diagonal for maximum efficiency
@@BryanLu0even other "shapes" work
00100
00000
10101
00000
00100
Works too
But I think the l shape, is the on, that Takes the longes time to reach every cell
The infected students thing is basically the same as filling up a hole in Minecraft such that the entire hole is filled with source blocks
still_lava115 ya you need water source on diagonal, iirc
functionally identical if you assume water cannot be picked back up and you only have 4 water buckets
No, no, they just close the schools at that point. Believe me, we know.
-The Future
Let's get 20 dislikes too, so it's also signed 2020.
At 14:56 you could also just teleport right away and start from the outside point and weave your way around. Both ways work because you're starting at an odd numbered node.
Yep! I actually saw that after having the animations done but didn’t want to change it lol
@@zachstar Lmao. It's all good. Always interesting to see new ways to view and approach a problem. Even ones where it's not complicated, but ideas can vary. Also, I would've done the same thing too lol. Love your videos. Keep up the good work, but dont overwork yourself.
The reasoning behind why it's possible with the teleport hole, is that it just makes the two "nodes" the same (since you can travel between them without restrictions), so you can just add their degree and get 14, which is even. Then you just start in one of the odd nodes :).
So at 14:38 is the puzzle solvable? Or do the odd nodes need to be adjacent to either each other or the worm hole?
Anyone who plays chess can solve this puzzle 8:56 easily, we have the same concept in our game. Sometimes enemy king (let's call it black) is in the worse position, but you can't pushem away or change the situation on a board with your king, because black repeats your moves and don't let your king walk in. This triangle in the right bottom is the key, in chess we call that triangulation, when black can repeatedly do only 2 moves to repeat the position, but you can do 3 and so make the black go away. Surprised that knowledge of chess helped me with some math problem, chess actually is a great game to sharpen your mind because quite often you've got to solve such puzzles on a board to win. Great video btw, it took me several hours to watch it and solve everything
My favourite TH-cam channel, you completely change the way I view mathematics and it’s applications
For the first problem, you can connect all dots in 3 lines. Just connect all the columns then tilt the lines and expand. The line will then intercept at some point
What do you mean?
As always, awesome video, man! I usually save your videos to watch them right after I wake up. Helps me wake up with the perspective I want for the day.
It has probably already been said, but something like the problem about the thief actually happens a lot in chess. When you reach the latest stages of the game with very few pieces (kings and pawns), you can find yourself in positions that would benefit one side or another depending on who has to move next. So the strategy involved is to "lose a tempo/move" by triangling with your king (using 3 moves to come back to the same spot, instead of 2) so you get the same position handing the turn to the other player and therefore forcing him to let you some space on the board. It's kind of hard to portrait really, but it is not that of an advanced tip you learn in chess, but still it made me instantly find the solution for this problem
I subbed three years ago when I decided to go into engineering and didn't know what to major in (ended up choosing electrical). Glad I stayed subbed because your videos are becoming more and more interesting. Happy to see that you have gained so many subscribers because you seriously have such high content quality.
Thank you for the comment and glad you've stuck around!
I came up with a slightly different method for the infected students. I noted that in order to completely span the length of a side, you need at least n/2 + 1/2 students, arranged in a X O X O X straight line. (I think for Even n, it's n/2 + 1, X O X O X X).
You need to do this for horizontal, and vertical, otherwise the infection won't be able to reach every row/column. While you can share an infected student between the two lines (e.x. lines are top row and leftmost column, sharing the top left corner infected student), that still leaves n/2 + 1/2 + n/2 - 1/2 = n. For evens, it works out to n+1.
So, it's not possible to infect all students with fewer than n infections.
This topic turned out to be very interesting, although it did not seem to me at first. Thank!
This came up in my recommended and the infected problem felt too real, very informative though good work!
Superb! Enjoyed that.
I love you, your content and your account, thank you so much for everything, hug from Portugal ❤
Is it possible? apparently sometimes it is and sometimes it isnt...
Who could have known
Great video btw
WHAAATTT THEEE *BEEP*
It's my first video on this channel and how come I not realise till 7:49, its Zach
This channel surprisingly gives me feels like the Vsauce used to do.
all hail vsauce
Watch Ted Ed
I'm pretty sure that I saw an video with these exactly same problems in one of the Michael's channels.
@@KenKelvinN he has spread his contents over so many channels and its so hard to pinpoint this in his contents.
The student infection question was very intuitive to me because its how water works in minecraft, if 2 water sources flow immedietly into an adjacent square, they become a water source block, and the most common way to fill a pool is to place water on 2 edges, which automatically fills the center and the opposite sides.
lol the chasing the thief problem is so obvious to anyone who's played roguelikes
Exactly what I was thinking lol
It's past midnight. I should be sleeping. But first, let's solve problems I will never encounter.
11:00 we litterally learned this in my a level further maths class today
1:53 This problem actually has a pretty interesting application in Minecraft:
Water blocks technically have two different states: Source water blocks that can be collected, and rushing water blocks that cannot.
Rushing water can be converted to source water using the same exact rules shown, and it is also proven there that the absolute minimum source water blocks needed to fill a pool of nxn size has to be n blocks.
The first one was a challenge our math teacher once gave us.
Me, not knowing the point of the riddle, just connected three points downwards, and then FOLDED the paper, so that the line actually connected all nine. I just thought of controlling space to get what I want, because I didn't got it right.
Big Brain or dumb failure of the rules? He said its good.
Sorry for the bad sentences, I usually don't speak english very often.
I guess you were allowed to lift the pencil off the paper then
You could also solve it by backtracking the pencil along existing line segments. You're not taking the pencil off of the paper and are not creating additional line segments, merely re-tracing existing segments.
@Tom Petitdidier Thank you for your help!
Couple of alternative solutions I came up with:
For the infection grid one, another way to solve it is to recognise that the infection can only be shared horizontally or vertically. Therefore, for any given student to become infected they must share a row of column with an already infected student. Some students that do share a row or column with an infected student may not become infected, but in any case where a student does not share a row nor a column with any other infected student, they can not become infected. Since there are five rows and five columns, we cannot place four infected points in such a way that this is satisfied.
And for the door one, my solution was very similar to yours, I just didn't know the technical mathematics for it. I first analysed a room with 5 doors. I noticed that if we do not start in that room, we would eventually become trapped there. This is because we would, at some point, enter through one door, and exit through another. Then, however later, we will again enter through a third door, then exit through a fourth
When we next enter that room, we would be trapped. However, this is not a problem if this is our final room. Nor is it a problem if it is our first room, because, we would exit the room first, rather than entering, making the pattern exit, enter, exit, enter, exit, and we can successfully leave. However, there are 3 rooms with 5 doors. We can only have one be the start and one be the finish, leaving the other the trap us forever. Then we starve to death.
when you intuitively know the proof around 4:00 because its the same as minecraft water physics
Another way to prove the infected students involves looking at efficiency.
The most efficient use of two tiles would be the max amount of squares two sources can infect. If you look at all possible arrangements of two sources, then the most efficient path would be the two squares connected by a corner, as this leads to two new infected. The new infections will always be generated inwards.
This means that for efficiency, you would need to start at any corner, and work your way to the other. Since there will never be an infection on the sides that do not lead to a corner with another infected, there can never be spread in that direction, meaning that the most efficient amount of starting tiles (the smallest amount tiles to fill in the square), must be n.
2:22 Ah the good old minecraft pool problem
There's a more intuitive way to resolve it though
The same goes with the thief problem
Did anyone else notice the infected student puzzle is basically just minecraft water source creation?
2:42 See also: Minecraft water source generation
3:20 the infection can never grow outward towards the edge without there being an initial infection, so you must have infected nodes on the edges, then work from there
"simple question, not so simple question"
Seems like exam questions
For the cop and robber one, I saw that bottom corner and realized I could catch him there, but gave up when i couldn’t think of a way to force him there, I didnt quite think of that
I believe the "9 dot game" is where the term "think outside the box" actually comes from, so you were actually right on point.
For the infection example: the perimeter solution is a nice mathematical proof for it, but you could also word it differently.
Given any n×n grid, the number of infected students must be equal to or greater than the number of diagonals in the grid. The diagonal configuration means that you'll start of with all sides of an infected square adjacent to a non-infected one, meaning no edges are shared between infected squares. It also means that at least 6 non-infected squares will be infected, then 4, then 2.
Funnily enough, the only reason I know of the solution I outlined above was due to a slightly different problem for someone I was tutoring in a math class geared towards teachers teaching math for K-5.
When watching the part at 4:10 i just kept thinking about infinite water sources in minecraft
The example at 10:00 is basically opposition in Chess. You triangulate with your King to end up 2 squares away from the opposing King to force them to either move back or move aside, similar to syncing up with the same type of city as the thief. The cornering of the thief is another Chess concept called zugzwang. The thief has no good moves to make since every move is a forced loss. You use opposition to force the opponent into zugzwang in both.
My mother once had me try to solve the 1st puzzle on this list and while it did take me like 20 minutes (I refused to give up!), the phrase "thinking outside the box" occurred to me as I was taking a bathroom break (lol), so I asked my mother if the dots were restrictive. As in, was I ONLY allowed to draw directly from dot to dot. She said no, I didn't have to... after which it pretty much solved itself xP. But yeah, that even occurring to me did take a long time...
When you add the hole in the house problem you are merging the 2 nodes, basically adding up the numbers on them. Thus you remove 2 odd vertices adding an even one and the graph is traversable.
There is actually a much simpler solution to the room problem as discussed in the video that I tried to work out while the video was paused. The rooms in the video can be put in two categories: Rooms with an odd number of doors and rooms with an even number of doors. Now, considerwhat would happen when you tried to 'complete' one of these rooms in isolation. Within each room, you MUST alternate between leaving the room and entering the room, because you cannot leave a room two times in a row. Now, consider the even-numbered room. There are two starting positions: inside and outside. Because you must always enter after you leave, or leave after you enter, you are either left inside the room if you started inside, or outside the room if you started outside. Now, however, consider the odd numbered room. If you start inside, your final move must always be to leave, and if you start outside, your final move must always be to enter, so this means that you always have to start inside the odd numbered room and outside the even numbered room if you don't want to get stuck inside each room. However, the problem contains 3 rooms with an odd number of doors, so the problem is impossible because you can only complete 2 odd-numbered rooms before getting stuck indefinitely, completing one by starting inside it and escaping and completing the other last by getting stuck inside. The hole fixes the problem because it allows you to 'transform' 2 odd-numbered rooms into even numbered rooms, while also allowing you to start outside those even-numbered rooms as to prevent getting stuck.
This is exactly the same logic as in the video. There can only be 2 odd numbered rooms, but the problem has 4. (Remember the outside counts as 1)
4:40 only 3*3 is infected but only with one recursion and the other is increase in log
the problem at 5:18 (infected students) felt realy easy to me. I could just tell how it would work for some reason. Then I realized, it is the same as water spreading mechanics in minecraft.
I knew solving it felt familiar. Making aquariums in Minecraft is useful.
It just surprised my brain that I already knew the answer of the infected students problems because that's how water sources spread in Minecraft
0:20 There can be thicc line wich connect all dots xD
Pls pls pls read this bcoz I took a long time to think it and literally, I am an 11yr old.
11:07
I can draw it
Without taking the pencil off the paper or without drawing the same edge twice
I will use a vertex twice bcoz it wasn't mentioned
It is a pentagon
Let the topmost point be A
The down left one B
The down right one C
The one in the middle of the pentagon D
The one between A and B will be E
And the one between A and C will be F
So start from F
Go to E
Go to A
Go to D
Go to B
Go to C
Again go to F
Then go to A
Ta da!!!!
Aww, you tried, but now you can't draw the line from B to E.
@@santoriomaker69 that's right sir
I tried my best though
9:48 is like triangulation of the king in chess
The infected students are like water source blocks in Minecraft. It seems that many Minecrafters don't know how many water source blocks they need to make a pool of water. A diagonal line of source blocks will fill a square pool, but when you want to extend the square to a rectangle, you only need 1 source block for every 2 new rows. So for an X by Y pool, with X being greater than Y, you only need X+(Y-X)/2 (rounded up) water source blocks. Although you can start with only 2 buckets of water and never need to worry about running out because you can scoop up water in the process of making the pool, you still waste time if you do more water collection and placement than necessary.
So he said no pun intended for the opening problem, but that's literally where "thinking outside the box" came from.
No it's not
@@Ten_Thousand_Locusts mate, yes it is.
@@evanbutler1396 it's literally not
@@Ten_Thousand_Locusts then what is the origin? I've presented a point, all you've said is I'm wrong without presenting any evidence to the contrary.
@@evanbutler1396 burden of proof isn't on me dumbass, you're the one making ridiculous statements out here not me.
Zach on the house problem: “this would be difficult to brute force
Me who gets it first try: “yes, this is big brain time”
2:08 everyone who knows how water spreads in minecraft has an intuitive understanding of this problem
I dont have Minecraft
L L uncultured
@@LL-pl2ek it's 25 dollars poor boy!!!! get outta here workin class bitch!!!!
@@wewladstbh
Did you think baby me could?
@@LL-pl2ek I didn't realise I was in contact with an infant! My apologies!
as a computer science major this is giving me so much painful flashbacks to regular expressions and algorithmic math then having to write proofs for them
3:35 beaver flu: a rare variant of bird flu that lasts forever
*uh oh*
Oh Yes
The way i see the box problem is this (sorry for my english)
Disease can spread only in a boundaries of your X-es, like if you could draw a rectangle in the box that includes all the X-es, disease cannot spread more than that, and so with 4 X-es you can cover maximum 4x4 square, and its not enough for spreading disease to all
You were just 5 months too early with the classroom example. It is now our reality.
Funny thing is in my mother tongue no one explained it to me as clear as you did :) I might even start to like maths now - thank you so much!
"It is impossible to start with 4 students and infect the whole class"!!!
Corona : " Hold my beer , Please 🍻🍻🍻
This puzzle was in Soviet magazine "Sciens and life" I think about 1976 - it's easy ,the lines must go a little bit further to form a triangle with a line in centre. But there was also a genius solution that if you imagine that these dots are lifting in the air and look at them above - only one line would be enough 😊
1:07 this is some weird minesweeper
The structure of the sick students puzzle is actually the same as the water source mechanics in Minecraft. And my gamer brain immediately went "No, I obviously need at least 5 original sources, down the diagonal" :D
I’m sorry, but… What did I just learn?
The problem at 8:43 reminded me of the opposition in chess and how often you need to triangulate the king in alot of pawn endgames
3:42 ... one of the instructors last name is " Van Dijk " !
0:34 You can impress your friends even more by following up with the question: "now, do it with THREE straight lines". Hint: you need large-enough paper or large-enough dots to do it.
I thought this video would be about an unlock pattern, lol
It’s funny because the infected problem is exactly like trying to create water sources in Minecraft. So I have a natural thought process on that already. You put them diagonally and that’s the most efficient way to cover a large area but it doesnt reach the end.
Yeah :D her ist the first english speaking person who pronounces Euler correctly :DDD 16:30
and I was watching the node-stuff asking myself the whole time: will he mention Euler at some point... wasn't disappointed but had to wait quite a while...
for the question with the students, an easy way to look at it is to imagine minecraft water sources, imagine a 5x5 water grid, you can't make the full grid source blocks with only 4 buckets of water.
0:34 Intend your puns, coward!
There's a version of the thief/police problem in one game community I'm in, called "parity". In the game Crypt of the Necrodancer you're on a square grid and can move once per beat, and there are some enemies that also move every beat after you do. So if you're on the same color tile as them on a checkerboard grid, you can't approach them safely, but if you're on the opposite color you always can. The solution is generally to do some action that lets you stay on the same tile for one move, so you can get next to them and attack. If anyone is interested, look up how to kill the red bat in necrodancer, there's a good video that shows the concept and explains it better than I could.
Who else first watched his skits and now finds it weird that this man is being smart? Lol.
Do the infected problem like you would fill a large area of water in minecraft
1) Put one in the corner
2) Expands from the two edges, but to save space, alternate between infected and non-infected
3) Profit
However, you need 5 infecteds to do this, therefore, it is impossible.