Is it possible? Simple questions, not so simple solutions 

If you get a pen thicc enough, you can connect all of them with 1 line

I have discrete math course flashbacks. All of this is fun and games, until you are graded for it.

"Is it possible to start with 4 students and infect the whole class?" 2020: Say no more.

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

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

1:53 the third problem is actually how water works in Minecraft

3:16 that moment when you've placed enough water sources in Minecraft to know the answer immediately.

Theory: whenever someone says "no pun intended" in a video/presentation... The pun was intended.

there is something unique about this channel

The classroom problem is the infinite water source problem in minecraft. Just sayin’...

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)

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!

3:06 and that's why you sit in the corner of the class room.

Can't believe you didn't mention triangulation in chess in the thief/cop problem.

love this video - good work Zach

I love you, your content and your account, thank you so much for everything, hug from Portugal ❤

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).

The police-theif game at 8:26 can win you a bet

2:27 *MINECRAFT WATER PHYSICS WANTS TO KNOW YOUR LOCATION*

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.

Ah, math that i dont understand. Perfect.

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

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.

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 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!

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.

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

No, no, they just close the schools at that point. Believe me, we know. -The Future

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.

Only be a robber in states with an even number of cities! Life hack...

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

This channel is awesome! I'm considering a career in statistics, and this channel is just feeding that idea. I love this channel!!!

My favourite RU-vid channel, you completely change the way I view mathematics and it’s applications

I CAN DO THE CONNECT THE 9 DOTS IN 1 LINE!!! *Uses a crayon without the wrapper and uses the side*

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

The problem of the infected students could be viewed as Minecraft water sources spreading. It's really the same concept.

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

Superb! Enjoyed 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.

I watch a lot of math videos, but this was the best I've seen so far!

This topic turned out to be very interesting, although it did not seem to me at first. Thank!

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.

I have could like to say that you have improved my way of my thinking

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

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.

Ima big fan of u und ur videos. Keep making great videos. 👍👍👍

13:31 ah yes, I didn't see the answer until you made that simple shape! Thank you

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

11:00 we litterally learned this in my a level further maths class today

Is it possible? apparently sometimes it is and sometimes it isnt... Who could have known Great video btw

2:42 See also: Minecraft water source generation

Wow..I loved the way you explained the topic which I find difficult to understand 👏👏

This channel surprisingly gives me feels like the Vsauce used to do.

This came up in my recommended and the infected problem felt too real, very informative though good work!

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 came up with a different (less elegant) solution to the nine dots problem. You can do it by drawing an m-shape with four line segments, with the only catch being that you would have to go back over already drawn lines.

My biggest problem is my inability to think outside the box, but almost all the ones that required logical thinking I was able to rationalise an answer. Nice video, btw.

"simple question, not so simple question" Seems like exam questions

4:40 only 3*3 is infected but only with one recursion and the other is increase in log

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

lol the chasing the thief problem is so obvious to anyone who's played roguelikes

2:22 Ah the good old minecraft pool problem There's a more intuitive way to resolve it though

I love how you changed the music when Euler's characteristic was revealed.

This was a nice video overall. We need more of graph theory problems.

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...

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.

Thank u so much of how to learn so much more and i was looking for a kind of website where it can teach me how to learn other equations and other lessons u inspire me to learn and solve problems the right way so thank u

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

These videos are so high quality it's amazing

Well, I gotta say, in the light of the first puzzle, where we have to use the space outside the square, I figured the answer to the second puzzle would for sure be to just let some of the rectangular tiles partly hanging outside the grid. That's perfectly doable.

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!

I LOVE problems like these. I hope to come up with my own someday

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.

WHAAATTT THEEE *BEEP* It's my first video on this channel and how come I not realise till 7:49, its Zach

Thank you. This really helped me fall asleep.

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.

6:57 rip that non infected guy

0:20 There can be thicc line wich connect all dots xD

8:30 - I solved this as soon as I saw it, because of a well-known triangulation concept in chess that goes the same way in king-pawn endings, with the same effect of "switching" the move order.

these type of videos are usually frustrating but i actually learned something

When watching the part at 4:10 i just kept thinking about infinite water sources in minecraft

when you intuitively know the proof around 4:00 because its the same as minecraft water physics

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.

we need a part 2!!!

3:42 ... one of the instructors last name is " Van Dijk " !

Do Sprouts theory! If it exists, I don’t know. This video just reminded me of Sprouts, maybe it’s the dots and lines...

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

Very well done.

9:48 is like triangulation of the king in chess

1:07 this is some weird minesweeper

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.

The infected students' problem reminds me of the segmentation problem in Image Processing (region growing and region splitting)

I believe the "9 dot game" is where the term "think outside the box" actually comes from, so you were actually right on point.

You were just 5 months too early with the classroom example. It is now our reality.

one thing I was thinking about during the 6 rooms problem as I was trying to solve it was that the graph of the nodes wasn't planar, so when he connected that to the problem being solvable on a taurus, i had this HUGE moment of ecstasy. like i used to just look at those problems and think, "oof unsolvable, big sad" but it's just impossible in 2D, it's doable in 3D, the problem just isn't presented that way so I never thought about that!

the cop and thief puzzle is like the opposition of kings in chess, you have to gain a tempo to get the opposition and once you do your opponent cant defend/ attack and thats what the diagonal does

"Is this possible to infect all of the students?" "Answer: *NO"* Coronavirus: Hold my Chinese..

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.

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.

According to the rules as stated, the first puzzle can also be solved in a number of configurations if back tracing is allowed and is not considered a new line segment. Meaning a new line segment can begins at some mid point (not only at end points) of a previous line segment. The final solution has many variations but will ultimately consist of 3 parallel lines and 1 line which intersects and is perpendicular to the 3 parallel lines.