As is being pointed out by some comments this very easily generalises to an n x n or even m x n grid. The exact same argument shows that k = m + n - 1 is maximal since k = m + n leads to a cycle. (Not sure how I never realised this but this has no effect on the actual argument).
@3Max
17 күн бұрын
Argh, spoiler in the pinned comment!
@strengthman600
19 күн бұрын
I hadn’t seen this trick before, representing a binary matrix as a graph whose rows and columns are vertices and whose edges are the true entries - and it’s immediately relevant to a projecteuler problem I’m working on :)
@Polyamathematics
19 күн бұрын
Amazing!
@Xaver_44
18 күн бұрын
it's actually really common in graph theory, it's called an adjacency matrix
@strengthman600
18 күн бұрын
@@Xaver_44 this is not an adjacency matrix
@TheArizus
18 күн бұрын
@@strengthman600It is an adjacency matrix if you replace the red squares by 1 and the blank squares by 0
@msolec2000
18 күн бұрын
@@strengthman600 It is. It's just that we're going from the matrix to the graph, instead of the usual way in the other direction
@Z_Inspector
19 күн бұрын
I think the difference between math explanation videos and textbooks goes unappreciated. Even things as subtle as coloring Odette's name red and Evan's blue can make the problem much easier to understand.
@Deus_Auto
5 күн бұрын
"Odette hates even numbers but enjoys painting, so she paints some of the squares of the grid." *paints 8 squares including even indices*
@britboy3456
19 күн бұрын
5:13 an easy counter-example to Evan always needing to find a rectangle is one you gave earlier: a diagonal staircase plus one square. No rectangle present, but Evan can win.
@ellabar252
19 күн бұрын
I really like this demonstration, but I feel like the airport sidetrack was very unmotivated. I feel like a more straightforward and motivated introduction of the representational graph is possible - e.g. "I noticed that for every successful coloring by Evan, I could make a path through the squared he colored where in each step I either stayed in the same row or stayed in the same column and landing back where I started. This seems kind of obvious in retrospect, since it's almost a property of there BEING even numbers of each. Actually, this is a property of Odette's coloring - if no such cycle exists, then it is a winning strategy." Indeed This feels (to me) like better motivation to introduce graphs and creating nodes for each row and column :) All in all, fantastic video and your delivery is clear and professional as always. Great work!
@Polyamathematics
19 күн бұрын
I agree that the airport analogy was a bit unmotivated but the idea of considering graphs and adjacency matrices is what motivated me and the idea of connecting places together really did go through my head (and maybe I was a bit biased since I enjoyed the idea of animating the map) I actually really like your concept that you can make a path staying in the same row or column which for some reason didn't cross my mind exactly in that way!
@ellabar252
17 күн бұрын
@@Polyamathematics it's kind of a generalized rectangle if you think about it which also really connects to one of the previous attempts :)
@soundsoflife9549
17 күн бұрын
I'm confused.
@andrewbuchanan5342
16 күн бұрын
Great video. (1) Revealing that the matrix can be viewed as an adjacency graph means that for k = 4n-1, Odette's winning strategies are exactly the trees. (2) Can have any rectangle sides m & n, not just 2n & 2n. (3) I was a little confused at the beginning. The words "game" and "strategy" normally suggest some alternation of play, but in fact Odette just picks a bunch of squares once. She only has one move. Her move *is* her entire strategy, so we don't get much mileage from the game metaphor, imho. But again it's great - thanks so much
@Noio.
17 күн бұрын
Hi! I recently discovered your channel through a comment from a editing video tutorial (something with 3d glow lines) and even tho i have no tangents with the content you make, i still find it incredibly interesting! What i m very intrigued by is your editing: it is high quality, and on point all throughout the video!! How do you edit like this? And also, how do you find the ideas to make this content, to then edit the videos? Thanks for being such awesome content creator aaannd you ll very soon sky-rocket in popularity!
@Polyamathematics
17 күн бұрын
I actually learnt all of my editing through youtube over 5+ years now on various softwares. Now I edit with davinci resolve and I think the quality of the videos just comes from years of experience not just with editing but with watching maths videos and seeing what looks visually appealing. I do plan at some point in the future on making a video about my editing process and perhaps even a video series.
@Noio.
17 күн бұрын
@@Polyamathematics that s impresive! Thanks for respoding and i m looking forward to seeing your next vids
@julianrosenfeld7177
18 күн бұрын
Fun observation which is that the first thought of trying to prove that Evan can always make a rectangle is actually not too far off. Since what he is actually trying to do is find some combination of rectangles that he can add, with the caveat being that placing a corner atop another corner turns it red again. Adding a rectangle in this way always preserves the “parity” of blue squares on each row and column so it will always remain even. Not sure how this would relate to your proof but I noticed it and thought it was interesting.
@marsh_prootogn
18 күн бұрын
Evan and Odette seem like great friends
@BrunoBarcelosAlves
8 күн бұрын
Evan actually wants them to be a couple but Odette prefers to remain single.
@marsh_prootogn
8 күн бұрын
@@BrunoBarcelosAlves awh
@Polyamathematics
8 күн бұрын
Sounds plausible
@benniboi7231
18 күн бұрын
These are really high quality. Please don't stop!
@Hankathan
19 күн бұрын
I love puzzles about colo(u)ring squares on grids!!! :))))
@devanshgupta794
19 күн бұрын
Its videos like these that keep me learning more and more
@juancappa3838
18 күн бұрын
Why not a (more general) m x n grid?
@TheArizus
18 күн бұрын
Hmm yeah it generalises pretty easily doesn't it
@jakeaustria5445
18 күн бұрын
Thank You
@stevejohnson1078
18 күн бұрын
Why did you pick 2n for the dimensions of the grid? n doesn't seem represent anything on its own, could you not have demonstrated this on an n by n grid instead?
@SgtSupaman
18 күн бұрын
2n makes it a board where the sides are specifically even. That didn't really matter for this, as this can be generalized to any n by n as well, but he probably just didn't think of that at the time.
@FunkyTurtle
18 күн бұрын
hey man i haven't watched the video yet, but i have to say, that thumbnail is beautiful. sleek and nicely colored. good job
@taranlee5203
3 күн бұрын
My guess is 11, trying to be odette and avoid making “rectangles” like the one at 2:10 I’m expecting to be wrong but I’m excited to find out! Edit: Sorry I was looking at the 6x6 when I paused, general idea n + n - 1 ok back to watching
@taranlee5203
3 күн бұрын
I love the graph theory approach, very elegant 😊
@GameJam230
15 күн бұрын
5:11 my guess at this point as to the reasoning behind 4n (if it’s correct) is related to the perimeter of a square with side length n. If you were to cut a 1x1 square chunk out of the corner of this square, creating an additional 2 sides and reducing the area of the shape, you’ll notice that the perimeter actually doesn’t change, as you can just move the indented sides back outwards and they reconnect to form the square again. This will only fail in a certain circumstance, but I’ll come back to it. The parallel here would be that the whole board is the square, and any removed chunks are the red squares. Thinking about it in this way reveals the problem. What if you remove 1x1 squares ALL THE WAY along two sides of the perimeter of the square? Obviously then you can’t just move all the sides out and reconnect them to have a larger square, the remaining line segments on the border of the resulting square are physically shorter, and this happens when you remove 2n-1 or more squares, but the perimeter decreases by TWO if you take squares off ALL the sides, which adds up to 4(n-1) squares
@f1reflam3
5 күн бұрын
0:30 I legit thought "this video looks like it's gonna start out with 'this video is sponsored by Brilliant!' " It's not a bad thing, it means Brilliant knows their audience
@AryanSingh-iu9vt
17 күн бұрын
Can you plzz upload Combinatorics problems from Russian Olympiads? They always have some really nice and elegant combi problems!
@jonathanlevy9635
17 күн бұрын
My approach was a bit more straightforward: if we have n*m grid, assume we have a maximal red coloring which achieves the goal, then it still works after row switching and column switching. we will choose one red cell arbitrarily and move it to the upper left corner by row and column switching. then, we can find another cell in the same row as it and move it by column switching to be adjacent (if theres no such, we will try finding one at the same column, if none as well, we will move one to be diagonally adjacent to it), we will keep doing it to the next one, making a string of adjacent cells. next we discover these facts: 1. no cell can be found above the chain or the left of it, or else we could have walked the chain from corner to corner, jumped to that cell and jumped back to where we started, maintaining a loop of even number of cells in every row and column. 2. the longest chain possible goes from the upper left corner to the lower right one, having a length of n+m-1. QED
@goesbymoon
18 күн бұрын
oh damn that shift in perspective was really cool wow
@rujon288
19 күн бұрын
abstract strategy!!
@Sjoerd-gk3wr
19 күн бұрын
nice problem and great video
@andy02q
2 күн бұрын
If Odette can get ahold of some blue color, then she can paint the entire square red except leave the very last column empty and then just put an odd number of blue squares in the last row.
@alexbanks9510
13 күн бұрын
Incredibly upsetting that the nodes of airports at 6:36 are roughly geographically accurate, except STN and LTN should be swapped
@skylardeslypere9909
18 күн бұрын
So for k at least 4n, Evan can always win, but can Odette always win for k at most 4n-1?
@SgtSupaman
18 күн бұрын
No, Odette cannot always win, even if she only colors in less than 4n squares. She must place her squares strategically in order to win. A simple way to see this is if, using a grid larger than n=1 (2x2), she puts only 4 squares but in a way that they form the corners of a square. Her 4 squares are less than 4n, but Evan still won.
@skylardeslypere9909
18 күн бұрын
@@SgtSupaman yes but I mean, does she always have a strategy to win? For any k up until 4n-1?
@SgtSupaman
18 күн бұрын
@@skylardeslypere9909 , sorry, I misunderstood what you were asking. Yes, Odette is always capable of winning with some arrangement of less than 4n squares. The trivial way to show this is just by filling in a single row and single column. For k=1, just fill one square in the bottom row; Odette wins. For k=2, put another square in that same row; Odette wins. Continue until k = 2n (so that row is filled), then start using a single column until you reach the version shown in the video, where the entire row and column is filled and Odette still wins.
@skylardeslypere9909
18 күн бұрын
@@SgtSupaman aahh, right. I forgot about the initial construction of 1 row and 1 column. Thanks!
@mertaliyigit3288
18 күн бұрын
A game about coloring squares on a grid? Isnt that go?
@josefuher6117
18 күн бұрын
why is the side lenght even(2n), it seem to me that this would work also for grids with odd side lenght with the general limit of 2L-1 wherw L is the side lenght, for the same reasons
@user-pr6ed3ri2k
18 күн бұрын
0:28 odd
@niraltahan7708
19 күн бұрын
Only 900 views?!?!?!
@lazghabkarim8031
19 күн бұрын
i'd like to know why this shift in prespective possible . the best i can think of is converting the table to an andjacency matrix.
@AubreyBarnard
18 күн бұрын
Converting the grid to an adjacency matrix is basically it. Let's index the rows of the grid / matrix with natural numbers i in I and index the columns with j in J. Then the space of matrix coordinates (i,j) is the Cartesian product I x J, but this is also the space of edges in a bipartite graph between I and J. So an entry in the matrix, either red or blue, is in 1-to-1 correspondence with colored edges in the bipartite graph. Then the parity of a row / col in the matrix corresponds to the parity of the degree of a node. Does that help?
@matthewleadbetter5580
18 күн бұрын
Unfortunately there's a horrible mistake in this video: You got Luton and Stansted the wrong way round. 😲
@infrabread
18 күн бұрын
There's a mobile game called 0hh1 which has a similar concept. So, for me, the video seemed logical.
@Fleecy_wurmple
18 күн бұрын
Why does this person reminds be of the blob guy
@jonathanlevy9635
17 күн бұрын
maybe Im wrong but the lemma you proved isnt relevant for this case. the lemma refers to any graph and hence dictates that the upper limit is 4n^2, much more than the goal of 4n
@TheArizus
17 күн бұрын
The number of vertices in the graph = no. Rows + no. Columns = 4n NOT the number squares (which is 4n^2)
@rodrigoqteixeira
16 күн бұрын
My proof: for that to happend either one row or one collumn has to have less than 2 cells. All the others have 2 because it's maximum number, that that one has 1 for the same reason. Just like that, you know the max value is 4k-1. Really, tha heck
@trevorclinton2573
18 күн бұрын
Just paint 0 squares in each row and column since 0 is even
@drdca8263
18 күн бұрын
His goal was specifically stated to be to color a non-zero even number of squares blue.
@bicycleshelter4818
18 күн бұрын
It has to be a non 0 number of new painted squares
@Oxygenationatom
5 күн бұрын
HOW ARE U DOING THIS AT AGE 17 😭😭😭😭, i cant even find time to do what I want with my high school classes at age 16, with less then a hour of distractions with 8 hours of work time. ?????
@Polyamathematics
5 күн бұрын
It's a lot easier when you enjoy what you're doing :)
@Oxygenationatom
4 күн бұрын
@@Polyamathematics you enjoy school??? Or making math videos (probably the latter)
@crillybafoon7730
18 күн бұрын
I got a different solution. For a grid of 2n by 2n, Odette 4n^2-2n+1 squares. This is done by colouring every square red in every single column except for one, and then colouring a single square red in the last column. Evan can obviously colour every single row so it has a non-zero, even number of blues, but as there is only a single red in the last column, this cannot be achieved
@magnuswahlstrom766
18 күн бұрын
No, Evan can leave the last column with 0 blue squares, which is even. He just needs to colour a 2x2 rectangle blue in your solution.
@SporeMystify
18 күн бұрын
3very column doesn't need a non zero even number of blues, thr non zero only applies to the total number of blues. In other words, an empty grid isn't a valid solution, but only rows or columns with blue squares in them need matching pairs
@jbrecken
19 күн бұрын
If zero is even, Evan can always win by coloring zero squares blue, regardless of how many red squares there are,.
@TheArizus
19 күн бұрын
"Attempts to recolour a NON-ZERO number of red squares..."
Пікірлер: 76