Day 12
day 12
1:54 am 4 October
lol so i just told myself; if you wokr every single moment up till smmc you will win it, ie. asia #1.
thats the deal. and this is gonna go past smmc too, smmc aint shit, its literally just the start. we gonna keep going. literally the whiz kid billionaire. lets go have fun then.
every single moment.
2:53 am
Ramsey implies schur:
coloring f:[N]->[r]
then for K_{n+1} use ramsey and colour edge {i,j} with f(j-i)
Then we'll find monochromatic triangle in it, say it's i,j,k and we have f(j-i)=f(k-j)=f(k-i)
Then just say j-i=a; k-j=b; k-i=c; then a, b , c have same colour in our schure colouring and clearly a+b=c
wow, fucking genius. hat bc
bruh.
ok i also thought, there can be two kinds of schur numbers; like N_s(r) and N_d(r) where in N_d(r) x=y is not allowed. N_s(r) is i guess the standard definition... but this is interesting, if you like for instance 2-color integers right, so you look at N_s(r) you'll pretty quickly see 5 is optimal , and like my algorithm for this was lexogrpahic dfs, like sort of look at the smallest possible ordering oyu can give while inductively not violating schur; like you'd do 01 (cos 00 violates schur) then 010 (011 would also work here, so note there's a choice) and then 010 and nomore possible, so go back to choice (dfs right) and look at 011 and then you can have 0110 and then nothing , and we've also run out of choices (first choice doesnt matter) so it's 5. and if you do this for N_d(2) you'll see the second optimal choice, ie.
00101110 is optimal and it gives N_d(2)=9... hmm interesting.
I've told chatgpt to make me a "Ramsey problem set"
I'll say more about what this means later, but yeah time to do some ramsey theroyyyy woohooo (genuinely crazy, you can go learn anyhting you want gng)
3:35
first i wanna prove schur implies flt mod p is always possible, lemme think about it.
btw one thing i wanna say is; your differentiatior is going to be your ability to solve genuinely diffiicult problems and not having balls and thinking big.
the last two is ok right, like already. the first one, there's so many kids who are out there just reading math books right, or just programming random shit. like a lottttt of kids. but there's very few kids who can like solve IMO 3's for instance. or like basically genuinley think deeply about difficult problems and solve difficult math problems. so get reallyyyyyy good at that and you'll be better than all the retards. and then among alll the nerds, most of them are actually just pussies, like they don't think big or have balls at all. so you'll be better than them too. thats the plan to go be the best ever. let's go.
3:58 am
wow i read this
thats crazy, this (the chat) is genuinely so inspiring.
it was so elementary, just a smart kid making smart not so difficult observations. literally the dream. lets keep working. literally the dream wow.
4:17
well, the schur implies fermat thingy was basically:
for any n, prove you can solve x^n+y^n=z^n (mod p) for all large enough p.
the proof is; look at A={x^n} where x ranges in Z_p (obv except 0)
A has size atleast about p/n (cos if you look at a generator g; g^n,g^2n... work atleast)
Then you sort of inductively build up other sets, like say a is smallest element not in A; then you look at B={ax^n} which is obv disjoint from A and so on and you keep going. so you basically get a partition of {1,..p-1} and choose a monochromatic sum soln ; and monochramitc means mod p it was ax^n and so on and you're done. this also works cos like size of A is p/n right, and each subgroup has that same size. so you have p/(p/n) ie. n colours. (wait does this only work for p that n divides p-1? wthelly???? cos it musst be an exact multiple right. ahhh nah i was tripping, it has atleast floor(p-1/n) elements, but it prolly has more. infact it'll have number of elements to be an exact divisor of p-1. damn yeah. ok so basically you have a n coloring of {1...p-1} and then choose three monochromatic that sum to each other, but monochromatic means they're x^n * constanct, remove the constant
i was spoiled on this problem cos i had heard the term normal subgroup when listening to this problem, and my proof i essentiallyrederive normal subgroups and the G/N thingy from napkin lmao. ok i really wanna study group theory tooo ahhh work work work so much interesting math man... think a lot and keep having fun.
Sagnik called me and told me this problem, i immediatly had a brilliant idea, here is the problem and my idea;
theres 6n points on a line, 4n blue, 2n green, prove there exist a line segment with 3n points such that exactly 2n of those are blue.
my proof is like we have a bunch of "sliding windows" of length 3n right, and each time you go left or right, the number of blue points in segment can either increase by 1 or decrease by 1 or remain same. (so like left or right means first consider the first 3n points line segment, [1,3n] ; go right means consider [2,3n+1]) and so on. then consider [1,3n] and [3n+1,6n]. one of these has >=2n blue things and the other has <=2n blue things. Assume one on left has less than 2n blue things. then as you slide to the end , at each poiint your blue is increaseing by at max one, but you reach more than 2n, ergo you must have crossed 2n... brilliant proof right
well anyways, the main thing again you gotta do is think veryyyyy deeply. right like all day just have fun enjoy math and think deeply bro. like this was fun. but now go think uninterrupted about ramsey problem set. focus. focus. focus.
6:03 am
yuh we lit. this is literallly going to make me the best ever. smmc is an ez dub now. this is actually so funnnnn, but you just gotta start and not stop at all , like for this to work, work non stop. literally non stop , thats the only way this shit can work man. like for reak theres os much math i wanna study. and like people have done so much math... i can literally do all of it. the world is mine. i just gotta work. what a simple deal ahahahah. all day tho, to start with lets just have fun all day and just not stop at all. you know what, lets do this, i am not going to use the internet at all tomorrow, thats it. like thats the deal with me. that includes whatsapp. literallyjust one day ,just say i wont stop and just dont stop. and remember that quote by illia topuria, once a boxer is knocked out, like once he hears the 10 count, its a lot easier to knock him out. so start and keep the momentum going, and remmeber if you stop even once this shits over with. all day, just do interesting math , or blog , or watch yufei zhao. but mostly do intersting math. 1 week. you alreayd know what you got to do. this shits getting fun now. new kid. just dont stop hahahahahhgut4hgpu245jhgpuh42. lets go
honestly man, just do not stop at all. just do not stop. this is litreally the dream. no negotiation.
prove you can two color integers s.t. no inifnite ap
i guess my N_d(r) thing is thought about, like in roths theorem , to avoid trivial (a,a,a) AP you change mantel's to : what is max edges you can have s.t. each edge is in exactly one triangle.
ok this should be inspiring, cos i believe i can do anything right. so someone solves these problesm. and no one in this world is better than me , ergo (i love this word now haha) i can solve all these problems. lets go math kid.
have fun, these next 7 days just have so much fun. go win smmc. but listen,before this year ends, i will solve atleast one unsolved math problem. along with doing a bunch of other interesting things ofcoures ^_^
7:07 am
looked at a lot of intersting math, durett, putnam and beyond etc.
i guess for the next week, since its oly, do more oly stuff. ill do more oly stuff after that also but like over the next weeek, do less "research" stuff
but this is soooo cool, im so excited about math right now, i havent been this excited in a long time ahaha. i was thinking about undeced ramsey number in the shower with my marker, literally lucknow man. from cengage quadractic numbers to research problems huh. lets go work. you gotta work, just have fun. this is literally how you get better. how did the little kid inlucknow go form nothing to this right. he just worked and did not stop no matter how frustrating.
now i have the opporunity to be the best in the world, no way you can;t take full advantage lesgoooooo
one last thing, i was talking to sagnik in the evening and we were talking about how id come to college graduation in a ferrari, and i would tell all the kids my story. hwo i skipepd all of college, didnt take no classes, dropped wtv courses lol just did things my way. sagnik said he was very excited for this. people would see the new meta. new inspiration and way for kids bruh. like this is literally going to happen. it has to.
7:44
nah fr, listen its arleady said:
if you work every single moment,i will win smmc. its literally written. if you work every single momeny, youre going to win. thats a given haha. like thats a fact.
i wanna feel so confident when going in, like ilia topuria man, if you do what youre supposed to the exam is just a bygone conclusion. like its already done. i am the best ever, and this is the perfect opportunity for me to start the journey. i wanna be arrogant. i wanna have fun. lets go get this shti man. every single moment. just work and im the best ever. im asia #1 by just working every single moment. leets fucking go man. all the way up. smmc really aint shit, i wanna dchange the world. but confidecne comes from making al the right decsions. i wanna go in so fucking confident hahahah. arrogant. all teh way up. sleep now, tomorrow is arelly big day.
4:20 pm, 4 october
yeah i just woke up. i really wanna fgo get this. im not taking any more L's at all. on 11 octoner, i will be so confident, cos i will have done what i said i was going to do, ie. work every moment, so ill go in wiht confidence and arrogance and win smmc. and then we're gonna keep this shit going. every single day working hard all day. being so arrogant. so confident. haveing so much fun every single day, just working hard non sotp and then we're gonna have so much fun being THE WHIZ KID BILLIONAIRE OF THSI GENERATION. All The Way Up. let's work.
4:51 pm
well i was watching yufei zhao's thingy while eating burger (khabib haha) lemme sketch out a couple proofs of mantel's theorem
max edges in triangle free graph?
whenever xy is an edge; d(x)+d(y) <= n (n is size of graph)
then consider $ \sum_{x \in V} d(x)^2 = \sum_{xy \in E} d(x)+d(y) \le m*n$ (m is number of edges)
but use cauchy schwarz on LHS to get its atleast $\frac{4m^2}{n}$ and you're done.
antoher proof, that acually gives us the maximal example right away is:
Let A be the maximum independent set in G, and let a be it's size.
Note, d(v) <= a ; cos the neighbours of v are themselves an independent set.
Let B= V/A
look at m. each edge e must pass through B.
so you get $m \le \sum_{b \in B} d(b) \le a*B$ and then you're done by am-gm or wtv.
the first inequality tells you it must be biparite (for maximal case) and then an-gm tells you sizes must be as close as possible .
I had indepently proved Turan's theorem , i dont know why i am not able to think of it. this is so fucking annoying ahhhh. wtv. yk what, i wanna go to jolly's, buy some stuff, come back shower and get right to it.
I am going to come back and start tthe ramsey. problem set. I will blog after that is done, atleast a min{4 hour, all problems done} session when i come back ^_^
just do not stop. 0 knockoutsmane.
6:49 pm
man i was feeling so fucking tired right now. jhust felt like sleeping. but i was like nah, if i work this moment, i will win smmc. i will be the best ever .i wanna be arrogant. itll be so fun to say best in asia, come suck my dick bitch lmao. anyways, we're finally doing the ramsey probelm set. 4 hours, lets go non stip
5:48 am