### tmwilliamlin168's blog

By tmwilliamlin168, history, 6 weeks ago,

Having had a ton of experience with CP, I have decided that I could share my knowledge the best with the world by making a course! This course currently has lessons designed for people of many levels, from newbies to low masters.

Currently, I have opened pre-registration. I'm currently just trying to get an idea of how many people would be interested in such a course (and also to catch possible bugs on the website).

Here is the course website: http://course.williamlin.io/

100 lucky winners who pre-register will be given beta access to the course for free! Pre-registration won't be open for long though, so be sure to do it quick!

Update:

Thanks to every who pre-registered! The pre-registration assessment proved to be tough, and we decided to offer spots for beta access even for those who did not pass the test! Look out for an email from course@williamlin.io for your decision letter, as well as solutions and more opportunities to gain beta access to the course!

Update 2:

Congrats to the people who were selected to be part of the beta course:

Spoiler

Here are the solutions:

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• +254

 » 6 weeks ago, # |   -14 Can you list the roadmap or syllabus?
 » 6 weeks ago, # | ← Rev. 2 →   +293 Happy April fool's to you too!
 » 6 weeks ago, # |   +238 Oops... No offense but seems like you lack a bit of practice...Solutions to this pre-registration test will be released soon.Come back again once you have learned the fundamentals!
•  » » 6 weeks ago, # ^ |   +47 Same result, can we practice together?
•  » » » 6 weeks ago, # ^ |   0 me to
 » 6 weeks ago, # |   +23 Problem 8 is broken, nothing happens after one stone is left.
•  » » 6 weeks ago, # ^ |   +3 I wasted a lot of time checking the HTML,JS and APIs and was surprised to found that there was no condition written for the time you Win the Test. There was only one thing "You lack fundamentals shit"LOL ! Tim you owe my my Time :'((
•  » » » 6 weeks ago, # ^ | ← Rev. 2 →   -7 Yes dirty culture of wasting someones time
•  » » » » 6 weeks ago, # ^ |   0 Not lying but I genuinely got very excited because I saw this blog as soon as it was posted and thought I can easily make it into the 100.Also, following tmwilliamlin168 from when he had only around 10K subs on YouTube, I am a fan of his screencast videos. But this joke disappointed me like hell :(
•  » » 6 weeks ago, # ^ |   0 Indeed. problem 8 is broken. It is not possible to get to one stone. You can't get to one stone if all piles have an odd number of stones. In the problem, he has [7,7,9,5,7]. To get to one stone, it's got to be [6,7,9.5.7]
 » 6 weeks ago, # |   +7 :(
•  » » 6 weeks ago, # ^ |   0 It is the maximum possible score . He may consider us if there is actually a course
 » 6 weeks ago, # |   +1 tmw orz
 » 6 weeks ago, # |   +22 Where can I practice this type of questions?
 » 6 weeks ago, # |   0 its seems that problem 8 is broken
•  » » 6 weeks ago, # ^ |   -9 last 3 questions are impossible to ace due to bugs in 6 and 8 , giving Wrong answer intentionally (or doesn't have any answer) in problem 7. In problem 7 i am getting 1,3 from brute forces but still it's giving WA .Probably April fool :) Others can save their 30 minutes .
•  » » » 6 weeks ago, # ^ |   +11 (1, 3) => (2, 6) => (4, 7) => (8, 8)
•  » » » » 6 weeks ago, # ^ | ← Rev. 2 →   0 is this your idea of april fool? you've got me man. I wasted 1 hour on this test.
•  » » » » 6 weeks ago, # ^ |   0 problem 8 is broken. I got one stone left and nothing happened.
•  » » » » » 6 weeks ago, # ^ |   0 My fault... the client is initialized with the wrong number of stones so the server doesn't think that the problem was solved.If you press reset though, it will sync with the server.
•  » » » » » » 6 weeks ago, # ^ |   0 i just tried,and it just start game from the begining and dont give me a accept like a correct answer
•  » » » » » » 6 weeks ago, # ^ | ← Rev. 2 →   -8 .....
•  » » » » » » 6 weeks ago, # ^ |   -28 tmwilliamlin168 It still does not work! I scored 5, and i can score 6 easily if this problem works please reply asap i have 10 minutes left
•  » » » » » » » 6 weeks ago, # ^ |   0 Congratulations on your solution. Maybe you would find it quicker if you notice that after every turn all piles are left with either an even or odd number of stones...
•  » » » » » » » » 6 weeks ago, # ^ |   0 Yeah man i realized that later, that we have 35 turns in total, and in each turn we lose 3 stones i.e it is impossible to have 1 stone left.
•  » » » » » » 6 weeks ago, # ^ |   -26 tmwilliamlin168 this is not fair you did wasted a lot human hours
•  » » » » 6 weeks ago, # ^ | ← Rev. 2 →   0 thanks so my brute force was wrong . I don't think answer for this problem exist .tmwilliamlin168 last problem still not working even if one stone is left
•  » » » » » 6 weeks ago, # ^ |   0 last one is pretty much impossible now!
 » 6 weeks ago, # |   +18 Wow i got free access to the course, Excited!
•  » » 6 weeks ago, # ^ |   0 What was your final score?
•  » » 6 weeks ago, # ^ |   0 He joked!
•  » » 6 weeks ago, # ^ |   0 Did you
 » 6 weeks ago, # |   0 Seems there's a bug in Matching Problem. I just selected random options and even after submitting the wrong answer, after I selected the correct one, it increased my score. I think that shouldn't be intended.
 » 6 weeks ago, # |   +57 Apparently I suck at games.
 » 6 weeks ago, # |   0 It seems wiiliam has changed last problem a bit to make it impossible .
•  » » 6 weeks ago, # ^ |   0 yes it is impossible now, I ran complete search and no solution was found.
 » 6 weeks ago, # |   +13 Had fun playing with Bob and losing ;(
 » 6 weeks ago, # | ← Rev. 2 →   +10 Thanks for the beta access, I'm very appreciate that!
 » 6 weeks ago, # | ← Rev. 2 →   +13 why you require people having prior knowledge if this course is from zero to hero ? Thanks but i am happy with content of Errichto and HealthyUG
•  » » 6 weeks ago, # ^ | ← Rev. 2 →   +8 Its preregistration he want to give access to only the good . But did anyone scored above 5 .
 » 6 weeks ago, # |   0 -_-
 » 6 weeks ago, # |   +51 Problem 9: Prove or disprove P=NP
 » 6 weeks ago, # |   +15 I thought this was a serious post at first. Got me william!
 » 6 weeks ago, # |   -24 Come on man tmwilliamlin168 , the questions were SO good to be true, but the online glitches were the issues. I request you to please take a similiar test in the near future, so that the people who were glitched out(me too) can be compensated. Thanks
 » 6 weeks ago, # |   +19 Best april fool prank ever! William orz!
 » 6 weeks ago, # |   +11 tmwilliamlin168 Nice April fool ! Problem 6 and 7 are not solvable (for 6 we can not make better than equal score) Last problem is glitched :/I have 2 stones everywhere but I can't move anymore haha Nice prank haha :p
 » 6 weeks ago, # |   +40 I knew I was gonna do it!
•  » » 6 weeks ago, # ^ |   +2 HTML?
•  » » 6 weeks ago, # ^ |   0 show off! -_-. I scored 3/8 lol
•  » » 6 weeks ago, # ^ |   0 Yo wtf !? What great knowledge in algorithms and data structures! has to do with this gaming test ?
 » 6 weeks ago, # |   0 Couldn't manage to get 6 or above(I scored 5). There is no better place for the people with weaker brains. :(
 » 6 weeks ago, # |   +9 How to prove that there is always a solution for problem 7?
•  » » 6 weeks ago, # ^ |   +19
•  » » 6 weeks ago, # ^ | ← Rev. 2 →   0 If you have a pair $(a, b)$, and you assume $a < b$, you can always make $a$ closer to $b$ by $1$ by doubling $a$ first, then $b$. Repeat this operation until they are equal.$(a, b) \rightarrow (2a, b + 1) \rightarrow (2a + 1, 2b + 2)$Edit: this solution is wrong
•  » » » 6 weeks ago, # ^ |   0 Doesn't this not work because $|(2a+1)-(2b+2)|=|2a-2b-1|$ isn't necessarily less than $|a-b|$.I think this is only true when $a-1 •  » » » » 6 weeks ago, # ^ | 0 Oh yeah you're right  » 6 weeks ago, # | 0 You are successful to make us fool in April fool's !!!  » 6 weeks ago, # | ← Rev. 3 → 0 Proof that no solution exists of problem 8 : https://ideone.com/XMlAcg •  » » 6 weeks ago, # ^ | +32 We can also see that overall we are adding 1 to the sum and subtracting 4 so apparently everytime we change the sum by 3.and the sum of given numbers is 35 which gives remainder as 2. It is only possible if sum%3 == 1  » 6 weeks ago, # | +16 After defeating Bob, you think that he's too easy of an opponent for you.Uhhh... what?  » 6 weeks ago, # | +10 wow! the course is rlly good tmw, im gonna spend more time using it now especially bcuz the open is this weekend.  » 6 weeks ago, # | +3 I've tried multiple times on Firefox and chrome to click pre-register. It pops up another small tab and then suddenly disappears with nothing happening. Anyone encountered that problem? •  » » 6 weeks ago, # ^ | 0 It happened with me as well but after few tries I was able to pre-register. •  » » 6 weeks ago, # ^ | 0 For me clicking 12 times did the job! •  » » 6 weeks ago, # ^ | 0 i have had this similar issue with some other website. disabling some extensions/add ons worked for me  » 6 weeks ago, # | 0 Was anyone able to solve anyone out of problem 6,7 and 8 ? •  » » 6 weeks ago, # ^ | 0 Were you able to attempt them? I tried to press things but the question was not interacting? •  » » » 6 weeks ago, # ^ | 0 It was interacting •  » » » » 6 weeks ago, # ^ | 0 The problems are impossible(8 for sure, 6 and 7 don't know) William Lin was just trying to fool us.It was fun and the work he put into this is a lot. Anyway better than the rickroll he did in a long challenge once. •  » » » » » 6 weeks ago, # ^ | +1 Yet he still manage to rickroll me this time (look at ur email). •  » » » » » » 6 weeks ago, # ^ | +3 yeah me too. Will we ever not fall for his traps :rofl: It was a well planned and perfectly executed rickroll though. •  » » 6 weeks ago, # ^ | ← Rev. 2 → 0 I think in game problem of a and b, no such pair exists where bob can lose. Not sure about others but since this a prank, I doubt there exists any solution for them. •  » » 6 weeks ago, # ^ | 0 No :( They seemed impossible •  » » 6 weeks ago, # ^ | ← Rev. 3 → 0 I think in problem 6, the best we could do is draw, since both we and opponent have half moves, so we can always divide blocks into equal partitions.In problem 7, I ran some searches, and found out that there were$81910$pairs$(x,y):\ \ x,y<=1000$possible with$final\ value<=1e9$, since this is increasing function, we might get all$499500\$ pairs on increasing upper bound, alas I don't have any proof!
•  » » 6 weeks ago, # ^ | ← Rev. 2 →   +8 For problem 6 you can show that the 2nd player has a strategy that can always at least tie the first player.The blocks are of length 3,5,7,9. The 2nd player strategy is as follows. If the player 1 paints in either middle of any block, the 2nd player can paint the the middle of any one of the remaining blocks. Otherwise the 2nd player can paint tile on the same block that the 1st player painted symmetrically on the other end, a reflection over the center.The 2nd player will always have a reply for the 1st player and this repeats until the game is over.You can show that the 1st and 2nd player will always end up with the same number of segments. Although this isn't the strategy that was actually used, the strategy the computer uses probably works similarly.
 » 6 weeks ago, # |   +16 finally!
•  » » 6 weeks ago, # ^ |   0 Did you get Correct answer though ?
•  » » » 6 weeks ago, # ^ |   +15 Yes
•  » » » » 6 weeks ago, # ^ |   +3 I think its a mathematical impossibility , the initial total was 35 and final expected total is 1 and in each step we reduce total by 3.
•  » » » » » 6 weeks ago, # ^ |   0 Yeah 2 stones will always remain
•  » » » » » 6 weeks ago, # ^ |   0 Not sure if the stones in number of piles have changed now. But last night when I tried, the initial stones were: 7 7 8 5 7 and I was able to get 0 0 1 0 0. Here are the steps: 7 7 8 5 7 8 6 7 4 6 9 5 6 3 5 8 6 5 2 4 7 7 4 1 3 6 6 5 0 2 5 5 4 1 1 4 4 3 2 0 3 3 2 1 1 2 2 1 2 0 1 1 0 1 1 0 0 1 0 0 
•  » » » » » » 6 weeks ago, # ^ |   0 it is now 7 7 9 5 7
•  » » » » » » 6 weeks ago, # ^ |   0 nater5247 so, what happened after u solved '6' ?
•  » » 6 weeks ago, # ^ |   0 HTML tricks?
•  » » » 6 weeks ago, # ^ |   0 Hehe of course
•  » » » 6 weeks ago, # ^ |   0 EvenX orz
 » 6 weeks ago, # |   0 Is it free??
 » 6 weeks ago, # |   0 Why it is restricted to solve those problem? It shows what I gained and don't let me upsolve other problems.
 » 6 weeks ago, # | ← Rev. 2 →   +16 Waah bete mauj kar li.
 » 6 weeks ago, # | ← Rev. 3 →   -26 Problem 6 Solution - Your Score: 10 Regions - Bob's Score: 12 Regions To win the game, you need to have your army separated into fewer regions than Bob  Hence, According to Problem Statement,You Win! [10<12]!Upvote if you found it Helpful!
•  » » 6 weeks ago, # ^ |   0 Well, I do agree that it will work, but on a condition that the simulation would work accordingly to the optimal moves we choose (as I tried the same strategy but didn't really work out for me).
•  » » » 6 weeks ago, # ^ |   0 The Program is Set in Such a Way that It's Impossible to Win! April Fool's Day's Prank by Author! 
•  » » 6 weeks ago, # ^ |   0 why you are try to ruin this!!
 » 6 weeks ago, # |   0 bruh i really thought there would be a course. Sadge
 » 6 weeks ago, # |   +17 I just like the fact that I knew it's April Fools and proved that the games can't be won but still didn't stop me for playing them for 30 minutes for no reason
 » 6 weeks ago, # |   +10 Am I wrong or is the stone game really impossible to win!!?Bob plays symmetrically unless you play center, in which case he plays center in one of the remaining centres (there are 4 so this is always possible).In the end all non-centre cells are coloured symmetrically. Therefore center colours won't change the number of regions, which will be the same for blue and red, thus you can't win.
•  » » 6 weeks ago, # ^ |   0 Ah wait lol, it really is an April Fools prank, well done!
•  » » 6 weeks ago, # ^ | ← Rev. 2 →   0 Were you able to solve the other two game problems (The minimizing block problem and equalizing two number problem) ? Well I guess all three games problems are unsolvable (maybe not). UPD: Also I think I have played a similar stone game which is impossible to win before
 » 6 weeks ago, # | ← Rev. 2 →   0 Test not working correctly except the timer! Not submitting on clicking after the 4th question.
 » 6 weeks ago, # |   0 Is it me or anyone else who cannot open http://course.williamlin.io/ this website?
•  » » 6 weeks ago, # ^ |   0 site is down for me too!
 » 6 weeks ago, # |   0 I can't do the test. I just press "start test" then it says Time's up: score 0 :c
 » 6 weeks ago, # |   0 editorial when?
 » 6 weeks ago, # |   0 Is hacking into the server the only way to pass this test??? Too bad I don't know how :'(.
 » 6 weeks ago, # |   0 Well played
 » 6 weeks ago, # |   +5 Auto comment: topic has been updated by tmwilliamlin168 (previous revision, new revision, compare).
•  » » 6 weeks ago, # ^ |   +16 DO NOT SELL THE PERSONAL DATA IN DARK WEB!
•  » » 6 weeks ago, # ^ |   +9 You know you are going to hell for pranking so many people for 3 days straight
•  » » 6 weeks ago, # ^ |   0
 » 6 weeks ago, # |   +16 The admissions committee has carefully reviewed your application. After much consideration, it is with great regret that we must inform you we are unable to offer you a spot in the beta course.The course was extremely popular and there was a large number of strong applicants; in light of this, we were unable to offer a spot to every worthy applicant.We recognize that this letter may come as a disappointment to you. Nevertheless, we encourage you to continue practicing competitive programming and wish you the best of luck in your future programming competitions.Lastly, we would like to thank you for the time and effort you took to submit your application. We encourage you to review the solutions to the pre-registration test and subscribe for future updates.
•  » » 6 weeks ago, # ^ |   +11 The solutions were quite interesting though
 » 6 weeks ago, # |   +14 Though u pranked,but u were successful in getting an estimate on how many users are interested in your course or any other project like this. Well played.
 » 6 weeks ago, # |   +16 It's funny how I managed to get rick-rolled twice
 » 6 weeks ago, # |   0 https://course.williamlin.io/story I decided it was time to send out decision letters to the 7000+ people who have pre-registered. pog, i wonder how many people got rickrolled
 » 6 weeks ago, # |   +1 You got me with the last one. I didn't notice that it wasn't possible until I saw that you aren't allowed to perform the operation if any piles are empty. I don't have a proof for questions 6 or 7 being impossible but I'm convinced 6 isn't, and 7 is just too mathy for me to want to try that badly.But fun joke, I enjoyed it!
 » 6 weeks ago, # |   0 Auto comment: topic has been updated by tmwilliamlin168 (previous revision, new revision, compare).
 » 4 weeks ago, # |   0 solution to no.5 seems kinda familiar. like I have seen this pattern somewhere else .can anyone tell me what it is?