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:
Here are the solutions:
Happy April fool's to you too!
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!
Same result, can we practice together?
Problem 8 is broken, nothing happens after one stone is left.
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 :'((
:(
tmw orz
Where can I practice this type of questions?
its seems that problem 8 is broken
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 .
(1, 3) => (2, 6) => (4, 7) => (8, 8)
is this your idea of april fool? you've got me man. I wasted 1 hour on this test.
problem 8 is broken. I got one stone left and nothing happened.
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.
i just tried,and it just start game from the begining and dont give me a accept like a correct answer
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
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...
Wow i got free access to the course, Excited!
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.
Apparently I suck at games.
It seems wiiliam has changed last problem a bit to make it impossible .
yes it is impossible now, I ran complete search and no solution was found.
Had fun playing with Bob and losing ;(
Thanks for the beta access, I'm very appreciate that!
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 demoralizer
Its preregistration he want to give access to only the good . But did anyone scored above 5 .
-_-
Problem 9: Prove or disprove P=NP
I thought this was a serious post at first. Got me william!
Best april fool prank ever! William orz!
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
I knew I was gonna do it!
HTML?
How to prove that there is always a solution for problem 7?
Proof by Lopsy
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
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 $$$a-1<b<a-1/3$$$, which has no integer solutions.
Oh yeah you're right
Proof that no solution exists of problem 8 : https://ideone.com/XMlAcg
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
Uhhh... what?
wow! the course is rlly good tmw, im gonna spend more time using it now especially bcuz the open is this weekend.
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?
Was anyone able to solve anyone out of problem 6,7 and 8 ?
Were you able to attempt them? I tried to press things but the question was not interacting?
It was interacting
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.
Yet he still manage to rickroll me this time (look at ur email).
yeah me too. Will we ever not fall for his traps :rofl: It was a well planned and perfectly executed rickroll though.
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.
finally!
Did you get Correct answer though ?
Yes
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.
HTML tricks?
EvenX orz
Waah bete mauj kar li.
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
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.
Ah wait lol, it really is an April Fools prank, well done!
editorial when?
Is hacking into the server the only way to pass this test??? Too bad I don't know how :'(.
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DO NOT SELL THE PERSONAL DATA IN DARK WEB!You know you are going to hell for pranking so many people for 3 days straight
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.
The solutions were quite interesting though
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.
It's funny how I managed to get rick-rolled twice
https://course.williamlin.io/story
pog, i wonder how many people got rickrolled
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!
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