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.↵
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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).↵
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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 [email protected] for your decision letter, as well as solutions and more opportunities to gain beta access to the course!↵
↵
**Update 2:**↵
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Congrats to the people who were selected to be part of the beta course:↵
↵
<spoiler summary="Spoiler">↵
Happy Extended April Fools'!↵
↵
There is no course but it doesn't matter; just keep practicing :)↵
</spoiler>↵
↵
Here are the solutions:↵
↵
<spoiler summary="1">↵
6↵
</spoiler>↵
↵
↵
<spoiler summary="2">↵
25↵
</spoiler>↵
↵
↵
<spoiler summary="3">↵
One possible solution is:↵
↵
~~~~~↵
.#.#.↵
.....↵
#.#.#↵
.....↵
.#.#.↵
~~~~~↵
</spoiler>↵
↵
↵
<spoiler summary="4">↵
0, 0, 0↵
</spoiler>↵
↵
↵
<spoiler summary="5">↵
~~~~~↵
.#.↵
.#.↵
###↵
~~~~~↵
</spoiler>↵
↵
↵
<spoiler summary="6">↵
Notice that you gain 0.5 extra regions if you capture a border cell and 0 otherwise. The strategy is then obvious, and it's also obvious that the best both of you can do is tie.↵
</spoiler>↵
↵
↵
<spoiler summary="7">↵
It's always possible for Bob to achieve the goal. See [this](https://www.reddit.com/r/mathriddles/comments/2v6eaj/doubling_and_adding_1/).↵
</spoiler>↵
↵
↵
<spoiler summary="8">↵
By a simple invariant argument, it's impossible to have 1 stone left.↵
</spoiler>↵
↵
↵
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 [email protected] 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 summary="Spoiler">↵
Happy Extended April Fools'!↵
↵
There is no course but it doesn't matter; just keep practicing :)↵
</spoiler>↵
↵
Here are the solutions:↵
↵
<spoiler summary="1">↵
6↵
</spoiler>↵
↵
↵
<spoiler summary="2">↵
25↵
</spoiler>↵
↵
↵
<spoiler summary="3">↵
One possible solution is:↵
↵
~~~~~↵
.#.#.↵
.....↵
#.#.#↵
.....↵
.#.#.↵
~~~~~↵
</spoiler>↵
↵
↵
<spoiler summary="4">↵
0, 0, 0↵
</spoiler>↵
↵
↵
<spoiler summary="5">↵
~~~~~↵
.#.↵
.#.↵
###↵
~~~~~↵
</spoiler>↵
↵
↵
<spoiler summary="6">↵
Notice that you gain 0.5 extra regions if you capture a border cell and 0 otherwise. The strategy is then obvious, and it's also obvious that the best both of you can do is tie.↵
</spoiler>↵
↵
↵
<spoiler summary="7">↵
It's always possible for Bob to achieve the goal. See [this](https://www.reddit.com/r/mathriddles/comments/2v6eaj/doubling_and_adding_1/).↵
</spoiler>↵
↵
↵
<spoiler summary="8">↵
By a simple invariant argument, it's impossible to have 1 stone left.↵
</spoiler>↵
↵
