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/↵
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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:**↵
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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 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 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>↵
↵
