Wow, it's week 8 already! If you haven't popped along to one of our Film Nights yet, then why not join us this Friday?

We've also got a talk about online communities, a solution to last fortnight's puzzle, and a new puzzle to get stuck in with!

Talk - "How To Build and Betray Online Communities: A Ravelry Case Study" by C Wringe - 19:15, Thu 19/11

We'll post a YouTube Premiere link in Slack (and on our calendar) shortly before the talk is scheduled to begin. Once the talk livestream has finished, join us in Discord for questions with the speaker!

Ravelry is a (relatively) small site for knitters and crocheters to share patterns, projects, and discussions. Being the main point for a (relatively) small community, there is some tea. At times, it has performed admirably - at others, failed miserably.

This makes it a great case study for those building platforms for online communities: What to adopt, where to improve, and what to never (oh god never) do.

Film Night - 19:00, Fri 20/11 - HackSoc Discord (The Pod)

We gather together in our Discord server to watch a film (or sometimes several). Apparently some of these might have a link to Computer Science, but we haven't spotted one yet.

Vote for the film you'd like to watch using our Film Night Portal!

We'll use our very own live-streaming platform to watch the film together. Hop into our Discord server at 7pm, and the film will begin after some time to catch up with your fellow members. (If you wouldn't like to hear any conversation during the film, then you're welcome to mute and deafen yourself!)

Split the coins into two equally sized piles, then flip all the coins in one of the piles. This is the solution as long as the number of coins with heads up is equal to the number of coins with tails up.

Follow up: can you find a solution with 100 coins, 10 with heads up and 90 with tails up? (If you think you've solved this one, get in touch and we'll let you know if you're right!)

Imagine a square board of N by N, such as the below board where N=3. On this board the integers 1 to N² must be placed in any order.

TODO: paste an image

A path is a series of N squares from the left to the right of the board moving only right, right-up, or right-down with each new square. The score of this path is the sum of the integers on the squares the path moves through.

For instance the below path scores 6 (path shown in emboldened italicised and underlined text):

TODO: paste an image

Alice claims that you can always find the path which has the best (lowest) score by working backwards from right to left picking the lowest lowest scoring square each time. Therefore finding the path in O(N) time.

Bob claims that the only way to find the path which has the best (lowest) score is by trying every possible route from left to right keeping track of the best scoring route. Therefore finding the path in O(3^N-1 × N) time.

Are either of them right? If not, can you find a method that does always find the best scoring route, and is it more or less complex than Bob’s approach?

Make sure you're in HackSoc's Slack workspace! We have over 200 members, and it's free to join the conversation, so sign up now!

If you're joining us for any of our virtual events, make sure you're also in our Discord server.

Have a great week,

Aaron 🤖