Description for today’s challenge is here.
The input for Day 9 looks like this:
AlphaCentauri to Snowdin = 66
AlphaCentauri to Tambi = 28
AlphaCentauri to Faerun = 60
Snowdin to Tambi = 22
Snowdin to Faerun = 12
there are similarities between this challenge and Day 7 in that they’re essentially a graph problem. In that case, let’s model this problem as a graph:
in our model of this problem, we have a set of cities, and the connections between them (where the key is the city to travel from, and the value is a map of the cities that you can travel to and the distance between the pair).
Like we did in Day 7, let’s add a helper function to help connect two cities – a and b – together, along with the distance between them.
Armed with the above, we can now go back to our input, parse each line into two cities and their distance, and then fold over them to build up a graph:
notice that the information we get from the input list is actually bi-directional, which is why in the fold we have to add the direction from a to b, as well as from b to a.
Next, starting with each cities, we want to find the routes that’ll visit each city once and only once.
aside: in hindsight, both the model and the path finding algorithm here can be simplified, I simply didn’t realise the distance data is bi-directional (it was implied through the example which I skimmed over and got started writing code…). For instance, the data is set up such that every city is connected to every other city, so the steps I have taken above to determine ‘which cities can I travel from here’ is redundant.
That said, I decided to keep the solution as it is because:
a) it can solve a more general problem where cities do not form a mesh network
b) it works, and pretty easy to understand, so why change
The last building block we need to compose our final solution is the ability to calculate the total distance we have to travel in a route:
and to answer the challenge:
in hindsight, it might be better to write the above as:
graph |> findPaths |> Seq.map calcDistance |> Seq.min
To work out the longest distance one can travel, simply flap the last line to look for the max distance instead, simple:
Enjoy what you’re reading? Subscribe to my newsletter and get more content on AWS and serverless technologies delivered straight to your inbox.
I’m an AWS Serverless Hero and the author of Production-Ready Serverless. I have run production workload at scale in AWS for nearly 10 years and I have been an architect or principal engineer with a variety of industries ranging from banking, e-commerce, sports streaming to mobile gaming. I currently work as an independent consultant focused on AWS and serverless.
In this course, we’ll cover everything you need to know to use AWS Step Functions service effectively. Including basic concepts, HTTP and event triggers, activities, design patterns and best practices.
Here is a complete list of all my posts on serverless and AWS Lambda. In the meantime, here are a few of my most popular blog posts.
- Lambda optimization tip – enable HTTP keep-alive
- You are thinking about serverless costs all wrong
- Many faced threats to Serverless security
- We can do better than percentile latencies
- I’m afraid you’re thinking about AWS Lambda cold starts all wrong
- Yubl’s road to Serverless
- AWS Lambda – should you have few monolithic functions or many single-purposed functions?
- AWS Lambda – compare coldstart time with different languages, memory and code sizes
- Guys, we’re doing pagination wrong