A ride that takes 10^20k years to complete in Roller Coaster Tycoon 2 [video]

Re:A ride that takes 10^20k years to complete in Roller Coas... by im3w1l on Monday August 03, 2020 @05:26PM (#24043663)

Let's pose the problem mathematically. Initial state is (0, forwards).

State (0, *) transitions to (1, forwards)

For k > 1, state (k, forwards) transitions to (k+1, forwards) with 5/8 probability and to (k-1, backwards) with probability 3/8.

For k > 1, state (k, backwards) transitions to (k+1, forwards) with 1/8 probability and to (k-1, backwards) with probability 7/8.

My intuition is that this very much resembles a biased walk so it should take exponential time. But there is the "momentum" to take into account.

Edit: It looks like we can use https://en.wikipedia.org/wiki/Absorbing_Markov_chain#Expecte...

[ Parent | on HN ]

Re:A ride that takes 10^20k years to complete in Roller Coas... by gbrown on Monday August 03, 2020 @05:41PM (#24043801)

Came here for this - I knew there must be an analytical solution involving Markov chains.

[ Parent | on HN ]

Re:A ride that takes 10^20k years to complete in Roller Coas... by tetha on Monday August 03, 2020 @06:21PM (#24044166)

I'm pretty sure this can be modeled with markov chains.

Your vertices would be (x, y, f), f being facing.

Edges are along the lines of (x, y, 0) - (x+1, y, {0,1,2,3}\1); (x, y, 1) - (x-1, y, {0,1,2,3}\0); (x, y, 2) - (x, y+1, {0,1,2,3}\3); (x, y, 3) - (x, y-1, {0,1,2,3}\2).

Probabilities are modeled in the video and need to be rotated 4 times for the four facings.

After that, I just don't recall my eigenvector calculation anymore. It feels like a good markov exercise, though.

[ Parent | on HN ]