Author Topic: "Desert Bus"  (Read 2738 times)

finka

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"Desert Bus"
« on: August 01, 2011, 07:54:52 PM »
This is a tiny, irritating, hopefully unfeaturable homage that sprang fully-formed into my mind after one death too many waiting on a conveyor between spikes. 

Anyone who knows more about VVVVVV frame timings care to figure out how long this actually takes?

cliff

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Re: "Desert Bus"
« Reply #1 on: August 02, 2011, 02:08:05 PM »
This was very cute, I like how you captured the bus veering right.

I guess those blocks are supposed to align at some time for you to be able to go through? Seems like that part could be really hard to time as a player.

As an alternate idea, I could imagine a massive level (which might be impossible) where you would constantly go in one direction but then need to press left to miss being thrown off onto spikes every once in a while.

Martze

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Re: "Desert Bus"
« Reply #2 on: August 02, 2011, 09:48:53 PM »
This is creative. I was expecting to load it up and find a big long empty path. It turned out to not be that.  You earn points for that.

Desert Bus is a game that is hilarious by concept but no fun at all to play. You captured that beautifully.

Let's work in 8-pixel units, for now. The columns are 18, 20, 22, 24, 26 squares high. Subtracting the width of the enemy--two squares--gives us the distance each enemy must travel between each bounce: 16, 18. 20, 22, and 24 squares. The least common multiple of that is 7920 squares.

I don't know the specific game timings, but the enemies go about 16 squares per second. That's about 500 seconds. Not bad, but you could make it a whole lot more infuriating by adding columns whose heights are odd.

finka

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Re: "Desert Bus"
« Reply #3 on: August 05, 2011, 11:17:16 PM »
It's worse than that, Martze, given that the enemies don't turn around instantaneously, so instead of a distance of 16 squares we should be counting 16+x, for some x which I don't know. 

Watching the shortest two columns until they line up again, unless I miscounted, took 38 periods of the smallest one to realign with 34 periods of the second-shortest.  So it appears x=1 and we really have lcm(17,19,21,23,25) = 3900225 (those are relatively prime) squares / the enemy speed. 
ETA: and it looks like you've omitted a factor of 2 corresponding to having to go up and down, so it's really 7800450 squares / the enemy speed, which is just short of 5 days 16 hours if it is 16 squares / second. 

ETA 2: though maybe it's still possible to make it through if some of the columns are one square off from their starting position.  I don't feel like Chinese remainder theoreming that up now...
« Last Edit: August 05, 2011, 11:42:02 PM by finka »