Fun with Compression Method 1

[Minim]

Hello everyone. This is another one of those Clam-type mathematical kind of threads. Unlike his thread which reveals the whole solution, I'm going to spoiler mine to see if you can figure this one out!

Last year, he focused on "We all fall down". This time, the focus is on one of my favourite Taxing levels, and the sequel to Fun 17's "Easy when you know how", Compression Method 1, because it's my idea of fun watching them get killed, but there's also a factor of how many would be killed depending on the position of the blocker when the lemmings are released.

The big question is, assuming that there's an infinite number of lemmings so that at least one lemming is covering a pixel, and that the blocker is placed on the far right of the brick closest to the steel, and the blocker is turned into a walker* how many lemmings would at most be killed by the four traps?

Spoiler

What we want to find out first is the distance in pixels the lemming has to travel by hitting the blocker twice. I used the unique NeoLemmix feature of clicking one frame at a time to help me do this. The answer: 59 pixels, quite surprisingly because the lemming takes two frames to jump up a four-pixel wall and one to drop down.

If I turned the blocker into a walker, the distance between the walker and the rest of the group is 6 pixels, making the total distance 65.

They will be walking across the trap now. Each trap's trigger area is four pixels long and takes one frame to kill a lemming and then waits another 14 frames before the next lemming is killed, meaning that when bunched together, 1 in 15 lemmings would be killed. There is also a 3 in 15 chance that depending on the distance, a lemming at the back of the bunch will be killed by the same trap just by walking on a different trigger area.

So let's put this walker formula into practice then shall we? So, walking across the first trap:

(59 + 6) / 15 + 3/15

65 / 15 + 3/15

= 4.33.. + 0.2
= 4.533.. (rounded up) = 5

The second to fourth traps are straightforward but come up with a very interesting answer with the absence of the walker and the length of the trigger area coming into play:

59 / 15 + 3/15
= 3.933... + 0.2
= 4.133... (rounded up) = 5

So, with an infinite number of lemmings and by turning the blocker into a walker you can lose up to 20 lemmings just by going
through the four traps, just like the original level's requirement.

*I say walker because what happens if I bomb the blocker? Well, if we try to work it out mathematically, this would probably make the formula
more complicated, because if a lemming is affected by the bomber's mask he would fall, and therefore his distance compared to
the rest of the bunch would change. If the problem was severe enough so that no lemming hits the trigger area by the time the
trap uses its last frame, the probability factor would alternate and I would have no choice but to test the level.

So, I hope all the mathematical formulae and details I've shown are clear. Any questions I'm happy to answer. Thanks for watching!