AA ID: f984d8 Feb. 22, 2018, 8:40 p.m. No.4852   🗄️.is 🔗kun   >>4853 >>4856 >>4857

>>4841

Personally I don't mind some irrelevant posts here and there. I'm talking about, like, a co-ordinated mass shilling, the board being filled with guro by someone who just wanted to annoy some people, the brown pill guy, that kind of thing.

 

>>4842

If we didn't go with a new bunker board and we tried staying here, one way to not only keep VQC looped in but to make sure he can say what he needs to would be to give him a mod account or a volunteer account so he can do what Q does on >>>/greatawakening/ with the locked thread. I was meant to give the board to him when we migrated here anyway. Otherwise, if we do create a bunker board, I guess we'll have to see if his Twitter account comes back. Obviously this has happened in the past.

 

>>4850

See >>4305

 

I think you're both right about factoring f. Is there any way to determine its parity and divisibility (i.e if it's prime there might be problems)? I'll put a few bitmaps together and see if anything comes out since I have some time. If we figure out a pattern to how f can be divided, it won't be a matter of trial and error trying to find a number f can be divided by for the f/(8*something) part.

AA ID: f984d8 Feb. 22, 2018, 9:03 p.m. No.4855   🗄️.is 🔗kun

I made images for nf and xf but they weren't very helpful. This one, however, is interesting. Obviously we don't need to know anything about gradients to calculate f from c, but what it does show is that for every chunk of however many fs there are in each of these lines, there's a gradient associated with that chunk. That means we could find the lowest point on each of these lines, take that number away from f, and then f would be divisible by its gradient. That makes it calculable, if we can figure out when each line starts, what each gradient is, and what the highest and lowest points are along each line. Of course, what it makes calculable is f minus something, but it looks like the lowest value is actually zero on every line, meaning we'd only need to work out the gradients and not subtract anything.