>>5887

>The final steps involve finding a base using (f-1)/8 for each of the eight triangles (some a bit bigger than others by a unit in length) which is larger than (n-1)/8 BUT smaller than (x+n)/8.

>larger than (n-1)/8 BUT smaller than (x+n)/8

For each given (x+n), there is a range of possible n values and a range of possible x values. Picture 2 shows this pretty well. For (x+n) = 9, n-1 is going to be in the range of 0 to 5 and x is going to be in the range of 3 to 8. Picture 1 was related to something else (I had found linear gradients in a bitmap but they didn't lead to anything because of the thing this picture explains), but it shows what VQC is talking about here: the range of n values and the range of x values overlap. If we don't know (x+n) but we want to set an n0 value based on our known f value, the minimum possible (x+n) and maximum possible n values for our given c also overlap in the same way (that's what the gradients were about, and I guess it's finally relevant). So n0 is currently a guess within an overlapping range. I could see if I can find the gradient picture and use it to make an equation for finding the range of n and (x+n) for a given c if that would help (it sounds like it would to me but I don't tend to get feedback about these things).

>>5904

Well thank you for the encouragement VA. I'm glad we all have good comradery here. You seem to be a big part of that. I think that first logo looks a lot less like a sigil compared to the other ones so I'd agree with your choice.

>>5960

>max n in a given x+n

I'm talking about the max n for a given c. That's what VQC's talking about. These pictures are the ones I was talking about from something I posted about that didn't go anywhere in February. I'll make bigger versions in a bit. It shows that there's a series of c values for which the corresponding (x+n) values follow the gradient along the bottom, and there's a series of c values for which the corresponding (x+n) values follow a slightly lower gradient but the gradient starts after the first gradient, and so on. The starting point of each gradient and the actual gradient are calculable, I'm pretty sure. What is means is that you can find the maximum and minimum (x+n) values for a given c. The same thing happens for n, as you can see in the second picture. What this means is that for a given c you can find the range of possible n values and the range of possible (x+n) values. That's exactly what VQC was talking about doing (finding the right f value (which we know for our given c) so that n0 is bigger than n-1 and smaller than (x+n)). The only problem is, for a given c, the range of possible n values and (x+n) values overlaps. n0 is somewhere in the overlap.

>I also looked for the formula for max n in a given x+n

It definitely seems possible to calculate, based on this graph I made. The x axis is (x+n) and the y axis is n. I added the black lines in Paint to show that it's a non-linear line made of increasingly-sized linear lines.

>>5961

I should maybe point out for anyone who might be confused (not necessarily you PMA) that the ranges I'm talking about in this post are the range of possible values but not every value is valid (it's only going to be one of them but if we were to guess it would be within this range), so it's different to the range of n values for a given (x+n) within which all the n values are valid.