Sorry I feel like a lot of this stuff is already known I'm just writing what I know for know about e+2n shifts. So when you shift with e+2n, you keep the same n and x values. Then your a and b both increase by one. Therefore your n-1 value stays the same as well as the nn value. f shifts down by 2(n-1) as you increase e by 2n. The (x+n) stays the same, so to compensate for the f shifts, the (2d-1)(n-1) also increases by 2(n-1). d increases by one each time also, and since n is the same, (d+n) increases by one each time. So the e = e+2n shifts to the right correlate with an increase in the (d+n) by one and keeping the (x+n) the same and also correlate with a=a+1, b=b+1. The opposite "move" would be to decrease increase (x+n) by one and keep (d+n) the same. To do this, you do a=a-1, b=b+1. To do this, you also need to ensure you have the same x value. For this, your n value is increasing by one each time. Then if you look at the f value it goes down then back up kind of parabolically as you do this and the vertex seems to be at the point where x=a=(n-1). I'm pretty sure that this gif of the xGrid (iterates through c values) is a visualization of the moves along this path, because both transformations involve keeping x the same. The x grid is also very easy to look at its pretty predictable I think that each little vertex is at (e,n) = (-x^2,0)