The page 6 of this paper says:
Condition 1: $p(0) = q(1)$
– Symmetry: $p(0) = q(1) = 1/6(\pi-2 + 4 \pi-1 + \pi)$
Condition 2: $p’(0) = q’(1)$
– Geometry: $p’(0) = q’(1) = 1/2 ((\pi – \pi-1) + (\pi-1 – \pi-2)) = 1/2 (\pi – \pi-2)$
...but I don't see what symmetry it's referring to and how it reached $1/6(\pi-2 + 4 \pi-1 + \pi)$ and $1/2 ((\pi – \pi-1) + (\pi-1 – \pi-2))$. Could someone explain that?
The article you refer to goes too fast. It should not be able to reach the conclusion $p(0)=\frac{1}{6}(p_{i-2}+4p_{i-1}+p_i)$ simply from positional continuity and symmetry. This article http://www2.cs.uregina.ca/~anima/408/Notes/Interpolation/UniformBSpline.htm provides a better explanation about how to derive the basis functions for uniform cubic B-spline. Open this article and scroll down to the section "Deriving the Weighting Functions". The weighting functions here mean the basis functions and you can rewrite them in matrix form to get what you want.