Lines Matching refs:neq

46 They are easy to handle so we won't cover them here. From now on, we assume $C_0 \neq C_1$ and $r_0
47 \neq r_1$.
49 As $r_0 \neq r_1$, we can find a focal point $C_f = (1-f) \cdot C_0 + f \cdot C_1$ where its
53 As $C_0 \neq C_1$, focal point $C_f$ is different from $C_1$ unless $r_1 = 0$. If $r_1 = 0$, we can
54 swap $C_0, r_0$ with $C_1, r_1$, compute swapped gradient $t_s$ as if $r_1 \neq 0$, and finally set
58 Assuming that we've done swapping if necessary so $C_1 \neq C_f$, we can then do a linear
68 always the bigger one (note that $f \neq 1$, otherwise we'll swap $C_0, r_0$ with $C_1, r_1$).
121     $x' = r_1 / (r_1^2 - 1) x,~ y' = \sqrt{|r_1^2 - 1|} / (r_1^2 - 1) y$ if $r_1 \neq 1$
148 * 2 additions, 3 multiplications, and 1 sqrt for $r_1 \neq 1$ (count substraction as addition;
150 * 1 more addition operation if $f \neq 0$
153 In comparison, for $r_1 \neq 1$ case, our current raster pipeline shading algorithm (which shall
322 As $r_1 \neq 1$, we can apply the similar transformation in Corollary 3. to get the two