A question about function dubins_LRL

[tinytinywell]

Hello, could you please tell me why p=mod(( 2pi - acos( tmp_lrl ) ), 2pi)? This is different from what I found in paper "Classification of the Dubins setq", and q is also different. Thank you! function param = dubins_LRL(alpha, beta, d) tmp_lrl = (6. - dd + 2cos(alpha - beta) + 2d(- sin(alpha) + sin(beta))) / 8.; if( abs(tmp_lrl) > 1) param = [-1, -1, -1]; return; else p = mod(( 2pi - acos( tmp_lrl ) ), 2pi); t = mod((-alpha - atan2( cos(alpha)-cos(beta), d+sin(alpha)-sin(beta) ) + p/2), 2pi); q = mod((mod(beta, 2pi) - alpha -t + mod(p, 2pi)), 2pi); param(1) = t; param(2) = p; param(3) = q; return; end end