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