Version 5.0 The LEDA User Manual

√
y v = b 1 c 2 + b 2 c 1 ± sign(r) 2c 1 c 2 ( √ N − D)
z v = √ N − a 1 a 2 − b 1 b 2
and
s = b 1 D 2 − b 2 D 1 , N = (a 2 1 + b2 1 )(a2 2 + b2 2 )
r = a 1 D 2 − a 2 D 1 , D = a 1 a 2 − b 1 b 2 .
Let us concentrate on one of these (say, we take the plus sign in both cases). The test
whether l 3 intersects, touches or misses the circle amounts to determining the sign of
E := dist 2 (v, l 3 ) − dist 2 (v, p) = (a 3x v + b 3 y v + c 3 ) 2
a 2 3 + b 2 3
− (x 2 v + y2 v ).
The following program computes the sign of Ẽ := (a2 3 + b2 3 ) · E using our data type real.
int incircle( real a 1 , real b 1 , real c 1 , real a 2 , real b 2 , real c 2 , real a 3 , real b 3 ,
real c 3 )
{
real RN = sqrt((a 1 ∗ a 1 + b 1 ∗ b 1 ) ∗ (a 2 ∗ a 2 + b 2 ∗ b 2 ));
real RN 1 = sqrt(a 1 ∗ a 1 + b 1 ∗ b 1 );
real RN 2 = sqrt(a 2 ∗ a 2 + b 2 ∗ b 2 );
real A = a 1 ∗ c 2 + a 2 ∗ c 1 ;
real B = b 1 ∗ c 2 + b 2 ∗ c 1 ;
real C = 2 ∗ c 1 ∗ c 2 ;
real D = a 1 ∗ a 2 − b 1 ∗ b 2 ;
real s = b 1 ∗ RN 2 − b 2 ∗ RN 1 ;
real r = a 1 ∗ RN 2 − a 2 ∗ RN 1 ;
int sign x = sign(s);
int sign y = sign(r);
real x v = A + sign x ∗ sqrt(C ∗ (RN + D));
real y v = B − sign y ∗ sqrt(C ∗ (RN − D));
real z v = RN − (a 1 ∗ a 2 + b 1 ∗ b 2 );
real P = a 3 ∗ x v + b 3 ∗ y v + c 3 ∗ z v ;
real D 2 3 = a 3 ∗ a 3 + b 3 ∗ b 3 ;
real R 2 = x v ∗ x v + y v ∗ y v ;
real E = P ∗ P − D 2 3 ∗ R2 ;
return sign(E);
}
We can make the above program more efficient if all coefficients a i , b i and c i , 1 ≤ i ≤ 3,
are k bit integers, i.e., integers whose absolute value is bounded by 2 k − 1. In [17, 14] we
showed that for Ẽ ≠ 0 we have |Ẽ| ≥ 2−24k−26 . Hence we may add a parameter int k in
the above program and replace the last line by
return E.sign(24 ∗ k + 26).