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@@ -4,17 +4,15 @@
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#define _USE_MATH_DEFINES
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#include <cmath>
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-#define TOO_LARGE_RATIO 1e12
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-
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namespace msdfgen {
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int solveQuadratic(double x[2], double a, double b, double c) {
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- // a = 0 -> linear equation
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- if (a == 0 || fabs(b)+fabs(c) > TOO_LARGE_RATIO*fabs(a)) {
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- // a, b = 0 -> no solution
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- if (b == 0 || fabs(c) > TOO_LARGE_RATIO*fabs(b)) {
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+ // a == 0 -> linear equation
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+ if (a == 0 || fabs(b) > 1e12*fabs(a)) {
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+ // a == 0, b == 0 -> no solution
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+ if (b == 0) {
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if (c == 0)
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- return -1; // 0 = 0
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+ return -1; // 0 == 0
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return 0;
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}
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x[0] = -c/b;
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@@ -35,41 +33,38 @@ int solveQuadratic(double x[2], double a, double b, double c) {
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static int solveCubicNormed(double x[3], double a, double b, double c) {
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double a2 = a*a;
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- double q = (a2 - 3*b)/9;
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- double r = (a*(2*a2-9*b) + 27*c)/54;
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+ double q = 1/9.*(a2-3*b);
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+ double r = 1/54.*(a*(2*a2-9*b)+27*c);
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double r2 = r*r;
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double q3 = q*q*q;
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- double A, B;
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+ a *= 1/3.;
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if (r2 < q3) {
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double t = r/sqrt(q3);
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if (t < -1) t = -1;
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if (t > 1) t = 1;
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t = acos(t);
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- a /= 3; q = -2*sqrt(q);
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- x[0] = q*cos(t/3)-a;
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- x[1] = q*cos((t+2*M_PI)/3)-a;
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- x[2] = q*cos((t-2*M_PI)/3)-a;
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+ q = -2*sqrt(q);
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+ x[0] = q*cos(1/3.*t)-a;
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+ x[1] = q*cos(1/3.*(t+2*M_PI))-a;
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+ x[2] = q*cos(1/3.*(t-2*M_PI))-a;
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return 3;
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} else {
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- A = -pow(fabs(r)+sqrt(r2-q3), 1/3.);
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- if (r < 0) A = -A;
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- B = A == 0 ? 0 : q/A;
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- a /= 3;
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- x[0] = (A+B)-a;
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- x[1] = -0.5*(A+B)-a;
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- x[2] = 0.5*sqrt(3.)*(A-B);
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- if (fabs(x[2]) < 1e-14)
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+ double u = (r < 0 ? 1 : -1)*pow(fabs(r)+sqrt(r2-q3), 1/3.);
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+ double v = u == 0 ? 0 : q/u;
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+ x[0] = (u+v)-a;
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+ if (u == v || fabs(u-v) < 1e-12*fabs(u+v)) {
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+ x[1] = -.5*(u+v)-a;
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return 2;
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+ }
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return 1;
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}
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}
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int solveCubic(double x[3], double a, double b, double c, double d) {
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if (a != 0) {
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- double bn = b/a, cn = c/a, dn = d/a;
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- // Check that a isn't "almost zero"
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- if (fabs(bn) < TOO_LARGE_RATIO && fabs(cn) < TOO_LARGE_RATIO && fabs(dn) < TOO_LARGE_RATIO)
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- return solveCubicNormed(x, bn, cn, dn);
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+ double bn = b/a;
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+ if (fabs(bn) < 1e6) // Above this ratio, the numerical error gets larger than if we treated a as zero
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+ return solveCubicNormed(x, bn, c/a, d/a);
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}
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return solveQuadratic(x, b, c, d);
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}
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