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      1 // from http://tog.acm.org/resources/GraphicsGems/gems/Roots3And4.c
      2 /*
      3  *  Roots3And4.c
      4  *
      5  *  Utility functions to find cubic and quartic roots,
      6  *  coefficients are passed like this:
      7  *
      8  *      c[0] + c[1]*x + c[2]*x^2 + c[3]*x^3 + c[4]*x^4 = 0
      9  *
     10  *  The functions return the number of non-complex roots and
     11  *  put the values into the s array.
     12  *
     13  *  Author:         Jochen Schwarze (schwarze (at) isa.de)
     14  *
     15  *  Jan 26, 1990    Version for Graphics Gems
     16  *  Oct 11, 1990    Fixed sign problem for negative q's in SolveQuartic
     17  *                  (reported by Mark Podlipec),
     18  *                  Old-style function definitions,
     19  *                  IsZero() as a macro
     20  *  Nov 23, 1990    Some systems do not declare acos() and cbrt() in
     21  *                  <math.h>, though the functions exist in the library.
     22  *                  If large coefficients are used, EQN_EPS should be
     23  *                  reduced considerably (e.g. to 1E-30), results will be
     24  *                  correct but multiple roots might be reported more
     25  *                  than once.
     26  */
     27 
     28 #include    <math.h>
     29 #include "CubicUtilities.h"
     30 #include "QuadraticUtilities.h"
     31 #include "QuarticRoot.h"
     32 
     33 int reducedQuarticRoots(const double t4, const double t3, const double t2, const double t1,
     34         const double t0, const bool oneHint, double roots[4]) {
     35 #ifdef SK_DEBUG
     36     // create a string mathematica understands
     37     // GDB set print repe 15 # if repeated digits is a bother
     38     //     set print elements 400 # if line doesn't fit
     39     char str[1024];
     40     bzero(str, sizeof(str));
     41     sprintf(str, "Solve[%1.19g x^4 + %1.19g x^3 + %1.19g x^2 + %1.19g x + %1.19g == 0, x]",
     42         t4, t3, t2, t1, t0);
     43     mathematica_ize(str, sizeof(str));
     44 #if ONE_OFF_DEBUG && ONE_OFF_DEBUG_MATHEMATICA
     45     SkDebugf("%s\n", str);
     46 #endif
     47 #endif
     48 #if 0 && SK_DEBUG
     49     bool t4Or = approximately_zero_when_compared_to(t4, t0) // 0 is one root
     50             || approximately_zero_when_compared_to(t4, t1)
     51             || approximately_zero_when_compared_to(t4, t2);
     52     bool t4And = approximately_zero_when_compared_to(t4, t0) // 0 is one root
     53             && approximately_zero_when_compared_to(t4, t1)
     54             && approximately_zero_when_compared_to(t4, t2);
     55     if (t4Or != t4And) {
     56         SkDebugf("%s t4 or and\n", __FUNCTION__);
     57     }
     58     bool t3Or = approximately_zero_when_compared_to(t3, t0)
     59             || approximately_zero_when_compared_to(t3, t1)
     60             || approximately_zero_when_compared_to(t3, t2);
     61     bool t3And = approximately_zero_when_compared_to(t3, t0)
     62             && approximately_zero_when_compared_to(t3, t1)
     63             && approximately_zero_when_compared_to(t3, t2);
     64     if (t3Or != t3And) {
     65         SkDebugf("%s t3 or and\n", __FUNCTION__);
     66     }
     67     bool t0Or = approximately_zero_when_compared_to(t0, t1) // 0 is one root
     68             && approximately_zero_when_compared_to(t0, t2)
     69             && approximately_zero_when_compared_to(t0, t3)
     70             && approximately_zero_when_compared_to(t0, t4);
     71     bool t0And = approximately_zero_when_compared_to(t0, t1) // 0 is one root
     72             && approximately_zero_when_compared_to(t0, t2)
     73             && approximately_zero_when_compared_to(t0, t3)
     74             && approximately_zero_when_compared_to(t0, t4);
     75     if (t0Or != t0And) {
     76         SkDebugf("%s t0 or and\n", __FUNCTION__);
     77     }
     78 #endif
     79     if (approximately_zero_when_compared_to(t4, t0) // 0 is one root
     80             && approximately_zero_when_compared_to(t4, t1)
     81             && approximately_zero_when_compared_to(t4, t2)) {
     82         if (approximately_zero_when_compared_to(t3, t0)
     83             && approximately_zero_when_compared_to(t3, t1)
     84             && approximately_zero_when_compared_to(t3, t2)) {
     85             return quadraticRootsReal(t2, t1, t0, roots);
     86         }
     87         if (approximately_zero_when_compared_to(t4, t3)) {
     88             return cubicRootsReal(t3, t2, t1, t0, roots);
     89         }
     90     }
     91     if ((approximately_zero_when_compared_to(t0, t1) || approximately_zero(t1))// 0 is one root
     92       //      && approximately_zero_when_compared_to(t0, t2)
     93             && approximately_zero_when_compared_to(t0, t3)
     94             && approximately_zero_when_compared_to(t0, t4)) {
     95         int num = cubicRootsReal(t4, t3, t2, t1, roots);
     96         for (int i = 0; i < num; ++i) {
     97             if (approximately_zero(roots[i])) {
     98                 return num;
     99             }
    100         }
    101         roots[num++] = 0;
    102         return num;
    103     }
    104     if (oneHint) {
    105         SkASSERT(approximately_zero(t4 + t3 + t2 + t1 + t0)); // 1 is one root
    106         int num = cubicRootsReal(t4, t4 + t3, -(t1 + t0), -t0, roots); // note that -C==A+B+D+E
    107         for (int i = 0; i < num; ++i) {
    108             if (approximately_equal(roots[i], 1)) {
    109                 return num;
    110             }
    111         }
    112         roots[num++] = 1;
    113         return num;
    114     }
    115     return -1;
    116 }
    117 
    118 int quarticRootsReal(int firstCubicRoot, const double A, const double B, const double C,
    119         const double D, const double E, double s[4]) {
    120     double  u, v;
    121     /* normal form: x^4 + Ax^3 + Bx^2 + Cx + D = 0 */
    122     const double invA = 1 / A;
    123     const double a = B * invA;
    124     const double b = C * invA;
    125     const double c = D * invA;
    126     const double d = E * invA;
    127     /*  substitute x = y - a/4 to eliminate cubic term:
    128     x^4 + px^2 + qx + r = 0 */
    129     const double a2 = a * a;
    130     const double p = -3 * a2 / 8 + b;
    131     const double q = a2 * a / 8 - a * b / 2 + c;
    132     const double r = -3 * a2 * a2 / 256 + a2 * b / 16 - a * c / 4 + d;
    133     int num;
    134     if (approximately_zero(r)) {
    135     /* no absolute term: y(y^3 + py + q) = 0 */
    136         num = cubicRootsReal(1, 0, p, q, s);
    137         s[num++] = 0;
    138     } else {
    139         /* solve the resolvent cubic ... */
    140         double cubicRoots[3];
    141         int roots = cubicRootsReal(1, -p / 2, -r, r * p / 2 - q * q / 8, cubicRoots);
    142         int index;
    143     #if 0 && SK_DEBUG // enable to verify that any cubic root is as good as any other
    144         double tries[3][4];
    145         int nums[3];
    146         for (index = 0; index < roots; ++index) {
    147             /* ... and take one real solution ... */
    148             const double z = cubicRoots[index];
    149             /* ... to build two quadric equations */
    150             u = z * z - r;
    151             v = 2 * z - p;
    152             if (approximately_zero_squared(u)) {
    153                 u = 0;
    154             } else if (u > 0) {
    155                 u = sqrt(u);
    156             } else {
    157                 SkDebugf("%s u=%1.9g <0\n", __FUNCTION__, u);
    158                 continue;
    159             }
    160             if (approximately_zero_squared(v)) {
    161                 v = 0;
    162             } else if (v > 0) {
    163                 v = sqrt(v);
    164             } else {
    165                 SkDebugf("%s v=%1.9g <0\n", __FUNCTION__, v);
    166                 continue;
    167             }
    168             nums[index] = quadraticRootsReal(1, q < 0 ? -v : v, z - u, tries[index]);
    169             nums[index] += quadraticRootsReal(1, q < 0 ? v : -v, z + u, tries[index] + nums[index]);
    170             /* resubstitute */
    171             const double sub = a / 4;
    172             for (int i = 0; i < nums[index]; ++i) {
    173                 tries[index][i] -= sub;
    174             }
    175         }
    176         for (index = 0; index < roots; ++index) {
    177             SkDebugf("%s", __FUNCTION__);
    178             for (int idx2 = 0; idx2 < nums[index]; ++idx2) {
    179                 SkDebugf(" %1.9g", tries[index][idx2]);
    180             }
    181             SkDebugf("\n");
    182         }
    183     #endif
    184         /* ... and take one real solution ... */
    185         double z;
    186         num = 0;
    187         int num2 = 0;
    188         for (index = firstCubicRoot; index < roots; ++index) {
    189             z = cubicRoots[index];
    190             /* ... to build two quadric equations */
    191             u = z * z - r;
    192             v = 2 * z - p;
    193             if (approximately_zero_squared(u)) {
    194                 u = 0;
    195             } else if (u > 0) {
    196                 u = sqrt(u);
    197             } else {
    198                 continue;
    199             }
    200             if (approximately_zero_squared(v)) {
    201                 v = 0;
    202             } else if (v > 0) {
    203                 v = sqrt(v);
    204             } else {
    205                 continue;
    206             }
    207             num = quadraticRootsReal(1, q < 0 ? -v : v, z - u, s);
    208             num2 = quadraticRootsReal(1, q < 0 ? v : -v, z + u, s + num);
    209             if (!((num | num2) & 1)) {
    210                 break; // prefer solutions without single quad roots
    211             }
    212         }
    213         num += num2;
    214         if (!num) {
    215             return 0; // no valid cubic root
    216         }
    217     }
    218     /* resubstitute */
    219     const double sub = a / 4;
    220     for (int i = 0; i < num; ++i) {
    221         s[i] -= sub;
    222     }
    223     // eliminate duplicates
    224     for (int i = 0; i < num - 1; ++i) {
    225         for (int j = i + 1; j < num; ) {
    226             if (AlmostEqualUlps(s[i], s[j])) {
    227                 if (j < --num) {
    228                     s[j] = s[num];
    229                 }
    230             } else {
    231                 ++j;
    232             }
    233         }
    234     }
    235     return num;
    236 }
    237