tasks a-s

Task/Problem-of-Apollonius/C/problem-of-apollonius.c

@@ -0,0 +1,110 @@
+#include <stdio.h>
+#include <tgmath.h>
+
+#define VERBOSE 0
+#define for3 for(int i = 0; i < 3; i++)
+
+typedef complex double vec;
+typedef struct { vec c; double r; } circ;
+
+#define re(x) creal(x)
+#define im(x) cimag(x)
+#define cp(x) re(x), im(x)
+#define CPLX "(%6.3f,%6.3f)"
+#define CPLX3 CPLX" "CPLX" "CPLX
+
+double cross(vec a, vec b) { return re(a) * im(b) - im(a) * re(b); }
+double abs2(vec a) { return a * conj(a); }
+
+int apollonius_in(circ aa[], int ss[], int flip, int divert)
+{
+	vec n[3], x[3], t[3], a, b, center;
+	int s[3], iter = 0, res = 0;
+	double diff = 1, diff_old = -1, axb, d, r;
+
+	for3 {
+		s[i] = ss[i] ? 1 : -1;
+		x[i] = aa[i].c;
+	}
+
+	while (diff > 1e-20) {
+		a = x[0] - x[2], b = x[1] - x[2];
+		diff = 0;
+		axb = -cross(a, b);
+		d = sqrt(abs2(a) * abs2(b) * abs2(a - b));
+
+		if (VERBOSE) {
+			const char *z = 1 + "-0+";
+			printf("%c%c%c|%c%c|",
+				z[s[0]], z[s[1]], z[s[2]], z[flip], z[divert]);
+			printf(CPLX3, cp(x[0]), cp(x[1]), cp(x[2]));
+		}
+
+		/* r and center represent an arc through points x[i].  Each step,
+		   we'll deform this arc by pushing or pulling some point on it
+		   towards the edge of each given circle. */
+		r = fabs(d / (2 * axb));
+		center = (abs2(a)*b - abs2(b)*a) / (2 * axb) * I + x[2];
+
+ 		/* maybe the "arc" is actually straight line; then we have two
+		   choices in defining "push" and "pull", so try both */
+		if (!axb && flip != -1 && !divert) {
+			if (!d) { /* generally means circle centers overlap */
+				printf("Given conditions confused me.\n");
+				return 0;
+			}
+
+			if (VERBOSE) puts("\n[divert]");
+			divert = 1;
+			res = apollonius_in(aa, ss, -1, 1);
+		}
+
+ 		/* if straight line, push dir is its norm; else it's away from center */
+		for3 n[i] = axb ? aa[i].c - center : a * I * flip;
+		for3 t[i] = aa[i].c + n[i] / cabs(n[i]) * aa[i].r * s[i];
+
+		/* diff: how much tangent points have moved since last iteration */
+		for3 diff += abs2(t[i] - x[i]), x[i] = t[i];
+
+		if (VERBOSE) printf(" %g\n", diff);
+
+ 		/* keep an eye on the total diff: failing to converge means no solution */
+		if (diff >= diff_old && diff_old >= 0)
+			if (iter++ > 20) return res;
+
+		diff_old = diff;
+	}
+
+	printf("found: ");
+	if (axb) printf("circle "CPLX", r = %f\n", cp(center), r);
+	else	 printf("line "CPLX3"\n", cp(x[0]), cp(x[1]), cp(x[2]));
+
+	return res + 1;
+}
+
+int apollonius(circ aa[])
+{
+	int s[3], i, sum = 0;
+	for (i = 0; i < 8; i++) {
+		s[0] = i & 1, s[1] = i & 2, s[2] = i & 4;
+
+		/* internal or external results of a zero-radius circle are the same */
+		if (s[0] && !aa[0].r) continue;
+		if (s[1] && !aa[1].r) continue;
+		if (s[2] && !aa[2].r) continue;
+		sum += apollonius_in(aa, s, 1, 0);
+	}
+	return sum;
+}
+
+int main()
+{
+	circ a[3] = {{0, 1}, {4, 1}, {2 + 4 * I, 1}};
+	circ b[3] = {{-3, 2}, {0, 1}, {3, 2}};
+	circ c[3] = {{-2, 1}, {0, 1}, {2 * I, 1}};
+	//circ c[3] = {{0, 1}, {0, 2}, {0, 3}}; <-- a fun one
+
+	puts("set 1"); apollonius(a);
+	puts("set 2"); apollonius(b);
+	puts("set 3"); apollonius(c);
+}