Compute the maximum number of runs possible for as large a string as possible
C
This does a recursive search for optimal solutions, heavily pruned using an upper bound on the number of runs that could be finished by the unknown remainder of the string. The upper bound computation uses a gigantic lookup table whose size is controlled by the constant L (L=11: 0.5 GiB, L=12: 2 GiB, L=13: 8 GiB).
On my laptop, this goes up through n = 50 in 100 seconds; the next line comes at 142 seconds.
#include <stdbool.h>
#include <stdint.h>
#include <stdio.h>
#include <string.h>
#define N (8*sizeof(unsigned long long))
#define L 13
static bool a[N], best_a[N];
static int start[N/2 + 1], best_runs;
static uint8_t end_runs[2 << 2*L][2], small[N + 1][2 << 2*L];
static inline unsigned next_state(unsigned state, int b)
{
state *= 2;
state += b;
if (state >= 2 << 2*L) {
state &= ~(2 << 2*L);
state |= 1 << 2*L;
}
return state;
}
static void search(int n, int i, int runs, unsigned state)
{
if (i == n) {
int r = runs;
unsigned long long m = 0;
for (int p = n / 2; p > 0; p--) {
if (i - start[p] >= 2*p && !(m & 1ULL << start[p])) {
m |= 1ULL << start[p];
r++;
}
}
if (r > best_runs) {
best_runs = r;
memcpy(best_a, a, n*sizeof(a[0]));
}
} else {
a[i] = false;
do {
int r = runs, bound = 0, saved = 0, save_p[N/2], save_start[N/2], p, s = next_state(state, a[i]);
unsigned long long m = 0;
for (p = n/2; p > i; p--)
if (p > L)
bound += (n - p + 1)/(p + 1);
for (; p > 0; p--) {
if (a[i] != a[i - p]) {
if (i - start[p] >= 2*p && !(m & 1ULL << start[p])) {
m |= 1ULL << start[p];
r++;
}
save_p[saved] = p;
save_start[saved] = start[p];
saved++;
start[p] = i + 1 - p;
if (p > L)
bound += (n - i)/(p + 1);
} else {
if (p > L)
bound += (n - (start[p] + p - 1 > i - p ? start[p] + p - 1 : i - p))/(p + 1);
}
}
bound += small[n - i - 1][s];
if (r + bound > best_runs)
search(n, i + 1, r, s);
while (saved--)
start[save_p[saved]] = save_start[saved];
} while ((a[i] = !a[i]));
}
}
int main()
{
for (int n = 0; n <= N; n++) {
if (n <= 2*L) {
for (unsigned state = 1U << n; state < 2U << n; state++) {
for (int b = 0; b < 2; b++) {
int r = 0;
unsigned long long m = 0;
for (int p = n / 2; p > 0; p--) {
if ((b ^ state >> (p - 1)) & 1) {
unsigned k = state ^ state >> p;
k &= -k;
k <<= p;
if (!(k & ~(~0U << n)))
k = 1U << n;
if (!((m | ~(~0U << 2*p)) & k)) {
m |= k;
r++;
}
}
}
end_runs[state][b] = r;
}
small[0][state] = end_runs[state][0] + end_runs[state][1];
}
}
for (int l = 2*L < n - 1 ? 2*L : n - 1; l >= 0; l--) {
for (unsigned state = 1U << l; state < 2U << l; state++) {
int r0 = small[n - l - 1][next_state(state, 0)] + end_runs[state][0],
r1 = small[n - l - 1][next_state(state, 1)] + end_runs[state][1];
small[n - l][state] = r0 > r1 ? r0 : r1;
}
}
if (n >= 2) {
search(n, 1, 0, 2U);
printf("%d %d ", n, best_runs);
for (int i = 0; i < n; i++)
printf("%d", best_a[i]);
printf("\n");
fflush(stdout);
best_runs--;
}
}
return 0;
}
Output:
$ gcc -mcmodel=medium -O2 runs.c -o runs
$ ./runs
2 1 00
3 1 000
4 2 0011
5 2 00011
6 3 001001
7 4 0010011
8 5 00110011
9 5 000110011
10 6 0010011001
11 7 00100110011
12 8 001001100100
13 8 0001001100100
14 10 00100110010011
15 10 000100110010011
16 11 0010011001001100
17 12 00100101101001011
18 13 001001100100110011
19 14 0010011001001100100
20 15 00101001011010010100
21 15 000101001011010010100
22 16 0010010100101101001011
23 17 00100101001011010010100
24 18 001001100100110110011011
25 19 0010011001000100110010011
26 20 00101001011010010100101101
27 21 001001010010110100101001011
28 22 0010100101101001010010110100
29 23 00101001011010010100101101011
30 24 001011010010110101101001011010
31 25 0010100101101001010010110100101
32 26 00101001011010010100101101001011
33 27 001010010110100101001011010010100
34 27 0001010010110100101001011010010100
35 28 00100101001011010010100101101001011
36 29 001001010010110100101001011010010100
37 30 0010011001001100100010011001001100100
38 30 00010011001001100100010011001001100100
39 31 001001010010110100101001011010010100100
40 32 0010010100101101001010010110100101001011
41 33 00100110010001001100100110010001001100100
42 35 001010010110100101001011010110100101101011
43 35 0001010010110100101001011010110100101101011
44 36 00101001011001010010110100101001011010010100
45 37 001001010010110100101001011010110100101101011
46 38 0010100101101001010010110100101001011010010100
47 39 00101101001010010110100101101001010010110100101
48 40 001010010110100101001011010010110101101001011010
49 41 0010100101101001010010110100101101001010010110100
50 42 00101001011010010100101101001011010110100101101011
51 43 001010010110100101001011010110100101001011010010100
Here are all optimal sequences for n ≤ 64 (not just the lexicographically first), generated by a modified version of this program and many hours of computation.
A remarkable nearly-optimal sequence
The prefixes of the infinite fractal sequence
1010010110100101001011010010110100101001011010010100101…
that is invariant under the transformation 101 ↦ 10100, 00 ↦ 101:
101 00 101 101 00 101 00 101 101 00 101 101 00 …
= 10100 101 10100 10100 101 10100 101 10100 10100 101 10100 10100 101 …
seem to have very nearly optimal numbers of runs—always within 2 of optimal for n ≤ 64. The number of runs in the first n characters divided by n approaches (13 − 5√5)/2 ≈ 0.90983. But it turns out this is not the optimal ratio—see comments.
Since it's not a race if there's only one horse, I'm submitting my solution though it's only a fraction the speed of Anders Kaseorg's and a only a third as cryptic. Compile with:
gcc -O2 run-count.c -o run-count
The heart of my algorithm is a simple shift and XOR scheme:
A run of zeros in the XOR result that's greater than or equal to the current period/shift indicates a run in the original string for this period. From this you can tell how long the run was and where it starts and ends. The rest of the code is overhead, setting up the situation and decoding the results.
I'm hoping it will make at least 28 after two minutes on Lembik's machine. (I wrote a pthread version, but only managed to make it run even slower.)
#include <stdio.h>
#include <stdlib.h>
#include <stdint.h>
#include <stdbool.h>
enum { START = 0, WIDTH } ;
// Compare and shuffle just one thing while storing two
typedef union {
uint16_t token;
uint8_t data[sizeof(uint16_t)];
} overlay_t;
#define SENTINAL (0) // marks the end of an array of overlay_t
#define NUMBER_OF_BITS (8 * sizeof(uint64_t))
void period_runs(uint64_t xor_bits, uint8_t nbits, uint8_t period, overlay_t *results) {
overlay_t *results_ptr = results;
uint8_t count = 0;
for (uint8_t position = 0; position < nbits; position++) {
if (xor_bits & 1ULL) {
if ((nbits - position) < period) {
break; // no room left to succeed further
}
if (count >= period) { // we found a run
results_ptr->data[START] = position - (count - 1);
results_ptr->data[WIDTH] = period + count;
results_ptr++;
}
count = 0;
} else {
count++;
}
xor_bits >>= 1;
}
if (count >= period) { // process the final run, if any
results_ptr->data[START] = 0;
results_ptr->data[WIDTH] = period + count;
results_ptr++;
}
results_ptr->token = SENTINAL;
}
void number_runs(uint64_t number, uint8_t bit_length, overlay_t *results) {
overlay_t sub_results[bit_length];
uint8_t limit = bit_length / 2 + 1;
uint64_t mask = (1ULL << (bit_length - 1)) - 1;
overlay_t *results_ptr = results;
results_ptr->token = SENTINAL;
for (uint8_t period = 1; period < limit; period++) {
uint64_t xor_bits = mask & (number ^ (number >> period)); // heart of the code
period_runs(xor_bits, bit_length - period, period, sub_results);
for (size_t i = 0; sub_results[i].token != SENTINAL; i++) {
bool stop = false; // combine previous and current results
for (size_t j = 0; !stop && results[j].token != SENTINAL; j++) {
// lower period result disqualifies higher period result over the same span
stop = (sub_results[i].token == results[j].token);
}
if (!stop) {
(results_ptr++)->token = sub_results[i].token;
results_ptr->token = SENTINAL;
}
}
mask >>= 1;
}
}
int main() {
overlay_t results[NUMBER_OF_BITS];
for (uint8_t bit_length = 2; bit_length < 25; bit_length++) {
int best_results = -1;
uint64_t best_number = 0;
for (uint64_t number = 1ULL << (bit_length - 1); number < (1ULL << bit_length); number++) {
// from the discussion comments, I should be able to solve this
// with just bit strings that begin "11...", so toss the rest
if ((number & (1ULL << (bit_length - 2))) == 0ULL) {
continue;
}
uint64_t reversed = 0;
for (uint8_t i = 0; i < bit_length; i++) {
if (number & (1ULL << i)) {
reversed |= (1ULL << ((bit_length - 1) - i));
}
}
if (reversed > number) {
continue; // ~ 1/4 of bit_strings are simply reversals, toss 'em
}
number_runs(number, bit_length, results);
overlay_t *results_ptr = results;
int count = 0;
while ((results_ptr++)->token != SENTINAL) {
count++;
}
if (count > best_results) {
best_results = count;
best_number = number;
}
}
char *best_string = malloc(bit_length + 1);
uint64_t number = best_number;
char *string_ptr = best_string;
for (int i = bit_length - 1; i >= 0; i--) {
*(string_ptr++) = (number & (1ULL << i)) ? '1' : '0';
}
*string_ptr = '\0';
printf("%u %d %s\n", bit_length, best_results, best_string);
free(best_string);
}
return 0;
}
Output:
> gcc -O2 run-count.c -o run-count
> ./run-count
2 1 11
3 1 110
4 2 1100
5 2 11000
6 3 110011
7 4 1100100
8 5 11001100
9 5 110010011
10 6 1100110011
11 7 11001100100
12 8 110110011011
13 8 1100100010011
14 10 11001001100100
15 10 110010011001000
16 11 1100100110010011
17 12 11001100100110011
18 13 110011001001100100
19 14 1101001011010010100
20 15 11010110100101101011
21 15 110010011001001100100
22 16 1100101001011010010100
23 17 11010010110100101001011
24 18 110100101001011010010100
25 19 1100100110010001001100100
26 20 11010010100101101001010010
27 21 110010011001000100110010011
28 22 1101001010010110100101001011