C++11, 28s
This also uses a dynamic programming approach based on rows. It took 28 seconds to run with argument 13 for me. My favorite trick is the next function which uses some bit bashing for finding the lexicographically next row arrangement satisfying a mask and the no-3-in-a-row rule.
Instructions
- Install the latest MinGW-w64 with SEH and Posix threads
- Compile the program with
g++ -std=c++11 -march=native -O3 <filename>.cpp -o <executable name> - Run with
<executable name> <n>
#include <vector> #include <stddef.h> #include <iostream> #include <string> #ifdef _MSC_VER #include <intrin.h> #define popcount32 _mm_popcnt_u32 #else #define popcount32 __builtin_popcount #endif using std::vector; using row = uint32_t; using xcount = uint8_t; uint16_t rev16(uint16_t x) { // slow static const uint8_t revbyte[] {0,128,64,192,32,160,96,224,16,144,80,208,48,176,112,240,8,136,72,200,40,168,104,232,24,152,88,216,56,184,120,248,4,132,68,196,36,164,100,228,20,148,84,212,52,180,116,244,12,140,76,204,44,172,108,236,28,156,92,220,60,188,124,252,2,130,66,194,34,162,98,226,18,146,82,210,50,178,114,242,10,138,74,202,42,170,106,234,26,154,90,218,58,186,122,250,6,134,70,198,38,166,102,230,22,150,86,214,54,182,118,246,14,142,78,206,46,174,110,238,30,158,94,222,62,190,126,254,1,129,65,193,33,161,97,225,17,145,81,209,49,177,113,241,9,137,73,201,41,169,105,233,25,153,89,217,57,185,121,249,5,133,69,197,37,165,101,229,21,149,85,213,53,181,117,245,13,141,77,205,45,173,109,237,29,157,93,221,61,189,125,253,3,131,67,195,35,163,99,227,19,147,83,211,51,179,115,243,11,139,75,203,43,171,107,235,27,155,91,219,59,187,123,251,7,135,71,199,39,167,103,231,23,151,87,215,55,183,119,247,15,143,79,207,47,175,111,239,31,159,95,223,63,191,127,255}; return uint16_t(revbyte[x >> 8]) | uint16_t(revbyte[x & 0xFF]) << 8; } // returns the next number after r that does not overlap the mask or have three 1's in a row row next(row r, uint32_t m) { m |= r >> 1 & r >> 2; uint32_t x = (r | m) + 1; uint32_t carry = x & -x; return (r | carry) & -carry; } template<typename T, typename U> void maxequals(T& m, U v) { if (v > m) m = v; } struct tictac { const int n; vector<row> rows; size_t nonpal, nrows_c; vector<int> irow; vector<row> revrows; tictac(int n) : n(n) { } row reverse(row r) { return rev16(r) >> (16 - n); } vector<int> sols_1row() { vector<int> v(1 << n); for (uint32_t m = 0; !(m >> n); m++) { auto m2 = m; int n0 = 0; int score = 0; for (int i = n; i--; m2 >>= 1) { if (m2 & 1) { n0 = 0; } else { if (++n0 % 3) score++; } } v[m] = score; } return v; } void gen_rows() { vector<row> pals; for (row r = 0; !(r >> n); r = next(r, 0)) { row rrev = reverse(r); if (r < rrev) { rows.push_back(r); } else if (r == rrev) { pals.push_back(r); } } nonpal = rows.size(); for (row r : pals) { rows.push_back(r); } nrows_c = rows.size(); for (int i = 0; i < nonpal; i++) { rows.push_back(reverse(rows[i])); } irow.resize(1 << n); for (int i = 0; i < rows.size(); i++) { irow[rows[i]] = i; } revrows.resize(1 << n); for (row r = 0; !(r >> n); r++) { revrows[r] = reverse(r); } } // find banned locations for 1's given 2 above rows uint32_t mask(row a, row b) { return ((a & b) | (a >> 1 & b) >> 1 | (a << 1 & b) << 1) /*& ((1 << n) - 1)*/; } int calc() { if (n < 3) { return n * n; } gen_rows(); int tdim = n < 5 ? n : (n + 3) / 2; size_t nrows = rows.size(); xcount* t = new xcount[2 * nrows * nrows_c]{}; #define tb(nr, i, j) t[nrows * (nrows_c * ((nr) & 1) + (i)) + (j)] // find optimal solutions given 2 rows for n x k grids where 3 <= k <= ceil(n/2) + 1 { auto s1 = sols_1row(); for (int i = 0; i < nrows_c; i++) { row a = rows[i]; for (int j = 0; j < nrows; j++) { row b = rows[j]; uint32_t m = mask(b, a) & ~(1 << n); tb(3, i, j) = s1[m] + popcount32(a << 16 | b); } } } for (int r = 4; r <= tdim; r++) { for (int i = 0; i < nrows_c; i++) { row a = rows[i]; for (int j = 0; j < nrows; j++) { row b = rows[j]; bool rev = j >= nrows_c; auto cj = rev ? j - nrows_c : j; uint32_t m = mask(a, b); for (row c = 0; !(c >> n); c = next(c, m)) { row cc = rev ? revrows[c] : c; int count = tb(r - 1, i, j) + popcount32(c); maxequals(tb(r, cj, irow[cc]), count); } } } } int ans = 0; if (tdim == n) { // small sizes for (int i = 0; i < nrows_c; i++) { for (int j = 0; j < nrows; j++) { maxequals(ans, tb(n, i, j)); } } } else { int tdim2 = n + 2 - tdim; // get final answer by joining two halves' solutions down the middle for (int i = 0; i < nrows_c; i++) { int apc = popcount32(rows[i]); for (int j = 0; j < nrows; j++) { row b = rows[j]; int top = tb(tdim2, i, j); int bottom = j < nrows_c ? tb(tdim, j, i) : tb(tdim, j - nrows_c, i < nonpal ? i + nrows_c : i); maxequals(ans, top + bottom - apc - popcount32(b)); } } } delete[] t; return ans; } }; int main(int argc, char** argv) { int n; if (argc < 2 || (n = std::stoi(argv[1])) < 0 || n > 16) { return 1; } std::cout << tictac{ n }.calc() << '\n'; return 0; }
