Jeff Somers's N Queens Solutions 最快的n皇后算法
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时间：2015年06月08日
1 /* Jeff Somers 2 * 3 * Copyright (c) 2002 4 * 5 * [email protected] 6 * or 7 * [email protected] 8 * 9 * April, 2002 10 * 11 * Program: nq 12 * 13 * Program to find number of solutions to the N queens problem. 14 * This program assumes a twos complement architecture. 15 * 16 * For example, you can arrange 4 queens on 4 x 4 chess so that 17 * none of the queens can attack each other: 18 * 19 * Two solutions: 20 * _ Q _ _ _ _ Q _ 21 * _ _ _ Q Q _ _ _ 22 * Q _ _ _ _ _ _ Q 23 * _ _ Q _ and _ Q _ _ 24 * 25 * Note that these are separate solutions, even though they 26 * are mirror images of each other. 27 * 28 * Likewise, a 8 x 8 chess board has 92 solutions to the 8 queens 29 * problem. 30 * 31 * Command Line Usage: 32 * 33 * nq N 34 * 35 * where N is the size of the N x N board. For example, 36 * nq 4 will find the 4 queen solution for the 4 x 4 chess 37 * board. 38 * 39 * By default, this program will only print the number of solutions, 40 * not board arrangements which are the solutions. To print the 41 * boards, uncomment the call to printtable in the Nqueen function. 42 * Note that printing the board arrangements slows down the program 43 * quite a bit, unless you pipe the output to a text file: 44 * 45 * nq 10 > output.txt 46 * 47 * 48 * The number of solutions for the N queens problems are known for 49 * boards up to 23 x 23. With this program, I‘ve calculated the 50 * results for boards up to 21 x 21, and that took over a week on 51 * an 800 MHz PC. The algorithm is approximated O(n!) (i.e. slow), 52 * and calculating the results for a 22 x 22 board will take about 8.5 53 * times the amount of time for the 21 x 21 board, or over 8 1/2 weeks. 54 * Even with a 10 GHz machine, calculating the results for a 23 x 23 55 * board would take over a month. Of course, setting up a cluster of 56 * machines (or a distributed client) would do the work in less time. 57 * 58 * (from Sloane‘s On-Line Encyclopedia of Integer Sequences, 59 * Sequence A000170 60 * http://www.research.att.com/cgi-bin/access.cgi/as/njas/sequences/eisA.cgi?Anum=000170 61 * ) 62 * 63 * Board Size: Number of Solutions to Time to calculate 64 * (length of one N queens problem: on 800MHz PC 65 * side of N x N (Hours:Mins:Secs) 66 * chessboard) 67 * 68 * 1 1 n/a 69 * 2 0 < 0 seconds 70 * 3 0 < 0 seconds 71 * 4 2 < 0 seconds 72 * 5 10 < 0 seconds 73 * 6 4 < 0 seconds 74 * 7 40 < 0 seconds 75 * 8 92 < 0 seconds 76 * 9 352 < 0 seconds 77 * 10 724 < 0 seconds 78 * 11 2680 < 0 seconds 79 * 12 14200 < 0 seconds 80 * 13 73712 < 0 seconds 81 * 14 365596 00:00:01 82 * 15 2279184 00:00:04 83 * 16 14772512 00:00:23 84 * 17 95815104 00:02:38 85 * 18 666090624 00:19:26 86 * 19 4968057848 02:31:24 87 * 20 39029188884 20:35:06 88 * 21 314666222712 174:53:45 89 * 22 2691008701644 ? 90 * 23 24233937684440 ? 91 * 24 ? ? 92 */ 93 94 #include <stdio.h> 95 #include <stdlib.h> 96 #include <time.h> 97 98 /* 99 Notes on MAX_BOARDSIZE: 100 101 A 32 bit unsigned long is sufficient to hold the results for an 18 x 18 102 board (666090624 solutions) but not for a 19 x 19 board (4968057848 solutions). 103 104 In Win32, I use a 64 bit variable to hold the results, and merely set the 105 MAX_BOARDSIZE to 21 because that‘s the largest board for which I‘ve 106 calculated a result. 107 108 Note: a 20x20 board will take over 20 hours to run on a Pentium III 800MHz, 109 while a 21x21 board will take over a week to run on the same PC. 110 111 On Unix, you could probably change the type of g_numsolutions from unsigned long 112 to unsigned long long, or change the code to use two 32 bit ints to store the 113 results for board sizes 19 x 19 and up. 114 */ 115 116 #ifdef WIN32 117 118 #define MAX_BOARDSIZE 21 119 typedef unsigned __int64 SOLUTIONTYPE; 120 121 #else 122 123 #define MAX_BOARDSIZE 18 124 typedef unsigned long SOLUTIONTYPE; 125 126 #endif 127 128 #define MIN_BOARDSIZE 2 129 130 SOLUTIONTYPE g_numsolutions = 0; 131 132 133 /* Print a chess table with queens positioned for a solution */ 134 /* This is not a critical path function & I didn‘t try to optimize it. */ 135 void printtable(int boardsize, int* aQueenBitRes, SOLUTIONTYPE numSolution) 136 { 137 int i, j, k, row; 138 139 /* We only calculated half the solutions, because we can derive 140 the other half by reflecting the solution across the "Y axis". */ 141 for (k = 0; k < 2; ++k) 142 { 143 #ifdef WIN32 144 printf("*** Solution #: %I64d ***\n", 2 * numSolution + k - 1); 145 #else 146 printf("*** Solution #: %d ***\n", 2 * numSolution + k - 1); 147 #endif 148 for ( i = 0; i < boardsize; i++) 149 { 150 unsigned int bitf; 151 /* 152 Get the column that was set (i.e. find the 153 first, least significant, bit set). 154 If aQueenBitRes[i] = 011010b, then 155 bitf = 000010b 156 */ 157 bitf = aQueenBitRes[i]; 158 159 row = bitf ^ (bitf & (bitf - 1)); /* get least significant bit */ 160 for ( j = 0; j < boardsize; j++) 161 { 162 /* keep shifting row over to the right until we find the one ‘1‘ in 163 the binary representation. There will only be one ‘1‘. */ 164 if (0 == k && ((row >> j) & 1)) 165 { 166 printf("Q "); 167 } 168 else if (1 == k && (row & (1 << (boardsize - j - 1)))) /* this is the board reflected across the "Y axis" */ 169 { 170 printf("Q "); 171 } 172 else 173 { 174 printf(". "); 175 } 176 } 177 printf("\n"); 178 } 179 printf("\n"); 180 } 181 } 182 183 184 /* The function which calculates the N queen solutions. 185 We calculate one-half the solutions, then flip the results over 186 the "Y axis" of the board. Every solution can be reflected that 187 way to generate another unique solution (assuming the board size 188 isn‘t 1 x 1). That‘s because a solution cannot be symmetrical 189 across the Y-axis (because you can‘t have two queens in the same 190 horizontal row). A solution also cannot consist of queens 191 down the middle column of a board with an odd number of columns, 192 since you can‘t have two queens in the same vertical row. 193 194 This is a backtracking algorithm. We place a queen in the top 195 row, then note the column and diagonals it occupies. We then 196 place a queen in the next row down, taking care not to place it 197 in the same column or diagonal. We then update the occupied 198 columns & diagonals & move on to the next row. If no position 199 is open in the next row, we back track to the previous row & move 200 the queen over to the next available spot in its row & the process 201 starts over again. 202 */ 203 void Nqueen(int board_size) 204 { 205 int aQueenBitRes[MAX_BOARDSIZE]; /* results */ 206 int aQueenBitCol[MAX_BOARDSIZE]; /* marks colummns which already have queens */ 207 int aQueenBitPosDiag[MAX_BOARDSIZE]; /* marks "positive diagonals" which already have queens */ 208 int aQueenBitNegDiag[MAX_BOARDSIZE]; /* marks "negative diagonals" which already have queens */ 209 int aStack[MAX_BOARDSIZE + 2]; /* we use a stack instead of recursion */ 210 register int* pnStack; 211 212 213 register int numrows = 0; /* numrows redundant - could use stack */ 214 register unsigned int lsb; /* least significant bit */ 215 register unsigned int bitfield; /* bits which are set mark possible positions for a queen */ 216 int i; 217 int odd = board_size & 1; /* 0 if board_size even, 1 if odd */ 218 int board_minus = board_size - 1; /* board size - 1 */ 219 int mask = (1 << board_size) - 1; /* if board size is N, mask consists of N 1‘s */ 220 221 222 /* Initialize stack */ 223 aStack[0] = -1; /* set sentinel -- signifies end of stack */ 224 225 /* NOTE: (board_size & 1) is true iff board_size is odd */ 226 /* We need to loop through 2x if board_size is odd */ 227 for (i = 0; i < (1 + odd); ++i) 228 { 229 /* We don‘t have to optimize this part; it ain‘t the 230 critical loop */ 231 bitfield = 0; 232 if (0 == i) 233 { 234 /* Handle half of the board, except the middle 235 column. So if the board is 5 x 5, the first 236 row will be: 00011, since we‘re not worrying 237 about placing a queen in the center column (yet). 238 */ 239 int half = board_size>>1; /* divide by two */ 240 /* fill in rightmost 1‘s in bitfield for half of board_size 241 If board_size is 7, half of that is 3 (we‘re discarding the remainder) 242 and bitfield will be set to 111 in binary. */ 243 bitfield = (1 << half) - 1; 244 pnStack = aStack + 1; /* stack pointer */ 245 246 aQueenBitRes[0] = 0; 247 aQueenBitCol[0] = aQueenBitPosDiag[0] = aQueenBitNegDiag[0] = 0; 248 } 249 else 250 { 251 /* Handle the middle column (of a odd-sized board). 252 Set middle column bit to 1, then set 253 half of next row. 254 So we‘re processing first row (one element) & half of next. 255 So if the board is 5 x 5, the first row will be: 00100, and 256 the next row will be 00011. 257 */ 258 bitfield = 1 << (board_size >> 1); 259 numrows = 1; /* prob. already 0 */ 260 261 /* The first row just has one queen (in the middle column).*/ 262 aQueenBitRes[0] = bitfield; 263 aQueenBitCol[0] = aQueenBitPosDiag[0] = aQueenBitNegDiag[0] = 0; 264 aQueenBitCol[1] = bitfield; 265 266 /* Now do the next row. Only set bits in half of it, because we‘ll 267 flip the results over the "Y-axis". */ 268 aQueenBitNegDiag[1] = (bitfield >> 1); 269 aQueenBitPosDiag[1] = (bitfield << 1); 270 pnStack = aStack + 1; /* stack pointer */ 271 *pnStack++ = 0; /* we‘re done w/ this row -- only 1 element & we‘ve done it */ 272 bitfield = (bitfield - 1) >> 1; /* bitfield -1 is all 1‘s to the left of the single 1 */ 273 } 274 275 /* this is the critical loop */ 276 for (;;) 277 { 278 /* could use 279 lsb = bitfield ^ (bitfield & (bitfield -1)); 280 to get first (least sig) "1" bit, but that‘s slower. */ 281 lsb = -((signed)bitfield) & bitfield; /* this assumes a 2‘s complement architecture */ 282 if (0 == bitfield) 283 { 284 bitfield = *--pnStack; /* get prev. bitfield from stack */ 285 if (pnStack == aStack) { /* if sentinel hit.... */ 286 break ; 287 } 288 --numrows; 289 continue; 290 } 291 bitfield &= ~lsb; /* toggle off this bit so we don‘t try it again */ 292 293 aQueenBitRes[numrows] = lsb; /* save the result */ 294 if (numrows < board_minus) /* we still have more rows to process? */ 295 { 296 int n = numrows++; 297 aQueenBitCol[numrows] = aQueenBitCol[n] | lsb; 298 aQueenBitNegDiag[numrows] = (aQueenBitNegDiag[n] | lsb) >> 1; 299 aQueenBitPosDiag[numrows] = (aQueenBitPosDiag[n] | lsb) << 1; 300 *pnStack++ = bitfield; 301 /* We can‘t consider positions for the queen which are in the same 302 column, same positive diagonal, or same negative diagonal as another 303 queen already on the board. */ 304 bitfield = mask & ~(aQueenBitCol[numrows] | aQueenBitNegDiag[numrows] | aQueenBitPosDiag[numrows]); 305 continue; 306 } 307 else 308 { 309 /* We have no more rows to process; we found a solution. */ 310 /* Comment out the call to printtable in order to print the solutions as board position*/ 311 /* printtable(board_size, aQueenBitRes, g_numsolutions + 1); */ 312 ++g_numsolutions; 313 bitfield = *--pnStack; 314 --numrows; 315 continue; 316 } 317 } 318 } 319 320 /* multiply solutions by two, to count mirror images */ 321 g_numsolutions *= 2; 322 } 323 324 /* Print the results at the end of the run */ 325 void printResults(time_t* pt1, time_t* pt2) 326 { 327 double secs; 328 int hours , mins, intsecs; 329 330 printf("End: \t%s", ctime(pt2)); 331 secs = difftime(*pt2, *pt1); 332 intsecs = (int)secs; 333 printf("Calculations took %d second%s.\n", intsecs, (intsecs == 1 ? "" : "s")); 334 335 /* Print hours, minutes, seconds */ 336 hours = intsecs/3600; 337 intsecs -= hours * 3600; 338 mins = intsecs/60; 339 intsecs -= mins * 60; 340 if (hours > 0 || mins > 0) 341 { 342 printf("Equals "); 343 if (hours > 0) 344 { 345 printf("%d hour%s, ", hours, (hours == 1) ? "" : "s"); 346 } 347 if (mins > 0) 348 { 349 printf("%d minute%s and ", mins, (mins == 1) ? "" : "s"); 350 } 351 printf("%d second%s.\n", intsecs, (intsecs == 1 ? "" : "s")); 352 353 } 354 } 355 356 /* main routine for N Queens program.*/ 357 int main(int argc, char** argv) 358 { 359 time_t t1, t2; 360 int boardsize; 361 362 if (argc != 2) { 363 printf("N Queens program by Jeff Somers.\n"); 364 printf("\[email protected] or [email protected]\n"); 365 printf("This program calculates the total number of solutions to the N Queens problem.\n"); 366 printf("Usage: nq <width of board>\n"); /* user must pass in size of board */ 367 return 0; 368 } 369 370 boardsize = atoi(argv[1]); 371 372 /* check size of board is within correct range */ 373 if (MIN_BOARDSIZE > boardsize || MAX_BOARDSIZE < boardsize) 374 { 375 printf("Width of board must be between %d and %d, inclusive.\n", 376 MIN_BOARDSIZE, MAX_BOARDSIZE ); 377 return 0; 378 } 379 380 time(&t1); 381 printf("N Queens program by Jeff Somers.\n"); 382 printf("\[email protected] or [email protected]\n"); 383 printf("Start: \t %s", ctime(&t1)); 384 385 Nqueen(boardsize); /* find solutions */ 386 time(&t2); 387 388 printResults(&t1, &t2); 389 390 if (g_numsolutions != 0) 391 { 392 #ifdef WIN32 393 printf("For board size %d, %I64d solution%s found.\n", boardsize, g_numsolutions, (g_numsolutions == 1 ? "" : "s")); 394 #else 395 printf("For board size %d, %d solution%s found.\n", boardsize, g_numsolutions, (g_numsolutions == 1 ? "" : "s")); 396 #endif 397 } 398 else 399 { 400 printf("No solutions found.\n"); 401 } 402 403 return 0; 404 }
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