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  • 爬山法、随机重启爬山法、模拟退火算法对八皇后问题和八数码问题的性能测试

    代码地址:https://github.com/laiy/AI/tree/master/awesome-search

    一些前提:

    1. 首先要明确这些算法并不是用于解决传统的搜索问题的(环境是可观察的,确定的,已知的,问题解是一个行动序列),这些算法适用于哪些关注解状态而不是路径代价的问题,我们讨论的搜索算法往往和现实世界的一些问题更加的契合。

    2. 为了便于测试我们选择了八皇后和八数码问题,不考虑它的一些特殊性质来作为一个搜索问题。(因为这两个问题本身就可以用经典搜索策略来解决,这里只是忽略掉问题的一些性质来模拟一个不可用经典搜索解决的问题)

     

    首先对八皇后问题和八数码问题做个简单的介绍:

    八皇后问题:

    The eight queens puzzle is the problem of placing eight chess queens on an 8×8 chessboard so that no two queens threaten each other. Thus, a solution requires that no two queens share the same row, column, or diagonal.

    八数码问题:

    对给的任意一个初始状态,找到回到目标状态的步骤。

    这两个问题都是经典的搜索问题,用DFS或BFS都可以找到解,这里我们探讨的是爬山法,随机重启爬山法和模拟退火算法对这两个问题的求解性能和能力。

     

    首先,编写测例生成器,我选择生成100k个初始状态的测例,生成思想:

    八皇后:生成八个1-8的随机数作为一个八皇后的初始状态,数字表示在每一列皇后的位置,例如4 6 8 2 7 1 3 5对应如下状态

    八数码:终点位置开始随机移动10k步达到一个有解初始状态

    测例生成器代码如下:

     1 /*
     2  * this program is used for generate the majority of 8 digits problems and 8 queens problems
     3  */
     4 
     5 #include <cstdio>
     6 #include <cstdlib>
     7 
     8 #define TESRCASE 100000
     9 #define STEP 10000
    10 #define UP 0
    11 #define DOWN 1
    12 #define LEFT 2
    13 #define RIGHT 3
    14 
    15 inline void swap(int *a, int *b) {
    16     *a ^= *b ^= *a ^= *b;
    17 }
    18 
    19 void generate_8_digits_problem() {
    20     FILE *fp = fopen("testcase_8_digits_problem", "w");
    21     int testcase = TESRCASE, direction, position, steps;
    22     while (testcase--) {
    23         int state[9] = {1, 2, 3, 4, 5, 6, 7, 8, 0};
    24         steps = STEP;
    25         position = 9;
    26         while (steps--) {
    27             direction = rand() % 4;
    28             switch (direction) {
    29                 case UP:
    30                     if (position <= 3)
    31                         break;
    32                     else {
    33                         swap(&state[position - 1], &state[position - 4]), position -= 3;
    34                         break;
    35                     }
    36                 case DOWN:
    37                     if (position >= 7)
    38                         break;
    39                     else {
    40                         swap(&state[position - 1], &state[position + 2]), position += 3;
    41                         break;
    42                     }
    43                 case LEFT:
    44                     if (position % 3 == 1)
    45                         break;
    46                     else {
    47                         swap(&state[position - 1], &state[position - 2]), position--;
    48                         break;
    49                     }
    50                 case RIGHT:
    51                     if (position % 3 == 0)
    52                         break;
    53                     else {
    54                         swap(&state[position - 1], &state[position]), position++;
    55                         break;
    56                     }
    57             }
    58         }
    59         fprintf(fp, "%d %d %d %d %d %d %d %d %d
    ", state[0], state[1], state[2], state[3], state[4], state[5], state[6], state[7], state[8]);
    60     }
    61     fclose(fp);
    62 }
    63 
    64 void generate_8_queens_problem() {
    65     FILE *fp = fopen("testcase_8_queens_problem", "w");
    66     int testcase = TESRCASE, state[8], position, i;
    67     while (testcase--) {
    68         for (i = 0; i < 8; i++)
    69             position = rand() % 8 + 1, state[i] = position;
    70         fprintf(fp, "%d %d %d %d %d %d %d %d
    ", state[0], state[1], state[2], state[3], state[4], state[5], state[6], state[7]);
    71     }
    72     fclose(fp);
    73 }
    74 
    75 int main() {
    76     generate_8_digits_problem();
    77     generate_8_queens_problem();
    78     return 0;
    79 }

    好,现在我们就得到了八皇后问题和八数码问题各100k个初始状态的测例了,分别保存在testcase_8_digits_problem和testcase_8_queens_problem中,每个测例占用1行。

    首先测试爬山法,我们约定用成功的测例的平均时间作为性能指标,用测例的解决率作为能力指标。

    爬山法的思想:对每一个状态,永远像最理想的状态前进(贪心)。

    对八皇后问题来说,我们用相互碰撞的皇后对作为状态的权重(weight),那么爬山法的做法就是对每一个后继状态计算weight,记录下最小weight的状态,前进。

    对八数码问题来说,我们用移动后曼哈顿距离作为状态权重(weight),最小weight的下一状态作为前进状态。

    如果周围的所有状态都不能达到“爬山”的效果,则算法达到一个山脊(极大值),如果未找到解则认为算法失败。

    爬山法测试代码如下:

      1 # ifndef CLK_TCK
      2 # define CLK_TCK CLOCKS_PER_SEC
      3 # endif
      4 
      5 #include <cstdio>
      6 #include <ctime>
      7 #include <cmath>
      8 #include <cstdlib>
      9 #include <vector>
     10 #include <algorithm>
     11 
     12 #define TESRCASE 100000
     13 #define UP 0
     14 #define DOWN 1
     15 #define LEFT 2
     16 #define RIGHT 3
     17 
     18 struct State {
     19     int direction, diff_manhattan;
     20     State(int i, int dis) {
     21         this->direction = i;
     22         this->diff_manhattan = dis;
     23     }
     24     bool operator<(const State &s) const {
     25         return diff_manhattan > s.diff_manhattan;
     26     }
     27 };
     28 
     29 double eight_digits_problem_time;
     30 double eight_queens_problem_time;
     31 int eight_digits_problem_failed_times;
     32 int eight_queens_problem_failed_times;
     33 int diff_manhattan_distance;
     34 
     35 inline void swap(int *a, int *b) {
     36     *a ^= *b ^= *a ^= *b;
     37 }
     38 
     39 inline bool solved(int *state) {
     40     for (int i = 0; i < 8; i++)
     41         if (state[i] != i + 1)
     42             return false;
     43     return true;
     44 }
     45 
     46 inline int manhattan_distance(int num, int position) {
     47     int dest_x = ceil(num / 3), dest_y = (num - 1) % 3 + 1, position_x = ceil(position / 3), position_y = (position - 1) % 3 + 1;
     48     return abs(dest_x - position_x) + abs(dest_y - position_y);
     49 }
     50 
     51 bool eight_digits_better(int *state, int position, int direction) {
     52     switch (direction) {
     53         case UP:
     54             if (position <= 3)
     55                 return false;
     56             else {
     57                 diff_manhattan_distance =  manhattan_distance(state[position - 4], position - 3) - manhattan_distance(state[position - 4], position);
     58                 return manhattan_distance(state[position - 4], position - 3) > manhattan_distance(state[position - 4], position);
     59             }
     60         case DOWN:
     61             if (position >= 7)
     62                 return false;
     63             else {
     64                 diff_manhattan_distance = manhattan_distance(state[position + 2], position + 3) - manhattan_distance(state[position + 2], position);
     65                 return manhattan_distance(state[position + 2], position + 3) > manhattan_distance(state[position + 2], position);
     66             }
     67         case LEFT:
     68             if (position % 3 == 1)
     69                 return false;
     70             else {
     71                 diff_manhattan_distance = manhattan_distance(state[position - 2], position - 1) - manhattan_distance(state[position - 2], position);
     72                 return manhattan_distance(state[position - 2], position - 1) > manhattan_distance(state[position - 2], position);
     73             }
     74         case RIGHT:
     75             if (position % 3 == 0)
     76                 return false;
     77             else {
     78                 diff_manhattan_distance =  manhattan_distance(state[position], position + 1) - manhattan_distance(state[position], position);
     79                 return manhattan_distance(state[position], position + 1) > manhattan_distance(state[position], position);
     80             }
     81     }
     82     return false;
     83 }
     84 
     85 void solve_one_case_of_8_digits_problem(int *state) {
     86     clock_t start_time = clock();
     87     int position, i;
     88     bool found;
     89     for (i = 0; i < 9; i++)
     90         if (state[i] == 0) {
     91             position = i + 1;
     92             break;
     93         }
     94     while (!solved(state)) {
     95         found = false;
     96         std::vector<State> v;
     97         for (i = 0; i < 4; i++) {
     98             if (eight_digits_better(state, position, i)) {
     99                 found = true;
    100                 v.push_back(State(i, diff_manhattan_distance));
    101             }
    102             if (i == 3 && found) {
    103                 std::sort(v.begin(), v.end());
    104                 switch (v[0].direction) {
    105                     case UP:
    106                         swap(&state[position - 1], &state[position - 4]), position -= 3;
    107                         break;
    108                     case DOWN:
    109                         swap(&state[position - 1], &state[position + 2]), position += 3;
    110                         break;
    111                     case LEFT:
    112                         swap(&state[position - 1], &state[position - 2]), position--;
    113                         break;
    114                     case RIGHT:
    115                         swap(&state[position - 1], &state[position]), position++;
    116                         break;
    117                 }
    118             }
    119         }
    120         if (!found) {
    121             eight_digits_problem_failed_times++;
    122             return;
    123         }
    124     }
    125     eight_digits_problem_time += (double)(clock() - start_time) / CLK_TCK;
    126 }
    127 
    128 void solve_8_digits_problem() {
    129     FILE *fp = fopen("testcase_8_digits_problem", "r");
    130     int original_state[9];
    131     while (fscanf(fp, "%d %d %d %d %d %d %d %d %d", original_state, original_state + 1, original_state + 2, original_state + 3, original_state + 4, original_state + 5, original_state + 6, original_state + 7, original_state + 8) != EOF)
    132         solve_one_case_of_8_digits_problem(original_state);
    133     fclose(fp);
    134 }
    135 
    136 int peers_of_attacking_queens(int *state) {
    137     int peers = 0, i, j, k;
    138     for (i = 0; i < 7; i++) {
    139         for (j = i + 1; j < 8; j++)
    140             if (state[j] == state[i])
    141                 peers++;
    142         for (j = i + 1, k = state[i] + 1; j < 8 && k <= 8; j++, k++)
    143             if (state[j] == k)
    144                 peers++;
    145         for (j = i + 1, k = state[i] - 1; j < 8 && k >= 1; j++, k--)
    146             if (state[j] == k)
    147                 peers++;
    148     }
    149     return peers;
    150 }
    151 
    152 void solve_one_case_of_8_queens_problem(int *state) {
    153     clock_t start_time = clock();
    154     int h = peers_of_attacking_queens(state), i, j, best_i, best_j, temp, record;
    155     while (h != 0) {
    156         best_i = -1;
    157         for (i = 1; i <= 8; i++) {
    158             record = state[i - 1];
    159             for (j = 1; j <= 8; j++) {
    160                 if (j != record) {
    161                     state[i - 1] = j;
    162                     temp = peers_of_attacking_queens(state);
    163                     if (temp < h)
    164                         h = temp, best_i = i, best_j = j;
    165                 }
    166             }
    167             state[i - 1] = record;
    168         }
    169         if (best_i == -1) {
    170             eight_queens_problem_failed_times++;
    171             return;
    172         }
    173         state[best_i - 1] = best_j;
    174     }
    175     eight_queens_problem_time += (double)(clock() - start_time) / CLK_TCK;
    176 }
    177 
    178 void solve_8_queens_problem() {
    179     FILE *fp = fopen("testcase_8_queens_problem", "r");
    180     int original_state[8];
    181     while (fscanf(fp, "%d %d %d %d %d %d %d %d", original_state, original_state + 1, original_state + 2, original_state + 3, original_state + 4, original_state + 5, original_state + 6, original_state + 7) != EOF)
    182         solve_one_case_of_8_queens_problem(original_state);
    183     fclose(fp);
    184 }
    185 
    186 void print_result() {
    187     printf("eight digits problem average solved times: %lf
    ", eight_digits_problem_time / (double)(TESRCASE - eight_digits_problem_failed_times));
    188     printf("eight digits problem solved rate: %lf
    ", 1 - double(eight_digits_problem_failed_times) / TESRCASE);
    189     printf("eight queens problem average solved times: %lf
    ", eight_queens_problem_time / (double)(TESRCASE - eight_queens_problem_failed_times));
    190     printf("eight queens problem solved rate: %lf
    ", 1 - double(eight_queens_problem_failed_times) / TESRCASE);
    191 }
    192 
    193 int main() {
    194     eight_digits_problem_time = 0;
    195     eight_queens_problem_time = 0;
    196     eight_digits_problem_failed_times = 0;
    197     eight_queens_problem_failed_times = 0;
    198     solve_8_digits_problem();
    199     solve_8_queens_problem();
    200     print_result();
    201     return 0;
    202 }

    测试结果如下:

    我们可以看到,用爬山法来解决八数码问题解决率是非常非常低的,99.9%+的测例用爬山法是无法得到解的。

    而对于八皇后问题,爬山法解决率也不是很理想,14.6%的测例是可以找到解的。

    优点是平均解决时间短,因为贪心并不需要回溯,找到底就结束了,可以理解为不回溯的启发式DFS。

     

    然后是随机重启爬山法

    随机重启爬山法的思想是,对于给定的初始状态如果不能找到解,自己随机生成一个初始状态重启求解直到找到最终的解为止。

    这个作为对于八皇后问题来说是合情合理的作为,对于八数码问题可能有人会认为是不妥当的,因为八数码问题的目标就是给定初始状态,找到到达最终状态的路线,初始状态是不能随意改变的。

    而八皇后问题不一样,八皇后问题目标就是找到最终状态,所有初始状态随机改变是可以允许的。

    提出这个问题的读者可以回到本文最开始理解一些我描述的前提。

    代码如下:

      1 # ifndef CLK_TCK
      2 # define CLK_TCK CLOCKS_PER_SEC
      3 # endif
      4 
      5 #include <cstdio>
      6 #include <ctime>
      7 #include <cmath>
      8 #include <cstdlib>
      9 #include <vector>
     10 #include <algorithm>
     11 
     12 #define TESRCASE 100000
     13 #define STEP 100
     14 #define UP 0
     15 #define DOWN 1
     16 #define LEFT 2
     17 #define RIGHT 3
     18 
     19 struct State {
     20     int direction, diff_manhattan;
     21     State(int i, int dis) {
     22         this->direction = i;
     23         this->diff_manhattan = dis;
     24     }
     25     bool operator<(const State &s) const {
     26         return diff_manhattan > s.diff_manhattan;
     27     }
     28 };
     29 
     30 double eight_digits_problem_time;
     31 double eight_queens_problem_time;
     32 int eight_digits_problem_failed_times;
     33 int eight_queens_problem_failed_times;
     34 int diff_manhattan_distance;
     35 
     36 inline void swap(int *a, int *b) {
     37     *a ^= *b ^= *a ^= *b;
     38 }
     39 
     40 inline bool solved(int *state) {
     41     for (int i = 0; i < 8; i++)
     42         if (state[i] != i + 1)
     43             return false;
     44     return true;
     45 }
     46 
     47 inline int manhattan_distance(int num, int position) {
     48     int dest_x = ceil(num / 3), dest_y = (num - 1) % 3 + 1, position_x = ceil(position / 3), position_y = (position - 1) % 3 + 1;
     49     return abs(dest_x - position_x) + abs(dest_y - position_y);
     50 }
     51 
     52 bool eight_digits_better(int *state, int position, int direction) {
     53     switch (direction) {
     54         case UP:
     55             if (position <= 3)
     56                 return false;
     57             else {
     58                 diff_manhattan_distance =  manhattan_distance(state[position - 4], position - 3) - manhattan_distance(state[position - 4], position);
     59                 return manhattan_distance(state[position - 4], position - 3) > manhattan_distance(state[position - 4], position);
     60             }
     61         case DOWN:
     62             if (position >= 7)
     63                 return false;
     64             else {
     65                 diff_manhattan_distance = manhattan_distance(state[position + 2], position + 3) - manhattan_distance(state[position + 2], position);
     66                 return manhattan_distance(state[position + 2], position + 3) > manhattan_distance(state[position + 2], position);
     67             }
     68         case LEFT:
     69             if (position % 3 == 1)
     70                 return false;
     71             else {
     72                 diff_manhattan_distance = manhattan_distance(state[position - 2], position - 1) - manhattan_distance(state[position - 2], position);
     73                 return manhattan_distance(state[position - 2], position - 1) > manhattan_distance(state[position - 2], position);
     74             }
     75         case RIGHT:
     76             if (position % 3 == 0)
     77                 return false;
     78             else {
     79                 diff_manhattan_distance =  manhattan_distance(state[position], position + 1) - manhattan_distance(state[position], position);
     80                 return manhattan_distance(state[position], position + 1) > manhattan_distance(state[position], position);
     81             }
     82     }
     83     return false;
     84 }
     85 
     86 void eight_digits_random_state(int *s) {
     87     int state[9] = {1, 2, 3, 4, 5, 6, 7, 8, 0}, steps, position, direction;
     88     steps = STEP;
     89     position = 9;
     90     while (steps--) {
     91         direction = rand() % 4;
     92         switch (direction) {
     93             case UP:
     94                 if (position <= 3)
     95                     break;
     96                 else {
     97                     swap(&state[position - 1], &state[position - 4]), position -= 3;
     98                     break;
     99                 }
    100             case DOWN:
    101                 if (position >= 7)
    102                     break;
    103                 else {
    104                     swap(&state[position - 1], &state[position + 2]), position += 3;
    105                     break;
    106                 }
    107             case LEFT:
    108                 if (position % 3 == 1)
    109                     break;
    110                 else {
    111                     swap(&state[position - 1], &state[position - 2]), position--;
    112                     break;
    113                 }
    114             case RIGHT:
    115                 if (position % 3 == 0)
    116                     break;
    117                 else {
    118                     swap(&state[position - 1], &state[position]), position++;
    119                     break;
    120                 }
    121         }
    122     }
    123     for (int i = 0; i < 9; i++)
    124         s[i] = state[i];
    125 }
    126 
    127 void solve_one_case_of_8_digits_problem(int *state) {
    128     clock_t start_time = clock();
    129     int position, i;
    130     bool found;
    131     for (i = 0; i < 9; i++)
    132         if (state[i] == 0) {
    133             position = i + 1;
    134             break;
    135         }
    136     while (!solved(state)) {
    137         found = false;
    138         std::vector<State> v;
    139         for (i = 0; i < 4; i++) {
    140             if (eight_digits_better(state, position, i)) {
    141                 found = true;
    142                 v.push_back(State(i, diff_manhattan_distance));
    143             }
    144             if (i == 3 && found) {
    145                 std::sort(v.begin(), v.end());
    146                 switch (v[0].direction) {
    147                     case UP:
    148                         swap(&state[position - 1], &state[position - 4]), position -= 3;
    149                         break;
    150                     case DOWN:
    151                         swap(&state[position - 1], &state[position + 2]), position += 3;
    152                         break;
    153                     case LEFT:
    154                         swap(&state[position - 1], &state[position - 2]), position--;
    155                         break;
    156                     case RIGHT:
    157                         swap(&state[position - 1], &state[position]), position++;
    158                         break;
    159                 }
    160             }
    161         }
    162         if (!found)
    163             eight_digits_random_state(state);
    164     }
    165     eight_digits_problem_time += (double)(clock() - start_time) / CLK_TCK;
    166 }
    167 
    168 void solve_8_digits_problem() {
    169     FILE *fp = fopen("testcase_8_digits_problem", "r");
    170     int original_state[9];
    171     while (fscanf(fp, "%d %d %d %d %d %d %d %d %d", original_state, original_state + 1, original_state + 2, original_state + 3, original_state + 4, original_state + 5, original_state + 6, original_state + 7, original_state + 8) != EOF)
    172         solve_one_case_of_8_digits_problem(original_state);
    173     fclose(fp);
    174 }
    175 
    176 int peers_of_attacking_queens(int *state) {
    177     int peers = 0, i, j, k;
    178     for (i = 0; i < 7; i++) {
    179         for (j = i + 1; j < 8; j++)
    180             if (state[j] == state[i])
    181                 peers++;
    182         for (j = i + 1, k = state[i] + 1; j < 8 && k <= 8; j++, k++)
    183             if (state[j] == k)
    184                 peers++;
    185         for (j = i + 1, k = state[i] - 1; j < 8 && k >= 1; j++, k--)
    186             if (state[j] == k)
    187                 peers++;
    188     }
    189     return peers;
    190 }
    191 
    192 inline void eight_queens_random_state(int *state) {
    193     for (int i = 0; i < 8; i++)
    194         state[i] = rand() % 8 + 1;
    195 }
    196 
    197 void solve_one_case_of_8_queens_problem(int *state) {
    198     clock_t start_time = clock();
    199     int h = peers_of_attacking_queens(state), i, j, best_i, best_j, temp, record;
    200     while (h != 0) {
    201         best_i = -1;
    202         for (i = 1; i <= 8; i++) {
    203             record = state[i - 1];
    204             for (j = 1; j <= 8; j++) {
    205                 if (j != record) {
    206                     state[i - 1] = j;
    207                     temp = peers_of_attacking_queens(state);
    208                     if (temp < h)
    209                         h = temp, best_i = i, best_j = j;
    210                 }
    211             }
    212             state[i - 1] = record;
    213         }
    214         if (best_i == -1)
    215             eight_queens_random_state(state), h = peers_of_attacking_queens(state);
    216         else
    217             state[best_i - 1] = best_j;
    218     }
    219     eight_queens_problem_time += (double)(clock() - start_time) / CLK_TCK;
    220 }
    221 
    222 void solve_8_queens_problem() {
    223     FILE *fp = fopen("testcase_8_queens_problem", "r");
    224     int original_state[8];
    225     while (fscanf(fp, "%d %d %d %d %d %d %d %d", original_state, original_state + 1, original_state + 2, original_state + 3, original_state + 4, original_state + 5, original_state + 6, original_state + 7) != EOF)
    226         solve_one_case_of_8_queens_problem(original_state);
    227     fclose(fp);
    228 }
    229 
    230 void print_result() {
    231     printf("eight digits problem average solved times: %lf
    ", eight_digits_problem_time / (double)(TESRCASE - eight_digits_problem_failed_times));
    232     printf("eight digits problem solved rate: %lf
    ", 1 - double(eight_digits_problem_failed_times) / TESRCASE);
    233     printf("eight queens problem average solved times: %lf
    ", eight_queens_problem_time / (double)(TESRCASE - eight_queens_problem_failed_times));
    234     printf("eight queens problem solved rate: %lf
    ", 1 - double(eight_queens_problem_failed_times) / TESRCASE);
    235 }
    236 
    237 int main() {
    238     eight_digits_problem_time = 0;
    239     eight_queens_problem_time = 0;
    240     eight_digits_problem_failed_times = 0;
    241     eight_queens_problem_failed_times = 0;
    242     solve_8_digits_problem();
    243     solve_8_queens_problem();
    244     print_result();
    245     return 0;
    246 }

    测试结果如下:

    我们可以看到随机重启爬山法的解决率是接近1的(不能说是1,因为确实存在无论怎么重启都找不到解的可能性)

    但是在每个测例解决的时间消耗上面就大打折扣了,对八数码问题,是爬山法单个测例解决时间的600倍,对八皇后问题单个测例平均解决时间大概是7倍。

    这个时间效率和爬山法的问题解决率有关系,解决率越大时间差距越小。

     

    现在测试最后一个算法:模拟退火算法

    模拟退火算法的思想:模拟炼金退火的步骤,温度随时间下降,在温度高的时候,分子是活跃的,有较大可能性随机运动。

    对应到具体问题上就是:对每一个后继状态,如果找到的状态不能使weight更好,也有exp(weight/temperature)的可能性去走这个状态,然后随着温度下降,这个概率越来越小,达到模拟退火的效果。

    我选择温度下降为比例下降,系数为0.8,每一个温度最多可以持续300次迭代,终止温度为0.000000001,八数码问题起始温度为10,八皇后问题起始温度为50。

    代码如下:

      1 # ifndef CLK_TCK
      2 # define CLK_TCK CLOCKS_PER_SEC
      3 # endif
      4 
      5 #include <cstdio>
      6 #include <ctime>
      7 #include <cmath>
      8 #include <cstdlib>
      9 #include <vector>
     10 #include <algorithm>
     11 
     12 #define TESRCASE 100000
     13 #define UP 0
     14 #define DOWN 1
     15 #define LEFT 2
     16 #define RIGHT 3
     17 #define EIGHT_DIGITS_ORIGINAL_TEMPERATURE 10
     18 #define EIGHT_QUEENS_ORIGINAL_TEMPERATURE 50
     19 #define MAX_TRIES_IN_ONE_TEMPERATURE 300
     20 #define STOP_TEMPERATURE 0.000000001
     21 #define COLD_DOWN_RATE 0.8
     22 
     23 double eight_digits_problem_time;
     24 double eight_queens_problem_time;
     25 int eight_digits_problem_failed_times;
     26 int eight_queens_problem_failed_times;
     27 double temperature;
     28 
     29 inline void swap(int *a, int *b) {
     30     *a ^= *b ^= *a ^= *b;
     31 }
     32 
     33 inline bool solved(int *state) {
     34     for (int i = 0; i < 8; i++)
     35         if (state[i] != i + 1)
     36             return false;
     37     return true;
     38 }
     39 
     40 inline int manhattan_distance(int num, int position) {
     41     int dest_x = ceil(num / 3), dest_y = (num - 1) % 3 + 1, position_x = ceil(position / 3), position_y = (position - 1) % 3 + 1;
     42     return abs(dest_x - position_x) + abs(dest_y - position_y);
     43 }
     44 
     45 bool eight_digits_better(int *state, int position, int direction) {
     46     int dis;
     47     switch (direction) {
     48         case UP:
     49             if (position <= 3)
     50                 return false;
     51             else {
     52                 dis = manhattan_distance(state[position - 4], position - 3) - manhattan_distance(state[position - 4], position);
     53                 return (dis > 0) ? true : ((double)(rand() % 1000) / 1000) < exp(dis / temperature);
     54             }
     55         case DOWN:
     56             if (position >= 7)
     57                 return false;
     58             else {
     59                 dis = manhattan_distance(state[position + 2], position + 3) - manhattan_distance(state[position + 2], position);
     60                 return (dis > 0) ? true : ((double)(rand() % 1000) / 1000) < exp(dis / temperature);
     61             }
     62         case LEFT:
     63             if (position % 3 == 1)
     64                 return false;
     65             else {
     66                 dis = manhattan_distance(state[position - 2], position - 1) - manhattan_distance(state[position - 2], position);
     67                 return (dis > 0) ? true : ((double)(rand() % 1000) / 1000) < exp(dis / temperature);
     68             }
     69         case RIGHT:
     70             if (position % 3 == 0)
     71                 return false;
     72             else {
     73                 dis = manhattan_distance(state[position], position + 1) - manhattan_distance(state[position], position);
     74                 return (dis > 0) ? true : ((double)(rand() % 1000) / 1000) < exp(dis / temperature);
     75             }
     76     }
     77     return false;
     78 }
     79 
     80 void solve_one_case_of_8_digits_problem(int *state) {
     81     clock_t start_time = clock();
     82     int position, i, tries_count;
     83     bool found;
     84     for (i = 0; i < 9; i++)
     85         if (state[i] == 0) {
     86             position = i + 1;
     87             break;
     88         }
     89     temperature = EIGHT_DIGITS_ORIGINAL_TEMPERATURE;
     90     tries_count = MAX_TRIES_IN_ONE_TEMPERATURE;
     91     while (!solved(state)) {
     92         found = false;
     93         for (i = 0; i < 4; i++)
     94             if (eight_digits_better(state, position, i)) {
     95                 found = true;
     96                 switch (i) {
     97                     case UP:
     98                         swap(&state[position - 1], &state[position - 4]), position -= 3;
     99                         break;
    100                     case DOWN:
    101                         swap(&state[position - 1], &state[position + 2]), position += 3;
    102                         break;
    103                     case LEFT:
    104                         swap(&state[position - 1], &state[position - 2]), position--;
    105                         break;
    106                     case RIGHT:
    107                         swap(&state[position - 1], &state[position]), position++;
    108                         break;
    109                 }
    110                 break;
    111             }
    112         if (--tries_count == 0)
    113             tries_count = MAX_TRIES_IN_ONE_TEMPERATURE, temperature *= COLD_DOWN_RATE;
    114         if (!found || temperature < STOP_TEMPERATURE) {
    115             eight_digits_problem_failed_times++;
    116             return;
    117         }
    118     }
    119     eight_digits_problem_time += (double)(clock() - start_time) / CLK_TCK;
    120 }
    121 
    122 void solve_8_digits_problem() {
    123     FILE *fp = fopen("testcase_8_digits_problem", "r");
    124     int original_state[9];
    125     while (fscanf(fp, "%d %d %d %d %d %d %d %d %d", original_state, original_state + 1, original_state + 2, original_state + 3, original_state + 4, original_state + 5, original_state + 6, original_state + 7, original_state + 8) != EOF)
    126         solve_one_case_of_8_digits_problem(original_state);
    127     fclose(fp);
    128 }
    129 
    130 int peers_of_attacking_queens(int *state) {
    131     int peers = 0, i, j, k;
    132     for (i = 0; i < 7; i++) {
    133         for (j = i + 1; j < 8; j++)
    134             if (state[j] == state[i])
    135                 peers++;
    136         for (j = i + 1, k = state[i] + 1; j < 8 && k <= 8; j++, k++)
    137             if (state[j] == k)
    138                 peers++;
    139         for (j = i + 1, k = state[i] - 1; j < 8 && k >= 1; j++, k--)
    140             if (state[j] == k)
    141                 peers++;
    142     }
    143     return peers;
    144 }
    145 
    146 void solve_one_case_of_8_queens_problem(int *state) {
    147     clock_t start_time = clock();
    148     int h = peers_of_attacking_queens(state), i, j, best_i, best_j, temp, record, tries_count, temp_h;
    149     temperature = EIGHT_QUEENS_ORIGINAL_TEMPERATURE;
    150     tries_count = MAX_TRIES_IN_ONE_TEMPERATURE;
    151     while (h != 0) {
    152         temp_h = 28;
    153         for (i = 1; i <= 8; i++) {
    154             record = state[i - 1];
    155             for (j = 1; j <= 8; j++) {
    156                 if (j != record) {
    157                     state[i - 1] = j;
    158                     temp = peers_of_attacking_queens(state);
    159                     if (temp < temp_h)
    160                         temp_h = temp, best_i = i, best_j = j;
    161                 }
    162             }
    163             state[i - 1] = record;
    164         }
    165         if (--tries_count == 0)
    166             tries_count = MAX_TRIES_IN_ONE_TEMPERATURE, temperature *= COLD_DOWN_RATE;
    167         if (!(temperature < STOP_TEMPERATURE) && (temp_h < h || ((double)(rand() % 1000) / 1000) < exp(temp_h / temperature)))
    168             state[best_i - 1] = best_j, h = temp_h;
    169         else {
    170             eight_queens_problem_failed_times++;
    171             return;
    172         }
    173     }
    174     eight_queens_problem_time += (double)(clock() - start_time) / CLK_TCK;
    175 }
    176 
    177 void solve_8_queens_problem() {
    178     FILE *fp = fopen("testcase_8_queens_problem", "r");
    179     int original_state[8];
    180     while (fscanf(fp, "%d %d %d %d %d %d %d %d", original_state, original_state + 1, original_state + 2, original_state + 3, original_state + 4, original_state + 5, original_state + 6, original_state + 7) != EOF)
    181         solve_one_case_of_8_queens_problem(original_state);
    182     fclose(fp);
    183 }
    184 
    185 void print_result() {
    186     printf("eight digits problem average solved times: %lf
    ", eight_digits_problem_time / (double)(TESRCASE - eight_digits_problem_failed_times));
    187     printf("eight digits problem solved rate: %lf
    ", 1 - double(eight_digits_problem_failed_times) / TESRCASE);
    188     printf("eight queens problem average solved times: %lf
    ", eight_queens_problem_time / (double)(TESRCASE - eight_queens_problem_failed_times));
    189     printf("eight queens problem solved rate: %lf
    ", 1 - double(eight_queens_problem_failed_times) / TESRCASE);
    190 }
    191 
    192 int main() {
    193     eight_digits_problem_time = 0;
    194     eight_queens_problem_time = 0;
    195     eight_digits_problem_failed_times = 0;
    196     eight_queens_problem_failed_times = 0;
    197     solve_8_digits_problem();
    198     solve_8_queens_problem();
    199     print_result();
    200     return 0;
    201 }

    这个测试时间比较久,题主跑了6个小时才跑出来,测试结果如下:

    可以看到,模拟退火算法的性能是比较优秀的,和爬山法的解决效率只相差0-1个数量级。

    而且解决率相比爬山法也大幅度提升,对于八数码问题问题解决率提升了10倍左右,对于八皇后问题解决率也提升了2倍多。

    结论:

    1. 解决NP问题用模拟退火算法在性能上和解决率上都有出色的表现。

    2. 随机重启爬山法来达到很高的解决率,但是时间效率非常低下。

    3. 爬山法有最高性能,但是问题解决率不堪入目。

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  • 原文地址:https://www.cnblogs.com/laiy/p/awesome_search.html
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