String 拼字块检查
对于拼字游戏中的平铺检查，您可以制作四个5x5的字母网格，总共100个平铺。我想做一个40个水平和垂直单词都有效的例子。可用分幅集包含：String 拼字块检查,string,algorithm,data-structures,scrabble,String,Algorithm,Data Structures,Scrabble,对于拼字游戏中的平铺检查，您可以制作四个5x5的字母网格，总共100个平铺。我想做一个40个水平和垂直单词都有效的例子。可用分幅集包含： 12 x E 9 x A，I 8 x O 6xN，R，T 4 x D，L，S，U 3 x G 2 x B、C、F、H、M、P、V、W、Y、空白磁贴（通配符） 1 x K，J，Q，x，Z 有效单词词典可用（700KB）。大约有12000个5个字母的有效单词 下面是一个示例，其中所有20个水平单词都有效： Z O W I E|P I N O T Y O G I
- 12 x E
- 9 x A，I
- 8 x O
- 6xN，R，T
- 4 x D，L，S，U
- 3 x G
- 2 x B、C、F、H、M、P、V、W、Y、空白磁贴（通配符）
- 1 x K，J，Q，x，Z
Z O W I E|P I N O T
Y O G I N|O C t A D <= blank being used as 't'
X E B E C|N A L E D
W A I T E|M E R L E
V I N E R|L U T E A
---------+---------
U S N E A|K N O S P
T A V E R|J O L E D
S O F T A|I A M B I
R I D G Y|H A I T h <= blank being used as 'h'
Q U R S H|G R O U F
Z O W I E|P I N O T
Y O G I N | O C t A D我会采取悲观的观点来处理这个问题（当然是天真的）。我试图证明没有5x5解决方案，因此肯定没有四个5x5解决方案。为了证明不存在5x5解决方案，我尝试从各种可能性中构建一个。如果我的猜想失败了，我能够构造一个5x5解，那么，我就有办法构造5x5解，我会尝试构造所有（独立的）5x5解。如果至少有4个，那么我将确定某个组合是否满足字母计数限制
[Edit]Null Set已确定存在“4430974 5x5解决方案”。这些有效吗？
我的意思是，我们可以使用的字母数量有限制。此限制可以表示为边界向量BV=[9，2，2，4，…]，对应于a，B，C等上的限制（您可以在Null集合的代码中看到此向量）。如果字母计数向量的每个项小于BV中的相应项，则5x5解有效。在创建5x5解决方案时，很容易检查它是否有效。也许4430974的数字可以减少，比如说减少到N
不管怎样，我们可以把这个问题表述为：在N中找到四个字母计数向量，其和等于BV。有（N，4）个可能的和（“N选择4”）。当N等于400万时，这个数字仍然在10^25左右——这不是一个令人鼓舞的数字。也许您可以搜索四个第一个项的总和为9的项，如果是，则检查第二个项的总和是否为2，以此类推
我要指出的是，在从N中选择4之后，计算是独立的，因此，如果你有一台多核机器，你可以通过并行解决方案加快计算速度
[Edit2]不过，并行化可能不会有多大区别。在这一点上，我可能会采取乐观的观点：肯定有比我预期的更多的5x5解决方案，因此最终的解决方案也可能比预期的更多。也许你不需要深入到10^25就可以找到一个。最终编辑：解决了这里有一个解决方案
GNAWN|jOULE
RACHE|EUROS
IDIOT|STEAN
PINOT|TRAvE
TRIPY|SOLES
-----+-----
HOWFF|ZEBRA
AGILE|EQUID
CIVIL|BUXOM
EVENT|RIOJA
KEDGY|ADMAN
这是一张用我的拼字游戏集制作的照片
一旦我找到了正确的方法，这个很容易找到，所以我打赌你可以通过这种方法找到更多。方法见下文
从每行和每列包含5个字母单词的字典中构造前缀树。递归地，如果给定的平铺放置为其列和行形成有效前缀，并且该平铺可用，并且下一个平铺放置有效，则该平铺放置有效。基本情况是，如果没有要放置的平铺，则它是有效的
正如格伦所说，找到所有有效的5x5板，看看其中是否有四块可以组合起来，这可能是有道理的。递归到100度听起来并不有趣
编辑：这是我的代码的第2版
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <stdbool.h>
typedef union node node;
union node {
node* child[26];
char string[6];
};
typedef struct snap snap;
struct snap {
node* rows[5];
node* cols[5];
char tiles[27];
snap* next;
};
node* root;
node* vtrie[5];
node* htrie[5];
snap* head;
char bag[27] = {9,2,2,4,12,2,3,2,9,1,1,4,2,6,8,2,1,6,4,6,4,2,2,1,2,1,2};
const char full_bag[27] = {9,2,2,4,12,2,3,2,9,1,1,4,2,6,8,2,1,6,4,6,4,2,2,1,2,1,2};
const char order[26] = {16,23,9,25,21,22,5,10,1,6,7,12,15,2,24,3,20,13,19,11,8,17,14,0,18,4};
void insert(char* string){
node* place = root;
int i;
for(i=0;i<5;i++){
if(place->child[string[i] - 'A'] == NULL){
int j;
place->child[string[i] - 'A'] = malloc(sizeof(node));
for(j=0;j<26;j++){
place->child[string[i] - 'A']->child[j] = NULL;
}
}
place = place->child[string[i] - 'A'];
}
memcpy(place->string, string, 6);
}
void check_four(){
snap *a, *b, *c, *d;
char two_total[27];
char three_total[27];
int i;
bool match;
a = head;
for(b = a->next; b != NULL; b = b->next){
for(i=0;i<27; i++)
two_total[i] = a->tiles[i] + b->tiles[i];
for(c = b->next; c != NULL; c = c->next){
for(i=0;i<27; i++)
three_total[i] = two_total[i] + c->tiles[i];
for(d = c->next; d != NULL; d = d->next){
match = true;
for(i=0; i<27; i++){
if(three_total[i] + d->tiles[i] != full_bag[i]){
match = false;
break;
}
}
if(match){
printf("\nBoard Found!\n\n");
for(i=0;i<5;i++){
printf("%s\n", a->rows[i]->string);
}
printf("\n");
for(i=0;i<5;i++){
printf("%s\n", b->rows[i]->string);
}
printf("\n");
for(i=0;i<5;i++){
printf("%s\n", c->rows[i]->string);
}
printf("\n");
for(i=0;i<5;i++){
printf("%s\n", d->rows[i]->string);
}
exit(0);
}
}
}
}
}
void snapshot(){
snap* shot = malloc(sizeof(snap));
int i;
for(i=0;i<5;i++){
printf("%s\n", htrie[i]->string);
shot->rows[i] = htrie[i];
shot->cols[i] = vtrie[i];
}
printf("\n");
for(i=0;i<27;i++){
shot->tiles[i] = full_bag[i] - bag[i];
}
bool transpose = false;
snap* place = head;
while(place != NULL && !transpose){
transpose = true;
for(i=0;i<5;i++){
if(shot->rows[i] != place->cols[i]){
transpose = false;
break;
}
}
place = place->next;
}
if(transpose){
free(shot);
}
else {
shot->next = head;
head = shot;
check_four();
}
}
void pick(x, y){
if(y==5){
snapshot();
return;
}
int i, tile,nextx, nexty, nextz;
node* oldv = vtrie[x];
node* oldh = htrie[y];
if(x+1==5){
nexty = y+1;
nextx = 0;
} else {
nextx = x+1;
nexty = y;
}
for(i=0;i<26;i++){
if(vtrie[x]->child[order[i]]!=NULL &&
htrie[y]->child[order[i]]!=NULL &&
(tile = bag[i] ? i : bag[26] ? 26 : -1) + 1) {
vtrie[x] = vtrie[x]->child[order[i]];
htrie[y] = htrie[y]->child[order[i]];
bag[tile]--;
pick(nextx, nexty);
vtrie[x] = oldv;
htrie[y] = oldh;
bag[tile]++;
}
}
}
int main(int argc, char** argv){
root = malloc(sizeof(node));
FILE* wordlist = fopen("sowpods5letters.txt", "r");
head = NULL;
int i;
for(i=0;i<26;i++){
root->child[i] = NULL;
}
for(i=0;i<5;i++){
vtrie[i] = root;
htrie[i] = root;
}
char* string = malloc(sizeof(char)*6);
while(fscanf(wordlist, "%s", string) != EOF){
insert(string);
}
free(string);
fclose(wordlist);
pick(0,0);
return 0;
}
编辑4：这是最终产品
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <stdbool.h>
typedef union node node;
union node {
node* child[26];
char string[6];
};
node* root;
node* vtrie[5];
node* htrie[5];
int score;
int max_score;
char block_1[27] = {4,2,0,2, 2,0,0,0,2,1,0,0,2,1,2,0,1,2,0,0,2,0,0,1,0,1,0};//ZEBRA EQUID BUXOM RIOJA ADMAN
char block_2[27] = {1,0,1,1, 4,2,2,1,3,0,1,2,0,1,1,0,0,0,0,1,0,2,1,0,1,0,0};//HOWFF AGILE CIVIL EVENT KEDGY
char block_3[27] = {2,0,1,1, 1,0,1,1,4,0,0,0,0,3,2,2,0,2,0,3,0,0,1,0,1,0,0};//GNAWN RACHE IDIOT PINOT TRIPY
//JOULE EUROS STEAN TRAVE SOLES
char bag[27] = {9,2,2,4,12,2,3,2,9,1,1,4,2,6,8,2,1,6,4,6,4,2,2,1,2,1,2};
const char full_bag[27] = {9,2,2,4,12,2,3,2,9,1,1,4,2,6,8,2,1,6,4,6,4,2,2,1,2,1,2};
const char order[26] = {16,23,9,25,21,22,5,10,1,6,7,12,15,2,24,3,20,13,19,11,8,17,14,0,18,4};
const int value[27] = {244,862,678,564,226,1309,844,765,363,4656,909,414,691,463,333,687,11998,329,218,423,536,1944,1244,4673,639,3363,0};
void insert(char* string){
node* place = root;
int i;
for(i=0;i<5;i++){
if(place->child[string[i] - 'A'] == NULL){
int j;
place->child[string[i] - 'A'] = malloc(sizeof(node));
for(j=0;j<26;j++){
place->child[string[i] - 'A']->child[j] = NULL;
}
}
place = place->child[string[i] - 'A'];
}
memcpy(place->string, string, 6);
}
void snapshot(){
static int count = 0;
int i;
for(i=0;i<5;i++){
printf("%s\n", htrie[i]->string);
}
for(i=0;i<27;i++){
printf("%c%d ", 'A'+i, bag[i]);
}
printf("\n");
if(++count>=1000){
exit(0);
}
}
void pick(x, y){
if(y==5){
if(score>max_score){
snapshot();
max_score = score;
}
return;
}
int i, tile,nextx, nexty;
node* oldv = vtrie[x];
node* oldh = htrie[y];
if(x+1==5){
nextx = 0;
nexty = y+1;
} else {
nextx = x+1;
nexty = y;
}
for(i=0;i<26;i++){
if(vtrie[x]->child[order[i]]!=NULL &&
htrie[y]->child[order[i]]!=NULL &&
(tile = bag[order[i]] ? order[i] : bag[26] ? 26 : -1) + 1) {
vtrie[x] = vtrie[x]->child[order[i]];
htrie[y] = htrie[y]->child[order[i]];
bag[tile]--;
score+=value[tile];
pick(nextx, nexty);
vtrie[x] = oldv;
htrie[y] = oldh;
bag[tile]++;
score-=value[tile];
}
}
}
int main(int argc, char** argv){
root = malloc(sizeof(node));
FILE* wordlist = fopen("sowpods5letters.txt", "r");
score = 0;
max_score = 0;
int i;
for(i=0;i<26;i++){
root->child[i] = NULL;
}
for(i=0;i<5;i++){
vtrie[i] = root;
htrie[i] = root;
}
for(i=0;i<27;i++){
bag[i] = bag[i] - block_1[i];
bag[i] = bag[i] - block_2[i];
bag[i] = bag[i] - block_3[i];
printf("%c%d ", 'A'+i, bag[i]);
}
char* string = malloc(sizeof(char)*6);
while(fscanf(wordlist, "%s", string) != EOF){
insert(string);
}
free(string);
fclose(wordlist);
pick(0,0);
return 0;
}
#包括
#包括
#包括
#包括
typedef联合节点；
联合节点{
节点*子节点[26]；
字符串[6]；
};
节点*根；
节点*vtrie[5]；
节点*htrie[5]；
智力得分；
int max_分数；
字符块_1[27]={4,2,0,2,0,0,2,1,0,0,2,1,2,0,1,2,2,0,0,2,0,0,0,0,0,1,0}//斑马马肉
字符块_2[27]={1,0,1,1,4,2,2,1,3,0,1,1,2,0,1,1,0,0,0,0,1,0,2,1,0,0,0}//HOWFF敏捷民事事件KEDGY
字符块_3[27]={2,0,1,1,1,1,4,0,0,0,0,3,2,0,2,0,3,0,0,0,1,0,0,0,0}//啃轴白痴比诺特里皮
//焦耳欧元硬脂底
炭袋[27]={9,2,2,4,12,2,3,2,9,1,1,4,2,6,8,2,1,6,4,4,2,2,1,2}；
const char full_bag[27]={9,2,2,4,12,2,3,2,9,1,1,4,2,6,8,2,1,6,4,4,2,2,1,2}；
常量字符顺序[26]={16,23,9,25,21,22,5,10,1,6,7,12,15,2,24,3,20,13,19,11,8,17,14,0,18,4}；
const int value[27]={2448626785642261309844765363465690941469146914633333687119983292184235361944124446736393363,0}；
空插入（字符*字符串）{
节点*位置=根；
int i；
对于（i=0；ichild[string[i]-'A']==NULL）{
int j；
place->child[string[i]-'A']=malloc（sizeof（node））；
对于（j=0；jchild[string[i]-'A']->child[j]=NULL；
}
}
place=place->child[string[i]-“A']；
}
memcpy（place->string，string，6）；
}
void快照（）{
静态整数计数=0；
int i；
对于（i=0；istring）；
}
对于（i=0；i=1000）{
出口（0）；
}
}
空拾取（x，y）{
如果（y==5）{
如果（分数>最大分数）{
快照（）；
最大分数=分数；
}
返回；
}
int i、tile、nextx、nexty；
节点*oldv=vtrie[x]；
节点*oldh=htrie[y]；
如果（x+1==5）{
nextx=0；
nexty=y+1；
}否则{
nextx=x+1；
nexty=y；
}
对于（i=0；ichild[order[i]]！=NULL&&
htrie[y]>child[order[i]！=NULL&&
（瓷砖=袋子[订单[i]]？订单[i]：袋子[26]？26:-1）+1）{
vtrie[x]=vtrie[x]>child[order[i]]；
htrie[y]=htrie[y]>child[order[i]]；
包[瓦]——；
分数+=价值[瓷砖]；
拾取（nextx、nexty）；
vtrie[x]=oldv；
htrie[y]=oldh；
包[瓷砖]+；
分数-=价值[瓷砖]；
}
}
}
int main（int argc，字符**argv）{
root=malloc（sizeof（node））；
文件*字表
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <stdbool.h>
typedef union node node;
union node {
node* child[26];
char string[6];
};
node* root;
node* vtrie[5];
node* htrie[5];
int score;
int max_score;
char block_1[27] = {4,2,0,2, 2,0,0,0,2,1,0,0,2,1,2,0,1,2,0,0,2,0,0,1,0,1,0};//ZEBRA EQUID BUXOM RIOJA ADMAN
char block_2[27] = {1,0,1,1, 4,2,2,1,3,0,1,2,0,1,1,0,0,0,0,1,0,2,1,0,1,0,0};//HOWFF AGILE CIVIL EVENT KEDGY
char block_3[27] = {2,0,1,1, 1,0,1,1,4,0,0,0,0,3,2,2,0,2,0,3,0,0,1,0,1,0,0};//GNAWN RACHE IDIOT PINOT TRIPY
//JOULE EUROS STEAN TRAVE SOLES
char bag[27] = {9,2,2,4,12,2,3,2,9,1,1,4,2,6,8,2,1,6,4,6,4,2,2,1,2,1,2};
const char full_bag[27] = {9,2,2,4,12,2,3,2,9,1,1,4,2,6,8,2,1,6,4,6,4,2,2,1,2,1,2};
const char order[26] = {16,23,9,25,21,22,5,10,1,6,7,12,15,2,24,3,20,13,19,11,8,17,14,0,18,4};
const int value[27] = {244,862,678,564,226,1309,844,765,363,4656,909,414,691,463,333,687,11998,329,218,423,536,1944,1244,4673,639,3363,0};
void insert(char* string){
node* place = root;
int i;
for(i=0;i<5;i++){
if(place->child[string[i] - 'A'] == NULL){
int j;
place->child[string[i] - 'A'] = malloc(sizeof(node));
for(j=0;j<26;j++){
place->child[string[i] - 'A']->child[j] = NULL;
}
}
place = place->child[string[i] - 'A'];
}
memcpy(place->string, string, 6);
}
void snapshot(){
static int count = 0;
int i;
for(i=0;i<5;i++){
printf("%s\n", htrie[i]->string);
}
for(i=0;i<27;i++){
printf("%c%d ", 'A'+i, bag[i]);
}
printf("\n");
if(++count>=1000){
exit(0);
}
}
void pick(x, y){
if(y==5){
if(score>max_score){
snapshot();
max_score = score;
}
return;
}
int i, tile,nextx, nexty;
node* oldv = vtrie[x];
node* oldh = htrie[y];
if(x+1==5){
nextx = 0;
nexty = y+1;
} else {
nextx = x+1;
nexty = y;
}
for(i=0;i<26;i++){
if(vtrie[x]->child[order[i]]!=NULL &&
htrie[y]->child[order[i]]!=NULL &&
(tile = bag[order[i]] ? order[i] : bag[26] ? 26 : -1) + 1) {
vtrie[x] = vtrie[x]->child[order[i]];
htrie[y] = htrie[y]->child[order[i]];
bag[tile]--;
score+=value[tile];
pick(nextx, nexty);
vtrie[x] = oldv;
htrie[y] = oldh;
bag[tile]++;
score-=value[tile];
}
}
}
int main(int argc, char** argv){
root = malloc(sizeof(node));
FILE* wordlist = fopen("sowpods5letters.txt", "r");
score = 0;
max_score = 0;
int i;
for(i=0;i<26;i++){
root->child[i] = NULL;
}
for(i=0;i<5;i++){
vtrie[i] = root;
htrie[i] = root;
}
for(i=0;i<27;i++){
bag[i] = bag[i] - block_1[i];
bag[i] = bag[i] - block_2[i];
bag[i] = bag[i] - block_3[i];
printf("%c%d ", 'A'+i, bag[i]);
}
char* string = malloc(sizeof(char)*6);
while(fscanf(wordlist, "%s", string) != EOF){
insert(string);
}
free(string);
fclose(wordlist);
pick(0,0);
return 0;
}
O(n * n/10 * n/10 * n/100 * n/100 * n/1000 * n/1000 ...
which for 12000 elements acts something like n^4 algorithm
AQFED, ZQABE, EDQDE, ELQUO
this means there are two valid positionings of those:
xZxxx
AQFED
xAxxx ---> this limits our search for words that contain [ABDEFZ] as the second letter
xBxxx
xExxx
same for the other
EDQDE ---> this limits our search for words that contain [EDLU] as the third letter
ELQUO
all appropriate words are in union of those two conditions
3736 2x2 solutions
8812672 3x3 solutions
The 1000th 4x4 solution is
A A H S
A C A I
L A I R
S I R E
The 1000th 5x5 solution is
A A H E D
A B U N A
H U R S T
E N S U E
D A T E D
The 1000th 2x4x4 solution is
A A H S | A A H S
A B A C | A B A C
H A I R | L E K U
S C R Y | S T E D
--------+--------
D E E D | D E E M
E I N E | I N T I
E N O L | O V E R
T E L T | L Y N E