Algorithm 纵横字谜回溯算法
我正在根据以下算法使用回溯实现纵横字谜生成器： 这是我的伪代码：Algorithm 纵横字谜回溯算法,algorithm,c#-4.0,graph-algorithm,backtracking,recursive-backtracking,Algorithm,C# 4.0,Graph Algorithm,Backtracking,Recursive Backtracking,我正在根据以下算法使用回溯实现纵横字谜生成器： 这是我的伪代码： > solve(words,grid): if words is empty: > if grid.isValudSol(): > return grid > else: > return None for each word in words: > possibleSol <- grid.fi
> solve(words,grid): if words is empty:
> if grid.isValudSol():
> return grid
> else:
> return None for each word in words:
> possibleSol <- grid.fillFirst(word)
> ret <- solve(words\{word},possibleSol)
> if (ret != None):
> return ret return None
>求解（单词，网格）：如果单词为空：
>如果grid.isValudSol（）：
>回流栅
>其他：
>对于words中的每个单词返回None：
>如果（ret！=无），则可能发生以下情况：
>返回ret返回无
Crossword.Crossword解决方案Crossword（列出单词，Crossword.Crossword板）{
如果（words.Count==0）
{
返回板；
}
//创建电路板的本地副本
Crossword.Crossword localBoard=新的Crossword.Crossword（board）；
foreach（单词中的var单词）
{
localBoard=新的纵横字谜。纵横字谜（board）；
int positions=localBoard.getPositionsNumber（字）；
如果（位置==0）{
返回null；
}
对于（int i=0；i public class Crossword
{
List<string> words = new List<string>();//for words
List<string> nomatchWords = new List<string>();//for words those have no match with others
List<string> usedWords = new List<string>();//for words we already used
char[,] renderResult;//your result as grid of words. You can draw it in WindowsFormsApp, WPF, etc.
//a grid.
//3 values of int[]:
//[0]==horizontal/vertical word(1/0)
//[1]==beginning (number of the first letter for horizontal axis)
//[2]==beginning(number of the first letter for vertical axis)
List<int[]> Grid = new List<int[]>();
int HorizontalSize=17;//size of renderResult. I left it as a default for this example. Its calculation depends on your algoritm for rendering
int VerticalSize=17;
public Crossword()
{
FillWords(this.words);//fill the list of words
}
//just for example; it can be reading from file etc.
void FillWords(List<string> words)
{
words.Add("alphabet");
words.Add("gramin");//may intersect the first at 'a'
words.Add("asturias");//may intersect the first or second at 'a'
words.Add("fitness");//may intersect all words
words.Add("shit");//? too difficult to calculate where it can place in the crossword. Lets do it by the algorithm
words.Add("programmer");//? you can watch it in debug how my recursion works
}
/// <summary>
/// the main method. Here we render all words in your words-list.
/// </summary>
/// <param name="crossword"></param>
public void Generate(Crossword crossword, DataGridView dgvr)
{
List<int[]> letterPosition=new List<int[]>();//can be empty because words may unmatch
//place EACH word in the crossword
while (words.Count != 0)//if words.Count=0, all words are placed (but there's also can be words with no match)
{
string currentWord = words[0];
Grid.Add(new int[3] { 1, 0, 0 });//new grid position for this word
if (usedWords.Count == 0)//this is the first word
{
usedWords.Add(currentWord);
crossword.Place(currentWord, Grid[0], letterPosition, 0);//place it as a new word
words.RemoveAt(0);
}
else//not the first word=> place the new word in position relative to any of used words
SolveCrossword(currentWord, letterPosition, 0);
PrintCrossword(????);//just for my app, do your logic here
}
//the step with nomatch words. Words that can be intersected with the others words properly will appear in crossword solution. Completely nomatch will not appear.
foreach(string currentWord in nomatchWords)
SolveCrossword(currentWord,letterPosition,0);//process NomatchWords like words
//HOW TO CHECK: try to place as word-list any list of words
}
/// <summary>
/// solution of crossword
/// </summary>
/// <param name="currentWord">the word that we try to place</param>
/// <param name="letterPosition">positions of letter where current word can intersect with the other</param>
/// <param name="intersectWord">the number of intersect word</param>
void SolveCrossword(string currentWord, List<int[]>letterPosition, int intersectWord)
{
if (usedWords.Count - intersectWord > 0)//we must check the intersection with all this words=>SolveCrossword checks the intersection from the last used word to first
//the other order will cause generation from the last words=>crossword will look like a tree instead of sharp
{
try
{
if (MatchWords(usedWords[intersectWord], currentWord, letterPosition))//no matches for its word with last word
{//usedWords.Count - 1 -
int positionCounter = 0;
if (this.Place(currentWord, Grid[intersectWord], letterPosition, positionCounter) == false)//no places for this word related to last placed word=> it's temporarily unmatched
{
letterPosition.Clear();
SolveCrossword(currentWord, letterPosition, ++positionCounter);//we're here because there's no valid intersections between current words=> lets seek the others
}
if (words.Count != 0)
words.RemoveAt(0);//backtrack will end when we'll find the correct position of word
return;
}
else
{
letterPosition.Clear();
SolveCrossword(currentWord, letterPosition, intersectWord + 1);
}
//if we're here, the intersection was not found=>it is nomatch word. It can be correct or completely nomatch word but we'll figure it out at last step
nomatchWords.Add(currentWord);
words.RemoveAt(0);
}
catch//we've got an exception if unchecked positions number->0
{
letterPosition.Clear();
SolveCrossword(currentWord, letterPosition, intersectWord + 1);
}
}
}
//how to find matches. Doesn't matter, you can use your own method to solve this
static bool MatchWords(string newword, string matchedword, List<int[]> letterPosition)
{
for (int i = 0; i < newword.Length; i++)
for (int j = 0; j < matchedword.Length; j++)
if (newword[i] == matchedword[j])
letterPosition.Add(new int[2]{i,j});//we need ALL matches, not only FIRST!<= it may cause problems with size of crossword and bug count of unmatched words
if (letterPosition.Count != 0)
return true;
else
return false;
}
/// <summary>
/// This is how to place words into crossword grid
/// </summary>
/// <param name="newword">our word</param>
/// <param name="letterPosition">Letter Positions list</param>
/// <param name="el">the number of element of LetterPositions. If current intersection is not valid, we move to next</param>
/// <returns>true if sucsess</returns>
bool Place(string newword, int[] intersectedWordGrid, List<int[]> letterPosition, int el)//
{
//placing newword related to lastword (setting up the last Grid element)
//try
//{
if (Grid.Count > 1)
{
if (intersectedWordGrid[0] == 1)//last word is horizontal=>we need vertical
{
Grid[Grid.Count - 1][0] = 0;
Grid[Grid.Count - 1][1] = intersectedWordGrid[1] - letterPosition[el][1];
Grid[Grid.Count - 1][2] = intersectedWordGrid[2] + letterPosition[el][0];
}
else//vertical=>horizontal
{
Grid[Grid.Count - 1][0] = 1;
Grid[Grid.Count - 1][1] = intersectedWordGrid[1] + letterPosition[el][0];
Grid[Grid.Count - 1][2] = intersectedWordGrid[2] - letterPosition[el][1];
}
usedWords.Add(newword);
}
//TODO: recalculate rendering grid for crossword (if you use another mechanic you're welcome)
//I use default boundaries of size 17x17
//try to render: if crossword with these Grid positions of words is valid=>good
if (Render())//renderResult now contains the current solution of crossword
{
letterPosition.Clear();
return true;
}
else//crossword is not valid with this grid
Place(newword, intersectedWordGrid, letterPosition, ++el);
return true;
}
//rendering your crossword
bool Render()
{
if (renderResult == null)
renderResult = new char[HorizontalSize, VerticalSize];
int[] renderword = Grid[Grid.Count - 1];//for each Grid coordinates
{
if (usedWords.Count != 0)
{
string currentword = usedWords[usedWords.Count-1];
//usedWords.RemoveAt(0);
int i=0, j = 0;//need i & j because changing grid coordinates may cause mistakes in other words positions
try
{
//rresultTempCopy = renderRes;
if (CheckWordsInSolution(currentword, renderResult, renderword))
for (int k = 0; k < currentword.Length; k++)//word were checked and it's valid=>write it to board
{
renderResult[HorizontalSize / 2 + renderword[1] + i, VerticalSize / 2 + renderword[2] + j] = currentword[k];
if (renderword[0] == 1)
j++;
else
i++;
}
else
{
usedWords.RemoveAt(usedWords.Count - 1);
return false;
}
}
catch
{
usedWords.RemoveAt(usedWords.Count - 1);
return false;
}
}
}
//we're here=>no mistakes, this is a valid result
return true;//sucsess
}
bool CheckWordsInSolution(string currentword, char[,] solutionTable, int[] wordCoordinates)
{
int i = 0, j = 0;
for (int k = 0; k < currentword.Length; k++)
//validation
if ((renderResult[HorizontalSize / 2 + wordCoordinates[1] + i, VerticalSize / 2 + wordCoordinates[2] + j] == '\0' ||//must be empty place for each letter
renderResult[HorizontalSize / 2 + wordCoordinates[1] + i, VerticalSize / 2 + wordCoordinates[2] + j] == currentword[k])//or the same letter
//we can use other rules, it depends on your type of crossword
//&& ((i == 0 && renderResult[HorizontalSize / 2 + wordCoordinates[1] + i - 1, VerticalSize / 2 + wordCoordinates[2] + j] == '\0') == false ||//before the beginning of word must be empty place
// (j == 0 && renderResult[HorizontalSize / 2 + wordCoordinates[1] + i, VerticalSize / 2 + wordCoordinates[2] + j - 1] == '\0') == false)
//&& ((i == k && renderResult[HorizontalSize / 2 + wordCoordinates[1] + i + 1, VerticalSize / 2 + wordCoordinates[2] + j] == '\0') == false ||//after the end of word must be empty place
// (j == k && renderResult[HorizontalSize / 2 + wordCoordinates[1] + i, VerticalSize / 2 + wordCoordinates[2] + j + 1] == '\0') == false)
)
{
if (wordCoordinates[0] == 1)
j++;
else
i++;
}
else
return false;
return true;
}
void PrintCrossword(???)
{
//draw your crossword here
}
}
公共类填字游戏
{
列表单词=新列表（）；//用于单词
List nomatchWords=new List（）；//用于与其他单词不匹配的单词
List usedWords=new List（）；//对于我们已经使用过的单词
char[，]renderResult；//您的结果是单词网格。您可以在WindowsFormsApp、WPF等中绘制它。
//网格。
//3 int[]的值：
//[0]==水平/垂直字（1/0）
//[1] ==开始（横轴第一个字母的编号）
//[2] ==开始（纵轴第一个字母的编号）
列表网格=新列表（）；
int HorizontalSize=17；//renderResult的大小。我将其作为本示例的默认值。其计算取决于渲染的算法
int垂直尺寸=17；
公共纵横字谜
{
FillWords（this.words）；//填写单词列表
}
//例如，它可以从文件等中读取。
无效填充词（列出词）
{
词语。添加（“字母表”）；
words.Add（“gramin”）；//可以在“a”处与第一个相交
words.Add（“asturias”）；//可以在“a”处与第一个或第二个相交
words.Add（“fitness”）；//可以与所有单词相交
words.Add（“shit”）；/？太难了，无法计算它在纵横字谜中的位置。让我们按照算法来做吧
words.Add（“程序员”）；/？您可以在调试中查看我的递归是如何工作的
}
///
///主要方法。这里我们呈现单词列表中的所有单词。
///
///
公共void生成（纵横字谜、DataGridView dgvr）
{
List letterPosition=new List（）；//可以为空，因为单词可能不匹配
//将每个单词放在纵横字谜中
while（words.Count！=0）//如果words.Count=0，则放置所有单词（但也可能存在不匹配的单词）
{
字符串currentWord=words[0]；
Add（newint[3]{1,0,0}）；//这个单词的新网格位置
if（usedWords.Count==0）//这是第一个字
{
usedWords.Add（currentWord）；
crossword.Place（currentWord，Grid[0]，letterPosition，0）；//将其作为新词放置
字。删除（0）；
}
else//not the first word=>将新词放在相对于任何已用单词的位置
解决纵横字谜（当前字，字母位置，0）；
PrintCrossword（？？？？；//仅针对我的应用程序，在此处执行您的逻辑
}
//使用nomatch单词的步骤。可以和其他单词正确相交的单词将出现在纵横填字游戏解决方案中。完全nomatch不会出现。
foreach（NOMATCHWORD中的字符串currentWord）
解决纵横字谜（当前字、字母位置）
public class Crossword
{
List<string> words = new List<string>();//for words
List<string> nomatchWords = new List<string>();//for words those have no match with others
List<string> usedWords = new List<string>();//for words we already used
char[,] renderResult;//your result as grid of words. You can draw it in WindowsFormsApp, WPF, etc.
//a grid.
//3 values of int[]:
//[0]==horizontal/vertical word(1/0)
//[1]==beginning (number of the first letter for horizontal axis)
//[2]==beginning(number of the first letter for vertical axis)
List<int[]> Grid = new List<int[]>();
int HorizontalSize=17;//size of renderResult. I left it as a default for this example. Its calculation depends on your algoritm for rendering
int VerticalSize=17;
public Crossword()
{
FillWords(this.words);//fill the list of words
}
//just for example; it can be reading from file etc.
void FillWords(List<string> words)
{
words.Add("alphabet");
words.Add("gramin");//may intersect the first at 'a'
words.Add("asturias");//may intersect the first or second at 'a'
words.Add("fitness");//may intersect all words
words.Add("shit");//? too difficult to calculate where it can place in the crossword. Lets do it by the algorithm
words.Add("programmer");//? you can watch it in debug how my recursion works
}
/// <summary>
/// the main method. Here we render all words in your words-list.
/// </summary>
/// <param name="crossword"></param>
public void Generate(Crossword crossword, DataGridView dgvr)
{
List<int[]> letterPosition=new List<int[]>();//can be empty because words may unmatch
//place EACH word in the crossword
while (words.Count != 0)//if words.Count=0, all words are placed (but there's also can be words with no match)
{
string currentWord = words[0];
Grid.Add(new int[3] { 1, 0, 0 });//new grid position for this word
if (usedWords.Count == 0)//this is the first word
{
usedWords.Add(currentWord);
crossword.Place(currentWord, Grid[0], letterPosition, 0);//place it as a new word
words.RemoveAt(0);
}
else//not the first word=> place the new word in position relative to any of used words
SolveCrossword(currentWord, letterPosition, 0);
PrintCrossword(????);//just for my app, do your logic here
}
//the step with nomatch words. Words that can be intersected with the others words properly will appear in crossword solution. Completely nomatch will not appear.
foreach(string currentWord in nomatchWords)
SolveCrossword(currentWord,letterPosition,0);//process NomatchWords like words
//HOW TO CHECK: try to place as word-list any list of words
}
/// <summary>
/// solution of crossword
/// </summary>
/// <param name="currentWord">the word that we try to place</param>
/// <param name="letterPosition">positions of letter where current word can intersect with the other</param>
/// <param name="intersectWord">the number of intersect word</param>
void SolveCrossword(string currentWord, List<int[]>letterPosition, int intersectWord)
{
if (usedWords.Count - intersectWord > 0)//we must check the intersection with all this words=>SolveCrossword checks the intersection from the last used word to first
//the other order will cause generation from the last words=>crossword will look like a tree instead of sharp
{
try
{
if (MatchWords(usedWords[intersectWord], currentWord, letterPosition))//no matches for its word with last word
{//usedWords.Count - 1 -
int positionCounter = 0;
if (this.Place(currentWord, Grid[intersectWord], letterPosition, positionCounter) == false)//no places for this word related to last placed word=> it's temporarily unmatched
{
letterPosition.Clear();
SolveCrossword(currentWord, letterPosition, ++positionCounter);//we're here because there's no valid intersections between current words=> lets seek the others
}
if (words.Count != 0)
words.RemoveAt(0);//backtrack will end when we'll find the correct position of word
return;
}
else
{
letterPosition.Clear();
SolveCrossword(currentWord, letterPosition, intersectWord + 1);
}
//if we're here, the intersection was not found=>it is nomatch word. It can be correct or completely nomatch word but we'll figure it out at last step
nomatchWords.Add(currentWord);
words.RemoveAt(0);
}
catch//we've got an exception if unchecked positions number->0
{
letterPosition.Clear();
SolveCrossword(currentWord, letterPosition, intersectWord + 1);
}
}
}
//how to find matches. Doesn't matter, you can use your own method to solve this
static bool MatchWords(string newword, string matchedword, List<int[]> letterPosition)
{
for (int i = 0; i < newword.Length; i++)
for (int j = 0; j < matchedword.Length; j++)
if (newword[i] == matchedword[j])
letterPosition.Add(new int[2]{i,j});//we need ALL matches, not only FIRST!<= it may cause problems with size of crossword and bug count of unmatched words
if (letterPosition.Count != 0)
return true;
else
return false;
}
/// <summary>
/// This is how to place words into crossword grid
/// </summary>
/// <param name="newword">our word</param>
/// <param name="letterPosition">Letter Positions list</param>
/// <param name="el">the number of element of LetterPositions. If current intersection is not valid, we move to next</param>
/// <returns>true if sucsess</returns>
bool Place(string newword, int[] intersectedWordGrid, List<int[]> letterPosition, int el)//
{
//placing newword related to lastword (setting up the last Grid element)
//try
//{
if (Grid.Count > 1)
{
if (intersectedWordGrid[0] == 1)//last word is horizontal=>we need vertical
{
Grid[Grid.Count - 1][0] = 0;
Grid[Grid.Count - 1][1] = intersectedWordGrid[1] - letterPosition[el][1];
Grid[Grid.Count - 1][2] = intersectedWordGrid[2] + letterPosition[el][0];
}
else//vertical=>horizontal
{
Grid[Grid.Count - 1][0] = 1;
Grid[Grid.Count - 1][1] = intersectedWordGrid[1] + letterPosition[el][0];
Grid[Grid.Count - 1][2] = intersectedWordGrid[2] - letterPosition[el][1];
}
usedWords.Add(newword);
}
//TODO: recalculate rendering grid for crossword (if you use another mechanic you're welcome)
//I use default boundaries of size 17x17
//try to render: if crossword with these Grid positions of words is valid=>good
if (Render())//renderResult now contains the current solution of crossword
{
letterPosition.Clear();
return true;
}
else//crossword is not valid with this grid
Place(newword, intersectedWordGrid, letterPosition, ++el);
return true;
}
//rendering your crossword
bool Render()
{
if (renderResult == null)
renderResult = new char[HorizontalSize, VerticalSize];
int[] renderword = Grid[Grid.Count - 1];//for each Grid coordinates
{
if (usedWords.Count != 0)
{
string currentword = usedWords[usedWords.Count-1];
//usedWords.RemoveAt(0);
int i=0, j = 0;//need i & j because changing grid coordinates may cause mistakes in other words positions
try
{
//rresultTempCopy = renderRes;
if (CheckWordsInSolution(currentword, renderResult, renderword))
for (int k = 0; k < currentword.Length; k++)//word were checked and it's valid=>write it to board
{
renderResult[HorizontalSize / 2 + renderword[1] + i, VerticalSize / 2 + renderword[2] + j] = currentword[k];
if (renderword[0] == 1)
j++;
else
i++;
}
else
{
usedWords.RemoveAt(usedWords.Count - 1);
return false;
}
}
catch
{
usedWords.RemoveAt(usedWords.Count - 1);
return false;
}
}
}
//we're here=>no mistakes, this is a valid result
return true;//sucsess
}
bool CheckWordsInSolution(string currentword, char[,] solutionTable, int[] wordCoordinates)
{
int i = 0, j = 0;
for (int k = 0; k < currentword.Length; k++)
//validation
if ((renderResult[HorizontalSize / 2 + wordCoordinates[1] + i, VerticalSize / 2 + wordCoordinates[2] + j] == '\0' ||//must be empty place for each letter
renderResult[HorizontalSize / 2 + wordCoordinates[1] + i, VerticalSize / 2 + wordCoordinates[2] + j] == currentword[k])//or the same letter
//we can use other rules, it depends on your type of crossword
//&& ((i == 0 && renderResult[HorizontalSize / 2 + wordCoordinates[1] + i - 1, VerticalSize / 2 + wordCoordinates[2] + j] == '\0') == false ||//before the beginning of word must be empty place
// (j == 0 && renderResult[HorizontalSize / 2 + wordCoordinates[1] + i, VerticalSize / 2 + wordCoordinates[2] + j - 1] == '\0') == false)
//&& ((i == k && renderResult[HorizontalSize / 2 + wordCoordinates[1] + i + 1, VerticalSize / 2 + wordCoordinates[2] + j] == '\0') == false ||//after the end of word must be empty place
// (j == k && renderResult[HorizontalSize / 2 + wordCoordinates[1] + i, VerticalSize / 2 + wordCoordinates[2] + j + 1] == '\0') == false)
)
{
if (wordCoordinates[0] == 1)
j++;
else
i++;
}
else
return false;
return true;
}
void PrintCrossword(???)
{
//draw your crossword here
}
}