public static void Search(ref string equ_str, out List <string> ans_str_list, int movMatch = 1, int max_ans_num = 10, double MAX_TIME = 3.0)
{
Equation root;
str2ssd(ref equ_str, out root);
List <Equation> ans_list;
Search(ref root, out ans_list, movMatch, max_ans_num, MAX_TIME);
ans_str_list = new List <string>(ans_list.Count);
for (int i = 0; i < ans_list.Count; i++)
{
Equation src = ans_list[i];
string dest;
if (ssd2str(ref src, out dest))
{
ans_str_list.Add(dest);
}
}
}
// Exist Function
private static bool exist(ref List <Equation> equ_list, ref Equation obj)
{
bool flag = true;
foreach (var equ in equ_list)
{
flag = true;
for (int i = 0; i < equ.ssds.Count; i++)
{
flag = flag & equ.ssds[i] == obj.ssds[i];
if (equ.ssds[i] != obj.ssds[i])
{
break;
}
}
if (flag == true)
{
return(true);
}
}
return(false);
}
// Check if the equation is valid
public static bool isValidEqu(string equ)
{
string left, right;
Equation.splitEqu(ref equ, out left, out right);
// no equal or no right expression
if (right == "")
{
return(false);
}
// right expression must be an integer
int _;
if (!int.TryParse(right, out _))
{
return(false);
}
// make sure the generate answer is not negative
if (_ < 0)
{
return(false);
}
// try solve the left
try
{
int ans = Expr.evaluate(left);
// if answer is negative, then cannot be handled
if (ans < 0)
{
return(false);
}
}
catch (Exception)
{ return(false); }
return(true);
}
public static void str2ssd(ref string src, out Equation dest)
{
dest = new Equation();
str2ssd(ref src, out dest.ssds, out dest.optr);
}
// Generate a puzzle, if found return true, else false
public static bool GenerateSearch(ref Equation root, out Equation puzzle, int movMatch = 1)
{
root.Attribute = Action.Remove;
puzzle = new Equation();
// Set up Open and Close Table
List <Equation> Open = new List <Equation>(), Close = new List <Equation>();
// mark if the puzzle is found
/* 深度优先搜索算法： */
//（1）把起始节点S 放到未扩展节点Open 表中。如果此节点为一目标节点，则得到一
//个解。
Open.Add(root);
//（2）如果Open 为一空表，或ans_list数量超过一定范围，则失败退出。
while (Open.Count > 0)
{
//（3）把Open 中第一个节点n 从Open 表移到Closed 表。
Equation cur_equ = Open[0];
Open.RemoveAt(0); Close.Add(cur_equ);
// Check if Open[0].depth < 2 * movMatch
if (cur_equ.get_depth() > 2 * movMatch)
{
continue;
}
//（4）扩展节点n ，产生其全部（时间太长，如何改进）后裔，对后裔生成其祖先、属性，(检查是否与前序节点一样)，并把它们放入Open 表的前端(末端的节点比前端
//的节点后移出)。如果没有后裔，则转向步骤（2）。
// 其中已经检查节点是否与祖父一样；
List <Equation> children;
randomExpand(cur_equ, out children);
Open.InsertRange(0, children);
//（5）如果后继节点中有任一个为目标节点，则求得一个解，判断其Attributes == Action.Remove && depth == 2*MovMatch，
// 成功退出；否则，转向步骤（2）。
for (int i = 0; i < children.Count; i++)
{
Equation son = children[i];
// if "son" needs to be placed one match, algorithm continues!
if (son.Attribute == Action.Place)
{
continue;
}
// if "son".depth != 2 * matches, then still not ending!
if (son.get_depth() != 2 * movMatch)
{
continue;
}
// now check!
if (isValidEqu(son))
{
// Optional: check if the result are "strictly" valid
if (matchDiff(ref root, ref son) != 2 * movMatch)
{
continue;
}
puzzle.ssds = new List <SSD>(son.ssds);
puzzle.Attribute = Action.Remove;
puzzle.optr = new Dictionary <int, char>(son.optr);
return(true);
}
}
}
return(false);
}
/* Search algorithm */
public static void Search(ref Equation root, out List <Equation> ans_list, int movMatch = 1, int max_ans_num = 10, double MAX_TIME = 3.0)
{
root.Attribute = Action.Remove;
ans_list = new List <Equation>();
//start watch
sw.Restart();
// Set up Open and Close Table
List <Equation> Open = new List <Equation>(), Close = new List <Equation>();
/* 深度优先搜索算法： */
//（1）把起始节点S 放到未扩展节点Open 表中。如果此节点为一目标节点，则得到一
//个解。
Open.Add(root);
if (isCorrect(root))
{
ans_list.Add(root);
}
//（2）如果Open 为一空表，或ans_list数量超过一定范围，则失败退出。
while (Open.Count > 0 && ans_list.Count < max_ans_num)
{
// check timer
if (sw.Elapsed.TotalSeconds > MAX_TIME)
{
sw.Stop(); return;
}
//（3）把Open 中第一个节点n 从Open 表移到Closed 表。
Equation cur_equ = Open[0];
Open.RemoveAt(0); Close.Add(cur_equ);
// Check if Open[0].depth < 2 * movMatch
if (cur_equ.get_depth() > 2 * movMatch)
{
continue;
}
//（4）扩展节点n ，产生其全部（时间太长，如何改进）后裔，对后裔生成其祖先、属性，(检查是否与前序节点一样)，并把它们放入Open 表的前端(末端的节点比前端
//的节点后移出)。如果没有后裔，则转向步骤（2）。
// 其中已经检查节点是否与祖父一样；
List <Equation> children;
expand(cur_equ, out children);
Open.InsertRange(0, children);
//（5）如果后继节点中有任一个为目标节点，则求得一个解，判断其Attributes == Action.Remove && depth == 2*MovMatch，
// 成功退出；否则，转向步骤（2）。
for (int i = 0; i < children.Count; i++)
{
Equation son = children[i];
// if "son" needs to be placed one match, algorithm continues!
if (son.Attribute == Action.Place)
{
continue;
}
// if "son".depth != 2 * matches, then still not ending!
if (son.get_depth() != 2 * movMatch)
{
continue;
}
// now check!
if (isCorrect(son) && !exist(ref ans_list, ref son))
{
ans_list.Add(son);
}
}
}
sw.Stop();
}
// expand with random seed
private static void randomExpand(Equation father, out List <Equation> children)
{
expand(father, out children);
Shuffle(ref children);
}
public static bool ssd2str(ref Equation src, out string dest)
{
return(ssd2str(ref src.ssds, ref src.optr, out dest));
}
