diff --git a/4.c b/4.c
new file mode 100644
index 0000000..bc26e66
--- /dev/null
+++ b/4.c
@@ -0,0 +1,255 @@
+#include<stdio.h>
+
+int main()
+{
+	//引用函数 
+	void Printf(char* p0);
+	int Sudoku(char* p0);
+	void Sudokuprint(char* p0);
+	void Sudokuback(int k,char* p0);
+	//给出矩阵 并打印 
+	char board[9][9]={
+	{'3','.','4','.','.','.','1','.','2'},
+	{'.','.','.','2','3','8','.','.','.'},
+	{'9','.','.','.','.','4','.','.','6'},
+	{'.','3','5','.','7','.','.','6','.'},
+	{'.','4','.','3','.','5','.','8','.'},
+	{'.','7','.','.','9','.','2','5','.'},
+	{'8','.','.','6','.','.','.','.','7'},
+	{'.','.','.','9','4','7','.','.','.'},
+	{'4','.','7','.','.','.','5','.','9'}};
+	char* p0 = board;
+	printf("The original Sudoku matrix:\n");
+	Printf(p0);   
+	int i,j,k;
+	//成立的情况 运用回溯函数
+	if(Sudoku(p0))
+	{
+		printf("True:Valid initial Sudoku matrix!\n");
+		Sudokuback(0, p0);
+	}	
+	//不成立的情况 
+	else
+	{
+		printf("False:Invalid initial Sudoku matrix!\n");
+		Sudokuprint(p0);
+		printf("No solution\n");
+	}
+	return 0;
+}
+//打印矩阵
+void Printf(char* p0)
+{
+	int i,j;
+	for(i=0;i<9;i++)
+	{
+		for(j=0;j<9;j++)
+		{
+			printf("%c ",p0[9*i+j]);
+		}
+		printf("\n");
+	}
+}
+//判断是数独但不打印
+int Sudoku(char* p0)
+{
+int i,j,number,s,q,p,t,g; 
+//此处引入1
+int c[10]={0,0,0,0,0,0,0,0,0,0};//用来储存数字出现的个数
+
+
+//下面小矩阵的判断
+int w[10]={1,2,3,4,5,6,7,8,9};
+int k=0;
+for(t=3;t<=9;t=t+3)
+{
+	for(g=3;g<=9;g=g+3)
+	{
+		for(i=t-3;i<t;i++)
+		{
+			for(j=g-3;j<g;j++)//分出来小矩阵
+			{
+				for(number=1;number<=9;number++)
+				{
+					if(p0[9*i+j] == number + '0')
+						{c[number]++;}
+				}
+			}
+		}
+		for(q=1;q<=9;q++)
+		{
+			if(c[q]>1)
+			{
+				return 0;
+			}
+		}
+		k++;
+		for(s=0;s<=9;s++)c[s]=0;
+	}
+
+}
+//完成行的判断
+for(i=0;i<9;i++)
+{
+
+	for(number=1;number<=9;number++)
+	{
+		for(j=0;j<9;j++)
+		{
+			if(p0[9*i+j] == number + '0')
+				c[number]++;
+		}
+	}
+	for(q=1;q<=9;q++)
+	{ 
+		if(c[q]>1)
+		{
+			return 0 ;
+		}
+}
+	for(p=0;p<10;p++)c[p]=0;
+}
+//同理列的判断
+for(i=0;i<9;i++)
+{
+	for(number=1;number<=9;number++)
+	{
+	for(j=0;j<9;j++)
+	{
+		if(p0[9*j+i] == number + '0')
+			c[number]++;}
+	}
+	for(q=1;q<=9;q++)
+	{
+		if(c[q]>1)
+		{
+			return 0;
+		}
+	}
+	for(p=0;p<10;p++)c[p]=0;
+}	
+return 1;
+}	
+//判断数独并打印
+void Sudokuprint(char* p0)
+{
+
+	int i, j, number, s, q, p, t, g;
+	//此处引入1
+	int c[10] = { 0,0,0,0,0,0,0,0,0,0 };//用来储存数字出现的个数
+
+
+	//下面小矩阵的判断
+	int w[10] = { 1,2,3,4,5,6,7,8,9 };
+	int k = 0;
+	for (t = 3;t <= 9;t = t + 3)
+	{
+		for (g = 3;g <= 9;g = g + 3)
+		{
+			for (i = t - 3;i < t;i++)
+			{
+				for (j = g - 3;j < g;j++)//分出来小矩阵
+				{
+					for (number = 1;number <= 9;number++)
+					{
+						if (p0[9 * i + j] == number + '0')
+						{
+							c[number]++;
+						}
+					}
+				}
+			}
+			for (q = 1;q <= 9;q++)
+			{
+				if (c[q] > 1)
+				{
+					printf("The number %d in the block %d has been used!\n", q, w[k]);
+					return ;
+				}
+			}
+			k++;
+			for (s = 0;s <= 9;s++)c[s] = 0;
+		}
+
+	}
+	//完成行的判断
+	for (i = 0;i < 9;i++)
+	{
+
+		for (number = 1;number <= 9;number++)
+		{
+			for (j = 0;j < 9;j++)
+			{
+				if (p0[9 * i + j] == number + '0')
+					c[number]++;
+			}
+		}
+		for (q = 1;q <= 9;q++)
+		{
+			if (c[q] > 1)
+			{
+				printf("The number %d in the col %d has been used!\n", q, i + 1);
+				return ;
+			}
+		}
+		for (p = 0;p < 10;p++)c[p] = 0;
+	}
+	//同理列的判断
+	for (i = 0;i < 9;i++)
+	{
+		for (number = 1;number <= 9;number++)
+		{
+			for (j = 0;j < 9;j++)
+			{
+				if (p0[9 * j + i] == number + '0')
+					c[number]++;
+			}
+		}
+		for (q = 1;q <= 9;q++)
+		{
+			if (c[q] > 1)
+			{
+				printf("The number %d in the column %d has been used!\n", q, i + 1);
+				return ;
+			}
+		}
+		for (p = 0;p < 10;p++)c[p] = 0;
+	}
+	return ;
+}
+//回溯法填空
+void Sudokuback(int k,char* p0)
+{
+	int m;
+	if (k == 81)
+	{
+		if (Sudoku(p0)) 
+		{ 
+			printf("The solution of Sudoku matrix:\n");
+			Printf(p0);
+		}
+		else printf("No sulotion\n");
+		return;
+	}
+	if (p0[k] == '.')
+	{
+		for (m = 1;m <= 9;m++)
+		{
+			p0[k] = m+'0';
+			if (Sudoku(p0))
+			{
+				Sudokuback(k + 1,p0);
+			}
+
+			else if (!Sudoku(p0)&&m==9)
+			{
+				p0[k] = '.';
+				return ;
+			}
+		}
+	}
+	else
+	{
+		Sudokuback(k + 1, p0);
+	}
+}
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