c program to check valid sudoku - c

Question

i'm writing c code to check whether given matrix is valid sudoku solution or not.
Input would be matrix of n*n size.
I have written code to check row and column but i'm not getting how to validate grids of sqrt(n)*sqrt(n) size.
my code is here
#include<stdio.h>
int main()
{
int i,j,count=0,sumrow;
int sumcol;
int n;
scanf("%d",&n);
int arr[n+1][n+1];
for(i=1;i<=n;i++)
for(j=1;j<=n;j++)
scanf("%d",&arr[i][j]);
for(i=1;i<=n;i++)
{
sumcol=0;
for(j=1;j<=n;j++)
sumcol+=arr[j][i];
if(sumcol!=(n*(n+1)/2))
count++;
}
for(i=1;i<=n;i++)
{
sumrow=0;
for(j=1;j<=n;j++)
{
sumrow+=arr[i][j];
}
// printf("%d\n",sumrow);
if(sumrow!=(n*(n+1)/2))
count++;
}
//printf("count%d ",count);
if(count==0)
printf("yes");
else
printf("no");
return 0;
}

Answer 1

Here i have a better solution. Instead of sum, we can use an integer flag.
//read sudoku
for(i=0;i<9;i++)
for(j=0;j<9;j++)
{
scanf("%c",&c);
a[i][j]=c-'0';
}
//checking rows
for(i=0;i<9;i++)
{
flag=0x0000;
for(j=0;j<9;j++)
flag|=1<<(a[i][j]-1);
if(flag!=0x01FF)
report("row",i,j-1);
}
//checking cols
for(j=0;j<9;j++)
{
flag=0x0000;
for(i=0;i<9;i++)
flag|=1<<(a[i][j]-1);
if(flag!=0x01FF)
report("col",i-1,j);
}
//checking Squares (3x3)
for(si=0;si<3;si++)
{
for(sj=0;sj<3;sj++)
{
flag=0x0000;
for(i=0;i<3;i++)
{
for(j=0;j<3;j++)
flag|=1<<(a[si*3+i][sj*3+j]-1);
}
if(flag!=0x01FF)
report("square",si*3+i-1,sj*3+j-1);
}
}
printf("\nThe sudoku is correct");
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Answer 2

You can check each of the smaller squares whether they are valid or not.For example if n=9 you have to check each of the 3 by 3 squares.To check each smaller square you can use an array of 10 elements and check whether any of the 1 to 9 value is repeated or not.
The algorithm is as follows
for each of the smaller grids
if(any of the digit repeats in a smaller grid)
return 0;
return 1;//valid
Following is some code to do the same
//A is the grid
int small=sqrt(n);
for(int i=0;i<small;i++)
{
for(int j=0;j<small;j++)
{
int row=small*i;//this will find the corresponding row
int col=small*j;//this will find the corresponding column
vector<int> used(10,0)
for(int p=row; p<row+small; p++) {
for(int q=col; q<col+small; q++)
{
if(A[p][q]=='0')//0 is not valid
return 0;
if(used[A[p][q]-'0']==1)//this digit has already been used
return 0;
if(used[A[p][q]-'0']) return 0;
used[A[p][q]-'0']=1;//now this particular digit has occurred and should not occur again
}
}
}
}
//if all's well return 1
return 1;
Note that the above code assumes that entire grid is filled and is not empty.If you want to check a partially filled grid,introduce a check.

Answer 3

int main() {
int a[10][10],i,j,n,k,sum,sum1,sum2,sum3,x,l;
printf("enter the size N of N*N sudoku\n");
scanf("%d",&n);
printf("enter the entries of sudoku row wise \n");
for (i=1;i<=n;i++) {
for (j=1;j<=n;j++) {
scanf("%d",&a[i][j]);
}
printf("\n");
}
printf("---------------------------------\n\n\n\n");
printf("the matrix you entered is \n");
for (i=1;i<=n;i++) {
for (j=1;j<=n;j++) {
printf("%d",a[i][j]);
printf("|");
}
printf("\n");
}
for (i=1;i<=n;i++) {
for (k=i;k==i;k++) {
sum=0;
for (j=1;j<=n;j++) {
sum = sum + a[i][j];
}
if(sum!=45)
x=1;
}
}
for (j=1;j<=n;j++) {
for(k=j;k==j;k++) {
sum=0;
for (i=1;i<=n;i++) {
sum = sum+a[i][j];
}
if (sum!=45)
x=1;
}
}
for (k=1;k<=3;k++) {
l = (1+(k-1)*n/3);
for (i=l;i<=k*n/3;i++) {
for(j=1;j<=3;j++) {
sum1 = sum1+a[i][j];
}
for (j=4;j<=6;j++) {
sum2 = sum2+a[i][j];
}
for (j=7;j<=9;j++) {
sum3 = sum3+a[i][j];
}
}
if (sum1!=45||sum2!=45||sum3!=45)
x=1;
}
if (x==1)
printf("sudoku not correct \n");
else
printf("correct sudoku");
return 0;
}

Answer 4

bool validateMatrix(int g_iMatrix1[][MAXCOLS],
int iROWS,
int iCOLS)
{
bool bRowUsed[MAXROWS][MAXCOLS] = {0};
bool bColUsed[MAXROWS][MAXCOLS] = {0};
bool bBlockUsed[MAXROWS][MAXCOLS] = {0};
//Matrix to keep record if current value is already set in current row..
memset(bRowUsed, false, (MAXROWS) * (MAXCOLS));
//Matrix to keep record if current value is already set in current column..
memset(bColUsed, false, (MAXCOLS) * (MAXCOLS));
//Matrix to keep record if current value is already set in current block of iSQRT * iSQRT..
//Lets assume the matrix is of size 9 * 9..
//So there will be 9 block of 3 * 3..
//Number the blocks from left to right as 0 to 8..
//We will be mapping 0 the block to 0th row, 1st block to 1st row, 2nd block to 2nd row and so on..
memset(bBlockUsed, false, (MAXROWS) * (MAXCOLS));
int iRows = 0,iCols = 0;
int iSQRT = int(sqrt(MAXCOLS));
for(iRows = 0;iRows < iROWS;iRows++)
{
for(iCols = 0;iCols < iCOLS;iCols++)
{
if(bRowUsed[iRows][g_iMatrix1[iRows][iCols] - 1] == true)
{
return false;
}
if(bColUsed[g_iMatrix1[iRows][iCols] - 1][iCols] == true)
{
return false;
}
//Number the blocks from left to right as 1 to 9..
//We will be mapping 0 the block to 0th row, 1st block to 1st row, 2nd block to 2nd row and so on..
//((iRows / iSQRT) * iSQRT) + (iCols / iSQRT) will map the block with above logic..
if(bBlockUsed[((iRows / iSQRT) * iSQRT) + (iCols / iSQRT)][g_iMatrix1[iRows][iCols] - 1] == true)
{
return false;
}
bRowUsed[iRows][g_iMatrix1[iRows][iCols] - 1] = true;
bColUsed[g_iMatrix1[iRows][iCols] - 1][iCols] = true;
bBlockUsed[((iRows / iSQRT) * iSQRT) + (iCols / iSQRT)][g_iMatrix1[iRows][iCols] - 1] = true;
}
}
return true;
}