Solving sudoku puzzle with c++ - c++

Question

I want to make a program for solving a 3*3 sudoku puzzle. I have made a program but it is only working for 50% problems and for the rest it gives 60% right solution. I do not know how to solve it in a finite number of steps for every possible problem. The technique I have used is that I am searching every individual element of array and check which no does not exist in the same row and column and then I put it in that unit and move to the next. But this is not the solution for every problem. The next thing that come to mind was that we should have to write each and every possible number for a unit and then proceed. But how will we decide that which number we should finally put in the unit. I just want that how to write a solution that will work for every problem. I hope you got the point.
The code I have written is
#include<iostream>
using namespace std;
int rowsearch(int l[9][9],int row,int num) // function to search a particular number from a row of array
{
int counter=0;
for (int c=0 ; c<9 ; c++)
{
if (l[row][c]==num)
counter=counter+1;
}
if (counter>0)
return 1;
else
return 0;
}
int colsearch(int l[9][9],int col,int num) // function to search a number from a column of an array
{
int counter=0;
for (int c=0 ; c<9 ; c++)
{
if (l[c][col]==num)
counter=counter+1;
}
if (counter>0)
return 1;
else
return 0;
}
int rowcolnotexist(int x[9][9],int row , int col) // to find a nuber which does not exists int a row and column
{
for (int c=1 ; c<=9 ; c++)
{
if ( rowsearch(x,row,c)!=1 && colsearch(x,col,c)!=1)
return c;
}
return 0;
}
int main()
{
int l[9][9]={};
// input of the list
for (int i=0 ; i<9 ; i++)
for (int j=0 ; j<9 ; j++)
{
cout<<"Enter "<<i+1<<"*"<<j+1<<"entry of the list(For not entering a number enter 0).";
cin>>l[i][j];
}
// operations
for (int i=0 ; i<9 ; i++)
{
for (int j=0 ; j<9 ; j++)
if (l[i][j]==0)
l[i][j]=rowcolnotexist(l,i,j);
}
// printing list
for (int i=0 ; i<9 ; i++)
{
for (int j=0 ; j<9 ; j++)
{
cout<<l[i][j];
if ((j+1)%3==0)
cout<<" ";
else
cout<<" ";
}
if ((i+1)%3==0)
cout<<"\n\n\n";
else
cout<<"\n\n";
}
return 0;
}

Answer 1

I'd recommend the same algorithm I use when solving them myself ;-)
Choose an arbitrary square and list all the valid values for it, based on what is present in all the other rows (there's probably a way to make a more efficient decision about which square to start with).
Then move on to a related empty square (there's probably a way to make a more efficient decision about which square to check next) and store all its possible values.
Rinse and repeat until you find a square that has only one valid value.
Then unzip.
This should sounds like a recursive problem, by now. (note: almost anything that can be done recursively can be done conventionally, but the idea is the same).
So, store up a list of partially solved squares, and when you get to a square that's been solved completely, go back through your list in reverse order re-evaluating your partial solutions with the new data (namely, the one you WERE able to solve).
Rinse and repeat for full body and volume.