void solve(Sudoku input){
int filledIn=input.getFilledInElements();
if(filledIn==81){
input.printGrid();
cout << endl;
return;
}
for(int i=0; i < 9; i++){
for(int j=0; j < 9; j++){
if(input.getCellElement(i,j)==0){
for(int k=1;k<=9; k++){
bool exists=false;
for(int l=0;l < 9;l++){
if(input.getCellElement(l,j)==k)
exists=true;
}
if(exists==false){
for(int l=0;l < 9;l++){
if(input.getCellElement(i,l)==k)
exists=true;
}
}
if(exists==false)
if((checkSquareConstraint(input,i,j,k))==true){
exists=true;
}
if(exists==false){
Sudoku temp2=new Sudoku(input);
temp2.setCellElement(i,j,k);
solve(temp2);
}
}
return;
}
}
}
}
//Function to check the only-one-number-per-square requirement
bool checkSquareConstraint(Sudoku input,int i,int j,int value){
int squareindex;
bool returned=false;
if(i < 3){
if(j < 3)
squareindex=1;
else if(j < 6)
squareindex=2;
else
squareindex=3;
}
else if(i < 6){
if(j < 3)
squareindex=4;
else if(j < 6)
squareindex=5;
else
squareindex=6;
}
else{
if(j < 3)
squareindex=7;
else if(j < 6)
squareindex=8;
else
squareindex=9;
}
int startk;
int endk;
int startl;
int endl;
/*The below can be written more compactly, but I thought it was easier
to understand this way*/
if(squareindex==1){
startk=0;
endk=3;
startl=0;
endl=3;
}
else if(squareindex==2){
startk=0;
endk=3;
startl=3;
endl=6;
}
if(squareindex==3){
startk=0;
endk=3;
startl=6;
endl=9;
}
else if(squareindex==4){
startk=3;
endk=6;
startl=0;
endl=3;
}
else if(squareindex==5){
startk=3;
endk=6;
startl=3;
endl=6;
}
else if(squareindex==6){
startk=3;
endk=6;
startl=6;
endl=9;
}
else if(squareindex==7){
startk=6;
endk=9;
startl=0;
endl=3;
}
else if(squareindex==8){
startk=6;
endk=9;
startl=3;
endl=6;
}
else if(squareindex==9){
startk=6;
endk=9;
startl=6;
endl=9;
}
for(int k=startk;k < endk;k++)
for(int l=startl;l < endl; l++)
if(input.getCellElement(k,l)==value)
returned=true;
return returned;
}