//Returns the minimal puzzle corresponding to this one
public Puzzle findMinimalPuzzle()
{
Boolean isSolutionUnique = this.isSolutionUnique();
//If this puzzle is already non-unique, just exit
if (!isSolutionUnique)
{
System.Console.WriteLine("This isn't unique!");
return(null);
}
//Tracks a list of potential candidates. i.e. the list of nodes the current node connects to
List <Puzzle> listOfCandidates = new List <Puzzle>();
//Represents the current node
Puzzle initialPuzzle = new Puzzle();
for (int i = 0; i < 9; i++)
{
for (int j = 0; j < 9; j++)
{
initialPuzzle.values[i, j] = this.values[i, j];
}
}
//Initially, the only node we see is the beginning puzzle
listOfCandidates.Add(initialPuzzle);
do
{
//Main logic loop
int len = listOfCandidates.Count;
for (int i = 0; i < len; i++)
{
//listOfCandidates.Count is constantly decreasing in length, so accessing listOfCandidates.ElementAt(i) is dangerous
if (i >= listOfCandidates.Count)
{
break;
}
Puzzle candidate = listOfCandidates.ElementAt(i);
for (int y = 0; y < 9; y++)
{
for (int x = 0; x < 9; x++)
{
//Check every location on the puzzle we're looking at to see if we can reduce it without making it non-unique
Puzzle nextPuzzle = candidate.findReducedPuzzle(x, y);
if (nextPuzzle != null)
{
if (!nextPuzzle.isContained(listOfCandidates))
{
//If we found a reduced non-unique non-duplicate version of the current puzzle, then add it to the list of candidates
listOfCandidates.Add(nextPuzzle);
}
}
}
}
if (listOfCandidates.Count > 1)
{
//Now we've added every node that this node connects to that is farther from the root
//(i.e. has less numbers given),so we don't need this node anymore
listOfCandidates.Remove(candidate);
}
//We're done with that candidate, move on to the next one in listOfCandidates
}
//We went through all the elements in listOfCandidates from before, but now that list has changed.
// -If there's more than one candidate puzzle left, then re-execute the for loop with the remaining candidates
// -If there's only one candidate puzzle left, exit and return that. It lasted through the most removals
// of numbers without being non-unique, so it must be the minimal puzzle
} while (!(listOfCandidates.Count == 1 && listOfCandidates.ElementAt(0).isSolutionUnique()));
return(listOfCandidates.ElementAt(0));
}