+from operator import attrgetter
+from weakref import ref
+
+class Cell(object):
+ """Represents a single cell/value within a sudoku grid."""
+
+ ### Accessors
+
+ def _get_solved(self):
+ """True iff this cell has been solved."""
+ return len(self._values) == 1
+ solved = property(_get_solved)
+
+ def _get_value(self):
+ """Returns this cell's value, if it has one known."""
+ if self.solved:
+ return self._values[0]
+ return None
+ value = property(_get_value)
+
+ grid = property(lambda self: self._grid())
+ constraints = property(attrgetter('_constraints'))
+
+ def __init__(self, grid, row, column):
+ self._grid = ref(grid)
+ self._row = row
+ self._col = column
+ self._values = range(self.grid.size)
+ self._constraints = []
+ self._normalized = False
+
+ def add_constraint(self, constraint):
+ self._constraints.append(constraint)
+
+ def set(self, value, normalize=True):
+ """Sets the value of this cell. If `normalize` is True or omitted, the
+ grid will be updated accordingly.
+ """
+ self._values = [value]
+ if normalize:
+ self._normalized = False
+ self.normalize()
+
+
+
+ def normalize(self):
+ """Checks to see if this cell has only one possible value left. If
+ so, sets that as its value and eliminates it from every related cell.
+ This method is exhaustive; that repeated calls should have no effect.
+ """
+
+ if self._normalized:
+ # Already done
+ return
+
+ # Set this now just in case of infinite looping
+ self._normalized = True
+
+ if not self.solved:
+ # Don't know the value yet
+ return
+
+ # Elimination time
+ for constraint in self.constraints:
+ for cell in constraint.cells:
+ if cell == self:
+ continue
+ cell.eliminate(self.value)
+
+
+ def eliminate(self, value):
+ """Eliminates the given value as a possibility for this cell."""
+ if value in self._values:
+ self._values.remove(value)
+
+ if len(self._values) == 0:
+ # XXX give me a real exception here
+ raise Exception
+
+ self._normalized = False
+ self.normalize()