# - - - P u z z l e . _ _ c l u e C h e c k def __clueCheck ( self, choiceList, slot, k ): '''Eliminate choices inconsistent with clue at position k. [ (choiceList is a set of Word instances) and (slot is a Slot instance) and (k is a position within slot) -> if the cell at slot[k] contains a clue letter -> choiceList := choiceList - (any word not having that letter at position [k]) else -> I ] '''
First we convert the index
k into a grid
position using Section 56, “
Slot.__getitem__(): Slot position to
puzzle position”. Then we
find the cell at that position using Section 32, “
Puzzle.whatCell(): Is there a cell at a
#-- 1 -- # [ coord := the Coord instance in self corresponding to # index [k] in slot ] coord = slot[k] #-- 2 -- # [ cell := the Cell instance in self at coord ] cell = self.whatCell ( coord )
cell has no clue, we are done.
Otherwise, examine each word in
and remove those whose
doesn't match the clue.
#-- 3 -- if cell.text == UNK_CELL: return else: clueLetter = cell.text
Removing the conflicts is just a little tricky. You might think that this code would do it:
# THIS DOES NOT WORK! for choice in choiceList: if choice[k] != clueLetter: choiceList.remove(choice)
Try it and you will get the message “RuntimeError:
Set changed size during iteration”. The correct
approach is to form a list from the set, then iterate over
the list, whose size will not change in the
for loop. By the way, we depend on the
Cell class to uppercase the clue letter, and the
Word class to uppercase the word list, so
that we don't have to a case-insensitive comparison here.
#-- 4 -- # [ choiceList := choiceList - (elements that do # not have (clueLetter) in position [k]) ] for choice in list(choiceList): if choice[k] != clueLetter: choiceList.remove ( choice )