In my CS 212 class on Udacity, the most complex lesson involved a crossword game program (for games such as Scrabble® and Words with Friends®). The program was developed incrementally. First I asked "what words can be made with a rack of seven letters?", then I asked "how can you place words onto a single row?", and finally, I, with the help of the students, developed a program to find the highest scoring play anywhere on the board. This approach made for a good sequence of exercises, each building on the previous one. But the code ended up being overly complicated—it accumlated technical debt—because it kept around ideas from each iteration.
In this notebook I will refactor the program to pay off the debt.
Our program uses these concepts:
s will stand for a square number, and sq for the contents of a square.)ACROSS or DOWN direction. (The variable dir stands for a direction.)A and 10 for Q) times the letter bonus score. The letter bonus is 2 when a tile is first placed on a double letter square (or the center star) and 3 when first placed on a triple letter square; it is 1 for a tile already on the board, or for a new tile played on a non-letter-bonus square. The letter score for a blank tile is always zero.This notebook uses these imports:
In [1]:
from __future__ import division, print_function
from collections import defaultdict, namedtuple
from IPython.display import HTML, display
import random
In [2]:
! [ -e enable1.txt ] || curl -O http://norvig.com/ngrams/enable1.txt
Now we can define a word and load the dictionary:
In [3]:
def Word(w) -> str: return w.strip().upper()
DICTIONARY = {Word(w) for w in open('enable1.txt')}
def is_word(word) -> bool:
"Is this a legal word in the dictionary?"
return word.upper() in DICTIONARY
In [4]:
len(DICTIONARY)
Out[4]:
In [5]:
list(DICTIONARY)[:10]
Out[5]:
In [6]:
'WORD' in DICTIONARY
Out[6]:
We'll represent a tile as a one-character string, like 'W'. We'll represent a rack as a string of tiles, usually of length 7, such as 'EELRTTS'. (I also considered a collections.Counter to represent a rack, but felt that str was simpler, and with the rack size limited to 7, efficiency was not a major issue.)
The blank tile causes some complications. We'll represent a blank in a player's rack as the underscore character, '_'. But once the blank is played on the board, it must be used as if it was a specific letter. However, it doesn't score the points of the letter. I chose to use the lowercase version of the letter to represent this. That way, we know what letter the blank is standing for, and we can distingush between scoring and non-scoring tiles. For example, 'EELRTT_' is a rack that contains a blank; and 'LETTERs' is a word played on the board that uses the blank to stand for the letter S.
We'll define letters to give all the distinct letters that can be made by a rack, and remove to remove letters from a rack (after they have been played).
In [7]:
BLANK = '_' # The blank tile (as it appears in the rack)
cat = ''.join # Function to concatenate strings
def letters(rack) -> str:
"All the distinct letters in a rack (including lowercase if there is a blank)."
if BLANK in rack:
return cat(set(rack.replace(BLANK, ''))) + 'abcdefghijklmnopqrstuvwxyz'
else:
return cat(set(rack))
def remove(tiles, rack) -> str:
"Return a copy of rack with the given tile(s) removed."
for tile in tiles:
if tile.islower(): tile = BLANK
rack = rack.replace(tile, '', 1)
return rack
In [8]:
is_word('LETTERs')
Out[8]:
In [9]:
letters('LETTERS')
Out[9]:
In [10]:
letters('EELRTT_')
Out[10]:
In [11]:
remove('SET', 'LETTERS')
Out[11]:
In [12]:
remove('TREaT', 'LETTER_')
Out[12]:
In the previous version of this program, the board was a two-dimensional matrix, and a square on the board was denoted by a (row, col) pair of indexes. There's nothing wrong with that representation, but for this version we will choose a different representation that is simpler in most ways:
ACROSS direction from one square to the next, increment the square index by 1.DOWN direction from one square to the next, increment the square index by 17.OFF, indicating that they are off the board.
The advantage of the border is that the code never has to check if it is at the edge of the board; it can always
look at the neighboring square without fear of indexing off the end of the board.How will we implement this? We'll define Board as a subclass of list and give it two additional attributes:
down: the increment to move in the down direction; 17 for a standard board.directions: the four increments to move to any neighboring square; (1, 17, -1, -17) in a standard board.Jupyter/Ipython notebooks have a special convention for displaying objects in HTML. We will adopt it as a method of Board:
_repr_html_: return a string of HTML that displays the board as a table.
In [13]:
ACROSS = 1 # The 'across' direction; 'down' depends on the size of the board
OFF = '#' # A square that is off the board
SL, DL, TL, STAR, DW, TW = EMPTY = '.:;*-=' # Single/double/triple letter; star, double/triple word bonuses
Square = int # Squares are implemented as integer indexes.
Direction = int # Directions are implemented as integer increments
class Board(list):
"""A Board is a (linear) list of squares, each a single character.
Note that board[s + down] is directly below board[s]."""
def __init__(self, squares):
list.__init__(self, squares)
down = int(len(squares)**0.5)
self.down = down
self.directions = (ACROSS, down, -ACROSS, -down)
def _repr_html_(self) -> str: return board_html(self)
We'll define WWF as the standard board for Words with Friends®.
In [14]:
WWF = Board("""
# # # # # # # # # # # # # # # # #
# . . . = . . ; . ; . . = . . . #
# . . : . . - . . . - . . : . . #
# . : . . : . . . . . : . . : . #
# = . . ; . . . - . . . ; . . = #
# . . : . . . : . : . . . : . . #
# . - . . . ; . . . ; . . . - . #
# ; . . . : . . . . . : . . . ; #
# . . . - . . . * . . . - . . . #
# ; . . . : . . . . . : . . . ; #
# . - . . . ; . . . ; . . . - . #
# . . : . . . : . : . . . : . . #
# = . . ; . . . - . . . ; . . = #
# . : . . : . . . . . : . . : . #
# . . : . . - . . . - . . : . . #
# . . . = . . ; . ; . . = . . . #
# # # # # # # # # # # # # # # # #
""".split())
In [15]:
assert len(WWF) == 17 * 17
In [60]:
def board_html(board) -> str:
"An HTML representation of the board."
size = board.down - 2
squares = [square_html(sq) for sq in board if sq != OFF]
row = ('<tr>' + '{}' * size)
return ('<table>' + row * size + '</table>').format(*squares)
board_colors = {
DL: ('lightblue', 66, 'DL'),
TL: ('lightgreen', 66, 'TL'),
DW: ('lightcoral', 66, 'DW'),
TW: ('orange', 66, 'TW'),
SL: ('whitesmoke', 66, ''),
STAR: ('violet', 100, '✭')}
def square_html(sq) -> str:
"An HTML representation of a square."
color, size, text = board_colors.get(sq, ('gold', 120, sq))
if text.isupper():
text = '<b>{}</b><sup style="font-size: 60%">{}</sup>'.format(text, POINTS.get(text, ''))
style = "background-color:{}; font-size:{}%; width:25px; height:25px; text-align:center; padding:0px"
return ('<td style="' + style + '">{}').format(color, size, text)
POINTS = defaultdict(int,
A=1, B=3, C=3, D=2, E=1, F=4, G=2, H=4, I=1, J=8, K=5, L=1, M=3,
N=1, O=1, P=3, Q=10, R=1, S=1, T=1, U=1, V=4, W=4, X=8, Y=4, Z=10)
In [61]:
WWF
Out[61]:
A Play describes the placement of tiles on the board. We will implement Play as a named tuple of four components:
start: the index number of the square that holds the first letter in the word.dir: the direction, with 1 indicating ACROSS and board.down (normally, 17) indicating DOWN.letters: the letters of the word, in order, as a str. Blanks are lowercase. Some letters are from the rack; some may have been on the board.rack: the letters that would remain in the player's rack after making this play. Not strictly necessary as part of the play, but useful information.The function make_play returns a new board with the play made on it. It does not do any checking to see if the play follows the rules.
In [18]:
Play = namedtuple('Play', 'start, dir, letters, rack')
def make_play(board, play) -> Board:
"Make the play on a copy of board and return the copy."
copy = Board(board)
end = play.start + len(play.letters) * play.dir
copy[play.start:end:play.dir] = play.letters
return copy
In [19]:
DOWN = WWF.down
plays = {Play(145, DOWN, 'ENTER', ''),
Play(144, ACROSS, 'BE', ''),
Play(138, DOWN, 'GAVE', ''),
Play(158, DOWN, 'MUSES', ''),
Play(172, ACROSS, 'VIRULeNT', ''),
Play(213, ACROSS, 'RED', ''),
Play(198, ACROSS, 'LYTHE', ''),
Play(147, DOWN, 'CHILDREN', ''),
Play(164, ACROSS, 'HEARD', ''),
Play(117, DOWN, 'BRIDLES', ''),
Play(131, ACROSS, 'TOUR', '')}
board = Board(WWF)
for play in plays:
board = make_play(board, play)
board
Out[19]:
This is our strategy for finding all possible legal plays on a board:
STAR square in the center counts as the only anchor square.)ABC, we find that B, BA, and BAC are all prefixes of the word BACK (and the rack contains other prefixes of other words as well).OFF square ahead), yield the play that made the word.So, each legal play will have a prefix of zero or more letters, followed by one letter from the rack covering an anchor square, followed by zero or more additional letters, which can be from the rack or already on the board.
In [20]:
def dict_prefixes(dictionary) -> set:
"The set of all prefixes of each word in a dictionary."
return {word[:i] for word in dictionary for i in range(len(word))}
PREFIXES = dict_prefixes(DICTIONARY)
In [21]:
len(PREFIXES)
Out[21]:
That's too many prefixes to look at; let's try a smaller example below. Note that the empty string is a prefix, and we include HELP because it is a prefix of HELPER, but we don't include HELPER, because there is nothing we can add to it to make a word in this dictionary:
In [22]:
dict_prefixes({'HELLO', 'HELP', 'HELPER'})
Out[22]:
The function rack_prefixes gives the set of prefixes that can be made just from the letters in the rack. Most of the work is done by extend_prefixes, which accumulates a set of prefixes into results:
In [23]:
def rack_prefixes(rack) -> set:
"All word prefixes that can be made by the rack."
return extend_prefixes('', rack, set())
def extend_prefixes(prefix, rack, results) -> set:
if prefix.upper() in PREFIXES:
results.add(prefix)
for L in letters(rack):
extend_prefixes(prefix+L, remove(L, rack), results)
return results
In [24]:
rack = 'ABC'
rack_prefixes(rack)
Out[24]:
The number of prefixes in a rack is usually on the order of a hundred, unless there is a blank in the rack:
In [25]:
len(rack_prefixes('LETTERS'))
Out[25]:
In [26]:
len(rack_prefixes('LETTER_'))
Out[26]:
In [27]:
def is_anchor(board, s) -> bool:
"Is this square next to a letter already on the board? (Or is it a '*')?"
return (board[s] == STAR or
board[s] in EMPTY and any(board[s + d].isalpha() for d in board.directions))
def all_anchors(board) -> list:
"A list of all anchor squares on the board."
return [s for s in range(len(board)) if is_anchor(board, s)]
In [28]:
all_anchors(WWF)
Out[28]:
In [29]:
anchors = all_anchors(board)
len(anchors)
Out[29]:
To visualize these anchors, we'll make each one be a star, on a copy of board:
In [62]:
board2 = Board(board)
for a in anchors:
board2[a] = STAR
board2
Out[62]:
Now we'll define a rack, and find all the prefixes for the rack:
In [31]:
rack = 'ABCHKNQ'
prefixes = rack_prefixes(rack)
len(prefixes)
Out[31]:
In [32]:
' '.join(prefixes)
Out[32]:
We wont go through all the anchor/prefix combinations; we'll just pick one: the anchor above the M in MUSES:
In [63]:
board3 = Board(board)
anchor = 141
board3[anchor] = STAR
board3
Out[63]:
There's only room for prefixes of length 0 or 1, because anything longer than that would hit the anchor to the right of the G in GAVE; to avoid duplication of effort, we only allow words to run into other anchors on the right, not the left. Let's try the 1-letter prefix B first:
In [64]:
board3[140] = 'B'
board3
Out[64]:
Now we can start to march forward. On the anchor square we can place any letter from the rack that makes a valid prefix, and that also turns .MUSES into a valid word. There's only one such letter, A:
In [35]:
board3[141] = 'A'
assert 'BA' in PREFIXES and is_word('A' + 'MUSES')
board3
Out[35]:
We can continue marching forward, trying letters from the rack that form valid prefixes. Let's try the combination CK:
In [36]:
board3[142:144] = 'CK'
assert 'BACKBE' in PREFIXES
board3
Out[36]:
We've spelled the word BACK, but we can't count it as a legal play, because we've hit two adjacent letters, BE, that are already on the board. We check that BACKBE froms a valid prefix, and continue to the next empty square, where we can choose an N:
In [37]:
board3[146] = 'N'
assert 'BACKBENC' in PREFIXES
board3
Out[37]:
We continue to the next square (a double word square), and place an H, which completes a word, BACKBENCH, and simultaneously makes a cross word, THE:
In [38]:
board3[148] = 'H'
assert is_word('BACKBENCH') and is_word('THE')
board3
Out[38]:
We would record this play, and backtrack to consider other letters for this and other prefix/anchor combinations. Now let's code this up!
The function all_plays generates all legal plays by first trying all prefix plays, and then trying to extend each one, one letter at a time. (Note that it also generates the empty play, because a player always has the option of passing.)
In [39]:
def all_plays(board, rack):
"""Generate all plays that can be played on board with this rack.
Try placing every possible prefix before every anchor point;
then extend one letter at a time, looking for valid plays."""
anchors = all_anchors(board)
prefixes = rack_prefixes(rack)
yield Play(0, 1, '', rack) # The empty play (no letters, no points)
for anchor in anchors:
for dir in (ACROSS, board.down):
for play in prefix_plays(prefixes, board, anchor, dir, rack):
yield from extend_play(board, play)
Note the syntax yield from, new in Python 3.3: "yield from c" is the same as "for x in c: yield x".
Now for the function prefix_plays, which returns a list of all partial plays consisting of a prefix placed before the anchor. Note that these are not legal plays; they are partial plays, some of which will end up being extended into legal plays.
There are two cases: if there are letters on the board immediately before the anchor, then those letters form the only allowable prefix. If not, we can use any prefix from the rack up to maxlen, which is the number of empty squares that do not run into another anchor, nor off the board.
In [40]:
def prefix_plays(prefixes, board, anchor, dir, rack) -> list:
"Return all Plays of a prefix to the left/above anchor."
if board[anchor-dir].isalpha(): # Prefix already on the board; only 1 prefix
start = scan_letters(board, anchor, -dir)
return [Play(start, dir, cat(board[start:anchor:dir]), rack)]
else: # Prefixes from rack fit in space before anchor
maxlen = (anchor - scan_to_anchor(board, anchor, -dir)) // dir
return [Play(anchor - len(prefix) * dir, dir, prefix, remove(prefix, rack))
for prefix in prefixes if len(prefix) <= maxlen]
Now extend_play takes a partial play, determines the square, s, that is one square past the end of the play, and tries all possible letters there. If adding a letter forms a valid prefix (and also does not form an invalid cross word), then we continue on (by calling extend_play recursively). If adding the letter forms a valid word, we yield the play.
In [41]:
def extend_play(board, play):
"Explore all ways of adding to end of play; return ones that form full words."
s = play.start + play.dir * len(play.letters)
if board[s] == OFF: return
cword = crossword(board, s, play.dir)
possible_letters = board[s].upper() if board[s].isalpha() else letters(play.rack)
for L in possible_letters:
prefix2 = play.letters + L
if prefix2.upper() in PREFIXES and valid_crossword(cword, L):
rack2 = play.rack if board[s].isalpha() else remove(L, play.rack)
play2 = Play(play.start, play.dir, prefix2, rack2)
if is_word(prefix2) and not board[s + play.dir].isalpha():
yield play2
yield from extend_play(board, play2)
def scan_letters(board, s, dir) -> Square:
"Return the last square number going from s in dir that is a letter."
while board[s + dir].isalpha():
s += dir
return s
def scan_to_anchor(board, s, dir) -> Square:
"Return the last square number going from s in dir that is not an anchor nor off board."
while board[s + dir] != OFF and not is_anchor(board, s + dir):
s += dir
return s
If adding a letter in, say, the ACROSS direction also adds on to a word in the DOWN direction, then we need to make sure that this cross word is also valid. The function crossword finds the cross word at square s and returns it with a '.' indicating the empty square where the new letter will be placed, so we would get '.MUSES' and 'T.E' for the two crosswords in the 'BACKBENCH' play.
In [42]:
def crossword(board, s, dir) -> str:
"""The word that intersects s in the other direction from dir.
Use '.' for the one square that is missing a letter."""
def canonical(L): return L if L.isalpha() else '.'
d = other(dir, board)
start = scan_letters(board, s, -d)
end = scan_letters(board, s, d)
return cat(canonical(board[s]) for s in range(start, end+d, d))
def valid_crossword(cword, L) -> bool:
"Is placing letter L valid (with respective to the crossword)?"
return len(cword) == 1 or cword.replace('.', L).upper() in DICTIONARY
def other(dir, board) -> Direction:
"The other direction (across/down) on the board."
return board.down if dir == ACROSS else ACROSS
In [43]:
crossword(board, 141, ACROSS)
Out[43]:
In [44]:
crossword(board, 148, ACROSS)
Out[44]:
The function valid_crossword checks if replacing the empty square with a specific letter will form a valid word:
In [45]:
valid_crossword('.MUSES', 'A')
Out[45]:
We can now see all the prefix plays for the anchor at 141 (just above MUSES):
In [46]:
prefix_plays(rack_prefixes(rack), board, 141, 1, rack)
Out[46]:
And we can see all the ways to extend the play of 'B' there:
In [47]:
set(extend_play(board, Play(start=140, dir=1, letters='B', rack='ACHKNQ')))
Out[47]:
Now we'll show how to count up the points made by a play. The score is the sum of the word score for the play, plus a bingo score if all seven letters are used, plus the sum of the word scores for any cross words. The word score is the sum of the letter scores (where each letter score may be doubled or tripled by a bonus square when the letter is first played on the square), all multiplied by any word bonus(es) encountered by the newly-placed letters.
In [48]:
def score(board, play) -> int:
"The number of points scored by making this play on the board."
return (word_score(board, play)
+ bingo(board, play)
+ sum(word_score(board, cplay)
for cplay in cross_plays(board, play)))
def word_score(board, play) -> int:
"Points for a single word, counting word- and letter-bonuses."
total, word_bonus = 0, 1
for (s, L) in enumerate_play(play):
sq = board[s]
word_bonus *= (3 if sq == TW else 2 if sq == DW else 1)
total += POINTS[L] * (3 if sq == TL else 2 if sq == DL else 1)
return word_bonus * total
def bingo(board, play) -> int:
"A bonus for using 7 letters from the rack."
return BINGO if (play.rack == '' and letters_played(board, play) == 7) else 0
BINGO = 35
Here are the various helper functions:
In [49]:
def letters_played(board, play) -> int:
"The number of letters played from the rack."
return sum(board[s] in EMPTY for (s, L) in enumerate_play(play))
def enumerate_play(play) -> list:
"List (square_number, letter) pairs for each tile in the play."
return [(play.start + i * play.dir, L)
for (i, L) in enumerate(play.letters)]
def cross_plays(board, play):
"Generate all plays for words that cross this play."
cross = other(play.dir, board)
for (s, L) in enumerate_play(play):
if board[s] in EMPTY and (board[s-cross].isalpha() or board[s+cross].isalpha()):
start, end = scan_letters(board, s, -cross), scan_letters(board, s, cross)
before, after = cat(board[start:s:cross]), cat(board[s+cross:end+cross:cross])
yield Play(start, cross, before + L + after, play.rack)
What should the BACKBENCH play score? The word covers two double-word bonuses, but no letter bonuses. The sum of the letter point values is 3+1+3+5+3+1+1+3+4 = 24, and 24×2×2 = 96. The cross word AMUSES scores 8, and THE is on a double word bonus, so it scores 6×2 = 12. There is one letter remaining in the rack, so no bingo, just a total score of 96 + 8 + 12 = 116.
In [50]:
score(board, Play(start=140, dir=1, letters='BACKBENCH', rack='Q'))
Out[50]:
We can find the highest scoring play by enumerating all plays and taking the one with the maximum score:
In [51]:
def highest_scoring_play(board, rack) -> Play:
"Return the Play that gives the most points."
return max(all_plays(board, rack), key=lambda play: score(board, play))
In [52]:
highest_scoring_play(board, rack)
Out[52]:
In [53]:
make_play(board, Play(start=140, dir=1, letters='BACKBENCH', rack='Q'))
Out[53]:
In [54]:
BAG = 'AAAAAAAAABBCCDDDDEEEEEEEEEEEEFFGGGHHIIIIIIIIIJKLLLLMMNNNNNNOOOOOOOOPPQRRRRRRSSSSTTTTTTUUUUVVWWXYYZ__'
len(BAG)
Out[54]:
Then the function play_game will take a list of player strategies as input, and play those strategies against each other over the course of a game. A strategy is a function that takes a board and a rack as input and returns a play. For example, highest_scoring_play is a strategy. If the optional argument verbose is true, then the board is displayed after each play.
In [55]:
def play_game(strategies=[highest_scoring_play, highest_scoring_play], verbose=True) -> list:
"A number of players play a game; return a list of their scores."
board = Board(WWF)
bag = list(BAG)
random.shuffle(bag)
scores = [0 for _ in strategies]
racks = [replenish('', bag) for _ in strategies]
while True:
old_board = board
for (p, strategy) in enumerate(strategies):
board = make_one_play(board, p, strategy, scores, racks, bag, verbose)
if racks[p] == '':
# Player p has gone out; game over
return subtract_remaining_tiles(racks, scores, p)
if old_board == board:
# No player has a move; game over
return scores
def make_one_play(board, p, strategy, scores, racks, bag, verbose) -> Board:
"""One player, player p, chooses a move according to the strategy.
We make the move, replenish the rack, update scores, and return the new Board."""
rack = racks[p]
play = strategy(board, racks[p])
racks[p] = replenish(play.rack, bag)
points = score(board, play)
scores[p] += points
board = make_play(board, play)
if verbose:
display(HTML('Player {} with rack {} makes {}<br>for {} points; draws: {}; scores: {}'
.format(p, rack, play, points, racks[p], scores)),
board)
return board
def subtract_remaining_tiles(racks, scores, p) -> list:
"Subtract point values from each player and give them to player p."
for i in range(len(racks)):
points = sum(POINTS[L] for L in racks[i])
scores[i] -= points
scores[p] += points
return scores
def replenish(rack, bag) -> str:
"Fill rack with 7 letters (as long as there are letters left in the bag)."
while len(rack) < 7 and bag:
rack += bag.pop()
return rack
In [56]:
%%javascript
IPython.OutputArea.auto_scroll_threshold = 9999;
In [57]:
play_game()
Out[57]:
That was an exciting game, with four bingos: TRICORNE, DINOSAUR, FISTULA, and IRRITATE. The FISTULA play garnered 99 points, from the bingo, the triple word score, and a sextuple bonus from the 4-point letter F (triple letter score in both directions).
But that was just one game; Let's get statistics for both players over, say, 10 games:
In [58]:
%%time
games = 10
scores = sorted(score for game in range(games)
for score in play_game(verbose=False))
print('min: {}, median: {}, mean: {}, max: {}'.format(
min(scores), scores[games], sum(scores)/(2*games), max(scores)))
In [59]:
def sames(A, B): return sorted(A) == sorted(B)
def test():
"Unit tests."
assert is_word('WORD')
assert is_word('LETTERs')
assert is_word('ETHyLENEDIAMINETETRAACETATES')
assert not is_word('ALFABET')
rack = 'ABCHKNQ'
assert sames(letters(rack), rack)
assert sames(letters('ABAC_'), 'ABCabcdefghijklmnopqrstuvwxyz')
assert dict_prefixes({'HELLO', 'HELP', 'HELPER'}) == {
'', 'H', 'HE', 'HEL', 'HELL', 'HELP', 'HELPE'}
assert rack_prefixes('ABC') == {'', 'A', 'AB', 'AC', 'B', 'BA', 'BAC', 'C', 'CA', 'CAB'}
assert len(rack_prefixes('LETTERS')) == 155
assert len(rack_prefixes('LETTER_')) == 1590
DOWN = WWF.down
plays = {
Play(145, DOWN, 'ENTER', ''),
Play(144, ACROSS, 'BE', ''),
Play(138, DOWN, 'GAVE', ''),
Play(158, DOWN, 'MUSES', ''),
Play(172, ACROSS, 'VIRULeNT', ''),
Play(213, ACROSS, 'RED', ''),
Play(198, ACROSS, 'LYTHE', ''),
Play(147, DOWN, 'CHILDREN', ''),
Play(164, ACROSS, 'HEARD', ''),
Play(117, DOWN, 'BRIDLES', ''),
Play(131, ACROSS, 'TOUR', '')}
board = Board(WWF)
for play in plays:
board = make_play(board, play)
assert len(WWF) == len(board) == 17 * 17
assert all_anchors(WWF) == [144]
assert all_anchors(board) == [
100, 114, 115, 116, 121, 127, 128, 130, 137, 139, 141, 143, 146, 148, 149, 150, 154, 156, 157, 159, 160,
161, 163, 171, 180, 182, 183, 184, 188, 190, 191, 193, 194, 195, 197, 206, 208, 210, 212, 216, 217, 218,
225, 227, 230, 231, 233, 236, 243, 248, 250, 265, 267]
assert crossword(board, 141, ACROSS) == '.MUSES'
assert crossword(board, 148, ACROSS) == 'T.E'
assert valid_crossword('.MUSES', 'A')
assert not valid_crossword('.MUSES', 'B')
assert sames(prefix_plays(rack_prefixes(rack), board, 141, 1, rack),
[Play(start=141, dir=1, letters='', rack='ABCHKNQ'),
Play(start=140, dir=1, letters='C', rack='ABHKNQ'),
Play(start=140, dir=1, letters='K', rack='ABCHNQ'),
Play(start=140, dir=1, letters='B', rack='ACHKNQ'),
Play(start=140, dir=1, letters='A', rack='BCHKNQ'),
Play(start=140, dir=1, letters='H', rack='ABCKNQ'),
Play(start=140, dir=1, letters='N', rack='ABCHKQ'),
Play(start=140, dir=1, letters='Q', rack='ABCHKN')])
assert sames(extend_play(board, Play(start=140, dir=1, letters='B', rack='ACHKNQ')),
{Play(start=140, dir=1, letters='BA', rack='CHKNQ'),
Play(start=140, dir=1, letters='BACKBENCH', rack='Q'),
Play(start=140, dir=1, letters='BAH', rack='CKNQ'),
Play(start=140, dir=1, letters='BAN', rack='CHKQ')})
assert len(BAG) == 100
assert replenish('RACK', ['X', 'B', 'A', 'G']) == 'RACKGAB'
assert replenish('RACK', []) == 'RACK'
assert replenish('RACK', ['A', 'B']) == 'RACKBA'
assert score(WWF, Play(144, ACROSS, 'BE', '')) == (3 + 1)
assert score(board, Play(140, ACROSS, 'BACKBENCH', 'Q')) == 116
return 'ok'
test()
Out[59]:
We can break that into three questions:
We can also ask: What's left to do?
Thanks to Markus Dobler for correcting one bug and making another useful suggestion.
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