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.rack is replenished with tiles until the player has 7 tiles or until the bag of tiles is empty.
with the rules of the game; it will be important in our algorithm that finds valid plays.
This notebook uses these imports:
from __future__ import division, print_function
from collections import defaultdict, namedtuple
from IPython.display import HTML, display
import random
We will represent the dictionary as a set of words. A word is an uppercase string of letters, like 'WORD'
. There are several standard dictionaries used by different communities of players; we will use the ENABLE dictionary—we can cache a local copy with this shell command:
! [ -e enable1.txt ] || curl -O http://norvig.com/ngrams/enable1.txt
Now we can define a word and load the dictionary:
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
len(DICTIONARY)
172820
list(DICTIONARY)[:10]
['DECIPHER', 'ROUGHLY', 'DULLSVILLES', 'INDEHISCENT', 'SAGEBRUSHES', 'COMES', 'WITCHERIES', 'HOMES', 'PECK', 'GEEZ']
'WORD' in DICTIONARY
True
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).
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
is_word('LETTERs')
True
letters('LETTERS')
'TLRSE'
letters('EELRTT_')
'TLREabcdefghijklmnopqrstuvwxyz'
remove('SET', 'LETTERS')
'LTER'
remove('TREaT', 'LETTER_')
'LE'
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:
we will include a border around the outside, making the board of size 17×17.
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.
the tile replaces the bonus value.
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.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®.
WWF = Board("""
# # # # # # # # # # # # # # # # #
# . . . = . . ; . ; . . = . . . #
# . . : . . - . . . - . . : . . #
# . : . . : . . . . . : . . : . #
# = . . ; . . . - . . . ; . . = #
# . . : . . . : . : . . . : . . #
# . - . . . ; . . . ; . . . - . #
# ; . . . : . . . . . : . . . ; #
# . . . - . . . * . . . - . . . #
# ; . . . : . . . . . : . . . ; #
# . - . . . ; . . . ; . . . - . #
# . . : . . . : . : . . . : . . #
# = . . ; . . . - . . . ; . . = #
# . : . . : . . . . . : . . : . #
# . . : . . - . . . - . . : . . #
# . . . = . . ; . ; . . = . . . #
# # # # # # # # # # # # # # # # #
""".split())
assert len(WWF) == 17 * 17
I want to diaplay the board in HTML, as a table with different background colors for the bonus squares; and gold-colored letter tiles. I also want to display the point values for each letter on the tiles; I'll use a defaultdict
of {letter: int}
named POINTS
for that.
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)
WWF
TW | TL | TL | TW | |||||||||||
DL | DW | DW | DL | |||||||||||
DL | DL | DL | DL | |||||||||||
TW | TL | DW | TL | TW | ||||||||||
DL | DL | DL | DL | |||||||||||
DW | TL | TL | DW | |||||||||||
TL | DL | DL | TL | |||||||||||
DW | ✭ | DW | ||||||||||||
TL | DL | DL | TL | |||||||||||
DW | TL | TL | DW | |||||||||||
DL | DL | DL | DL | |||||||||||
TW | TL | DW | TL | TW | ||||||||||
DL | DL | DL | DL | |||||||||||
DL | DW | DW | DL | |||||||||||
TW | TL | TL | TW |
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.
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
Let's test out what we've done so far. I'll put some words on a board, which I will call board
:
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
TW | TL | TL | TW | |||||||||||
DL | DW | DW | DL | |||||||||||
DL | DL | DL | DL | |||||||||||
TW | TL | DW | TL | TW | ||||||||||
DL | DL | DL | DL | |||||||||||
DW | TL | TL | DW | B3 | ||||||||||
TL | DL | DL | T1 | O1 | U1 | R1 | ||||||||
G2 | DW | B3 | E1 | C3 | DW | I1 | ||||||||
TL | A1 | M3 | N1 | H4 | E1 | A1 | R1 | D2 | ||||||
V4 | I1 | R1 | U1 | L1 | e | N1 | T1 | TL | I1 | DW | L1 | |||
E1 | DL | S1 | DL | E1 | L1 | Y4 | T1 | H4 | E1 | |||||
TW | TL | E1 | DW | R1 | E1 | D2 | TL | S1 | ||||||
DL | S1 | R1 | DL | |||||||||||
DL | DW | DW | E1 | DL | ||||||||||
TW | TL | TL | N1 | TW |
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.
Here we define the set of all prefixes of all words in the dictionary:
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)
len(PREFIXES)
276374
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:
dict_prefixes({'HELLO', 'HELP', 'HELPER'})
{'', 'H', 'HE', 'HEL', 'HELL', 'HELP', 'HELPE'}
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
:
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
rack = 'ABC'
rack_prefixes(rack)
{'', 'A', 'AB', 'AC', 'B', 'BA', 'BAC', 'C', 'CA', 'CAB'}
The number of prefixes in a rack is usually on the order of a hundred, unless there is a blank in the rack:
len(rack_prefixes('LETTERS'))
155
len(rack_prefixes('LETTER_'))
1590
An anchor square is either the star in the middle of the board, or an empty square that is adjacent to a letter:
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)]
all_anchors(WWF)
[144]
Let's work through the process of finding plays on the example board
. First, we'll find all the anchors:
anchors = all_anchors(board)
len(anchors)
53
To visualize these anchors, we'll make each one be a star, on a copy of board
:
board2 = Board(board)
for a in anchors:
board2[a] = STAR
board2
TW | TL | TL | TW | |||||||||||
DL | DW | DW | DL | |||||||||||
DL | DL | DL | DL | |||||||||||
TW | TL | DW | TL | TW | ||||||||||
DL | DL | DL | DL | ✭ | ||||||||||
DW | TL | TL | ✭ | ✭ | ✭ | B3 | ||||||||
TL | ✭ | DL | ✭ | ✭ | ✭ | T1 | O1 | U1 | R1 | |||||
✭ | G2 | ✭ | DW | ✭ | ✭ | B3 | E1 | ✭ | C3 | ✭ | ✭ | ✭ | I1 | |
✭ | A1 | ✭ | ✭ | M3 | ✭ | ✭ | ✭ | N1 | ✭ | H4 | E1 | A1 | R1 | D2 |
✭ | V4 | I1 | R1 | U1 | L1 | e | N1 | T1 | ✭ | I1 | ✭ | ✭ | ✭ | L1 |
✭ | E1 | ✭ | ✭ | S1 | ✭ | ✭ | ✭ | E1 | ✭ | L1 | Y4 | T1 | H4 | E1 |
TW | ✭ | ✭ | E1 | ✭ | ✭ | R1 | E1 | D2 | ✭ | ✭ | ✭ | S1 | ||
DL | ✭ | S1 | ✭ | ✭ | ✭ | R1 | ✭ | DL | ✭ | |||||
DL | ✭ | DW | ✭ | E1 | ✭ | DL | ||||||||
TW | TL | TL | ✭ | N1 | ✭ |
Now we'll define a rack, and find all the prefixes for the rack:
rack = 'ABCHKNQ'
prefixes = rack_prefixes(rack)
len(prefixes)
88
' '.join(prefixes)
' ANKH ACN CANK BHA ACK HA HACKN A HACKB AN ABN KAC AQ AK AC BH CAN BAK BAC HANC CN N NAB H ANC C ANK CHAN CHAK KH KHAN NAK HAK BA QAN HANK CAH Q BACK CAB KHA BACH HAC BAH BANK CAK CA NACH AHC KAN NAC KAH KACH AH HACK BANKC CHAQ BANQ ANCH BHAK KANB BAN KAB ACKN NA K AB BACKH ANH KN KB HAB KNA KNAC B KA ABH CHA CHAB HAN ACQ CH BANC KBA ACH QA BHAN'
We wont go through all the anchor/prefix combinations; we'll just pick one: the anchor above the M
in MUSES
:
board3 = Board(board)
anchor = 141
board3[anchor] = STAR
board3
TW | TL | TL | TW | |||||||||||
DL | DW | DW | DL | |||||||||||
DL | DL | DL | DL | |||||||||||
TW | TL | DW | TL | TW | ||||||||||
DL | DL | DL | DL | |||||||||||
DW | TL | TL | DW | B3 | ||||||||||
TL | DL | DL | T1 | O1 | U1 | R1 | ||||||||
G2 | DW | ✭ | B3 | E1 | C3 | DW | I1 | |||||||
TL | A1 | M3 | N1 | H4 | E1 | A1 | R1 | D2 | ||||||
V4 | I1 | R1 | U1 | L1 | e | N1 | T1 | TL | I1 | DW | L1 | |||
E1 | DL | S1 | DL | E1 | L1 | Y4 | T1 | H4 | E1 | |||||
TW | TL | E1 | DW | R1 | E1 | D2 | TL | S1 | ||||||
DL | S1 | R1 | DL | |||||||||||
DL | DW | DW | E1 | DL | ||||||||||
TW | TL | TL | N1 | TW |
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:
board3[140] = 'B'
board3
TW | TL | TL | TW | |||||||||||
DL | DW | DW | DL | |||||||||||
DL | DL | DL | DL | |||||||||||
TW | TL | DW | TL | TW | ||||||||||
DL | DL | DL | DL | |||||||||||
DW | TL | TL | DW | B3 | ||||||||||
TL | DL | DL | T1 | O1 | U1 | R1 | ||||||||
G2 | B3 | ✭ | B3 | E1 | C3 | DW | I1 | |||||||
TL | A1 | M3 | N1 | H4 | E1 | A1 | R1 | D2 | ||||||
V4 | I1 | R1 | U1 | L1 | e | N1 | T1 | TL | I1 | DW | L1 | |||
E1 | DL | S1 | DL | E1 | L1 | Y4 | T1 | H4 | E1 | |||||
TW | TL | E1 | DW | R1 | E1 | D2 | TL | S1 | ||||||
DL | S1 | R1 | DL | |||||||||||
DL | DW | DW | E1 | DL | ||||||||||
TW | TL | TL | N1 | TW |
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
:
board3[141] = 'A'
assert 'BA' in PREFIXES and is_word('A' + 'MUSES')
board3
TW | TL | TL | TW | |||||||||||
DL | DW | DW | DL | |||||||||||
DL | DL | DL | DL | |||||||||||
TW | TL | DW | TL | TW | ||||||||||
DL | DL | DL | DL | |||||||||||
DW | TL | TL | DW | B3 | ||||||||||
TL | DL | DL | T1 | O1 | U1 | R1 | ||||||||
G2 | B3 | A1 | B3 | E1 | C3 | DW | I1 | |||||||
TL | A1 | M3 | N1 | H4 | E1 | A1 | R1 | D2 | ||||||
V4 | I1 | R1 | U1 | L1 | e | N1 | T1 | TL | I1 | DW | L1 | |||
E1 | DL | S1 | DL | E1 | L1 | Y4 | T1 | H4 | E1 | |||||
TW | TL | E1 | DW | R1 | E1 | D2 | TL | S1 | ||||||
DL | S1 | R1 | DL | |||||||||||
DL | DW | DW | E1 | DL | ||||||||||
TW | TL | TL | N1 | TW |
We can continue marching forward, trying letters from the rack that form valid prefixes. Let's try the combination CK
:
board3[142:144] = 'CK'
assert 'BACKBE' in PREFIXES
board3
TW | TL | TL | TW | |||||||||||
DL | DW | DW | DL | |||||||||||
DL | DL | DL | DL | |||||||||||
TW | TL | DW | TL | TW | ||||||||||
DL | DL | DL | DL | |||||||||||
DW | TL | TL | DW | B3 | ||||||||||
TL | DL | DL | T1 | O1 | U1 | R1 | ||||||||
G2 | B3 | A1 | C3 | K5 | B3 | E1 | C3 | DW | I1 | |||||
TL | A1 | M3 | N1 | H4 | E1 | A1 | R1 | D2 | ||||||
V4 | I1 | R1 | U1 | L1 | e | N1 | T1 | TL | I1 | DW | L1 | |||
E1 | DL | S1 | DL | E1 | L1 | Y4 | T1 | H4 | E1 | |||||
TW | TL | E1 | DW | R1 | E1 | D2 | TL | S1 | ||||||
DL | S1 | R1 | DL | |||||||||||
DL | DW | DW | E1 | DL | ||||||||||
TW | TL | TL | N1 | TW |
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
:
board3[146] = 'N'
assert 'BACKBENC' in PREFIXES
board3
TW | TL | TL | TW | |||||||||||
DL | DW | DW | DL | |||||||||||
DL | DL | DL | DL | |||||||||||
TW | TL | DW | TL | TW | ||||||||||
DL | DL | DL | DL | |||||||||||
DW | TL | TL | DW | B3 | ||||||||||
TL | DL | DL | T1 | O1 | U1 | R1 | ||||||||
G2 | B3 | A1 | C3 | K5 | B3 | E1 | N1 | C3 | DW | I1 | ||||
TL | A1 | M3 | N1 | H4 | E1 | A1 | R1 | D2 | ||||||
V4 | I1 | R1 | U1 | L1 | e | N1 | T1 | TL | I1 | DW | L1 | |||
E1 | DL | S1 | DL | E1 | L1 | Y4 | T1 | H4 | E1 | |||||
TW | TL | E1 | DW | R1 | E1 | D2 | TL | S1 | ||||||
DL | S1 | R1 | DL | |||||||||||
DL | DW | DW | E1 | DL | ||||||||||
TW | TL | TL | N1 | TW |
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
:
board3[148] = 'H'
assert is_word('BACKBENCH') and is_word('THE')
board3
TW | TL | TL | TW | |||||||||||
DL | DW | DW | DL | |||||||||||
DL | DL | DL | DL | |||||||||||
TW | TL | DW | TL | TW | ||||||||||
DL | DL | DL | DL | |||||||||||
DW | TL | TL | DW | B3 | ||||||||||
TL | DL | DL | T1 | O1 | U1 | R1 | ||||||||
G2 | B3 | A1 | C3 | K5 | B3 | E1 | N1 | C3 | H4 | I1 | ||||
TL | A1 | M3 | N1 | H4 | E1 | A1 | R1 | D2 | ||||||
V4 | I1 | R1 | U1 | L1 | e | N1 | T1 | TL | I1 | DW | L1 | |||
E1 | DL | S1 | DL | E1 | L1 | Y4 | T1 | H4 | E1 | |||||
TW | TL | E1 | DW | R1 | E1 | D2 | TL | S1 | ||||||
DL | S1 | R1 | DL | |||||||||||
DL | DW | DW | E1 | DL | ||||||||||
TW | TL | TL | N1 | TW |
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.)
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.
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.
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.
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
crossword(board, 141, ACROSS)
'.MUSES'
crossword(board, 148, ACROSS)
'T.E'
The function valid_crossword
checks if replacing the empty square with a specific letter will form a valid word:
valid_crossword('.MUSES', 'A')
True
We can now see all the prefix plays for the anchor at 141 (just above MUSES
):
prefix_plays(rack_prefixes(rack), board, 141, 1, rack)
[Play(start=141, dir=1, letters='', rack='ABCHKNQ'), Play(start=140, dir=1, letters='A', rack='BCHKNQ'), Play(start=140, dir=1, letters='N', rack='ABCHKQ'), Play(start=140, dir=1, letters='H', rack='ABCKNQ'), Play(start=140, dir=1, letters='C', rack='ABHKNQ'), Play(start=140, dir=1, letters='Q', rack='ABCHKN'), Play(start=140, dir=1, letters='K', rack='ABCHNQ'), Play(start=140, dir=1, letters='B', rack='ACHKNQ')]
And we can see all the ways to extend the play of 'B'
there:
set(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')}
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.
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:
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.
score(board, Play(start=140, dir=1, letters='BACKBENCH', rack='Q'))
116
We can find the highest scoring play by enumerating all plays and taking the one with the maximum score:
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))
highest_scoring_play(board, rack)
Play(start=140, dir=1, letters='BACKBENCH', rack='Q')
make_play(board, Play(start=140, dir=1, letters='BACKBENCH', rack='Q'))
TW | TL | TL | TW | |||||||||||
DL | DW | DW | DL | |||||||||||
DL | DL | DL | DL | |||||||||||
TW | TL | DW | TL | TW | ||||||||||
DL | DL | DL | DL | |||||||||||
DW | TL | TL | DW | B3 | ||||||||||
TL | DL | DL | T1 | O1 | U1 | R1 | ||||||||
G2 | B3 | A1 | C3 | K5 | B3 | E1 | N1 | C3 | H4 | I1 | ||||
TL | A1 | M3 | N1 | H4 | E1 | A1 | R1 | D2 | ||||||
V4 | I1 | R1 | U1 | L1 | e | N1 | T1 | TL | I1 | DW | L1 | |||
E1 | DL | S1 | DL | E1 | L1 | Y4 | T1 | H4 | E1 | |||||
TW | TL | E1 | DW | R1 | E1 | D2 | TL | S1 | ||||||
DL | S1 | R1 | DL | |||||||||||
DL | DW | DW | E1 | DL | ||||||||||
TW | TL | TL | N1 | TW |
Now let's play a complete game. We start with a bag of tiles:
BAG = 'AAAAAAAAABBCCDDDDEEEEEEEEEEEEFFGGGHHIIIIIIIIIJKLLLLMMNNNNNNOOOOOOOOPPQRRRRRRSSSSTTTTTTUUUUVVWWXYYZ__'
len(BAG)
100
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.
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
%%javascript
IPython.OutputArea.auto_scroll_threshold = 9999;
play_game()
TW | TL | TL | TW | |||||||||||
DL | DW | DW | DL | |||||||||||
DL | DL | DL | DL | |||||||||||
TW | TL | DW | TL | TW | ||||||||||
DL | DL | DL | DL | |||||||||||
DW | TL | TL | DW | |||||||||||
TL | DL | DL | TL | |||||||||||
DW | D2 | O1 | G2 | E1 | Y4 | |||||||||
TL | DL | DL | TL | |||||||||||
DW | TL | TL | DW | |||||||||||
DL | DL | DL | DL | |||||||||||
TW | TL |