Boggle


random trip report
Want to know all the words in a Boggle board? Enter the board in the field below, one row per line. It can be any square size (4x4, 5x5, etc.).

Standard dictionary Jumbo dictionary

Min word length

Python source:

import math, sys
from dict import *

# return True if the num1 and num2 positions in the grid are adjacent
#
def adjacent(num1, num2, len):
    num1_row = num1//len
    num1_col = num1 % len
    num2_row = num2//len
    num2_col = num2 % len
    if abs(num1_row - num2_row) >= 2:
        return False
    if abs(num1_col - num2_col) >= 2:
        return False
    return True

# return True is word is in board
# "used" says which letters have been used so far
#
def find_word(board, word, used, n, edge):
    nused = len(used)
    if nused:
        last = used[nused-1]
    for i in range(n):
        if nused and i == last:
            continue
        if nused and not adjacent(i, last, edge):
            continue
        if board[i] != word[0]:
            continue
        if nused and i in used:
            continue
        used.append(i)
        if len(word) == 1:
            return True
        return find_word(board, word[1:], used, n, edge)
    return False
 
min_length = 4

# print list of all words in board
#
def solve(board, jumbo):
    n = len(board)
    edge = int(math.sqrt(n))
    count = 0
    for w in get_dictionary(jumbo, True):
        if len(w) < min_length:
            continue
        if find_word(board, w, [], n, edge):
            print(w)
            count += 1
    print 'count: ', count
    
b = ['a', 'n', 'e', 'r',
     's', 'e', 'h', 't',
     'm', 'o', 'l', 'i',
     'w', 'b', 'c', 'v']

x = list(sys.argv[1])
min_length = int(sys.argv[2])
solve(x, len(sys.argv)>3)

Copyright 2017 © David P. Anderson