Python柏忌游戏

Question

我正在努力完成这项针对学校的Python挑战游戏，我创建了一个绘图板函数，可以创建一个包含16个随机字母的绘图板。如何使用def函数要求用户输入一个单词，然后根据单词的长度对其进行评分？一旦完成，游戏应该可以正常运行。任何帮助都将不胜感激：)

import random
def loadwords ():
print "Loading word list from file.. "
wordList = []
inFile = open ('words.txt','r')
for line in inFile:
wordList.append(line.strip().lower())
#inFile : locates file in the folder, opens it
#wordlist: list of the words (strings)
print " " , len(wordList), "words loaded"
inFile.close()
return wordList
def spellCheck (word, wordList):
if (word in wordList) == True:
return True
else:
return False
def drawBoard (randomLetters):
'''Takes a randomList of 16 characters
Prints out a 4 x 4 grid'''
print " %s %s %s %s " %(randomLetters [0], randomLetters [1], randomLetters [2], randomLetters [3])
print " %s %s %s %s " %(randomLetters [4], randomLetters [5], randomLetters [6], randomLetters [7])
print " %s %s %s %s " %(randomLetters [8], randomLetters [9], randomLetters [10], randomLetters [11])
print " %s %s %s %s " %(randomLetters [12], randomLetters [13], randomLetters [14], randomLetters [15])
def wordinput ():
#asks user to input the longest word they can from grid
wordinput = raw_input ("Enter a word made up of the letters in the 4x4 table")
for letters in wordinput:
letters == randomLetters
def randomLetters ():
letters = []
alphabet = ['a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','v','w','x','y','z']
for i in range (0,16,1):
letters.append(random.choice(alphabet))
return letters
dictionary = loadwords()
letterList = randomLetters()
drawBoard(letterList)
wordinput(randomLetters)

Answer 1

raw_input(在Python 2中)或input(在Python 3中)允许读取字符串。

然后你需要检查字母表中是否包含所有字母。。。

def wordInLetter(word, letters):
for i in range(word):
if word[i] not in letters:
return False
return True

或更短：

def wordInLetter(word, letters):
return all(letter in letters for letter in word)

但是等一下，你不想让一个字母用几次！

让我们使用Counter来跟踪每个字母在信袋中的次数，并从中计算：

from collections import Counter
def wordInLetter(word, letters):
available = Counter(letters)
for letter in word:
if available[letter] == 0:
return False
available[letter] -= 1
return True

这似乎很有效。我希望这正是你所需要的。

In [3]: wordInLetter('kos','akos')
Out[3]: True
In [4]: wordInLetter('kos','ako')
Out[4]: False
In [5]: wordInLetter('koss','akos')
Out[5]: False

---

编辑

因此，我们不仅对word是否可以在letters中组合感兴趣，而且还想知道是否可以通过匹配相邻的字母来实现。所以另一个尝试：

import math
def wordInLetterSearch(word, letters, startX, startY):
# Assume: letters is a list of X letters, where X is a k*k square
k = int(len(letters) ** 0.5)
if len(word) == 0:
return True
def letter(x, y):
return letters[x + y*k]
def adjacent(x, y):
if x > 0:
yield x-1, y
if x < k-1:
yield x+1, y
if y > 0:
yield x, y-1
if y < k-1:
yield x, y+1
# try to move in all 4 directions
return any(letter(x2, y2) == word[0] and wordInLetterSearch(word[1:], letters, x2, y2)
for x2, y2 in adjacent(startX, startY))
def wordInLetter(word, letters):
k = int(len(letters) ** 0.5)
def coords(i):
return i%k, i/k
# look for a starting point
return any(letter == word[0]
and wordInLetterSearch(word[1:], letters,
coords(i)[0], coords(i)[1])
for i, letter in enumerate(letters)) coords(i)[1])
for i, letter in enumerate(letters))

这有点复杂。基本上，这是一个两步搜索：首先我们寻找一个起点(letters中字母与单词第一个字符匹配的位置)，然后我们尽可能递归地将蛇形移动到相邻的字段。

只需少量修改即可完全匹配Boggle规则：

• 修改CCD_ 7以允许对角线太(容易)
• 防止两次访问同一地点(中等；提示：向wordInLetterSearch添加参数，使用set)

我把这些留作练习。

Answer 2

这里是另一个变体。它应该更容易阅读，但想法是一样的：简单的回溯。有时，在调用方中止迭代更容易(不，不执行下一步)，有时在被调用方中止(在每次迭代中，检查前置条件，如果无效则中止)。

def wordInBoardIter(letters, word, x, y):
n = int(len(letters)**0.5)
if word == "": return True # empty - all letters are found
if x<0 or y<0 or x>=n or y>= n: return False #outside of board
if letters[x+y*n] != word[0]: return False # we are looking at the wrong position
# one step further:
return any(wordInBoardIter(letters, word[1:], x+dx,y+dy) for dx,dy in [(1,0), (0,1), (-1,0), (0,-1)])

def wordInBoard(letters, word):
n = int(len(letters)**0.5)
return any(any(wordInBoardIter(letters, word, x,y) for x in range(n)) for y in range(n))
if __name__ == '__main__':
letters = ['a', 'h', 'e', 'l',
'x', 'd', 'l', 'l',
'y', 'v', 'r', 'o',
'z', 'w', 'o', 'w']
print "hello     : %s" % ("found" if wordInBoard(letters, "hello") else "not found")
print "helloworld: %s" % ("found" if wordInBoard(letters, "helloworld") else "not found")
print "foobar    : %s" % ("found" if wordInBoard(letters, "foobar") else "not found")

这个版本中有相同的练习(对角平铺，不允许重复使用相同的字母两次)。

Answer 3

我还没有玩过太多的boggle myslef，但可以使用的解决方案是获取用户输入的单词，并使用len()命令返回单词的长度。然后取这个长度并打分。下面是一个基本示例(根据游戏规则进行修改)：

def wordinput ():
#asks user to input the longest word they can from grid
wordinput = raw_input ("Enter a word made up of the letters in the 4x4 table")
for letters in wordinput:
letters == randomLetters
scoreWord(wordInput)
def scoreWord(word)
#finds the amount of characters in the word
wordLength = len(word)
#multiplies the maunt of letters in the word by two to get the score
#(not sure how boggle scoring works)
score = wordLength * 2