这是模拟游戏Cows and Bulls的三位数字
我试图获得两个数字之间的牛和公牛数量。其中一个是由计算机生成的，另一个是由用户猜测的。我已经解析了我拥有的两个数字，现在我有两个列表，每个列表包含三个元素，每个元素都是数字中的一个数字。所以：
237将列出[2,3,7]列表。我确保保持相对指数。一般模式是：(hundreds, tens, units)。
这两个列表存储在两个列表中：machine和person。
所以，我编写了以下代码，最直观的算法：
在此循环开始之前， cows和bulls被初始化为0.
for x in person:
if x in machine:
if machine.index(x) == person.index(x):
bulls += 1
print x,' in correct place'
else:
print x,' in wrong place'
cows += 1
我开始使用计算机猜测的不同类型的数字来测试它。
很随机，我决定277.我猜到了447.这里，我得到了第一个线索，这个算法可能不起作用。我有1头牛和0头公牛。虽然我应该有1头牛和1头牛。
这是第一个算法的输出表：
Guess Output Expected Output
447 0 bull, 1 cow 1 bull, 0 cow
477 2 bulls, 0 cows 2 bulls, 0 cows
777 0 bulls, 3 cows 2 bulls, 0 cows
很明显，当计算机随机选择的数字中有重复数字时，此算法无效。
我试图理解为什么会发生这些错误，但我不能。我已经尝试了很多，但我在算法中看不到任何错误（可能是因为我写了它！）
在考虑这几天后，我尝试了这个：
在此循环开始之前， cows和bulls被初始化为0.
for x in range(3):
for y in range(3):
if x == y and machine[x] == person[y]:
bulls += 1
if not (x == y) and machine[x] == person[y]:
cows += 1
我对这个更有希望。但是当我测试这个时，这就是我得到的：
Guess Output Expected Output
447 1 bull, 1 cow 1 bull, 0 cow
477 2 bulls, 2 cows 2 bulls, 0 cows
777 2 bulls, 4 cows 2 bulls, 0 cows
我所犯的错误在这里很清楚，我明白这些数字一次又一次被计算在内。
即：277对477
当你计算公牛时，那么2只公牛就会出现，那就好了。但是当你算上奶牛时：
这里的匹配完全正确，因为我按照这个编写了代码。但这不是我想要的。而且我不知道在此之后要做什么。
我想强调的是，如果计算机选择的数字中没有重复数字，两种算法都能正常工作。
答案 0 :(得分：5)
def digits(number):
return [int(d) for d in str(number)]
def bulls_and_cows(guess, target):
guess, target = digits(guess), digits(target)
bulls = [d1 == d2 for d1, d2 in zip(guess, target)].count(True)
bovine = 0
for digit in set(guess):
bovine += min(guess.count(digit), target.count(digit))
return bulls, bovine - bulls
请注意bulls_and_cows(277, 447)将返回1头牛和0头奶牛。这就是我个人所期望的：为什么277中的前7个算作牛，因为已经有447的7牛？
答案 1 :(得分：1)
.index()返回给定输入的第一次出现的索引：
>>> [1, 5, 5].index(5)
1
您应该使用enumerate()来获取所有可能的索引：
>>> for i, j in enumerate([1, 5, 5]):
... if j == 5:
... print i
...
1
2
但是，似乎可以使用zip()来完成，除非我弄错了：
for x, y in enumerate(zip(player,computer)):
if y[0] in computer and not y[0] == y[1]:
print y[0], "is in the number"
cows += 1
elif y[0] == y[1]:
print y[0], "in the correct place"
bulls += 1
else:
print y[0], "is not in the number"
使用player = [4, 4, 7]：
4 is not in the number
4 is not in the number
7 in the correct place
>>> cows
0
>>> bulls
1
使用player = [4, 7, 7]：
4 is not in the number
7 in the correct place
7 in the correct place
>>> cows
0
>>> bulls
2
答案 2 :(得分：1)
Here is an algorithm which compares digits and positions.
I use it in a program for finding the target in a 4-digit
version of the game when the target can begin with "0".
It works with or without repeating digits equally well.
In the table below are shown all 9 results/values of variable
ED (Equal Digits) when comparing digits in each position of
the guess 447 with each digit of the target 447, for the
special case when the guess G$(3) equals the target T$(3)*.
For any other case the table is different and the values for
B and C will change accordingly.
It's a fact that for us, humans it's easier to count
Bulls and Cows without repeating digits.
With repeating digits I need to see the table below.
!!!Important!!!
Do not judge the values for B and C. Just use them.
4 4 7 3 Bulls 3 Cows
-------- (I=J) (I<>J)
4| 1 1 0 1 . . . 1 .
|
4| 1 1 1 . 1 . 1 . 1
|
7| 0 0 1 . . 1 . . .
Here is the algorithm in Liberty BASIC:
'-------------------------------------
[calculate_Bulls_and_Cows]
B = 0: C = 0
FOR I=1 TO 3
FOR J=1 TO 3
ED=(T$(I)=G$(J)) 'finding Equal Digits makes ED=1
B = B+ED*(I=J) 'Bulls increment when ED*(I=J)=1
C = C+ED*(I<>J) 'Cows increment when ED*(I<>J)=1
NEXT J
NEXT I
return
'------------------------------------
_________
*In this case I decided Digits to be considered
text variables because of some benefits when
it wasn't restricted the target to begin with "0".
答案 3 :(得分：0)
def game(a,b,c,d):
from random import randint
m=randint(0,9)
n=randint(0,9)
o=randint(0,9)
p=randint(0,9)
cow=bull=0
A={a:m,b:n,c:o,d:p}
k=A.keys()
v=A.values()
bull=cow=0
for i in k:
if i==A[i]:
bull+=1
elif i in v:
cow+=1
if (cow>0 or bull>0):
print(bull,"Bull and ",cow,"Cow")
print("Correct No. : ",m,n,o,p)
print("You've Entered : ",a,b,c,d)
答案 4 :(得分：-1)
嘿，制作牛牛计划的最佳方法是使用字符和字符串而不是数字。那么你就不必费心去打扰一百或十几个地方了。您可以在python：http://funpythonprojects.blogspot.in/2013/07/cow-bull-game-aka-guess-number-in-python.html
中按照我的4位牛牛游戏教程进行操作答案 5 :(得分：-1)
import random
def num_checker(guess_num,answer):
guess_num=list(str(guess_num))
ans=list(str(answer))
cow=0
bull=0
for a in range(0,4):
if guess_num[a]==ans[a]:
bull+=1
ans[a]=10#changed values to counter reccuring digits
guess_num[a]=20#changed values to counter reccuring digits
for a in range(0,4):
for b in range (0,4):
if guess_num[a]==ans[b]:
cow+=1
ans[b]=30#changed values to counter reccuring digits
break
final=[bull,cow]
return final
#..................................................
ans=random.randrange(1000,10000)
print("this is the program to gues a four digit number")
#......................................................#continue asking until you get it right
while True:
print("just for reference answer is:",ans)
num_typed=int(input("please guess a four digit the number?\n "))
reply=num_checker(num_typed,ans)
if reply==[4,0]:
print("correct")
print(reply[0],"bull",reply[1],"cows and the ans is",ans)
break
else:
print(reply[0],"bulls",reply[1],"cows")