Question

我是一名初学程序员，并认为创建一个程序来循环使用＃34; all＆＃34;可能的密码。在1-10的范围内，您对以下＆＃34;黑客程序的评价效率如何？＆＃34;有没有办法让以下代码更有效？您可能知道的任何算法或技巧？

我考虑过移动＆＃39; e＆＃39;和更常见的字母到我的字符数组的开头，因为这些字母更常用。

import sys

password = raw_input("Enter a password: ")

characters = []
lower = ["a","b","c","d","e","f","g","h","i","j","k","l","m","n","o","p","q","r","s","t","u","v","w","x","y","z"]
upper = ["A","B","C","D","E","F","G","H","I","J","K","L","M","N","O","P","Q","R","S","T","U","V","W","X","Y","Z"]
number = ["1","2","3","4","5","6","7","8","9","0"]
symbol = ["!","@","#","$","%","^","&","*","(",")","_","+","-","="]

#Creates array with most keyboard characters
for i in range(len(lower)):
    characters.append(lower[i])
for i in range(len(upper)):
    characters.append(upper[i])
for i in range(len(number)):
    characters.append(number[i])
for i in range(len(symbol)):
    characters.append(symbol[i])

found = False

#For 1 character passwords
for a in range(len(characters)):
    if (characters[a] == password):
        print (characters[a])
        found = True
if (found):
    sys.exit()

#For 2 character passwords
for a in range(len(characters)):
    for b in range(len(characters)):
        if (characters[a] + characters[b] == password):
            print (characters[a] + characters[b])
            found = True
if (found):
    sys.exit()

for a in range(len(characters)):
    for b in range(len(characters)):
        for c in range(len(characters)):
            if (characters[a] + characters[b] + characters[c] == password):
                print (characters[a] + characters[b] + characters[c])
                found = True
if (found):
    sys.exit()

for a in range(len(characters)):
    for b in range(len(characters)):
        for c in range(len(characters)):
            for d in range(len(characters)):
                if (characters[a] + characters[b] + characters[c] + characters[d] == password):
                    print (characters[a] + characters[b] + characters[c] + characters[d])
                    found = True
if (found):
    sys.exit()

for a in range(len(characters)):
    for b in range(len(characters)):
        for c in range(len(characters)):
            for d in range(len(characters)):
                for e in range(len(characters)):
                    if (characters[a] + characters[b] + characters[c] + characters[d] + characters[e] == password):
                        print (characters[a] + characters[b] + characters[c] + characters[d] + characters[e])
                        found = True
if (found):
    sys.exit()

Answer 1

首先，您不必输入所有这些符号

>>> import string
>>> string.uppercase
'ABCDEFGHIJKLMNOPQRSTUVWXYZ'
>>> string.lowercase
'abcdefghijklmnopqrstuvwxyz'
>>> string.digits
'0123456789'
>>>

所以你可以这样做：

lower = string.lowercase

等等

这可以消失：

#Creates array with most keyboard characters
for i in range(len(lower)):
    characters.append(lower[i])
for i in range(len(upper)):
    characters.append(upper[i])
for i in range(len(number)):
    characters.append(number[i])
for i in range(len(symbol)):
    characters.append(symbol[i])

并替换为：

characters = "".join(upper, lower, number, symbol)

这可以简化：

#For 1 character passwords
for a in range(len(characters)):
    if (characters[a] == password):
        print (characters[a])
        found = True
if (found):
    sys.exit()

为：

for a in characters:
    print a
    sys.exit()

多字符检查需要循环，但尝试编写循环，就像我在上面的单个字符代码中所显示的那样

Answer 2

在这方面，必须提及itertools。这个答案并没有提供答案，这就是为什么你的编程效率高或不高的原因＆＃34;但它提醒你如果你很快就想要一个设置/脚本，你可以通过它来强制某些东西，那里有一个很好的lib。

import itertools
import string

for guess in itertools.product(string.lowercase, repeat=10): 
     password = ''.join(guess) #put the array elements together
     print password

itertools.product为您提供给定数组元素的所有可能组合（笛卡尔交叉积），在本例中为小写alpahabet，您可以使用生成所有3个行的任意长度密码的代码代码。

现在将repeat=10重写为变量表达式，并从for执行1循环到您喜欢的任何密码长度。

Answer 3

最有效的代码是

password = raw_input("Enter a password: ")
guess = password
print guess

：）

好吧开玩笑说，你的解决方案是一个简单的暴力攻击来找到密码，如果我们没有关于密码的额外信息（除了法律字母表是什么）那么暴力攻击是最好的攻击（就其而言）算法复杂度）。

但是你的代码有改进实现的余地。

而不是

for i in range(len(lower)):
  characters.append(lower[i])
for i in range(len(upper)):
  characters.append(upper[i])
for i in range(len(number)):
  characters.append(number[i])
for i in range(len(symbol)):
  characters.append(symbol[i])

这样做

characters = lower + upper + number + symbol

而不是有太多的案例来处理密码可能是什么，让我们使用一些数学来简化实现：）

请注意，如果该字符列表的大小为n，并且密码的长度为l，那么使用一些基本的Discrete Math / Combinatorics，我们可以使用给定的字母表集合进行n ^ l种可能的组合。我们可以简单地将每个字符组合映射到以base n表示的单个数字。例如，如果n = 2并且字母表仅包括＆＃39; a＆＃39;并且＆＃39; b＆＃39;然后基数2中的数字串10将对应于字符串＆＃34; ba＆＃34;。 Similalry字符串00100将压缩到＆＃34; aabaa＆＃34;

因此，我们的任务减少为生成所有可能的字符串。

这可以通过递归来完成（它有一点开销，但它减少了实现的负担）

def recurse(s, l, n):
    if n == l:
        if s == password: return True
        return False:
    for c in characters:
        if recurse(s + c, l, n+1) == True: return True:
    return False



 recurse("", len(password), 0)

Python密码程序

3 个答案: