在我的课程中,我的任务是创建一个Caesar Cipher解码器,它接受一串输入并使用字母频率找到最好的字符串。如果不确定有多大意义,请发布问题:
编写执行以下操作的程序。首先,它应该读取一行输入,即编码消息,并且将由大写字母和空格组成。您的程序必须尝试使用shift S的所有26个可能值来解码消息;在这26条可能的原始信息中,打印出具有最高优点的信息。 为方便起见,我们将为您预先定义变量letterGoodness,长度为26的列表,等于上面频率表中的值
到目前为止我有这个代码:
x = input()
NUM_LETTERS = 26 #Can't import modules I'm using a web based grader/compiler
def SpyCoder(S, N):
y = ""
for i in S:
x = ord(i)
x += N
if x > ord('Z'):
x -= NUM_LETTERS
elif x < ord('A'):
x += NUM_LETTERS
y += chr(x)
return y
def GoodnessFinder(S):
y = 0
for i in S:
if x != 32:
x = ord(i)
x -= ord('A')
y += letterGoodness[x]
return y
def GoodnessComparer(S):
goodnesstocompare = GoodnessFinder(S)
goodness = 0
v = ''
for i in range(0, 26):
v = SpyCoder(S, i)
goodness = GoodnessFinder(v)
if goodness > goodnesstocompare:
goodnesstocompare = goodness
return v
y = x.split()
z = ''
for i in range(0, len(y)):
if i == len(y) - 1:
z += GoodnessComparer(y[i])
print(z)
编辑:Cristian Ciupitu建议做出修改
请忽略缩进错误,当我复制代码时,它们可能会出现。
该程序的工作方式如下:
我不太确定问题出在哪里,第一次测试:LQKP OG CV GKIJV DA VJG BQQ
打印正确的信息:由ZOO加入我的
然而下一个测试:UIJT JT B TBNQMF MJOF PG UFYU GPS EFDSZQUJOH
给出一个垃圾串:SGHR HR Z RZLOKD KHMD NE SDWS ENQ CDBQXOSHMF
当它应该是:这是一个用于解密的文本样本行
我知道我必须:
尝试每个班次值
获得单词
的“善”
返回最高优点的字符串。
我希望我的解释有意义,因为我现在很困惑。
答案 0 :(得分:3)
这是我的实现工作正常。
您应该打印每条可能消息的优点,并查看程序输出的原因。
letterGoodness = dict(zip(string.ascii_uppercase,
[.0817,.0149,.0278,.0425,.1270,.0223,.0202,
.0609,.0697,.0015,.0077,.0402,.0241,.0675,
.0751,.0193,.0009,.0599,.0633,.0906,.0276,
.0098,.0236,.0015,.0197,.0007]))
trans_tables = [ str.maketrans(string.ascii_uppercase,
string.ascii_uppercase[i:]+string.ascii_uppercase[:i])
for i in range(26)]
def goodness(msg):
return sum(letterGoodness.get(char, 0) for char in msg)
def all_shifts(msg):
msg = msg.upper()
for trans_table in trans_tables:
txt = msg.translate(trans_table)
yield goodness(txt), txt
print(max(all_shifts(input())))
答案 1 :(得分:2)
我的最终解决方案有效,感谢精彩的Cristian Ciupitu。
x = input()
NUM_LETTERS = 26 #Can't import modules I'm using a web based grader/compiler
def SpyCoder(S, N):
y = ""
for i in S:
if(i.isupper()):
x = ord(i)
x += N
if x > ord('Z'):
x -= NUM_LETTERS
elif x < ord('A'):
x += NUM_LETTERS
y += chr(x)
else:
y += " "
return y
def GoodnessFinder(S):
y = 0
for i in S:
if i.isupper():
x = ord(i)
x -= ord('A')
y += letterGoodness[x]
else:
y += 1
return y
def GoodnessComparer(S):
goodnesstocompare = GoodnessFinder(S)
goodness = 0
v = ''
best_v = S
for i in range(0, 26):
v = SpyCoder(S, i)
goodness = GoodnessFinder(v)
if goodness > goodnesstocompare:
best_v = v
goodnesstocompare = goodness
return best_v
print(GoodnessComparer(x))
感谢您的帮助!
答案 2 :(得分:0)
我正在使用相同的教程并使用稍微不同的方法。这避免了创建和调用函数:
inp = input() #to hold code text
code = list(inp) #store code as a list
soln = [] #store the 'Goodness' for each of 26 possible answers
y=0 #variable to hold total goodness during calculations
clear = [] #will hold decoded text
pos=0 #position marker for a list
#for every possible value of shift
#note range as 0 to 25 are valid shifts and shift 26 = shift 0
for shift in range(0,26):
for i in code: #loop through each letter in code
if i == " ": #spaces have no score so omit them
continue
else: #if it's a letter
x = ord(i)-shift #apply the test shift
if x < 65: #prevent shifting outside A-Z range
x = x + 26
x = x - 64 #turn ord into character position in A-Z with A=1
x = letterGoodness[x-1] #turn this into the Goodness score
y = y + x #add this to a running total
soln.insert(shift-1,y) #AFTER decoding all letters in code, add total(y) to list of scores
y = 0 #reset y before next test value
bestSoln=max(soln) #find highest possible score
for i in range(0,26): #check the list of solutions for this score
if soln[i]==bestSoln: #the position in this list is the shift we need
bestShift = i+1 #+1 as the first solution is 0
for i in code: #now decode the original text using our best solution
if i == " ": #spaces are not encoded so just add these to the string
clear.insert(pos," ") #pos used to track next position for final string
pos = pos + 1
continue
else:
x = ord(i)-bestShift #same operation as before
if x < 65:
x = x + 26
z = chr(x)
clear.insert(pos,z) #add the decoded letter to the clear text
pos = pos + 1
print("".join(clear)) #join the list of clear text into one string and print it
请注意,此代码的许多部分都可以(并且应该)进行压缩,例如
x = x - 64
x = letterGoodness[x-1]
y = y + x
他们被扩展为“显示我的工作”以进行辅导练习。