Which printable ASCII characters will usually appear in an english text?_问答_开发者

I have been trying to solve Project Euler's problem #59 for a while, and I am having trouble because some of it seems somewhat more ambiguous than previous problems.

As background, the problem says that the given t开发者_Python百科ext file is encrypted text with the ASCII codes saved as numbers. The encryption method is to XOR 3 lowercase letters cyclically with the plaintext (so it is reversible). The problem asks for the key that decrypts the file to English text. How should I restrict the character set of my output to get the answer, without trying to sift through all possible plaintexts (26^3)?

I have tried restricting to letters, spaces, and punctuation, and that did not work.

To clarify: I want to determine, out of all printable ASCII characters, which ones I can probably discard and which ones I can expect to be in the plaintext string.

Have you tried two of the most "basic" and common tools in analyzing the algorithm used?

Analyze the frequency of the characters and try to match it against English letter frequency
Bruteforce using keys from a wordlist, most often common words are used as keys by "dumb" users

To analyze the frequency for this particular problem you would have to split the string every third element since the key is of length 3, you should now be able to produce three columns:

you have to analyse the frequency for each column, since now they are independent of the key.

Ok, actually took my time and constructed the 3 complete columns and counted the frequency for each of the columns and got the two most frequent item or each column:

Col1  Col2  Col3
71    79    68
2     1     1

Now if you check for instance: http://en.wikipedia.org/wiki/Letter_frequency You have the most frequent letters, and don't forget you have spaces and other characters which is not present on that page, but I think you can assume that space is the most frequent character.

So now it is just a matter of xor:ing the most frequent characters in the table I provided with the most frequent characters in English language, and see if you get any lowercase characters, I found a three letter word which I think is the answer with only this data.

Good luck and by the way, it was a nice problem!

A possible solution is to simply assume the presence of a given three-character sequence in the encrypted text. You can use a three-letter word, or a three letter sequence which is likely to appear in English text (e.g. " a ": the letter 'a' enclosed between two spaces). Then simply try all possible positions of that sequence in the encrypted text. Each position allows you to simply recompute the key, then decrypt the whole text into a file.

Since the original text has length 1201, you get 1199 files to skim through. At that point it is only a matter of patience, but you can make it much faster by using a simple text search utility on another frequent sequence in English (e.g. "are"), for instance with the Unix tool grep.

I did just that, and got the decrypted text in less than five minutes.

I'll admit upfront I'm not familiar with an XOR cipher.

However, it seems very similar to the concept of the vigenere cipher. Escpecially in the line where they mention for unbreakable encryption the keylength equals the message length. That screams Vernam Cipher.

As mentioned in the other answer, the strategical approach to breaking a vigenere cipher involves a probabilistic approach. I will not go into detail because most of the theory I learned was relatively complicated, but it can be found here keeping in mind that vignere is a series of caesar ciphers.

The problem makes it easy for you though because you already know the keylength. Because of that, as you mentioned, you can simply bruteforce by trying every single 3 letter combination.

Here's what I would do: take a reasonably sized chunk of the ciphertext, say maybe 10-20 characters, and try the brute force approach on that. Keep track of all the keys that seem to create understandable sequences of letters and then use those on the whole ciphertext. That way we can employ the obvious brute forcing method, but without bruteforcing the entire problem, so I don't think you'll have to worry about limiting your output.

That said, I agree that as you're creating the output, if you ever get a non printable character, you could probably break your loop and move on to the next key. I wouldn't try anything more specific than that because who knows what the original message could have, never make assumptions about the data you're dealing with. Short circuiting logic like that is always a good idea, especially when implementing a brute force solution.

Split the ciphertext into 3.

Ciphertext1 comprises the 1st, 4th, 7th, 10th...numbers Ciphertext2 comprises the 2nd, 5th, 8th, 11th...numbers Ciphertext3 comprises the 3rd, 6th, 9th, 12th...numbers

Now you know that each cyphertext is encrypted with the same key letter. Now do a standard frequency analysis on it. That should give you enough clues as to what the letter is.

I just solved this problem a few days ago. Without spoiling it for you, I want to describe my approach to this problem. Some of what I say may be redundant to what you already knew, but was part of my approach.

First I assumed that the key is exactly as described, three lowercase ASCII letters. So I began brute forcing at 'aaa' and went to 'zzz'. While decrypting, if any resulting byte was a value lower than 32 (the ASCII value of space, the lowest "printable" ASCII value) or higher than 126 (the ASCII value of the tilde '~' which is the highest printable character in ASCII) than I assumed the key was invalid because any value outside 32 and 126 would be an invalid character for a plain text stretch of English. As soon as a single byte is outside of this range, I stopped decrypting and went to the next possible key.

Once I decrypted the entire message using a particular key (after passing the first test of all bytes being printable characters), I needed a way to verify it as a valid decryption. I was expecting the result to be a simple list of words with no particular order or meaning. Through other cryptography experience, I thought back to letter frequency, and most simply that your average English word in text is 5 characters long. The file contains 1201 input bytes. So that would mean that there would be (on average) 240 words. After decrypting, I counted how many spaces were in the resulting output string. Since Project Euler is anything but average, I compared the number of spaces to 200 accounting for longer, more obscure words. When an output had more than 200 spaces in it, I printed out the key it was decrypted with and the output text. The one and only output that has more than 200 spaces is the answer. Let me tell you that it's more than obvious that you have the answer when you see it.

Something to point out is that the answer to the question is NOT the key. It is the sum of all the ASCII values of the output string. This approach will also solve the equation under the one minute mark, in fact, it times in around 3 or 4 seconds.