Java：解密位置未知的凯撒密码

Question

这里是一个加密的消息，首先使用3个位置的凯撒密码，然后5个位置。

WKH surjudp zdv krvwhg eb dfwru Slhufh Eurvqdq dqg kdg frpphqwdub iurp pdqb Kroobzrrg dfwruv dqg iloppdnhuv Prylh txrwdwlrqv wkdw ylhzhuv xvh lq wkhlu rzq olyhv dqg VLWXDWLRQV

YMJ uwtlwfr bfx mtxyji gd fhytw Unjwhj Gwtxsfs fsi mfi htrrjsyfwd kwtr rfsd Mtqqdbtti fhytwx fsi knqrrfpjwx Rtanj vztyfyntsx ymfy anjbjwx zxj ns ymjnw tbs qnajx fsi XNYZFYNTSX

解密时应该说:该节目由演员皮尔斯·布鲁斯南主持，有许多好莱坞演员和电影制作人的评论，观众在自己的生活和处境中使用的电影语录

好的，这是我写的类（您将需要所有的导入），我想感谢任何提前帮助的人：

public class CodeBreaker {
    public final int NUMBER_OF_LETTERS = 26;

    private double[] knownFrequencies = new double[NUMBER_OF_LETTERS];

    public double[] getKnownFrequencies() {
        return knownFrequencies;
    }

    public void setKnownFrequencies(double[] knownFrequencies) {

        this.knownFrequencies = knownFrequencies;
    }

    /**
     * Method reads in a file with a lot of text in it and
     * then use that to figure out the frequencies of each character
     *
     * @param trainingFileName
     */
    public void train(String trainingFileName) {
        try {
            Scanner fileIO = new Scanner(new File(trainingFileName));

            int total = 0;
            String temp = "";

            while (fileIO.hasNext()) {
                //reading into file and storing it into a string called temp
                temp += fileIO.next().toLowerCase().replaceAll("[ -,!?';:.]+", "");
                //converting temp string into a char array
            }

            char[] c = temp.toCharArray();
            total += c.length; // how many characters are in text
            int k = (int) 'a'; // int value of lowercase letter 'a'
            int[] counter = new int[NUMBER_OF_LETTERS];

            for (int j = 0; j < total; j++) {
                for (int i = k - k; i < knownFrequencies.length; i++) {

                    char[] d = new char[knownFrequencies.length];

                    d[i] = (char) (k + i);

                    if (c[j] == d[i]) {//checking to see if char in text equals char in d array

                        counter[i]++;

                        knownFrequencies[i] = (double) counter[i] / total;
                    }
                }
            }

            fileIO.close();
        } catch (FileNotFoundException e) {
            // TODO Auto-generated catch block
            System.err.println(e);
            System.exit(0);
        }
    }


    /**
     * Main decryption method used to take coded text from a file, figure out the positions in the CaesarCipher
     * and then decode it onto another file.
     *
     * @param cipherTextFileName
     * @param outputFileName
     * @return
     */
    public int decrypt(String cipherTextFileName, String outputFileName) {
        Scanner fileIO;
        int numberOfPositions = 0;
        double distance = 0.000000;

        try {
            fileIO = new Scanner(new File(cipherTextFileName));

            PrintWriter writer = new PrintWriter(new File(outputFileName));

            String temp = "";

            while (fileIO.hasNext()) {
                //reading into file and storing it into a string called temp
                temp += fileIO.next().toLowerCase().replaceAll(" ", "");
            }

            fileIO.close();

            do {
                distance = 0.0;
                int total = 0;
                double[] observedFreq = new double[NUMBER_OF_LETTERS];
                temp = decrypt(temp, numberOfPositions);
                char[] c = temp.toCharArray(); //store decrypted chars into an array
                total += c.length; // how many characters are in text

                int k = (int) 'a'; // int value of lowercase letter 'a'
                int[] counter = new int[NUMBER_OF_LETTERS]; //use to count the number of characters in text

                for (int j = 0; j < total; j++) {
                    for (int i = k - k; i < observedFreq.length; i++) {
                        char[] d = new char[observedFreq.length];
                        d[i] = (char) (k + i);
                        if (c[j] == d[i]) { //checking to see if char in text equals char in d array
                            counter[i]++;
                            observedFreq[i] = (double) counter[i] / total;
                        }
                    }
                }

                //Formula for finding distance that will determine the numberOfPositions in CaesarCipher
                for (int j = 0; j < knownFrequencies.length; j++) {
                    distance += Math.abs(knownFrequencies[j] - observedFreq[j]); //This is the part of the code I am having trouble with
                }

                numberOfPositions = numberOfPositions + 1;

            } while (distance > 0.6); //This is the part of the code I am having trouble with

            Scanner fileIO2 = new Scanner(new File(cipherTextFileName));

            while (fileIO2.hasNextLine()) {

                //reading into file and storing it into a string called temp
                temp = fileIO2.nextLine();

                writer.println(decrypt(temp, numberOfPositions));

            }
            writer.close();
            fileIO2.close();
            System.out.println(distance);

        } catch (FileNotFoundException e) {
            // TODO Auto-generated catch block
            System.err.println(e);
            System.exit(0);
        }
        return numberOfPositions;

    }

    /**
     * CaesarCipher decrypt and encrypt methods
     *
     * @param ciphertext
     * @param numberOfPositions
     * @return
     */
    public String decrypt(String ciphertext, int numberOfPositions) {
        // TODO Auto-generated method stub
        return encrypt(ciphertext, -numberOfPositions);
    }

    public String encrypt(String msg, int offset) {
        offset = offset % 26 + 26;
        StringBuilder encoded = new StringBuilder();
        for (char i : msg.toCharArray()) {
            if (Character.isLowerCase(i)) {
                int j = (i - 'a' + offset) % 26;
                encoded.append((char) (j + 'a'));
            } else if (Character.isUpperCase(i)) {
                int h = (i - 'A' + offset) % 26;
                encoded.append((char) (h + 'A'));
            } else {
                encoded.append(i);
            }
        }
        return encoded.toString();
    }

    // barebones main method to test your code
    public static void main(String[] args) {
        // args[0] contains the filename of the training file
        // args[1] contains the filename of the cipher text file
        // args[2] contains the filename of the output file
        CodeBreaker cb = new CodeBreaker();
        cb.train(args[0]);
        System.out.println(cb.decrypt(args[1], args[2]));

    }
}

Answer 1

解码凯撒密码的标准方法被称为“顺着字母表跑下来”。本质上是一种强力解决方案；你尝试了所有的可能性。因为只有26个可能的键，所以没有那么难。

举你的例子：

WKH SURJUDP ZDV KRVWHG ...

wkh surjudp zdv krvwhg ...
xli tvskveq aew lswxih ...
ymj uwtlwfr bfx mtxyji ...
znk vxumxgs cgy nuyzkj ...
aol wyvnyht dhz ovzalk ...
bpm xzwoziu eia pwabml ...
cqn yaxpajv fjb qxbcnm ...
dro zbyqbkw gkc rycdon ...
esp aczrclx hld szdepo ...
ftq bdasdmy ime taefqp ...
gur cebtenz jnf ubfgrq ...
hvs dfcufoa kog vcghsr ...
iwt egdvgpb lph wdhits ...
jxu fhewhqc mqi xeijut ...
kyv gifxird nrj yfjkvu ...
lzw hjgyjse osk zgklwv ...
max ikhzktf ptl ahlmxw ...
nby jlialug qum bimnyx ...
ocz kmjbmvh rvn cjnozy ...
pda lnkcnwi swo dkopaz ...
qeb moldoxj txp elpqba ...
rfc npmepyk uyq fmqrcb ...
sgd oqnfqzl vzr gnrsdc ...
the program was hosted ...
uif qsphsbn xbt iptufe ...
vjg rtqitco ycu jquvgf ...
wkh surjudp zdv krvwhg ...

对于一个人来说，只要26条线就能找到正确的线是很简单的。对于计算机来说，这就更难了。你数字母频率的主意很好。您还可以标记字母对，如“qx”，标记字母对，如“th”。计算所有26个可能结果的得分，并选择最高的得分结果。只要你调整好了你的得分方法，那么你就有很好的机会找到正确的解决方案。