Question

class VigenèreCipher {
    constructor(key, abc) {
        this.encode = function (str) {
            //split the string into an array
            let arr = abc.split("");
            let keyValue = [];

            for (let i = 0; i < key.length; i++) {
                //finding the letter(key) value and push it to key value
                keyValue.push(arr.indexOf(key[i]));
            }
            let redacted = "";
            let pointer = 0;
            for (let i = 0; i < str.length; i++) {
                if (arr.includes(str[i])) {
                    let shift = arr.indexOf(str[i]) + keyValue[pointer];

                    if (shift >= 26) {
                        redacted += arr[shift - 26];
                    } else {
                        redacted += arr[shift];
                    }
                    //debugging code
                    console.log(
                        `string: ${str[i]}`,
                        `shift: ${shift - 26}`,
                        `shiftTo ${arr[shift]}`,
                        `pointer: ${pointer}`,
                        `KeyValue: ${keyValue[pointer]}`
                    );
                    pointer += 1;
                } else {
                    pointer = 0;
                    redacted += str[i];
                }

                if (pointer >= keyValue.length) {
                    pointer = 0;
                }
            }

            //console.log(keyValue);
            console.log(`redacted: ${redacted}`);
            return redacted
        };
        this.decode = function (str) {
            let arr = abc.split("");
            let keyValue = [];

            for (let i = 0; i < key.length; i++) {
                //finding the letter(key) value and push it to key value
                keyValue.push(arr.indexOf(key[i]));
            }
            let decoded = "";
            let pointer = 0;
            for (let i = 0; i < str.length; i++) {
                if (arr.includes(str[i])) {
                    let shift = arr.indexOf(str[i]) - keyValue[pointer];
                    if (shift < 0) {
                        decoded += arr[26 - Math.abs((arr.indexOf(str[i]) - keyValue[pointer]))];
                    } else {
                        decoded += arr[Math.abs((arr.indexOf(str[i]) - keyValue[pointer]))];
                    }

                    //debugging code
                    console.log(
                        `string: ${str[i]} `,
                        `shift: ${arr.indexOf(str[i]) - keyValue[pointer]} `,
                        `shiftTo ${arr[arr.indexOf(str[i]) - keyValue[pointer]]} `,
                        `pointer: ${pointer} `,
                        `KeyValue: ${keyValue[pointer]} `,
                        `arrindex: ${arr.indexOf(str[i])} `,

                    );
                    pointer += 1;
                } else {
                    pointer = 0;
                    decoded += str[i];
                }

                if (pointer >= keyValue.length) {
                    pointer = 0;
                }
            }

            //console.log(keyValue);
            console.log(`decode: ${decoded}`);
            return decoded
        };
    };
}

Vigenère密码是一种通过使用一系列不同的密码来加密字母文本的方法基于关键字字母的凯撒密码。它是多字母替换的一种简单形式。在凯撒密码中，字母表中的每个字母都沿一定数量的位置移动；例如，在移位3的凯撒密码中，A变成D，B变成E，Y变成B，依此类推。 Vigenère密码由多个具有不同移位值的凯撒密码组成。假设在每个字符的字符长度上重复键。请注意，某些实现仅在字符是字母的一部分时才在字符上重复键-此处不是这种情况。通过将凯撒（Caesar）移位应用于具有键在字母表中的相应索引的字符来得出移位。 / strong>

**ar abc, key; abc = "abcdefghijklmnopqrstuvwxyz"; key = "password" c = new VigenèreCipher(key, abc); Test.assertEquals(c.encode('codewars'), 'rovwsoiv'); Test.assertEquals(c.decode('rovwsoiv'), 'codewars'); Test.assertEquals(c.encode('waffles'), 'laxxhsj'); Test.assertEquals(c.decode('laxxhsj'), 'waffles'); Test.assertEquals(c.encode('CODEWARS'), 'CODEWARS'); Test.assertEquals(c.decode('CODEWARS'), 'CODEWARS');** **Expected: 'xt\'k o vwixl qzswej!', instead got: 'xt\'h p hhaxp rihzaf!' Expected: 'it\'s a shift cipher!', instead got: 'it\'v z gwqfp bzaeiv!'** **Expected: 'ドオカセガヨゴザキアニ', instead got: 'ドテタヒガォゴザキノイ' Expected: 'ドモアリガトゴザイマス', instead got: 'ドタシェガホゴザイィス'** **The code solve most problem but won't solve those any idea whats going on or how to fix it?**

Answer 1

当您检测到不在字母表中的字符时，会将else中的指针重置为零。通常，只有在加密时才增加密钥中的“指针”，否则您将其保留，就不会重置它。

Answer 2

帖子中给出的信息

... 请注意，某些实现仅在字符是字母的一部分时才在字符上重复键-此处不是这种情况。 ...

要实现此目的，您需要在处理完每个每个字符后递增指针和/或将其设置为零。您可以在循环结束前立即进行pointer的无条件更新，删除其他指针更新，例如

    ...
    pointer += 1;
    if( pointer >= keyValue.length) {
        pointer = 0
    }
} // end of for loop

或通过直接将其计算为

    pointer = (pointer + 1) % keyValue.length;

这将使代码按预期方式将"it's a shift cipher!"编码为"xt'k o vwixl qzswej!"，并将结果解码回原始数据而不会出错（在encode和{{1}中都修复了指针更新之后}。）

VigenèreCipher大部分测试都通过了编码，但是有些不起作用

2 个答案: