用盐加扰和解扰javascript数组

Question

所以我用 md5 做了一个 javascript 字符串加扰器。

var MD5 = function(d){var r = M(V(Y(X(d),8*d.length)));return r.toLowerCase()};function M(d){for(var _,m="0123456789ABCDEF",f="",r=0;r<d.length;r++)_=d.charCodeAt(r),f+=m.charAt(_>>>4&15)+m.charAt(15&_);return f}function X(d){for(var _=Array(d.length>>2),m=0;m<_.length;m++)_[m]=0;for(m=0;m<8*d.length;m+=8)_[m>>5]|=(255&d.charCodeAt(m/8))<<m%32;return _}function V(d){for(var _="",m=0;m<32*d.length;m+=8)_+=String.fromCharCode(d[m>>5]>>>m%32&255);return _}function Y(d,_){d[_>>5]|=128<<_%32,d[14+(_+64>>>9<<4)]=_;for(var m=1732584193,f=-271733879,r=-1732584194,i=271733878,n=0;n<d.length;n+=16){var h=m,t=f,g=r,e=i;f=md5_ii(f=md5_ii(f=md5_ii(f=md5_ii(f=md5_hh(f=md5_hh(f=md5_hh(f=md5_hh(f=md5_gg(f=md5_gg(f=md5_gg(f=md5_gg(f=md5_ff(f=md5_ff(f=md5_ff(f=md5_ff(f,r=md5_ff(r,i=md5_ff(i,m=md5_ff(m,f,r,i,d[n+0],7,-680876936),f,r,d[n+1],12,-389564586),m,f,d[n+2],17,606105819),i,m,d[n+3],22,-1044525330),r=md5_ff(r,i=md5_ff(i,m=md5_ff(m,f,r,i,d[n+4],7,-176418897),f,r,d[n+5],12,1200080426),m,f,d[n+6],17,-1473231341),i,m,d[n+7],22,-45705983),r=md5_ff(r,i=md5_ff(i,m=md5_ff(m,f,r,i,d[n+8],7,1770035416),f,r,d[n+9],12,-1958414417),m,f,d[n+10],17,-42063),i,m,d[n+11],22,-1990404162),r=md5_ff(r,i=md5_ff(i,m=md5_ff(m,f,r,i,d[n+12],7,1804603682),f,r,d[n+13],12,-40341101),m,f,d[n+14],17,-1502002290),i,m,d[n+15],22,1236535329),r=md5_gg(r,i=md5_gg(i,m=md5_gg(m,f,r,i,d[n+1],5,-165796510),f,r,d[n+6],9,-1069501632),m,f,d[n+11],14,643717713),i,m,d[n+0],20,-373897302),r=md5_gg(r,i=md5_gg(i,m=md5_gg(m,f,r,i,d[n+5],5,-701558691),f,r,d[n+10],9,38016083),m,f,d[n+15],14,-660478335),i,m,d[n+4],20,-405537848),r=md5_gg(r,i=md5_gg(i,m=md5_gg(m,f,r,i,d[n+9],5,568446438),f,r,d[n+14],9,-1019803690),m,f,d[n+3],14,-187363961),i,m,d[n+8],20,1163531501),r=md5_gg(r,i=md5_gg(i,m=md5_gg(m,f,r,i,d[n+13],5,-1444681467),f,r,d[n+2],9,-51403784),m,f,d[n+7],14,1735328473),i,m,d[n+12],20,-1926607734),r=md5_hh(r,i=md5_hh(i,m=md5_hh(m,f,r,i,d[n+5],4,-378558),f,r,d[n+8],11,-2022574463),m,f,d[n+11],16,1839030562),i,m,d[n+14],23,-35309556),r=md5_hh(r,i=md5_hh(i,m=md5_hh(m,f,r,i,d[n+1],4,-1530992060),f,r,d[n+4],11,1272893353),m,f,d[n+7],16,-155497632),i,m,d[n+10],23,-1094730640),r=md5_hh(r,i=md5_hh(i,m=md5_hh(m,f,r,i,d[n+13],4,681279174),f,r,d[n+0],11,-358537222),m,f,d[n+3],16,-722521979),i,m,d[n+6],23,76029189),r=md5_hh(r,i=md5_hh(i,m=md5_hh(m,f,r,i,d[n+9],4,-640364487),f,r,d[n+12],11,-421815835),m,f,d[n+15],16,530742520),i,m,d[n+2],23,-995338651),r=md5_ii(r,i=md5_ii(i,m=md5_ii(m,f,r,i,d[n+0],6,-198630844),f,r,d[n+7],10,1126891415),m,f,d[n+14],15,-1416354905),i,m,d[n+5],21,-57434055),r=md5_ii(r,i=md5_ii(i,m=md5_ii(m,f,r,i,d[n+12],6,1700485571),f,r,d[n+3],10,-1894986606),m,f,d[n+10],15,-1051523),i,m,d[n+1],21,-2054922799),r=md5_ii(r,i=md5_ii(i,m=md5_ii(m,f,r,i,d[n+8],6,1873313359),f,r,d[n+15],10,-30611744),m,f,d[n+6],15,-1560198380),i,m,d[n+13],21,1309151649),r=md5_ii(r,i=md5_ii(i,m=md5_ii(m,f,r,i,d[n+4],6,-145523070),f,r,d[n+11],10,-1120210379),m,f,d[n+2],15,718787259),i,m,d[n+9],21,-343485551),m=safe_add(m,h),f=safe_add(f,t),r=safe_add(r,g),i=safe_add(i,e)}return Array(m,f,r,i)}function md5_cmn(d,_,m,f,r,i){return safe_add(bit_rol(safe_add(safe_add(_,d),safe_add(f,i)),r),m)}function md5_ff(d,_,m,f,r,i,n){return md5_cmn(_&m|~_&f,d,_,r,i,n)}function md5_gg(d,_,m,f,r,i,n){return md5_cmn(_&f|m&~f,d,_,r,i,n)}function md5_hh(d,_,m,f,r,i,n){return md5_cmn(_^m^f,d,_,r,i,n)}function md5_ii(d,_,m,f,r,i,n){return md5_cmn(m^(_|~f),d,_,r,i,n)}function safe_add(d,_){var m=(65535&d)+(65535&_);return(d>>16)+(_>>16)+(m>>16)<<16|65535&m}function bit_rol(d,_){return d<<_|d>>>32-_}

var hex = {
    a: 10,
    b: 11,
    c: 13,
    d: 14,
    e: 15,
    f: 16
}

function scrambleText(text, salt) {
    if(text.length > 32) throw "No more than 32 chars.";

    var strs = text.split("");
    var final = "";

    var hash = MD5(salt);
    var hashlength = hash.length, i = 0;

    while(i < hashlength) {
        if(!strs.length) break;
        var num = (hex[hash[i]] || parseInt(hash[i])) * 2 % strs.length;
        final += strs[num];
        strs.splice(num, 1);
        ++i;
    }
    return final
}

我可以打乱字符串，但是如果我知道盐，我该如何解乱打乱的文本？

我尝试过，但效果不佳

function unscrambleText(text, salt) {
    var strs = text.split("");

    var hash = MD5(salt);
    var hashlength = hash.length, i = strs.length - 1, c = 0;;

    var final = [];

    while(i >= 0) {
        var num = (hex[hash[i]] || parseInt(hash[i])) * 2 % (strs.length - i);
        final[num] = strs[c];

        final += strs[num];
        strs.splice(num, 1);
        --i;
        ++c;
    }
    return final
}

你们能帮我解决这个问题吗？我已经尝试了几个小时。

我取名为“salt”的字符串的 md5，然后将其用作对字符串进行打乱的参考。我只使用字符串来简化问题。在我的实际代码中，我有一组图像数据要加扰，然后能够对其进行解扰 example of image

Answer 1

这是一个用于加密/解密文本的愚蠢脚本。

哈希函数来自这里：Simple (non-secure) hash function for JavaScript? 显然，它是 Java 哈希函数的 JavaScript 版本。

脚本获取“salt”，对其进行散列，并使用散列中的每个字符作为偏移量，以在 ascii 表中向上或向下移动字母。所以它本质上是ROT n，其中 n 是散列盐的值之一。 “烂”可能是对这个打乱功能的一个很好的描述：D。

const hash = (str) =>
  str.split("").reduce((hash, char) => {
    hash = (hash << 5) - hash + char.charCodeAt(0);
    return hash & hash;
  }, 0);

const offset = (num, i) =>
  parseInt(num.toString().charAt(i % num.toString().length));

const scramble = (st, sa, d = 1) =>
  st
    .split("")
    .map((c, i) => String.fromCharCode(c.charCodeAt(0) + offset(sa, i) * d))
    .join("");

const unscramble = (str, salt) => scramble(str, salt, -1);

const salt = "This is my salt";
const str = "My text to 'scramble'.";

const scram = scramble(str, hash(salt));
const unscram = unscramble(scram, hash(salt));

console.log("scrambled:", scram);
console.log("unscrambled:", unscram);