基于超奇偶性的高速对称加密算法
郝伟 2020/05/27
对称加密指加密和解密所使用的密钥都是是一把解密，而非对称加密指加密和解密使用的不是同一种密钥，性能上要比对称加密弱很多。对称加密：AES(AdvancedEncryptionStandard)算法作为加密的标准算法[]因其可靠、高效、安全性能高得到了广泛应用。然而，大量用户容易形成对服务器突发大量的访问请求，而且传统的AES串行方式的加密效率低下，不能满足流式加密的速度和服务质量需求。本文研究是一种高效率的对称加密，具有运行速度级快，算法简单，安全性好等特点，并在数学上证明在没有密钥（包括间接）的情况下，实现数据的绝对安全。
本公开的实施例提供了一种对称加密方法，此加密方法包括获取待加密明文二进制数组input；使用密钥二进制数组key对所述待加密明文二进制数组input进行异或加密，生成密文二进制数组output；根据所述待加密明文二进制数组input得到随机数b；根据所述随机数b对所述密文二进制数组output进行奇偶性随机混合，得到奇偶性随机混合加密结果二进制数组output′。以此方式，利用了大容量密钥、异或数据加密和基于异或奇偶性的数据乱序算法，能够在保证数据安全性的前提下，提高加解密运算速度。
package test; import java.io.FileInputStream; import java.io.FileOutputStream; public class FileEncryption { static String input = "D:\\Research\\20200522_ImageEncrpytion\\WindowsFormsApp1\\bin\\Debug\\GameOfThroneS01E01.mkv"; // 可以是任何大文件 static String output = "D:\\Research\\20200522_ImageEncrpytion\\WindowsFormsApp1\\bin\\Debug\\GameOfThroneS01E01_out.mkv"; static String output1 = "D:\\Research\\20200522_ImageEncrpytion\\WindowsFormsApp1\\bin\\Debug\\GameOfThroneS01E01_out1.mkv"; static String key = "SunBeach.jpg"; public static void main(String[] args) throws Exception { // 使用密钥对指定的文件加密。 Encrypt(input, output, key, 0); Encrypt(output, output1, key, 1); } public static String Info = null; public static int maxsize = 1048576; static int[] ones = { 0, 1, 1, 2, 1, 2, 2, 3, 1, 2, 2, 3, 2, 3, 3, 4, 1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5, 1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5, 2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6, 1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5, 2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6, 2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6, 3, 4, 4, 5, 4, 5, 5, 6, 4, 5, 5, 6, 5, 6, 6, 7, 1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5, 2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6, 2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6, 3, 4, 4, 5, 4, 5, 5, 6, 4, 5, 5, 6, 5, 6, 6, 7, 2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6, 3, 4, 4, 5, 4, 5, 5, 6, 4, 5, 5, 6, 5, 6, 6, 7, 3, 4, 4, 5, 4, 5, 5, 6, 4, 5, 5, 6, 5, 6, 6, 7, 4, 5, 5, 6, 5, 6, 6, 7, 5, 6, 6, 7, 6, 7, 7, 8 }; public static byte[] readKey(String file, int length){ byte[] key; return key; } public static boolean Encrypt(String infile, String outfile, String keyfile, int type) throws Exception { byte[] key = readBytes(keyfile, maxsize); FileInputStream fin = new FileInputStream(infile); FileOutputStream fout = new FileOutputStream(outfile); long total = fin.available(); byte[] bts = new byte[maxsize]; double t1 = 0, t2 = 0, t3 = 0; int readCount = maxsize; byte bx = 0b1010110; while (readCount >= maxsize) { long dt1 = System.nanoTime(); readCount = fin.read(bts, 0, maxsize); long dt2 = System.nanoTime(); bx = getBX(bts, key, type); // System.out.println("Bx=" + (bx & 0xff)); for (int i = 0; i < readCount; i++) bts[i] ^= key[i] ^ bx; long dt3 = System.nanoTime(); fout.write(bts, 0, readCount); long dt4 = System.nanoTime(); t1 += dt2 - dt1; t2 += dt3 - dt2; t3 += dt4 - dt3; } fin.close(); fout.close(); t1 /= 1000000000.0; t2 /= 1000000000.0; t3 /= 1000000000.0; String speed_reading = String.format("%.6f", total / t1 / 1073741824); String speed_encryption = String.format("%.6f", total / t2 / 1073741824); String speed_writing = String.format("%.6f", total / t3 / 1073741824); Info = String.format("总用时：%.4f s, 总容量：%.4f GB.\n", t1 + t2 + t3, total / 1024.0 / 1024 / 1024); Info += String.format("读取用时 %4f s, 速度：%s GB/s\n", t1, speed_reading); Info += String.format("加密用时 %4f s, 速度：%s GB/s\n", t2, speed_encryption); Info += String.format("写入用时 %4f s, 速度：%s GB/s\n", t3, speed_writing); System.out.println(Info); return true; } private static byte getBX(byte[] bts, byte[] key, int etype) { return etype == 0 ? encrypt(bts, key) : decrypt(bts, key); } private static byte decrypt(byte[] bts, byte[] key) { if (bts.length <= 1024) return 0; int offset = 0, pos = 0; for (int i = 0, w = 1; i < 8; i++, w *= 2) offset += ((ones[bts[i] & 0xff] + 100 - ones[key[i] & 0xff]) % 2) * w; // System.out.println("offset2: " + (offset & 0xff)); for (int i = 0, w = 1; i < 8; i++, w *= 2) { int even = (ones[bts[(i + offset) % bts.length] & 0xff] + 100 - ones[key[i + offset] & 0xff]) % 2; pos += even == 0 ? 0 : w; } byte b = (byte) (pos & 0xff); while (ones[b & 0xff] % 2 > 0) b = (byte) (b << 1); return b; } private static byte encrypt(byte[] bts, byte[] key) { if (bts.length <= 1024) return (byte) 0xff; int offset = 0, pos = 0; for (int i = 0, w = 1; i < 8; i++, w *= 2) offset += ones[bts[i] & 0xff] % 2 == 0 ? 0 : w; for (int i = 0, w = 1; i < 8; i++, w *= 2) { int odd = (ones[bts[(i + offset) % bts.length] & 0xff]) % 2; pos += odd == 0 ? 0 : w; } byte b = (byte) (pos & 0xff); while (ones[b & 0xff] % 2 > 0) b = (byte) (b << 1); return b; } public static byte[] readBytes(String keyfile, int maxsize) throws Exception { byte[] bts = new byte[maxsize]; FileInputStream fs = new FileInputStream(keyfile); int left = maxsize, p = 0; while (left > 0) { p += fs.read(bts, maxsize - left, left); left -= p; } fs.close(); return bts; } }
总用时：2.0012 s, 总容量：0.7685 GB. 读取用时 1.228805 s, 速度：0.625438 GB/s 加密用时 0.126546 s, 速度：6.073212 GB/s 写入用时 0.645808 s, 速度：1.190045 GB/s 总用时：1.1109 s, 总容量：0.7685 GB. 读取用时 0.445182 s, 速度：1.726352 GB/s 加密用时 0.055458 s, 速度：13.858152 GB/s 写入用时 0.610306 s, 速度：1.259273 GB/s
【悬赏】
任务目标：在非破解程序的前提下，通过观察输入和输出分析出加密方法或密钥。
悬赏金额：200元
运行方式：在安装了JDK1.8及以上版本后，在程序文件目录下输入
for /L %i in (1,1,10) do java FastEncryption str
注意事项：此为非重要任务，不要影响正常工作。
推荐指令： for /L %i in (1,1,10) do java FastEncryption str%i