AES算法

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求S Box

求s盒的过程是对于每一个有限域DF(2^8)中，256个数，每个数一个字节，对每个数求逆元，然后在用矩阵转置

算出来的s盒是固定的，如下

   | 0  1  2  3  4  5  6  7  8  9  a  b  c  d  e  f---|--|--|--|--|--|--|--|--|--|--|--|--|--|--|--|--|00 |63 7c 77 7b f2 6b 6f c5 30 01 67 2b fe d7 ab 76 10 |ca 82 c9 7d fa 59 47 f0 ad d4 a2 af 9c a4 72 c0 20 |b7 fd 93 26 36 3f f7 cc 34 a5 e5 f1 71 d8 31 15 30 |04 c7 23 c3 18 96 05 9a 07 12 80 e2 eb 27 b2 75 40 |09 83 2c 1a 1b 6e 5a a0 52 3b d6 b3 29 e3 2f 84 50 |53 d1 00 ed 20 fc b1 5b 6a cb be 39 4a 4c 58 cf 60 |d0 ef aa fb 43 4d 33 85 45 f9 02 7f 50 3c 9f a8 70 |51 a3 40 8f 92 9d 38 f5 bc b6 da 21 10 ff f3 d2 80 |cd 0c 13 ec 5f 97 44 17 c4 a7 7e 3d 64 5d 19 73 90 |60 81 4f dc 22 2a 90 88 46 ee b8 14 de 5e 0b db a0 |e0 32 3a 0a 49 06 24 5c c2 d3 ac 62 91 95 e4 79 b0 |e7 c8 37 6d 8d d5 4e a9 6c 56 f4 ea 65 7a ae 08 c0 |ba 78 25 2e 1c a6 b4 c6 e8 dd 74 1f 4b bd 8b 8a d0 |70 3e b5 66 48 03 f6 0e 61 35 57 b9 86 c1 1d 9e e0 |e1 f8 98 11 69 d9 8e 94 9b 1e 87 e9 ce 55 28 df f0 |8c a1 89 0d bf e6 42 68 41 99 2d 0f b0 54 bb 16

代码是参考维基百科，然后改成Java的，但是还是没搞明白是怎么求逆元的！！！！

private static void generateSbox() {    byte p = 1, q = 1;    do {        p = (byte)(p ^ (p << 1) ^ ((p & 0x80) > 0 ? 0x1B : 0));        System.out.println("p: " + p);        q ^= q << 1;        q ^= q << 2;        q ^= q << 4;        q ^= (q & 0x80) > 0 ? 0x09 : 0;        sbox[(int)p&0xFF] = (byte)(q ^ (q << 4 | q >> 4)^(q << 3 | q >> 5) ^ (q << 2 | q >> 6) ^ (q << 1 | q >> 7) ^ 0x63);    } while (p != 1);    sbox[0] = (byte)0x63;}

待写

https://github.com/dhuertas/AES/blob/master/aes.c

package com.zheyue.encrypt.coder;import org.apache.commons.codec.binary.Base64;import org.springframework.context.annotation.Configuration;import javax.crypto.Cipher;import javax.crypto.SecretKey;import javax.crypto.spec.SecretKeySpec;import java.security.Key;import java.util.Arrays;/** * @author FD * @date 2016/11/30 */@Configurationpublic class AESCoder {    private static byte[] sbox = new byte[256];    public static final String KET_ALGORITHM = "AES";    public static final String CIPHER_ALGORITHM = "AES/ECB/PKCS5Padding";    private static Key toKey(byte[] key) {        SecretKey secretKey = new SecretKeySpec(key, KET_ALGORITHM);        return secretKey;    }    public static byte[] decrypt(byte[] data, byte[] key) throws Exception{        Key k = toKey(key);//        Cipher.getInstance(CIPHER_ALGORITHM, "BC");        Cipher cipher = Cipher.getInstance(CIPHER_ALGORITHM);        cipher.init(Cipher.DECRYPT_MODE, k);        return cipher.doFinal(data);    }    public static byte[] encrypt(byte[] data, byte[] key) throws Exception{        Key k = toKey(key);//        Cipher.getInstance(CIPHER_ALGORITHM, "BC");        Cipher cipher = Cipher.getInstance(CIPHER_ALGORITHM);        cipher.init(Cipher.ENCRYPT_MODE, k);        return cipher.doFinal(data);    }//    public static void main(String[] args) throws Exception{//        String data = "你是谁";//        System.out.println(data.length());//        System.out.println(data.codePointCount(0, data.length()));//        int index = data.offsetByCodePoints(0, 1);//        System.out.println(data.codePointAt(index));//////        String key = "ffffffffdddddddd";////        System.out.println("密钥  "+key);////        System.out.println("原文  "+data);////        byte[] data4 = Arrays.copyOf(data.getBytes(), 16);////        System.out.println("加密前字节长度 "+ data.getBytes().length);////        byte[] data2 = encrypt(data.getBytes(), key.getBytes());////////        System.out.println("加密后字节长度 "+ data2.length);////        for (int i = 0; i < data2.length; i++) {////            System.out.print(data2[i]+" ");////        }////        System.out.println();////        System.out.println("base64密文  " + Base64.encodeBase64String(data2));////////        byte[] data3 = decrypt(data2, key.getBytes());////        System.out.println("原文  " + new String(data3));////        generateSbox();////        for (int i = 0; i < 256; i++) {////                System.out.format("%x", sbox[i]);////                System.out.println();////        }////    }    //乘以2    private byte xtime (byte x) {        int tmp = x << 1;        if((x&0x80) > 0) {            tmp ^= 0x1B;        }        return (byte)tmp;    }    //得到s盒 每一个字节计算逆元    //得到逆元计算yi=xi + x(i+4)mod8 + x(i+5)mod8 + x(i+6)mod8 + x(i+7)mod8 + ci    //假设输入字节的值为a=a7a6a5a4a3a2a1a0，则输出值为S[a7a6a5a4][a3a2a1a0]，S-1的变换也同理。    //这里mod不是整数 而是多项式x^8+x^4+x^3+x+1 -> 100011011 -> 283    private static void generateSbox() {        byte p = 1, q = 1;        do {            p = (byte)(p ^ (p << 1) ^ ((p & 0x80) > 0 ? 0x1B : 0));//            System.out.println("p: " + ((int)p&0xFF));            q ^= q << 1;            q ^= q << 2;            q ^= q << 4;            q ^= (q & 0x80) > 0 ? 0x09 : 0;//            System.out.println("q: " + (byte)(q ^ (q << 4 | q >> 4)^(q << 3 | q >> 5) ^ (q << 2 | q >> 6) ^ (q << 1 | q >> 7) ^ 0x63));            sbox[(int)p&0xFF] = (byte)(q ^ (q << 4 | q >> 4)^(q << 3 | q >> 5) ^ (q << 2 | q >> 6) ^ (q << 1 | q >> 7) ^ 0x63);        } while (p != 1);        sbox[0] = (byte)0x63;    }    //费马小定理求逆元 求 ab mod n = 1的逆元b//    private static int getInverse(int a, int n) {//        return pow_mod(a, n-2, n);//    }    //行移位变换    public void shiftRows(byte[][] data) {        byte tmp = data[1][0];        for (int i = 1; i < 4; i++) {            data[1][i-1] = data[1][i];        }        data[1][3] = tmp;        tmp = data[2][0];        data[2][0] = data[2][2];        data[2][2] = tmp;        tmp = data[2][1];        data[2][1] = data[2][3];        data[2][3] = tmp;        tmp = data[3][3];        for (int i = 1; i < 4; i++) {            data[3][i] = data[3][i-1];        }        data[3][0] = tmp;    }    public void mixColumns(byte[][] data) {        for (int i = 0; i < 4; i++) {            for (int j = 0; j < 4; j++) {            }        }    }    /**     * (a*a^-1 = 1) mod n -> (a*a^-1+ k*n) mod n = 1     * 扩展欧几里得求a的逆元 (a*x+by)mod n = 1 其中gcd(a, b) = 1 所以x就是a的逆元 b就是n y就是k(y这里求出来没啥用)     * @param a 求a的逆元 已知     * @param b 就是n 已知     */    private static void gcd(int a, int b, Int d, Int x, Int y) {        //当b==0 gcd(a, b) = a, 所以(a*1+0*y)mod n = a -> x=1 y=0 d=a 然后递归回去逆推回去        if(b == 0) {            d.setX(a);            x.setX(1);            y.setX(0);        }        else {            gcd(b, a % b, d, y, x);            y.setX(y.getX() - x.getX() * (a/b));        }    }    //计算模n下a的逆。如果不存在逆， 返回-1    private static int inv(int a, int n) {        Int d = new Int(0), x = new Int(0), y = new Int(0);        gcd(a, n, d, x, y);        return d.getX() == 1 ? (x.getX()+n)%n : -1;    }    //快速幂 求a^b mod n    private static int pow_mod(int a, int b, int n) {        int ans = 1;        while (b > 0) {            if ((b&1) == 1) {                ans = ans*a;                ans %= n;            }            a *= a;            a %= n;        }        return ans;    }    static class Int {        int x;        public Int(int x) {            this.x = x;        }        public int getX() {            return x;        }        public void setX(int x) {            this.x = x;        }    }}

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