c - Speck Cipher - C 中的解密问题

Question

我正在尝试通过添加解密模块来扩展 Speck 密码 ( https://github.com/inmcm/Simon_Speck_Ciphers ) 的以下实现。我按照 NSA 原始论文中关于 Speck 和 Simon (https://eprint.iacr.org/2013/404) 的说明实现了解密算法。

目前我有点卡住了，因为我无法正确破译从加密例程生成的密文。我已经看过在 SO 上发布的其他类似问题，但无济于事。

通过检查我的代码和标准输出，我注意到从第一次迭代开始，在解密阶段存储在 x_word 中的值是不正确的。因此，以下指令可能存在问题:

x_word = rotate_left( (sub_mod((x_word ^ *(round_key_ptr + 21 - i)), y_word, 65535)), 7);

我在这里发布我的代码的相关部分:

#define rotate_left(x,n) (x >> (word_size - n)) | (x << n)
#define rotate_right(x,n) (x << (word_size - n)) | (x >> n) 

void Speck_Encrypt_32(uint8_t *key_schedule, uint8_t *plaintext, uint8_t *ciphertext) {

    const uint8_t word_size = 16;
    uint16_t y_word = *(uint16_t *)plaintext;
    uint16_t x_word = *(((uint16_t *)plaintext) + 1);
    uint16_t *round_key_ptr = (uint16_t *)key_schedule;
    uint16_t * word_ptr = (uint16_t *)ciphertext;

    for(uint8_t i = 0; i < 22; i++) {  // Block size 32 has only one round number option        

        x_word = ((rotate_right(x_word, 7)) + y_word) ^ *(round_key_ptr + i);
        y_word = (rotate_left(y_word, 2)) ^ x_word;

        printf("y_word - (%d) - %u\n", i, (unsigned int)y_word);
        printf("x_word - (%d) - %u\n", i, (unsigned int)x_word);
    }
    // Assemble Ciphertext Output Array   
    *word_ptr = y_word;
    word_ptr += 1;
    *word_ptr = x_word;
    return;
}

static inline uint16_t sub_mod(uint16_t a, uint16_t b, uint16_t m)
{
  if ( a>=b )
    return a - b;
  else
    return m - b + a;
}


void Speck_Decrypt_32(uint8_t *key_schedule, uint8_t *plaintext, uint8_t *ciphertext) {

    const uint8_t word_size = 16;
    // Swapping cipher text words
    uint16_t y_word = *(uint16_t *)ciphertext;
    uint16_t x_word = *(((uint16_t *)ciphertext) + 1);
    uint16_t *round_key_ptr = (uint16_t *)key_schedule;
    uint16_t * word_ptr = (uint16_t *)plaintext;

    for(int i=0; i < 4; i++) {
        printf("Ciphertext Byte %02d: %02x \n",i,ciphertext[i]);
        printf("Plaintext Byte %02d: %02x \n",i,plaintext[i]);
    } 

    // Reading round keys in reverse order
    for(uint8_t i = 0; i < 22; i++) {  // Block size 32 has only one round number option

        //printf("y_word - (%d) - %u\n", i, (unsigned int)y_word);
        //printf("x_word - (%d) - %u\n", i, (unsigned int)x_word);
        // Inverting rotations and using custom modular subtraction
        y_word = rotate_right((x_word ^ y_word), 2);
        x_word = rotate_left( (sub_mod((x_word ^ *(round_key_ptr + 21 - i)), y_word, 65535)), 7);

    }
    // Assemble Plaintext - Swapping plaintext words  
    *word_ptr = y_word;
    word_ptr += 1;
    *word_ptr = x_word;
    return;
}

我有一段时间没有用 C 编写代码了，可能会有一些错误。

Answer 1

鉴于 repo 协议(protocol)的作者没有绕过编写 C 解密例程，我首先担心 C 设置或加密加密例程的有效性。

与其从那里开始，不如看看其中一个有效的 C 实现:

https://github.com/tomasaazevedo/Speck_8-bit_C

https://github.com/yuehann/Simon_Speck

https://github.com/mysummergit/myssl/blob/master/demos/mycipher/speck.c