用Java中的BigIntegers进行位移 [英] BitShifting with BigIntegers in Java

问题描述

我正在使用BigIntegers在Java中实现DES加密。

我通过执行BigInteger.leftShift（int n）方法将Java BigIntegers左移二进制键。 N（Kn）的关键取决于Kn-1移位的结果。我得到的问题是，我在每个密钥生成后打印出结果，并且转移不是预期的。关键在2 Cn和Dn（左和右）分开。

我具体试图这样做：
要做左移，移动每个位于左边的一个位置，除了第一个位，循环到块的末尾。

似乎根据O的结果取决于调动，转移。

结果：

c0：11110101010100110011000011110

d0：11110001111001100110101010100

c1：111101010101001100110000111100

d1：111100011110011001101010101000

c2：11110101010100110011000011110000

d2：11110001111001100110101010100000

c3 ：1111010101010011001100001111000000

d3：1111000111100110011010101010000000

c4：111101010101001100110000111100000000

d4：111100011110011001101010101000000000

c5：11110101010100110011000011110000000000

d5：11110001111001100110101010100000000000

c6：1111010101010011001100001111000000000000

d6：1111000111100110011010101010000000000000

c7：111101010101001100110000111100000000000000

d7：111100011110011001101010101000000000000000

c8：1111010101010011001100001111000000000000000

d8：1111000111100110011010101010000000000000000

c9 ：111101010101001100110000111100000000000000000

d9：111100011110011001101010101000000000000000000

c10：11110101010100110011000011110000000000000000000

d10：11110001111001100110101010100000000000000000000

c11：1111010101010011001100001111000000000000000000000

d11：1111000111100110011010101010000000000000000000000

c12：111101010101001100110000111100000000000000000000000

d12：111100011110011001101010101000000000000000000000000

c13：11110101010100110011000011110000000000000000000000000

d13：11110001111001100110101010100000000000000000000000000

c14：1111010101010011001100001111000000000000000000000000000000

d14：111100011 11001100110101010100000000000000000000000000000000

c15：11110101010100110011000011110000000000000000000000000000

d15：11110001111001100110101010100000000000000000000000000000

解决方案

BigInteger实现无限精度整数，所以向左移动将不会向左添加零。您需要旋转：

  private static BigInteger rotateLeft（BigInteger bi）{
 BigInteger ret = bi.shiftLeft（1 ）; 
 if（ret.testBit（32））{
 ret = ret.clearBit（32）.setBit（0）; 
} 
 return ret; 
} 
  

对于32位数字来说这将是非常低效的，所以你可能只需使用原语来旋转DES的28位半部分。我不熟悉DES算法，所以我假设你需要BigInteger的其他东西。

  private static BigInteger rotateLeftPrimitive （BigInteger bi）{
 int value = bi.intValue（）; 
返回BigInteger.valueOf（（（value< 1）& 0xffffffe）|（（value>>> 27）& 1））; 
} 
  

I am implementing DES Encryption in Java with use of BigIntegers.

I am left shifting binary keys with Java BigIntegers by doing the BigInteger.leftShift(int n) method. Key of N (Kn) is dependent on the result of the shift of Kn-1. The problem I am getting is that I am printing out the results after each key is generated and the shifting is not the expected out put. The key is split in 2 Cn and Dn (left and right respectively).

I am specifically attempting this: "To do a left shift, move each bit one place to the left, except for the first bit, which is cycled to the end of the block. "

It seems to tack on O's on the end depending on the shift. Not sure how to go about correcting this.

Results:

c0: 11110101010100110011000011110

d0: 11110001111001100110101010100

c1: 111101010101001100110000111100

d1: 111100011110011001101010101000

c2: 11110101010100110011000011110000

d2: 11110001111001100110101010100000

c3: 1111010101010011001100001111000000

d3: 1111000111100110011010101010000000

c4: 111101010101001100110000111100000000

d4: 111100011110011001101010101000000000

c5: 11110101010100110011000011110000000000

d5: 11110001111001100110101010100000000000

c6: 1111010101010011001100001111000000000000

d6: 1111000111100110011010101010000000000000

c7: 111101010101001100110000111100000000000000

d7: 111100011110011001101010101000000000000000

c8: 1111010101010011001100001111000000000000000

d8: 1111000111100110011010101010000000000000000

c9: 111101010101001100110000111100000000000000000

d9: 111100011110011001101010101000000000000000000

c10: 11110101010100110011000011110000000000000000000

d10: 11110001111001100110101010100000000000000000000

c11: 1111010101010011001100001111000000000000000000000

d11: 1111000111100110011010101010000000000000000000000

c12: 111101010101001100110000111100000000000000000000000

d12: 111100011110011001101010101000000000000000000000000

c13: 11110101010100110011000011110000000000000000000000000

d13: 11110001111001100110101010100000000000000000000000000

c14: 1111010101010011001100001111000000000000000000000000000

d14: 1111000111100110011010101010000000000000000000000000000

c15: 11110101010100110011000011110000000000000000000000000000

d15: 11110001111001100110101010100000000000000000000000000000

解决方案

BigInteger implements infinite-precision integers, so shifting to the left will keep adding zeros to the left. You need a rotate:

private static BigInteger rotateLeft(BigInteger bi) {
    BigInteger ret = bi.shiftLeft(1);
    if (ret.testBit(32)) {
        ret = ret.clearBit(32).setBit(0);
    }
    return ret;
}

That's going to be rather inefficient for 32-bit numbers, so you might as well just use primitives to rotate DES' 28-bit halves. I'm not familiar with the DES algorithm, so I'll assume you need BigInteger for something else.

private static BigInteger rotateLeftPrimitive(BigInteger bi) {
    int value = bi.intValue();
    return BigInteger.valueOf(((value << 1) & 0xffffffe) | ((value >>> 27) & 1));
}

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