鉴于异或放大器;两个数字的总和。如何找到号码？ [英] Given XOR & SUM of two numbers. How to find the numbers?

问题描述

由于异或放大器;两个数字的总和。如何找到号码？
例如x = A + B，Y = A ^ B;如果X，Y给出，如何让A，B？
如果不能，放弃的原因。

Given XOR & SUM of two numbers. How to find the numbers? For example, x = a+b, y = a^b; if x,y are given, how to get a, b? And if can't, give the reason.

推荐答案

无法完成。一个反例是0/100和4/96。这两项之和为100，异或为100。

Can't be done. One counter example is 0/100 and 4/96. Both these sum to 100 and xor to 100.

因此​​给予100之和的100的XOR结果，你可以不知道的其中的可能性的产生这两个数字。

Hence given a sum of 100 and an xor result of 100, you cannot know which of the possibilities generated those two numbers.

有关它的价值，这个程序会检查只数 0..255 的可能性：

For what it's worth, this program checks the possibilities with just the number 0..255:

#include <stdio.h>
static void output (unsigned int a, unsigned int b) {
    printf ("%u:%u=%u %u\n", a+b, a^b, a, b);
}
int main (void) {
    unsigned int limit = 256;
    unsigned int a, b;
    output (0, 0);
    for (b = 1; b != limit; b++)
        output (0, b);
    for (a = 1; a != limit; a++)
        for (b = 1; b != limit; b++)
            output (a, b);
    return 0;
}

您可以然后采取输出和按摩它来给你所有的重复可能性：

You can then take that output and massage it to give you all the repeated possibilities:

testprog | sed 's/=.*$//' | sort | uniq -c | grep -v ' 1 '

这给：

  7 100:100
  2 100:16
  4 100:18
  2 100:2
  4 100:20
  4 100:24
  8 100:26
  2 100:4
  2 100:64
  4 100:66
  4 100:68
  4 100:80
  8 100:82
  8 100:84
  8 100:88
 16 100:90

等。

即使在减少集，还有一些产生相同的总和/异或相当多的组合，最糟糕的是大量产生的255/255总和/异或可能性，其中一些是：

Even in that reduced set, there are quite a few combinations which generate the same sum/xor, the worst being the large number of possibilities that generate a sum/xor of 255/255, some of which are:

255:255=0 255
255:255=1 254
255:255=2 253
255:255=3 252
255:255=4 251
255:255=5 250
255:255=6 249
255:255=7 248
255:255=8 247
255:255=9 246
255:255=10 245
255:255=11 244
255:255=12 243
255:255=13 242
255:255=14 241
255:255=15 240
255:255=16 239
255:255=17 238
255:255=18 237

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