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Comment by pedrocr | Hacker News Reader

>note that m XOR d is basically m + d without the carry bit

This isn't true. XOR is addition without any carry bits so there are several solutions that have the same XOR and difference

There are at most 2 solutions: if m XOR d = a then either m+d = a or m+d = a+N.

With the combination of m-d = b, you will find m and d.

If both of them satisfied ( remember m and d are bounded by 1 and N) then we actually have two solutions.

Here's a counterexample.

Sequence is [1, 2, 3, 4, 6, 6, 7, 8, 9, 10] so m = 5 and d = 6

  diff = 5-6 = -1
  xor = 5^6 = 3

  1-2 = -1
  1^2 = 3

  5-6 = -1
  5^6 = 3

  9-10 = -1
  9^10 = 3

So three different combinations, within N, that obey the two constraints

OK. I think I find a way around.

Assume d XOR m = a. Find k = position of the first bit (from MSB to LSB) in a. Then d + m can be in [a + 2^(k+1),a + 2^(k+2),...,a + 2^63].

We will have less than 63 cases. In each case, after we find d and m, make sure they are in range and then linear test.

So in the end, we may need to run at most 63+3 = 66 N. But it still linear though.

Make sense!

That fails my solution!