An XOR value between n integers of 1 to N (O (1) algorithm)

problem: Ask 1 to the N it N an XOR value between integers, i.e. 1 XOR 2 xor 3 ... xor n

Record f (x, Y) as the XOR value of all integers x through Y.

To f (2^k, 2^ (k+1)-1) (note that the ^ in the article represents "power", Xor means "XOR", or "or"):

2^k to 2^ (k+1)-1 This 2^k number, the highest bit (+k bit) of 1 number is 2^k,

If K >= 1, then the 2^k is even, the highest bit of the 2^k number (+k bit) is removed, the difference or value is unchanged.

Thus F (2^k, 2^ (k+1)-1) = f (2^k-2^k, 2^ (k+1) -1-2^k) = f (0, 2^k-1)

Thus f (0, 2^ (k+1)-1) = f (0, 2^k-1) XOR F (2^k, 2^ (k+1)-1) = 0 (k >= 1)

i.e. f (0, 2^k-1) = 0 (k >= 2)

The highest bit 1 for f (0, N) (n >= 4) is set at the +k bit (k >= 2),

F (0, N) = f (0, 2^k-1) XOR F (2^k, N) = f (2^k, N)

2^k to n this n+1-2^k number, the highest bit (+k bit) Total M = n+1-2^k 1, remove the highest bit of 1

When n is odd, M is an even number, thus f (0, N) = f (2^k, N) = f (0, N-2^k)

Due to the n-2^k and N parity, recursion above formula, can get: F (0, N) = f (0, n% 4)

When n% 4 = = 1 o'clock, f (0, N) = f (0, 1) = 1

When n% 4 = = 3 o'clock, f (0, N) = f (0, 3) = 0

When n is even, M is odd, thus f (0, N) = f (2^k, N) = f (0, n-2^k) or 2^k

That is, the highest bit 1 remains the same, due to the n-2^k and N parity, recursion above the formula,

Available: F (0, N) = NN or f (0, N 4) (minimum 2 position of nn n = 0)

When n% 4 = = 0 o'clock, f (0, n) = n

When n% 4 = = 2 o'clock, f (0, N) = NN or 3 = n + 1 (Formula pair n = 2 is still true)

Sum up:

F (1, N) = f (0, N) =

n N% 4 = = 0

1 n% 4 = = 1

N +1 n% 4 = = 2

0 n% 4 = = 3

Code:

unsigned xor_n(unsigned n)

{

unsigned t = n & 3;

if (t & 1) return t / 2u ^ 1;

return t / 2u ^ n;

}

An XOR value between n integers of 1 to N (O (1) algorithm)