Evaluate the variance or value between the N integers 1 to n (O (1) algorithm)

Question: Please1ToNThisNThe exclusive OR value between integers, that is1 XOR 2 XOR 3... XOR n

 

NoteF (x, y)IsXToYAll integers.

 

PairF (2 ^ K, 2 ^ (k + 1)-1 )(Note:ArticleIn^Indicates the "power ",XORIt indicates "different or ",Or"Or"):

2 ^ KTo2 ^ (k + 1)-1This2 ^ KNumber, the highest bit (+ KBit)1The number is2 ^ K,

IfK> = 1, Then2 ^ KIs an even number.2 ^ KMaximum number of digits(+ KBit)Remove the value. The value or value remains unchanged.

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

ThereforeF (0, 2 ^ (k + 1)-1) = f (0, 2 ^ k-1) xor f (2 ^ K, 2 ^ (k + 1) -1) = 0 (k> = 1)

That is F (0, 2 ^ k-1) = 0 (k> = 2)

 

PairF (0, n)(N> = 4)SetNThe highest bit1Yes+ KBit(K> = 2),

F (0, n) = f (0, 2 ^ k-1) xor f (2 ^ k, n) = F (2 ^ k, n)

Pair2 ^ KToNThisN + 1-2 ^ KNumber, highest bit(+ KBit)TotalM = n + 1-2 ^ KItems1,Remove the highest bit1

 

WhenNIf it is an odd number,MIs an even number, soF (0, n) = F (2 ^ k, n) = f (0, N-2 ^ K)

BecauseN-2 ^ KAndNFor parity, the formula above is recursive and can be obtained as follows:F (0, n) = f (0, N % 4)

WhenN % 4 = 1, F (0, n) = f (0, 1) = 1

WhenN % 4 = 3,F (0, n) = f (0, 3) = 0

 

WhenNIf it is an even number,MIs an odd number, soF (0, n) = F (2 ^ k, n) = f (0, N-2 ^ K) or 2 ^ K

That is to say, the highest bit1Remain unchanged becauseN-2 ^ KAndNSame as parity, recursive formula above,

Available:F (0, n) = nn or F (0, N % 4) (NNIsNThe lowest2Location0)

WhenN % 4 = 0,F (0, n) = N

WhenN % 4 = 2,F (0, n) = nn or 3 = n + 1(Formula pairN = 2Still valid)

 

To sum up:

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

N % 4 = 0

1 N % 4 = 1

N + 1 N % 4 = 2

0 n % 4 = 3

 

Code:

UnsignedXor_n(UnsignedN)

{

unsigned T = n & 3 ;

If ( T & 1 ) return T / 2u ^ 1 ;

 ReturnT/ 2u ^N;

}