3

Problem : Given a range $[n,m]$, find the xor of all even numbers in this range. Constraints : $1 \le n \le m \le 10^{18}$

How do we solve this problem?

P.S. I asked this question at stackoverflow but that was not the appropriate place to ask this question. So I re-asked it here.

gt6989b
  • 55,125
Daga
  • 291

3 Answers3

6

Let $$f(n) = \oplus_{\text{$i$ even}}^n i$$ and wlog $n$ is even (replace it with $n-1$ if it's odd, without any change to the answer).

Then

  • if $n \equiv 0 \pmod{8}$, have $f(n) = n$;
  • if $n \equiv 2 \pmod{8}$, have $f(n) = 2$;
  • if $n \equiv 4 \pmod{8}$, have $f(n) = n+2$;
  • if $n \equiv 6 \pmod{8}$, have $f(n) = 0$.

These are easily proved by considering the binary expansion of $f(n)$: in all cases, we want $f(n-2) \oplus n$.

  • If $n \equiv 2 \pmod{8}$, then we are taking the previous sum and XORing it with $n$; that is $(n-2) \oplus n$ inductively, which is $2$.
  • If $n \equiv 4 \pmod{8}$, then we are taking the previous sum and XORing it with $n$; that is $2 \oplus n$, which (since $n$ ends with two zeros in its expansion) is $n+2$.
  • If $n \equiv 6 \pmod{8}$, then likewise it is $n \oplus n = 0$.
  • If $n \equiv 0 \pmod{8}$, then it is $0 \oplus n = n$.

Base case is easy.

Then the xor in range $[n, m]$ is $f(m) \oplus f(n-1)$.

Patrick Stevens
  • 36,995
  • Your answer appears to be right, as I've lazily verified with a foetus of a program: http://www.tutorialspoint.com/viewproject.php?URL=execute_python_online.php&PID=0Bw_CjBb95KQMSVVaSmNoekNITTQ. But it would be lovely if you could furnish it with a proof. – Colm Bhandal Dec 03 '15 at 18:18
  • @ColmBhandal Done. – Patrick Stevens Dec 03 '15 at 18:23
  • I think I am misunderstanding the final part. If I want the xor on $[4,6]$, I think the answer would be $2$. But $f(6) = 0$ and $f(3)=f(2)=2$ so I would arrive at $-2$ by the formula suggested? – Jason Knapp Dec 03 '15 at 18:37
  • @JasonKnapp Oh yes, thanks for the spot. I meant xor, not minus. – Patrick Stevens Dec 03 '15 at 18:48
  • Beautiful! That last part was getting to me also. I never thought of induction, duh! Very nicely done sir. – Colm Bhandal Dec 03 '15 at 18:53
  • 1
    @ColmBhandal tips hat Induction is usually correct for proving piecewise things, in my experience. – Patrick Stevens Dec 03 '15 at 18:56
2

If n is odd, change n to n-1.

Starting with n, find the next number by adding 2 to the previous number.

f(n) = $2 \oplus 4 \oplus ... \oplus (n-2) \oplus n$

Converting the numbers to binary, its easy to see that the least significant bit of all the numbers is same (because all are even numbers). Adding 2 to previous number, say x, to get (x+2) its equivalent to ((x/2) + 1)*2.

Converting the numbers in binary and using the above observation we can easily infer the following:

f(n) = 2*g(n), where g(n) = $1 \oplus 2 \oplus ... \oplus (n-2)/2 \oplus (n/2)$

So it reduces the problem to find the XOR of all numbers from 1 to k where k = n/2. This can be solved as follows:

Then

  • if $k \equiv 0 \pmod{4}$, have $f(n) = k$;
  • if $k \equiv 1 \pmod{4}$, have $f(n) = 1$;
  • if $k \equiv 2 \pmod{4}$, have $f(n) = k+1$;
  • if $k \equiv 3 \pmod{4}$, have $f(n) = 0$.

This can be observed from the pattern:

0000 <- 0  [k]
0001 <- 1  [1]
0010 <- 3  [k+1]
0011 <- 0  [0]
0100 <- 4  [a]
0101 <- 1  [1]
0110 <- 7  [k+1]
0111 <- 0  [0]
1000 <- 8  [k]
1001 <- 1  [1]
1010 <- 11 [k+1]
1011 <- 0  [0]
1100 <- 12 [k]
1101 <- 1  [1]
1110 <- 15 [k+1]
1111 <- 0  [0]

So for the range [n,m] the ans is $f(m) \oplus f(n-1)$

P.S. This is how I solved it and got it right.

Daga
  • 291
0

Something like this:

For bit $k$, where $k=0$ is the units bit, blocks of length $2^{k+1}$ xor to zero, since there are $2^k$ with a $0$ and $2^k$ with a $1$.

Therefore, we can xor the $k$-th bit from $i= n+\lfloor \frac{m-n}{2^{k+1}} \rfloor $ to $m$ to get that bit, since the blocks of length $2^{k+1}$ xor to zero.

Do this until $2^{k} > m$.

marty cohen
  • 110,916