如何查找给定范围内的整数数甚至设置了位

Question

I want to know how many numbers say x, in a given range lets say l and r, l < r, present where number of 1's in binary representation of x is even. Is there any efficient way to find that?

Answer 1

Here's a key fact:

• If n is a non-negative even number, then exactly half of the non-negative integers less than n have even parity and the other half have odd parity. ("Even parity" means that the number of bits in the binary representation is even.)

Suppose we need to count the number of integers in the range [ l , r ) with some property P , and we know how to solve this problem for any range where l is 0. ( "[ l , r )" is a "half-open range": all integers n where l ≤ n < r . That makes the arithmetic easier.) Then we just need to subtract in order to solve the general problem where l ≠0: COUNT[ l , r ) = COUNT[0, r ) − COUNT[0, l ).

The first fact doesn't quite tell us all we need to know, since it only works for even n . But if n is odd, n-1 is even, and all we need to do is check the parity of n-1 itself, which is the extra number not in the range [0, n -1).

Putting all that together, if we have the range [ l , r ), we compute the count as follows:

• COUNT(0, l ) is l /2 if l is even, and ( l -1)/2+PARITY( l ) if l is odd
• COUNT(0, r ) is r /2 if r is even, and ( r -1)/2+PARITY( r ) if r is odd
• COUNT( l , r ) is COUNT(0, r ) − COUNT(0, l )

That last computation requires at most two parity computations, regardless of how large the range is, as well as a couple of divisions and a subtraction.

If this were a mathematics site, I might feel compelled to prove the assertion in the key fact at the beginning, but since this is CS I will content myself with a proof outline. We first note that if i is even, then PARITY( i ) and PARITY( i +1) are different (since the binary representations only differ in the last bit). Conversely, if i is odd, then PARITY( i ) and PARITY( i -1) are different. Now, take all the integers in [0, n ) and divide them into the set of integers with odd parity and the set with even parity, and consider the homomorphism

f ( i )⇒ i +1 if i is even;
f ( i )⇒ i −1 if i is odd;

The image of f over one of the two subsets of [0, n ) is the other subset, since the parity of f ( i ) is different from the parity of i . So the two subsets are the same size.

Answer 2

Assuming that your range falls in positive integer domain, I wrote the following C++ code

#include <iostream> 
#include <cmath> 
using namespace std;

int find_even_bits_helper(int upper, int level, bool sign) {
    // find even-bit number starting from 0
    // level and sign just for printing purpose 
    for (int i = 0; i <= level; ++i) cout << ' ';
    cout << "[0," << upper << "] -> " << '\n';
    int answer;
    if (upper == -1) {
        answer = 0;
    }
    else if (upper == 0) {
        answer = 1;
    }
    else if ((upper & (upper + 1)) == 0) {
        answer = (upper + 1) / 2;
    }
    else {
        int threshold = pow(2, floor(log2(upper)));
        int base_part = find_even_bits_helper(threshold - 1, level + 1, sign);
        int extra_part = (upper - threshold + 1) \
                       - find_even_bits_helper(upper - threshold, level + 1, !sign);
        // Essentially remove the highest 1 bit, thus all odd-bit becomes even-bit 
        answer = base_part + extra_part;
    }
    for (int i = 0; i <= level; ++i) cout << ' ';
    if (sign) cout << "+";
    else cout << "-";
    cout << answer << '\n';
    return answer;
}

int find_even_bits(int lower, int upper) {
    return find_even_bits_helper(upper, 0, true) \
         - find_even_bits_helper(lower - 1, 0, false);
}

int main() {
    cout << "Result is: " << find_even_bits(7,34) << '\n';
}

Testing range [7, 34] outputs:

 [0,34] ->
  [0,31] ->
  +16
  [0,2] ->
   [0,1] ->
   -1
   [0,0] ->
   +1
  -1
 +18
 [0,6] ->
  [0,3] ->
  -2
  [0,2] ->
   [0,1] ->
   +1
   [0,0] ->
   -1
  +1
 -4
Result is: 14

It shows that [0, 34] has 18 even-bit integers and [0, 6] has 4 even-bit integers.

Therefore, the answer is 18 - 4 = 14 .