Updating partial sums with Fenwick tree

The two straightforward ways to code updating and retrieval of partial sums of sequences are: (a) maintain the array of sequence elements - on each update of a sequence element update just the sequence element, and each time you need a partial sum you compute it from scratch, (b) maintain the array of partial sums - on each update of a sequence element update all the partial sums this element contributes to. In the case (a) update takes [math]\displaystyle{ O(1) }[/math] time and retrieval takes [math]\displaystyle{ O(N) }[/math], in the case (b) update takes [math]\displaystyle{ O(N) }[/math] and retrieval [math]\displaystyle{ O(1) }[/math] time. [math]\displaystyle{ N }[/math] is the length of the sequence. The Fenwick tree data structure (although it's not really a tree) has both update and retrieval taking [math]\displaystyle{ O(\log N) }[/math] time in the worst case, while still using the same [math]\displaystyle{ O(N) }[/math] amount of space.

The idea is to have two actions on positive integers, say [math]\displaystyle{ A }[/math] and [math]\displaystyle{ B }[/math], so that orbit of [math]\displaystyle{ x }[/math] under action of [math]\displaystyle{ A }[/math], and orbit of [math]\displaystyle{ y }[/math] under action of [math]\displaystyle{ B }[/math] intersect in exactly one point if [math]\displaystyle{ x \leq y }[/math], and none otherwise. We will maintain an array of size [math]\displaystyle{ N }[/math] with all elements initialized to 0. On each update of a sequence element [math]\displaystyle{ i }[/math], update all the elements of the array with indices in the [math]\displaystyle{ A }[/math] orbit of [math]\displaystyle{ i }[/math] by that same amount. To retrieve the partial sum of elements [math]\displaystyle{ 1 }[/math] thru [math]\displaystyle{ j }[/math], sum up all the elements from the array with indices in the [math]\displaystyle{ B }[/math] orbit of [math]\displaystyle{ j }[/math]. In the case (a) above [math]\displaystyle{ A }[/math] is identity and [math]\displaystyle{ B(j) = j-1 }[/math], for (b) it's [math]\displaystyle{ A(i) = i+1 }[/math] and [math]\displaystyle{ B }[/math] is the identity.

Here's the punchline: for a positive integer [math]\displaystyle{ n }[/math], if we call [math]\displaystyle{ r(n) }[/math] the integer we get when we isolate the rightmost, i.e. the least significant, set bit in the binary representation of [math]\displaystyle{ n }[/math] (so e.g. [math]\displaystyle{ r(6) = 2 }[/math] and [math]\displaystyle{ r(8) = 8 }[/math]), then [math]\displaystyle{ A(i) = i + r(i) }[/math] and [math]\displaystyle{ B(i) = i - r(i) }[/math] are two such actions. What's more, in C and its ilk, [math]\displaystyle{ r(i) }[/math] is coded as (-i)&i.

Proof: [math]\displaystyle{ A }[/math] increases its argument while [math]\displaystyle{ B }[/math] decreases it, so if [math]\displaystyle{ x=y }[/math] then [math]\displaystyle{ A^0(x) = B^0(y) }[/math], and that's the only intersection. If [math]\displaystyle{ x \lt y }[/math], then binary representations of [math]\displaystyle{ x }[/math] and [math]\displaystyle{ y }[/math] are [math]\displaystyle{ y = a1b_y }[/math] and [math]\displaystyle{ x = a0b_x }[/math] where [math]\displaystyle{ a }[/math], [math]\displaystyle{ b_x }[/math], [math]\displaystyle{ b_y }[/math] are (possibly empty) binary sequences, and [math]\displaystyle{ b_x }[/math] and [math]\displaystyle{ b_y }[/math] are of same length [math]\displaystyle{ |b| }[/math]. We may have had to pad the representation of [math]\displaystyle{ x }[/math] by zeros to the left as needed for this. The action [math]\displaystyle{ B }[/math] unsets the rightmost set bit, so eventually [math]\displaystyle{ y }[/math] will under action of [math]\displaystyle{ B }[/math] arrive at [math]\displaystyle{ a10^{|b|} }[/math], if it's not of that form to begin with. If [math]\displaystyle{ b_x }[/math] has no set bits, then one more action of [math]\displaystyle{ B }[/math] on [math]\displaystyle{ y }[/math] will make it equal to [math]\displaystyle{ x }[/math]. If [math]\displaystyle{ b_x }[/math] has set bits, then [math]\displaystyle{ A }[/math], which turns a number of the form [math]\displaystyle{ z01^{m}0^{n} }[/math] into [math]\displaystyle{ z10^{m+n} }[/math] will eventually bring [math]\displaystyle{ x }[/math] to [math]\displaystyle{ a10^{|b|} }[/math], which is where we left [math]\displaystyle{ y }[/math]. By the argument from the beginning, this is the only place the two orbits intersect. QED

Here's a simple implementation in C++, add error checking as needed

#include <vector>

class fenwick_tree {
    std::vector<int> v_;

  public:
    fenwick_tree(int N) : v_(N+1) {}

    // neither add() nor get() will work 
    // for k outside of [1, N] range

    // add val to the k-th element
    void add(int k, int val) {
        for (int i=k; i<int(v_.size()); i += (-i&i))
            v_[i] += val;
    }

    // get sum of elements 1 thru k
    int get(int k) const {
        int r=0;
        for (int i=k; i>0; i -= (-i&i))
            r += v_[i];
        return r;
    }
};