How to adapt Fenwick tree to answer range minimum queries

Question

Fenwick tree is a data-structure that gives an efficient way to answer to main queries:

• add an element to a particular index of an array update(index, value)
• find sum of elements from 1 to N find(n)

both operations are done in O(log(n)) time and I understand the logic and implementation. It is not hard to implement a bunch of other operations like find a sum from N to M.

I wanted to understand how to adapt Fenwick tree for RMQ. It is obvious to change Fenwick tree for first two operations. But I am failing to figure out how to find minimum on the range from N to M.

After searching for solutions majority of people think that this is not possible and a small minority claims that it actually can be done (approach1, approach2).

The first approach (written in Russian, based on my google translate has 0 explanation and only two functions) relies on three arrays (initial, left and right) upon my testing was not working correctly for all possible test cases.

The second approach requires only one array and based on the claims runs in O(log^2(n)) and also has close to no explanation of why and how should it work. I have not tried to test it.

---

In light of controversial claims, I wanted to find out whether it is possible to augment Fenwick tree to answer update(index, value) and findMin(from, to).

If it is possible, I would be happy to hear how it works.

Credenza

In first approach there is an explicit statement that if you're searching for maximum, you are allowed to increase values in a cell, but not decrease, or else the values returned won't match the array. Since you're searching for a minimum, you are prohibited from increasing stored values if you are using this approach. The reason is that if you move array values towards the extremum you're seeking, you can just renew stored data, this takes O(log n), but if you're decreasing from extremum, the worst case is recalculating the entire array, this is way too expensive and not implemented. – Nostology

Answer 1

Yes, you can adapt Fenwick Trees (Binary Indexed Trees) to

• Update value at a given index in O(log n)
• Query minimum value for a range in O(log n) (amortized)

We need 2 Fenwick trees and an additional array holding the real values for nodes.

Suppose we have the following array:

index 1  2  3  4  5  6  7  8  9 10 11 12 13 14 15
value 1  0  2  1  1  3  0  4  2  5  2  2  3  1  0

We wave a magic wand and the following trees appear:

Note that in both trees each node represents the minimum value for all nodes within that subtree. For example, in BIT2 node 12 has value 0, which is the minimum value for nodes 12,13,14,15.

Queries

We can efficiently query the minimum value for any range by calculating the minimum of several subtree values and one additional real node value. For example, the minimum value for range [2,7] can be determined by taking the minimum value of BIT2_Node2 (representing nodes 2,3) and BIT1_Node7 (representing node 7), BIT1_Node6 (representing nodes 5,6) and REAL_4 - therefore covering all nodes in [2,7]. But how do we know which sub trees we want to look at?

Query(int a, int b) {
  int val = infinity // always holds the known min value for our range

  // Start traversing the first tree, BIT1, from the beginning of range, a
  int i = a
  while (parentOf(i, BIT1) <= b) {
    val = min(val, BIT2[i]) // Note: traversing BIT1, yet looking up values in BIT2
    i = parentOf(i, BIT1)
  }

  // Start traversing the second tree, BIT2, from the end of range, b
  i = b
  while (parentOf(i, BIT2) >= a) {
    val = min(val, BIT1[i]) // Note: traversing BIT2, yet looking up values in BIT1
    i = parentOf(i, BIT2)
  }

  val = min(val, REAL[i]) // Explained below
  return val
}

It can be mathematically proven that both traversals will end in the same node. That node is a part of our range, yet it is not a part of any subtrees we have looked at. Imagine a case where the (unique) smallest value of our range is in that special node. If we didn't look it up our algorithm would give incorrect results. This is why we have to do that one lookup into the real values array.

To help understand the algorithm I suggest you simulate it with pen & paper, looking up data in the example trees above. For example, a query for range [4,14] would return the minimum of values BIT2_4 (rep. 4,5,6,7), BIT1_14 (rep. 13,14), BIT1_12 (rep. 9,10,11,12) and REAL_8, therefore covering all possible values [4,14].

Updates

Since a node represents the minimum value of itself and its children, changing a node will affect its parents, but not its children. Therefore, to update a tree we start from the node we are modifying and move up all the way to the fictional root node (0 or N+1 depending on which tree).

Suppose we are updating some node in some tree:

• If new value < old value, we will always overwrite the value and move up
• If new value == old value, we can stop since there will be no more changes cascading upwards

• If new value > old value, things get interesting.

  • If the old value still exists somewhere within that subtree, we are done
  • If not, we have to find the new minimum value between real[node] and each tree[child_of_node], change tree[node] and move up

Pseudocode for updating node with value v in a tree:

while (node <= n+1) {
  if (v > tree[node]) {
    if (oldValue == tree[node]) {
      v = min(v, real[node])
      for-each child {
        v = min(v, tree[child])
      }
    } else break
  }
  if (v == tree[node]) break
  tree[node] = v
  node = parentOf(node, tree)
}

Note that oldValue is the original value we replaced, whereas v may be reassigned multiple times as we move up the tree.

Binary Indexing

In my experiments Range Minimum Queries were about twice as fast as a Segment Tree implementation and updates were marginally faster. The main reason for this is using super efficient bitwise operations for moving between nodes. They are very well explained here. Segment Trees are really simple to code so think about is the performance advantage really worth it? The update method of my Fenwick RMQ is 40 lines and took a while to debug. If anyone wants my code I can put it on github. I also produced a brute and test generators to make sure everything works.

I had help understanding this subject & implementing it from the Finnish algorithm community. Source of the image is http://ioinformatics.org/oi/pdf/v9_2015_39_44.pdf, but they credit Fenwick's 1994 paper for it.

Jen

Interesting, is it possible to implement range update here (range decrement, actually): "RangeUpdate (index, delta)" that would decrement all the values in array on interval [0 .. index] (i.e. from the start of array to index element) faster than O(index * log n)? – Flown

Hmmh, I don't think it's possible with this data structure. – Jen

So this would basically be the same as a segment tree. You're storing the same intervals, but in a different format. – Reformism

@Reformism Segment tree is a different data structure entirely. Both data structures can be used to solve the same problem with ~same time complexity. – Jen

That's true, but if you look at the segments which are covered by a segment tree and double fenwick tree, you'll notice that they are the same – Reformism

Answer 2

The Fenwick tree structure works for addition because addition is invertible. It doesn't work for minimum, because as soon as you have a cell that's supposed to be the minimum of two or more inputs, you've lost information potentially.

If you're willing to double your storage requirements, you can support RMQ with a segment tree that is constructed implicitly, like a binary heap. For an RMQ with n values, store the n values at locations [n, 2n) of an array. Locations [1, n) are aggregates, with the formula A(k) = min(A(2k), A(2k+1)). Location 2n is an infinite sentinel. The update routine should look something like this.

def update(n, a, i, x):  # value[i] = x
    i += n
    a[i] = x
    # update the aggregates
    while i > 1:
        i //= 2
        a[i] = min(a[2*i], a[2*i+1])

The multiplies and divides here can be replaced by shifts for efficiency.

The RMQ pseudocode is more delicate. Here's another untested and unoptimized routine.

def rmq(n, a, i, j):  # min(value[i:j])
    i += n
    j += n
    x = inf
    while i < j:
        if i%2 == 0:
            i //= 2
        else:
            x = min(x, a[i])
            i = i//2 + 1
        if j%2 == 0:
            j //= 2
        else:
            x = min(x, a[j-1])
            j //= 2
    return x

Radish

have you looked at approach 1 in my link. Based on my testing it works for majority of the cases. – Credenza

@SalvadorDali I haven't. – Radish

Thanks, @DavidEisenstat. Although this does not answer the question, I think this is the simplest solution for RMQ. I implemented your idea in gist.github.com/tuanchauict/e2c873fdceb7fb3018ebd47f395cbf32 – Thunderous

Answer 3

I've just implemented code for @atte-juvonen's approach. I'd highly recommend try to implement it on your own, but if it's a bit hard for someone, here is the code:

#include<bits/stdc++.h>
using namespace std;
#define int long long
#define maxn 200002
vector<int>t1(maxn, INT_MAX), t2(maxn,INT_MAX),vec(maxn, INT_MAX);

void update(int k, int v, int old){
    int node=k;
    int vv=v;
    while(node<maxn){
        if(vv>t1[node]){
            vv=min(vv, vec[node]);
            int lsb=(node & -node);
            int child=node-lsb;
            while(lsb/2>0){
                child+=lsb/2;
                lsb/=2;
                vv=min(vv,t1[child]);
            }
        }
        if(vv==t1[node])break;
        t1[node]=vv;
        node+=(node & -node);
    }
    node=k;
    while(node>0){
        if(v > t2[node]) {
            v = min(v,vec[node]);
            int ax=1;
            while(ax!=(node & -node)){
                int child=node+ax;
                ax<<=1;
                v=min(v, t2[child]);
            }
        }
        if(v == t2[node])break;
        t2[node] = v;
        node -= (node & -node);
    }
}
int query(int a, int b){
    // traversing BIT1, yet looking up values in BIT2
    int node=a;
    int val=INT_MAX;
    while(node+(node&-node)<=b){
        val=min(val, t2[node]);
        node+=(node&-node);
    }
    // traversing BIT2, yet looking up values in BIT1
    node=b;
    while(node-(node&-node)>=a){
        val=min(val, t1[node]);
        node-=(node&-node);
    }
    val=min(val, vec[node]);
    return val;
}

int32_t main(){
    int n,q,x,a,b; 
    cin>>n>>q;
    for(int i=1;i<=n;i++){
        cin>>vec[i];
        update(i,vec[i], INT_MAX);
    }
    while(q--){
        cin>>x>>a>>b;
        if(x==1){
            int old=vec[a];
            vec[a]=b;
            update(a,b, old);
        }
        else
            cout<<query(a,b)<<endl;
    }
}

Sola