A Segment Tree for Christmas

Posted on 25th December 201524th October 2020 by parth

Prerequisites:
Firm grounding in the Basics of Data Structures
Recursion
Algorithmic complexity

Welcome to a festive new installment of Algosaurus! ‘Tis the season of Christmas trees and holly, so today, we’ll be talking about a special tree for the occasion, called the Segment Tree.

Mind you, this isn’t such an easy-to-understand data structure, but is indeed one of the most elegant and useful ones out there. It’s a long-ish post, so we’ve kept a few checkpoints to keep track of where you are:

Introduction
Operations on a Segment Tree
Lazy Propagation

Peep the duck and her family is decorating their Christmas tree with shiny new baubles today.

She has some mild OCD, so she divides her tree into different sections and wants to calculate the number of baubles between any 2 sections. At the same time, her babies keep adding baubles to any section whenever they like.

She decides to keep track of the number of baubles in each section through an array A.

Let’s call the operation of finding the number of baubles between section x and section y, query x y; the operation of Peep’s babies adding y baubles to section x, add x y

Basically, query x y asks the sum of the values $A_x$ to $A_y$ , and add x y asks to add the value $y$ at the index $x$ .

Of course, that doesn’t sound hard at all! Peep can just iterate from $A_i$ to $A_j$ each time she decides to query, and updates are even simpler, she can just set $A_x = A_x + y$ .

Let us analyze her solution. She takes $O(N)$ time per query, and $O(1)$ time per update.

But what if there are a lot of queries? We’d need to iterate over a range multiple times, so that doesn’t sound so good in terms of complexity.

Enter, the prefix sum array. Instead of merely storing A, we store an array B, which stores the *cumulative sum* up till that index of A, like so:

Despite the super fancy name and definition, prefix sums are simple creatures. Calculating them is easy-peasy.

def prefix_calc(A):
    B = []
    curr = 0
    for i in A:
        curr += i
        B.append(curr)
    return B

Note how we’re calculating each element of B in $O(1)$ , for a total run-time of $O(N)$ . Now, we can answer queries in $O(1)$ , by simply returning $B_y - B_{x-1}$ .

Yay! But wait, what happens when her babies add some new baubles, ie. we get an update?

For every update in A, we need to update the elements in B as well, since the sum changes accordingly. We need to run over $O(N)$ elements of B, so updates are $O(N)$ .

When her tree is tiny, she can still afford to count the baubles individually. But what if she has to decorate the Christmas tree at Rockefeller Center?

Hmm, looks like we need to think a bit about this.

So the prefix sum is very quick for queries, and the generic array is very quick for updates. An important question to ask at this point is, why?

It’s because in the prefix sum, any given query can be broken into 2 values in the structure we have. And in the generic array, each update is to change a single element. These are all constant values which don’t change with the *indexes* of the queries or the updates.

What if we found a middle ground, one that does both things reasonably quickly?

The beauty of the prefix sum array B, is that it functions using a running sum broken into intervals, but the downside is that you have to update every element of B when even a single element of the original array A is updated.

In other words, the cumulative sum of every element up till the given index in B, is strongly dependent on the elements of array A. We need to decouple intervals from updates as much as we can, because we’re lazy and don’t want to update our interval values unless we *really* need to.

Why? Because that speeds up updates of course! Let’s play around with array B a bit.

Hmm, this isn’t particularly helpful. We’ll still have to loop over all the elements in A to produce this value, in case of an update. Let’s split this into 2 and see what happens.

Slightly better. If any update happens between indices 1…4 of A, we don’t have to calculate the running sum of indices 5…8 anymore. We can simply recalculate the running sum of 1…4 and add to it the run sum of 5..8.

Now we’re talking. Any change in the interval, say, 3…4 won’t affect the run sums of 1..2 and 5…8. Simply recalculate the run sum of 3…4 and let the change in 3…4 ‘percolate’ to 1…4 and then to 1…8.

And finally…

The last level of ‘decoupling’ updates from intervals, and reducing the dependence of pre-calculated run sums from the elements of A. Any change in a single element won’t affect the rest of the elements.

This is called a segment tree.

Since this kind of structure is formed by the *union* of intervals or segments, it supports any function which supports the union operator.

That means Peep can find the greatest number of baubles between any 2 sections as well! (Since max supports the union operator as well, ie. if max(a, b) = a and max(c, d) = d, then max(a, b, c, d) = max(a, d))

Why is this nice? Note that each index in the original array appears in $ceil(\log N) + 1$ nodes (one on each level), which is $O(\log N)$ . This means that if we change the value of a single element, $O(\log N)$ nodes in the segment tree change. Which implies that we can do updates quickly.

Note that any interval can be expressed as a union of some nodes of the tree. As an example look at the nodes we need for $1\dots 7$ .

We claim that any interval can be represented by $O(\log N)$ nodes in the tree, so we can conceivably do queries quickly too. We’ll prove this after talking about the query function itself right here, so stay put!

Let’s now try to count the number of nodes. There’s $N$ leaves (leaves are nodes with no children), $N/2$ nodes in the level above, $N/4$ above that, and so on, until $1$ . This gives us the sum $1 + 2 + 4 + \dots + N$ , which sums to $2\cdot N - 1 = O(N)$

Summing up, this means we can do updates and queries in $O(\log N)$ , in $O(N)$ space.

Let’s concretely make our functions now.

Now, for each node indexed $v$ , if $v$ isn’t a leaf node, its children are indexed $2\cdot v$ and $2\cdot v + 1$ . These observations suffice to describe a build method, with which we can initialize the tree. Note that we’ve used an array of size $4 \cdot N$ to represent the tree. This is necessary because of the full binary-heap like representation of the segment tree we’re using.

max_N = (1 << 20) #This is the maximum size of array our tree will support
seg = [0] * (4 * max_N) #Initialises the ’tree’ to all zeroes.

def build(someList, v, L, R):
    global seg
    if (L == R):
        seg[v] = someList[L]
    else:
        mid = (L + R) / 2 #Integer division is equivalent to flooring.
        build(someList, 2 * v, L, mid)
        build(someList, 2 * v + 1, mid + 1, R)
        seg[v] = seg[2 * v] + seg[2 * v + 1]

For the update procedure, the simplest way is to travel down to the lowest level, make the necessary update, and let it travel back up the tree as we add the left and right children for every node.

Let’s use an example!

This procedure can be described very cutely with a recursive function.

def update(v, L, R, x, y):
    global seg
    if (L <= x <= R): #the index to be updated is part of this node
        if (L == R): #this is a leaf node
            seg[v] += y
        else:
            mid = (L + R) / 2
            update(2 * v, L, mid, x, y)
            update(2 * v + 1, mid + 1, R, x, y)

            #update node using new values of children
            seg[v] = seg[2 * v] + seg[2 * v + 1]

Consider a query. Say that the query asks for the sum of values in the interval $x \dots y$ . Say, further, that we are on node $v$ , which represents the interval $l \dots r$ . There are three possibilities:

1. $l \dots r$ and $x \dots y$ do not intersect. In this case, we are not interested in the value of this node.
2. $x \leq l \leq r \leq y$ . In this case, we will use the value of the node as is.
3. There is a part of $l \dots r$ , which resides outside $x \dots y$ . Here, it makes sense to examine the children of $v$ , to find the intervals that fit better, or not at all (see cases 1 and 2).

Let’s look at a concrete example!

So the answer to the query is stored in run sum (or whatever you call it).

This can be represented in a recursive method as follows.

def query(v, L, R, x, y):
    if (R < x or y < L):
        #case 1
        return 0
    elif (x <= L <= R <= y):
        #case 2
        return seg[v]
    else:
        #case 3
        mid = (L + R) / 2
        return query(2 * v, L, mid, x, y) + query(2 * v + 1, mid + 1, R, x, y)

We’ve put up a global list-based version and a class-based version of the segment tree code on Gist, so go check them out!

Now, some interesting stuff. Let’s say that instead of the sum function, we had some function $f$ , which takes as input a list of values and outputs a single real number. What properties should it have so that our segment tree can support it?

We are combining intervals, in $O(1)$ for the sum function, which gives us a runtime of $O(\log N)$ per operation, so if combining intervals takes time $T$ , then each operation is $O(T\cdot\log N)$ . So we can use this structure for any $f$ whose value of $T$ is low enough to be acceptable for the given situation.

For example, for $f$ = max function, $T = O(1)$ , so the segment tree works quickly for that.

We’ll now prove that the query function runs in $O(\log N)$ for any arbitrary interval.

At the heart of the proof is the fact that the query function visits at most $4$ nodes in each level.

We always visit the root level, and it has only $1$ node, so we have a base case. Also, let’s assume that the hypothesis holds true for all levels $\leq k$ .

Now, if we visit $1$ node on the $k^{\text{th}}$ level, then we can visit only $2$ in the next (since a single node has $2$ children). Note how the function only visits continuous nodes in the next level.

If we visit $2$ , then they have only $4$ children, which means we’re still fine visiting all $4$ .

We will never visit $3$ nodes on a level. This is because if an interval does not fit well, we visit both of its children, which implies that we never visit an odd number of nodes on a level (except on root).

If we visit $4$ nodes on a level, then by the previous argument, they must be continuous. We claim, by way of contradiction, that we will need to visit all $8$ children from these nodes. This implies that we have case $3$ for every one of these nodes. Note, however, that if any two nodes have case $3$ , the query interval spans every node between them, which contradicts our supposition. So we’ll only visit the children of at-most $2$ nodes on this level, and hence at most $4$ nodes on the next level.

Now, since the hypothesis holds for all levels $\leq k$ implies it holds for level $k + 1$ , by strong induction, it holds for every level.

Now, since there are $O(\log N)$ levels, the query function will use at most $4\cdot\log N = O(\log N)$ nodes per query.

And we’re done!

What if we had updates of the type:

add x y z

where one has to add $z$ to every element between $x$ and $y$ (both inclusive)?

We could do $y - x + 1$ updates on our tree, but that would be $O(N \log N)$ , which is atrocious.

Alternatively, note that we can break up the update interval into some nodes (exactly like the query function).

But instead of reporting the sums of all the nodes which exhibit case 2, we will update all of them. If $z$ is added to every element of a node of size $R - L + 1$ , its sum increases by $(R - L + 1) \cdot z$ , which means we can update each node in $O(1)$ , for an overall runtime of $O(\log N)$ .

But *wait*.

The children of every non-leaf node we changed do not have the new value. What if we get a query that uses those nodes? Let’s create an additional variable for every node, called lazy, so that $lazy_v$ stores the pending addition to node $v$ .

We’ll only do additions to a node when we *need* to use the node, since we’re lazy folks and don’t want to do any work unless we need to.

What do we mean by pending addition? The idea is that when a node falls in case 2 (from the cases we defined around the query function), we update it, and add $z$ to the pending values of its children. This adds an additional procedure to both the update and the query functions.

Before we do any work on a node, we need to apply any pending add updates, and propagate these updates to its children, by adding the updates to their lazy values.

This part is magical! All the pending changes for every update are squashed into a single update. No multiple updates are required anymore and we can make do with only 1, that too only when we need to.

Since we’re only doing an additional constant amount of work per node, everything is still $O(\log N)$ !

Let’s make this procedure clearer with an example.

Now, let’s see how the lazy values come into play. Say that we call get a query 4 5.

The idea is hopefully precise enough to translate to code now. We’ve put up the code right here, so go take a look at it!

This concludes our article on segment tree. The segment tree is a rather wonderful data structure, so we hope we’ve done justice to it. As usual, we love hearing from you, so do send us your feedback at rawrr@algosaur.us!

Acknowledgements:
Stack Exchange thread on proof of time complexity of query operation on segment tree

A Heap of (less) Complexity: Binomial Heaps

Posted on 5th November 201512th January 2016 by radhika

Prerequisites:
Firm grounding in the Basics of Data Structures
Recursion
Algorithmic complexity

Hi, welcome to another installment of Algosaurus, this time on Binomial Heaps. They are a more efficient form of heaps than the simple binary heaps we looked at earlier, not to mention they have a beautiful intuition to them. Even though they aren’t frequently used, they’re still good for expanding your mind.

Let’s compare the complexities of various operations on Binary, Binomial, and Fibonacci Heaps respectively.

Need I say anything further?

As a disclaimer, this article is definitely not aimed at beginners to programming. The Binomial Heap concept is somewhat complex and you might look like this after your first reading:

That said, let’s begin!

First, a quick recap of binary heaps. In my last article, I implemented binary heaps as max heaps, where the largest element was stored at the root of the tree.

Today, we’re going to show them as min-heaps, where the root element is the smallest.

Heaps were created to improve the complexity of graph algorithms like Dijkstra’s Algorithm, by executing the algorithm using a heap. Notably, binary min-heaps are used as priority queues there.

Thing is, two more operations are often used in graph algorithms.

One, merging two heaps together to form a new heap.

Two, decreasing the value stored in a node, called decrease-key.

There’s no easy way to merge regular binary heaps apart from reconstructing the heap from scratch, making the complexity $O(m \log n)$ . That’s no good.

Once again, the time complexities for operations on binary heaps are as follows:

We can perform most operations in $O(\log n)$ time or less.

Question is, can we do better?

Don’t worry Algosaurus, we can make that reality.

How do we merge nicely then?

Algosaurus then remembers that game he used to love playing on the phone, which merged tiles of the same power of 2 together.

You’ve played 2048 too, right?

Let’s get some intuition using binary arithmetic now.

What if we could adapt this approach to store our elements in ‘packets’ whose sizes are powers of 2?

With this, we have three properties fixed:

Sizes must be powers of two
Can efficiently fuse packets of the same size
Can efficiently find the minimum element of each packet

Let’s get down to deletion now.

Relax Algosaurus. Remember how any number can be expressed in terms of powers of 2?

Let’s ‘fracture’ the packet from which we deleted the element.

Then put it all back together.

With that, our four properties are:

Sizes must be powers of two
Can efficiently fuse packets of the same size
Can efficiently find the minimum element of each packet
Can efficiently ‘fracture’ a packet of $2^k$ elements into similar packets of smaller powers of 2

In what form should we express our packets in?

With binomial trees of course!

Quoting directly from the slides referenced in the Acknowledgements:

“A binomial tree of order k is a type of tree recursively defined as follows:
A binomial tree of order k is a single node whose children are binomial trees of order 0, 1, 2, …, k – 1.”

Let’s apply the heap property to the binomial trees.

A binomial heap is basically a forest of heap-ordered binomial trees. Now, let’s check whether our binomial trees adhere to the 4 properties we set earlier.

Sizes must be powers of two – Pretty obvious.
Can efficiently fuse packets of the same size – As shown below, it’s a trivial operation!

Can efficiently find the minimum element of each packet – Each tree has a pointer to its minimum element, so it can be retrieved in $O(1)$ time.
Can efficiently ‘fracture’ a packet of $2^k$ elements into similar packets of smaller powers of 2

Since we’ve finally got an intuitive understanding of merging and deletion for the ‘packets’, let’s step through doing so in Binomial Heaps.

insert():

extractMin():

For now, I’m just going to say that the amortized time of insertion into a binomial heap is $O(1)$ . Feel free to take my word, or check them out in the slides given in the Acknowledgements.

But we still have a couple of problems…

When we intermix insert() and extractMin(), we force expensive insertions to happen repeatedly, wrecking our amortized time for insertion from $O(1)$ to $O(\log n)$ , because we have to merge $\log n$ binomial trees together at the same time.

Hmm. How do we make the speed of our insertion operations independent of whether we do deletions in the middle?

In Bengali we have a saying, “If you don’t have a head, you won’t have a headache”.

What if we just don’t merge the $O(\log n)$ binomial trees together, avoiding the step altogether?

That is, just add the isolated, unmerged binomial tree to the forest with every insertion and do nothing else.

We’ll only coalesce the trees together when we call extractMin(), only when we need to. The number of trees required to coalesce all the disjoint trees, is $\log n$ , just like the number of bits required to represent a decimal number $n$ , is $\log n$ .

We can’t just ‘merge’ all the mini-heaps together because the ‘merge’ operation assumes that all the trees in the binomial heap are in ascending order, and that isn’t the case here.

Let’s go back to our 2048-analogy for some intuition.

And we’re done!

Here are the final amortized complexities of our ‘lazy’ binomial heap:
insert(): $O(1)$
merge(): $O(1)$
findMin(): $O(1)$
extractMin(): $O(\log n)$

I’m not proving them here, in interest of keeping this proof-free. Feel free to check them out in the slides given in the Acknowledgements.

This concludes my article on Binomial Heaps. Although this may not have been as fun as my other articles, I hope it managed to demystify Binomial Heaps for you.

Acknowledgements:
This article simply wouldn’t have been possible without this series of slides CS166 from Stanford. These slides were the template for my article, so full props to them.
http://web.stanford.edu/class/archive/cs/cs166/cs166.1146/lectures/06/Slides06.pdf

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A Segment Tree for Christmas

A Heap of (less) Complexity: Binomial Heaps