Introduction to Li Chao Tree

A Li Chao tree (also known as a Dynamic Convex Hull or Segment Tree with lazy propagations) is a data structure that allows for efficient dynamic maintenance of the convex hull of a set of points in a 2D plane. The Li Chao tree allows for dynamic insertion, deletion, and query operations on the set of points, and can be used in a variety of geometric problems such as line segment intersection, visibility graphs, and polygon triangulation.

The structure of a Li Chao tree is a segment tree, a balanced binary tree data structure where each leaf node represents a single point, and each internal node represents a convex hull of the points represented by its child nodes.
The tree is balanced to ensure that each leaf node is at roughly the same depth, which allows for logarithmic time complexity for operations such as insertion and deletion.

To go further one should begin with some geometric utility functions as given below:

C++

struct line {
    long double m, b;

    // This overloads the () operator,
    // allowing the struct to be used as a
    // function that takes in a value x and
    // returns the result of m * x + b.
    long double operator()(long double x)
    {

        return m * x + b;
    }
};

vector<line> a(N * 4);

Java

class Line {
    double m, b;
    
    public Line(double m, double b) {
        this.m = m;
        this.b = b;
    }
    
    public double evaluate(double x) {
        return m * x + b;
    }
}

List<Line> a = new ArrayList<>(N * 4);

Python3

class Line:
    def __init__(self, m, b):
        self.m = m
        self.b = b
    def __call__(self, x):
        return self.m * x + self.b

a = [Line(0,0) for i in range(N * 4)]

public class Line {
    private double m;
    private double b;
    public Line(double m, double b)
    {
        this.m = m;
        this.b = b;
    }

    public double Evaluate(double x) { 
          return m * x + b; 
    }
}

List<Line> a = new List<Line>(N * 4);

JavaScript

class Line {
  constructor(m, b) {
    this.m = m;
    this.b = b;
  }

  // This overloads the () operator,
  // allowing the struct to be used as a
  // function that takes in a value x and
  // returns the result of m * x + b.
  call(x) {
    return this.m * x + this.b;
  }
}

const a = new Array(N * 4).fill(null).map(() => new Line(0, 0));

On every node of the segment tree, we are storing the line that maximizes (or minimizes) the value of the mid. If the interval of the node is [L, R), then the line stored in it will maximize(or minimize). \frac{L+R}{2}

Insert: Let's say we are inserting a new line to the node which is corresponding to Interval [L, R).To make it easy let's assume that the line on the node has a smaller slope. So, mid =\ frac{L + R}{2}. Now we have two cases to think about.

Note: Red is the original line in the node

Case 1: red(mid) < blue(mid)

In this Case, We have to replace red line with blue.Now, should we remove red? The answer is No, because there is need of red in segment [L, mid). Because of this reason we should pass red to the node with interval [L, mid), which is its left son.

representation of case 1
Case 2: red(mid) > blue(mid)

In similar way, we should pass blue to its right son and keep red in this node, whose interval is [mid, R).

representation of case 2

Below is the code for the above approach:

Code block

C++

void insert(int l, int r, line segment, int idx = 0)
{
    // if the range represented by the current node is just
    // one element
    if (l + 1 == r) {
        // if the segmentment to be inserted has a greater
        // slope than the one in the current node, replace
        // it
        if (segment(l) > a[idx](l))
            a[idx] = segment;
        return;
    }
    // find the middle of the range
    int mid = (l + r) / 2;
    // calculate the indices of the left and right child
    // nodes
    int leftson = idx * 2 + 1, rightson = idx * 2 + 2;
    // if the segmentment to be inserted has a smaller slope
    // than the one in the current node, swap them
    if (a[idx].m > segment.m)
        swap(a[idx], segment);
    // if the segmentment in the current node has a smaller
    // value at the middle point than the segmentment to be
    // inserted
    if (a[idx](mid) < segment(mid)) {
        // swap the segmentments in the current node and the
        // left child node
        swap(a[idx], segment);
        // insert the segmentment into the left child node
        insert(l, mid, segment, leftson);
    }
    // otherwise, insert the segmentment into the right
    // child node
    else
        insert(mid, r, segment, rightson);
}

Java

void insert(int l, int r, Line segment, int idx)
{
    // if the range represented by the current node is just
    // one element
    if (l + 1 == r) {
        // if the segment to be inserted has a greater slope
        // than the one in the current node, replace it
        if (segment.get(l) > a[idx].get(l)) {
            a[idx] = segment;
        }
        return;
    }
    // find the middle of the range
    int mid = (l + r) / 2;
    // calculate the indices of the left and right child
    // nodes
    int leftson = idx * 2 + 1, rightson = idx * 2 + 2;
    // if the segment in the current node has a smaller
    // slope than the segment to be inserted, swap them
    if (a[idx].m > segment.m)
        swap(a[idx], segment);
    // if the segment in the current node has a smaller
    // value at the middle point than the segment to be
    // inserted
    if (a[idx].get(mid) < segment.get(mid)) {
        // swap the segments in the current node and the
        // left child node
        swap(a[idx], segment);
        // insert the segment into the left child node
        insert(l, mid, segment, leftson);
    }
    // otherwise, insert the segment into the right child
    // node
    else {
        insert(mid, r, segment, rightson);
    }
}

Python3

def insert(l, r, segment, idx=0):
    # if the range represented by the current node is just
    # one element
    if l + 1 == r:
        # if the segmentment to be inserted has a greater
        # slope than the one in the current node, replace
        # it
        if segment(l) > a[idx](l):
            a[idx] = segment
        return
    # find the middle of the range
    mid = (l + r) / 2
    # calculate the indices of the left and right child
    # nodes
    leftson = idx * 2 + 1
    rightson = idx * 2 + 2
    # if the segmentment to be inserted has a smaller slope
    # than the one in the current node, swap them
    if a[idx].m > segment.m:
        swap(a[idx], segment)
    # if the segmentment in the current node has a smaller
    # value at the middle point than the segmentment to be
    # inserted
    if a[idx](mid) < segment(mid):
        # swap the segmentments in the current node and the
        # left child node
        swap(a[idx], segment)
        # insert the segmentment into the left child node
        insert(l, mid, segment, leftson)
    # otherwise, insert the segmentment into the right
    # child node
    else:
        insert(mid, r, segment, rightson)

# This code is contribtued by Tapesh(tapeshdua420)

void Insert(int l, int r, Line segment, int idx)
{
    // if the range represented by the current node is just
    // one element
    if (l + 1 == r)
    {
        // if the segment to be inserted has a greater slope
        // than the one in the current node, replace it
        if (segment.Get(l) > a[idx].Get(l))
        {
            a[idx] = segment;
        }
        return;
    }
    // find the middle of the range
    int mid = (l + r) / 2;
    // calculate the indices of the left and right child
    // nodes
    int leftson = idx * 2 + 1, rightson = idx * 2 + 2;
    // if the segment in the current node has a smaller
    // slope than the segment to be inserted, swap them
    if (a[idx].m > segment.m)
        Swap(a[idx], segment);
    // if the segment in the current node has a smaller
    // value at the middle point than the segment to be
    // inserted
    if (a[idx].Get(mid) < segment.Get(mid))
    {
        // swap the segments in the current node and the
        // left child node
        Swap(a[idx], segment);
        // insert the segment into the left child node
        Insert(l, mid, segment, leftson);
    }
    // otherwise, insert the segment into the right child
    // node
    else
    {
        Insert(mid, r, segment, rightson);
    }
}
// This code is contributed by Tapesh(tapeshdua420)

JavaScript

function insert(l, r, segment, idx=0) {
    // if the range represented by the current node is just
    // one element
    if (l + 1 === r) {
        // if the segmentment to be inserted has a greater
        // slope than the one in the current node, replace
        // it
        if (segment(l) > a[idx](l)) {
            a[idx] = segment;
        }
        return;
    }
    
    // find the middle of the range
    let mid = (l + r) / 2;
    // calculate the indices of the left and right child
    // nodes
    let leftson = idx * 2 + 1;
    let rightson = idx * 2 + 2;
    // if the segmentment to be inserted has a smaller slope
    // than the one in the current node, swap them
    if (a[idx].m > segment.m) {
        swap(a[idx], segment);
    }
    // if the segmentment in the current node has a smaller
    // value at the middle point than the segmentment to be
    // inserted
    if (a[idx](mid) < segment(mid)) {
        // swap the segmentments in the current node and the
        // left child node
        swap(a[idx], segment);
        // insert the segmentment into the left child node
        insert(l, mid, segment, leftson);
    } 
    // otherwise, insert the segmentment into the right
    // child node
    else {
        insert(mid, r, segment, rightson);
    }
}

// This code is contribtued by Tapesh(tapeshdua420)

Code for queries, as we know we only need to consider the intervals that contain the point we need to ask from the way we inserted lines.

C++

long double query(int l, int r, int x, int idx = 0)
{
    // If the range of the query is only
    // one element, return the result of
    // the function a at index o applied to x
    if (l + 1 == r)
        return a[idx](x);

    // Find the middle of the range
    int mid = (l + r) / 2;
    // Find the index of the left and right
    // children of the current node
    int leftson = idx * 2 + 1;
    int rightson = idx * 2 + 2;

    // If the value being queried is less
    // than the middle of the range,
    // recursively call query on
    // the left child
    if (x < mid)
        return max(a[idx](x), query(l, mid, x, leftson));

    // Otherwise, recursively call query
    // on the right child
    else
        return max(a[idx](x), query(mid, r, x, rightson));
}

Java

public double query(int l, int r, int x, int idx) {
    // If the range of the query is only
    // one element, return the result of
    // the function a at index o applied to x
    if (l + 1 == r) {
        return a[idx](x);
    }

    // Find the middle of the range
    int mid = (l + r) / 2;
    // Find the index of the left and right
    // children of the current node
    int leftson = idx * 2 + 1;
    int rightson = idx * 2 + 2;

    // If the value being queried is less
    // than the middle of the range,
    // recursively call query on
    // the left child
    if (x < mid) {
        return Math.max(a[idx](x), query(l, mid, x, leftson));
    }

    // Otherwise, recursively call query
    // on the right child
    else {
        return Math.max(a[idx](x), query(mid, r, x, rightson));
    }
}
// This code is contributed by Tapesh(tapeshdua420)

Python3

def query(l, r, x, idx = 0):
    # If the range of the query is only
    # one element, return the result of
    # the function a at index o applied to x
    if l + 1 == r:
        return a[idx](x)

    # Find the middle of the range
    mid = (l + r) // 2
    
    # Find the index of the left and right
    # children of the current node
    leftson = idx * 2 + 1
    rightson = idx * 2 + 2

    # If the value being queried is less
    # than the middle of the range,
    # recursively call query on
    # the left child
    if x < mid:
        return max(a[idx](x), query(l, mid, x, leftson))

    # Otherwise, recursively call query
    # on the right child
    else:
        return max(a[idx](x), query(mid, r, x, rightson))
      
# This code is contributed by ik_9

using System;

public class Program
{
    // Define the delegate type for the functions in the 'a' array
    public delegate long double Function(int x);

    // Define the 'a' array of functions
    public static Function[] a = new Function[10]; // Adjust the size as needed

    public static long double Query(int l, int r, int x, int idx = 0)
    {
        // If the range of the query is only one element, return the result of
        // the function 'a' at index 'idx' applied to 'x'
        if (l + 1 == r)
        {
            return a[idx](x);
        }

        // Find the middle of the range
        int mid = (l + r) / 2;

        // Find the index of the left and right children of the current node
        int leftSon = idx * 2 + 1;
        int rightSon = idx * 2 + 2;

        // If the value being queried is less than the middle of the range,
        // recursively call Query on the left child
        if (x < mid)
        {
            return Math.Max(a[idx](x), Query(l, mid, x, leftSon));
        }
        // Otherwise, recursively call Query on the right child
        else
        {
            return Math.Max(a[idx](x), Query(mid, r, x, rightSon));
        }
    }

    public static void Main(string[] args)
    {
        // Example usage
        // Initialize the 'a' array with functions
        a[0] = (int x) => (long double)x; // Replace with your own functions
        a[1] = (int x) => (long double)(x * x);
        a[2] = (int x) => (long double)(x + 1);

        long double result = Query(0, 10, 5); // Example query
        Console.WriteLine("Result: " + result);
    }
}

JavaScript

function query(l, r, x, idx = 0) {
    // If the range of the query is only
    // one element, return the result of
    // the function a at index o applied to x
    if (l + 1 === r) {
        return a[idx](x);
    }

    // Find the middle of the range
    let mid = Math.floor((l + r) / 2);
    
    // Find the index of the left and right
    // children of the current node
    let leftson = idx * 2 + 1;
    let rightson = idx * 2 + 2;

    // If the value being queried is less
    // than the middle of the range,
    // recursively call query on
    // the left child
    if (x < mid) {
        return Math.max(a[idx](x), query(l, mid, x, leftson));
    } 
    // Otherwise, recursively call query
    // on the right child
    else {
        return Math.max(a[idx](x), query(mid, r, x, rightson));
    }
}

// This code is contributed by Tapesh(tapeshdua420)

Time Complexity: O(NLogN), for the construction of the tree and O(LogN) for each Query
Auxiliary Space: O(N)

Advantages of Li Chao tree:

Dynamic Convex Hull: The Li Chao tree is a dynamic data structure that can handle line segments and can be used to maintain the convex hull of a set of lines in real-time.
Fast Queries: Li Chao tree has a logarithmic time complexity for querying the minimum value on a given line segment, making it efficient for large datasets.
Space-efficient: Li Chao tree uses a balanced binary tree structure, which is space-efficient.

Disadvantages of the Li Chao tree:

Complexity: The construction of the Li Chao tree takes O(n log n) time and O(n) space, which can be computationally expensive for large datasets.
Limited to lines: Li Chao tree can only handle line segments and not other types of geometric shapes.
Limited to 2D: Li Chao tree is limited to 2-dimensional space and cannot be used for 3D or higher-dimensional problems.
Limited to linear functions: Li Chao tree is limited to linear functions and cannot be used for non-linear functions.

Introduction to Li Chao Tree

Advantages of Li Chao tree:

Disadvantages of the Li Chao tree:

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