# Rectangle Geometry

Authors: Darren Yao, Michael Cao, Benjamin Qi, Ben Dodge

Contributors: Allen Li, Andrew Wang, Dong Liu, Ryan Chou

Problems involving rectangles whose sides are parallel to the coordinate axes.

### Prerequisites

Resources | ||||
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IUSACO | module is based off this |

Most problems in this category include only two or three squares or rectangles, in which case you can simply draw out cases on paper. This should logically lead to a solution.

## Example - Fence Painting

Focus Problem – try your best to solve this problem before continuing!

### Slow Solution

Since all the intervals lie between the range, $[0, 100]$, we can mark each interval of length $1$ contained within each interval as painted using a loop. Then the answer will be the number of marked intervals.

**Time Complexity:** $\mathcal{O}(\text{max coordinate})$

C++

#include <bits/stdc++.h>using namespace std;const int MAX_POS = 100;int main() {freopen("paint.in", "r", stdin);freopen("paint.out", "w", stdout);int a, b, c, d;

Java

import java.io.*;import java.util.*;public class Paint {static final int MAX_POS = 100;public static void main(String[] args) throws IOException {Kattio io = new Kattio("paint");int a = io.nextInt();int b = io.nextInt();

Python

import sysMAX_POS = 100sys.stdin = open("paint.in", "r")sys.stdout = open("paint.out", "w")a, b = map(int, input().split())c, d = map(int, input().split())

However, this solution would not work for higher constraints (ex. if the coordinates were up to $10^9$).

### Fast Solution

Calculate the answer by adding the original lengths and subtracting the intersection length.

The official analysis splits computing the intersection length into several cases. However, we can do it in a simpler way. An interval $[x,x+1]$ is contained within both $[a,b]$ and $[c,d]$ if $a\le x$, $c\le x$, $x<b$, and $x<d$, or in other words if, $\max(a,c)\le x$ and $x<\min(b,d)$. So the length of the intersection is $\min(b,d)-\max(a,c)$ if this quantity is positive and zero otherwise!

**Time Complexity:** $\mathcal{O}(1)$

C++

#include <bits/stdc++.h>using namespace std;int main() {freopen("paint.in", "r", stdin);freopen("paint.out", "w", stdout);int a, b, c, d;cin >> a >> b >> c >> d;

Java

import java.io.*;import java.util.*;public class Paint {public static void main(String[] args) throws IOException {Kattio io = new Kattio("paint");int a = io.nextInt();int b = io.nextInt();int c = io.nextInt();int d = io.nextInt();

Python

import syssys.stdin = open("paint.in", "r")sys.stdout = open("paint.out", "w")a, b = map(int, input().split())c, d = map(int, input().split())total = (b - a) + (d - c) # the sum of the two intervalsintersection = max(min(b, d) - max(a, c), 0) # subtract the intersectionunion = total - intersectionprint(union)

## Example - Blocked Billboard

Think of this as the 2D analog of the previous example.

Focus Problem – try your best to solve this problem before continuing!

### Slow Solution

**Time Complexity:** $\mathcal{O}((\text{max coordinate})^2)$

Since all coordinates are in the range $[-1000,1000]$, we can simply go through each of the $2000^2$ possible visible squares and check which ones are visible using nested for loops.

C++

#include <bits/stdc++.h>using namespace std;const int MAX_POS = 2000;bool visible[MAX_POS][MAX_POS];int main() {freopen("billboard.in", "r", stdin);freopen("billboard.out", "w", stdout);for (int i = 0; i < 3; ++i) {

Java

import java.io.*;import java.util.*;public class Billboard {private static final int MAX_POS = 2000;public static void main(String[] args) throws IOException {Kattio io = new Kattio("billboard");int visible[][] = new int[MAX_POS][MAX_POS];

Python

Code runs marginally faster in Python when placed in a function, so we can use this to get all 10 test cases. When taken out of the function, the code passes only 9 test cases.

import sysMAX_POS = 2000def main():sys.stdin = open("billboard.in", "r")sys.stdout = open("billboard.out", "w")visible = [[False for _ in range(MAX_POS)] for _ in range(MAX_POS)]

Of course, this wouldn't suffice if the coordinates were changed to be up to $10^9$.

### Fast Solution

**Time Complexity:** $\mathcal{O}(1)$

Java

Creating a class `Rect`

to represent a rectangle makes the code easier
to understand.

import java.io.*;import java.util.*;class Rect {int x1, y1, x2, y2;int area() { return (y2 - y1) * (x2 - x1); }}public class Billboard {public static void main(String[] args) throws IOException {

Alternative Implementation

C++

Note how creating a struct `Rect`

to represent a rectangle makes the code easier
to understand.

#include <bits/stdc++.h>using namespace std;struct Rect {int x1, y1, x2, y2;int area() { return (y2 - y1) * (x2 - x1); }};int intersect(Rect p, Rect q) {int xOverlap = max(0, min(p.x2, q.x2) - max(p.x1, q.x1));

Python

Note how creating a class `Rect`

to represent a rectangle makes the code easier
to understand.

import sysclass Rect:def __init__(self):self.x1, self.y1, self.x2, self.y2 = map(int, input().split())def area(self):return (self.y2 - self.y1) * (self.x2 - self.x1)

## Common Formulas

Certain tasks show up often in rectangle geometry problems. For example, many problems involve finding the overlapping area of two or more rectangles based on their coordinate points, or determining whether two rectangles intersect. Here, we'll discuss these formulas.

Note that these formulas only apply to rectangles which have sides parallel to the coordinate axes.

### Coordinates

A rectangle can be represented with $2$ points: its top right corner and bottom left corner. We'll label these points $tr$ (top right) and $bl$ (bottom left).

In this module, we'll assume that increasing $x$ moves to the right and increasing $y$ moves up.

### Finding area

The formula for finding the area of an individual rectangle is $w \cdot l$.

$\texttt{length}$ is the length of the vertical sides, and $\texttt{width}$ is the length of the horizontal sides.

- $\texttt{width} = \texttt{tr}_x - \texttt{bl}_x$
- $\texttt{length} = \texttt{tr}_y - \texttt{bl}_y$
- $\texttt{area} = \texttt{width} \cdot \texttt{length}$

### Implementation

C++

long long area(int bl_x, int bl_y, int tr_x, int tr_y) {long long length = tr_y - bl_y;long long width = tr_x - bl_x;return length * width;}

Java

int area(int bl_x, int bl_y, int tr_x, int tr_y) {int length = tr_y - bl_y;int width = tr_x - bl_x;return length * width;}

Python

def area(bl_x: int, bl_y: int, tr_x: int, tr_y: int) -> int:length = tr_y - bl_ywidth = tr_x - bl_xreturn length * width

### Checking if two rectangles intersect

Given two rectangles $a$ and $b$, there are only two cases where they do not intersect:

- $\texttt{tr}_{a_y}$ $\leq$ $\texttt{bl}_{b_y}$ or $\texttt{bl}_{a_y}$ $\geq$ $\texttt{tr}_{b_y}$.
- $\texttt{bl}_{a_x}$ $\geq$ $\texttt{tr}_{b_x}$ or $\texttt{tr}_{a_x}$ $\leq$ $\texttt{bl}_{b_x}$.

In all other cases, the rectangles intersect.

### Implementation

C++

bool intersect(vector<int> s1, vector<int> s2) {int bl_a_x = s1[0], bl_a_y = s1[1], tr_a_x = s1[2], tr_a_y = s1[3];int bl_b_x = s2[0], bl_b_y = s2[1], tr_b_x = s2[2], tr_b_y = s2[3];// no overlapif (bl_a_x >= tr_b_x || tr_a_x <= bl_b_x || bl_a_y >= tr_b_y ||tr_a_y <= bl_b_y) {return false;} else {return true;}}

Java

boolean intersect(int[] s1, int[] s2) {int bl_a_x = s1[0], bl_a_y = s1[1], tr_a_x = s1[2], tr_a_y = s1[3];int bl_b_x = s2[0], bl_b_y = s2[1], tr_b_x = s2[2], tr_b_y = s2[3];// no overlapif (bl_a_x >= tr_b_x || tr_a_x <= bl_b_x || bl_a_y >= tr_b_y ||tr_a_y <= bl_b_y) {return false;} else {return true;}}

Python

def intersect(s1, s2) -> bool:bl_a_x, bl_a_y, tr_a_x, tr_a_y = s1[0], s1[1], s1[2], s1[3]bl_b_x, bl_b_y, tr_b_x, tr_b_y = s2[0], s2[1], s2[2], s2[3]# no overlapif bl_a_x >= tr_b_x or tr_a_x <= bl_b_x or bl_a_y >= tr_b_y or tr_a_y <= bl_b_y:return Falseelse:return True

### Finding area of intersection

We'll assume that the shape formed by the intersection of two rectangles is itself a rectangle.

First, we'll find this rectangle's length and width. $\texttt{width} = \min(\texttt{tr}_{a_x}, \texttt{tr}_{b_x}) - \max(\texttt{bl}_{a_x}, \texttt{bl}_{b_x})$. $\texttt{length} = \min(\texttt{tr}_{a_y}, \texttt{tr}_{b_y}) - \max(\texttt{bl}_{a_y}, \texttt{bl}_{b_y})$.

If either of these values are negative, the rectangles do not intersect. If they are zero, the rectangles intersect at a single point. Multiply the length and width to find the overlapping area.

### Implementation

C++

int inter_area(vector<int> s1, vector<int> s2) {int bl_a_x = s1[0], bl_a_y = s1[1], tr_a_x = s1[2], tr_a_y = s1[3];int bl_b_x = s2[0], bl_b_y = s2[1], tr_b_x = s2[2], tr_b_y = s2[3];return ((min(tr_a_x, tr_b_x) - max(bl_a_x, bl_b_x)) *(min(tr_a_y, tr_b_y) - max(bl_a_y, bl_b_y)));}

Python

def inter_area(s1, s2) -> int:bl_a_x, bl_a_y, tr_a_x, tr_a_y = s1[0], s1[1], s1[2], s1[3]bl_b_x, bl_b_y, tr_b_x, tr_b_y = s2[0], s2[1], s2[2], s2[3]return (min(tr_a_x, tr_b_x) - max(bl_a_x, bl_b_x)) * (min(tr_a_y, tr_b_y) - max(bl_a_y, bl_b_y))

Java

int interArea(int[] s1, int[] s2) {int bl_a_x = s1[0], bl_a_y = s1[1], tr_a_x = s1[2], tr_a_y = s1[3];int bl_b_x = s2[0], bl_b_y = s2[1], tr_b_x = s2[2], tr_b_y = s2[3];return ((Math.min(tr_a_x, tr_b_x) - Math.max(bl_a_x, bl_b_x)) *(Math.min(tr_a_y, tr_b_y) - Math.max(bl_a_y, bl_b_y)));}

## Problems

Status | Source | Problem Name | Difficulty | Tags | |
---|---|---|---|---|---|

Bronze | Very Easy | ## Show TagsRectangle | |||

Bronze | Easy | ## Show TagsRectangle | |||

CF | Normal | ## Show TagsRectangle | |||

CF | Normal | ## Show TagsRectangle |

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