# find the number of 11x11 squares in a 55x55 square grid?

### Mohammed

Guys, does anyone know the answer?

get find the number of 11x11 squares in a 55x55 square grid? from screen.

## The Number of Lattice Squares

What is the total number of squares that can fit into an n x n grid? *Lattice squares are the squares whose vertices are on the grid points. There are two types of lattice squares, grid ones and the tilted ones.

Eda Aydemir Jun 25, 2021 4 min read

## The Number of Lattice Squares*

**There are many puzzles about the number squares you can draw by using the grid points ( lattice points) on a given grid. Here is an**

**example**

**;**

**The correct answer is not 9 (the number of 1x1 squares). There are many other squares you can create using the given points.**

**These hard to catch tilted squares makes these puzzles interesting!**

**Now we have a harder puzzle to work on!**

## What is the total number of squares that can fit into an n x n grid?

***Lattice squares are the squares whose vertices are on the grid points.**

**There are two types of lattice squares, grid ones and the tilted ones.**

**Let’s define a "grid square" as a square whose vertices are lattice points and sides are along the axis. (vertical squares). They are easy to create and have square number areas.**

**A "tilted square" is a square whose vertices are still lattice points, but its sides are not along the axis.**

**Tilted squares have whole number areas. The side length of a tilted square can easily be found by using the Pythagorean Theorem.**

**Now, let’s have a look at a 3 x 3 squares and find the total number of grid and tilted squares that can be drawn using the lattice points.**

**The number of grid squares that can be drawn is 9 +4 +1 = 14**

**Now, let’s find the number of tilted squares**

**The number of tilted squares that can be drawn is 4 + 2 = 6.**

**Then, the total number of lattice squares is 14 + 6 = 20 by using the points of a 3 x 3 grid.**

**One may wonder if there is a short way of finding the number of squares for an n x n square.**

**The questions we need to answer are;**

**The number of grid squares in a n x n square**

**The side length of the biggest tilted square that can be drawn in an n x n square**

**The number of tilted squares in a n x n square**

**The total number of lattice squares in an n x n square.**

**Any relation among the number of tilted squares and grid squares**

**We need to investigate all the possible squares carefully and record our findings systematically to be able to find answers to these questions.**

**Here is a**

**Polypad file**

**you can work on to make drawings;**

**You may need more grids to highlight to create different squares. Good luck!**

**------ ***------**

**SOLUTION**

**We can start solving this puzzle by remembering another one! Famous"**

**Checkerboard Puzzle**

**".**

**The answer of the Checkerboard Problem gives us the number of grid squares.**

**To be able to find the total number of squares on a checkerboard, we need to consider that the board has 2 x 2 squares, 3 x 3 squares, 4 x 4 squares and so on other than 64 unit squares.**

**If we organize our findings in a table. We may easily see that they follow the pattern of square numbers.**

**Number of Grid squares in a n x n square;**

## Find the number of squares inside the given square grid

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## Find the number of squares inside the given square grid

Last Updated : 08 Mar, 2022

Read Discuss Courses Practice Video

Given a grid of side **N * N**, the task is to find the total number of squares that exist inside it. All squares selected can be of any length.

**Examples:**

**Input:**N = 1

**Output:**1

**Input:**N = 2

**Output:**5

**Input:**N = 4

**Output:**30

Recommended: Please try your approach on first, before moving on to the solution.

**Approach 1:**Taking a few examples, it can be observed that for a grid on size

**N * N**, the number of squares inside it will be

**12 + 22 + 32 + … + N2**

Below is the implementation of the above approach:

## C++

// C++ implementation of the approach

#includeusing namespace std;

// Function to return the number

// of squares inside an n*n grid

int cntSquares(int n)

{ int squares = 0;

for (int i = 1; i <= n; i++) {

squares += pow(i, 2);

} return squares; } // Driver code int main() { int n = 4;

cout << cntSquares(4);

return 0; }

## Java

// Java implementation of the approach

class GFG {

// Function to return the number

// of squares inside an n*n grid

static int cntSquares(int n)

{ int squares = 0;

for (int i = 1; i <= n; i++) {

squares += Math.pow(i, 2);

} return squares; } // Driver code

public static void main(String args[])

{ int n = 4;

System.out.print(cntSquares(4));

} }

## Python3

# Python3 implementation of the approach

# Function to return the number

# of squares inside an n*n grid

def cntSquares(n) : squares = 0;

for i in range(1, n + 1) :

squares += i ** 2; return squares; # Driver code

if __name__ == "__main__" :

n = 4;

print(cntSquares(4));

# This code is contributed by AnkitRai01

## C#

// C# implementation of the approach

using System; class GFG {

// Function to return the number

// of squares inside an n*n grid

static int cntSquares(int n)

{ int squares = 0;

for (int i = 1; i <= n; i++)

{

squares += (int)Math.Pow(i, 2);

} return squares; } // Driver code

public static void Main(String []args)

{ int n = 4;

Console.Write(cntSquares(n));

} }

// This code is contributed by 29AjayKumar

## Javascript

**Output:**

30

Time Complexity: O(n)

Auxiliary Space: O(1)

**Approach 2:**By the use of direct formula.

However, the sum has the closed form (direct formula) . Hence, we can employ this to calculate the sum in time.

Below is the implementation of the above approach:

## C++

// C++ implementation of the approach

#includeusing namespace std;

int cnt_squares (int n)

{

/* Function to return the number

of squares inside an n*n grid */

return n * (n + 1) * (2 * n + 1) / 6;

} // Driver code int main() {

cout << cnt_squares (4) << endl;

return 0; }

## Java

// Java implementation of the approach

class GFG {

static int cntSquares (int n) {

/* Function to return the number

of squares inside an n*n grid */

return n * (n + 1) * (2 * n + 1) / 6;

} // Driver code

public static void main(String args[]) {

System.out.println (cntSquares(4));

} }

## Python3

# Python3 implementation of the approach

"""

Function to return the number

of squares inside an n*n grid

""" def cntSquares(n) :

return int (n * (n + 1) * (2 * n + 1) / 6)

# Driver code

if __name__ == "__main__" :

print (cntSquares (4));

## C#

// C# implementation of the approach

using System; class GFG {

/* Function to return the number

of squares inside an n*n grid */

static int cntSquares (int n)

{

return n * (n + 1) * (2 * n + 1) / 6;

} // Driver code

public static void Main (String[] args)

{

Console.Write (cntSquares (4));

} }

## Javascript

## Find number of squares inside a given square grid in C++

Here we will check out how to find the number of squares inside a given square grid in C++ using simple formula with the corresponding code.

## Find number of squares inside a given square grid in C++

By HARISHWAR REDDY MUNUKUNTLA

In this tutorial, we will check out how to find the number of squares inside a given square grid in C++ with the corresponding code.

If the square grid contains a side of N*N, then we are going to find the total number of squares present in it.

## calculate the number of squares inside a square grid in C++

If we observe the number of squares that exist in smaller square grids then we can draw a pattern that will guide us to construct a simple formula to find the number of squares in a square grid of any length.

The total number of squares in a square grid of side 1 =1.

Total number of squares in a square grid of side 2 =5. ⇒ 4 small squares, and 1 2×2 square.

The total number of squares in a square grid of side 3 =14. ⇒ 9 small squares, 4 2×2 squares and 1 3×3 square.

The total number of squares in a square grid of side 4 =30. ⇒ 16 small squares, 9 3×3 squares, 4 2×2 squares and 1 4×4 square.

**By observing above sentences we can draw a pattern**

1×1 :-1^2 = 1.

2×2 :- 2^2 + 1^2 = 5.

3×3 :- 3^2 + 2^2 + 1^2 = 14.

4×4 :- 4^2 + 3^2 + 2^2 + 1^2 = 30.

Likewise, for nxn is n^2 + (n-1)^2 + (n-2)^2…………….(n-n+1)^2.

Then N(s)n = n^2 + (n-1)^2 + (n-2)^2................(n-n+1)^2 = n(n + 1)(2n + 1)/6

**1.By using the formula N(s)n = n^2 + (n-1)^2 + (n-2)^2…………….(n-n+1)^2:**#include

using namespace std;

int main() { int n,squares=0;

cout<<"enter the length of a square:";

cin>>n; while(n>0) {

squares += pow(n, 2);

cout<<"no of squares in a square grid:"<n--; }

return 0; }

**Output:**

enter the length of a square:5

no of squares in a square grid:55

2.**By using the formula N(s)n =n(n + 1)(2n + 1)/6:**

using namespace std;

int squares(int n) {

return n * (n + 1) * (2 * n + 1) / 6;

} int main() { int m;

cout<<"enter length of a square:";

cin>>m;

cout <<"no of squares in a square grid:"<< squares(m);

return 0; }

**Output:**

enter length of a square:10

no of squares in a square grid:385

Similarly, you can find the number of squares inside a given square grid of any length by using these programs.

Thus, I hope this tutorial will help you.

Guys, does anyone know the answer?