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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 = 14Now, 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 squareThe side length of the biggest tilted square that can be drawn in an n x n squareThe number of tilted squares in a n x n squareThe total number of lattice squares in an n x n square.Any relation among the number of tilted squares and grid squaresWe 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;

स्रोत : www.funmathfan.com

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 = 1Output: 1

Input: N = 2Output: 5

Input: N = 4Output: 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

#include

using 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

#include

using 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

स्रोत : www.geeksforgeeks.org

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:

#include

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.

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