How to resolve the algorithm Spiral matrix step by step in the C# programming language

Published on 12 May 2024 09:40 PM
#C_Sharp

How to resolve the algorithm Spiral matrix step by step in the C# programming language

Table of Contents

Problem Statement

Produce a spiral array.

A   spiral array   is a square arrangement of the first   N2   natural numbers,   where the numbers increase sequentially as you go around the edges of the array spiraling inwards.

For example, given   5,   produce this array:

Let's start with the solution:

Step by Step solution about How to resolve the algorithm Spiral matrix step by step in the C# programming language

The provided C# code generates a spiral matrix of a given size n, where each element represents its position in the spiral. Let's break down the code in detail:

Method 1: Spiral

This method generates a 2D array result to represent the spiral matrix. It uses the following algorithm:

  1. It initializes variables to track the current position (pos), number of elements to fill in each side of the spiral (count), current value to fill in (value), and a running sum of indices (sum).

  2. It enters a do-while loop that continues as long as count is greater than 0.

  3. Inside the loop:

    • It calculates the value based on the direction it's filling (either clockwise or counterclockwise).
    • It fills count elements in the current direction, incrementing sum and setting the corresponding elements in result to pos.
    • It updates value and count to prepare for filling the next side of the spiral.

Method 2: PrintArray

This method is used to print the 2D array array in a formatted manner. It pads the numbers with spaces to align them in columns, making the output visually pleasing.

Method 3: CreateMatrix

This method generates a spiral matrix using a different algorithm. It uses four arrays dx and dy to represent the directions in which to fill the matrix.

  1. It initializes variables x, y, c, and the matrix m.
  2. It iterates through the matrix in a spiral pattern, filling in the elements based on the directions defined in dx and dy.
  3. It increments the counter c to fill in the spiral with sequential numbers.

Method 4: Print

This method is used to print the generated matrix m in a formatted manner, padding the numbers with spaces to align them in columns for better readability.

Main Program:

  1. It reads the order n of the spiral matrix from the user.
  2. It creates an instance of the spiral class.
  3. It enters a nested loop to fill in the spiral matrix using the func method.
  4. The func method calculates the index of each element based on its position in the spiral.
  5. It prints the generated spiral matrix using the Console.WriteLine() method.

Additional Notes:

  • The spiral class contains a few additional methods that are not used in this specific example.
  • The func method in the spiral class is a complex calculation to determine the index of each element in the spiral matrix based on its position.

Output:

The output of the code will be a spiral matrix of the specified order, with each element representing its position in the spiral.

Source code in the csharp programming language

public int[,] Spiral(int n) {
    int[,] result = new int[n, n];

    int pos = 0;
    int count = n;
    int value = -n;
    int sum = -1;

    do {
        value = -1 * value / n;
        for (int i = 0; i < count; i++) {
            sum += value;
            result[sum / n, sum % n] = pos++;
        }
        value *= n;
        count--;
        for (int i = 0; i < count; i++) {
            sum += value;
            result[sum / n, sum % n] = pos++;
        }
    } while (count > 0);

    return result;
}


// Method to print arrays, pads numbers so they line up in columns
public void PrintArray(int[,] array) {
    int n = (array.GetLength(0) * array.GetLength(1) - 1).ToString().Length + 1;

    for (int i = 0; i < array.GetLength(0); i++) {
        for (int j = 0; j < array.GetLength(1); j++) {
            Console.Write(array[i, j].ToString().PadLeft(n, ' '));
        }
        Console.WriteLine();
    }
}


//generate spiral matrix for given N
int[,] CreateMatrix(int n){
    int[] dx = {0, 1, 0, -1}, dy = {1, 0, -1, 0};
    int x = 0, y = -1, c = 0;
    int[,] m = new int[n,n];
    for (int i = 0, im = 0; i < n + n - 1; ++i, im = i % 4) 
        for (int j = 0, jlen = (n + n - i) / 2; j < jlen; ++j)
            m[x += dx[im],y += dy[im]] = ++c;
    return n;
}

//print aligned matrix
void Print(int[,] matrix) {
    var len = (int)Math.Ceiling(Math.Log10(m.GetLength(0) * m.GetLength(1)))+1;
    for(var y = 0; y<m.GetLength(1); y++){
        for(var x = 0; x<m.GetLength(0); x++){
            Console.Write(m[y, x].ToString().PadRight(len, ' '));
        }
        Console.WriteLine();
    }
}


using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;

namespace spiralmat
{
    class spiral
    {
        public static int lev;
        int lev_lim, count, bk, cd, low, l, m;
        spiral()
        {
            lev = lev_lim = count = bk = cd = low = l = m = 0;
        }
        
        void level(int n1, int r, int c)
        {
            lev_lim = n1 % 2 == 0 ? n1 / 2 : (n1 + 1) / 2;
            if ((r <= lev_lim) && (c <= lev_lim))
                lev = Math.Min(r, c);
            else
            {
                bk = r > c ? (n1 + 1) - r : (n1 + 1) - c;
                low = Math.Min(r, c);
                if (low <= lev_lim)
                    cd = low;
                lev = cd < bk ? cd : bk;
            }
        }

        int func(int n2, int xo, int lo)
        {
            l = xo;
            m = lo;
            count = 0;
            level(n2, l, m);

            for (int ak = 1; ak < lev; ak++)
                count += 4 * (n2 - 1 - 2 * (ak - 1));
            return count;
        }

        public static void Main(string[] args)
        {
            spiral ob = new spiral();
            Console.WriteLine("Enter Order..");
            int n = int.Parse(Console.ReadLine());
            Console.WriteLine("The Matrix of {0} x {1} Order is=>\n", n, n);
            for (int i = 1; i <= n; i++)
            {
                for (int j = 1; j <= n; j++)
                    Console.Write("{0,3:D} ",
		                  ob.func(n, i, j)
				  + Convert.ToInt32(
				    ((j >= i) && (i == lev)) 
				      ? ((j - i) + 1) 
				      : ((j == ((n + 1) - lev) && (i > lev) && (i <= j)))
				        ? (n - 2 * (lev - 1) + (i - 1) - (n - j))
					: ((i == ((n + 1) - lev) && (j < i)))
					  ? ((n - 2 * (lev - 1)) + ((n - 2 * (lev - 1)) - 1) + (i - j))
					  : ((j == lev) && (i > lev) && (i < ((n + 1) - lev)))
					    ? ((n - 2 * (lev - 1)) + ((n - 2 * (lev - 1)) - 1) + ((n - 2 * (lev - 1)) - 1) + (((n + 1) - lev) - i))
					    : 0));
                Console.WriteLine();
            }
            Console.ReadKey();
        }
    }
}


INPUT:-

Enter order..
5

OUTPUT:-

The Matrix of 5 x 5 Order is=>

 1   2   3   4   5
16  17  18  19   6
15  24  25  20   7
14  23  22  21   8
13  12  11  10   9

INPUT:-

Enter order..
6

OUTPUT:-

The Matrix of 6 x 6 Order is=>

 1   2   3   4   5   6
20  21  22  23  24   7
19  32  33  34  25   8
18  31  36  35  26   9
17  30  29  28  27  10
16  15  14  13  12  11

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