Printing a Hollow Square Star Pattern
Dive into coordinate-based pattern printing by utilizing conditional statements to carve out the center of a square matrix.
Carving the Grid
Printing a solid square is easy (two nested loops). Printing a hollow square introduces a new concept: conditional printing based on matrix coordinates.
The Problem
Print a hollow square of size N. If N = 4:
****
* *
* *
****
The Coordinate Check
To solve this, we generate a full N x N grid using two loops, but we use an if statement to decide whether to print a star * or a blank space .
When does a star appear?
- On the very first row (
row == 1) - On the very last row (
row == n) - On the very first column (
col == 1) - On the very last column (
col == n)
The Implementation
int n = 4;
for (int row = 1; row <= n; row++) {
for (int col = 1; col <= n; col++) {
// If it's a boundary coordinate, print a star
if (row == 1 || row == n || col == 1 || col == n) {
System.out.print("*");
} else {
// Otherwise, print empty space
System.out.print(" ");
}
}
System.out.println();
}
Why This Matters
This is the exact logic used in 2D array boundary traversal. If you need to write an algorithm that only processes the exterior walls of a maze or a matrix, you use this exact boolean logic.
The Takeaway
Hollow shapes cannot be solved by simply adjusting the loop boundaries. They require full grid traversal with internal conditional logic acting as a filter for the specific coordinates you care about.
Generate an N x N grid and print a star only if the current row or column is at the absolute boundary (1 or N). Otherwise, print a space.
You could, but it is incredibly messy. Coordinate-based boolean logic is the standard and most readable way to handle hollow shapes.
Because terminal fonts are usually taller than they are wide. A star (*) takes up more vertical space than horizontal space. You can fix this by printing '* ' (with a space) to widen the shape.
This exact boundary logic (checking if row==0 or col==max) is used in graph traversal algorithms (like BFS/DFS in a matrix) to detect when you hit a wall.
The logic is identical, just replace 'row == n' with 'row == height' and 'col == n' with 'col == width'.