A mirror diagonal pattern prints digit i on the main diagonal and again on a mirrored diagonal. Spaces fill every other column, forming a V-shape that meets at the bottom tip.
Remember
Rule: left j = 1..rows → print i if i == j else space
right k = rows-1..1 → print i if i == k else space
1 1
2 2
3 3
4 4
5 ← rows = 5
In C# use two inner loops: left half with Console.Write(i == j ? j.ToString() : " "), right half with i == k while k counts down from rows - 1, then WriteLine().
Approach
How to Solve It
Use one outer loop and two inner loops. Left half tests i == j; right half tests i == k while counting k down from rows - 1 (skip the center column).
Method
Idea
Best for
Two halves + conditions
Left i == j, right i == k, else space
Learning, interviews, exams
Rows input
Same logic with a user-chosen height
Practice / demos
Pseudocode
Pseudocode
for i from 1 to rows:
for j from 1 to rows:
print i if i == j else a space
for k from (rows - 1) down to 1:
print i if i == k else a space
print newline
for (k = rows - 1; k >= 1; k--) Console.Write(i == k ? k.ToString() : " ");
Skip center twice
Right loop starts at rows - 1, not rows
Chars per row
2 * rows - 1
Write vs WriteLine
API
Effect
Use for
Console.Write
Stays on the same line
Each column (digit or space) in both halves
Console.WriteLine
Ends the current line
After both inner loops finish a row
Try it
Live Preview
Change the row count and the V-shape updates instantly — capped at 9 so every digit stays a single character.
Whole numbers from 1 to 9. Tap a chip or type a value — the preview redraws as you go.
Live resultrows = 5 · 5×9 cells
1 1
2 2
3 3
4 4
5
Trace
Worked Walkthrough — Row i = 3, rows = 5
Trace where digit 3 lands on both diagonals.
Half
Loop
When digit prints
Left
j = 1..5
At j = 3 → 3 (spaces elsewhere)
Right
k = 4..1
At k = 3 → 3 (spaces elsewhere)
Full row: 3 3 (9 characters). Each of n rows prints 2n - 1 chars → O(n²).
Code
C# Programs
Three complete programs: fixed rows = 5, user-input rows, and a compact rows = 3 demo. Use View Output for sample results.
Example 1 — Fixed rows = 5
Hard-coded height — print digit when i == j or i == k, otherwise a space.
C#
using System;
namespace MyApp
{
class Program
{
static void Main(string[] args)
{
int rows = 5;
int i, j, k;
for (i = 1; i <= rows; i++)
{
for (j = 1; j <= rows; j++)
Console.Write(i == j ? j.ToString() : " ");
for (k = rows - 1; k >= 1; k--)
Console.Write(i == k ? k.ToString() : " ");
Console.WriteLine();
}
}
}
}
Output
1 1
2 2
3 3
4 4
5
How It Works
1. Left half. Scan j = 1..rows; print the digit only when i == j (main diagonal).
2. Right half. Scan k = rows-1..1; print when i == k — starting at rows - 1 skips the center column.
3. Bottom tip. On the last row, both diagonals meet — only one digit appears in the middle.
Example 2 — User Input (rows)
Read the row count and draw the same V-shaped mirror diagonals.
C#
using System;
namespace MyApp
{
class Program
{
static void Main(string[] args)
{
int rows, i, j, k;
Console.Write("Enter the number of rows: ");
if (!int.TryParse(Console.ReadLine(), out rows) || rows < 1)
{
Console.WriteLine("Please enter a positive whole number.");
return;
}
for (i = 1; i <= rows; i++)
{
for (j = 1; j <= rows; j++)
Console.Write(i == j ? j.ToString() : " ");
for (k = rows - 1; k >= 1; k--)
Console.Write(i == k ? k.ToString() : " ");
Console.WriteLine();
}
}
}
}
2. Same core. Only the loop limits change — both diagonal conditions stay identical.
3. Single-digit tip. Cap demos at rows ≤ 9 so every digit stays one character wide.
Example 3 — Compact rows = 3
Same two-loop structure with a smaller height for quick paper tracing.
C#
using System;
namespace MyApp
{
class Program
{
static void Main(string[] args)
{
int rows = 3;
int i, j, k;
for (i = 1; i <= rows; i++)
{
for (j = 1; j <= rows; j++)
Console.Write(i == j ? j.ToString() : " ");
for (k = rows - 1; k >= 1; k--)
Console.Write(i == k ? k.ToString() : " ");
Console.WriteLine();
}
}
}
}
Output
1 1
2 2
3
How It Works
1. Three rows. Wide V on row 1, closer digits on row 2, single tip on row 3.
2. Trace on paper. Confirm the right loop starts at 2 (not 3) so the center is not doubled.
Edge Cases & Pitfalls
Check these before calling the solution done.
k = rows
Doubled center
If the right loop starts at k = rows, the middle column prints twice. Keep k = rows - 1.
print i always
No spaces
Printing i on every column fills the row with digits. Use the ternary: digit only when i == j / i == k.
WriteLine
Broken rows
If WriteLine sits inside either half, each column lands on its own line. Call it only after both loops.
rows = 1
Single tip
Output is just 1 — the right loop does not run. A good sanity check.
rows > 9
Multi-digit values
Past 9, values like 10 break column alignment. Cap demos at 9 or use fixed-width formatting.
Bad input
Convert.ToInt32 throws
Prefer int.TryParse so non-numeric input does not crash the program.
Analysis
Time and Space Complexity
Program
Time
Extra space
Fixed / compact (Examples 1, 3)
O(n²)
O(1)
User input (Example 2)
O(n²)
O(1)
Each of n rows prints 2n - 1 characters → about 2n² prints. For n = 5 that is 45 characters.
Remember
Key Takeaways
Two diagonals: left uses i == j, right uses i == k, else space.
Skip the center twice: right loop starts at rows - 1 so the middle column is not doubled.
Write vs WriteLine: columns stay on the line; WriteLine advances after both halves.
Next step: Program 54 mirrors this V downward into a full diamond.
One line: for each i, print digit on i == j and i == k, spaces elsewhere, then WriteLine().
Frequently Asked Questions
The mirrored half prints rows-1 positions to avoid duplicating the center column. On the final row, only the main diagonal digit remains.
It prints the row number on the main diagonal (left) and on a mirrored diagonal (right), creating a symmetric V-shape like 1..5 on both sides.
The first loop prints the left half across rows columns. The second prints the right mirrored half across rows-1 columns in reverse.
Skipping the center column prevents printing the middle digit twice on rows where i equals the center index.
Change rows or read it from user input with TryParse — see Example 2.
O(n²) for n rows because each row prints about 2n-1 characters using nested loops.
Program 52 builds palindromic digit rows with m++ and m--. Program 53 uses spacing and i == j / i == k to place digits on mirrored diagonals.
Yes. Print when i == j or i + j == rows + 1 in a single column loop.
Prefer int.TryParse(Console.ReadLine(), out rows) so bad input does not throw FormatException.
One row prints a single 1 — the left loop prints at j = 1 and the right loop does not run.
🤔
Did you know?
Each row prints the row number on the main diagonal (left) and on a mirrored diagonal (right) using i == j and i == k. Row 3 shows 3 on both sides — about 2n-1 characters per row, so O(n²) total.
