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
Follows the increasing-decreasing pyramid in Program 52; next is the full mirror diagonal diamond in Program 54.
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
Compact trace
Dry-run with rows = 3 before coding rows = 5
Paper tracing and labs
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--) printf(i == k ? "%d" : " ", k);
Skip center twice
Right loop starts at rows - 1, not rows
Chars per row
2 * rows - 1
Printing Numbers vs Starting a New Line
API
Effect
Use for
printf("%d", j) / printf(" ")
Stays on the same line
Each column in both halves
printf("\n")
Ends the current line
After both inner loops
Print all columns without a newline, then end the row once.
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, scanf input, and a compact rows = 3 demo. Use View Output to reveal sample results.
Example 1 — Fixed rows = 5
Hard-coded height — print digit when i == j or i == k, otherwise a space.
C
#include <stdio.h>
int main(void)
{
int rows = 5;
int i, j, k;
for (i = 1; i <= rows; i++)
{
for (j = 1; j <= rows; j++)
printf(i == j ? "%d" : " ", j);
for (k = rows - 1; k >= 1; k--)
printf(i == k ? "%d" : " ", k);
printf("\n");
}
return 0;
}
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 rows with scanf and reject non-positive values.
C
#include <stdio.h>
int main(void)
{
int rows;
int i, j, k;
printf("Enter the number of rows: ");
if (scanf("%d", &rows) != 1 || rows <= 0)
{
printf("Please enter a positive integer.\n");
return 1;
}
for (i = 1; i <= rows; i++)
{
for (j = 1; j <= rows; j++)
printf(i == j ? "%d" : " ", j);
for (k = rows - 1; k >= 1; k--)
printf(i == k ? "%d" : " ", k);
printf("\n");
}
return 0;
}
Output (when user enters 4)
Enter the number of rows: 4
1 1
2 2
3 3
4
How It Works
1. Prompt and validate. Reject bad input with a clear message.
2. Same core. Only the loop limits come from the user — both diagonal conditions stay identical.
3. Safer input tip. Cap demos to single-digit rows:
Safer input
if (scanf("%d", &rows) != 1 || rows < 1 || rows > 9)
{
printf("Enter a whole number from 1 to 9.\n");
return 1;
}
Example 3 — Compact rows = 3
Same two-loop structure with a smaller height for quick paper tracing.
C
#include <stdio.h>
int main(void)
{
int rows = 3;
int i, j, k;
for (i = 1; i <= rows; i++)
{
for (j = 1; j <= rows; j++)
printf(i == j ? "%d" : " ", j);
for (k = rows - 1; k >= 1; k--)
printf(i == k ? "%d" : " ", k);
printf("\n");
}
return 0;
}
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.
3. Scale up next. Once the small demo is clear, use Examples 1–2 for five rows or user input.
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.
\n inside
Broken rows
If printf("\n") sits inside either half, each column lands on its own line. Call the newline 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.
scanf
Check the return value
If scanf fails, rows may be uninitialized — always test scanf(...) == 1.
Analysis
Time and Space Complexity
Program
Time
Extra space
Fixed / input (Examples 1–2)
O(n²)
O(1)
Compact rows = 3 (Example 3)
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.
Break the row: call printf("\n") only after both halves.
Complexity:O(n²) time from n × (2n - 1) characters; O(1) extra space.
One line: for each i, print digit on i == j and i == k, spaces elsewhere, then printf("\n").
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 scanf — 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.
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.
