A diagonal mirror number diamond is Program 57’s inverse-V pyramid plus a mirrored bottom half — digit i only on the left and right diagonals, spaces everywhere else.
Remember
Rule: top i = 1..rows, bottom i = rows−1..1
same left/right diagonal print as Program 57
1
2 2
3 3
4 4
5 5
4 4
3 3
2 2
1 ← rows = 5 (9 lines)
Follows the diagonal mirror pyramid in Program 57; next is the hollow square border in Program 59.
Approach
How to Solve It
Reuse Program 57’s diagonal row logic twice: climb 1..rows, then descend rows-1..1 so the peak line appears once.
Method
Idea
Best for
Two outer loops
Top 1..rows, bottom rows-1..1
Learning, interviews, exams
Shared row body
Same left/right diagonal loops in both halves
Avoiding duplicated logic bugs
Pseudocode
Pseudocode
function print_row(i, rows):
for j from rows down to 1:
if i == j: print i else print space
for k from 2 to rows:
if i == k: print i else print space
print newline
for i from 1 to rows:
print_row(i, rows)
for i from rows - 1 down to 1:
print_row(i, rows)
for (k = 2; k <= rows; k++) printf(i == k ? "%d" : " ", k);
Total lines
2 * rows - 1
Printing Numbers vs Starting a New Line
API
Effect
Use for
printf("%d", i) / printf(" ")
Stays on the same line
Each diagonal digit or filler space
printf("\n")
Ends the current line
After both left and right loops
Print digits and spaces without a newline, then end the row once.
Try it
Live Preview
Change the half-height and the hollow diamond updates instantly — capped at 9 for readable single-digit demos.
Whole numbers from 1 to 9. Tap a chip or type a value — the preview redraws as you go.
Live resultrows = 5 · 9 lines
1
2 2
3 3
4 4
5 5
4 4
3 3
2 2
1
Trace
Worked Walkthrough — rows = 5
See how the two outer loops share one row body and why the bottom starts at rows - 1.
Half
i values
Lines produced
Top
1..5
1 → 2 2 → … → 5
Bottom
4..1
4 → 3 3 → … → 1
Total lines = 2×5−1 = 9. Starting the bottom at 5 would print the peak twice.
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 half-height — top pyramid, then mirrored bottom with the same diagonal logic.
C
#include <stdio.h>
int main(void)
{
int rows = 5;
int i, j, k;
for (i = 1; i <= rows; i++)
{
for (j = rows; j >= 1; j--)
printf(i == j ? "%d" : " ", j);
for (k = 2; k <= rows; k++)
printf(i == k ? "%d" : " ", k);
printf("\n");
}
for (i = rows - 1; i >= 1; i--)
{
for (j = rows; j >= 1; j--)
printf(i == j ? "%d" : " ", j);
for (k = 2; k <= rows; k++)
printf(i == k ? "%d" : " ", k);
printf("\n");
}
return 0;
}
Output
1
2 2
3 3
4 4
5 5
4 4
3 3
2 2
1
How It Works
1. Top half. Same as Program 57: left j = rows..1, right k = 2..rows.
2. Bottom half. Replay the same row body for i = rows - 1 down to 1.
3. One peak. Skipping i = rows on the way down keeps the middle line unique.
Example 2 — User Input Rows
Read half-height with scanf, validate, then run both outer loops.
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 = rows; j >= 1; j--)
printf(i == j ? "%d" : " ", j);
for (k = 2; k <= rows; k++)
printf(i == k ? "%d" : " ", k);
printf("\n");
}
for (i = rows - 1; i >= 1; i--)
{
for (j = rows; j >= 1; j--)
printf(i == j ? "%d" : " ", j);
for (k = 2; k <= rows; k++)
printf(i == k ? "%d" : " ", k);
printf("\n");
}
return 0;
}
Output (when user enters 4)
Enter the number of rows: 4
1
2 2
3 3
4 4
3 3
2 2
1
How It Works
1. Prompt and validate. Reject bad input before printing.
2. Same core. Top and bottom loops match Example 1 — only rows comes from the user.
3. Safer input tip. Cap demos for readable console output:
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
Five-line diamond — easy to confirm the bottom starts at 2, not 3.
C
#include <stdio.h>
int main(void)
{
int rows = 3;
int i, j, k;
for (i = 1; i <= rows; i++)
{
for (j = rows; j >= 1; j--)
printf(i == j ? "%d" : " ", j);
for (k = 2; k <= rows; k++)
printf(i == k ? "%d" : " ", k);
printf("\n");
}
for (i = rows - 1; i >= 1; i--)
{
for (j = rows; j >= 1; j--)
printf(i == j ? "%d" : " ", j);
for (k = 2; k <= rows; k++)
printf(i == k ? "%d" : " ", k);
printf("\n");
}
return 0;
}
Output
1
2 2
3 3
2 2
1
How It Works
1. Five lines. Top prints 1, 2 2, 3 3; bottom reprints 2 2 and 1.
2. Trace on paper. Confirm the peak 3 3 appears once — bottom starts at rows - 1.
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.
i = rows..1
Double peak
If the bottom loop starts at rows, the widest line prints twice. Always start at rows - 1.
k = 1
Triple digit on a row
If the right loop starts at 1, the center column can print a third digit. Keep k = 2.
\n inside
Broken rows
If printf("\n") sits inside an inner loop, each character lands on its own line. Call it only after both diagonal loops finish.
rows = 1
Single line
Output is just 1 — the bottom loop (rows-1..1) does not run.
rows > 9
Multi-digit width
Digits 10+ break neat column alignment with %d. Cap demos at 9 or use a fixed field width.
scanf
Check the return value
Validate scanf before looping — a failed read leaves rows uninitialized.
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)
You print 2n−1 lines, each scanning about 2n−1 positions → O(n²) time. Only a few loop variables are needed.
Remember
Key Takeaways
Pyramid + mirror: top 1..rows, bottom rows-1..1, same diagonal row body.
Skip the peak twice: bottom starts at rows - 1.
Two diagonals per row: print i when i == j or i == k; spaces elsewhere.
Complexity:O(n²) time from 2n−1 lines × ~2n positions; O(1) extra space.
One line: Program 57 once upward, then again from rows-1 down to 1.
Frequently Asked Questions
Each row prints the same digit on the left diagonal and the right diagonal; spaces fill the remaining positions.
The top half already printed the peak row — starting at rows-1 avoids duplicating the middle line.
An inverse-V pyramid from 1 to rows, then mirrored back down to 1 — 2*rows-1 lines total.
Program 57 prints only the top pyramid half. Program 58 adds a second outer loop from rows-1 down to 1 for the bottom half.
Each column prints either the row digit or a space — those conditions pick exactly the two diagonal positions.
Change rows or read it from user input with scanf — see Example 2.
O(n²) for n rows because you print 2n-1 lines, each scanning about 2n positions.
Yes. Replace the digit printf with printf("*") in both diagonal print branches.
🤔
Did you know?
Print the Program 57 pyramid for the top half, then mirror with for (i = rows-1; i >= 1; i--). Total lines = 2×rows-1 — each row scans about 2×rows-1 positions.
