An odd-length descending number triangle prints digits 1..i on each row, but the outer loop steps by 2 so only odd widths appear.
Remember
Rule: for i = max..1 step −2, print 1..i
1234567
12345
123
1 ← max = 7 (odd widths only)
In C you solve it with for (i = max; i >= 1; i -= 2) and an inner loop that prints 1..i with printf("%d", j), then printf("\n") ends the row. Change the step to i-- if you want every width (Program 1 style).
Approach
How to Solve It
Outer loop steps by 2 for odd widths only. Inner loop prints 1..i. End each row with printf("\n").
Method
Idea
Best for
Step by 2
Outer visits odd widths only; inner prints 1..i
Learning loop steps, this pattern
Step by 1
Same inner loop; outer uses i-- for every width
Contrast / full descending triangle
Pseudocode
Pseudocode
for i from max down to 1 step −2:
for j from 1 to i:
print j (no newline)
print newline
Cheat sheet
Goal
Pattern
Odd widths only
for (i = max; i >= 1; i -= 2)
Print digits 1..i
for (j = 1; j <= i; j++) printf("%d", j);
End the row
printf("\n");
Force odd max
if (max % 2 == 0) max -= 1;
Every width
for (i = max; i >= 1; i--)
Printing Numbers vs Starting a New Line
API
Effect
Use for
printf("%d", j)
Stays on the same line
Each digit j
printf("\n")
Ends the current line
After the inner loop
Print all digits without a newline, then end the row once.
Try it
Live Preview
Change the maximum width and the odd-length triangle updates instantly. Even values are adjusted down to the nearest odd number.
Whole numbers from 1 to 9. Even input becomes the next lower odd max (e.g. 6 → 5).
Live resultmax 7 · 16 digits
1234567
12345
123
1
Trace
Worked Walkthrough — max = 7
Trace each outer-loop value of i (stepping by 2) and the ascending digits printed.
i
Inner j
Printed row
Digits
7
1..7
1234567
7
5
1..5
12345
5
3
1..3
123
3
1
1..1
1
1
Total digits are 7+5+3+1 = 16 — the square ((max+1)/2)² for odd max.
Code
C Programs
Three complete programs: fixed odd max, scanf input with odd forcing, and a step-by-1 contrast. Use View Output for sample results.
Example 1 — Fixed max = 7
Outer loop steps by 2 for odd widths; inner loop prints digits 1..i.
C
#include <stdio.h>
int main(void)
{
int i, j;
for (i = 7; i >= 1; i -= 2)
{
for (j = 1; j <= i; j++)
printf("%d", j);
printf("\n");
}
return 0;
}
Output
1234567
12345
123
1
How It Works
1. Step by 2.for (i = 7; i >= 1; i -= 2) yields 7, 5, 3, 1 — even widths never run.
2. Print digits. When i = 7, the inner loop prints 1234567; when i = 1, just 1.
3. Newline.printf("\n") after the inner loop starts the next shorter odd-length row.
Example 2 — User Input
Read the maximum with scanf; force it odd if the user enters an even number.
C
#include <stdio.h>
int main(void)
{
int max;
int i, j;
printf("Enter an odd maximum (e.g., 9): ");
if (scanf("%d", &max) != 1 || max < 1)
{
printf("Please enter a positive integer.\n");
return 1;
}
if (max % 2 == 0)
max -= 1;
for (i = max; i >= 1; i -= 2)
{
for (j = 1; j <= i; j++)
printf("%d", j);
printf("\n");
}
return 0;
}
Output (when user enters 6)
Enter an odd maximum (e.g., 9): 6
12345
123
1
How It Works
1. Validate and force odd. Require scanf success; if (max % 2 == 0) max -= 1 turns 6 into 5.
2. Same nested loops. With max = 5, rows are 12345, 123, and 1.
3. Safer demos. Cap at single-digit max if you want a tight look:
Safer input
if (scanf("%d", &max) != 1 || max < 1 || max > 9)
{
printf("Enter a whole number from 1 to 9.\n");
return 1;
}
Example 3 — Step by 1 (Contrast)
Change i -= 2 to i-- to print every width — a full descending triangle.
C
#include <stdio.h>
int main(void)
{
int max = 5;
int i, j;
for (i = max; i >= 1; i--)
{
for (j = 1; j <= i; j++)
printf("%d", j);
printf("\n");
}
return 0;
}
Output
12345
1234
123
12
1
How It Works
1. Every width.i-- visits 5, 4, 3, 2, 1 — including even lengths that Example 1 skips.
2. Same inner loop. Digits are still 1..i; only the outer step changed. Keep i -= 2 for the odd-only lesson.
Edge Cases & Pitfalls
Check these before calling the solution done.
wrong step
i-- instead of i -= 2
That prints every width (Example 3). Keep stepping by 2 for odd-only rows.
even max
Starting at an even maximum
With max = 6 and no adjustment, the first row is even-length. Force odd with max -= 1 when needed.
\n inside
printf("\n") inside the inner loop
That puts each digit on its own line. Print digits without a newline; call printf("\n") only after the row finishes.
max = 1
Single digit
Output is just 1 — a good sanity check.
max ≤ 0
Empty output
The outer loop never runs. Validate and prompt again for clearer UX.
multi-digit
max > 9
printf("%d", 10) prints two characters, so columns drift. Labs usually stick to odd max values up to 9.
Analysis
Time and Space Complexity
Program
Time
Extra space
Odd-only loops (Examples 1–2)
O(n²)
O(1)
Step by 1 (Example 3)
O(n²)
O(1)
Odd widths print 1+3+…+n = ((n+1)/2)² digits for odd n — still quadratic in the maximum width.
Remember
Key Takeaways
Rule: for odd i from max down to 1 stepping by 2, print 1..i.
Step matters:i -= 2 skips even widths; i-- prints every width.
printf vs newline: digits stay on the line; printf("\n") advances after each row.
Next step: Program 15 builds an alternating binary number triangle.
One line: for each odd width from max down to 1, print 1..i with printf, then printf("\n").
Frequently Asked Questions
The outer loop uses i -= 2, so it visits only odd widths: 7, 5, 3, 1. Even lengths like 6, 4, 2 are skipped.
Because i -= 2 skips even row lengths. After 1234567 (7 digits), the next row is 5 digits (12345), not 6.
printf("%d", j) stays on the same line. printf("\n") ends the current line. Digits use printf; the row break uses printf("\n") after the inner loop.
Change the outer loop step to i-- (or i -= 1). Then you print 7, 6, 5, 4, 3, 2, 1 — every width.
Program 13 alternates ascending/descending direction with i % 2 (12345, 4321, 123, 21, 1). Program 14 always prints 1..i but only for odd row lengths using i -= 2.
O(n²) where n is the maximum row width. You print about 1+3+5+…+n odd-width digits, which is still O(n²).
Subtract 1 to make it odd (e.g. if max % 2 == 0 then max -= 1) so the first row stays odd-length, or document that even input shifts the pattern.
Check scanf's return value: if (scanf("%d", &max) != 1) handle bad input. Unchecked scanf leaves max uninitialized on failure.
The outer loop never runs, so nothing is printed. Validate and prompt again if you want a clear user message.
🤔
Did you know?
Only odd-length rows print. The outer loop uses i −= 2 (7, 5, 3, 1) and the inner loop prints 1..i — still O(n²) total digit prints for maximum width n.
