Shape Rule
1..i digits on row i
Row rows prints 1..rows, then one fewer digit each line, ending with a single 1.

The descending number triangle pattern is the classic first console pattern: nested loops, cout vs newline, and a clear visual result. This tutorial covers the shape rule, loop structure, a live preview, algorithm steps, worked C++ examples, edge cases, and complexity.
1..i digits on row i
Row rows prints 1..rows, then one fewer digit each line, ending with a single 1.
Rows
for (i = rows; i >= 1; i--) walks each line from the longest down to one digit.
Digits
for (j = 1; j <= i; j++) prints digits 1 through i on that row.
Same line / next line
Digits use cout << j; end each row with cout << "\n".
1–20 rows
Pick a row count and draw the descending number triangle instantly in the browser.
Complexity
Total digit prints = n(n+1)/2; extra memory stays O(1).
A descending number triangle pattern starts with the longest row and loses one digit per line. Each row prints consecutive digits from 1 up to the current row length, shrinking from top to bottom.
In C++ you usually solve it with two nested for loops: the outer loop picks the row, the inner loop prints digits 1..i on that row, then cout << "\n" moves to the next line.
It is the standard first pattern exercise in C++ courses. Once nested loops and cout click, pyramids, diamonds, and hollow shapes become much easier.
On row i, print digits 1 through i.
Outer counts down rows; inner prints digits.
cout << j in the inner loop; cout << "\n" after.
Gateway to left-shifted, pyramid, and hollow patterns.
In short: for each row i from rows down to 1, print digits 1..i with cout << j, then call cout << "\n".
Given a positive integer rows, print a descending number triangle: each row i shows digits 1 through i, with the outer loop counting from rows down to 1.
// First 5 rows (conceptual shape)
// 12345
// 1234
// 123
// 12
// 1 | Item | Type | Description |
|---|---|---|
rows | int | Number of triangle lines to print (typically ≥ 1). |
| Printed output | text | Each row prints 1..i; the top row has rows digits, the bottom row has one digit. |
for i from rows down to 1:
for j from 1 to i:
print j (no newline)
print newline | Approach | Idea | Best for |
|---|---|---|
| Nested loops | Outer rows + inner digits | Learning and interviews |
row += char('0' + j) loop | Build a whole row in one call | Shorter production-style demos |
| Goal | Pattern |
|---|---|
| Walk each row | for (i = rows; i >= 1; i--) |
Print digits 1..i | for (j = 1; j <= i; j++) cout << j; |
| End the row | cout << "\n"; |
| One-line row shortcut | row += char('0' + j); cout << row; |
| Ascending variant | for (i = 1; i <= rows; i++) (grows upward) |
Same triangle — different ways to emit characters.
same linePrints a digit without moving to the next line
new lineEnds the current row after all digits are printed
whole rowBuilds digits 1..i in a string, then prints once
loops firstMaster nested loops before the string shortcut
Reach for this triangle when teaching or testing nested-loop basics.
Most C++ pattern series start here before pyramids and diamonds.
Outer/inner bound practice with an immediate visual check.
Combine loops with cin for a flexible row count.
Try left-shifted starts (Program 2), pyramids, or spaced digits next.
This is a console teaching pattern — not how you build modern app screens.
Key benefit: one small program that locks in nested loops, output sequencing, and O(n²) thinking.
Choose a row count between 1 and 20 and draw the descending number triangle in the browser.
Three complete C++ programs — fixed row count, cin input, and a string row shortcut. Click View Output to reveal sample console results.
Print five rows of the descending number triangle with nested loops.
rows = 5Hard-coded height — ideal for first demos and screenshots.
#include <iostream>
using namespace std;
int main() {
int rows = 5;
int i, j;
for (i = rows; i >= 1; --i) {
for (j = 1; j <= i; ++j) {
cout << j;
}
cout << "\n";
}
return 0;
} When i = 5, the inner loop prints 12345. When i = 4, it prints 1234, and so on until i = 1 prints 1. cout << "\n" after the inner loop starts the next row.
Let the user choose the height at runtime.
Read the row count with cin >> rows (check cin.fail() in real apps).
#include <iostream>
using namespace std;
int main() {
int rows;
int i, j;
cout << "Enter the number of rows: ";
cin >> rows;
for (i = rows; i >= 1; --i) {
for (j = 1; j <= i; ++j) {
cout << j;
}
cout << "\n";
}
return 0;
} Same nested-loop core as Example 1; only the source of rows changes. Non-numeric input sets cin’s fail bit if you ignore errors — always validate in safer labs.
Same shape by building each row in a string before printing.
string row builderAppend each digit to a string, then print the full row with cout.
#include <iostream>
#include <string>
using namespace std;
int main() {
int rows = 5;
int i, j;
for (i = rows; i >= 1; --i) {
string row;
for (j = 1; j <= i; ++j) {
row += char('0' + j);
}
cout << row << "\n";
}
return 0;
} row += char('0' + j) appends one digit when j is 1..9. Same nested-loop structure as Example 1; a handy alternative that builds each line before printing. Keep either style for exams that want both loop bounds visible.
#include <iostream> brings in cout / cin. Set rows (fixed or from input).
for (i = rows; i >= 1; i--) selects the current line, starting at the widest.
for (j = 1; j <= i; j++) prints digits 1..i with cout << j.
cout << "\n" ends the row so the next outer iteration starts fresh.
Total digit prints: 1+2+…+n = n(n+1)/2 — O(n²) time, O(1) extra memory.
rows = 4Trace each outer-loop value of i (counting down) and count how many digits the inner loop prints.
i | Inner j range | Printed row | Digits this row |
|---|---|---|---|
4 | 1..4 | 1234 | 4 |
3 | 1..3 | 123 | 3 |
2 | 1..2 | 12 | 2 |
1 | 1..1 | 1 | 1 |
Total digit prints: 1 + 2 + 3 + 4 = 10 = 4×5/2.
Where this tiny pattern (and its loop structure) shows up beyond the homework prompt.
Clearest visual proof that outer and inner bounds interact.
Example: change j <= i and watch the shape change.
Foundation for inverted, pyramid, diamond, and hollow variants.
Example: Program 2 changes the inner start value each row.
Practice cout vs row newline without complex math.
Example: put cout << "\n" inside the inner loop by mistake.
Swap digits for letters, stars, or spaced output once the loop works.
Example: print j + " " for spaced digits on each row.
Triangular totals make O(n²) concrete for beginners.
Example: count printed digits for n = 10 → 55.
Pair the pattern with cin and positive-row checks.
Example: reject rows <= 0 and re-prompt.
Pro Tip: when an interviewer asks for patterns, explain the outer/inner roles first — then write the loops. The story matters as much as the code.
Why this pattern earns a permanent spot in beginner C courses.
Wrong bounds show up immediately as a broken staircase.
Only loops and console output — no arrays or math libraries.
Invert, center, hollow, or change the fill character with small edits.
Streaming output needs no storage beyond loop counters.
Pro Tip: learn the nested-loop version first; treat row += char('0' + j) loop as a polish shortcut afterward.
Small habits that keep number-pattern code clean.
Use rows (or n) and keep i/j for row/column — or rename to row/col.
cin Return ValueAvoid using uninitialized rows when the user types letters instead of a number.
Only call cout << "\n" after the inner loop finishes the row.
rows..1 on the outer loop with j <= i matches “row i prints digits 1..i” naturally.
Trace rows = 3 on paper before coding larger demos.
Pro Tip: if the output is a vertical list of single digits, you almost certainly put cout << "\n" inside the inner loop.
Mistakes that commonly break descending number patterns.
Each digit lands on its own line — you get a column, not a triangle.
→ Use cout << j for digits; cout << "\n" only after the inner loop.
j <= rows prints a rectangle; wrong outer bounds flatten or invert the shape.
→ For this shape, keep j <= i.
Omitting cout << "\n" glues every digit onto one endless line.
→ Always end the row after the inner loop.
Letters or empty input leave rows unset if you ignore cin errors.
→ Check cin.fail() after cin >> rows and re-prompt on failure.
Switching to i = 0 without adjusting the inner bound prints an empty first row or wrong counts.
→ If 0-based, print digits 1..i+1 (e.g. j <= i + 1).
Check these inputs before calling the solution done.
Output is just 1 on one line.
Outer loop never runs — print nothing or show a message.
rows < 0Treat as invalid; re-prompt instead of silent empty output.
Output grows as n²/2 characters — fine for labs, noisy for huge n.
Unchecked cin leaves rows unset — check cin.fail().
Try cout << j << " " for spaces between numbers.
Try these variations to lock in the pattern.
for (i = 1; i <= rows; i++) as outer loopcin until rows >= 1cout << j << " " between digitsn(n+1)/2 — hence O(n²) time.cout << j stays on the line; cout << "\n" advances — use both deliberately.rows > 0 for interactive programs; rows = 1 should print a single 1.Quick Takeaway: outer loop picks the row, inner loop prints digits 1..i, then break the line — that is the whole pattern.
| Program | Time | Extra space |
|---|---|---|
| Nested loops (Examples 1–2) | O(rows²) | O(1) |
row += char('0' + j) loop (Example 3) | O(rows²) | O(rows) per row string (temporary) |
The descending number triangle pattern is a small nested-loop exercise with lasting payoff: row/column thinking, cout vs row newline, and O(n²) intuition. Master the classic two-loop version, then optionally shorten rows with row += char('0' + j) loop.
Practice the three examples above, then continue to Program 2 for a left-shifted descending triangle.
Row i prints 1..i — keep cout << j for digits and cout << "\n" for the break, and validate row counts when reading input.
cout << j for digits and cout << "\n" after each rowrows ≥ 1 for interactive programscin.fail() before using rowscout << "\n" inside the inner digit looprows = 1 edge casePrint the triangle the beginner-friendly way.
Row i prints 1..i
DefinitionControls each row
CodePrints digits with cout << j
CodeEnds each row
I/OO(n²) time
AnalysisRow i prints digits 1 through i. The outer loop counts down from rows, so the first line is longest and each row shortens by one digit — still O(n²) total prints.
Shift the start digit each row — 12345, 2345, 345, 45, 5.
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