A left-aligned descending number triangle starts every row at rows and counts down to the current i, so each line gets shorter from the right.
Rule: print rows..i (no leading spaces)
54321
5432
543
54
5 ← 5 rowsFollows the reverse descending triangle in Program 3; next is the ascending number triangle in Program 5.
One nested-loop idea: ascending outer bound, descending inner digits — tight digits with end="", then bare print().
| Method | Idea | Best for |
|---|---|---|
rows..i loops | Start each row at rows; stop at i | Learning, interviews, exams |
| User-input rows | Same logic with a variable height | Practice / demos |
for i from 1 to rows:
for j from rows down to i:
print j
print newline| Goal | Pattern |
|---|---|
| Raise the stop value | for i in range(1, rows + 1): |
Print rows..i | for j in range(rows, i - 1, -1): print(j, end="") |
| Spaced digits | print(j, end=" ") |
| End the row | print() |
| API | Effect | Use for |
|---|---|---|
print(j, end="") | Stays on the same line | Each digit |
print() | Ends the current line | After the inner loop |
Print digits without a newline, then end the row once.
Change the row count and the shrinking descending triangle updates instantly.
rows = 4Trace how the stop value i shortens each row while every line still starts at 4.
i | Digits j | Printed row |
|---|---|---|
1 | 4 3 2 1 | 4321 |
2 | 4 3 2 | 432 |
3 | 4 3 | 43 |
4 | 4 | 4 |
Row length is rows - i + 1. The first digit is always rows.
Three complete programs: fixed rows = 5, input() variant, and a compact rows = 3 demo. Use View Output to reveal sample results.
rows = 5Ascending outer loop; each row prints from 5 down to the current i.
rows = 5
for i in range(1, rows + 1):
for j in range(rows, i - 1, -1):
print(j, end="")
print()1. Outer loop. i grows from 1 to 5 — the inner stop value rises, so rows get shorter.
2. Inner loop. j always starts at 5 and counts down to i — every row begins with 5.
3. Newline. Bare print() after the digits starts the next shorter row.
Read rows with input(), validate, then use the same rows..i logic.
try:
rows = int(input("Enter rows: "))
except ValueError:
print("Please enter a positive whole number.")
raise SystemExit(1)
if rows < 1:
print("Please enter a positive whole number.")
raise SystemExit(1)
for i in range(1, rows + 1):
for j in range(rows, i - 1, -1):
print(j, end="")
print()1. Prompt and validate. Catch ValueError; require rows >= 1 before printing.
2. Same core. rows..i matches Example 1 — only rows comes from the user.
3. Safer input tip. Cap demos for readable single-digit output:
if rows < 1 or rows > 9:
print("Enter a whole number from 1 to 9.")
raise SystemExit(1)rows = 3Same structure with only three rows — easy to confirm every line still starts at rows.
rows = 3
for i in range(1, rows + 1):
for j in range(rows, i - 1, -1):
print(j, end="")
print()1. Same rules. Outer raises i; inner prints from 3 down to i.
2. Quick check. The last row is a single 3.
3. Scale up next. Once the small demo is clear, use Examples 1–2 for five rows or user input.
Check these before calling the solution done.
for i in range(rows, 0, -1) with j = i..1 prints 4321, 321… Keep ascending i and inner rows..i.
j down to 1 every rowThat reprints 54321 five times. Stop at range(rows, i - 1, -1).
If bare print() sits inside the inner loop, you get one digit per line. Call it only after the loop.
Default print(j) adds a newline. Use end="" for tight digits (or end=" " only if you want spaces).
Output is just 1. A good sanity check for input validation.
ValueErrorBare int(input()) crashes on non-numeric text — wrap it in try/except ValueError.
| Program | Time | Extra space |
|---|---|---|
| Fixed / input (Examples 1–2) | O(n²) | O(1) |
| Compact rows = 3 (Example 3) | O(n²) | O(1) |
Row lengths are n + (n−1) + … + 1 = n(n+1)/2 → O(n²) time. Only a few loop variables are needed.
rows..i — always starts at rows.i each row so the stop value climbs and the line shortens.print() advances after each row.O(n²) time; O(1) extra space.One line: for each i from 1 to rows, print rows..i with no leading spaces.
Each row starts at rows and counts down to i. Row i prints rows - i + 1 digits — total prints = n(n+1)/2; output is left-aligned with no leading spaces.
Continue with a triangle that counts upward on each growing row.
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