An increasing number triangle from 1 prints i values on row i, each computed as i + j - 1 — so each row starts with its own row number.
Rule: for i from 1 to rows
for j from 1 to i
print (i + j - 1) and a space
1
2 3
3 4 5
4 5 6 7
5 6 7 8 9 ← rows = 5Follows the triangle from 11 in Program 32; next is the variant from 0 in Program 34.
Outer loop grows i. Inner loop prints i values with the formula i + j - 1, separated by spaces.
| Method | Idea | Best for |
|---|---|---|
Formula i + j - 1 | When j = 1, value equals i — row starts at its index | Learning, interviews |
| input() rows | Same loops; read rows at runtime | Interactive practice |
| Compact rows = 3 | Same structure; easier to trace by hand | Paper walkthrough |
for i from 1 to rows:
for j from 1 to i:
print (i + j - 1) and a space
print newline| Goal | Pattern |
|---|---|
| Grow each row | for i in range(1, rows + 1): |
| Print i values | for j in range(1, i + 1): |
| Start at 1 | print(i + j - 1, end=" ") |
| Row starts at i | When j = 1, i + j - 1 = i |
| End of row | print() |
| API | Effect | Use for |
|---|---|---|
print(value, end=" ") | Stays on the same line | Each number |
print() | Ends the current line | After the inner loop |
Print each value with a trailing space, then break once with print(). Putting print() inside the inner loop prints one number per line.
Change the row count and the increasing triangle updates instantly.
rows = 5Trace how each (i, j) pair maps to i + j - 1.
i | Values | Count | Prints |
|---|---|---|---|
1 | 1 | 1 | 1 |
2 | 2, 3 | 2 | 2 3 |
3 | 3, 4, 5 | 3 | 3 4 5 |
4 | 4..7 | 4 | 4 5 6 7 |
5 | 5..9 | 5 | 5 6 7 8 9 |
Total numbers = 1 + 2 + … + n = n(n+1)/2 → O(n²).
Three complete programs: fixed rows = 5, input() rows, and compact rows = 3. Use View Output for sample results.
rows = 5Hard-coded height — print i + j - 1 with a trailing space on each cell.
for i in range(1, 6):
for j in range(1, i + 1):
print(i + j - 1, end=" ")
print()1. Outer grows the width. Row i prints exactly i numbers.
2. Formula fills the cells. i + j - 1 makes row i start at i and count upward.
3. Newline once. Call print() only after the inner loop finishes.
Read rows at runtime, then use the same formula core.
rows = int(input("Enter rows: "))
if rows < 1:
print("Please enter a positive integer.")
else:
for i in range(1, rows + 1):
for j in range(1, i + 1):
print(i + j - 1, end=" ")
print()1. Read and validate. Parse the answer; reject non-positive values before looping.
2. Same core. i + j - 1 matches Example 1 — only rows comes from the user.
3. Safer input tip. Bare int(input()) raises ValueError on letters. Prefer:
raw = input("Enter rows: ").strip()
try:
rows = int(raw)
except ValueError:
print("Please enter a positive integer.")
raise SystemExit(1)
if rows < 1:
print("Please enter a positive integer.")
raise SystemExit(1)rows = 3Same structure with only three rows — easy to confirm the formula on paper.
rows = 3
for i in range(1, rows + 1):
for j in range(1, i + 1):
print(i + j - 1, end=" ")
print()1. Three rows. i = 1 → 1; i = 2 → 2 3; i = 3 → 3 4 5.
2. Trace on paper. If you use i + j instead of i + j - 1, every value shifts up by one.
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.
Using i + j or 9 + i + j shifts every value — rows no longer start at the row number. Keep i + j - 1.
Inner bound range(1, rows + 1) makes every row the same width. Use range(1, i + 1).
Without end=" ", values run together (23). Always print a trailing space.
If print() sits inside the inner loop, you get one number per line. Call it only after the loop.
Output is just 1. A good sanity check for input validation.
Letters crash bare int(input()) — wrap in try/except and require rows >= 1.
| Program | Time | Extra space |
|---|---|---|
| Examples 1–3 | O(n²) | O(1) |
Row i prints i numbers. Total = 1 + 2 + … + n = n(n+1)/2 → O(n²) time. Only a few loop variables are needed.
i + j - 1 — row i starts at i.j = 1..i so row i has i numbers.print() advances after each row.O(n²) time from n(n+1)/2 prints; O(1) extra space.One line: for each row i, print i + j - 1 for j = 1..i with spaces, then print().
Each printed value is computed as i + j - 1. Row i = 1 prints 1; row i = 4 prints 4, 5, 6, 7 — a left-shifted increasing triangle starting at 1.
Continue with the same shape using i + j with i starting at 0.
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