An increasing number triangle from 0 uses zero-based loops: row i prints i + 1 values of i + j — so the first cell is 0.
Rule: for i from 0 to max_n
for j from 0 to i
print (i + j) and a space
0
1 2
2 3 4
3 4 5 6
4 5 6 7 8
5 6 7 8 9 10 ← max_n = 5Follows the triangle from 1 in Program 33; next is the right-aligned incremental triangle in Program 35.
Outer loop runs i = 0..max_n. Inner loop prints j = 0..i with the formula i + j, separated by spaces.
| Method | Idea | Best for |
|---|---|---|
Formula i + j | When j = 0, value equals i — row starts at its index | Learning, interviews, exams |
| Compare with Program 33 | Same shape; zero-based loops replace i + j - 1 | Seeing how indexing changes the start |
for i from 0 to max_n:
for j from 0 to i:
print (i + j) and a space
print newline| Goal | Pattern |
|---|---|
| Grow each row | for i in range(0, max_n + 1): |
| Print i+1 values | for j in range(0, i + 1): |
| Start at 0 | print(i + j, end=" ") |
| Row starts at i | When j = 0, i + j = i |
| End of row | print() |
| API | Effect | Use for |
|---|---|---|
print(value, end=" ") | Stays on the same line | Each number on the row |
print() | Ends the current line | After the inner loop finishes |
Print values without a newline, then end the row once.
Change the max i and the zero-based triangle updates instantly — capped at 9 for readable demos.
max_n = 5Trace how each (i, j) pair maps to i + j.
i | Values | Count | Prints |
|---|---|---|---|
0 | 0 | 1 | 0 |
1 | 1, 2 | 2 | 1 2 |
2 | 2, 3, 4 | 3 | 2 3 4 |
3 | 3..6 | 4 | 3 4 5 6 |
4 | 4..8 | 5 | 4 5 6 7 8 |
5 | 5..10 | 6 | 5 6 7 8 9 10 |
Total numbers = 1 + 2 + … + (max_n+1) = (max_n+1)(max_n+2)/2 → O(n²).
Three complete programs: fixed max_n = 5, input() variant, and a compact max_n = 2 demo. Use View Output to reveal sample results.
max_n = 5Hard-coded max — zero-based loops print i + j with a trailing space.
for i in range(0, 6):
for j in range(0, i + 1):
print(i + j, end=" ")
print()1. Outer grows from 0. Row i prints exactly i + 1 numbers.
2. Formula fills the cells. i + j makes row i start at i and count upward.
3. Newline once. Call bare print() only after the inner loop finishes.
Read max_n with input(), validate, then use the same formula core.
try:
max_n = int(input("Enter max i: "))
except ValueError:
print("Please enter a non-negative integer.")
raise SystemExit(1)
if max_n < 0:
print("Please enter a non-negative integer.")
raise SystemExit(1)
for i in range(0, max_n + 1):
for j in range(0, i + 1):
print(i + j, end=" ")
print()1. Prompt and validate. Catch ValueError; reject negative values before printing.
2. Same core. i + j matches Example 1 — only max_n comes from the user.
3. Safer input tip. Cap demos for readable output:
if max_n < 0 or max_n > 9:
print("Enter a whole number from 0 to 9.")
raise SystemExit(1)max_n = 2Same structure with only three rows — easy to confirm zero-based indexing on paper.
max_n = 2
for i in range(0, max_n + 1):
for j in range(0, i + 1):
print(i + j, end=" ")
print()1. Three rows. i = 0 → 0; i = 1 → 1 2; i = 2 → 2 3 4.
2. Trace on paper. If you start at i = 1, the leading 0 disappears and the shape matches Program 33 less cleanly.
3. Scale up next. Once the small demo is clear, use Examples 1–2 for max 5 or user input.
Check these before calling the solution done.
Starting at i = 1 or using i + j - 1 drops the leading 0. Keep range(0, max_n + 1) with i + j.
Inner range(0, max_n + 1) makes every row the same width. Use range(0, i + 1).
Without end=" ", values run together (12). Always print a trailing space.
If bare print() sits inside the inner loop, you get one number per line. Call it only after the loop.
Output is just 0. 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 max = 2 (Example 3) | O(n²) | O(1) |
Row i prints i + 1 numbers. Total = 1 + 2 + … + (n+1) = (n+1)(n+2)/2 for max_n = n → O(n²) time. Only a few loop variables are needed.
i + j — first cell is 0 when both loops start at 0.j = 0..i so row i has i + 1 numbers.end=" " in the inner loop; bare print() after it finishes.O(n²) time from (n+1)(n+2)/2 prints; O(1) extra space.One line: for each i from 0 to max_n, print i + j for j = 0..i with spaces, then print().
Each printed value is computed as i + j. With i = 0 and j = 0 the first row prints 0; row i = 2 prints 2, 3, 4 — a zero-based left-shifted increasing triangle.
Continue with a right-aligned triangle that counts continuously across rows.
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