A left-shifted odd number triangle prints only odd values from the row start up to max_n, stepping by 2 — so you see 13579, 3579, 579, 79, 9.
Rule: for i = 1..max step 2, print j = i..max step 2
13579
3579
579
79
9 ← max = 10In Python both loops use range(..., 2) from an odd start: print(j, end="") prints each odd digit, then bare print() ends the row. Program 18 alternates odd and even rows next.
Outer loop from 1 to max_n stepping by 2. Inner loop from current i to max_n stepping by 2. End each row with print().
| Method | Idea | Best for |
|---|---|---|
| Odd + step 2 | Both loops use range(..., 2) from an odd start | Learning, interviews, exams |
| Even mirror | Start both loops at 2 | Comparing odd-only vs even-only |
for i from 1 to max step 2:
for j from i to max step 2:
print j (no newline)
print newline| Goal | Pattern |
|---|---|
| Outer (odd starts) | for i in range(1, max_n + 1, 2): |
| Inner (odds to max) | for j in range(i, max_n + 1, 2): print(j, end="") |
| End the row | print() |
| Force odd max | if max_n % 2 == 0: max_n -= 1 |
| Even mirror | for i in range(2, max_n + 1, 2): |
| API | Effect | Use for |
|---|---|---|
print(j, end="") | Stays on the same line | Each odd digit |
print() | Ends the current line | After the inner loop |
Print all digits without a newline, then end the row once.
Change the maximum and the left-shifted odd triangle updates instantly.
Trace three outer values when max_n = 10 — watch the start shift while only odds print.
i | Inner j | Length | Printed row |
|---|---|---|---|
1 | 1, 3, 5, 7, 9 | 5 | 13579 |
3 | 3, 5, 7, 9 | 4 | 3579 |
5 | 5, 7, 9 | 3 | 579 |
9 | 9 | 1 | 9 |
10 never prints because the loops stay on odd values. Raising i by 2 drops one odd digit from the left each row.
Three complete programs: fixed max, input() with odd adjust, and an even-number mirror. Use View Output for sample results.
max_n = 10Hard-coded maximum — both loops step by 2 from an odd start.
max_n = 10
for i in range(1, max_n + 1, 2):
for j in range(i, max_n + 1, 2):
print(j, end="")
print()1. Outer picks the start. i runs 1, 3, 5, 7, 9 — that value is where each row begins.
2. Inner prints odds to max. j steps from i to max_n by 2 with print(j, end="").
3. End the row. Call bare print() only after the inner loop finishes.
Read the maximum from the keyboard. Subtract 1 when the value is even so the bound is odd.
max_n = int(input("Enter the maximum value: "))
if max_n % 2 == 0:
max_n -= 1
for i in range(1, max_n + 1, 2):
for j in range(i, max_n + 1, 2):
print(j, end="")
print()1. Adjust even input. 8 becomes 7 so the last odd matches a clear odd bound.
2. Same loop core. The nested step-2 loops match Example 1.
3. Safer input tip. Validate before adjusting:
raw = input("Enter the maximum value: ").strip()
try:
max_n = int(raw)
except ValueError:
print("Enter a positive whole number.")
raise SystemExit(1)
if max_n < 1:
print("Enter a positive whole number.")
raise SystemExit(1)
if max_n % 2 == 0:
max_n -= 1Start both loops at 2 to print only even digits with the same left-shift shape.
max_n = 10
for i in range(2, max_n + 1, 2):
for j in range(i, max_n + 1, 2):
print(j, end="")
print()1. Same left-shift structure. Only the start value changes from 1 to 2.
2. Even values only. i and j visit 2, 4, 6, 8, 10.
3. Compare side by side. Seeing both odd and even mirrors makes the step-2 idea easy to remember.
Check these before calling the solution done.
2, not step 1range(i, max_n + 1) reprints consecutive integers, not an odd-only triangle.
if max_n % 2 == 0: max_n -= 1 keeps the last odd clear when reading input.
print(j) inside the inner loopThat puts each digit on its own line. Use print(j, end="") for digits; call bare print() only after the row finishes.
Output is just 1 — a good sanity check.
The outer loop never runs. Validate and prompt again for clearer UX.
max_n > 9print(11, end="") prints two characters. Labs often stick to max ≤ 10 for a clean single-digit look.
| Program | Time | Extra space |
|---|---|---|
| Odd triangle (Examples 1–2) | O(n²) | O(1) |
| Even mirror (Example 3) | O(n²) | O(1) |
About half the integers up to n are visited, but nested loops still make total work O(n²).
i, print odds from i to max_n with step 2.print() advances after each row.One line: start at an odd i, print odds through max_n with step 2, then raise the start.
Both loops step by 2 with range(..., 2), so only odd numbers print. Each row starts at a larger odd value, so the triangle shifts left — still O(n²) for maximum n.
Continue with rows that flip between odd and even number sequences.
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