Shape Rule
j % 2 alternates 0 and 1
Row 1 prints 1, row 2 prints 10, row 3 prints 101, and so on as width grows.

The alternating binary number triangle with an ascending inner loop prints each row starting with 1 and alternating 0/1 as width grows. This tutorial covers the shape rule, loop structure, a live preview, algorithm steps, worked C examples, edge cases, and complexity.
j % 2 alternates 0 and 1
Row 1 prints 1, row 2 prints 10, row 3 prints 101, and so on as width grows.
Rows
for (i = 1; i <= rows; i++) makes each new row one digit longer than the previous.
1..i ascending
for (j = 1; j <= i; j++) prints j % 2 while counting up, so every row starts with 1.
Same line / next line
Binary digits use printf("%d", j % 2); end each row with printf("\n").
1–20 rows
Pick a row count and draw the ascending-inner binary triangle instantly in the browser.
Complexity
Total digit prints still = n(n+1)/2; extra memory stays O(1).
An alternating binary number triangle (starting with 1) grows each row by one digit while alternating between 0 and 1 using the modulo operator. With rows = 5, the output is 1, 10, 101, 1010, 10101.
In C you solve it with an ascending outer loop and an ascending inner loop: for (j = 1; j <= i; j++) prints j % 2, then printf("\n") ends each row.
It is a natural follow-up to Program 15 — same modulo idea, different inner-loop direction.
j % 2 yields 0 for even j, 1 for odd j.
j = 1 up to i makes every row start with 1.
printf("%d", j % 2) in the inner loop; printf("\n") after.
Follow Program 15 (descending inner); continue to Program 17 (left-shifted odd numbers).
In short: for each row i from 1 to rows, print j % 2 for j from 1 up to i, then call printf("\n").
Given a positive integer rows, print an alternating binary number triangle starting with 1: row i has i digits from j % 2 as j counts up from 1 to i.
// rows = 5 (conceptual shape)
// 1
// 10
// 101
// 1010
// 10101 | Item | Type | Description |
|---|---|---|
rows | int | Number of triangle lines to print (typically ≥ 1). |
| Printed output | text | Each row has i alternating binary digits from j % 2. |
for i from 1 to rows:
for j from 1 to i:
print j % 2 (no newline)
print newline | Approach | Idea | Best for |
|---|---|---|
Ascending inner + j % 2 | 1, 10, 101, … | Learning and interviews |
Flip with 1 - (j % 2) | Start rows with 0 instead of 1 | Parity inversion variant |
| Program 15 variant | Descending inner loop | Produces 1, 01, 101, … |
| Goal | Pattern |
|---|---|
| Walk each row | for (i = 1; i <= rows; i++) |
| Print binary digit | for (j = 1; j <= i; j++) printf("%d", j % 2); |
| End the row | printf("\n"); |
| Flip parity | printf("%d", 1 - (j % 2)); |
| Program 15 variant | for (j = i; j >= 1; j--) printf("%d", j % 2) (descending inner) |
| Row + column parity | printf("%d", (i + j) % 2); |
Same binary triangle family — inner-loop direction changes the row shape.
parityEven j → 0, odd j → 1
flippedInverts every digit — row 1 starts with 0
asc innerThis page — produces 1, 10, 101, …
compare 15Try Program 15’s descending inner loop next
Reach for this pattern when teaching the modulo operator inside nested loops.
Natural follow-up after Program 15 — same modulo, different inner-loop direction.
Outer/inner bound practice with an immediate visual check.
Combine loops with scanf for a flexible row count.
Compare Program 15 (descending inner loop) and Program 17 (left-shifted odd numbers) 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 alternating binary triangle in the browser.
Three complete C programs — fixed rows, a flip variant, and a user-input version. Click View Output to reveal sample console results.
Print five rows of the ascending-inner binary triangle with j % 2.
rows = 5Hard-coded height — ideal for first demos and screenshots.
#include <stdio.h>
int main() {
int rows = 5;
int i, j;
for (i = 1; i <= rows; ++i) {
for (j = 1; j <= i; ++j) {
printf("%d", j % 2);
}
printf("\n");
}
return 0;
} When i = 1, the inner loop prints 1 % 2 = 1. When i = 3, it prints 1%2=1, 2%2=0, 3%2=1 as 101, and so on as row width grows. printf("\n") after the inner loop starts the next row.
Invert parity so the first row starts with 0 instead of 1.
1 - (j % 2)Start each row with 0 instead of 1 by inverting the parity output.
#include <stdio.h>
int main() {
int rows = 5;
int i, j;
for (i = 1; i <= rows; ++i) {
for (j = 1; j <= i; ++j) {
printf("%d", 1 - (j % 2));
}
printf("\n");
}
return 0;
} 1 - (j % 2) flips every digit: where j % 2 was 1 it prints 0, and vice versa. Row 1 becomes 0 instead of 1.
Read the row count at runtime and scale the binary triangle.
Read rows with scanf("%d", &rows) and apply the same j % 2 logic.
#include <stdio.h>
int main() {
int rows;
int i, j;
printf("Enter the number of rows: ");
scanf("%d", &rows);
for (i = 1; i <= rows; ++i) {
for (j = 1; j <= i; ++j) {
printf("%d", j % 2);
}
printf("\n");
}
return 0;
} Same nested-loop core as Example 1; only the source of rows changes. Non-numeric input leaves rows unset if you ignore scanf’s return value — always check it in safer labs.
#include <stdio.h> brings in printf / scanf. Set rows (fixed or from input).
for (i = 1; i <= rows; i++) makes each row one digit longer than the previous.
for (j = 1; j <= i; j++) prints j % 2 with printf to alternate 0 and 1.
printf("\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 and note the j % 2 values printed as j counts up.
i | Inner j order | j % 2 values | Printed row |
|---|---|---|---|
1 | 1 | 1 | 1 |
2 | 1, 2 | 1, 0 | 10 |
3 | 1, 2, 3 | 1, 0, 1 | 101 |
4 | 1, 2, 3, 4 | 1, 0, 1, 0 | 1010 |
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: use (i + j) % 2 for row+column parity grids.
Practice printf vs row newline without complex math.
Example: put printf("\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 still → 55.
Pair the pattern with scanf return checks 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 ascending inner loop first; compare with Program 15’s descending inner loop to see how direction changes each row.
Small habits that keep number-pattern code clean.
Use rows (or n) and keep i/j for row/column — or rename to row/col.
scanfCheck the return value so bad input does not leave rows uninitialized.
Only call printf("\n") after the inner loop finishes the row.
Run both pages with the same rows to see how inner-loop direction changes output.
Trace rows = 5 on paper before coding larger demos.
Pro Tip: if the output is a vertical list of single digits per line, you almost certainly put printf("\n") inside the inner loop.
Mistakes that commonly break alternating binary number patterns.
Each digit lands on its own line — you get a column, not a triangle.
→ Use printf("%d", j % 2) for binary digits; printf("\n") only after the inner loop.
Counting j down instead of up produces Program 15’s shape (01 on row 2).
→ For this shape, keep for (j = 1; j <= i; j++).
Omitting printf("\n") glues every digit onto one endless line.
→ Always end the row after the inner loop.
Letters or empty input leave rows uninitialized.
→ Check scanf return value 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 i with wrong inner bound (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 scanf leaves rows unset — check the return value.
printf("%d", j) prints 1,2,3… — use printf("%d", j % 2) for binary output.
Try these variations to lock in the pattern.
j = i down to 11 - (j % 2) so row 1 starts with 0printf("%d", (i + j) % 2) for a checkerboard-style gridn rows.printf("%d", j % 2) stays on the line; printf("\n") advances — mix them carefully.rows > 0 for interactive programs; rows = 1 should print a single 1.Quick Takeaway: outer loop grows row length, inner loop prints j % 2 ascending, then break the line.
| Program | Time | Extra space |
|---|---|---|
| Nested loops (Examples 1–2) | O(rows²) | O(1) |
| User input (Example 3) | O(rows²) | O(1) |
The alternating binary number triangle with an ascending inner loop is a compact lesson in how loop direction changes output. Master the j % 2 version, then compare with Program 15’s descending inner loop.
Practice the three examples above, then continue to Program 17 for the left-shifted odd number triangle.
Use for (j = 1; j <= i; j++) with printf("%d", j % 2) — keep printf("\n") for the break, and validate row counts when reading input.
j = 1..i before codingprintf("%d", j % 2) for digits and printf("\n") after each rowrows ≥ 1 for interactive programsscanf return value before using rowsprintf("\n") inside the inner digit loopj directly instead of j % 2j down when you meant this page’s ascending inner looprows = 1 edge casePrint the pattern the beginner-friendly way.
j % 2 alternates 0 and 1
DefinitionCounts j up to i
Codej % 2 picks digit
LogicEnds each row
I/OO(n²) time
AnalysisEach row prints alternating 0 and 1 using j % 2. The inner loop counts up from 1 to i, so every row starts with 1 — still O(n²) total prints.
Move on to the left-shifted odd number triangle in the C number-pattern series.
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