0-Centered Descending Mirror Number Pattern in C

What You’ll Learn

The 0-centered descending mirror pattern prints 0, 909, 89098, … up to 1234567890987654321 — a natural step after the palindrome triangle in Program 27. This tutorial covers three nested loops, a fixed zero center, a live preview, algorithm steps, worked C examples, edge cases, and complexity.

📝 Problem & Approach

Given max digit 9, print 10 rows of a 0-centered mirror: for each descending i, print i..9, then 0, then 9..i on the same line.

// max = 9 (conceptual shape)
// 0
// 909
// 89098
// 7890987
// ...
// 1234567890987654321

Inputs & Outputs

Minimal workflow

for i from max+1 down to 1:
    for j from i to max:
        print j
    print 0
    for k from max down to i:
        print k
    print newline

Approach comparison

🔮 Live Preview

Choose a max digit between 3 and 9 and draw the 0-centered mirror pattern in the browser.

Press "Draw pattern".

Examples Gallery

Three complete C programs — fixed max digit, custom max input, and spaced output variant. Click View Output to reveal sample console results.

#include <stdio.h>

int main() {
    int i, j, k;

    for (i = 10; i >= 1; --i) {
        for (j = i; j < 10; ++j)
            printf("%d", j);

        printf("0");

        for (k = 9; k >= i; --k)
            printf("%d", k);

        printf("\n");
    }

    return 0;
}

0
909
89098
7890987
678909876
56789098765
4567890987654
345678909876543
23456789098765432
1234567890987654321

#include <stdio.h>

int main() {
    int max;
    int i, j, k;

    printf("Enter max digit (1-9): ");
    scanf("%d", &max);
    if (max < 1) max = 1;
    if (max > 9) max = 9;

    for (i = max + 1; i >= 1; --i) {
        for (j = i; j <= max; ++j)
            printf("%d", j);

        printf("0");

        for (k = max; k >= i; --k)
            printf("%d", k);

        printf("\n");
    }

    return 0;
}

Enter max digit (1-9): 4
0
404
34043
2340432
123404321

#include <stdio.h>

int main() {
    int max = 9;
    int i, j, k;

    for (i = max + 1; i >= 1; --i) {
        for (j = i; j <= max; ++j)
            printf("%d ", j);

        printf("0 ");

        for (k = max; k >= i; --k)
            printf("%d ", k);

        printf("\n");
    }

    return 0;
}

0 
9 0 9 
8 9 0 9 8 
7 8 9 0 9 8 7 

🔎 Worked Walkthrough — max = 9 (selected rows)

Trace selected outer-loop values of i, the left half, center, right half, and full row output.

When i = max + 1, both inner loops are empty — only 0 prints. Each row grows as i decreases.

🎯 Practice Problems

Try these variations to lock in the pattern.

Notes

⏱️ Time and Space Complexity

💡 Best Practices