Number-Star Diamond Pattern in Python

What You’ll Learn

The number-star diamond prints 1, 2*2, 3*3*3, … 5*5*5*5*5, then mirrors back down — a natural step after the right-aligned triangle in Program 30. This tutorial covers two outer loops, modulus alternation, a live preview, worked Python examples, edge cases, and complexity.

📝 Problem & Approach

Given n = 5, print a number-star diamond: top half i = 1..n, bottom half i = n-1..1, each row alternating digit i and * via j % 2.

# n = 5 (conceptual shape)
for i in range(1, n + 1):
    for j in range(1, i * 2):
        if j % 2 == 0:
            print("*", end="")
        else:
            print(i, end="")
    print()

for i in range(n - 1, 0, -1):
    for j in range(1, i * 2):
        if j % 2 == 0:
            print("*", end="")
        else:
            print(i, end="")
    print()

Inputs & Outputs

Minimal workflow

for i from 1 to n:
    for j from 1 to i*2 - 1:
        if j % 2 == 0: print *
        else: print i
    print newline
for i from n-1 down to 1:
    for j from 1 to i*2 - 1:
        if j % 2 == 0: print *
        else: print i
    print newline

Approach comparison

🔮 Live Preview

Choose a height between 3 and 7 and draw the number-star diamond in the browser.

Press "Draw pattern".

Examples Gallery

Three complete Python programs — fixed height, user input with conditional expression form, and a smaller trace demo. Click View Output to reveal sample console results.

for i in range(1, 6):
    for j in range(1, i * 2):
        if j % 2 == 0:
            print("*", end="")
        else:
            print(i, end="")
    print()

for i in range(4, 0, -1):
    for j in range(1, i * 2):
        if j % 2 == 0:
            print("*", end="")
        else:
            print(i, end="")
    print()

1
2*2
3*3*3
4*4*4*4
5*5*5*5*5
4*4*4*4
3*3*3
2*2
1

n = int(input("Enter n: "))
if n < 1:
    raise SystemExit

for i in range(1, n + 1):
    for j in range(1, i * 2):
        print("*" if j % 2 == 0 else i, end="")
    print()

for i in range(n - 1, 0, -1):
    for j in range(1, i * 2):
        print("*" if j % 2 == 0 else i, end="")
    print()

Enter n: 3
1
2*2
3*3*3
2*2
1

n = 3

for i in range(1, n + 1):
    for j in range(1, i * 2):
        if j % 2 == 0:
            print("*", end="")
        else:
            print(i, end="")
    print()

for i in range(n - 1, 0, -1):
    for j in range(1, i * 2):
        if j % 2 == 0:
            print("*", end="")
        else:
            print(i, end="")
    print()

1
2*2
3*3*3
2*2
1

🔎 Worked Walkthrough — top half n = 5

Trace each outer-loop value of i, inner-loop range, character count, and full row output.

Characters per row = 2*i-1. Bottom half repeats rows 4, 3, 2, 1 in reverse.

🎯 Practice Problems

Try these variations to lock in the pattern.

Notes

⏱️ Time and Space Complexity

💡 Best Practices