Python Pattern Programming Part-2

 

Various types of python pattern programming using for loop

Pattern 6:

This pattern will print a diamond of numbers with a width equal to the input odd number.

python
n = int(input("Enter the odd number of rows: "))

for i in range(n):
    if i <= n//2:
        print(" "*(n//2-i) + "".join(str(j) for j in range(i+1)) + "".join(str(j) for j in range(i, 0, -1)))
    else:
        print(" "*(i-n//2) + "".join(str(j) for j in range(n-i, 0, -1)) + "".join(str(j) for j in range(1, n-i+1)))

Output for n = 7:

   1
  121
 12321
1234321
 12321
  121
   1

Pattern 7:

This pattern will print a spiral of numbers with a size equal to the input number.

python
n = int(input("Enter the size: "))

matrix = [[0 for i in range(n)] for j in range(n)]

i, j, di, dj = 0, 0, 0, 1

for k in range(n**2):
    matrix[i][j] = k+1
    if matrix[(i+di)%n][(j+dj)%n]:
        di, dj = dj, -di
    i, j = i+di, j+dj

for i in range(n):
    for j in range(n):
        print(matrix[i][j], end="\t")
    print()

Output for n = 4:

1	2	3	4	
12	13	14	5	
11	16	15	6	
10	9	8	7	

Pattern 8:

This pattern will print a diamond of characters with a width equal to the input uppercase letter.

python
import string

letter = input("Enter an uppercase letter: ")

n = string.ascii_uppercase.index(letter) + 1

for i in range(n):
    print(" "*(n-i-1) + string.ascii_uppercase[i] + " "*(2*i-1) + (string.ascii_uppercase[i] if i != 0 else ""))
    
for i in range(n-2, -1, -1):
    print(" "*(n-i-1) + string.ascii_uppercase[i] + " "*(2*i-1) + (string.ascii_uppercase[i] if i != 0 else ""))

Output for letter = "E":

mathematica
    A    
   B B   
  C   C  
 D     D 
E       E
 D     D 
  C   C  
   B B   
    A    

Pattern 9:

This pattern will print a right-angled triangle of characters with a height equal to the input uppercase letter.

python
import string

letter = input("Enter an uppercase letter: ")

n = string.ascii_uppercase.index(letter) + 1

for i in range(n):
    print(string.ascii_uppercase[i]*(i+1))

Output for letter = "D":

python
A
BB
CCC
DDDD

Pattern 10:

This pattern will print a checkerboard of characters with a size equal to the input number.

python
n = int(input("Enter the size: "))

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

Output for n = 5:

python
* * *
 * * 
* * *
 * * 
* * *

I hope you find these additional examples helpful!