Сформировать двумерный массив по правилу

def generate():
    last = list('00000012')
    while last.index('2') > 0:
        copy_ = last.copy()
        copy_[-3] = '.'
        yield ''.join(copy_)
        new_position = last.index('2') - 1
        last = list('00000000')
        last[new_position] = '2'
        last[new_position - 1] = '1'
        last[new_position + 1] = '1'
    yield '21000.00'
 
 
for line in generate():
    print(line)

n = int(input('N = '))
 
matrix = []
 
for i in range(n):
    row = [0] * n
    row[i] = 2
    if i == 0:
        row[1] = 1
    elif i == n - 1:
        row[n - 2] = 1
    else:
        row[i-1] = row[i + 1] = 1
    row.reverse()
    matrix.append(row)
 
for row in matrix:
    print(*row)

n = int(input())
arr = [[0] * (n - 2) + [1, 2], [2, 1] + [0] * (n - 2)]
 
for i in range(1, n - 1):
    line = [0] * n
    line[i - 1: i + 2] = [1, 2, 1]
    arr.insert(1, line)
 
print(*arr, sep="\n")

import numpy as np
n = int(input())
a = np.ones((n, n), dtype=int)
a[np.diag_indices(n)] = 2
a = np.rot90(np.tril(np.triu(a, k = -1), k = 1))
print(a)

n = int(input())
[print(*[x if (x := 2-abs(n-i-j-1)) > 0 else 0 for i in range(n)]) for j in range(n)]

n = int(input())
a = tuple(tuple((i + j + 1 == n) + (abs(n - i - j - 1) <= 1) for j in range(n)) for i in range(n))
for row in a:
    print(*row)