Python:實現返回 Collatz 序列及其任意正整數的長度算法
from __future__ import annotations
def n31(a: int) -> tuple[list[int], int]:
if not isinstance(a, int):
raise TypeError(f"Must be int, not {
type(a).__name__}")
if a < 1:
raise ValueError(f"Given integer must be greater than 1, not {
a}")
path = [a]
while a != 1:
if a % 2 == 0:
a = a // 2
else:
a = 3 * a + 1
path += [a]
return path, len(path)
def test_n31():
""" >>> test_n31() """
assert n31(4) == ([4, 2, 1], 3)
assert n31(11) == ([11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1], 15)
assert n31(31) == (
[
31,
94,
47,
142,
71,
214,
107,
322,
161,
484,
242,
121,
364,
182,
91,
274,
137,
412,
206,
103,
310,
155,
466,
233,
700,
350,
175,
526,
263,
790,
395,
1186,
593,
1780,
890,
445,
1336,
668,
334,
167,
502,
251,
754,
377,
1132,
566,
283,
850,
425,
1276,
638,
319,
958,
479,
1438,
719,
2158,
1079,
3238,
1619,
4858,
2429,
7288,
3644,
1822,
911,
2734,
1367,
4102,
2051,
6154,
3077,
9232,
4616,
2308,
1154,
577,
1732,
866,
433,
1300,
650,
325,
976,
488,
244,
122,
61,
184,
92,
46,
23,
70,
35,
106,
53,
160,
80,
40,
20,
10,
5,
16,
8,
4,
2,
1,
],
107,
)
if __name__ == "__main__":
num = 4
path, length = n31(num)
print(f"The Collatz sequence of {
num} took {
length} steps. \nPath: {
path}")
