No. Different speeds guarantee a meeting in a finite cycle; equal-speed phase two locates its entrance.

Fast’s extra distance at the meeting is a whole number of cycle lengths. That congruence makes distance from head to entrance match the remaining cycle distance from meeting.

What the function promises: Return the entrance node of a stable singly linked list cycle, or None.

class Node:
    def __init__(self, value, next_node=None):
        self.value, self.next = value, next_node

def build(values):
    head = None
    for value in reversed(values):
        head = Node(value, head)
    return head

def to_list(head):
    answer = []
    while head is not None:
        answer.append(head.value)
        head = head.next
    return answer

def cycle_entrance(head):
    slow = fast = head
    while fast is not None and fast.next is not None:
        slow, fast = slow.next, fast.next.next
        if slow is fast:
            seeker = head
            while seeker is not slow:
                seeker, slow = seeker.next, slow.next
            return seeker
    return None

a, b, c = Node(1), Node(2), Node(3)
a.next, b.next, c.next = b, c, b
assert cycle_entrance(a) is b
assert cycle_entrance(build([1,2])) is None

Work and storage: O(n) time and O(1) extra space.

When to recognize it: Different speeds guarantee a meeting in a finite cycle; equal-speed phase two locates its entrance.

Check before applying it: The arithmetic assumes deterministic next references that do not change during traversal.

After the meeting, walk once around the cycle while counting edges.