Linked List loop detection

Proof:
---------o
suppose circle distance=C
suppose entry of LL to cycle starting point=x
first collision happened point=K
distance bw k-> entry of the cycle=y
2*slow=fast
2*(c-y+x)=x+c+c-y (One circle already completed by the fast pointer)
2c-2y+2x=x+2c-y
x=y so after the first collision of slow & fast, both starts moving one step so x=y & x is the distance of starting of LL to the cycle entry point.
***Move fast only when you can move two nodes***
class Solution:
    def hasCycle(self, head: Optional[ListNode]) -> bool:
        slow=head
        fast=head
        while fast and fast.next: #move the fast only if you can move two nodes
            slow=slow.next
            fast=fast.next.next
            if slow==fast:
                break
        if not fast or fast.next==None: #check if fast==None or fast.next==None because in previous while, even fast.next=None, we considered not a cycle and didn't move
            return False
        slow=head
        while fast and slow!=fast:
            slow=slow.next
            fast=fast.next
        return slow.val