slow and fast, the first one moves forward one step at a time, and the second one moves forward two steps at a time. Eventually, the two pointers will meet somewhere. In the above example, they meet at number 9 (or G).
G is where two-pointers fast and slow meet.E is the entrance of the loop.S is the first node of the linked list.F is the distance from the chain’s beginning to the loop’s entrance (from S to E). (eg: 1 – 3 – 5)a is the distance from the loop’s entrance to the point where two pointers meet (from E to G). (eg: 5 – 6 – 7 – 8 – 9)C is the length of the loop. (eg: the loop is 5 – 6 – 7 – 8 – 9 – 10, the length is 6 )
The total distance that the slow pointer has moved is $F + a + xC$. $x$ is the number of times the slow pointer completes the loop.
The total distance that the fast pointer has moved is $F + a + yC$. Likewise, $y$ is the number of times the fast pointer completes the loop.
Since the fast pointer moves two steps at a time, its distance will be double the slow pointer’s distance.
\[2*(F + a + xC) = F + a + yC\]
\[2F + 2a + 2xC = F + a + yC\]
\[F + a + 2xC = yC\]
\[F + a = yC – 2xC\]
\[F + a = (y – 2x)C\]
Denote $n = y – 2x, n \ge 0$
\[F + a = nC\]
$nC$ means to complete the loop (length is C) n times
In other words, if we have two pointers a and b, a starts at beginning of the chain (node S), b starts at where slow and fast pointers meet (node G), and both two new pointers move one step at a time. The first pointer moves a distance equal to $F + a$, and the second pointer moves a distance equal to $nC$, they will meet at G.
\[F + a= (n – 1)C + C\] \[F + a= (n – 1)C + (C – a) + a\] \[F = (n – 1)C + (C – a)\]
This means, that if pointer a starts from the beginning of the chain (node S), pointer b starts from where two pointers slow and fast meet (node G). They eventually meet at node E, the entrance of the loop.