# List Cycle Detection

Linked list cycle detection problem is a very instructive and fun problem to reason about. This article will state the problem and explains how we can solve it efficiently while giving some insight on the underlying maths.

Well, list can gets corrupted and some node can be linked by one or more nodes (as in the following figure). This could lead to never ending traversal of the list. So it make sense to solve the following problem:

Cicular List

## List Cycle Detection - Problem Statement

1. Given  a linked list detect (if any) check if the list is circular i.e. contains a cycle
2. Find the starting node of the cycle (the node with two inward arrows in the figure)

The problem is easily solvable in $O(n^2)$ time and$O(n)$ space considering that we can visit the list from the head and store in a separate list the visited nodes. As the visit continues we check if the node we are examining was previously visited (is that node contained in the support list?). If yes the list is circular and that node is the start point of the cycle. If we reach the tail of the list then the list is not circular. We can lower the complexity of this approach down to$O(nlog(n))$ time using a more efficient support (set) data structure (like a tree set). But we can do much better, and the rest of the article will show how to obtain $O(n)$ time and $O(1)$.

## List Cycle Detection - Floyd’s algorithm

We use the fact that like clock's hands things iterating on a cicle will meet at some point in the future. Consider two runner $R_1,R_2$ with velocities$V_1,V_2=2V_1$ starting running from the same point in a circular stadium. They will meet again when the slower runner reach the starting point for the second time (Why? By the time the slower one has completed half circle the faster has completed a complete cycle and by the time the slower finishes his run, arriving at the starting point again, the faster has completed a second entire cycle).

Things are a bit more complicated in the List Cycle Detection problems because the iterators do not necessarily start from the same point. Consider two iterators $p,q$ with velocities $v_p=1,v_q=2v_p = 2$  respectively. Suppose the cycle has length $n$ and that it starts at node number $A < n$ When the slower iterator reaches $A$ the faster is at location $2A$. How many iteration $k$ it will take before they meet? And at which node?

The situation is described by the following congruence

• $A + kv_p \equiv 2A + k2v_p mod(n)$
• $\Rightarrow 2A + k2v_p \equiv A + kv_p \; mod(n)$
• $\Rightarrow A + k2v_p \equiv kv_p \; mod(n)$
• $\Rightarrow A +kv_p \equiv 0 \;mod(n)$
• $\Rightarrow A +k \equiv 0 \;mod(n)$

which has solution $k = n-A$. This means that they will meet after $k=n-A$ iteration of the slower iterator and since it starts from $A$ this means that they will meet at $A$ nodes before the beginning of the cycle. We can use that fact to count $A$ nodes from the beginning of the list. Once the iterators matche in the cycle, we can move the fast iterator back to the beginning of the list and iterate forward one node per stepo with both iterators until they match again (They are far away $A$ nodes from the beginning of the cycle, so they will meet exaclty in that point.

Let's consider now the case when $A \geq n$. This means that by the time the slower iterator reaches the beginning of the cycle the faster one has completed more that a cycle. What will be the starting point for the faster one? We argue that once p reaches A q is at node 2A but since A > n this means that some it will be at position A + (A mod(n)). We can now use similar argument to the previous case and write:

• $A + kv_p \equiv A + (A mod(n)) + k2v_p \;mod(n)$
• $A + (A mod(n)) + k2v_p \equiv A + kv_p\;mod(n)$
• $(A mod(n)) + kv_p mod(n) \equiv 0mod(n)$
• $(A mod(n)) + k mod(n) \equiv 0 mod(n)$ since $v_p =1$

which has solution $k = n-(A mod(n))$. This means that the meetpoint is $A mod(n)$ nodes before the beginning of the cycle. If we do the sape operation as the previous case we obtain the sme result. Itrators wil meet at the beginning of the cycle. Why? Well advancing $q$ makes $p$ cycles possibly several times ( remember that $A \geq n$  ) and it will clearly stops at $A+(n-A mod(n)) + A mod(n) = A +n \;(mod (n))= A$.

In other words the slower pointer is at first  at node number $A+(n-A mod(n))$. We can write$A = bn + r$ where $r = A \;mod(n)$. After $A$ advancing steps it will be at location  $A+(n-A \;mod(n)) +bn +r (mod n)$ Since $bn \; mod (n)=0$ the result follows.

As an example consider in which we have a list with a cycle of length $n=4$ and starts at node number $10$. The first part of the algorithm tells us that the nodes will meet at node $10 + 4 - 10 \: mod(4) = 12$. Moving the fast pointer back to te head of the list and iterating one node per time both iterators will lead the slower point to node:

• 12 again after advancing of 4 nodes
• 12 again after advancing of 4 nodes
• 10 advancing of the remaining 2 nodes.