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The tortoise and the hare

Recently, I start to work on leetcode's problems. My goal is to solve two problems per day (mission possible, right?). The problems I'm looking at are 142. Linked List Cycle II and 287. Find the Duplicate Number, which both can be solved by Folyd's Tortoise and Hare algorithm.

This post will try to take a deeper look at the correctness of the algorithm and how to apply it to solve problems.


Floyd's Tortoise and Hare algorithm is used with three purposes under the context of linked list:

  1. Detect whether there is a cycle in the list
  2. Find the starting point of the cycle (i.e. list 1->4->3->4, starting point is 4)
  3. Decide the length of the cycle (i.e. 2 for above example)

The algorithm idea is following:

  1. We use two pointers: tortoise and hare. Both start at the beginning of the list. Hare runs twice as fas the tortoise.
  2. If there is no-cycle, then hare will reach the finish line before the tortoise.
  3. If there is a cycle, then hare will always be ahead and eventually he would so far ahead that he laps the tortoise. That's the place we know we have a cycle in the list.
  4. Once we detect the cycle, we send hare back to the beginning and advance both of them at the same speed until they meet again. The second meeting place, which we'll prove immediately, is the entry point of the cycle.
  5. Then, one of them will keep moving to finish the victory lap to find the period of the cycle.

The key difference when the list has a cycle is that at some point on the track, the hare will be at the same spot as the tortoise ...

Proof of correctness

Let \(\mu\) be the index of the start of the cycle, and let \(\lambda\) be period of the cycle. Let \(i\) be the distance (i.e number of nodes) that tortoise travels and let \(x_i\) denotes the index of the node at which both tortoise and hare meet. \(x_0\) is the first node in the list.


The notation here is similar to the concept in physics: distance vs. displacement. \(i\) is the "distance" or the number of nodes that our character (tortoise or hare) has travelled since the beginning of the list and \(x_i\) is the "displacement" between the first node and the current node that our characters are at.

The key observation for showing the correctness of the algorithm lies in the following fact:

$$ \begin{equation} x_{j+k\lambda} = x_j \text{ for all integers }j \ge \mu \text{ and } k \ge 0 \label{eqn:1} \end{equation} $$

This statement says that going around the loop any number of times takes you back to the same places as long as you start somewhere on the loop. Let's define the following set of notation here for future use:

  • \(y\) be the displacement between \(x_{\mu}\) and \(x_i\)
  • \(m\) be the number of laps that tortoise have travelled before he meets with hare at \(x_i\)
  • \(n\) be the number of laps that hare have travelled before he meets with tortoise at \(x_i\)

Since the hare runs twice as fast as the tortoise, then the distance hare travelled when he meets with tortoise is \(2i\) 1. Then, we have the following set of equations

$$ \begin{eqnarray} i = \mu + y + m \cdot \lambda \label{eqn:2} \\ 2i = \mu + y + n \cdot \lambda \label{eqn:3} \end{eqnarray} $$

Now we subtract \ref{eqn:2} from \ref{eqn:3} and we have

$$ \begin{equation} i = (n-m) \cdot \lambda \label {eqn:4} \end{equation} $$

Let's revisit our key observation \ref{eqn:1} and set \(j = \mu\) and \(k = (n-m)\), we have \(x_{\mu + (n-m)\lambda} = x_{\mu}\). Then, by \ref{eqn:4}, we have \(x_{\mu + i} = x_{\mu}\), which can be rewritten as \(x_{i+\mu} = x_{\mu}\)!!! This equation tells us that the node at which the cycle begins (i.e \(x_{\mu}\)) is exactly the same node as the node that is \(\mu\) nodes away from the index at which tortoise and hare meet (i.e. \(x_i\)).


The proof can be much shorter once we have \(x_{2i} = x_i\). By \ref{eqn:1}, we also have \(x_{i+k\lambda} = x_{2i}\), which leads to \(k\lambda = i\). Since \(x_\mu\) also meets the condition of \ref{eqn:1}, we have \(x_{\mu + k\lambda} = x_\mu\). Substitutes \(i = k\lambda\) in and get \(x_{\mu+i} = x_\mu\). The conclusion follows.

Two problems

142. Linked List Cycle II

The first problem is a straightforward application of the algorithm: Given a linked list, return the node where the cycle begins. If there is no cycle, return NULL. The code is following

 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
struct ListNode *detectCycle(struct ListNode *head) {
    if(head == NULL || head->next == NULL)
        return NULL;

    struct ListNode *tortoise;
    struct ListNode *hare;
    struct ListNode *curr;

    tortoise = hare = curr = head;

    while(hare != NULL && hare->next != NULL)
        hare = hare->next->next;
        tortoise = tortoise->next;
        if (hare == tortoise)  // there is a cycle
            while(curr != tortoise)
                curr = curr->next;
                tortoise = tortoise->next;
            return curr;  // find the entry location
    return NULL; // there is no cycle

287. Find the Duplicate Number

This problem is a lot tricker than a previous one: we need identify this problem can also be solved by the Floyd's Tortoise and Hare algorithm, which is not obvious at first glance

Given an array nums containing \(n + 1\) integers where each integer is between \(1\) and \(n\) (inclusive), prove that at least one duplicate number must exist. Assume that there is only one duplicate number, find the duplicate one. note: There is only one duplicate number in the array, but it could be repeated more than once.

The key point is to identify that the problem description is another way of describing a linked list, which requires somewhat deeper understanding of the algorithm itself.

The algorithm is, in fact, used to find a cycle in a sequence of iterated function values:

$$ x_0, x1 = f(x_0), x_2 = f(x_1), \dots, x_i = f(x_{i-1}), \dots $$

For example, the sequence \(1,3,4,2,1\) can be considered as a sequence of iterated function values with \(x_0 = 1, x_1 = f(1) = 3, x_2 = f(3) = 4, x_3 = f(4) = 2, x_4 = f(2) = 1\). Let's try another representation:

index 0 1 2 3 4
value 1 3 4 2 1

Surprisingly, the function \(f\) simply map the index to the corresponding values. With this table, the above sequence can be converted as a linked list:

0 -> 1 -> 3 -> 2 -> 4
     ^              |

This list is constructed by the definition of a sequence of iterated function values. The arrow (->) is the function \(f\). Then, we can apply our algorithm to solve this problem:

int findDuplicate(int* nums, int numsSize) {
  int tortoise;
  int hare;
  tortoise = nums[0];
  hare = nums[nums[0]];
  while (tortoise != hare)
    tortoise = nums[tortoise];
    hare = nums[nums[hare]];
  hare = 0;
  while (tortoise != hare)
    tortoise = nums[tortoise];
    hare = nums[hare];
  return hare;

This implementation slightly deviates from the algorithm description above:

  1. Tortoise and hare don't start from the same place at the beginning. This doesn't matter really because eventually they will be in the loop.

  2. We use tortoise = nums[tortoise] instead of tortoise++ for advancing tortoise, for example. This is the place where "a sequence of iterative function values" idea appears. In fact, this is also how we constructed our linked list.

  3. hare = 0 not hare = nums[0] can be confusing. We can think about this from our linked list representation: our list starts from \(0\) (required by \(f\), which maps index to value) and if we starts from hare = nums[0], that violates our algorithm.

  1. \(x_{2i} = x_i\) immediately follows this statement. Then, if we let
    \(l\) be the number of laps by which hare is ahead, then \(2i = i + l \cdot \lambda\) and we have \(i = l\lambda\). Then we set \(k=l\) in \ref{eqn:1} and reach the same conclusion. This way we don't need to define \(y\),\(m\),\(n\), which can make proof a little simpler notation-wise.  

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