Why increase pointer by two while finding loop in linked list, why not 3,4,5?

Question

I had a look at question already which talk about algorithm to find loop in a linked list. I have read Floyd's cycle-finding algorithm solution, mentioned at lot of places that we have to take two pointers. One pointer( slower/tortoise ) is increased by one and other pointer( faster/hare ) is increased by 2. When they are equal we find the loop and if faster pointer reaches null there is no loop in the linked list.

Now my question is why we increase faster pointer by 2. Why not something else? Increasing by 2 is necessary or we can increase it by X to get the result. Is it necessary that we will find a loop if we increment faster pointer by 2 or there can be the case where we need to increment by 3 or 5 or x.

Answer 1

From a correctness perspective, there is no reason that you need to use the number two. Any choice of step size will work (except for one, of course). However, choosing a step of size two maximizes efficiency.

To see this, let's take a look at why Floyd's algorithm works in the first place. The idea is to think about the sequence x₀, x₁, x₂, ..., x_n, ... of the elements of the linked list that you'll visit if you start at the beginning of the list and then keep on walking down it until you reach the end. If the list does not contain a cycle, then all these values are distinct. If it does contain a cycle, though, then this sequence will repeat endlessly.

Here's the theorem that makes Floyd's algorithm work:

The linked list contains a cycle if and only if there is a positive integer j such that for any positive integer k, x_j = x_jk.

Let's go prove this; it's not that hard. For the "if" case, if such a j exists, pick k = 2. Then we have that for some positive j, x_j = x_2j and j ≠ 2j, and so the list contains a cycle.

For the other direction, assume that the list contains a cycle of length l starting at position s. Let j be the smallest multiple of l greater than s. Then for any k, if we consider x_j and x_jk, since j is a multiple of the loop length, we can think of x_jk as the element formed by starting at position j in the list, then taking j steps k-1 times. But each of these times you take j steps, you end up right back where you started in the list because j is a multiple of the loop length. Consequently, x_j = x_jk.

This proof guarantees you that if you take any constant number of steps on each iteration, you will indeed hit the slow pointer. More precisely, if you're taking k steps on each iteration, then you will eventually find the points x_j and x_kj and will detect the cycle. Intuitively, people tend to pick k = 2 to minimize the runtime, since you take the fewest number of steps on each iteration.

We can analyze the runtime more formally as follows. If the list does not contain a cycle, then the fast pointer will hit the end of the list after n steps for O(n) time, where n is the number of elements in the list. Otherwise, the two pointers will meet after the slow pointer has taken j steps. Remember that j is the smallest multiple of l greater than s. If s ≤ l, then j = l; otherwise if s > l, then j will be at most 2s, and so the value of j is O(s + l). Since l and s can be no greater than the number of elements in the list, this means than j = O(n). However, after the slow pointer has taken j steps, the fast pointer will have taken k steps for each of the j steps taken by the slower pointer so it will have taken O(kj) steps. Since j = O(n), the net runtime is at most O(nk). Notice that this says that the more steps we take with the fast pointer, the longer the algorithm takes to finish (though only proportionally so). Picking k = 2 thus minimizes the overall runtime of the algorithm.

Hope this helps!

Answer 2

Let us suppose the length of the list which does not contain the loop be s, length of the loop be t and the ratio of fast_pointer_speed to slow_pointer_speed be k.

Let the two pointers meet at a distance j from the start of the loop.

So, the distance slow pointer travels = s + j. Distance the fast pointer travels = s + j + m * t (where m is the number of times the fast pointer has completed the loop). But, the fast pointer would also have traveled a distance k * (s + j) (k times the distance of the slow pointer).

Therefore, we get k * (s + j) = s + j + m * t.

s + j = (m / k-1)t.

Hence, from the above equation, length the slow pointer travels is an integer multiple of the loop length.

For greatest efficiency , (m / k-1) = 1 (the slow pointer shouldn't have traveled the loop more than once.)

therefore , m = k - 1 => k = m + 1

Since m is the no.of times the fast pointer has completed the loop , m >= 1 . For greatest efficiency , m = 1.

therefore k = 2.

if we take a value of k > 2 , more the distance the two pointers would have to travel.

Hope the above explanation helps.