## Detecting a loop in a linked list

I was recently asked about Floyd’s cycle-finding algorithm, see here for details. In particular the person wanted to know what speeds the two pointers could move at for the algorithm to work. In Floyd’s algorithm the pointers move 1 step and 2 steps at a time respectively.

It’s an interesting question which touches on an element of number theory called modulo arithmetic and the study of congruence relations.

In Floyd’s algorithm the “hare” and “tortoise” both start at the beginning of the linked list at time, t = 0. At each step the “tortoise” moves 1 place and the hare moves 2 places. Given an arbitrary linked list with a cycle then it is clear the “tortoise” and “hare” won’t meet until they are both in the cycle. For this reason we don’t need to know how many steps it took for both of them to be in the cycle, and we only care from the point that both are in the cycle and we will actually call this t = 0. We will call and the positions that the “tortoise” and “hare” are in on the earliest step that both are in the cycle.

We don’t know how long the cycle is in length, but we will call it, n. We label the first element in the cycle 0, the second 1, the third 2, …., the n-th will be labelled n-1.

If we are at the position labelled n-1 and we move one place then we are at the position labelled 0.

The above diagram demonstrates what the values of t, n, , will be for the above linked list.

To give a few concrete examples we will pretend to be working with a cycle of length 12 as this gives us the familiar intuition that we have from using our watches.

however we will relabel the 12 position to be 0.

Now consider the “tortoise” and “hare” starting at positions 0 and 1 respectively. If the “tortoise” and “hare” move 1 place and 2 places respectively at a time, will they ever meet?

A quick bit of working with some pen and paper will yield the result that after 11 steps the “tortoise” will be at position 11 and the “hare” would have gone round the watch once and reached position 11 again.

It is true that with the “tortoise” moving 1 place and the “hare” moving 2 places at a time that they will always meet eventually regardless of their start positions (, ). The reasons for this will be described later.

What we would like to know is whether other values that the “tortoise” and “hare” can move at each step yields the same result of them meeting eventually.

Consider the start positions as we have above but this time the “tortoise” moves 3 places and the “hare” moves 5 places at a time. Will they meet then?

The positions at each step for the first 15 steps is shown below.

Step | Tortoise | Hare |
---|---|---|

0 | 0 | 1 |

1 | 3 | 6 |

2 | 6 | 11 |

3 | 9 | 4 |

4 | 0 | 9 |

5 | 3 | 2 |

6 | 6 | 7 |

7 | 9 | 0 |

8 | 0 | 5 |

9 | 3 | 10 |

10 | 6 | 3 |

11 | 9 | 8 |

12 | 0 | 1 |

13 | 3 | 6 |

14 | 6 | 11 |

15 | 9 | 4 |

You may have noticed that at step 12 we have started to repeat the pattern of step 0. The “tortoise” and “hare” didn’t meet in the first 12 steps and so will never meet. Clearly steps of 3 and 5 don’t work for Floyd’s algorithm and these start positions. However if the “tortoise” started at position 0 and the “hare” at position 2 then it is easy to check that they will meet, so the start positions do matter as well!

Before we write out the problem more formally, let’s first formalise this “clock” arithmetic we have been using implicitly.

We say that if (b – a) is a multiple of n.

The concrete examples we will give will use our familiar clock.

— that is to say that 1pm and 1am look the same on a 12 hour clock. They both use the 1 position.

— that is to say if we start at 1pm and add 24 hours then it is still 1pm but just a day later

In both examples 13 – 1 = 12 and 25 – 1 = 24 are both multiples of 12.

Given the start positions and and the places moved at each step and and a cycle of length n then we are looking for a solution to the following equation:

This can be rewritten to:

This is a standard equation in number theory and it has a solution if and only if

The gcd is the greatest common divisor. More information can be found at gcd.

Given that the length of the cycle n and the start positions are arbitrary the only way to guarantee that the above equation has a solution is to ensure that and the only way to guarantee this for all n is to make . In other words the places that the “tortoise” and “hare” move at a time must differ by 1.

This explains why the choice of 1 and 2 work for all linked lists with a cycle, however values such as 4 and 5 will also work. 99 and 100 or even the values 121 and 120!