How can we find the starting node of a loop in link list?

Let's see.

You position the hare and the tortoise at 1, and let them run, the hare twice as fast as the tortoise.

At the zeroth step, both are at 1. At the first step, the tortoise moves to 2 and the hare moves to 3, and so on.

1 1
2 3
3 5
4 7
5 3
6 5
7 7 

So the hare and the tortoise meet at 7.

Now place the tortoise at the beginning, and let them run again, now with the same speed.

1 7
2 8
3 3 

So they indeed have met at the first element of the cycle.

And that's how it works for the given situation.

OK, let's keep it simple.

Say you have two runners A and B. A move forward by 1 node each step, B moves by 2 nodes.

If it's a cyclic list, they will finally meet each other.

At that time

Say the distance A has moved so far is m, so for B it's 2m

Also note that

 m = a + b
2m = a + b + k * lengthOfLoop

Because for B, it has moved whatever distance A has moved, plus k(some number we don't care) of loops. a is the distance before the loop point, b is the distance A moved after the loop point.

Then we have (after some math)

a = k * lengthOfLoop - b

Now we put B back to the head of the list, and reduce his speed to 1.

For B, it's a nodes away from the loop point. For A, he already passed the loop point by b, and according to the equation above, he is also a nodes away from the loop point.

So, after a more steps, A and B will meet again at the loop point.

Okay, direct answer: The [EDIT: linked list, not sequence] you gave contains no cycles. Here's what will happen. In the first part of the algorithm, the tortoise and hair will start out at x1=2 and x2=3, respectively. Then they will advance to x2=3 and x4=5. Then to x3=4 and x6=7. Then to x4=5 and x8=3. Then the hare will cease to advance, since there isn't anything beyond x8, and the algorithm will yield that no cycles were found.

Below I compiled a little GIF that shows Floyd's cycle-finding applied to a different sequence, that does contain cycles.