for in range cycle_len (slow))cycle_len_advanced_iter cycle_len_advanced_iter next it head both iterators advance in tandem while it is not cycle_len_advanced_iter it it next cycle_len_advanced_iter cycle_len_advanced_iter next return it iter is the start of cycle return none no cycle let be the number of nodes to the start of the cyclec the number of nodes on the cycleand the total number of nodes then the time complexity is (fo(co( )-- (ffor both pointers to reach the cycleand (cfor them to overlap once the slower one enters the cycle variantthe following program purports to compute the beginning of the cycle without determining the length of the cycleit has the benefit of being more succinct than the code listed above is the program correctdef has_cycle (head)fast slow head while fast and fast next and fast next nextslow fast slow next fast next next if slow is fastthere is cycle tries to find the start of the cycle slow head both pointers advance at the same time while slow is not fastslow fast slow next fast next return slow slow is the start of cycle return none no cycle