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 FAST algorithm of checkSamecircularList(x, y) without L CircularList ref but find ref
My book has this exercise:

R-7.6 Suppose that x and y are references to nodes of circularly linked lists,
although not necessarily the same list. Describe a fast algorithm for telling
if x and y belong to the same list.

I wonder if there is the FAST algorithm of checkSamecircularList(x, y) without L CircularList ref but find ref to L from the first element x... can do this??? after check if y is in this List of x... well ?

checkSamecircularList(L, x, y):

RESTART: F:/DOWNLOADS/Textbooks - Source Code/Python DS Tamassia/ch07/exe/
same list x, y of L: True
same list x1, y of L: False
same list x1, y1 of L: False
same list x1, y1 of L1: True

if __name__ == '__main__':

    # Function to find sum of the given  
    # Circular linked list      
    def checkSamecircularList(L, x, y): 
      temp = L._tail   # head
      # tsum = 
      temp = temp._next

      xyes = False
      yyes = False

      while(temp is not L._tail):
        if temp is x:
          xyes = True
        if temp is y:
          yyes = True        
        temp = temp._next
      return xyes and yyes

    L = CircularQueue()

    L1 = CircularQueue()

    x = L.first()
    y = (L.first())._next

    x1 = L1.first()
    y1 = (L1.first())._next

    print("same list x, y of L: " + str(checkSamecircularList(L, x, y)) )
    print("same list x1, y of L: " + str(checkSamecircularList(L, x1, y)) )
    print("same list x1, y1 of L: " + str(checkSamecircularList(L, x1, y1)) )
    print("same list x1, y1 of L1: " + str(checkSamecircularList(L1, x1,y1)) )

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