Q:

Most efficient way to find spatial order from a list of tuples?

I have a circle-growth algorithm (line-growth with closed links) where new points are added between existing points at each iteration.
The linkage information of each point is stored as a tuple in a list. That list is updated iteratively.

QUESTIONS:

What would be the most efficient way to return the spatial order of these points as a list ?
Do I need to compute the whole order at each iteration or is there a way to cumulatively insert the new points in a orderly manner into that list ?

All I could come up with is the following:
tuples = [(1, 4), (2, 5), (3, 6), (1, 6), (0, 7), (3, 7), (0, 8), (2, 8), (5, 9), (4, 9)]

starting_tuple = [e for e in tuples if e[0] == 0 or e[1] == 0][0]
## note: 'starting_tuple' could be either (0, 7) or (0, 8), starting direction doesn't matter

order = list(starting_tuple) if starting_tuple[0] == 0 else [starting_tuple[1], starting_tuple[0]]
## order will always start from point 0

idx = tuples.index(starting_tuple)
## index of the starting tuple

def findNext():
    global idx
    for i, e in enumerate(tuples):
        if order[-1] in e and i != idx:
            ind = e.index(order[-1])
            c = 0 if ind == 1 else 1
            order.append(e[c])
            idx = tuples.index(e)

for i in range(len(tuples)/2):
    findNext()

print order

It is working but it is neither elegant (non pythonic) nor efficient. 
It seems to me that a recursive algorithm may be more suitable but unfortunately I don't know how to implement such solution.
Also, please note that I'm using Python 2 and can have access to full python packages only (no numpy)

A:

Rather than recursion, this seems more like a dictionary and generator problem to me:
from collections import defaultdict

def findNext(tuples):
    previous = 0
    yield previous  # our first result

    dictionary = defaultdict(list)

    # [(1, 4), (2, 5), (3, 6), ...] -> {0: [7, 8], 1: [4, 6], 2: [5, 8], ...}
    for a, b in tuples:
        dictionary[a].append(b)
        dictionary[b].append(a)

    current = dictionary[0][0]  # dictionary[0][1] should also work
    yield current  # our second result

    while True:
        a, b = dictionary[current]  # possible connections

        following = a if a != previous else b  # only one will move us forward

        if following == 0:  # have we come full circle?
            break

        yield following  # our next result

        previous, current = current, following  # reset for next iteration

tuples = [(1, 4), (2, 5), (3, 6), (1, 6), (7, 0), (3, 7), (8, 0), (2, 8), (5, 9), (4, 9)]

generator = findNext(tuples)

for n in generator:
    print n

OUTPUT
% python test.py
0
7
3
6
1
4
9
5
2
8
% 

Algorithm currently assumes we have more than two nodes.

