Find a chain that fill range without gaps - range

Question

Given a range f.e. :
range : 1 4
there are 3 lists of pairs that 'fill' this range w/o gaps. Call it a 'chain'
chain1: 1 2, 2 3, 3 4
chain2: 1 2, 2 4
chain3: 1 3, 3 4
the following lists of pairs fail to fill the range, so it is not a chain :
1 2 <= missing 2 3, 3 4 OR 2 4
1 3 <= missing 3 4
1 2, 3 4 <= missing 2 3
1 2, 2 3 <= missing 3 4
2 3, 3 4 <= missing 1 2
Now the question is given a random list of pairs and a range how can I find is there a combination of pairs that makes a CHAIN.
Pairs do not repeat, are always (lower-value, higher-value) and never overlap i.e. this is never an input [ 1 3, 2 4 ]
In addition I can pre-filter only the pairs that are within the range, so f.e. you dont have to worry about pairs like : 4 6, 7 9, 0 1 ...
update ... this is also valid input
chain: 1 2, 1 3, 3 4
chain: 1 2, 1 3, 1 4
chain: 1 2, 2 4, 3 4
i.e a pair can subsume another .... this breaks the sort-and-loop idea

Answer 1

implemented the Union-Find algo from here in python ..
q=QFUF(10)
In [131]: q.is_cycle([(0,1),(1,3),(2,3),(0,3)])
Out[131]: True
In [132]: q.is_cycle([0,1,4,3,1])
Out[132]: True
only cares about first two elements of the tuple
In [134]: q.is_cycle([(0,1,0.3),(1,3,0.7),(2,3,0.4),(0,3,0.2)])
Out[134]: True
here
import numpy as np
class QFUF:
""" Detect cycles using Union-Find algorithm """
def __init__(self, n):
self.n = n
self.reset()
def reset(self): self.ids = np.arange(self.n)
def find(self, a): return self.ids[a]
def union(self, a, b):
#assign cause can be updated in the for loop
aid = self.ids[a]
bid = self.ids[b]
if aid == bid : return
for x in range(self.n) :
if self.ids[x] == aid : self.ids[x] = bid
#given next ~link/pair check if it forms a cycle
def step(self, a, b):
# print(f'{a} => {b}')
if self.find(a) == self.find(b) : return True
self.union(a, b)
return False
def is_cycle(self, seq):
self.reset()
#if not seq of pairs, make them
if not isinstance(seq[0], tuple) :
seq = zip(seq[:-1], seq[1:])
for tpl in seq :
a,b = tpl[0], tpl[1]
if self.step(a, b) : return True
return False