map cities by rotated letters

cities_match.py

from itertools import combinations

# quick and dirty implementation, not optimized
cities = ['Tokyo', 'London', 'Rome', 'Donlon', 'Kyoto', 'Paris']
normalized = [(c, ''.join(sorted(c.lower()))) for c in cities]
cartesian = list(combinations(normalized, 2))
matches = [[l[0], r[0]] for l, r in cartesian if l[1] == r[1]]
matches.extend([c for c in cities if c not in [m[0] for m in matches] + [m[1] for m in matches]])
matches
# output [['Tokyo', 'Kyoto'], ['London', 'Donlon'], 'Rome', 'Paris']

cities_match_rot.py

# improved version by only rotating, not sorting
# -- rotate i times to the left
rot = lambda s, i: s[i:] + s[0:(i)]
# -- keep unrotated with i == 0, rotate every possible way
rotated = [(i, c, rot(c.lower(), i)) for c in cities for i in range(len(c))]
# -- find matches where left is equal to right and right is the original name
matches = [(l[1], r[1]) for l, r in combinations(rotated, 2) if l[2] == r[2] and r[0] == 0]
matches.extend([c for c in cities if c not in [m[0] for m in matches] + [m[1] for m in matches]])
matches
# output [('Tokyo', 'Kyoto'), ('London', 'Donlon'), 'Rome', 'Paris']

cities_match_rot_slim.py

cities = ['Tokyo', 'London', 'Rome', 'Donlon', 'Kyoto', 'Paris']

# actually, can be done in 3 statements, without combinatorics. ok it starts to get somewhat involved 
rot = lambda s, i: s[i:] + s[0:(i)]
matches = [(c, rot(c.lower(), i).capitalize()) 
           for j, c in enumerate(cities) 
           for i in range(len(c)) 
           if rot(c.lower(), i) in [cx.lower() for cx in cities[j:]] and i > 0]
matches.extend([c for c in cities if c not in [m[0] for m in matches] + [m[1] for m in matches]])
matches
# output [('Tokyo', 'Kyoto'), ('London', 'Donlon'), 'Rome', 'Paris']

Python solution to https://hackernoon.com/how-to-lose-an-it-job-in-10-minutes-3d63213c8370