Bron to Clique

As The Historians wander around a secure area at Easter Bunny HQ, you come across posters for a LAN party scheduled for today! 
Maybe you can find it; you connect to a nearby datalink port and download a map of the local network (your puzzle input).

The network map provides a list of every connection between two computers. For example:
kh-tc
qp-kh
de-cg
ka-co
Each line of text in the network map represents a single connection; the line kh-tc represents a connection between the 
computer named kh and the computer named tc. Connections aren't directional; tc-kh would mean exactly the same thing.

LAN parties typically involve multiplayer games, so maybe you can locate it by finding groups of connected computers. 
Start by looking for sets of three computers where each computer in the set is connected to the other two computers.

If the Chief Historian is here, and he's at the LAN party, it would be best to know that right away. You're pretty 
sure his computer's name starts with t, so consider only sets of three computers where at least one computer's name 
starts with t. That narrows the list down to 7 sets of three inter-connected computers:
co,de,ta
co,ka,ta
de,ka,ta
qp,td,wh
tb,vc,wq
tc,td,wh
td,wh,yn
Find all the sets of three inter-connected computers. How many contain at least one computer with a name that starts 
with t?

def is_name_valid(a: str, b: str, c: str) -> bool:
    return a.startswith('t') or b.startswith('t') or c.startswith('t')

def find_trios(pairs: dict[str, set]) -> set:
    relationships = defaultdict(set)
    for pair in pairs:
        relationships[pair[0]].add(pair[1])
        relationships[pair[1]].add(pair[0])

    trios = set()
    for origin in relationships.keys():
        for related_computer in relationships[origin]:
            if related_computer > origin: # removes redundant checks A-B / B-A
                common = relationships[related_computer] & relationships[origin] 
                for computer in common:
                    if is_name_valid(origin, related_computer, computer):
                        trio = frozenset([origin, related_computer, computer])
                        trios.add(trio)
    return trios

There are still way too many results to go through them all. You'll have to find the LAN party another way and go there yourself.

Since it doesn't seem like any employees are around, you figure they must all be at the LAN party. If that's true, the 
LAN party will be the largest set of computers that are all connected to each other. That is, for each computer at the LAN party, 
that computer will have a connection to every other computer at the LAN party.

In the above example, the largest set of computers that are all connected to each other is made up of co, de, ka, and ta. 
Each computer in this set has a connection to every other computer in the set:
ka-co
ta-co
de-co
ta-ka
de-ta
ka-de
The LAN party posters say that the password to get into the LAN party is the name of every computer at the LAN party, 
sorted alphabetically, then joined together with commas. (The people running the LAN party are clearly a bunch of nerds.)
 In this example, the password would be co,de,ka,ta.

What is the password to get into the LAN party?

def get_lan_password(names: list[str]) -> str:
    return ",".join(sorted(names))

BronKerbosch(R, P, X):
    if P is empty and X is empty:
        report R as a maximal clique
        return

    for each node in P:
        BronKerbosch(R ∪ {node}, P ∩ N(node), X ∩ N(node))
        P = P \ {node}
        X = X ∪ {node}

     aa
    / \
  bb---cc
  |
  dd

bron_kerbosch(R={}, P={aa, bb, cc, dd}, X={})
├── node = aa
│   └── bron_kerbosch(R={aa}, P={bb, cc}, X={})
│       ├── node = bb
│       │   └── bron_kerbosch(R={aa, bb}, P={cc}, X={})
│       │       └── node = cc
│       │           └── bron_kerbosch(R={aa, bb, cc}, P={}, X={}) ✅ Maximal clique!
│       └── node = cc
│           └── bron_kerbosch(R={aa, cc}, P={}, X={bb}) ✖ Not maximal
├── node = bb
│   └── bron_kerbosch(R={bb}, P={cc, dd}, X={aa})
│       ├── node = cc
│       │   └── bron_kerbosch(R={bb, cc}, P={}, X={aa}) ✖ Not maximal
│       └── node = dd
│           └── bron_kerbosch(R={bb, dd}, P={}, X={}) ✅ Maximal clique!
├── node = cc
│   └── bron_kerbosch(R={cc}, P={}, X={aa, bb}) ✖ Not maximal
└── node = dd
    └── bron_kerbosch(R={dd}, P={}, X={bb}) ✖ Not maximal

def bron_kerbosch(r: set[str], p: set[str], x: set[str], graph: dict[str, set], cliques: list[set]):
    if not p and not x:
        cliques.append(r)
        return
    for vertex in list(p):
        bron_kerbosch( r | {vertex}, p & graph[vertex], x & graph[vertex], graph, cliques)
        p.remove(vertex)
        x.add(vertex)

def find_largest_clique(graph: dict[str, set]) -> str:
    cliques_found = []
    bron_kerbosch(set(), set(graph.keys()), set(), graph, cliques_found)
    largest_clique = max(cliques_found, key=len)
    return get_lan_password(list(largest_clique))

def bron_kerbosch(r: set[str], p: set[str], x: set[str], graph: dict[str, set], cliques: list[set]):
    if not p and not x:
        cliques.append(r)
        return
    # pivot
    pivot_candidates = p | x
    # choose the pivot from the candidates with the highest number of connexions
    u = max(pivot_candidates, key=lambda vertex: len(graph[vertex] & p))
    candidate_neighbours = graph[u]
    # reduce search space by avoiding direct neighbours of the pivot
    for vertex in p - (candidate_neighbours):
        bron_kerbosch( r | {vertex}, p & graph[vertex], x & graph[vertex], graph, cliques)
        p.remove(vertex)
        x.add(vertex)