Lightbulbs in Two Separate Rooms

Question

You are in a room with six light switches. Down the hallway, there are two other rooms, each with three incandescent lightbulbs.

Each lightbulb is uniquely controlled by exactly one of the switches. At the start, all of the lightbulbs are switched off.

If you can only visit one of the two rooms at a time, what is the minimum number of visits required to be able to distinguish which switch corresponds to which bulb?

Answer 1

With the allowed conditions light on/off-cold/off-warm, then the minimal number of visits is:

3

Strategy

For the first visit in a room 4 switches are turned on (s1-s4) and after some time 2 switches are turned off (s3-s4) before visiting a room. Two possible results: All three conditions are in the room (v1) or two different lightbulbs are in the same condition (v2)

For v1 the second visit is in the same room with s1, s3, s5 turned on. After this visit we know all three switches for this room and the remaining three switches for the second room. Two of this three switches are turned on and after some time one of the switches is turned off. With all different conditions on all three lightbulbs the puzzle is solved.

For v2 the second visit is in the same room. The two switches for the known two lightbulbs in the room are turned in the states on/off. The two switches for the last known lightbulb are also turned in different states on/off. After the visit we know all three switches for this room and the remaining three switches for the second room. Two of this three switches are turned on and after some time one of the switches is turned off. With all different conditions on all three lightbulbs the puzzle is solved.

Answer 2

If we are only allowed to look at the lights, then the minimal number of visits is:

5

Proof by code:

import collections
import itertools
 
matchings = [(p, (0,) * 6) for p in itertools.permutations(range(6))]
 
#by symmetry, on first pull we only care about number pulled
pull_sets0 = [(1,) * k + (0,) * (6 - k) for k in range(1,4)]
 
#only makes sense to pull 3 or fewer switches in 1 go
pull_sets = [x for x in itertools.product(range(2), repeat=6) if sum(x) <= 3]
 
def go(matchings, bound):
    if len(matchings) == 1:
        return 1
    if bound == 0:
        return 0
    for pull_set in (pull_sets0 if len(matchings) == 720 else pull_sets):
        new_matchings = []
        for m, s in matchings:
            s = list(s)
            for i, j in enumerate(m):
                if pull_set[i]:
                    s[j] ^= 1
            new_matchings.append((m, tuple(s)))
        match_groups_1 = collections.defaultdict(lambda: [])
        match_groups_2 = collections.defaultdict(lambda: [])
        for m, s in new_matchings:
            match_groups_1[s[:3]].append((m, s))
            match_groups_2[s[3:]].append((m, s))
        if len(matchings) == 720:
            match_groups_2 = {}
        for match_groups in (match_groups_1, match_groups_2):
            if len(match_groups) > 1:
                if all(go(ms, bound - 1) for ms in match_groups.values()):
                    return 1
    return 0
 
for bound in itertools.count(1):
    print(f"trying {bound}...")
    if go(matchings, bound):
        break
print("success!!")