Aid for logical puzzle

Question

I am looking for some help to solve this logical puzzle. The problem I am asked to solve is:

"You’re a famous detective and you’re trying to solve a second murder. You know that the murderer was one of exactly 6 suspects. Among the suspects, one was actually a witness of the crime, but you don’t know who this is. The witness is afraid of the murderer and will not say anything if the murderer will be able to hear it. What you can do is repeatedly select a group of people to go to a separate room. There you can ask them if anyone knows who the murderer is. If the witness is in the room and the murderer is not, they will speak up. You could take all 6 people aside one-by-one, but that will take a long time. How small of a number of groups can you take aside to guarantee that the witness speaks up?"

After studying the proposed solutions from the community, the general agreement was: one needs 4 groups. I have found a 3 groups solution.

Let’s call N1, N2, N3, N4 the suspects who don’t know who the murderer is, and W and M the witness and the murderer.

We start by splitting the suspects into two groups of three.

First step:

N1 N2 N3 N4 M W

Second step: N4 N2 N3 N1 M W

Nobody speaks up, but we know that N1 is neither the witness nor the murderer.

Third step: N4 M N3 N1 N2 W

Now witness and murderer are separated and can be identified.

Answer 1

This is the Fano Plane:

There are 7 points and 7 lines. Every point is connected to every other point by exactly one line. Each line has 3 points on it.

You only have 6 suspects, so you can throw away one point, and the lines through that point.

This will leave 4 lines. Interrogate the group on each line. This will mean that each witness is interrogated twice, but there will be different people in the room each time.

Answer 2

To show that 3 groups is not sufficient:

• We want to identify the (witness, murderer) ordered pair.
• There are $ 6 \times 5 = 30 $ pairs to check.
• Each grouping allows us to say that (person in group, person not in group) is NOT the pairing.
• Each grouping allows us to check at most $ x(6-x) \leq 9 $ pairs.
• Hence, to check all 30 pairs, we need at least $ \lceil \frac{30}{9} \rceil = 4$ groups.
• User317176 provided such an example of 4 groups, namely checking $EGB, CGF, EDC, BDF$ (we drop point A)