Knights vs. Knaves

Cathy says there are at most 3 Knights, while Dave says there are at most 3 Knights. Clearly, at least one of these two people is telling the truth. Let's take a look at the cases:

• There are less than 3 Knights. Then, Dave is a Knight. Since Alice, Evelyn, and Frank claimed there were exactly 0, 2, and 1 Knights respectively, exactly one of them must be a Knight. This makes 2 Knights (the truth-teller and Dave), so there are 2 Knights: Bob and Dave.

• There are exactly 3 Knights. Then, Cathy and Dave are Knights. However, the third Knight would have to have said there were exactly 3 Knights, but no one did, so this is impossible.

• There are more than 3 Knights. Then, Cathy is a Knight. But only one of Alice/Bob/Evelyn/Frank could be telling the truth, so there are at most 2 Knights, thus causing a contradiction.

Hence, the only possibility is that there are 2 Knights: Bob and Dave.

Just a little semantic to be sorted out, you never explicitly stated that all the people in the group were either knights or knaves, so, it could be argued that some of them are not knights but are still telling the truth.

Alice is obviously not a Knight because if she was it would be a paradox.

Bob, Eve and Frank all claim there to be an exact number, so 2 are liars. Eve is lying because if she was telling the truth then all 3 would be telling the truth, which can't happen.

We know that there's at most 3 Knights. We know that Cathy is lying because that means either Frank or Bob are right, but since they all claim there are less than 3 knights they contradict Cathy, meaning she's lying.

We now know for 100% sure that Dave is telling the truth.

This is where it gets a little tricky. Bob can't be lying, because if he was that would mean that Frank is lying. However that leaves Dave as the only Knight, which means that Frank is right, which causes a paradox. Therefore Bob is telling the truth and Frank must be lying.

2 Knights, Bob and Dave

I'm not a fan of the wording. What does "At most, three of us are Knights" mean? It should be "Three or less of us are Knights." Here's why: Suppose Frank is a Knight. Then at most there is one Knight, so Dave's statement is incorrect. Then A-E are all telling lies, thus are Knaves, and Frank's statement is true, and consistent with him being a Knight.

If Dave said "Three or less of us are Knights" then Frank cannot be a Knight, as the problem has as the solution.

Alice - "None of us are Knights." Obviously this cannot be true because if it were true, then she would be a knight, meaning that her claim would have to be Exactly one of us is a Knight.

Bob - "Exactly two of us are Knights."
This cannot be determined yet.

Cathy - "At least three of us are Knights This cannot be determined either.

Dave - "At most three of us are Knights." This cannot be determined either.

Evelyn - "Exactly five of us are Knights." Again, this cannot be determined yet.

Frank - "Exactly one of us is a Knight."
If this one is indeed true, then both he and Dave would be correct since Dave is claiming that there are less than three Knights present. Since both claims would be true, there would in fact be two Knights present, meaning Frank is lying.

At this point, Bob, Cathy, Dave, and Evelyn remain.

Since there are only four remaining who can be Knights, Evelyn must be lying since her claim is that there are exactly five present.

Now, there are only the three of them left: Bob, Cathy, and Dave. Bob is claiming that there are only two Knights. Cathy claims that there are three. Since both of these are less than or equal to three Knights, Dave's claim must be true, automatically making him a Knight.

However, since Cathy is claiming that there are at least three Knights, she is automatically counting Bob out if she were telling the truth because Bob states that there are in fact only two Knights present. Therefore, Cathy must be lying.

This leaves Bob and Dave. Both of their claims are truthful, proving them to be the two Knights present in the group of six.

The above graph helps visualize the solution. Each vertex represents one of the six people in the group. Verticies are connected with an edge if their statements are compatible i.e. both statements can be true at the same time. E.g. Dave's statement is compatible with Cathy's as there could be three knights exactly.

Since the graph is connected , there exists at least two knights.

However, a quick inspection of the graph reveals that the largest complete subgraph has two verticies (K2). This means that at most, two nodes from the graph are compatible with each other, or at most two people are knights. If there were three or more knights, there would be a triangle shaped cycle in the graph but there isn't...

So... there are two knights.

Alice, Evelyn and Frank are Knaves since their statements can't be true in any situation leaving 3 left that could be knights, it's obvious that either Cathy or Dave is a Knight meaning there is at least one Knight, but Cathy says there must be three meaning Bob would be a Knight but if Bob is a Knight only two can be Knights meaning Cathy is a Knave thus making Dave a Knight by default and since Bob does not contradict Dave they are both the only Knights available.

Ifill out the following grid which has the claims of each person in each row and the Truth or Falsness of that claim if the actual number is the one at the top of the column. The only column that has the number of T's that matches the actual is 2. Hence there are 2 Knights. We don't need to know who they are: -

My thought process:

Alice is not a Knight because the statement "none of us are knights" contradicts her being a knight, so I only thought about the other cases.

I noticed that the 2 knight, 5 knight, and 1 knight scenarios are mutually exclusive (bob, evelyn, frank respectively). Therefore out of the 3 of them there is at most 1 knight.

Only one of Cathy and Dave will back up any of those scenarios, therefore boosting the potential knight count to 2. Cathy and Dave both back the 3 knight scenario but not any other person, therefore at most there are only 2 knights.

I was wondering if this thought process is the quickest, or if there are better ways of grouping/eliminating possibilities.

Alice says, "None of us are Knights."
Bob says, "Exactly two of us are Knights."
Cathy says, "At least three of us are Knights."
Dave says, "At most three of us are Knights."
Evelyn says, "Exactly five of us are Knights."
Frank says, "Exactly one of us is a Knight."

There are 6 statements made between Alice, Bob, Cathy, Dave, Evelyn and Frank, 4 in direct equalities and 2 inequalities. Among the group of the 4 using equalities, there should be AT MOST only one Knight there, while for the rest who used inequalities, there should be AT LEAST one Knight there though they might be both telling the truth only if the number of Knights equal 3. But if there really were 3 Knights, then they need an equality which would have said exactly that as the last Knight around the table.

Rechecking, we only have A = 0, B = 2, C ≥ 3, D ≤ 3, E = 5 & F = 1, and ? = 3 is nowhere to be found. Thus, only one of the inequalities told the truth (and that's Dave) but there are those who claimed of 1 or 2 Knight(s) from equalities, too. Since the number spoken have to take the speaker's self into account in addition to the established facts of AT LEAST one from the inequality group, then the real Knights must be D ≤ 3 & B = 2, an assortment of 2 person.

If Alice was telling the truth, that would make her a knight, contradicting her own statement. Therefore she is a knave, and we can eliminate her.

If Bob was a knight, he and Cathy/Dave/Evelyn or Frank are knights. However Cathy states there are 3 or more knights, not exactly 2. Evelyn says exactly 5 and Frank says exactly 1, so they are not knights. However, Dave says at most 3 of them are knights, meaning there could be 1, 2 or 3. This is the only statement that fits with Bob stating there are 2 knights, so they are both telling the truth. So Bob and Dave are the 2 knights.

Alice can't be right, for her statement would contradict itself. Evelyn can't be right either, for there are at least 3 "exactly" statements. Cathy can't be right, for hers would cancel out Bob's statement and make hers untrue. Neither Bob's nor Dave's statements contradict each other or themselves, so they are the two Knights.

Case 1: None of us are knights: Contradiction, knaves lie

Case 2:Exactly two of us our knights: If true then, No contradiction

Alice:Lie-Knave

Bob: Truth-Knight

Cathy:Lie- Knave

Dave: Truth- Knight

Evelyn: Lie knave

Frank:Lie-knave

Case 3:"At least three of us are Knights." If true leads to contradictory solution of only two knights possible Alice:Lie-Knave

Bob: Lie-Knave

Cathy:True- Knight

Dave: Truth/lie- Knight if Evelyn is false

Evelyn: Truth/lie- knight is Dave is false

Frank:Lie-knave

Case 4 "At most three of us are Knights." If true leads to contradiction

Alice:Lie-Knave

Bob: Lie-Knave

Cathy:Lie- knave

Dave: Truth- Knight

Evelyn:Lie knave

Frank:Lie-knave

Case 5:Exactly five of us are Knights If true: leads to contradiction as others indicate different values simultaneously

Case 6:Exactly one of us is a Knight If true: contradiction found in Dave's statement being true.