Murder Mysteries at Masochist's Cove

The Game Room
1

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Normally it would seem too, well, sadistic to post Murders at Masochist’s Cove. But with Chronos asking about the strange spear-like thing made of dry yellow wood, this seems fair game.

Hercule Poirot is awakened one morning by a call from his old friend, the Constable of Masochist’s Cove. “We need you at once, Hercule. Please!”
– Oh no; I’d need double my usual fee to contend with the chicanery at Masochist’s Cove.
“Then you’re in luck! We have two murders for you to solve.”

Inspector Poirot hopped on a jetfoil.

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Background

There are four Islands in Philosopher’s Archipelago:
[ul][li] Truther’s Island, whose inhabitants speak only the truth.[/li][li] Moe’s-Leetroo Island, whose inhabitants utter mostly true statements, but always accompanied by a lie.[/li][li] Mosely-Fib Island, whose inhabitants utter mostly false statements, but always accompanied by a truth.[/li][li] Liar’s Island, whose inhabitants speak only lies.[/li][/ul]
On Halloween, the natives of all four islands meet for fun and games at Masochist’s Cove. Of course any native can easily let the others know where he comes from; for example he can say “Two plus Two is Four. Two Plus Two is Four. Two plus Two is Five.” and it will be obvious he comes from Moe’s-Leetroo. But they find it more fun to keep others guessing!

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Unfortunately there were two murders at yesterday’s Halloween celebration. In each case, there were three witnesses; exactly one of the witnesses fired the fatal bullet. All three witnesses know who the murderer is, and what Islands the others come from. The Constable has managed to glean this much, and has taken down some statements, but all the suspects are refusing to give any further statements. You know that 3, 2, 1 or 0 of each suspect’s sentences are true, depending on which island he comes from, but you don’t know which islands these suspects come from.

In the first interrogation room, you find Abel, Bera and Cain, one of whom murdered Pandora. All you have to go on are their original statements.

Abel says
(1) Exactly two of us three are from Mosely-Fib Island.
(2) I murdered Pandora.
(3) None of us are from Liar’s Island.

Bera says
(4) Exactly one of us is from Mosely-Fib Island.
(5) An odd number of us are from Moe’s-Leetroo.
(6) Cain murdered Pandora.

Cain says
(7) Either we’re all from Mosely-Fib Island or none of us are.
(8) At most one of us is from Moe’s-Leetroo.
(9) Abel murdered Pandora.

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In the second interrogation room, three suspects are being quizzed about the murder of Delilah. The circumstances are similar; in fact you’ll see that the statements are all analogous to the first murder except for Rob’s (1).

Rob says
(1) An odd number of us are from Mosely-Fib Island.
(2) I murdered Delilah.
(3) None of us are from Liar’s Island.

Sam says
(4) Exactly one of us is from Mosely-Fib Island.
(5) An odd number of us are from Moe’s-Leetroo.
(6) Tom murdered Delilah.

Tom says
(7) Either we’re all from Mosely-Fib Island or none of us are.
(8) At most one of us is from Moe’s-Leetroo.
(9) Rob murdered Delilah.

Can you help Poirot? Who killed Pandora? Who killed Delilah?

2

To be clear, when the suspects say things like “all of us”, they just mean the three suspects for that particular murder, not all six of them? That is to say, this is two separate puzzles, not one big puzzle?

3

Correct.

4

OK, tackling the first one. I think that the first thing to figure out is how many of them are from Mosly-True. Can they all be from there? No, because in that case both (1) and (4) are false, and (2) and (6) are inconsistent, so A and C can’t both be from Mosly-True.

Can there be exactly two from Mosly-True? In that case, (1), (5), and (8) are all false, meaning that for the two MT-ers, their other statements must all be true. (6) is inconsistent with both (2) and (9), meaning that B can’t be the MT-er, so A and C both are. In this case, A would be the murderer, and (3) and (7) must both be true, which means that there are no liars or MF-ers. But B has at least two lies in this case, so he must be a liar or MF-er, an inconsistency. Thus, there are not exactly two MT-ers.

Between those two statements, we have that there is at most one MT-er. But that is exactly what (8) says, and therefore, Statement (8) is definitely true.

OK, now how many can there be from Liar’s Island? Clearly not all three, since C spoke a truth. Can there be exactly two? Then they must be A and B. From the falsehood of 5, there must be an even number of MT-ers, hence zero. Neither C nor A can be the murderer, so it must be B, which makes (9) false. C cannot be an MTer, a truther, or a liar, so he must be an MF. But that would mean exactly one MF, so (4) would be true, a contradiction. So there cannot be exactly two liars, either.

How about one liar? If there is one liar and it’s A, then…

Hm, wait, I seem to have found a consistent solution. If A is a liar, B is a truther, and C is a mostly-fib and the murderer, then we have no inconsistencies. But that doesn’t mean that that’s the only consistent answer.

Let’s see if I can find any situation where C is not the murderer. In that case, (6) is false.

…I don’t have time for any more right now.

5

It’s easy to get the placenames Moe’s-Leetroo and Mosely-Fib confused. I should have used more distinctive names, perhaps Fibber’s Island for the latter.

6

Yes, but I don’t think I fell afoul of that. The only place I crossed them was that if there are exactly two MTs, then the statement that there are exactly two MFs is false.

Oh, and I don’t expect that the case I found is the only consistent one. The puzzle only asks who the murderer is, not the islands of the witnesses, and so it’s possible that there are other configurations with the same murderer. It’s also possible that there are consistent configurations with different murderers, of course, and while this would make it a poor puzzle, that possibility needs to be logically ruled out, not just discarded on the grounds of poor puzzling.

7

With zero from Moe’s-Leetroo, B’s (5) is false and B is not a truther.

Blame me. The “Mosely-Fib” / “Moe’s-Leetroo” were very poor name choices.

8

Answers:

Cain killed Pandora
Tom killed Delilah

9

By my reckoning you’ve caught one murderer, but have let the other murderer go free.

10

Hmmm… so that means 0 mostly-truthers, but I’ve always assumed that 0 was an even number, making B’s statement #5 false.
Just looking at the first puzzle, I think I decided that none of them can be truthers. But I haven’t worked out what they are.

11

I hope the word “Masochist’s” in OP title gave fair warning. I’m somewhat relieved that others found the puzzle difficult; I was afraid it was just me.

Correct. I’ll spoiler a slight Hint.

[SPOILER]By considering each of his responses in turn, 8, 7, 9, you can determine, with much difficulty, the truth/falsity of those statements and deduce that Cain is from Mosely-Fib Island.

But that’s the end of it! As Chronos guessed, Abel and Bera can each be from Mosely-Fib, Liar’s or Moe’s-Leetroo Island. (But not from Truther’s Island as TexCat points out.) There are three different solutions, but in each case the murderer is the same.

Similarly, in the 2nd puzzle you can deduce that Tom is from Mosely-Fib Island. There are three different cases again (though this time a Truther is possible), all with the same murderer, though not the murderer corresponding in puzzle 1.

Even when these hints the puzzles may be very hard.[/SPOILER]

12

I am curious, septimus – Did you write this puzzle yourself? It is really impressive!

13

Yes, I constructed the two puzzles the same day I posted them. I also wrote a computer program in C to verify solution uniqueness. (In fact I’ve been half-expecting a programmer to show up and demonstrate the ease with which a Prolog program could solve these puzzles.)

Whether it’s “impressive” is another matter. I think it’s interesting that the logical implications among just nine simple statements can be so complicated and tedious to untangle. In the first puzzle, exactly one of #1, #4, #7 must be true – I hoped that might lead to a shortcut some clever Doper would find that I didn’t.

Maybe I’ll hunt through my old folders and see if I can find a fun-looking puzzle that hasn’t already been published under my real name.

14

I’m still on the first puzzle and can find only one answer that Bera is the murderer. Are you sure there are three answers?

A is mostly true (T F T). B is mostly false (F T F). And C is mostly false (F T F).

15

I did find 2 solutions to the second puzzle and both of them point to Tom.

16

There is another solution to Puzzle 1, with Abel a Liar; and a third solution with Bera a Liar. In each case Bera is the murderer.

(Again, I hope the “Masochist” gave warning these were obscenely difficult. When I contributed to Logic Puzzles I asked whether puzzles could be too difficult and was told No; indeed they had to be extremely difficult to earn the maximum, $125 stipend. Since solution proof outlines were required, it was not a high-wage job. :stuck_out_tongue: )