This one took a while to get online because I just could not figure out the last part… then the answer was explained to me, and it just bothered me more. *Anyway*, here’s the deal: You only get one shot at it, so make it count. You’re trying to make it through a labyrinth of sorts. Behind some doors are certain death, and behind others is the next challenge [“Safe”]. For the slower ones out there, you’re looking to *avoid* certain death. So here’s how it works: You start at trial 1. Get it right (by picking the right door) and you get a point, and move on to trial 2. Get it wrong and you are done. So if you get all of them right except trial 3, you get 2 points (for trial 1 and trial 2). Make sense? Good. If you get past all 10 trials [10 points] and solve The Final Trial, you’ll get 5 more points [Total of 15 points]. Then there’s always the 5 point bonus for being first. Off you go:

### The 10 Trials

- One sign is true and the other is false. It is possible that both rooms are Safe, or both rooms are certain death.

**Door 1**

This room is safe, and the other room is certain death.**Door 2**

One room is safe, and one room is certain death. - Both Signs are true or both signs are false.

**Door 1**

At least one of these rooms is safe.**Door 2**

The other room is certain death. - Both Signs are true or both signs are false.

**Door 1**

Either this room is certain death, or the other room is safe.**Door 2**

The other room is safe. - If Room 1 is safe, the sign on the door is true, but if it is certain death, the sign is false. In Room 2, the situation is reversed. Both rooms might be safe, or both rooms might be certain death.

**Door 1**

Both rooms are safe.**Door 2**

Both rooms are safe. - If Room 1 is safe, the sign on the door is true, but if it is certain death, the sign is false. In Room 2, the situation is reversed. Both rooms might be safe, or both rooms might be certain death.

**Door 1**

At least one room is safe.**Door 2**

The other room is safe. - If Room 1 is safe, the sign on the door is true, but if it is certain death, the sign is false. In Room 2, the situation is reversed. Both rooms might be safe, or both rooms might be certain death.

**Door 1**

It makes no difference what door you pick.**Door 2**

The other room is safe. - If Room 1 is safe, the sign on the door is true, but if it is certain death, the sign is false. In Room 2, the situation is reversed. Both rooms might be safe, or both rooms might be certain death.

**Door 1**

It*does*matter what door you pick.**Door 2**

You should pick the other door. - If Room 1 is safe, the sign on the door is true, but if it is certain death, the sign is false. In Room 2, the situation is reversed. Both rooms might be safe, or both rooms might be certain death. Neither sign is assigned to a door, they simply sit on the ground, with no indication of which door they belong to. One reads: “This room is certain death.” The other reads “Both rooms lead to certain death.”

**Door 1****Door 2** - Now 3 doors exist. One door is safe, and the other lead to certain death. At most, one sign is true.

**Door 1**

This room is certain death.**Door 2**

This room is safe.**Door 3**

Door 2 leads to certain death. - 3 doors still exist. The sign for the “Safe” room is true, and at least one of the other signs is false.

**Door 1**

Door 2 leads to certain death.**Door 2**

This door leads to certain death**Door 3**

Door 1 leads to certain death

### The Final Trial

One room leads to safety, the rest are either empty or lead to certain death. The sign on the door leading to safety is true. The signs on the doors leading to certain death are false. The signs leading to empty rooms can be either true or false.

**Room 1**: Safety is in an odd numbered room.

**Room 2**: This room is empty.

**Room 3**: Either sign 5 is right or sign 7 is wrong.

**Room 4**: Sign 1 is wrong.

**Room 5**: Either sign 2 or sign 4 is right.

**Room 6**: Sign 3 is wrong.

**Room 7**: Safety is *not* in room 1.

**Room 8**: This room is certain death, and room 9 is empty.

**Room 9**: This room is certain death, and sign 6 is wrong.

As is sits now, this riddle is impossible, but it would not be if I told you if room 8 was empty or not. That having been said, what door do you choose?

My thanks to Josh and Serita

1. Door 2

2. Door 2

3. Door 1

4. Door 2

5. Door 1

6. Door 2

7. Door 1

8. Door 2

9. Door 1

10. Door 1

Final. Door 7

2

2

2

2

1

2

1

1

1

1

Please explain what you mean by “the situation is reversed”. and could you give an example.

Thanks, it would be helpful.

The situation is reversed.

The given phrase is “If Room 1 is safe, the sign on the door is true, but if it is certain death, the sign is false.”

thus the reverse situation would be:

“If Room 2 is safe, the sign on the door is false, but if it is certain death, the sign is true.”

Nevermind the explanation…

1. #2 safe

2. #2 safe

3. #1 and 2 safe

4. #2 safe

5. #1 safe

6. #2 safe

7. #1 safe

8. #2 safe

9. #1 safe

10. #1 safe

Final Trial…..I choose room number 7.

final

room 7

I dug up my symbolic logic from the back of my mind for some of these.

Which is not to say that I got them right, just that I had fun doing them.

Here goes:

1. Door 2

2. Door 2

3. Door 2

4. Door 2

5. Door 1

6. Door 2

7. Door 1

8. Door 2

9. Door 1

10. Door 1

Final trial: room 8

1. Door 2

2. Door 2

3. Door 1 (and 2)

4. Door 2

5. Door 1

6. Door 2

7. Door 1

8. Door 2

9. Door 1

10. Door 1

Final Trial : Door 7

Can you explain the final trial? I guess I should have looked at it longer, since everyone else got it right, but now that I can’t get any points for it I’d just hear it explained…

Klaas: I had to have Serita explain it to me, so I will use her explanation. Let me know if you desire more information:

First, just from the statements on doors, we know that room 2 is either empty or has a tiger, and that the lady is not in rooms 8 or 9 since then their door statements would not be true.

Ok, so then, if we are told that 8 is not empty, then we know it must contain a tiger since it can’t contain the lady. But with a tiger, its statement must be made false. Thus, room 9 is not empty. But then, if 9 is not empty, 9 must contain a tiger, and therefore 6 must be correct in order for 9’s sign to be false.

Now, since 6 is correct, sign 3 is wrong (and therefore room 3 does not contain the lady). Using deMorgan’s law, we have that 5 is wrong and 7 is right from sign 3. Since 5 is wrong, no lady in 5, and since 7 is right, no lady in 1.

Ok, but if 5 is wrong, we know 4 is wrong. This is because the negation of statement 5 is [{wrong(2) or wrong(4)} and {wrong(2) or right(4)} and {right(2) or wrong(4)}]. If 2 is wrong, then four must be wrong, and if 2 is right, four must be wrong in order for the statement to be true. Thus 4 is wrong.

BUt if four is wrong, 1 is right. In which case lady is in odd-numbered room. But we know lady is not in 1, 3, 5 or 9.

Therefore lady is in 7.

If we were told that room 8 IS empty, then we can’t narrow it down to a specific room.. and since we’re told we can based on information given, we know room 8 must be empty.

[Thanks Serita!]

Ok, now I’m going to have to go through the whole thing myself. But in the meantime, I have two objections to the answer above.

“… sign 3 is wrong…. Using deMorgan’s law, we have that 5 is wrong and 7 is right from sign 3.”

“Sign 3: Either sign 5 is right or sign 7 is wrong.”

Either/or is exclusive, so deMorgan’s law doesn’t apply.

“Ok, but if 5 is wrong, we know 4 is wrong. This is because the negation of statement 5 is [{wrong(2) or wrong(4)} and {wrong(2) or right(4)} and {right(2) or wrong(4)}].”

“Sign 5: Either sign 2 or sign 4 is right.”

Looks like an identical either/or construction has now been negated differently. Also, it’s not true that 2>~4. In fact, 2>4 and ~2>~4.

I have to go home now, but I’ll keep working on this. Should I post here, or email Serita directly?

You’ve hit on the same problem that I had with it… and I complained to Serita about it… So if Serita wanted to jump in and say here what she said to me earlier, that’d be neat

From what I understand and remember, it all has to do with how you take the word “either”. You and I read it one way, where Serita [and logicians, apparently] read it another way. I think it goes like this:

If we have a stament “Either A is true or B is False”;

You and I would read that as “A is true and B is True” or “A is false and B is false” in order for that statement to be true. [Right?]

But Serita [from what I’m told is a “logic” definition] “either” would mean this:

“A is true, and B can be anything” or “A is false, and B is false” in order for that statement to be true.

If that’s how either is used, then DeMorgan’s law works on sign 3 to make it “5 is wrong and 7 is right”

As for sign 5, the construction is slightly different .. but I’m not sure if that matters. If sign 5 is right, then as far as I understand it [keep in mind I’m on your side with this whole “either” thing.. I have a full IM log of me railing against the English language and its ambiguity] neither 2 or 4 can be right. If just 2 is right, then sign 5 is true, if just 4 is right, then sign 5 is true, and if 2 and 4 are both right, then by that magic Either usage sign 5 is made true. Thus, 2 and 4 are both false. I don’t get how 2 can be right in this mix, but hopefully Serita does. My understanding is that neither 2 nor 4 can be right.. 2 ends up not matter, but 4 being wrong is important.

How much of that made sense? Yeah.. I guess all I’m really sure of is that there’s no way it could be in 8, 9, or 2.

Ok, upon further reflection, I see that “{wrong(2) or wrong(4)} and {wrong(2) or right(4)} and

{right(2) or wrong(4)}” is just a fancy way of saying that both 2 and 4 are wrong. Which, if we’re using an inclusive or, is true. In fact, it’s deMorgan’s law again.

So yeah, turns out both of my objections are due to treating either/or as inclusive (and obfuscating the second time).

It’s true that the logical or is inclusive, but I guess I thought that when you’re using words, adding the “either” changes the logical structure. I.e. “A or B” mean (A v B) but “Either A or B” means [(A v B) . ~(A . B)].

So I’ll believe that Serita’s answer works if you take “either/or” to mean “one or both”, but I don’t think it should be taken that way. And the question remains whether it’s solvable, and if so whether the answer is the same, if “either/or” is taken to mean “one but not both”. Probably I won’t be able to stop myself from tackling that, but at the moment I’m going to eat dinner.