@killer He's on the moon.

Ok, so my new best idea to the spider one is to not give them a strategy at all. Then as time approaches infinity, the chance of a spider pursuing a fly perfectly approaches 1 and since the spiders are faster, the fly will eventually be caught (or, at least, its probability of not being caught approaches 0). You said that it was a spatial reasoning question, though, and this is almost certainly not an ideal solution, so I'll keep thinking.

Diplo, can I ask two questions to the same god if I desire?

They have to be given deterministic orders, you could argue that any pseudo-RNG is deterministic, but the fly then knows exactly what the results of the RNG are and can plan accordingly.

I must say, diplo, the puzzle really seems like it should be impossible.

Let's say you ask the first question. OK. Now, you ask the second. Choosing it based on the answer to the first would not make that much sense, since you don't understand the first anyway, but let's say you can. Then you can get yes or no answers to either of two questions. We now have four possibilities -- DJ, DD, JJ, JD. You get one more question. The problem is, no matter how you're choosing, you _might_ end up asking the third question to the Random god, in which case, the answer would be completely worthless. It wouldn't tell you anything.
So there are only 4 results you can count on, yet there are six possible arrangements of the gods.

I'm sure I'm missing something here. I just wanted to gripe. lol.

just keep the riddle thread alive

There is a solution. And many debates on it. You can find them online ( I will not be so pompous to claim i invented such a good riddle).

OK thanks. :-) Just making sure. I hope you won't be offended if I say it had occurred to me as a possibility that you might have posted an impossible riddle as a joke and wanted to make sure before spending more time.

just to tempt and mess with you there is a whole wikipedia page to this one riddle

Nice, but I want to think longer first.

Here's one:

What starts with 'T' and ends in 'ru Ninja'?

A bad diplomacy player :-)

@ghug I think I may have it now, but I havent fully thought it through so maybe not.

Have two of the spiders be "guards" of the edges ac and bd while the last spider goeas around in a abcdabcd pattern forever. How the guards work is that the are always at a vertex if the last spider is at that vertex. So the algorithm could be:
Spider A starts at a then goes abcdabcd... forever.
Spider B starts at b, waits the amount of time it takes a spider to go travel an edge, then goes to d, then waits, then goes to b, and so on.
Spider C starts at a, then waits, then goes to c, then waits, then goes to a, and so on.

Nice job, that might be the fastest anyone's ever solved it.

now we need a new riddle

gotta go. Don't let this thread die! It must live on!

Ok, I can give two, both pretty doable (and I've seen them before, so they might not be particularly obscure). I'll look at the other one that seems impossible later:

You have 25 horses and can have 5 horses in a race at a time. How many races do you have to have to identify the top 3 horses?

Clarifications: We are assuming that if 1 horse is better than another, it will beat that horse every single time. Further, if A is better than B and B is better than C, A is better than C. Last, you can't time the horses and compare their results across races (in other words, the answer isn't 5). Each race only directly gives you information about the relationships between the 5 horses in that race.

Next one:

You have 60 meters of rope and are at the top of an 80 meter cliff. Halfway down, there is a ledge on which you can stand. Both the top of the cliff and the ledge have something stable you can tie a rope to. The rope can just barely hold your weight so you can't modify it to make it longer. How do you get down? Note: The solution doesn't involve saying something like "and the you could survive falling the last 10 meters".

I guess maybe you need a knife for the second one.

Answer to the first one is 7, I believe. Starting the second now.

The second one's stumping me, and I have an essay to procrastinate on, I'll try it again tomorrow if nobody else has solved it.

Glad I got you with one of them. :)

I think I got it.

Cut the rope intoi a 20 meter and 40 meter section. Tie off one end of the 20 meter to the top and tie a loop in the bottom of it. Then feed the 40 meter rope through the loop so it is half (20 meters) on each side. Climb down and when you get down you can pull the 40 meter back through the loop to tie off and climb the rest of the way down.

Nice. Now that I have nothing to distract me I can leave in peace.

OK diplo, you didn't answer my question, but I'm going to assume that I can ask two of my questions to the same god. If so, then I have a solution.

First: I will make extensive use of "xor," and for shorthand, that's what I'll say. If the gods didn't recognize that as an English word, I would spell it out, saying, "Is it true that precisely one of p and q is true, where p is the proposition that...." etc.

Second: it turns out the language thing doesn't matter at all. If instead of the question Q, I ask (Q xor (Da means no)), then it turns out that I can act, with respect to True and False, as though Da means yes and Ja means no. So from now on, I'll just act as if they are saying yes or no, for ease.

There are six possibilities. Labeling in order ABC, these are TFR, TRF, FTR, FRT, RTF, RFT. The first question (Q1) I would ask, to A, is "Is C Random xor You False?" If the answer was Yes (Da), the remaining possibilities would be RTF, RFT, TFR, FTR. In particular, notice that I now know that B is either T or F. So now, ask Q2 "Is 2 + 2 = 4?" If he said yes, then I know I have RTF or FTR, so I now just ask B, "Is A Random?" I know I can trust his answer, so that determines everything. If the answer to Q2 is false, I know B is False, so I can just do the same but take the negative.

Similarly, if the answer to Q1 is no, then I have RTF, RFT, TRF, or FRT. This time, I know that C is either True or False, so I now do the same trick, but with him.

And that is all.

Nice job Draug!

@semck The language thing is brilliant! The second part I had but figuring out that the language didn't matter at all was something I didn't even consider.

Thanks ucla! I was very happy when I thought of that.

So here's one, mainly for US people. John, who lives in a pacific coast state, was talking to Barbara, who lives in an Atlantic coast state, and the clocks on each of their walls read (correctly) exactly the same time. How was this possible?

There's a bit of Oregon in MST and a bit of Florida in CST. When daylight savings time falls back, CST does it first, and thus shares the time with MST for an hour.

Good job ghug.

Alright, I'm actually done for the night now. I''ll leave one last one before I go:

There are 150 people on an island, 50 of them have blue eyes and 100 of them have brown eyes, none of them know the eye color distributions beforehand, only that brown and blue are the only possible eye colors. They are not allowed to communicate, and there are no reflective surfaces on the island. Every day, anyone who has determined their eye color with complete certainty leaves the island, and everyone else remains. Who leaves, and when?