Escape the Black Ship - Win a birdy!

Kharnor1 · Posted: Tue Apr 08, 2008 3:03 am Post subject:

Yay! Someone else finally got it!

Congrats Sante! Very Happy

I'll give the rest of you, hmm... exactly 5 days to try and solve it before I post the solution.

Sante, I'll try and catch you ingame sometime. Unless you don't actually want the birdy either? Razz

Sante · Posted: Tue Apr 08, 2008 12:17 pm Post subject:

Yay! Erm, I mean, Yarrrrr!

(/me wants birdy. But let me know who was next in line, in case I get over it.)

Thanks Kharnor.

Sante
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Sante
~~Ultimate~~ Variable sailor, crew poet,
As lost as ever.

Carrol · Posted: Tue Apr 08, 2008 1:45 pm Post subject:

Congrats Sante!

I'm still perplexed. Sad

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Spacehog · Posted: Tue Apr 08, 2008 3:24 pm Post subject:

My final answer was going to be....."The skellies, at dawn". But somehow I doubt that would have been right, or would the joke been appreciated!!!!!

Kusje has tried, oh my how he has tried to explain the answer to me. But I just can't get my head around it. No matter how he explained, and tried different examples, I just can't accept the logic of it.

I do believe the answer is right Khar.....I just at the same time can't accept that it it truly does work that way. It's an enigma, wrapped in mysetry and surrounded in complete insanity!!!!

(Now to work out how to bribe Sante!!!!). Hmm limericks, poems...cold hard cash!!!!!!!

Well done Sante, maybe it's because I have a scientific mind, or maybe it's because I am thick, but I am so glad someone lese did solve it!
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Kharnor1 · Posted: Tue Apr 08, 2008 6:47 pm Post subject:

Bullseye Bud already tried that, Space Very Happy

Sante now has her birdy. Kusje, you can bug her for that portrait. Smile

Kharnor1 · Posted: Sun Apr 13, 2008 6:54 am Post subject:

Bah. Had a long post typed up and lost it somehow.

Anyway, here's the original problem:
http://www.xkcd.com/blue_eyes.html

Here's the solution:
http://xkcd.com/solution.html

Here's a 300+ post forum discussion attacking the problem from all possible angles, including a lot of great explanations:
http://forums.xkcd.com/viewtopic.php?f=3&t=3

Basically, the 50 red skellies leave on the 50th day. The reasoning includes a lot of long chains of "I know that A knows that B knows that.... ....knows that Z sees at least x red skellies". It's very hard to wrap your head around it, and even once you know the answer it's hard to accept it. But I assure you that it works!

There's a reason it's described as 'the hardest logic puzzle in the world'.

You may all stab me now. ^__^

ara · Posted: Sun Apr 13, 2008 12:04 pm Post subject:

/me stabs kharnor repeatedly
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Sante · Posted: Sun Apr 13, 2008 1:20 pm Post subject:

The easy way for me to get my head around it is by each skellie thinking "okay, the only way this is going to work is if everyone counts the number of red skellies they can see, and waits that many days plus one until they leave. If the red skellies all leave before that day, then I know I'm not red."

This works because all the red skellies will only see 49 red skellies, so they all leave on the 50th day. All the blue and white skellies see 50 red skellies, so they would leave on the 51st day, except that the red skellies have already left the day before, so they know not to after all.

Actually, you could almost argue, though I didn't include it in my original answer, that the blue and white skellies, having seen what the red skellies did, would do the same, (since there is more than one skellie of each colour). Obviously, the smaller number would go first, so the 26 white skellies would leave on the 76th day, and the 34 blue skellies on the 110th day. No more black ship! If there had been a lone skeleton in some other colour, they wouldn't have escaped, and there's probably some exception where the numbers of blue and white skeletons are similar, that I need to figure out too. EDIT: No, there's not. It still works even if there are the same number of skellies, or if they are one number different.
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Sante
~~Ultimate~~ Variable sailor, crew poet,
As lost as ever.

kusje · Solid Forum Tart Joined: 20 Jan 2006 Posts: 188

You can't really argue that though. Because they don't know that there are only blue and white skellies left. For all you know, there might be a pink skellie somewhere in the mix.

There's another argument against that but I can't seem to remember it. Will try to recall and post more later.

Bullseye Bud · Posted: Sun Apr 13, 2008 3:04 pm Post subject:

No, they are still on the hunt and will replentish thier fold before they all escape...

hehe

Kharnor1 · Posted: Sun Apr 13, 2008 4:52 pm Post subject:

You can't use the "they just have to use the same strategy" approach, because they need to be SURE of their colour. In order to be totally sure, they must know that the strategy works, and they have to know that everyone else knows it works, and they have to know that everyone knows that everyone knows... etc.

Which they don't, because it doesn't work in every case.

Start from the bottom and work up:
If theres one white, he won't see any others and he will never leave.
If there are two whites, they will each see one, know that the other one will not leave no matter what he sees, so on the second day they have no more information. They can't just leave because it fits the strategy, because it's possible that one of them is a different colour and the other guy just didn't leave because he doesn't know there are any whites at all.
If there are three, they each see two whites going through the same process, and therefore THEY can't leave either, because the two whites won't have a reason to leave on day 2.
And so on.

This is one theory that is brought up and bashed down repeatedly in the thread I linked to.

Sante · Posted: Sun Apr 13, 2008 5:20 pm Post subject:

Apologies Kharnor, for not reading the other link. And yes, the "needing to be sure of their colour" bit is probably the crux of it. I'll have to go read it now!
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Sante
~~Ultimate~~ Variable sailor, crew poet,
As lost as ever.