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[Qvist]

Why don't they just look into the water of the ocean around them? 
But this is an interesting riddle and pretty similar to one I enjoyed the most.


There are 3 Logic students having their final exam. The professor says:
"Instead of writing an exam I just let you having a final test.
There are 5 hats, 3 of them are red and 2 of them are white.
Please stand in a row, so that one of you can see the other two, another one can only see one and the front one can see no-one.
I will put a hat on each of you and then removing the 2 leftover hats out of the room.
You aren't allowed to talk or use any help or items.
If anyone of you can tell me what color the hat has he's wearing, all of you pass this test."
After an hour of silence the student on the front who couldn't see any other student gives the right answer.
--
How did he do that and what color had his hat?

[soulnet]

Alternate solution: They will all return to the deck when the game ends.

I guess this being in the non-Dominion related subforum, this solution is incorrect, though.

[GendoIkari]

Nope, still dont getit.

Start with 3 people with blue and 3 with green.
B1
B2
B3
G1
G2
G3

Now obviously they can all see each other, and the Blues can see 2 other blues.

So when the stranger turns up and says I can see someone with blue eyes.
If I am B1, he can possibly be talking about B2, B3 or even me. But also i know that B3 can see B2 (and vice versa), so they could assume the stranger is talking about the other one.

Before the stranger's comment... if you are B1, you know that everyone knows there's at least 1 blue-eyes. However... for all you know, you have green eyes. So you assume for the moment that you do have green eyes. And if you do, then B2 would only be seeing 1 blue-eyes person (B3), and so B2 has no way of knowing whether or not B3 sees a blue-eyes person. So for all you know, B2 doesn't know if B3 knows that there's a blue-eyes person.

[Ozle]

I can understand it with 98 green and 2 blue obviously, as the other blue person must only be able to see me

[Ozle]

Nope, still dont getit.

Start with 3 people with blue and 3 with green.
B1
B2
B3
G1
G2
G3

Now obviously they can all see each other, and the Blues can see 2 other blues.

So when the stranger turns up and says I can see someone with blue eyes.
If I am B1, he can possibly be talking about B2, B3 or even me. But also i know that B3 can see B2 (and vice versa), so they could assume the stranger is talking about the other one.

Before the stranger's comment... if you are B1, you know that everyone knows there's at least 1 blue-eyes. However... for all you know, you have green eyes. So you assume for the moment that you do have green eyes. And if you do, then B2 would only be seeing 1 blue-eyes person (B3), and so B2 has no way of knowing whether or not B3 sees a blue-eyes person. So for all you know, B2 doesn't know if B3 knows that there's a blue-eyes person.

Perfect, got it

[Ozle]

Its very much like the Unexpected Hanging Paradox

[Reyk]

Why don't they just look into the water of the ocean around them? 
But this is an interesting riddle and pretty similar to one I enjoyed the most.


There are 3 Logic students having their final exam. The professor says:
"Instead of writing an exam I just let you having a final test.
There are 5 hats, 3 of them are red and 2 of them are white.
Please stand in a row, so that one of you can see the other two, another one can only see one and the front one can see no-one.
I will put a hat on each of you and then removing the 2 leftover hats out of the room.
You aren't allowed to talk or use any help or items.
If anyone of you can tell me what color the hat has he's wearing, all of you pass this test."
After an hour of silence the student on the front who couldn't see any other student gives the right answer.
--
How did he do that and what color had his hat?

I assume his (nr. 1) hat is red. If hats of nr. 2 and nr. 1 were white nr. 3 - who can see both - would announce his red hat.
If he doesn't do this nr. 2 can announce his red hat if nr. 1 has a white one.
If he doesn't do this either nr. 1 can announce his red hat regardless of the color of the other two.

[GendoIkari]

Its very much like the Unexpected Hanging Paradox

Hah, I just spend some time a few days ago trying to explain that one.... I host a game night, and we were talking about switching which night we have it. We couldn't decide on a night, so I joked that game night would be on a random night each week; and no one would be able to expect which night it happens on; until I send the email saying it's time for game night... the paradox of the unexpected game night.

[Galzria]

Logic Problems? Ehhh... This is close enough:

Here's a sequence of numbers. Can you tell me what the next number is and why?:

4, 6, 12, 18, 30, 42, 60, X

[Ozle]

Nope, still dont getit.

Start with 3 people with blue and 3 with green.
B1
B2
B3
G1
G2
G3

Now obviously they can all see each other, and the Blues can see 2 other blues.

So when the stranger turns up and says I can see someone with blue eyes.
If I am B1, he can possibly be talking about B2, B3 or even me. But also i know that B3 can see B2 (and vice versa), so they could assume the stranger is talking about the other one.

Before the stranger's comment... if you are B1, you know that everyone knows there's at least 1 blue-eyes. However... for all you know, you have green eyes. So you assume for the moment that you do have green eyes. And if you do, then B2 would only be seeing 1 blue-eyes person (B3), and so B2 has no way of knowing whether or not B3 sees a blue-eyes person. So for all you know, B2 doesn't know if B3 knows that there's a blue-eyes person.

Hang on, lost it again!
Haha

B1
B2
B3
B4
GX (assume the number of green people doesnt matter?)

I am B1, I dont know whether I have green eyes or not, but now we are back to where I was before taking meout of the equation, that B2 can clearly see two people with Blue eyes, so could be talking about either of them, B3 or B4.

[Galzria]

Logic Problems? Ehhh... This is close enough:

Here's a sequence of numbers. Can you tell me what the next number is and why?:

4, 6, 12, 18, 30, 42, 60, X

I'll even be nice, and tell you that the number that comes after the X is 102.

[Ozle]

Logic Problems? Ehhh... This is close enough:

Here's a sequence of numbers. Can you tell me what the next number is and why?:

4, 6, 12, 18, 30, 42, 60, X

78?

[Galzria]

Logic Problems? Ehhh... This is close enough:

Here's a sequence of numbers. Can you tell me what the next number is and why?:

4, 6, 12, 18, 30, 42, 60, X

78?

No!

[theory]

The blue eyes puzzle was nicely laid out by xkcd a while ago: http://xkcd.com/blue_eyes.html (no spoilers)

if you want the solution, replace "blue_eyes.html" with "solution.html"

[GendoIkari]

Nope, still dont getit.

Start with 3 people with blue and 3 with green.
B1
B2
B3
G1
G2
G3

Now obviously they can all see each other, and the Blues can see 2 other blues.

So when the stranger turns up and says I can see someone with blue eyes.
If I am B1, he can possibly be talking about B2, B3 or even me. But also i know that B3 can see B2 (and vice versa), so they could assume the stranger is talking about the other one.

Before the stranger's comment... if you are B1, you know that everyone knows there's at least 1 blue-eyes. However... for all you know, you have green eyes. So you assume for the moment that you do have green eyes. And if you do, then B2 would only be seeing 1 blue-eyes person (B3), and so B2 has no way of knowing whether or not B3 sees a blue-eyes person. So for all you know, B2 doesn't know if B3 knows that there's a blue-eyes person.

Hang on, lost it again!
Haha

B1
B2
B3
B4
GX (assume the number of green people doesnt matter?)

I am B1, I dont know whether I have green eyes or not, but now we are back to where I was before taking meout of the equation, that B2 can clearly see two people with Blue eyes, so could be talking about either of them, B3 or B4.

When you go to B4 instead of B3, you just have to extend the logic out one step further. As B1... assume for the moment you have green instead of blue eyes. Think about what B2 now knows... Well, B2 is now in the exact same position that B1 was in earlier, back when it only went up to 3. So for all you know, B2 could be following all the same logic that you would follow if you were B1 in the situation where it only went up to 3 instead of 4.

[Reyk]

Nope, still dont getit.

Start with 3 people with blue and 3 with green.
B1
B2
B3
G1
G2
G3

Now obviously they can all see each other, and the Blues can see 2 other blues.

So when the stranger turns up and says I can see someone with blue eyes.
If I am B1, he can possibly be talking about B2, B3 or even me. But also i know that B3 can see B2 (and vice versa), so they could assume the stranger is talking about the other one.

Before the stranger's comment... if you are B1, you know that everyone knows there's at least 1 blue-eyes. However... for all you know, you have green eyes. So you assume for the moment that you do have green eyes. And if you do, then B2 would only be seeing 1 blue-eyes person (B3), and so B2 has no way of knowing whether or not B3 sees a blue-eyes person. So for all you know, B2 doesn't know if B3 knows that there's a blue-eyes person.

Hang on, lost it again!
Haha

B1
B2
B3
B4
GX (assume the number of green people doesnt matter?)

I am B1, I dont know whether I have green eyes or not, but now we are back to where I was before taking meout of the equation, that B2 can clearly see two people with Blue eyes, so could be talking about either of them, B3 or B4.

That's why they all leave the island on 100th day and not earlier.

[Drab Emordnilap]

Logic Problems? Ehhh... This is close enough:

Here's a sequence of numbers. Can you tell me what the next number is and why?:

4, 6, 12, 18, 30, 42, 60, X

72

x + 1 and x - 1 are both prime.

[Galzria]

Logic Problems? Ehhh... This is close enough:

Here's a sequence of numbers. Can you tell me what the next number is and why?:

4, 6, 12, 18, 30, 42, 60, X

72

x + 1 and x - 1 are both prime.

Yes!

[Ozle]

Haha I only got that wrong because I cant add up!

[Qvist]

Why don't they just look into the water of the ocean around them? 
But this is an interesting riddle and pretty similar to one I enjoyed the most.


There are 3 Logic students having their final exam. The professor says:
"Instead of writing an exam I just let you having a final test.
There are 5 hats, 3 of them are red and 2 of them are white.
Please stand in a row, so that one of you can see the other two, another one can only see one and the front one can see no-one.
I will put a hat on each of you and then removing the 2 leftover hats out of the room.
You aren't allowed to talk or use any help or items.
If anyone of you can tell me what color the hat has he's wearing, all of you pass this test."
After an hour of silence the student on the front who couldn't see any other student gives the right answer.
--
How did he do that and what color had his hat?

I assume his (nr. 1) hat is red. If hats of nr. 2 and nr. 1 were white nr. 3 - who can see both - would announce his red hat.
If he doesn't do this nr. 2 can announce his red hat if nr. 1 has a white one.
If he doesn't do this either nr. 1 can announce his red hat regardless of the color of the other two.

Exactly. This might be not so complicated as Galzria's, but I like it a lot because it's hard enough for an "average guy" without being so complicated to not understand the solution.

[Ozle]

OK, finally got the eyed people day one, i missed that they all left on the 100th day! (Thought they left on that day!)

[GendoIkari]

Ok, another favorite, definitely difficult.

A witch captures 50 maidens. She puts them all in a dungeon and tells them: "Tomorrow morning I will separate you into 50 separate cells, so you will not be able to communicate or see each other in any way after tonight. Then every once in a while, a few times a day or so, I will randomly choose one of you and take you to my tower. There you will find 2 levers that each can be set to either up or down. You must flip exactly one lever (either from down to up or up to down), then I will return you to your cell. I won't tell you what position the levers are in to begin with, but I won't ever touch the levers. If at some point you think that every single one of you have visited the tower at least once, then let me know. If you are correct, and everyone has visited the tower at least once, I will release all of you. But if you are wrong, then you will stay here forever."

What plan do the maidens come up with that night?

[jonts26]

I'll just leave this here

http://en.wikipedia.org/wiki/The_Hardest_Logic_Puzzle_Ever

[DStu]

What plan do the maidens come up with that night?

Probably not optimal:

You could elect some master maiden.  Use switch 1 as dummyswitch which does not carry any information.  If usual maiden enters the room and switch 2 is "down", and she hasn't switched it already, switch it
if switch 2 is "up", or this maiden has already switched switch 2, switch switch 1.
if master maiden enters room and switch 2 is "up", switch it to "down"
otherwise use switch 1

Now if the master maiden has turned the switch2 down for 50 times, all other maiden have entered the room.


Estimated maiden in rooms ~N^2 log N I think.

[Galzria]

I'll just leave this here

http://en.wikipedia.org/wiki/The_Hardest_Logic_Puzzle_Ever

The simpler version that I know of that is:

There are two computers, and two doors. Behind one door is a lion that will eat you, and behind the other is freedom and riches that will leave you happy for life.

One computer is programmed to answer any query made of it truthfully.
The other is programmed to answer any query made of it falsely.

You do not know which computer is which.

You may input one question only into one of the two computers. What question do you ask, and which computer do you ask it of?