Puzzle challenges!

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Jarom
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Re: Puzzle challenges!

Post by Jarom »

Solution:
Spoiler:
Exactly! I didn't expected you to figure this out so quickly. So, you're next.
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Inky
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Re: Puzzle challenges!

Post by Inky »

Oh, cool! :D

This is a variation on the classic hat riddle posted earlier.

Hats Again

The evil sorcerer with the colored hat collecting hobby is back, and he's managed to capture another 100 dwarves (they just don't learn...). He gives each dwarf either a black hat or a white hat at random (probably reusing the same hats as last time... disgusting!) This time the dwarves are standing in a circle so each can see every other dwarf's hat, but now they are all required to guess at the same time, so they can't communicate using their answers like before.

The sorcerer does want to give the dwarves a fair chance though (after all, he'd have no more victims to play hat games with if all the dwarves went extinct), so he'll let them all go if at least half the dwarves guess correctly. As usual the dwarves can formulate a strategy together beforehand, but there's no communicating after that.

What strategy can the 100 dwarves use to guarantee that at least 50 of them guess correctly?

(Like the previous hat riddle this one also has a generalization: if you have kN dwarves randomly given N different colors of hats, how can they guarantee that at least k guess correctly? The basic approach is similar to the previous hat riddle.)
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WTrawi
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Re: Puzzle challenges!

Post by WTrawi »

I've got a question that might be a spoiler:
Spoiler:
Not very active on the forum anymore, but I still read it, and I also still play Wesnoth and draw a bit.
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Inky
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Re: Puzzle challenges!

Post by Inky »

WTrawi wrote:Can the dwarves communicate in other ways e.g. secret moves? Since this time they can see each other.
They don't want to try that since the evil sorcerer takes his hat games very seriously and cheaters will be met with the most gruesome of deaths :twisted:
(There is a guaranteed strategy that does not involve any kind of cheating.)
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Samonella
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Re: Puzzle challenges!

Post by Samonella »

After wasting hours trying to make an almost-perfect strategy be always-perfect, I finally threw it out and now I think I 've got the right one:
Spoiler:
How to generalize it? Ug, I'll think about that later, my brain hurts right now. Or better yet, somebody else finish it up for me.
The last few months have been nothing but one big, painful reminder that TIMTLTW.

Creator of Armory Mod, The Rising Underworld, and Voyage of a Drake: an RPG
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Inky
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Re: Puzzle challenges!

Post by Inky »

Yeah! That's right, looks like the dwarves get to live to see another day (and now the 100 hats have been used twice and are even more gross :P )

Maybe an easier way to look at it:
2 hat colors
Suppose you had only 2 dwarves and needed one of them to be right. One dwarf assumes they both have the same color hat (guesses the color he sees) and the other assumes they have different colored hats (guesses the opposite of what he sees).
Then for 100 dwarves, you can just have them pair off and do that for each of the 50 pairs.

Equivalently, if you assigned white hats = 0 and black = 1, 50 dwarves would assume the sum of the 100 hats is even and the other 50 would assume the sum is odd. (The word for even/oddness is parity by the way.) Thinking of it this way makes it easier to generalize to N colors (though this method is much less practical since you just know some of those dwarves are going to be drunk and add wrong :doh: )
Here's the answer to the general problem too, but you can always try to figure it out on your own before reading:
N hat colors
This time the dwarves divide into N groups of k dwarves each and assign each hat color a number from 0 to N-1.

Group 1 assumes the sum of all kN hats is divisible by N, group 2 assumes the sum has remainder 1 when divided by N,.....,Group N assumes the sum has remainder N-1 when divided by N. The remainder when you divide the sum by N is a number from 0 to N-1, so one of the groups has to be right!
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Samonella
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Re: Puzzle challenges!

Post by Samonella »

I just found out from PM that apparently people are waiting for me, since I answered the last riddle. Sorry, I don't have a new one to post right now.
The last few months have been nothing but one big, painful reminder that TIMTLTW.

Creator of Armory Mod, The Rising Underworld, and Voyage of a Drake: an RPG
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WTrawi
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Re: Puzzle challenges!

Post by WTrawi »

Hey everyone, wake up! :D

Here's a good one (and really hard, so you can get some help in PM's if it's really hopeless): "How can this following addition be true? 873+381+1184+3=2560 2562"
Last edited by WTrawi on November 19th, 2017, 12:48 pm, edited 2 times in total.
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GunChleoc
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Re: Puzzle challenges!

Post by GunChleoc »

Well, this strategy did not quite work:
Spoiler:
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Ravana
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Re: Puzzle challenges!

Post by Ravana »

GunChleoc wrote:Well, this strategy did not quite work:
Spoiler:
The Black Sword
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Re: Puzzle challenges!

Post by The Black Sword »

Spoiler:
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WTrawi
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Re: Puzzle challenges!

Post by WTrawi »

Ravana wrote:Casual testing with bases 8..12 showed no progress.
It should, did I type it wrong then? :shock:
The Black Sword wrote:Hi, btw. :)
Hi :)
Not very active on the forum anymore, but I still read it, and I also still play Wesnoth and draw a bit.
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Samonella
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Re: Puzzle challenges!

Post by Samonella »

WTrawi wrote:
Ravana wrote:Casual testing with bases 8..12 showed no progress.
It should, did I type it wrong then? :shock:
It doesn't work for any base. Looking at the last digits, 3+1+4+3=(some multiple of 10) is only true for base 11, but the second to last digit doesn't work in that base.

The Black Sword's answer is pretty clever though.
The last few months have been nothing but one big, painful reminder that TIMTLTW.

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WTrawi
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Re: Puzzle challenges!

Post by WTrawi »

I checked the book where I found this riddle years ago and the addition goes like this: 873+381+1184+3=2562 (not 2560).
I'm really sorry for mistyping it! :augh: :lol:
Spoiler:
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GunChleoc
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Re: Puzzle challenges!

Post by GunChleoc »

Here's one from the internet that I Wesnothified up a bit:

One sunny day, a footpad, a mage and a lord wake up together in a courtyard. Confused, they look around them and find three horses grazing nearby: a Frisian, a Lusitano and a Mongolian horse. They also find a rack with saddles and bridles, and each has a sign with a name on it: Fred, Limerick and Majestix. Next to the rack, they find a note saying:
  1. The Lusitano belongs to Majestix if and only if the Frisian belongs to a mage.
  2. The Mongolian belongs to Fred if and only if the Lusitano belongs to a mage.
  3. Majestix is a lord if and only if the Lusitano belongs to a footpad.
  4. Limerick is a footpad if and only if the Lusitano belongs to Majestix.
  5. The Frisian belongs to a mage if and only if Fred is a lord.
  6. Majestix is a lord if and only if the Mongolian belongs to Fred.
They don't remember how they got there, nor do they remember their own names – it seems like somebody with an odd sense of humor cast both a teleport and an amnesia spell on them. How ever will each of them find their own horse?
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