Puzzle challenges!

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Ephraim
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Re: Puzzle challenges!

Post by Ephraim » September 1st, 2017, 1:14 pm

WTrawi wrote:It seems no one can solve this puzzle :D (I know it's against the rules, but I know a really-really good one if someone wants a new riddle)
Really? Well... it seems it is time to give a hint... Caladbolg was very near from the answer.
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Deciton_Reven
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Re: Puzzle challenges!

Post by Deciton_Reven » September 1st, 2017, 10:01 pm

I mean, unless it's accidental, the guard says things in a very odd way, which would lead me to believe it's the structure of the spy's sentence that's wrong. I couldn't tell you how it wrong though so I haven't tried. And if it's not improper structure you've got me.

Well unless it's the really weak-sauce answer the spy should have said 4 and a half as he would have been the ninth merchant and by answering half of 8 instead of half of 9 gave away he was no merchant.

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Ephraim
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Re: Puzzle challenges!

Post by Ephraim » September 1st, 2017, 10:08 pm

Deciton_Reven wrote:I mean, unless it's accidental, the guard says things in a very odd way, which would lead me to believe it's the structure of the spy's sentence that's wrong. I couldn't tell you how it wrong though so I haven't tried. And if it's not improper structure you've got me.

Well unless it's the really weak-sauce answer the spy should have said 4 and a half as he would have been the ninth merchant and by answering half of 8 instead of half of 9 gave away he was no merchant.
Wrong. The structure of the sentence doesn't matter. About the "4 and a half" answer... no. It is incorrect too.
The Great Library of WML is huge and extense. I must find the book who will lead me to create a perfect campaign.
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Deciton_Reven
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Re: Puzzle challenges!

Post by Deciton_Reven » September 3rd, 2017, 1:59 pm

One of my friends took a shot at it and this was his answer:

Okay this time I got it for real. The password isn't half of the number the guard gives you. The sum of all of the numbers you say must be one less than the sum of all the numbers he says. This includes homophones. When addressing the knight, the guard said "for" and "six". The knight responded with "to", "three", and "for". When talking to the beggar, the guard only said "twelve". The beggar responded with "five" and "six".

It follows then, that the spy should have said something more along the lines of "Of course! Of the four cities I've been trading in, this one is my favorite to visit!"

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Zap-Zarap
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Re: Puzzle challenges!

Post by Zap-Zarap » September 3rd, 2017, 2:02 pm

Wow.. cool answer. I hope it is the right one.
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Ephraim
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Re: Puzzle challenges!

Post by Ephraim » September 3rd, 2017, 6:05 pm

Deciton_Reven wrote:One of my friends took a shot at it and this was his answer:

Okay this time I got it for real. The password isn't half of the number the guard gives you. The sum of all of the numbers you say must be one less than the sum of all the numbers he says. This includes homophones. When addressing the knight, the guard said "for" and "six". The knight responded with "to", "three", and "for". When talking to the beggar, the guard only said "twelve". The beggar responded with "five" and "six".

It follows then, that the spy should have said something more along the lines of "Of course! Of the four cities I've been trading in, this one is my favorite to visit!"
Good answer, but it is incorrect. It's time to say the correct one...
Correct answer:
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Zap-Zarap
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Re: Puzzle challenges!

Post by Zap-Zarap » September 3rd, 2017, 6:09 pm

:lol: Great Riddle. Thanks for entertaining us that much.
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Samonella
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Re: Puzzle challenges!

Post by Samonella » September 3rd, 2017, 9:33 pm

This isn't a new puzzle, just a short comic i happened to come across and thought i'd share since it's so applicable. :lol:
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Samonella
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Re: Puzzle challenges!

Post by Samonella » September 7th, 2017, 4:07 am

I just came across a pretty good one! In front of you is a table with 100 coins. 10 of them are laying heads up, and the other 90 are tails up. However, the room is dark and you can't see which are which, and they are smooth enough that you also can't tell them apart by touch or any other method. Your goal is to sort the coins into two piles following these two rules:
1) Each coin must be laying with one face down (ie, balancing them on their side is not allowed).
2) The two piles must have an equal number of coins facing heads up.

Edit: Winner is Skeptical_troll!
Last edited by Samonella on September 7th, 2017, 4:48 pm, edited 1 time in total.
The last few months have been nothing but one big, painful reminder that TIMTLTW.

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skeptical_troll
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Re: Puzzle challenges!

Post by skeptical_troll » September 7th, 2017, 4:06 pm

I think I got it!
Spoiler:
My turn:

101 dwarves are captured by an evil sorcerer and imprisoned in a dungeon. The sorcerer wants to execute them, but agrees to concede a chance of survival.

He tells the Dwarves "You will have to form a queue, and I will put either a black or a white hat on your head. Each of you will only be able to see the color of the hat of all Dwarves ahead in the queue, so the last will see the next 100, the second last the next 99 etc.

You can talk among yourself now and elaborate a strategy, but when it will happen I will ask to each of you the color of your own hat, starting from the last one. all the dwarves who guess wrong will be killed".

What is the best strategy the dwarves can elaborate? How many can be sure to survive? Note that, once they are in the queue, they will not be allowed to say anything other than their guess about the color.

BONUS VERSION: (this one is quite hard, but if you fancy math go for it): what would be the solution if instead of just black and white the hats could be of N colours. In that case the dwarves would know which N colours.

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Re: Puzzle challenges!

Post by Samonella » September 7th, 2017, 4:46 pm

That's right!

I don't think I understand your puzzle very well. Is the sorcerer waiting at the front of the queue, giving them hats one by one and immediately making each one guess?
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Caladbolg
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Re: Puzzle challenges!

Post by Caladbolg » September 7th, 2017, 5:41 pm

skeptical_troll wrote: 101 dwarves are captured by an evil sorcerer and imprisoned in a dungeon. The sorcerer wants to execute them, but agrees to concede a chance of survival.

He tells the Dwarves "You will have to form a queue, and I will put either a black or a white hat on your head. Each of you will only be able to see the color of the hat of all Dwarves ahead in the queue, so the last will see the next 100, the second last the next 99 etc.

You can talk among yourself now and elaborate a strategy, but when it will happen I will ask to each of you the color of your own hat, starting from the last one. all the dwarves who guess wrong will be killed".

What is the best strategy the dwarves can elaborate? How many can be sure to survive? Note that, once they are in the queue, they will not be allowed to say anything other than their guess about the color.
My reasoning:
Spoiler:
BONUS VERSION: (this one is quite hard, but if you fancy math go for it): what would be the solution if instead of just black and white the hats could be of N colours. In that case the dwarves would know which N colours.
Now, for this one, I'm a bit less certain, but the logic makes sense. I also tried applying it to a random sequence of hats generated in excel and it worked fine.
Spoiler:

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WTrawi
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Re: Puzzle challenges!

Post by WTrawi » September 9th, 2017, 7:47 am

Haha, I know this riddle! :D Not sure if I have the right to tell if your answer is correct so I'll put it in a spoiler section.
Spoiler:
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skeptical_troll
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Re: Puzzle challenges!

Post by skeptical_troll » September 9th, 2017, 9:35 am

Indeed, Caladbolg's solution is correct, good job! :)

@Samonella: sorry if my formulation was a bit ambiguous. No, the dwarves are all in the queue with their hat on, only then the sorcerer starts interrogating them, starting from the last one.

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Re: Puzzle challenges!

Post by Caladbolg » September 9th, 2017, 7:52 pm

My turn then:

You have a balancing scale with two plates. You also have 10 coins, all identical in shape and in weight, except for one, which is heavier than the others. What is the least number of weighings needed to determine which coin is the heavy one?

For a more general riddle, the same question, but with N coins (all same weight except one that is heavier).

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