Puzzle challenges!

[Ephraim]

It seems no one can solve this puzzle (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.

[Deciton_Reven]

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.

[Ephraim]

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.

[Deciton_Reven]

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!"

[Zap-Zarap]

Wow.. cool answer. I hope it is the right one.

[Ephraim]

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:

First, the guard says a number:
(Knight's case): Six
(Beggar's case): Twelve
(Spy's case): Eight

Then, the other person says another number:
(Knight's answer): Three
(Beggar's answer): Six
(Spy's answer): Four

What did the spy think? He thought that the password is: "The guard will give me a number. I must include in my answer the half of that number. Eight - Four."

But he was wrong, because the correct password is...


You must include in your answer the number of letters that the number given by the guard has.

Yes. It is. Let's check if it is correct:
(Knight's case): "Six" has got "three" letters
(Beggar's case): "Twelve" has got "six" letters
(Spy's case): "Eight" has got four "five" letters.

He he he... It seems that this was the most difficult riddle!

[Zap-Zarap]

Great Riddle. Thanks for entertaining us that much.

[Samonella]

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.

[Samonella]

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!

[skeptical_troll]

I think I got it!

Spoiler:

make two piles, one of 90 the other of 10 coins. If the 90-coins pile contains X heads u, the other one will contain 10-X heads, cause they are 10 in total. So flip upside down the small pile and you will have X heads up in both piles.

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.

[Samonella]

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?

[Caladbolg]

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:

The first dwarf that is asked (the last one in the queue) cannot possibly know the color of his hat. The strategy would be to use up his answer to convey information to others. As there are black and white hats (2 options) and the problem deals with natural numbers, I'm led to believe it can be solved by partitioning the naturals in 2 sets: evens and odds. Mod 2 arithmetic is even better so I'll use that.

Count 'black' as 0, 'white' as 1. In order of being asked (from the end), the first dwarf adds up all of the numbers/hats he sees in front of him (mod 2). In other words, if the sum is even, the result is 0, if odd it is 1. He gives that sum as his answer (says 'white' if sum=1, 'black' if 0). Whether he lives or dies is a coin-toss, but he provided the others with the needed info.

Next dwarf heard the answer. Now he calculates the sum in front of him the same way. His sum doesn't include his own number. If his number was 1, the sum he gets will be different from the one he heard; otherwise it will be the same. So by comparing his sum with the one he heard, he can deduce his own number/color.

All of the other dwarves do the same, refering to the answer of the first dwarf, but they must also take into account that all of the answers before them were correct and they need to subtract them from the answer they initially heard.

That way, only 1 dwarf (at most) will have to die. It's a bit hard to explain it without drawing though so sorry if it makes little sense.

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:

With 2 colors we used mod 2 arithmetic, here I tried using mod N. The situation is completely analogous. Code the N colors with numbers 0,1,..., N-1. The first dwarf takes the risk and sums the numbers he sees in front of him (mod N). All of those in front of him refer to his answer. They subtract previous answers (which they know were correct) and compare that with the sum of numbers in front of them. The difference will be a number from 0 to N-1, and it will correspond to the color of their hat.

In other words, at most 1 dwarf needs to die here, too.

[WTrawi]

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

Spoiler:

First I thought no one will answer this but you were right! Congrats!

[skeptical_troll]

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.

[Caladbolg]

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).