Puzzle challenges!

[Inky]

I love this logic puzzle! I like how your version has a fantasy theme with dragons The ending was kind of a downer though...wouldn't it be nicer if something positive happened instead? (I feel like in the current situation the best thing to do is just not think about it )

As for a hint, I hope this is helpful without giving away too much:

Spoiler:

Think about what would happen if there were only 2 dragons with green eyes on the island.

[Ephraim]

Thinking about Inky's hint...

Spoiler:

If there were just two dragons, they will think: "the other one has green eyes, so I have not reason to worry, he will turn to stone next dawn". And, the next dawn, the other dragon won't be petrified, because he thinked the same that the other one.
Then, at the second day, one will think: "Wait... if he didn't get petrified, is because he was pretty sure that he hasn't green eyes just why... He was pretty sure that I have green eyes!" And the other one, think the same.
The next dawn, both were petrified, because they known they have green eyes.

That just happens with 3 dragons. Let's call them A, B and C.
C will think: A and B have got green eyes, so A will think that B will get petrified because A thinks that he hasn't got green eyes. They can't include me in this math because I haven't got green eyes.
Next dawn, anyone have turned into stone. C will think: Yeah, as I said yesterday, A and B didn't turn into stone because both think that they haven't got green eyes, but now they know that they have got green eyes, because if the other didn't turn into stone was because he was pretty sure that the other one had green eyes... Next dawn, them will turn into stone."
Dawn arrived, and anyone turned into stone. C says: "Wait! If they haven't turned into stone was because they were pretty sure that... I have got green eyes, as another one! Both them think the same that me! So... I will get petrified!
Next dawn, the three dragons get petrified.

That just happens with 4, 5 and... 100!

Is it the correct answer?

[Elder2]

Ok, I think I have figured it out, if there are 2 dragons, then, knowing that at least one of them has green eyes, dragon A is looking at dragon B, he sees that dragon B has green eyes, therefore either dragon B is the only dragon with green eyes or they both have green eyes. If only dragon B had green eyes, dragon B looking at dragon A would see that dragon A doesn't have green eyes and thus he would infer that he has green eyes himself and he would turn into stone in the morning.
In the morning of the first day, dragon A doesn't see dragon B turning into stone and so dragon B doesn't see dragon A turning into stone since he follows the same train of thought, therefore both dragons conclude that they both must have green eyes and they turn into stone in the morning the second day.

If we extrapolate that to have 50 pairs of dragons we have the solution.

[Inky]

Yes, Ephraim's answer and reasoning is correct! Nice!

If we extrapolate that to have 50 pairs of dragons we have the solution.

Unfortunately the 50 pairs thing would only work if the sailor divided the 100 dragons into 50 pairs and told each pair "at least one of you has green eyes." But your reasoning for 2 dragons is exactly correct - you just need to think about 3 dragons next.

[Ephraim]

Yes, Ephraim's answer and reasoning is correct! Nice!

Hurray! Well, here goes the next one:

The two sages and the evil king:

"Two sages of a town were imprisoned by an evil king. In order to test the intelligence of the sages, he locked them in cells separated from a tower: one looked eastward and the other westward, so that they could not communicate with the other one. Between the two, they could see all the cities that made up the kingdom, but no city was visible at the same time by both. The king told them that the cities of the kingdom were five or eight, and that they would be released immediately after one of them told the jailer that every morning he would bring them food, how many cities were part of the kingdom. In addition, the king told them they had a week or they would end up hanging. But on the third morning, the two sages were released after one of them discovered through a logical procedure how many cities the kingdom was made up of. What logical process led them to solve their problem? How many cities make up the kingdom? "

Warning! It's very difficult! Maybe you should cooperate to get the answer.

Solution and Winner:

There were many answer, but... Ravana said the correct one first. Anyway, I post here Inky's answer, because is easier to understand:

Day 1: If either sage saw 6,7, or 8 cities they would report 8 on Day 1.

Day 2: Since no one said anything on Day 1, both sages now know that both see 0-5 cities. Therefore if either sage saw 0,1, or 2 cities they would know that since the other sage saw less than 6 cities, then there cannot be 8 cities, and would report 5.

Day 3: Now both sages know that both see 3-5 cities, and since any sums of these are at least 6 then there must be 8 cities.

[Elder2]

Hmmm, well, right, thinking about it the dragon thing is correct, but for this to work for large group of dragons, like the 100 dragons here it also requires each dragon to have perfect information about every other dragon of the 100, whether he got petrified or not, otherwise I think it wouldn't work. Also I think they would need to have perfect memory.

[Ravana]

8.

Cases

Code: Select all

I see 0-2 cities.
Skip day.
d2 say 5, or he would have returned 8 on d1.

I see 3 cities.
He sees 2 or 5.
If he sees 2, he knows I see 3 or 6, and if 6, would return 8 on d1.
So at d2+ he knows I see 3, and returns 5 on d2.

I see 4 cities.
He sees 1 or 4.
If he sees 1, he knows I see 4 or 7, and if 7, would return 8 on d1.
	So at d2+ he knows I see 4, and returns 5 on d2.
If he sees 4, he knows I see 1 or 4, and if 1, then on d2 would return 5.
	So at d3+ he knows I see 4, and returns 8 on d3.

[Elder2]

@Ephraim i think I got it, the answer is: 8

Each of the sages can see a number of cities from 0 to 8. There are sages A and B. Neither of the sages (probably assuming they are perfect logicians XD) reported the number of cities on the first morning or the second morning.


If sage A saw 8 cities the sage B must see 0 cities and on the second day sage A would report 8 cities on the first morning, because it is the only possibility.

If sage A saw 7 cities, sage B must see 1 cities so sage A would report 8 cities on the first morning.

If sage A saw 6 cities, sage B must see 2 cities so sage A would report 8 cities on the first morning.

If sage A saw 5 cities, sage B can see either 3 or 0 cities. If sage B saw 0 cities then, sage B would conclude sage A must see either 8 or 5 cities, If sage A saw 8 cities he would report it on the first morning so A must see 5 cities, and sage B would report 5 cities on the second morning, but he didn't.
Therefore sage B must see 3 cities, so sage A would report 8 cities on the third morning

If sage A saw 4 cities, sage B must see either 4 or 1 cities, if sage B saw 1 cities he would conclude that sage A must see either 7 or 4 cities, if sage A saw 7 he would report 8 cities on the first morning, if he saw 4, then sage B would report 5 cities on the second morning.
Therefore sage B must see 4 cities, and sage A reports 8 cities on the third morning

If sage A saw 3 cities, sage B can see either 2 cities or 5 cities. If sage B saw 2 cities he would conclude sage A can see either 6 or 3, he can't see 6 because he would report it on the first morning, so he must see 3 and would report 5 on the second morning. Because he didn't sage B must see 5 cities, so sage A would report 8 cities on the third morning

If sage A saw 2 cities, sage B can see either 6 or 3, he can't see 6 because he would report it, so sage A would report 5 cities on the second morning, he didn't

If sage A saw 1 city then sage B can see either 7 or 4 cities, if he saw 7 he would report it on the first morning, so he must see 4 cities and sage A would report 5 cities on the second morning, he didn't


PS. Ravana's approach is better. XD

[Inky]

Hmmm, well, right, thinking about it the dragon thing is correct, but for this to work for large group of dragons, like the 100 dragons here it also requires each dragon to have perfect information about every other dragon of the 100, whether he got petrified or not, otherwise I think it wouldn't work. Also I think they would need to have perfect memory.

Yes, each dragon can see the 99 other dragons so he'll know whether any other dragon got petrified. However there's no need for perfect memory because on days 1-99 no one gets petrified so there's nothing to remember.

---
Cities problem: I think this could be done a little more elegantly by considering both sages at once:

Day 1: If either sage saw 6,7, or 8 cities they would report 8 on Day 1.

Day 2: Since no one said anything on Day 1, both sages now know that both see 0-5 cities. Therefore if either sage saw 0,1, or 2 cities they would know that since the other sage saw less than 6 cities, then there cannot be 8 cities, and would report 5.

Day 3: Now both sages know that both see 3-5 cities, and since any sums of these are at least 6 then there must be 8 cities.

[Samonella]

Wow, I disappear for a few hours, and suddenly my riddle has been hinted at, solved, a new one proposed and it sure looks like it has been solved as well! Props to Inky, that's the exact hint I was planning to give.

[Ephraim]

Whoa! You are so clever. Ravana, ElderofZion and Inky's answer were correct. Honestly, I prefer Inky's solution, is better explained, but Ravana posted the answer first, so... let's say that Ravana won.
Do you know other puzzles, Ravana?

[Ravana]

I let someone else post.

[Velensk]

If you're looking for an interesting one, there's one I remember reading in a book that was quite fun though I don't remember the answer. There were actually many of this style of question in the book of which this was but one however this is the one that stuck with me for some reason.
>As a note I'm somewhat throwing you into the deep end, as in the book, you've gone through a lot of problems that show the small jumps of logic needed to make this problem simpler but there's no reason one couldn't work their way through them without previous problems if so inclined. The problem is actually not that difficult, or at least, not compared to some of the questions that came after it in the book.

-In the Land of Transylvania there are Humans, Vampires, Faerie, and Alien Infiltrators. These groups are all aware of the others and understand how each other work. The Vampires, Faerie, and Alien Infiltrators all are masters of disguise and visually indistinguishable from humans.

-Humans will always tell the truth (as they perceive it)
-Vampires and Faerie will both always lie (as they perceive it)
-The Alien Infiltrators when they first arrived were puzzled by the behavior of those around them and have developed an odd habit where they always tell the truth (as they perceive it) every other question and lie the rest of the time but there is no way to know how one responded to the last question they were asked when you first meet them.

-Humans and Vampires may be sane or insane.
-Faerie are always insane
-Alien Infiltrators are always sane

Beings who are sane will perceive the truth as the truth and falsehood as falsehood. Beings who are insane perceive all things that are false as being true and all things that are true as being false.

So for instance, all Faerie believe that they are a Human, a Vampire, and an Alien Infiltrator (and also all possible subsets of those three) but if you ask them if they are a Faerie, they will say yes because Faerie always lie.

>The question is, what is the maximum number of yes/no style questions one must ask any random inhabitant of the land to determine what race they are and whether or not they are sane?

EDIT: After actually typing the question, I remembered the answer so I can confirm for you.

[Samonella]

Nice puzzle! I think I figured out how to determine what they are in max 4 questions, though it may be possible to do it in less.

Possible solution:

First, ask it whether it is a faerie, twice. If the answers are different, then it obviously is a sane alien infiltrator.

If it answered no both times:

It cannot be a faerie, because faeries are insane but always say the opposite of what they think is true, so they end up telling the truth. Therefore, since it is neither alien nor faerie, it is either human or vampire. Additionally, we know it is telling the truth (real truth, not truth by what it thinks) because it truthfully said it was not a faerie. Therefore we can ask one more question, "are you human" and it will answer yes if it's a sane human and no if it's an insane vampire.

If it answered yes both times:

Now the possibilities are a faerie or a human/vampire that tells lies (real lies, not lies by what it thinks). Therefore, ask "are you a vampire;" it only answers yes if it is an insane human. Otherwise, ask whether it is human; an insane faerie will still say no, and a sane vampire will say yes.

EDIT: About the cities problem... did the king threaten that if they guessed wrong they'd be killed or something? I can think of a way to be released by day 2, and it's a much simpler logical process... there are only two possibilities for the number of cities, after all...

[Velensk]

Spoiler:

Your answer is actually incorrect. There's a way to always get the answer within 3 questions.

To give a hint, I will tell you that the vast majority of the information given (although it would be useful in a typical knights and knaves), is actually mostly irrelevant to the easiest way to solve this problem.