Logic Riddles - Page 6

On November 05 2012 04:30 gymnast wrote: so the prisoners know the order or not?

On October 28 2012 21:32 Art.Cascade wrote: 100 damn smart and organised prisoners. The guards set up a room, with 100 boxes, and place the (unique) name of all prisoner in the boxes. The guard collect one prisoner at a time from his cell to the room with boxes, and let the prisoner open 50 of the 100 boxes, one after the other. If any prisoner fails to find his own name in any of the 50 boxes he opens, then all the prisoners are killed. If all prisoners manage to find their name in the 50 tries, then they all go free. They have time to prepare a strategy before the box-opening starts, but once the first prisoner is in the room opening boxes, no more communication between the prisoners is possible. The prisoners do not know in which order they will be called in the room, and the guards set up the room in the exact same way for each prisoner. How can the prisoners maximise their chances to get free? hint: they can get well above 10%.

I finally gave up and looked at the solution. It's absolutely brilliant. Kind of simple (as most brilliant things are), but still very hard to come up with. No silly tricks or anything though. Art.Cascade has explained it really well.

It's kind of funny how people find different things difficult. The hints didn't help me at all because those were the things that I realized almost instantly. It's also fairly obvious that the order in which the prisoners are let in doesn't matter, since they have to assume that the other guys have found/will find their names, or else they're all screwed anyway. Still, I couldn't come up with the solution. I guess I lack the creativity or something ^^

You should definitely not feel bad for not solving the prisoner problem. I don't think many percent manage to find the solution to this riddle themselves. Actually, one of the reasons that I like this problem is that I was the only one (or at least first, don't know what's happened since) solving it in the thread on teamliquid, and it took me over 24 hours. Most of the other puzzles posted at that time (it was a bit of a riddle frenzy at TL for some time) were solved within minutes.

More problems!!!
A) Santas raindeers and helpers
1) Santa has 100 raindeers. Or any large number, maybe he has infinite number of raindeers, who knows? Anyway, not important. A lot of raindeers.
2) They are in booths in the stable. Let's say for simplicity that the booths are numbered 1, 2, ...
3) The booths start all closed.
4) The first of santas little helper (helper number 1) opens all the booths.
5) The second helper (helper number 2) closes all even numbered booths.
6) Helper 3 goes to every booth with a number divisible with 3 (ie, booth 3, 6, 9, ...) and change the state of the booth. Ie, if it's open, he closes it, and if it's closed, he opens it. So booth 3 will be open, so he closes it. Booth 6 is closed, so he opens it. Booth 9 is open, so he closes it. etc.
7) And it continues like this. Helper number n will go to every booth divisible with n, and change the state of the booth. There are as many helpers as booths, so the last helper will just open/close the very last booth.
8) When all the helpers are done, which booths are open, and which are closed?

B) Eye colours at island (hard)
1) An island with 100 inhabitants, all of them very intelligent, logical, rational, and aware that all the other inhabitants are the same.
2) All the inhabitants have either brown or blue eyes, and they are all aware of that.
3) All inhabitants knows the eye colour of all other inhabitants, but no inhabitant knows their own eye colour.
4) If an inhabitant somehow figures out his own eye colour by sunset, he/she will leave the island during the night. The inhabitants see each other daily, so all other will notice before sunset if someone leaves the island.
5) 50 of the inhabitants have blue eyes, 50 have brown. The inhabitants don't know these numbers though. (So for example a inhabitant with brown eyes will see 50 with blue eyes, and 49 with brown eyes, but doesn't know if he is the 51:st blue, or 50:th brown.)
6) The inhabitants never talk about eye colour. Seeing how they are all very intelligent etc, and everyone likes the island, they make absolutely sure to not communicate in any way whatsoever about even the faintest clue regarding eye colours.
7) One day (let's call it day 0 for simplicity) a stranger arrives at the island, gathers all the inhabitants, and loudly declares (everyone hears, everyone knows that everyone else hears, everyone knows that everyone else also knows that everyone hears, etc) that there is at least one person has brown eyes. All the inhabitants trust him (everyone knows that he cannot lie or something).
8) The stranger leaves. Assuming that nothing else disturbs the inhabitants, what will happen when?

I think I got the questions right, but please correct me or ask if anything is unclear.

On November 06 2012 05:33 Art.Cascade wrote: More problems!!! A) Santas raindeers and helpers 8) When all the helpers are done, which booths are open, and which are closed?

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I don't understand something in the second one.
You say in 5) that there are 50 brown eyes ones, and 50 blue eyes ones, and in 3) that all inhabitants know the eye colour of the other inhabitants.
So that should imply, that every inhabitant, knows his own eye colour : I see 49 brown eyes inhabitants, so I have blue eyes, and thus I leave.

A classical one that I will try to put in other words, so that it is not easily found with google.

There are 5 kids who just got 12 candies for halloween and they have to share it. Youngest one has to propose a way to distribute the candies among the kids. Then, they proceed to vote, and if the strict majority rejects the proposition, the youngest kid is evicted and leaves with nothing. And the 2nd younger kid will then have to propose a distribution, and they will again vote for it, and may or not accept or evict this kid. And this goes on, until the distribution is accepted.

Assuming that there are all loving candies and are not friends (they don't matter if a kid is evicted), what is the distribution the youngest have to make in order to get some candies ?

On November 06 2012 06:42 kingpowa wrote: Show nested quote + On November 06 2012 05:33 Art.Cascade wrote: More problems!!! A) Santas raindeers and helpers 8) When all the helpers are done, which booths are open, and which are closed? + Show Spoiler + The closed ones are all x² with x²<n and n.

nope

+ Show Spoiler +

On November 06 2012 06:47 kingpowa wrote: I don't understand something in the second one. You say in 5) that there are 50 brown eyes ones, and 50 blue eyes ones, and in 3) that all inhabitants know the eye colour of the other inhabitants. So that should imply, that every inhabitant, knows his own eye colour : I see 49 brown eyes inhabitants, so I have blue eyes, and thus I leave.

I've heard this one before so I won't answer, but the point is that they don't know there are 50 people with brown eyes and 50 with blue eyes.

spoiler because maybe hint:
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On November 06 2012 08:21 KoeBawlt wrote: Show nested quote + On November 06 2012 06:42 kingpowa wrote:   On November 06 2012 05:33 Art.Cascade wrote: More problems!!! A) Santas raindeers and helpers 8) When all the helpers are done, which booths are open, and which are closed? + Show Spoiler + The closed ones are all x² with x²<n and n. nope + Show Spoiler + door n is closed if it is divisible by an even number of numbers between 1 and n.

Your answer is not wrong, but I prefer mine.
go further.

On November 06 2012 05:33 Art.Cascade wrote: More problems!!! A) Santas raindeers and helpers 1) Santa has 100 raindeers. Or any large number, maybe he has infinite number of raindeers, who knows? Anyway, not important. A lot of raindeers. 2) They are in booths in the stable. Let's say for simplicity that the booths are numbered 1, 2, ... 3) The booths start all closed. 4) The first of santas little helper (helper number 1) opens all the booths. 5) The second helper (helper number 2) closes all even numbered booths. 6) Helper 3 goes to every booth with a number divisible with 3 (ie, booth 3, 6, 9, ...) and change the state of the booth. Ie, if it's open, he closes it, and if it's closed, he opens it. So booth 3 will be open, so he closes it. Booth 6 is closed, so he opens it. Booth 9 is open, so he closes it. etc. 7) And it continues like this. Helper number n will go to every booth divisible with n, and change the state of the booth. There are as many helpers as booths, so the last helper will just open/close the very last booth. 8) When all the helpers are done, which booths are open, and which are closed? .

+ Show Spoiler +

On November 06 2012 08:21 KoeBawlt wrote: Show nested quote + On November 06 2012 06:42 kingpowa wrote:   On November 06 2012 05:33 Art.Cascade wrote: More problems!!! A) Santas raindeers and helpers 8) When all the helpers are done, which booths are open, and which are closed? + Show Spoiler + The closed ones are all x² with x²<n and n. nope + Show Spoiler + door n is closed if it is divisible by an even number of numbers between 1 and n.

You are both right.

Show nested quote + On November 06 2012 06:47 kingpowa wrote: I don't understand something in the second one. You say in 5) that there are 50 brown eyes ones, and 50 blue eyes ones, and in 3) that all inhabitants know the eye colour of the other inhabitants. So that should imply, that every inhabitant, knows his own eye colour : I see 49 brown eyes inhabitants, so I have blue eyes, and thus I leave. I've heard this one before so I won't answer, but the point is that they don't know there are 50 people with brown eyes and 50 with blue eyes. spoiler because maybe hint: + Show Spoiler + So when a brown eyed person sees 50 blue eyed people and 49 brown eyed people she doesn't know whether there are 51 blue eyed people and 49 brown eyed people or 50 of each.

Exactly, thanks. The inhabitants don't know they are 50-50, that's just a piece of information I give to you as the problem solver. I'll edit my post to clarify.

On November 06 2012 15:27 HotChip wrote: Show nested quote + On November 06 2012 05:33 Art.Cascade wrote: More problems!!! A) Santas raindeers and helpers 1) Santa has 100 raindeers. Or any large number, maybe he has infinite number of raindeers, who knows? Anyway, not important. A lot of raindeers. 2) They are in booths in the stable. Let's say for simplicity that the booths are numbered 1, 2, ... 3) The booths start all closed. 4) The first of santas little helper (helper number 1) opens all the booths. 5) The second helper (helper number 2) closes all even numbered booths. 6) Helper 3 goes to every booth with a number divisible with 3 (ie, booth 3, 6, 9, ...) and change the state of the booth. Ie, if it's open, he closes it, and if it's closed, he opens it. So booth 3 will be open, so he closes it. Booth 6 is closed, so he opens it. Booth 9 is open, so he closes it. etc. 7) And it continues like this. Helper number n will go to every booth divisible with n, and change the state of the booth. There are as many helpers as booths, so the last helper will just open/close the very last booth. 8) When all the helpers are done, which booths are open, and which are closed? . + Show Spoiler + It's a sequence wich goes like this: open - 2xClosed - open - 4xClosed - open - 6xClosed - open - 8xClosed - open 10xClosed...

close, but you must have made a mistake somewhere. Go through the first few booths manually and you will see that your solution isn't accurate, and I think you will be able to fix it. The principle seems right though, as the answer is almost right.

see post 17 and 18. already asked.

thank you for pointing this out, didn't notice that ^^

On November 06 2012 07:03 kingpowa wrote: A classical one that I will try to put in other words, so that it is not easily found with google. There are 5 kids who just got 12 candies for halloween and they have to share it. Youngest one has to propose a way to distribute the candies among the kids. Then, they proceed to vote, and if the strict majority rejects the proposition, the youngest kid is evicted and leaves with nothing. And the 2nd younger kid will then have to propose a distribution, and they will again vote for it, and may or not accept or evict this kid. And this goes on, until the distribution is accepted. Assuming that there are all loving candies and are not friends (they don't matter if a kid is evicted), what is the distribution the youngest have to make in order to get some candies ?

+ Show Spoiler +

For the reindeer,

+ Show Spoiler +

I hope they are not really question to hire brokers cuz damn those are retarded.

Mixing exact and precise question with "creative" bullshit is absurd.


What odds i lay on the flip? 9999999999:1... cuz why not? its more profitable and it doesnt say you will reject Ev- offers and we are already guessing the height of the mother fucking elevator guy, those questions are are useful to determine somebodys intelligence as much as a coloring book, shame on any company who uses these.



rant off... carry on

Why absurd? It is incredibly important to be able to identify a problem that is impossible to solve with a direct mathematical approach, that has to be worked around. Many very intelligent people have wasted huge amount of time trying to solve impossible problems.

In real life you will not know if the problem you have in front of you is solvable with mathematics, or if you need "creative bullshit".

On November 10 2012 11:16 waga wrote: Show nested quote + On November 06 2012 07:03 kingpowa wrote: A classical one that I will try to put in other words, so that it is not easily found with google. There are 5 kids who just got 12 candies for halloween and they have to share it. Youngest one has to propose a way to distribute the candies among the kids. Then, they proceed to vote, and if the strict majority rejects the proposition, the youngest kid is evicted and leaves with nothing. And the 2nd younger kid will then have to propose a distribution, and they will again vote for it, and may or not accept or evict this kid. And this goes on, until the distribution is accepted. Assuming that there are all loving candies and are not friends (they don't matter if a kid is evicted), what is the distribution the youngest have to make in order to get some candies ? + Show Spoiler + 2 + 4 x 2.5 = 12

nop. Can't break the candy.
When a kid proposes a sharing, the other ones when voting, will think "If I evict him, will I get more candies ?"
And it is actually about optimal choices.

On November 11 2012 08:19 kingpowa wrote: Show nested quote + On November 10 2012 11:16 waga wrote:   On November 06 2012 07:03 kingpowa wrote: A classical one that I will try to put in other words, so that it is not easily found with google. There are 5 kids who just got 12 candies for halloween and they have to share it. Youngest one has to propose a way to distribute the candies among the kids. Then, they proceed to vote, and if the strict majority rejects the proposition, the youngest kid is evicted and leaves with nothing. And the 2nd younger kid will then have to propose a distribution, and they will again vote for it, and may or not accept or evict this kid. And this goes on, until the distribution is accepted. Assuming that there are all loving candies and are not friends (they don't matter if a kid is evicted), what is the distribution the youngest have to make in order to get some candies ? + Show Spoiler + 2 + 4 x 2.5 = 12 nop. Can't break the candy. When a kid proposes a sharing, the other ones when voting, will think "If I evict him, will I get more candies ?" And it is actually about optimal choices.

+ Show Spoiler +