Riddles

[tanuki_chau]

I'm sorry ahead of time as I just came up with this on the spot:

In North Korea there are several internment camps dotted across the nation. Anyone who has found themselves on the wrong side of the political power will soon find themselves in such a prison. The Prisoners are forced to perform hard and dangerous slave labor with no possibility of release. However in Hwasong Political Prison, there is a gang famous even outside the fences and mine fields surrounding the camp. The Salamdeul Kkangpae were practically untouchable. The gang started as a close knit group of villagers in a small town in Samjiyon-gun. As the gang grew they stuck to their roots. All the members knew the others surname and given name. After growing in power, they attempted an unsuccessful coup to bring democracy to the troubled peninsula. Even though only a handful of the gang members participated in the coup. As an act of loyalty all members surrendered themselves to the state police as a form of political protest. Despite Pyongyang’s attempts to control any information about the gang, somehow information makes its way to the masses. They are the Robin Hoods’ of North Korea. Their popularity even rivals the Supreme Leader Kim Jong-un. This left the ruling class with a problem. If they killed the gang members, news would break out and a mass revolt would occur, potentially bringing forth the very coup Salamdeul had hoped for. On the other hand they couldn't simply leave them be. They needed to be made an example of. They needed to break Salamdeul. The only method of coercion possible in this situation was The Tower.

The Tower was feared by all, even the Salamdeul. The Tower itself was not an instrument of death. This is what made it feared. When prisoners left The Tower, they were no longer like themselves. It was as if something inside The Tower took away what made them who they were, leaving a shell of a person. This is why The Tower was so feared.

Unfortunately for the warden of the camp, only one prisoner can go into the Tower at a time. So how can the warden threaten the gang into obedience?

[Lord_Potato]

Ask who wants to go?

[Helix935]

hmm i will have a go but i only i have one i can think that is sadly very short

also i should have seen that the old riddle was faith but i had never thought there may be a religious aspect to it

also wait is tannuki_chau saying a riddle or not? Now i am not sure if i should put my riddle unless someone points it out

if it is a riddle then i say that they would send either to leader of the gang first or worse their children

[vexx32]

My initial thought would be to study the gang for some time to figure out who was perceived as the most strong-willed of the lot. Break him, and the rest will be unlikely to want to chance it. However, it is not certain. Hm. I guess you could play a sort of mind game with them all, feeding them misinformation as to who actually goes into the Tower and when, give them reason to fear it well before anyone actually goes anywhere near the Tower. Break one or two at a time, and make sure none of the other prisoners can find out for sure who, but that they can easily make guesses which are almost certainly wrong. Break them using the Tower as a source of fear, but you have to make them fear it to start with.

Interesting riddle, indeed!

[Starwhip]

You separate them into different rooms. You tell each of them that they are going in the Tower, and then directly after that you blindfold them and restrict their hearing and walk them in the direction of the Tower. They cannot see or hear one another, so they would each submit and be taken back. They don't know that every one of them had the same treatment.

I like logic puzzles, maybe that is what we should do next!

[tanuki_chau]

Nope for all accounts. I wrote this very badly and I'm sorry for that. I'm better at solving them than making them. If you want I can do a fun logic game one I got when I interviewed at Google?

[vexx32]

Hm? Well, if you want I'm still curious as to the answer for this one, though xD

It does sound like you were going for a specific answer, but I can't see it xD

[tanuki_chau]

The answer is: the warden sends them in alphabetical order, the idea being he only has to threaten the first one and he will cease and so on. In a pure universe it would work however people are too complex. I digress. Here is a better one.

Inside a room with no windows and one door there are three light bulbs. Outside the door there are two light switches. Those light switches are connected to the light bulbs. If the door is opened the lights shut off regardless of light switch position. How would you find out which switch is connected to what bulb?

Another one is there are 1000 two sided coins. 980 coins are in one direction and 20 are in another. You are blindfolded and have gloves preventing you from feeling the characteristics of the coins. You are required to separate the coins into two groups so that each group has the same number of coins heads up as the other group.

[bigyihsuan]

For riddle 1, have one switch on for ten seconds, one for 5 seconds, and one that stays off. Fell the bulbs afterwards. The warmest is the ten second switch, the second warmest five, and coldest off.

[KvickFlygarn87]

Number 1, turn on one light switch, open door, enter room, close door. Check what light bulb is on. Open door, exit room, close door. Flip both switches (the one that was on before is off, vice versa). Open door, enter room, close door. Check what light bulb is on. The one that hasn't been on is the un-wired one.

Number 2, I would use my forehead to feel which is on which side.

[tanuki_chau]

bigy gets the first one. The second is harder though and no that doesn't count kvick, you can't feel with your forehead.

[vexx32]

I'm thinking there might be some math involved in that one. I am way too tired to be doing it now, though xD Something to do with probabilities, perhaps? My initial reaction is to simply shuffle them as best and randomly as humanly possible, then just divide into two piles.

[Specialist290]

Another one is there are 1000 two sided coins. 980 coins are in one direction and 20 are in another. You are blindfolded and have gloves preventing you from feeling the characteristics of the coins. You are required to separate the coins into two groups so that each group has the same number of coins heads up as the other group.

Count twenty coins into one pile, flipping each one over on the other side as you do so. That is your first group. The rest are the second group. No matter which coins you choose, the number of coins facing heads-up in both piles will be equal.

[Starwhip]

For the coin one, you have to separate them into groups of 500 piles, right? Hmm.

Nevermind. I reread the riddle.

Let's see...

980 O

20 X

Required: 10X......10X and any number of O.

One moment...

"Count twenty coins into one pile, flipping each one over on the other side as you do so. That is your first group. The rest are the second group. No matter which coins you choose, the number of coins facing heads-up in both piles will be equal."

Well damn, you're right. Next riddle.

Edited December 5, 2013 by Starwhip

[tanuki_chau]

Bingo Specialist got it. It's a math riddle.

[Tex_NL]

For the coin one, you have to separate them into groups of 500 piles, right? Hmm.

Nevermind. I reread the riddle.

Let's see...

980 O

20 X

Required: 10X......10X and any number of O.

One moment...

"Count twenty coins into one pile, flipping each one over on the other side as you do so. That is your first group. The rest are the second group. No matter which coins you choose, the number of coins facing heads-up in both piles will be equal."

Well damn, you're right. Next riddle.

Normally I'm not too bad at these puzzles but I still don't get the logic in this one.

If you have 980 tail / 20 head you would indeed get equal numbers of head, as was required, in both piles doing it this way. But what if you have 980 head / 20 tail? Oops, now I have equal numbers of tail in both piles but the riddle asks for head.

Since the original riddle did not state the position of the mixed coins but does ask for heads up in the split piles there is no definitive answer.

[tanuki_chau]

Normally I'm not too bad at these puzzles but I still don't get the logic in this one.

If you have 980 tail / 20 head you would indeed get equal numbers of head, as was required, in both piles doing it this way. But what if you have 980 head / 20 tail? Oops, now I have equal numbers of tail in both piles but the riddle asks for head.

Since the original riddle did not state the position of the mixed coins but does ask for heads up in the split piles there is no definitive answer.

It works against all logic, which is why math is awesome. Lets cut the numbers down for simplicity and say there are 20 coins. When you randomly take 10 coins for one group, let's say it contained eight tails and two heads. There were ten tails and ten heads to begin with. This means the ten coins that are left consist of the opposite: two tails and eight heads! They will always complement each other like that. So when you flip over the group of eight tails and two heads, it will become a group of two tails and eight heads  exactly the same as the other group.

[bigyihsuan]

I'm only good at solving riddles, and I'm not that good at making riddles, but here we go:

Let's say that you are a road builder between Cities A, B, C, and D. They all are points in a 1 by 1 mile square.

You need to build the shortest road between all of the cities. You cannot build bridges due to their expense. What is the most efficient road configuration that goes between all of the cities?

[briansun1]

stright/use a wormhole

[TechnicalK3rbal]

Uhhhh....abcd?

No! There all in the square, you didn't say that they were the points on the border of the square, so they can all be right next to each other and you don't need roads!

[Tex_NL]

It works against all logic, which is why math is awesome. Lets cut the numbers down for simplicity and say there are 20 coins. When you randomly take 10 coins for one group, let's say it contained eight tails and two heads. There were ten tails and ten heads to begin with. This means the ten coins that are left consist of the opposite: two tails and eight heads! They will always complement each other like that. So when you flip over the group of eight tails and two heads, it will become a group of two tails and eight heads  exactly the same as the other group.

True, but this is NOT the same as Specialists answer. He split the 1000 into a pile of 980 and a pile of 20, flipping the 20.

Doing this will result into an equal number of coins in both piles with the same face as the 20 in the original pile. Since you never stated if the original 20 where heads or tails there is NO WAY to predict the outcome. The only thing you know for sure you'll end up with an equal number of either hears or tails.

Edit:

Strike the part where I 'True' as it is NOT! Your example only works with equal numbers in the original pile.

Lets say we have 20 coins, by random 16 are heads, 4 are tails. I take 10: 9 heads and 1 tails. The original pile now consists of the remaining 7 heads and 3 tails. I flip the 10 I took: 9 tails and 1 heads.

I don't know what universe you come from or what you've been smoking but 7 heads and 3 tails is NOT the same as 9 tails and 1 heads. Not even close.

Edited December 5, 2013 by Tex_NL

[vexx32]

Okay, so Specialist's turn. Damn, I really should have thought of that xD

(For those still arguing... If there are 20 coins in the original bunch which are reversed, and you need to make another pile from this original bunch that will afterwards have an identical number of heads to those remaining in the other pile, taking 20 coins at random and flipping them will yield identical numbers of heads. If in the 20 coins I take, six of them were heads in the original pile, the number of heads remaining in the initial pile is now 14, whilst there are now 6/20 coins in the new pile which are tails instead of heads, leaving 14 heads in both piles.)

Edited December 6, 2013 by vexx32

[Mr Tegu]

http://www.bbc.co.uk/iplayer/episode/b03k6ypz/The_Joy_of_Logic/

If you enjoyed that last one, you might find this quite interesting.

[Helix935]

i would have just rearranged them by size

[Specialist290]

Cool. Alright then, the new riddle (no poetry, this time -- couldn't think of a good way to make it rhyme):

It has neither eye nor lens, yet you can use it to see distant places.

It has neither mouth nor voice, yet through it a dead man can speak to a living one.

It can hold entire worlds, yet children can hold it in their hands.

No matter how much you take out of it, it is always as full as it was when you opened it.

What is it?