There is a king and there are his n prisoners. The king has a dungeon in his castle that is shaped like a circle, and has n cell doors around the perimeter, each leading to a separate, utterly sound proof room. When within the cells, the prisoners have absolutely no means of communicating with each other.
The king sits in his central room and the n prisoners are all locked in their sound proof cells. In the king’s central chamber is a table with a single chalice sitting atop it. Now, the king opens up a door to one of the prisoners’ rooms and lets him into the room, but always only one prisoner at a time! So he lets in just one of the prisoners, any one he chooses, and then asks him a question, “Since I first locked you and the other prisoners into your rooms, have all of you been in this room yet?” The prisoner only has two possible answers. “Yes,” or, “I’m not sure.” If any prisoner answers “yes” but is wrong, they all will be beheaded. If a prisoner answers “yes,” however, and is correct, all prisoners are granted full pardons and freed. After being asked that question and answering, the prisoner is then given an opportunity to turn the chalice upside down or right side up. If when he enters the room it is right side up, he can choose to leave it right side up or to turn it upside down, it’s his choice. The same thing goes for if it is upside down when he enters the room. He can either choose to turn it upright or to leave it upside down. After the prisoner manipulates the chalice (or not, by his choice), he is sent back to his own cell and securely locked in.
The king will call the prisoners in any order he pleases, and he can call and recall each prisoner as many times as he wants, as many times in a row as he wants. The only rule the king has to obey is that eventually he has to call every prisoner in an arbitrary number of times. So maybe he will call the first prisoner in a million times before ever calling in the second prisoner twice, we just don’t know. But eventually we may be certain that each prisoner will be called in ten times, or twenty times, or any number you choose.
Here’s one last monkey wrench to toss in the gears, though. The king is allowed to manipulate the cup himself, k times, out of the view of any of the prisoners. That means the king may turn an upright cup upside down or vice versa up to k times, as he chooses, without the prisoners knowing about it. This does not mean the king must manipulate the cup any number of times at all, only that he may.
Assume that both the king and the prisoners have a complete understanding of the game as I have just explained it to you, and that the prisoners were given time beforehand to come up with a strategy. The king was able to hear the prisoners discuss, however, so also assume that if there is a way to foil a strategy, the king will know it and exploit the weakness. The prisoners must utilize a strategy that works in absolutely every single possible case.
Now you must figure out not only how to keep the prisoners alive, but how to also ensure their eventual freedom. When can any one of them be certain they’ve all been in the central chamber of the dungeon at least once? And how? Don’t try to imagine any trickery like scratching messages in the soft gold of the chalice. The problem is as simple as it sounds. The prisoners have absolutely no way of communicating with each other except through the two orientations of the chalice. If any of them attempts any trickery at all they will all be beheaded. All the prisoners can do is turn the chalice upside down or right side up, as they choose, whenever they are called into the chamber.
Princes, Peasants, and Merchants
In the land of Gog, princes always lie, peasants always tell the truth, while merchants sometimes tell the truth and soemtimes lie. A tourist is enjoying an afternoon refreshment in one of the local pubs when the bartender (who always tells the truth) says to her: “Do you see those three men over there? One is a peasant, another a prince, and the third a merchant. You may ask them three yes/no questions, always indicating which man you wish should answer. If, after asking these three questions, you correctly identify the peasant, prince, and merchant, they will buy you a drink.” The tourist is indeed very thirsty. What questions should she ask?
Let’s say I have a stack of sticks: all identical, inflexible, unbreakable. Sticks can touch only at their ends, not in between.
If I give you 3 sticks, you can make one triangle. If I give you 2 more sticks (5), you can make 2 triangles. If I give you another stick (6), how can you make 4 triangles?
18 Heads Up
You’re in a dark room with 50 quarters, 18 of which are heads up. You are allowed to move around the coins or flip some or all of them, if you wish. Problem is, it’s too dark to tell what you’re moving or flipping (no, you can’t figure it out by touch either). Your job is to split the coins into two groups, each of which has the same number of heads up coins. How do you accomplish this?
Where’s the father?
The mother is 21 years older than the child. In 6 years from now, the mother will be 5 times as old as the child. Question: Where’s the father?
When I solve these, I’ll put my solution in the comments section.
For those that like solving math/logic riddles, you might wanna visit Your Favorite Math/Logic Riddles?
Almost forgot about my favorite riddles site: [ wu :: riddles]
Where’s the father?
M = mother’s age
C = child’s age
M = C + 21 //mother is 21 years older than the child
M + 6 = 5(C+6) //in 6 years, mother will be 5x older than the child
Now the math:
C + 21 + 6 = 5C + 30
C + 27 = 5C + 30
4C = -3
C = -3/4 = -.75yrs
3/4 or .75 = 9months
so at this moment, the baby isn’t technically born yet. in fact, it’s 9 months before the baby is born, so where is the father? he’s having sex with this child’s mom.
The King and the Chalice
So I was thinking along the same lines and was discussion with ungsunghero, but it seems I was a bit off (chose 2k+1 instead of 2k+2) and my calculations fell throught. It’s 3:30am and I sorta gave up at that point, but someone posted the solution on /.
http://ask.slashdot.org/comments.pl?sid=165444&cid=13802056
let me first reword the problem to the way I think of it. Let us think of all the prisoner having “tokens” that they are passing around via the cup. Flipping the cup up means you leave one of your tokens in the center room. Flipping it down means you take the token. Obviously only one token can be left in the center room at a time.
Now the problem just becomes one of having the leader collect enough tokens. The cup can start up or down, so there may be an extra token in play. Allowing the king to flip the cup out of sight of the prisoners allows him to add or remove a token from the system with each flip. Thus if the prisoners start with a total of x tokens among all of them, over the course of the game that total may change to between x – k and x + k + 1 (+ 1 there because the cup might start sitting upright).
Now, the solution is that all prisoners except the leader start with 2k + 2 tokens (so the prisoners start with a total of (n – 1)(2k + 2) tokens. The leader will say yes once he has collected (n – 1)(2k + 2) – k tokens. We need to show that the leader can always collect this many tokens, and that he cannot get this many tokens unless ever prisoner has left at least one token in the center room.
We can see that the leader can always collect this many tokens because the king can take at most k tokens out of play leaving only (n – 1)(2k + 2) – k tokens. Since all players will be called an arbitrary number of times, they will all get a chance to leave all their tokens in the center room and the leader will get a chance to pick them all up (the ones that the king doesn’t take).
Now consider the case where one prisoner is not called out for a very long time and all other tokens are allowed to be transferred to the leader first. In addition we will assume that the cup starts up and that the king adds k tokens to the system. In this case, there are (n – 2)(2k + 2) + k + 1 = (n – 1)(2k + 2) – 2k – 2 + k + 1 = (n – 1)(2k + 2) – k – 1 tokens in play. This is one less that the required number of tokens for the leader to say yes, so he can’t possibly say yes until that last prisoner comes out and gets to transfer one of his tokens to the leader.
Sticky Triangles
Create a tetrahedron (triangular pyramid) which has 4 triangles and has 6 sides.
http://www.fact-archive.com/encyclopedia/upload/thumb/e/eb/240px-Tetrahedron.jpg
18 Heads Up
I sorta cheated on this one and ruined it for ungsunghero. Someone had posted the solution on /. and it didn’t make sense to me, so I posted it to ungsunghero for verification and then about 30 secs later after rereading it, it made sense.
First of all, you might want to know that the piles DO NOT have to be the same size. If you think this hint will help you, you might wanna give it another shot.
To solve this, all you have to do is split the coins into 2 piles. One pile of 18 and another pile of 32. The rules just say 2 piles of the same # of heads up. Flip all coins from the 18-coin pile. You will now have equal # of heads in both piles. To explain the reasoning, let’s say of those 18 heads up coins, x coins went into the 18-coin pile and (18-x) coins went into the the 32-coin pile. If you flip all coins from the 18-coin pile, you’ll end up with (18-x) coins with heads up which equals the (18-x) heads up in the 32-coin pile.
The wording of this riddle was definitely tricky!