You have just purchased a small company called Company X. Company X has N employees, and everyone is either an engineer or a manager. You know for sure that there are more engineers than managers at the company.
Everyone at Company X knows everyone else's position, and you are able to ask any employee about the position of any other employee. For example, you could approach employee A and ask "Is employee B an engineer or a manager?" You can only direct your question to one employee at a time, and can only ask about one other employee at a time. You're allowed to ask the same employee multiple questions if you want.
Your goal is to find at least one engineer to solve a huge problem that has just hit the company's factory. The problem is so urgent that you only have time to ask N-1 total questions.
The major problem with questioning the employees, however, is that while the engineers will always tell you the truth about other employees' roles, the managers may lie to you if they like. You can assume that the managers will do their best to confuse you.
How can you find at least one engineer by asking at most N-1 questions?
You can find at least one engineer using the following process:
Put all of the employees in a conference room. If there happen to be an even number of employees, pick one at random and send him home for the day so that we start with an odd number of employees. Note that there will still be more engineers than managers after we send this employee home.
Then call them out one at a time in any order. You will be forming them into a line as follows:
If there is nobody currently in the line, put the employee you just called out in the line.
Otherwise, if there is anybody in the line, then we do the following. Let's call the employee currently at the front of the line Employee_Front, and call the employee who we just called out of the conference room Employee_Next.
So ask Employee_Front if Employee_Next is a manager or an engineer.
If Employee_Front says "manager", then send both Employee_Front and Employee_Next home for the day.
However, if Employee_Front says "engineer", then put Employee_Next at the front of the line.
Keep doing this until you've called everyone out of the conference room. Notice that at this point, you'll have asked N-1 or less questions (you asked at most one question each time you called an employee out except for the first employee, when you didn't ask a question, so that's at most N-1 questions).
When you're done calling everyone out of the conference room, the person at the front of the line is an engineer. So you've found your engineer!
But the real question: how does this work?
We can prove this works by showing a few things.
First, let's show that if there are any engineers in the line, then they must be in front of any managers.
We'll show this with a proof by contradiction. Assume that there is a manager in front of an engineer somewhere in the line. Then it must have been the case that at some point, that engineer was Employee_Front and that manager was Employee_Next. But then Employee_Front would have said "manager" (since he is an engineer and always tells the truth), and we would have sent them both home. This contradicts their being in the line at all, and thus we know that there can never be a manager in front of an engineer in the line.
So now we know that after the process is done, if there are any engineers in the line, then they will be at the front of the line. That means that all we have to prove now is that there will be at least one engineer in the line at the end of the process, and we'll know that there will be an engineer at the front.
So let's show that there will be at least one engineer in the line. To see why, consider what happens when we ask Employee_Front about Employee_Next, and Employee_Front says "manager". We know for sure that in this case, Employee_Front and Employee_Next are not both engineers, because if this were the case, then Employee_Front would have definitely says "engineer". Put another way, at least one of Employee_Front and Employee_Next is a manager. So by sending them both home, we know we are sending home at least one manager, and thus, we are keeping the balance in the remaining employees that there are more engineers than managers.
Thus, once the process is over, there will be more engineers than managers in the line (this is also sufficient to show that there will be at least one person in the line once the process is over). And so, there must be at least one engineer in the line.
Put altogether, we proved that at the end of the process, there will be at least one engineer in the line and that any engineers in the line must be in front of any managers, and so we know that the person at the front of the line will be an engineer.
The Pope, Beyonce, POTUS, and Bill Gates are on the same plane.
There are only 3 parachutes left for the 4 of them.
POTUS says: "As the President, I think I should have the right to have a parachute, because I rule millions of people in the greatest nation of all."
Beyonce says: "As one of the greatest singers of all-time, I think I should deserve to be safe. I bring tears and laughter to millions of people, and I'm an important contributor to pop music."
Bill Gates says: "As one of the richest successful company owners, I think I should live because I'm on top of the economics cycle, creating jobs and incomes for millions of people. I am a wealthy and intelligent man."
Finally, the Pope says: "I'm an old, religious man. I lived a life that's full, I helped millions of people find their way through God, I'm ready to let go of a parachute and to face my fate."
Which one of them will abandon the parachute and die?
Did I ever mention that the plane was crashing? No one's gonna die.
100 men are in a room, each wearing either a white or black hat. Nobody knows the color of his own hat, although everyone can see everyone else's hat. The men are not allowed to communicate with each other at all (and thus nobody will ever be able to figure out the color of his own hat).
The men need to line up against the wall such that all the men with black hats are next to each other, and all the men with white hats are next to each other. How can they do this without communicating? You can assume they came up with a shared strategy before coming into the room.
The men go to stand agains the wall one at a time. If a man goes to stand against the wall and all of the men already against the wall have the same color hat, then he just goes and stands at either end of the line. However, if a man goes to stand against the wall and there are men with both black and white hats already against the wall, he goes and stands between the two men with different colored hats. This will maintain the state that the line contains men with one colored hats on one side, and men with the other colored hats on the other side, and when the last man goes and stands against the wall, we'll still have the desired outcome.
There are five vowels in English language (a, e, i, o, u, and sometimes y).
Easy version: Can you tell us a word contains of these 5 vowels?
Hard: Can you tell us a word contains of all these 6 vowels with y?
Very difficult: Can you tell us a word contains of all these vowels with y in their alphabetical order?
In their alphabetical order:
This guy living on the 20th floor in an apartment building got up early each morning to go to work in a downtown store. He always went into the elevator on the 20th floor and rode down to the entrance (1st floor). When he came home he always rode the elevator from the entrance and up to the 8th floor. He walked out of the elevator and walked the stairs up to his apartment on the 20th floor. Why didn't he take the elevator all the way up to his apartment?
This guy is midget and can only reach to the 8th floor button.
It was a Pink Island. There were 201 individuals (perfect logicians) lived in the island. Among them 100 people were blue eyed people, 100 were green eyed people and the leader was a black eyed one.
Except the leader, nobody knew how many individuals lived in the island. Neither have they known about the color of the eyes. The leader was a very strict person. Those people can never communicate with others. They even cannot make gestures to communicate. They can only talk and communicate with the leader. It was a prison for those 200 individuals.
However, the leader provided an opportunity to leave the island forever but on one condition. Every morning he questions the individuals about the color of the eyes! If any of the individuals say the right color, he would be released. Since they were unaware about the color of the eyes, all 200 individuals remained silent. When they say wrong color, they were eaten alive to death. Afraid of punishment, they remained silent.
One day, the leader announced that "at least 1 of you has green eyes! If you say you are the one, come and say, I will let you go if you are correct! But only one of you can come and tell me!"
How many green eyed individuals leave the island and in how many days?
All 100 green eyed individuals will leave on the 100th night.
Consider, there is only one green eyed individual lived in the island. He will look at all the remaining individuals who have blue eyes. So, he can get assured that he has green eyes!
Now consider 2 people with green eyes. Only reason the other green-eyed person wouldn't leave on the first night is because he sees another person with green eyes. Seeing no one else with green eyes, each of these two people realize it must be them. So both leaves on second night.
This is the same for any number. Five people with green eyes would leave on the fifth night and 100 on the 100th, all at once.
Search: Monty Hall problem
Why it's important for the solution that the leader said the new information "at least 1 of you has green eyes", when they must knew from the beginning, that there are no less than 99 green-eyed people on the island? Because they cannot depart the island without being certain, they cannot begin the process of leaving until the guru speaks, and common knowledge is attained.
Search: Common knowledge (logic)