## Hold without touching it

What can you hold without ever touching it?

A conversation.

What can you hold without ever touching it?

A conversation.

See also best riddles or new riddles.

Two words are anagrams if and only if they contain the exact same letters with the exact same frequency (for example, "name" and "mean" are anagrams, but "red" and "deer" are not).
Given two strings S1 and S2, which each only contain the lowercase letters a through z, write a program to determine if S1 and S2 are anagrams. The program must have a running time of O(n + m), where n and m are the lengths of S1 and S2, respectively, and it must have O(1) (constant) space usage.

First create an array A of length 26, representing the counts of each letter of the alphabet, with each value initialized to 0. Iterate through each character in S1 and add 1 to the corresponding entry in A. Once this iteration is complete, A will contain the counts for the letters in S1. Then, iterate through each character in S2, and subtract 1 from each corresponding entry in A. Now, if the each entry in A is 0, then S1 and S2 are anagrams; otherwise, S1 and S2 aren't anagrams.
Here is pseudocode for the procedure that was described:
def areAnagrams(S1, S2)
A = new Array(26)
A.initializeValues(0)
for each character in S1
arrayIndex = mapCharacterToNumber(character) //maps "a" to 0, "b" to 1, "c" to 2, etc...
A[arrayIndex] += 1
end
for each character in S2
arrayIndex = mapCharacterToNumber(character)
A[arrayIndex] -= 1
end
for (i = 0; i < 26; i++)
if A[i] != 0
return false
end
end
return true
end

What runs around the house, but does not move?

The fence.

What is the similarity between your eyes and your school?

Both have pupils.

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 beginning of eternity,
The end of time and space.
The beginning of every end,
And the end of every place.

E.

Mr. Black, Mr. Gray, and Mr. White are fighting in a truel. They each get a gun and take turns shooting at each other until only one person is left. Mr. Black, who hits his shot 1/3 of the time, gets to shoot first. Mr. Gray, who hits his shot 2/3 of the time, gets to shoot next, assuming he is still alive. Mr. White, who hits his shot all the time, shoots next, assuming he is also alive. The cycle repeats. All three competitors know one another's shooting odds. If you are Mr. Black, where should you shoot first for the highest chance of survival?

He should shoot at the ground. If Mr. Black shoots the ground, it is Mr. Gray's turn. Mr. Gray would rather shoot at Mr. White than Mr. Black, because he is better. If Mr. Gray kills Mr. White, it is just Mr. Black and Mr. Gray left, giving Mr. Black a fair chance of winning. If Mr. Gray does not kill Mr. White, it is Mr. White's turn. He would rather shoot at Mr. Gray and will definitely kill him. Even though it is now Mr. Black against Mr. White, Mr. Black has a better chance of winning than before.

2+3=8,
3+7=27,
4+5=32,
5+8=60,
6+7=72,
7+8=?
Solve it?

98
2+3=2*[3+(2-1)]=8
3+7=3*[7+(3-1)]=27
4+5=4*[5+(4-1)]=32
5+8=5*[8+(5-1)]=60
6+7=6*[7+(6-1)]=72
therefore
7+8=7*[8+(7-1)]=98
x+y=x[y+(x-1)]=x^2+xy-x

There are several chickens and rabbits in a cage (with no other types of animals). There are 72 heads and 200 feet inside the cage. How many chickens are there, and how many rabbits?

There are 44 chickens and 28 rabbits in the cage.
Let c be the number of chickens, and r be the number of rabbits.
r + c = 72
4r + 2c = 200
To solve the equations, we multiply the first by two, then subtract the second.
2r + 2c = 144
2r = 56
r = 28
c = 44

It makes more noise than a dog in your house. What is that?

Two dogs.

What do you get when you cross a chicken and a chihuahua?

Pooched eggs.