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Was curious what the probably was for at least one song in a shuffled playlist of n songs to be in its proper spot. I derived this formula that definitely works. I’m betting there’s a way to simplify it down further, but I’m not good at math anymore. (alternating +/– signs in the series)

Was curious what the probably was for at least one song in a shuffled playlist of n songs to be in its proper spot. I derived this formula that definitely works. I’m betting there’s a way to simplify it down further, but I’m not good at math anymore. (alternating +/– signs in the series)

If one remembers this particular episode from the popular sitcom ‘Friends’ where Ross is trying to carry a sofa to his apartment, it seems that moving a sofa up the stairs is ridiculously hard.

But life shouldn’t be that hard now should it?

The mathematician Leo Moser posed in 1966 the following curious mathematical problem: what is the shape of largest area in the plane that can be moved around a right-angled corner in a two-dimensional hallway of width 1? This question became known as the moving sofa problem, and is still unsolved fifty years after it was first asked.

The most common shape to move around a tight right angled corner is a square.

And another common shape that would satisfy this criterion is a semi-circle.

But what is the largest area that can be moved around?

Well, it has been conjectured that the shape with the largest area that one can move around a corner is known as “Gerver’s sofa”. And it looks like so:

Wait.. Hang on a second

This sofa would only be effective for right handed turns. One can clearly see that if we have to turn left somewhere we would be kind of in a tough spot.

Prof.Romik from the University of California, Davis has proposed this shape popularly know as Romik’s ambidextrous sofa that solves this problem.

Although Prof.Romik’s sofa may/may not be the not the optimal solution, it is definitely is a breakthrough since this can pave the way for more complex ideas in mathematical analysis and more importantly sofa design.

(15+15=30

Three

Five

Nine

Formally, we define an even integer as such:

Suppose n is an integer. Then we say n is even if there exists an integer k such that n = 2k.

In other words an integer n is even if it can be rewritten as 2 multiplied by another integer k. Looking at some examples we can see that 2 is even because it can be written as 2(1). 4 can be written as 2(2). 6 can be written as 2(3). And so on.

So now we look at 0. Is there any way we can write it in the form 0 = 2k? Yes we can! 0 itself is an integer and we know that 0 multiplied by anything is 0. So 0 = 2(0). Therefore, 0 is even.

We can also show that its not odd. Odd integers are defined similarly to even integers.

Suppose n is an integer. Then we say n is odd if there exists an integer k such that n = 2k + 1.

Examples: 1 = 2(0) + 1, 3 = 2(1) + 1, 5 = 2(4) + 1, …

We can prove that 0 is not odd by contradiction.

First we assume that 0 IS odd. This means that there exists an integer k such that 0 = 2k + 1. We can determine this k algebraically and come to the conclusion that k = -½. This is where the contradiction lies. By the definition of an odd integer, k must also be an integer. However -½ is not an integer. Therefore, 0 is not odd.

Are you sure about that?

(15+15=30

Three

Five

Nine

Formally, we define an even integer as such:

Suppose n is an integer. Then we say n is even if there exists an integer k such that n = 2k.

In other words an integer n is even if it can be rewritten as 2 multiplied by another integer k. Looking at some examples we can see that 2 is even because it can be written as 2(1). 4 can be written as 2(2). 6 can be written as 2(3). And so on.

So now we look at 0. Is there any way we can write it in the form 0 = 2k? Yes we can! 0 itself is an integer and we know that 0 multiplied by anything is 0. So 0 = 2(0). Therefore, 0 is even.

We can also show that its not odd. Odd integers are defined similarly to even integers.

Suppose n is an integer. Then we say n is odd if there exists an integer k such that n = 2k + 1.

Examples: 1 = 2(0) + 1, 3 = 2(1) + 1, 5 = 2(4) + 1, ...

We can prove that 0 is not odd by contradiction.

First we assume that 0 IS odd. This means that there exists an integer k such that 0 = 2k + 1. We can determine this k algebraically and come to the conclusion that k = -1/2. This is where the contradiction lies. By the definition of an odd integer, k must also be an integer. However -1/2 is not an integer. Therefore, 0 is not odd.

(15+15=30

Three

Five

Nine