## Class Meeting for April 17

We reviewed the final bits of our exploration of social media and then started to consider Chapter 23 in the text. I gave an introduction to public-key encryption and also discussed the nature of software development when it comes to free/open-source versus proprietary options.

Everyone should have a schedule at this point as to when they are going to perform their presentation.

We started chapter 23 today. Read all of Chapter 23 by Friday. I’ll be updating some info on the website on Thursday, April 18.

If you have not planned a day/time for your presentation, we will schedule you on Friday. There are some that I already know are going to be absent on Friday and I will communicate with them individually.

## Class Meeting for April 12

Today, we talked mostly about social media. Given that there are, on average, 340 million tweets per day on Twitter right now, we calculated how much it costs to store tweets on Twitter per day. We found that there are at least 47GB worth of tweets per day, which from one perspective sounds large. However, if we consider that this value represents most significant events in the entire world for the day in question (plus a lot of junk), it actually becomes a bit amazing that 47GB can store all of this data.

I gave examples for North Korea and some other “trending” topics.

See this blog post that relates to what I spent a good chunk of my spring break doing, along with my brother at Room214 in Boulder.

## Class Meeting for April 10

Here is information about recycling in Boulder County.

Today, we discussed some of the numbers related to plastics recycling. Related information can be found here. We also discussed some of the guidelines for recycling in Boulder, what can be recycled and what can’t, and what the recycling center explicitly asks with regards to plastic bags and so on.

Oddly, the Safeway at 28th and Iris in Boulder does not have any recycling bins for plastics, but does have a recycling container specifically for people to deposit plastic bags in.

## Class Meetings for April 3 & 5

The electrical/computational world around us is completely encoded in binary, or base-2. Between Wednesday and Friday, we did the following.

• Converted numbers from base 2 to base 10.
• Converted numbers from base 10 to base 2 (a bit more tricky).
• Added numbers in binary. (See this tutorial)
• Multiplied numbers by 2 in binary. (See my in my office for questions on this.)
• Considered a famous problem in mathematics that was formed to be computed in binary.
• Considered parity check functions for detecting when an evil scrambler or environmental factors swap digits in a binary stream.

## A Quick Parity Check Function

I am going to be sending you a message that is 6 bits long, representing a number. A problem that can and does occur in this process is that a bit or two may get scrambled during transmission. So, I will append a “parity check digit” to the front of my message. If the message has the label $m$, then the parity check function will be $p(m)$ and it will have the value

$p(m)=\begin{cases}0 & \text{if the number of 1s in m is even}\cr 1 & \text{if the number of 1s in m is odd}\end{cases}.$

So, the message I am sending now looks something like $p(m)\ |\ m$. For instance, if I wanted to send you the number 5, my initial message would be $m=000101$, representing the number 5 in binary. Now, since there are an even number of 1s in this message, we have $p(m)=0$, and so our seven bit message now reads $0000101$. When the person on the other end receives this message, they will be able to detect if any single bit has been changed.

## The Limitations of Computers

Most modern computers are 64-bit machines, meaning that each block of data they consider is 64-bits in length. This page contains information far beyond the scope of this class, but the table near the top of the page shows how large various types of integers can be based on how many bits are used to store the integer. When we are working in a computer system that isn’t capable of working with big enough integers, we have a situation called “overflow.” When programmers make an error in not knowing the limitations of their computers, big things can go wrong as a result. A classic example is the Ariane 5 rocket explosion.

## Class Meeting for April 1

We reviewed a bit about the Warlpiri and considered how quickly one can have a descendant in a specific section number. Then, we started to look at error correcting codes by looking at base-2 numbers. See this web page for a discussion about converting from base-2 to base-10, and vice-versa.

On Wednesday, we’ll look more at base-2 numbers and consider sending messages and how we might retrieve a scrambled message.

## Class Meetings for March 18-22

We finished looking at the Brower-Cousteau Model and moved on to our next topic: genetics. We started with a look at the Warlpiri people and Chapter 15 in the text. The main idea is that they have a specific set of social rules dealing with 8 sets of people. Depending on which set you’re in, you can only marry those in a specific other set. And, based on the set number of your parents, you will have a specific set number.

On Friday, we looked at how we could determine, using graph theory, how long it would possibly take to get from one specific set number to another specific set number using the rules in the Warlpiri society. I’ll review this briefly on Monday.

## Class Meeting for March 15

Today we went over an example of the Brower-Cousteau model, which can be found in chapter 7 of the text.

## The Sun’s Power on Earth Measured in Paper

We have discussed for some time the magnitude of solar power reaching the surface of the earth, and the point of today’s lecture is to put that in perspective a bit. The question is: We know that the earth receives roughly 81,000 terawatts of energy from the sun, that is, $8.1\times 10^{16}$ joules of energy every single second … but how much energy is that?

100 joules per second (100 watts) will power a standard incandescent light bulb. If we were to measure each watt as a single piece of paper, which averages about 0.1mm thick, then we’d have a stack of paper 1cm thick representing the power being used by that light bulb, since 100 times 0.1mm is 10mm, or 1cm. If we were to view the total energy coming in from the sun, that 81,000 terawatts, how tall would the corresponding stack of paper be? Here, we’re looking at $8.1\times 10^{16}$ times $1.0\times 10^{-3}$, where the first number represents the number of joules hitting the earth each second and the second is the thickness of a single sheet of paper in meters. The result is in meters, $m$. We calculate that as follows.

\begin{align}\left(8.1\times 10^{16}\right)\times\left(1.0\times 10^{-3}m\right)&=8.1\times 1.0\times 10^{16}\times 10^{-3}m\cr &=8.1\times 10^{13}m\cr &=8.1\times 10^{10}km\end{align}

where now the answer has been reduced to kilometers. How big is that? It’s significantly bigger than the distance from the earth to the sun.

## The Brower-Cousteau Model

The Brower-Cousteau model says that we can make something too big or too small easier to intuitively grasp by scaling it appropriately to a human scale. The example we looks at in class was looking at how thick the earth’s atmosphere is. If we were to scale to earth down to the size of a basketball, how thick would the atmosphere be?

The earth has a radius of $6.38\times 10^6$ meters (at the equator anyway), while as basketball has a radius of $1.193\times 10^{-1}$ meters, as we calculated in class from the circumference. That means that proportionally speaking, the ration of the basketball to the earth is roughly

$\frac{1.193\times 10^{-1}m}{6.38\times 10^6m}\approx 1.8699\times 10^{-8}.$

This means that if we know a distance on the earth, we can get the corresponding distance on the basketball by multiplying it by $1.8699\times 10^{-8}$. For instance, according to NASA, roughly 50% of the earth’s atmosphere is located below 5.6km (18,000ft) in elevation. How big is that on the basketball? Let’s compute it.

\begin{align}5.6km \times 1.8699\times 10^{-8} &= 5.6\times 10^3m\times 1.8699\times 10^{-8}\cr &= 5.6\times 1.8699\times 10^3\times 10^{-8}m\cr &=10.47 \times 10^{-5}m \cr &= 10.47 \times 10^{-2}mm\cr &=1.47\times 10^{-1}mm\end{align}

In other words, if the earth were scaled down to the size of a basketball, 50% of the atmosphere would be within the first 2 tenths of a millimeter from the surface of the basketball.

## Class Meeting for March 13

Some videos for today.

A response to the idea that solar power only provides power while the sun is shining.

Capacitors charge and discharge quickly, but overall hold very little charge. So, what if you put a whole bunch of them together? You get a supercapacitor.

## Class Meetings for March 8 and 11

Most of everything we covered in class on Friday and Monday can be found in chapter 6. If you have any specific questions, as there is one short calculation problem on the homework, please email me or come to office hours.