# #MeasureEarth: a world-wide experiment to determine whether Earth is flat or round

Just want to draw some attention to this cool world-side data collection project happening on Oct 24. I am hoping there will be a lot of involvement because I am planning on making this part of a laboratory class I teach here at Merrimack College. The more people who participate, the more data we will have to play with!

A group of people are claiming that the Earth is flat.  Meanwhile, much of humanity believes the Earth is round.  Who is right, and who is wrong?

Folks: we live in the Internet age, which means we can test these ideas with an EXPERIMENT.  To participate, all you need are three one-meter measuring sticks, a sunny day, some flat ground, and an Internet connection.  I hereby declare October 24th, 2017 as the day we #MeasureEarth.

What to measure

The goal is to determine the length of shadow your stick creates when you hold it straight up on a specific date at a specific time of day.  Then compare the length of shadow you measured with what people in other parts of the world measured.

Wait, why?

Do you expect people at other latitudes to measure the same length of shadow that you measure?  Why or why not?  What do you…

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# My Rubix Cube Solution

In this post I am going to present my solution for the Rubix cube. I came up with it during my last year as an undergrad (at Zoomass!), when I was enrolled in an abstract algebra class. The final topic of the class was the Rubix cube, since it can be used to study groups, group actions, and other such things. We were not taught exactly how to solve the cube in the class, but the class gave me some notations (which you see below) and some of the concepts which helped me find my own solution. From there it was just a matter of spending every day with the thing trying to figure it out – If you lived in the Amherst-Northampton area circa Spring 2005, you were almost certainty in danger of being hit by a distracted driver in a Subaru with a cube.

This solution is not meant to be fast – it’s supposed to be easy. I wanted to make it as easy to solve as possible, so I would have to remember the fewest number of moves in order to do it. It takes me around 5 minutes to solve the cube using these techniques. The most interesting thing I find about my method is that is only requires you to know 3 moves. Memorizing these three moves allows you to solve any initial cube orientation. First some notation:

These basic moves should be pretty obvious; $D$ turns the bottom row of the cube to the right. When I write $D^{-1}$, it means do the opposite; turn the bottom row to the left.

Step 1: Get the top. I won’t go through this, getting the top is not hard, and at worst requires a set of three simple moves to get it. If you need help, try getting the cross first, and then the corners. And make sure that the middles of the second row are solved too. It’s easy to ensure that happens while you’re getting the top, but don’t forget to do it!

Step 2: The following set of basic moves we will call M1: $DF^{-1}D^{-1}FD^{-2}F^{-1}DFD^{-2}F^{-1}DF$ will the following thing to your cube:

I hope that diagram is understandable. It shows the front, left, and bottom faces of the cube. You can see that it takes an edge from the bottom and moves it to the middle layer, and keeps the rest of the middle layer fixed. It screws up the bottom all over the place, but don’t worry about that. I should also come clean about this; you need the opposite of this move too in order to get the bottom edge to the right edge on the second layer. This requires some mind acrobats, but just do the inverse of each basic move in M1 (in the same order – this is not an inverse of M1, but rather the inverse transposition).

Step 3: Next we have M2, which goes like this: $FLDL^{-1}D^{-1}F^{-1}D$. This move does the following:

Focus on the bottom edges. It flips the one in front and exchanges the 2 on the bottom. It also fixes the last edge on the bottom (not numbered). You can use this to solve the bottom cross, since it exchanges two edges. You can always solve the bottom cross in this manner.

Step 4a: (assuming the corners are not in the right place) The third move M3 is $F^{-1}D^{-2}FDLDL^{-1}F^{-1}DL$. It screws up quite a few things, but most importantly it exchanges the corners on the right bottom (345 and 8910).

This move can always be used to get the corners into the right position; they don’t need to be in the right orientation, that’s what the next move is for.

Step 4b: (assuming the corners are in the right locations, just twisted) Now this is really clever; if you do M2 four times in a row, it does the following:

The key here is that it fixes everything except what is notated. So once you have the bottom cross solved, this twists those three corners. If all four corners are twisted, you can always use this move to twist three, then move the cube to twist a different three.

You will find that you may need to do the last few moves over and over again. There is a particular orientation (bottom cross solved, bottom corners opposite where they should) which can occur and is very frustrating. You will have to use M3 several times, screwing up the cross, in order to get the corners in the right orientation.

There is nothing too special about solving the Rubix cube, and like I have said, this is not the fastest method. The reason I find it interesting that you really only need 3 moves in order to solve the cube from any position. Another thing that might possibly be interesting is each of these moves is order 12; this means that when you do it 12 times, the cube returns to it’s original orientation. I don’t know if there is anything special about this or just a coincidence.