# Rubik’s Cube

The Rubik’s Cube is a 3D combination puzzle. It is the brainchild of the Hungarian professor of architecture Ernő Rubik, who invented in it 1974. Since then, over 300 million Rubik’s Cubes have been sold worldwide. If they were stacked on top of each other, they would reach the top of Mount Everest, twice.

It is estimated that in the mid-eighties about a fifth of the world’s population had, at some point, handled a Rubik’s Cube. And because of its simple design, people continue to be astounded by its devilish complexity. The percentage of people that has ever solved the Cube, is more somewhat more difficult to ascertain.

Obviously, there is only one solution in which all six sides of the Cube have the same colour; as for all the different unsolved states, the original 3x3x3 Cube has 43252003274489856000 (that is to say, over 43 quintillion) possible Cubes. If there was a Cube for every permutation, they would cover the Earth with 273 layers – a sea of Cubes 15,5 meters (50 foot) deep. If there was a cube scrambled for every permutation and they were laid end to end then they would stretch approximately 261 light years – from Earth to Alpha Columbae.

Because of the vast amount different Cubes, algorithms (a sequence of moves that has a desired effect) are used to solve the Cube. Without algorithms to solve the Cube, it could take ages: if you made a single turn of one of the Cube’s faces every second, it would take you 1371,51 billion years to go through all the possible configurations. The universe is only 13,82 billion years old. If you had started this project during the Big Bang, you still would not be done yet.

Amazingly, the best speed cubers (people who take part in speed cubing – a sport where competitors try and solve the cube as quickly as possible) can solve the cube in under six seconds. At the time of writing, the world record is 4,90 seconds; the record for blindfolded solving (including memorization beforehand) is 21,05 seconds.

Mike: Look I don’t know, ok; it’s like a fucking Rubik’s Cube! I mean, it’s impossible at this point.
Selina: What? Mike, a Rubik’s cube is not impossible to solve.
Gary: Yeah, I saw an Asian kid do it in like ten seconds.
Selina: Ten seconds Mike.
– Veep (2012) Season 1, Episode 3; “Catherine” [No. 3]

# Tautology

A tautology is a needless repetition of an idea, especially in words other than those of the immediate context, without imparting additional force or clearness, as in ‘Morning sunrise.’

In logic, a tautology is a compound propositional form all of whose instances are true, as A or not A. An instance of such a form, as ‘This candidate will win or will not win.’

In other words, tautology is unnecessary repetition. For example ‘They spoke in turn, one after the other’ is considered a tautology because ‘in turn’ and ‘one after the other’ mean the same thing.

“At a certain point talk about ‘essence’ and ‘oneness’ and the universal becomes more tautological than inquisitive.” – Christopher Hitchens

# A Comic Book Excerpt

‘As you journey along the path you meet an old man. He tells you that modern science has proved that all our actions and decisions are merely the machinations of a predetermined universe and that our concept of ‘free will’ is naught but a comforting illusion.

If you agree with this hypothesis turn to page 72.

If you disagree, turn to page 72.’

Unknown Source

# Graham’s Number

Graham’s number is a very, very big number that was discovered by a man called Ronald Graham. It is the answer to a problem in an area of mathematics called Ramsey theory, and is one of the biggest number ever used in a mathematics study.

Even if every number in the number was written in the tiniest writing possible, it would still be too big to fit in all the universe that scientists have seen so far – in other words: the universe is just too small a place to be able to write this number in.

Graham’s number is connected to the following problem in the branch of mathematics known as Ramsey theory: (Note that the symbol ^ is used to denote “to the power”)

Consider an n-dimensional hypercube, and connect each pair of vertices to obtain a complete graph on 2^n vertices. Then colour each of the edges of this graph either red or blue. What is the smallest value of n for which every such colouring contains at least one single-coloured 4-vertex planar complete subgraph?

Graham and Rothschild proved in 1971 that this problem has a solution, N*, and gave as a bounding estimate 6 ≤ N* ≤ N, with the upper bound N a particular, explicitly defined, very large number. In other words, the smallest possible value of N was thought to be 6 in 1971. However, this answer has been debunked. A possible solution points to the smallest value of N being at least 11, it may well be 12, but the answer lies between 11 and Graham’s Number (G).

To convey the difficulty of appreciating the enormous size of Graham’s number, it may be helpful to express—in terms of exponentiation alone—just the first term (g1) of the rapidly growing 64-term sequence.

(I) 3x3x3 is 3^3 is 27.

(II) 3^^3 is 7625597484987 you can think of this as 3 mutiplied by its self 3^3 times so 3x3x3x3x3….27 times.

(III.a) 3^^^3 is so huge, its digits would fill up the universe and beyond. it has 3638334640025 decimal digits – and this is only the start.

(III.b) 3^^^3 can also be represented as g1.

(IV) g2 is equal to 3^^^^^^^^^^^^^^^^^^^^^^…3 the number of arrows (^) in this number is g1 this means there are 3^^^3, arrows, or levels of exponents, in g2.

(V) g3 is equal to 3^^^^^^^^^^…3 the number of arrows is the value of g2 and so on.

(VI) g64 is equal to G, graham’s number.

Because of the Knuth up-arrow notation described here we know that the last ten numbers in Graham’s number are […] 2464195387. But the actual entire number (the remaining cyphers preceding these last ten numbers put together in one large number) is virtually infinitely longer.

The point of Graham’s number (G) is that it is the answer to the upper bound of N in this particular hypercube problem. Whatever the smallest value of N must be in this particular problem remains unclear. The answer probably lies between 11 and G. There is quite a margin of error to render faulty.

# Ramsey and The Pigeonhole Principle

Ramsey theory, named after the British mathematician and philosopher Frank P. Ramsey, is a branch of mathematics that studies the conditions under which order must appear. Problems in Ramsey theory typically ask a question of the form: ‘how many elements of some structure must there be to guarantee that a particular property will hold?’

Illustration of The Pigeonhole Principle

Suppose, for example, that we know that n pigeons have been housed in m pigeonholes. How big must n be before we can be sure that at least one pigeonhole houses at least two pigeons? The answer is the pigeonhole principle: if n > m, then at least one pigeonhole will have at least two pigeons in it.

The photograph on the right shows a number of pigeons in holes. Here there are n = 10 pigeons in m = 9 holes, so by the pigeonhole principle, at least one hole has more than one pigeon: in this case, both of the top corner holes contain two pigeons. The principle says nothing about which holes are empty: for n = 10 pigeons in m = 9 holes, it simply says that at least one hole here will be over-full; in this case, the bottom-left hole is empty. Ramsey’s theory generalizes this principle as explained below.

A typical result in Ramsey theory starts with some mathematical structure that is then cut into pieces. How big must the original structure be in order to ensure that at least one of the pieces has a given interesting property?

For example, consider a complete graph of order n; that is, there are n vertices and each vertex is connected to every other vertex by an edge. A complete graph of order 3 is called a triangle. Now colour every edge red or blue. How large must n be in order to ensure that there is either a blue triangle or a red triangle? It turns out that the answer is 6.

Another way to express this result is as follows: at any party with at least six people, there are three people who are all either mutual acquaintances (each one knows the other two) or mutual strangers (each one does not know either of the other two).

A comic result of the pigeonhole principle is the “proof” that in the city of New York (or any other city with a population over a million) at least two people have the same number of hairs on their head. The reasoning is as follows: an average human being has about 150.000 hairs on the scalp; it is reasonable to assume that no human being has more than 1.000.000 hairs on the scalp. Over a million people live in New York. The population of n people (exceeding a million) has to be arranged in m (1.000.000 or less) collections; one collection is possible for each number of hairs on the scalp. Because n population > m different numbers of hairs on the scalp, there are at least two people in one of these collections – at least two people with the same number of hairs.