## Terminal Hexagonal Lattice

Here: have a script to generate plaintext hexagonal lattices for you when you’re feeling blue.

import itertools
pattern1 = ". ."
pattern0 = "   "
def main(repeat,length):
for _ in itertools.repeat(None, length):
print (pattern1+pattern0)*repeat
print (pattern0+pattern1)*repeat
if __name__ == "__main__":
main(10,10)

. .   . .   . .   . .   . .   . .   . .   . .   . .   . .
. .   . .   . .   . .   . .   . .   . .   . .   . .   . .
. .   . .   . .   . .   . .   . .   . .   . .   . .   . .
. .   . .   . .   . .   . .   . .   . .   . .   . .   . .
. .   . .   . .   . .   . .   . .   . .   . .   . .   . .
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. .   . .   . .   . .   . .   . .   . .   . .   . .   . .
. .   . .   . .   . .   . .   . .   . .   . .   . .   . .
. .   . .   . .   . .   . .   . .   . .   . .   . .   . .
. .   . .   . .   . .   . .   . .   . .   . .   . .   . .
. .   . .   . .   . .   . .   . .   . .   . .   . .   . .
. .   . .   . .   . .   . .   . .   . .   . .   . .   . .
. .   . .   . .   . .   . .   . .   . .   . .   . .   . .
. .   . .   . .   . .   . .   . .   . .   . .   . .   . .
. .   . .   . .   . .   . .   . .   . .   . .   . .   . .
. .   . .   . .   . .   . .   . .   . .   . .   . .   . .
. .   . .   . .   . .   . .   . .   . .   . .   . .   . .
. .   . .   . .   . .   . .   . .   . .   . .   . .   . .
. .   . .   . .   . .   . .   . .   . .   . .   . .   . .
. .   . .   . .   . .   . .   . .   . .   . .   . .   . .

## Quick Basics of Enumerative Combinatorics

While going through past notebooks, I came across a table I’d compiled which covered basic enumerative combinatorics.

choose $$r$$ from $$n$$ ordered unordered
no repititions $$\frac{n!}{(n-r)!}$$ $${n \choose r}$$
repetitions $$n^r$$ $${n+r-1 \choose r}$$

Note that $${n \choose r} = \frac{n!}{r!(n-r)!}$$ and is pronounced “$$n$$ choose $$r$$”.

## SPOILERS: Using Simple Combinatorics

DISCLAIMER: This is the solution to Project Euler’s problem 15. Please attempt to solve the problem yourself before reading my solution.

Starting in the top left corner of a 2×2 grid, and only being able to move to the right and down, there are exactly 6 routes to the bottom right corner.

How many such routes are there through a 20×20 grid?

I like to use this problem to demonstrate the efficacy of using simple maths to improve code.

from itertools import permutations
def unique(iterable):
seen = set()
for x in iterable:
if x in seen:
continue
yield x
options = [1]*20 + [0]*20
counter = 0
for a in unique(permutations(options)):
counter = counter+1
print counter

Use simple combinatorics!

To find the number of unique routes through a 20×20 grid, use our friend: the concept of permutations with repeated elements:

$$\frac{\text{number of elements}!}{\text{repetitions of character}!*\text{repetitions of other character}!*…}$$

import math
print math.factorial(40)/(math.factorial(20)*math.factorial(20))

Even better – one line in Haskell:

product [1..40] div product[1..20]^2

## Controlling Fear

Having a reliable method to control your fear and achieve focus quickly is indispensable. During one sleepless night reading Frank Herbert, I discovered my solution.

A bit of backstory:
During undergrad, it was common before and during tests for me to be unnecessarily nervous. This degenerated my ability to focus and fully apply myself to the important task at hand: acing the test creatively in a reasonable time frame.

While meditating, one of my many calming methods is the recitation of poems, such as the Jabberwocky, Litany Against Fear, and If. I find that of all poems, the Litany calms me the most quickly.

To immediately calm myself in risky situations, I use the same techniques. My favoured methods are solving simple integrations or reciting the Litany Against Fear.

This Litany, from the Original Dune, an “incantation spoken by many highly educated people who faced danger or fear during their everyday lives. The incantation helped focus their minds in times of peril.”

The content of the Litany:
I must not fear.
Fear is the mind-killer.
Fear is the little-death that brings total obliteration.
I will face my fear.
I will permit it to pass over me and through me.
And when it has gone past I will turn the inner eye to see its path.
Where the fear has gone there will be nothing.
Only I will remain.

If you aren’t sure how to begin searching for a reliable and personalized method: think of things that make you happy and calm and read while simultaneously reading the Litany aloud to yourself.

## Force Quit in Ubuntu

A surprising amount of Linux users are unfamiliar with how to force quit programs via the command-line. It is often that your only option to escape a process gone awry is using the term window.

I was coding with some friends yesterday and to my surprise, they didn’t know how to quit an unresponsive process. I fear that this ignorance is widespread. The procedure is fairly simple. Use the following command to list ongoing processes:

ps x

After you’ve located the UID of the problem process (usually a 4 or 5-digit natural number), kill it.

Let’s say our UID is 27182.

kill -9 27182

Done!

## Doodling with Painter’s Tape

Large scale doodling with tape is an interesting experience.

## Penrose Triangle

I wanted wall art for my office $$\rightarrow$$ I made myself wall art with painter’s tape during a 30min break from working on my simulation. The triangle is fairly large, $$h \approx$$ 130 cm $$\approx$$ 4’4″

## Introductory Science Kit

Katriona Guthrie-Honea, Austin Russell, and I recently brainstormed to come up with an introductory science kit. The goal of the kit is to create wonder and light the flame of curiosity; revealing that science is a series of fun puzzles and that magic is more beautiful when you understand how it is done.

Prism/raingutter
Magnetic dust / buckyballs
Low power 532nm/650nm
Ultrasound sensor
Use Lego mindstorms
Strings/ropes – harmonics
Paints / subtractive colors
Mirrors
Diffraction
Different refractive indices
Gears / basic motor
Paperclip motor
Budget microscope w/ samples
Pictures of mathematical phenomena, ex. Fractals

## A Method to Not Prove the Collatz Conjecture

Number theory is the cocaine of mathematics. The following resulted not from my intention to prove the conjecture, but from the drive to deeply understand its underlying structure.

After obsessively thinking about it for a few weeks, I came up with a possible approach out of the blue.

I scribbled the proof down at the breakfast table on Christmas day, and this rushed, unedited transcription of thought to paper led to assumptions and oversights that I could not justify when I revisited my scribblings.

As I was transcribing my proof from messy scribbles in my notebook into LaTeX, I found errors (described below). I encourage the reader to attempt to find the errors before they are explained!

In the spirit of How Not to Prove the Poincare Conjecture, I present the following:

Claim:

$$\forall n \in \mathbb{N}$$ the number sequence

n_{i+1} =
\begin{cases}
n_i/2, & \text{if }n_i\text{ is even} \\
3n_i+1, & \text{if }n_i\text{ is odd}
\end{cases}

will always contain the closed chain $$4 \rightarrow 2 \rightarrow 1 \rightarrow 4$$.

Proof:
Let $$f \equiv$$ the product factor between $$n_i$$ and $$n_{i+1}$$

If $$n_i$$ is odd:

$$n_{i+1} = 3n_i+1$$

$$n_{i+1} = fn_i$$

$$f_{i, i+1} = \frac{3n_{i+1}}{n_i} \Leftrightarrow f_{i+1, i} = \frac{n_i}{3n_i+1}$$ [Eq. 1]

Let $$n_1 \rightarrow n_2 \rightarrow \ldots \rightarrow n_{g-1} \rightarrow n_g$$ be any Collatz sequence, started by an arbitrary number $$n_1 \in \mathbb{N}$$ and ended after $$g-1$$ steps on a number $$n_g \in \mathbb{N}$$.

Every even number has an odd number somewhere in it’s sequence, so we will assume that $$n_1$$ is always odd.

A number sequence is an open chain when $$n_i \neq n_j, i \neq j$$ for $$g-1$$ steps. (In other words, there are no repeated numbers in the number sequence.) The chain has $$e$$ even and $$o$$ odd steps.

For the total factor between $$n_1$$ and $$n_g$$ after $$g-1$$ steps for an open chain, it follows from Eq. 1 that

$$2^e\prod_{j=1}^o\frac{n_j}{3n_j+1} = \frac{n_1}{n_g}$$ where $$j$$ is the $$j^{th}$$ odd number in the sequence. [Eq. 2]

A number sequence is a closed chain if $$n_g = n_1$$ and $$n_i \neq n_j, i \neq j$$; $$i, j \in [2, g-1]$$ for $$g$$ steps.

The total factor between $$n_i$$ and $$n_g$$ for closed chains is thus

$$2^e\prod_{j=1}^o\frac{n_j}{3n_j+1} = 1$$

This proof will demonstrate that no other closed chain exists and no open chain exists with endless length.

Let’s reverse engineer it. We start at the last element of the sequence with $$g$$ steps and show that given $$n_g = 1$$ we can reach all odd numbers > 1 by the inverse Collatz sequence.

Using the definition [Eq. 2] of an open chain

$$2^e\prod_{j=1}^o\frac{n_j}{3n_j+1} = \frac{n_1}{n_g} = n_1$$

Let’s do a proof by exception. If we have no other odd numbers apart from 1 in our sequence ($$o =0$$), it follows that

$$2^{e_{g}} = n_1$$

$$o = 0$$ is the only open chain with only even steps. We will address this abnormality later in the [attempted] proof.

Let $$o =1$$

$$2^{e_{0}}\frac{n_1}{3n_1 + 1} = n_1 \Rightarrow n_1 = \frac{2^{e_{1}}-1}{3}$$

Let $$o = 2$$

$$2^{e_{2}}\frac{n_1}{3n_1+1}\frac{n_2}{3n_2+1} \Rightarrow n_1 = \frac{2^{e_{2} – e_{1}}(n_2-1)}{3}$$.

Let $$o=3$$

$$2^{e_{3}}\frac{n_1}{3n_1+1}\frac{n_2}{3n_2+1}\frac{n_3}{3n_3+1} = n_1$$

$$2^{e_{3} – (e_{2} – e_{g-1})}\frac{n_1}{3n_1+1}n_2 = n_1$$

and so on.

It follows in general

$$n_1 = \frac{2^{x}n_2-1}{3}, x \in \mathbb{N}$$ [Eq.3]

ERROR EXPLANATION:

The $$o=2$$ simplification doesn’t work. I use the $$o=1$$ equation to substitute in a simpler form for $$n_2$$, but that equation only holds when $$n_1$$ is the end of a cycle. In this case $$n_2$$ would not be the end of a cycle, thus the $$o=1$$ equation for $$n_1$$ cannot be plugged in for $$n_2$$.The same follows for $$o=3$$ etc.

I did the following:
$$2^{e_2}[\frac{n_1}{(3n_1 + 1)}][\frac{n_2}{(3n_2 + 1)}] = n_1$$
$$2^{e_2}[\frac{n_2}{(3n_2 + 1)}] = 3n_1 + 1$$
$$n_1 = \frac{(2^{e_2}[\frac{n_2}{(3n_2 + 1)}] – 1)}{3}$$

from $$o=1$$:

$$2^{e_1}[\frac{n_1}{(3n_1 + 1)}] = n_1$$
$$\frac{n_1}{(3n_1 + 1)} = \frac{n_1}{2^{e_1}}$$
sub in 2 for 1 $$\leftarrow$$ DOES NOT FOLLOW
$$\frac{n_2}{(3n_2 + 1)} = \frac{n_2}{2^{e_1}}$$ $$\leftarrow$$ DOES NOT FOLLOW

sub in that mess:

$$n_1 = \frac{(2^{e_2}[\frac{n_2}{(3n_2 + 1)}] – 1)}{3}$$
$$n_1 = \frac{(2^{e_2}[{n_2}{2^{e_1}}] – 1)}{3}$$ $$\leftarrow$$ DOES NOT FOLLOW
$$n_1 = \frac{[2^{(e_2-e_1)}n_2 – 1]}{3}$$

Even if that did work, I have several arithmetic errors. If we plug in 2’s for 1’s in the $$o=1$$ equation and solve for $$\frac{n_2}{(3n_2 + 1)}$$, the result should be $$\frac{n_2}{2^{e_2}}$$ rather than $$\frac{n_2}{2^{e_1}}$$.

Further, even if $$\frac{n_2}{2^{e_1}}$$ was the right thing to plug in for $$\frac{n_2}{(3n_2 + 1)}$$ the result in $$o=2$$ should be $$\frac{[2^{(e_2-e_1)}n_2 – 1}{3}$$ (i.e., no parentheses around “$$n_2 – 1$$”). I correct this in $$o=3$$. The rest of the proof depends on finding solutions to [Eq. 3], which I’ve just shown to be moot. Nevertheless, back to the attempted proof…

So when do integer solutions to [Eq. 3] exist? Let’s partition the set of odd natural number into 3 disjoint subsets. $$1^{st}$$ subset is set of all odds which are multiples of 3. There are no solutions for this subset, since

$$n_1 = \frac{2^x3_j-1}{3} = 2^xj – \frac{1}{3}$$, $$j \in \mathbb{N}$$

The second subset is given by $$n_{3ep} = 6i_{ep} + 1$$, $$i_{ep} \in \mathbb{N}$$ which solves the recurrence formula [Eq. 3] for even powers $$x_{ep}$$.

The third subset is given by $$n_{3op} = 6i_{op} – 1$$, $$i_{op} \in \mathbb{N}$$ which solves the recurrence formula [Eq. 3] for odd powers $$x_{op}$$.

With these two subsets satisfying the recursion formula, we can get all start numbers $$n_1$$ of Collatz sequences generated by their direct follower. But, addressing the previous anomaly, we have one exception to our recurrence formula. The case $$n_2,n_1 = 1,1$$ gives us $$4 \rightarrow 2 \rightarrow 1 \rightarrow 4$$. Here, the only odd number (1) in the sequence is the factor between $$n_1$$ and $$n_g$$, for $$4 \rightarrow 2 \rightarrow 1 \rightarrow 4$$ is a closed chain. This sequence continues to iterate on itself and is an exception to the recurrence relations.

ERROR EXPLANATION:

My 3 subsets are ill defined and not mutually exhaustive.
Subset 1 is clear because $$n_2\mod 3 = 0$$ will never satisfy [Eq. 3].
For the second and third subsets, I state that $$2^x\mod 3 = 1$$ when $$x$$ is even and $$2^x\mod 3 = 2$$ when $$x$$ is odd. Which is sound.
Subset 2 ($$x$$ even): $$n_2 = (6i + 1)$$, therefore $$[(2^x)(n_2) – 1]\mod 3 = [2^x\mod 3][(6i + 1) \mod 3] – (1 \mod 3)$$
$$= (1 \mod 3)(1 \mod 3) – (1 \mod 3)$$
$$= (1 \mod 3) – (1 \mod 3) = 0 \mod 3$$ which satisfies [Eq. 3].

Subset 3 ($$x$$ odd): $$n_2 = (6i – 1)$$, therefore $$[(2^x)(n_2) – 1] \mod 3 = [2^x \mod 3][(6i – 1) \mod 3] – (1 mod 3)$$
$$= (2 \mod 3)(2 \mod 3) – (1 \mod 3)$$
$$= (1 \mod 3) – (1 \mod 3) = 0 \mod 3$$ which satisfies [Eq. 3].

It was pointed out to me that I could have used $$3i + 1$$ and $$3i – 1$$ instead of $$6i + 1$$ and $$6i – 1$$ to convey the same result (it’s just mod 3).

In my approach, we find not 3 but 5 disjoint subsets:
1. $$n_2 \mod 3 = 0$$, no solution
2. $$n_2 \mod 3 = 1$$ and $$x$$ even, solution
3. $$n_2 \mod 3 = 1$$ and $$x$$ odd, no solution
4. $$n_2 \mod 3 = 2$$ and $$x$$ even, no solution
5. $$n_2 \mod 3 = 2$$ and $$x$$ odd, solution

These subsets are both mutually exclusive and mutually exhaustive.

I wrapped up the [incorrect] proof as follows:

The conjecture is true for all odd numbers. Every even number has an odd number somewhere in its sequence. $$\Rightarrow$$ The conjecture is true for all evens. $$\Rightarrow$$ The conjecture is true for all $$n \in \mathbb{N}.$$ QED

## Doodle Painting: Tesseract

The equation defines the points that are contained by a tesseract in Euclidean 4-space.
If you look closely, there is an etching that says “sp2” which refers to graphene’s bonds.