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## Dragon curves II

This post appeared in issue 05 of

*Chalkdust*. I strongly recommend reading the rest of*Chalkdust*.Take a long strip of paper. Fold it in half in the same direction a few times. Unfold it and look at the shape the edge of the paper
makes. If you folded the paper \(n\) times, then the edge will make an order \(n\) dragon curve, so called because it faintly resembles a
dragon. Each of the curves shown on the cover of issue 05 of

*Chalkdust*is an order 10 dragon curve.The dragon curves on the cover show that it is possible to tile the entire plane with copies of dragon curves of the same order. If any
readers are looking for an excellent way to tile a bathroom, I recommend getting some dragon curve-shaped tiles made.

An order \(n\) dragon curve can be made by joining two order \(n-1\) dragon curves with a 90° angle between their tails. Therefore, by
taking the cover's tiling of the plane with order 10 dragon curves, we may join them into pairs to get a tiling with order 11 dragon
curves. We could repeat this to get tilings with order 12, 13, and so on... If we were to repeat this

*ad infinitum*we would arrive at the conclusion that an order \(\infty\) dragon curve will cover the entire plane without crossing itself. In other words, an order \(\infty\) dragon curve is a space-filling curve.Like so many other interesting bits of recreational maths, dragon curves were popularised by Martin Gardner in one of his

*Mathematical Games*columns in*Scientific American*. In this column, it was noted that the endpoints of dragon curves of different orders (all starting at the same point) lie on a logarithmic spiral. This can be seen in the diagram below. Although many of their properties have been known for a long time and are well studied, dragon curves continue to appear in new and
interesting places. At last year's Maths Jam conference, Paul Taylor gave a talk about my favourite surprise occurrence of
a dragon.

Normally when we write numbers, we write them in base ten, with the digits in the number representing (from right to left) ones, tens,
hundreds, thousands, etc. Many readers will be familiar with binary numbers (base two), where the powers of two are used in the place of
powers of ten, so the digits represent ones, twos, fours, eights, etc.

In his talk, Paul suggested looking at numbers in base -1+i (where i is the square root of -1; you can find more adventures of i here) using the digits 0 and 1. From right to left, the columns of numbers in this
base have values 1, -1+i, -2i, 2+2i, -4, etc. The first 11 numbers in this base are shown below.

Number in base -1+i | Complex number |

0 | 0 |

1 | 1 |

10 | -1+i |

11 | (-1+i)+(1)=i |

100 | -2i |

101 | (-2i)+(1)=1-2i |

110 | (-2i)+(-1+i)=-1-i |

111 | (-2i)+(-1+i)+(1)=-i |

1000 | 2+2i |

1001 | (2+2i)+(1)=3+2i |

1010 | (2+2i)+(-1+i)=1+3i |

Complex numbers are often drawn on an Argand diagram: the real part of the number is plotted on the horizontal axis and the imaginary part
on the vertical axis. The diagram to the left shows the numbers of ten digits or less in base -1+i on an Argand diagram. The points form
an order 10 dragon curve! In fact, plotting numbers of \(n\) digits or less will draw an order \(n\) dragon curve.

Brilliantly, we may now use known properties of dragon curves to discover properties of base -1+i. A level \(\infty\) dragon curve covers
the entire plane without intersecting itself: therefore every Gaussian integer (a number of the form \(a+\text{i} b\) where \(a\) and
\(b\) are integers) has a unique representation in base -1+i. The endpoints of dragon curves lie on a logarithmic spiral: therefore
numbers of the form \((-1+\text{i})^n\), where \(n\) is an integer, lie on a logarithmic spiral in the complex plane.

If you'd like to play with some dragon curves, you can download the Python code used
to make the pictures here.

### Similar posts

Dragon curves | MENACE at Manchester Science Festival | The Mathematical Games of Martin Gardner | MENACE |

### Comments

Comments in green were written by me. Comments in blue were not written by me.

**2016-10-08**

## Logical contradictions

During my Electromagnetic Field talk this year, I spoke about @mathslogicbot, my Twitter bot that is working its way through the tautologies in propositional calculus. My talk included my conjecture that the number of tautologies of length \(n\) is an increasing sequence (except when \(n=8\)). After my talk, Henry Segerman suggested that I also look at the number of contradictions of length \(n\) to look for insights.

A contradiction is the opposite of a tautology: it is a formula that is False for every assignment of truth values to the variables. For example, here are a few contradictions:

$$\neg(a\leftrightarrow a)$$
$$\neg(a\rightarrow a)$$
$$(\neg a\wedge a)$$
$$(\neg a\leftrightarrow a)$$
The first eleven terms of the sequence whose \(n\)

$$0, 0, 0, 0, 0, 6, 2, 20, 6, 127, 154$$
^{th}term is the number of contradictions of length \(n\) are:This sequence is A277275 on OEIS. A list of contractions can be found here.

For the same reasons as the sequence of tautologies, I would expect this sequence to be increasing. Surprisingly, it is not increasing for small values of \(n\), but I again conjecture that it is increasing after a certain point.

### Properties of the sequences

There are some properties of the two sequences that we can show. Let \(a(n)\) be the number of tautolgies of length \(n\) and let \(b(n)\) be the number of contradictions of length \(n\).

First, the number of tautologies and contradictions, \(a(n)+b(n)\), (A277276) is an increasing sequence. This is due to the facts that \(a(n+1)\geq b(n)\) and \(b(n+1)\geq a(n)\), as every tautology of length \(n\) becomes a contraction of length \(n+1\) by appending a \(\neg\) to be start and vice versa.

This implies that for each \(n\), at most one of \(a\) and \(b\) can be decreasing at \(n\), as if both were decreasing, then \(a+b\) would be decreasing. Sadly, this doesn't seem to give us a way to prove the conjectures, but it is a small amount of progress towards them.

### Similar posts

Logic bot, pt. 2 | Logic bot | How OEISbot works | Raspberry Pi weather station |

### Comments

Comments in green were written by me. Comments in blue were not written by me.

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**2016-10-06**

## Palindromic dates

Thanks to Marc, I noticed that today's date is a palindrome in two different date formats—DMMYY (61016) and DMMYYYY (6102016).

This made me wonder when there will be another date that is palindromic in multiple date formats, so I wrote a Python script to find out.

Turns out there's not too long to wait: 10 July 2017 will be palindromic in two date formats (MDDYY and MDDYYYY). But before that, there's 1 July 2017, which is palindromic in three date formats (YYMMD, YYMD and MDYY). Most exciting of all, however, is 2 February 2020, which is palindromic in 7 different formats!

The next palindromic dates are shown in the following table. It will update as the dates pass.

\(n\) | Next day with \(\geq n\) palindromic formats | Formats |

1 | 1 September 2019 | YYMMD,YYMD,MDYY |

2 | 1 September 2019 | YYMMD,YYMD,MDYY |

3 | 1 September 2019 | YYMMD,YYMD,MDYY |

4 | 2 February 2020 | YYYYMMDD,DDMMYYYY,MMDDYYYY,YYYYMDD,YYMDD,DDMYY,MMDYY |

5 | 2 February 2020 | YYYYMMDD,DDMMYYYY,MMDDYYYY,YYYYMDD,YYMDD,DDMYY,MMDYY |

6 | 2 February 2020 | YYYYMMDD,DDMMYYYY,MMDDYYYY,YYYYMDD,YYMDD,DDMYY,MMDYY |

7 | 2 February 2020 | YYYYMMDD,DDMMYYYY,MMDDYYYY,YYYYMDD,YYMDD,DDMYY,MMDYY |

A full list of future palindromic dates can be found here.

### Similar posts

Dragon curves II | Logical contradictions | Dragon curves | How OEISbot works |

### Comments

Comments in green were written by me. Comments in blue were not written by me.

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**2016-03-30**

## Dragon curves

Take a piece of paper. Fold it in half in the same direction many times. Now unfold it. What pattern will the folds make?

I first found this question in one of Martin Gardner's books. At first, you might that the answer will be simple, but if you look at the shapes made for a few folds, you will see otherwise:

The curves formed are called

*dragon curves*as they allegedly look like dragons with smoke rising from their nostrils. I'm not sure I see the resemblance:As you increase the order of the curve (the number of times the paper was folded), the dragon curve squiggles across more of the plane, while never crossing itself. In fact, if the process was continued forever, an order infinity dragon curve would cover the whole plane, never crossing itself.

This is not the only way to cover a plane with dragon curves: the curves tessellate.

Dragon curves of different orders can also fit together:

To generate digital dragon curves, first notice that an order \(n\) curve can be made from two order \(n-1\) curves:

This can easily be seen to be true if you consider folding paper: If you fold a strip of paper in half once, then \(n-1\) times, each half of the strip will have made an order \(n-1\) dragon curve. But the whole strip has been folded \(n\) times, so is an order \(n\) dragon curve.

Because of this, higher order dragons can be thought of as lots of lower order dragons tiled together. An the infinite dragon curve is actually equivalent to tiling the plane with a infinite number of dragons.

If you would like to create your own dragon curves, you can download the Python code I used to draw them from GitHub. If you are more of a thinker, then you might like to ponder what difference it would make if the folds used to make the dragon were in different directions.

### Similar posts

Dragon curves II | MENACE in fiction | Building MENACEs for other games | Origins of World War I |

### Comments

Comments in green were written by me. Comments in blue were not written by me.

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**2015-08-29**

## How OEISbot works

A few weeks ago, I made OEISbot, a Reddit bot which posts information whenever an OEIS sequence is mentioned.

This post explains how OEISbot works. The full code can be found on GitHub.

### Getting started

OEISbot is made in Python using PRAW (Python Reddit Api Wrapper). PRAW can be installed with:

**bash**

Before making a bot, you will need to make a Reddit account for your bot, create a Reddit app and obtain API keys. This python script can be used to obtain the necessary keys.

Once you have your API keys saved in your praw.ini file, you are ready to make a bot.

### Writing the bot

First, the necessary imports are made, and test mode is activated if the script is run with test as an argument. We also define an exception that will be used later to kill the script once it makes a comment.

**python**

import re

import urllib

import json

from praw.objects import MoreComments

import sys

test = False

if len(sys.argv) > 1 and sys.argv[1] == "test":

test = True

print("TEST MODE")

class FoundOne(BaseException):

pass

To prevent OEISbot from posting multiple links to the same sequence in a thread, lists of sequences linked to in each thread can be loaded and saved using the following functions.

**python**

print(seen)

with open("/home/pi/OEISbot/seen/"+_id, "w") as f:

return json.dump(seen, f)

def open_list(_id):

try:

with open("/home/pi/OEISbot/seen/" + _id) as f:

return json.load(f)

except:

return []

The following function will search a post for a mention of an OEIS sequence number.

**python**

seen = open_list(id_)

re_s = re.findall("A([0-9]{6})", text)

re_s += re.findall("oeis\.org/A([0-9]{6})", url)

if test:

print(re_s)

post_me = []

for seq_n in re_s:

if seq_n not in seen:

post_me.append(markup(seq_n))

seen.append(seq_n)

if len(post_me) > 0:

post_me.append(me())

comment(joiner().join(post_me))

save_list(seen, id_)

raise FoundOne

The following function will search a post for a comma-separated list of numbers, then search for it on the OEIS. If there are 14 sequences or less found, it will reply. If it finds a list with no matches on the OEIS, it will message /u/PeteOK, as he likes hearing about possibly new sequences.

**python**

seen = open_list(id_)

if test:

print(text)

re_s = re.findall("([0-9]+\, *(?:[0-9]+\, *)+[0-9]+)", text)

if len(re_s) > 0:

for terms in ["".join(i.split(" ")) for i in re_s]:

if test:

print(terms)

if terms not in seen:

seen.append(terms)

first10, total = load_search(terms)

if test:

print(first10)

if len(first10)>0 and total <= 14:

if total == 1:

intro = "Your sequence (" + terms \

+ ") looks like the following OEIS sequence."

else:

intro = "Your sequence (" + terms + \

+ ") may be one of the following OEIS sequences."

if total > 4:

intro += " Or, it may be one of the " + str(total-4) \

+ " other sequences listed [here]" \

"(http://oeis.org/search?q=" + terms + ")."

post_me = [intro]

if test:

print(first10)

for seq_n in first10[:4]:

post_me.append(markup(seq_n))

seen.append(seq_n)

post_me.append(me())

comment(joiner().join(post_me))

save_list(seen, id_)

raise FoundOne

elif len(first10) == 0:

post_me = ["I couldn't find your sequence (" + terms \

+ ") in the [OEIS](http://oeis.org). "

"You should add it!"]

message("PeteOK",

"Sequence not in OEIS",

"Hi Peter, I've just found a new sequence (" \

+ terms + ") in [this thread](link). " \

"Please shout at /u/mscroggs to turn the " \

"feature off if its spamming you!")

post_me.append(me())

comment(joiner().join(post_me))

save_list(seen, id_)

raise FoundOne

def load_search(terms):

src = urllib.urlopen("http://oeis.org/search?fmt=data&q="+terms).read()

ls = re.findall("href=(?:'|\")/A([0-9]{6})(?:'|\")", src)

try:

tot = int(re.findall("of ([0-9]+) results found", src)[0])

except:

tot = 0

return ls, tot

The markup function loads the necessary information from OEIS and formats it. Each comment will end with the output of the me function. The ouput of joiner will be used between sequences which are mentioned.

**python**

pattern = re.compile("%N (.*?)<", re.DOTALL|re.M)

desc = urllib.urlopen("http://oeis.org/A" + seq_n + "/internal").read()

desc = pattern.findall(desc)[0].strip("\n")

pattern = re.compile("%S (.*?)<", re.DOTALL|re.M)

seq = urllib.urlopen("http://oeis.org/A" + seq_n + "/internal").read()

seq = pattern.findall(seq)[0].strip("\n")

new_com = "[A" + seq_n + "](http://oeis.org/A" + seq_n + "/): "

new_com += desc + "\n\n"

new_com += seq + "..."

return new_com

def me():

return "I am OEISbot. I was programmed by /u/mscroggs. " \

"[How I work](http://mscroggs.co.uk/blog/20). " \

"You can test me and suggest new features at /r/TestingOEISbot/."

def joiner():

return "\n\n- - - -\n\n"

Next, OEISbot logs into Reddit.

**python**

access_i = r.refresh_access_information(refresh_token=r.refresh_token)

r.set_access_credentials(**access_i)

auth = r.get_me()

The subs which OEISbot will search through are listed. I have used all the math(s) subs which I know about, as these will be the ones mentioning sequences.

**python**

"learnmath","mathbooks","cheatatmathhomework","matheducation",

"puremathematics","mathpics","mathriddles","askmath",

"recreationalmath","OEIS","mathclubs","maths"]

if test:

subs = ["TestingOEISbot"]

For each sub OEISbot is monitoring, the hottest 10 posts are searched through for mentions of sequences. If a mention is found, a reply is generated and posted, then the FoundOne exception will be raised to end the code.

**python**

for sub in subs:

print(sub)

subreddit = r.get_subreddit(sub)

for submission in subreddit.get_hot(limit = 10):

if test:

print(submission.title)

look_for_A(submission.id,

submission.title + "|" + submission.selftext,

submission.url,

submission.add_comment)

look_for_ls(submission.id,

submission.title + "|" + submission.selftext,

submission.add_comment,

submission.url,

r.send_message)

flat_comments = praw.helpers.flatten_tree(submission.comments)

for comment in flat_comments:

if ( not isinstance(comment, MoreComments)

and comment.author is not None

and comment.author.name != "OEISbot" ):

look_for_A(submission.id,

re.sub("\[[^\]]*\]\([^\)*]\)","",comment.body),

comment.body,

comment.reply)

look_for_ls(submission.id,

re.sub("\[[^\]]*\]\([^\)*]\)","",comment.body),

comment.reply,

submission.url,

r.send_message)

except FoundOne:

pass

### Running the code

I put this script on a Raspberry Pi which runs it every 10 minutes (to prevent OEISbot from getting refusals for posting too often). This is achieved with a cron job.

**bash**

### Making your own bot

The full OEISbot code is available on GitHub. Feel free to use it as a starting point to make your own bot! If your bot is successful, let me know about it in the comments below or on Twitter.

Edit: Updated to describe the latest version of OEISbot.

### Similar posts

Logic bot | Raspberry Pi weather station | Logical contradictions | Logic bot, pt. 2 |

### Comments

Comments in green were written by me. Comments in blue were not written by me.

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