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2016-10-08

During my EMF 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 $$n$$th term is the number of contradictions of length $$n$$ are:
$$0, 0, 0, 0, 0, 6, 2, 20, 6, 127, 154$$
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.

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2016-06-05

## Making Names in Life

The Game of Life is a cellular automaton invented by John Conway in 1970, and popularised by Martin Gardner.
In Life, cells on a square grid are either alive or dead. It begins at generation 0 with some cells alive and some dead. The cells' aliveness in the following generations are defined by the following rules:
• Any live cell with four or more live neighbours dies of overcrowding.
• Any live cell with one or fewer live neighbours dies of loneliness.
• Any dead cell with exactly three live neighbours comes to life.
Starting positions can be found which lead to all kinds of behaviour: from making gliders to generating prime numbers. The following starting position is one of my favourites:
It looks boring enough, but in the next generation, it will look like this:
If you want to confirm that I'm not lying, I recommend the free Game of Life Software Golly.

### Going Backwards

You may be wondering how I designed the starting pattern above. A first, it looks like a difficult task: each cell can be dead or alive, so I need to check every possible combination until I find one. The number of combinations will be $$2^\text{number of cells}$$. This will be a very large number.
There are simplifications that can be made, however. Each of the letters above (ignoring the gs) is in a 3×3 block, surrounded by dead cells. Only the cells in the 5×5 block around this can affect the letter. These 5×5 blocks do no overlap, so can be calculated seperately. I doesn't take too long to try all the possibilities for these 5×5 blocks. The gs were then made by starting with an o and trying adding cells below.

### Can I Make My Name?

Yes, you can make your name.
I continued the search and found a 5×5 block for each letter. Simply Enter your name in the box below and these will be combined to make a pattern leading to your name!

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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
pip install praw
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 praw
import re
import urllib
import json

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
def save_list(seen, _id):
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:
except:
return []
The following function will search a post for a mention of an OEIS sequence number.
python
def look_for_A(id_, text, url, comment):
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
def look_for_ls(id_, text, comment, link, message):
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)
if test:
print(first10)
if len(first10)>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). "
message("PeteOK",
"Sequence not in OEIS",
"Hi Peter, I've just found a new sequence (" \
"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

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
def markup(seq_n):
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
= praw.Reddit("OEIS link and description poster by /u/mscroggs.")

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
subs = ["TestingOEISbot","math","mathpuzzles","casualmath","theydidthemath",
"learnmath","mathbooks","cheatatmathhomework","matheducation",
"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
try:
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,
look_for_ls(submission.id,
submission.title + "|" + submission.selftext,
submission.url,
r.send_message)

and comment.author is not None
and comment.author.name != "OEISbot" ):
look_for_A(submission.id,
re.sub("$[^$]*\]$$[^$$*]\)","",comment.body),
comment.body,
look_for_ls(submission.id,
re.sub("$[^$]*\]$$[^$$*]\)","",comment.body),
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
*/10 * * * * python /path/to/bot.py

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.

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2015-08-27

## MENACE

### Machine Educable Noughts And Crosses Engine

In 1961, Donald Michie build MENACE (Machine Educable Noughts And Crosses Engine), a machine capable of learning to be a better player of Noughts and Crosses (or Tic-Tac-Toe if you're American). As computers were less widely available at the time, MENACE was built from from 304 matchboxes.
Taken from Trial and Error by Donald Michie [2]
The original MENACE.
To save you from the long task of building a copy of MENACE, I have written a JavaScript version of MENACE, which you can play against here.

### How To Play Against MENACE

To reduce the number of matchboxes required to build it, MENACE aways plays first. Each possible game position which MENACE could face is drawn on a matchbox. A range of coloured beads are placed in each box. Each colour corresponds to a possible move which MENACE could make from that position.
To make a move using MENACE, the box with the current board position must be found. The operator then shakes the box and opens it. MENACE plays in the position corresponding to the colour of the bead at the front of the box.
For example, in this game, the first matchbox is opened to reveal a red bead at its front. This means that MENACE (O) plays in the corner. The human player (X) then plays in the centre. To make its next move, MENACE's operator finds the matchbox with the current position on, then opens it. This time it gives a blue bead which means MENACE plays in the bottom middle.
The human player then plays bottom right. Again MENACE's operator finds the box for the current position, it gives an orange bead and MENACE plays in the left middle. Finally the human player wins by playing top right.
MENACE has been beaten, but all is not lost. MENACE can now learn from its mistakes to stop the happening again.

### How MENACE Learns

MENACE lost the game above, so the beads that were chosen are removed from the boxes. This means that MENACE will be less likely to pick the same colours again and has learned. If MENACE had won, three beads of the chosen colour would have been added to each box, encouraging MENACE to do the same again. If a game is a draw, one bead is added to each box.
Initially, MENACE begins with four beads of each colour in the first move box, three in the third move boxes, two in the fifth move boxes and one in the final move boxes. Removing one bead from each box on losing means that later moves are more heavily discouraged. This helps MENACE learn more quickly, as the later moves are more likely to have led to the loss.
After a few games have been played, it is possible that some boxes may end up empty. If one of these boxes is to be used, then MENACE resigns. When playing against skilled players, it is possible that the first move box runs out of beads. In this case, MENACE should be reset with more beads in the earlier boxes to give it more time to learn before it starts resigning.

### How MENACE Performs

In Donald Michie's original tournament against MENACE, which lasted 220 games and 16 hours, MENACE drew consistently after 20 games.
Taken from Trial and Error by Donald Michie [2]
Graph showing MENACE's performance in the original tournament.
After a while, Michie tried playing some more unusual games. For a while he was able to defeat MENACE, but MENACE quickly learnt to stop losing. You can read more about the original MENACE in A Matchbox Game Learning-Machine by Martin Gardner [1] and Trial and Error by Donald Michie [2].
You may like to experiment with different tactics against MENACE yourself.

### Play Against MENACE

I have written a JavaScript implemenation of MENACE for you to play against. The source code for this implementation is available on GitHub.
When playing this version of MENACE, the contents of the matchboxes are shown on the right hand side of the page. The numbers shown on the boxes show how many beads corresponding to that move remain in the box. The red numbers show which beads have been picked in the current game.
The initial numbers of beads in the boxes and the incentives can be adjusted by clicking Adjust MENACE's settings above the matchboxes. My version of MENACE starts with more beads in each box than the original MENACE to prevent the early boxes from running out of beads, causing MENACE to resign.
Additionally, next to the board, you can set MENACE to play against random, or a player 2 version of MENACE.
Edit: After hearing me do a lightning talk about MENACE at CCC, Oliver Child built a copy of MENACE. Here are some pictures he sent me:
Edit: Oliver has written about MENACE and the version he built in issue 03 of Chalkdust Magazine.
Edit: Inspired by Oliver, I have built my own MENACE. I took it to the MathsJam Conference 2016. It looks like this:

#### References

A Matchbox Game Learning-Machine by Martin Gardner. Scientific American, March 1962. [link]
Trial and Error by Donald Michie. Penguin Science Survey, 1961.

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2017-09-11
영준이가~ 좋아하는~ 토론~게임
유동훈
2016-12-07
I've also made a physical MENACE since writing this. I plan to make a video at some point: I'll let you know when I do...
Matthew
2016-12-07
Oh, I just read further and saw that Oliver made the physical MENACE. I shall menace him instead!
Steve Paget
2016-12-07
I would love to see a game being played with this machine. Could you make a Youtube video of a game in progress, so we can see it in action? How long does a typical game last with a trained engine?
Steve Paget
2015-12-14
There may be a mistake in the source code somewhere. I will look into it.
Matthew

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2015-03-15

## Logic Bot, pt. 2

A few months ago, I set @mathslogicbot going on the long task of tweeting all the tautologies (containing 140 characters or less) in propositional calculus with the symbols $$\neg$$ (not), $$\rightarrow$$ (implies), $$\leftrightarrow$$ (if and only if), $$\wedge$$ (and) and $$\vee$$ (or). My first post on logic bot contains a full explanation of propositional calculus, formulae and tautologies.

### An Alternative Method

Since writing the original post, I have written an alternative script to generate all the tautologies. In this new method, I run through all possible strings of length 1 made with character in the logical language, then strings of length 2, 3 and so on. The script then checks if they are valid formulae and, if so, if they are tautologies.
In the new script, only formulae where the first appearances of variables are in alphabetical order are considered. This means that duplicate tautologies are removed. For example, $$(b\rightarrow(b\wedge a))$$ will not be counted as it is the same as $$(a\rightarrow(a\wedge b))$$.
You can view or download this alternative code on github. All the terms of the sequence that I have calculated so far can be viewed here and the tautologies for these terms are here.

### Sequence

One advantage of this method is that it generates the tautologies sorted by the number of symbols they contain, meaning we can generate the sequence whose $$n$$th term is the number of tautologies of length $$n$$.
The first ten terms of this sequence are
$$0, 0, 0, 0, 2, 2, 12, 6, 57, 88$$
as there are no tautologies of length less than 5; and, for example two tautologies of length 6 ($$(\neg a\vee a)$$ and $$(a\vee \neg a)$$).
This sequence is listed as A256120 on OEIS.

#### Properties

There are a few properties of this sequence that can easily be shown. Throughout this section I will use $$a_n$$ to represent the $$n$$th term of the sequence.
Firstly, $$a_{n+2}\geq a_n$$. This can be explained as follows: let $$A$$ be a tautology of length $$n$$. $$\neg\neg A$$ will be of length $$n+2$$ and is logically equivalent to $$A$$.
Another property is $$a_{n+4}\geq 2a_n$$: given a tautology $$A$$ of length $$n$$, both $$(a\vee A)$$ and $$(A\vee a)$$ will be tautologies of length $$n+4$$. Similar properties could be shown for $$\rightarrow$$, $$\leftrightarrow$$ and $$\wedge$$.
Given properties like this, one might predict that the sequence will be increasing ($$a_{n+1}\geq a_n$$). However this is not true as $$a_7$$ is 12 and $$a_8$$ is only 6. It would be interesting to know at how many points in the sequence there is a term that is less than the previous one. Given the properties above it is reasonable to conjecture that this is the only one.
Edit: The sequence has been published on OEIS!

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