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 2015-08-27 
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. Edit: Added the redrawn graph on the left.
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:

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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Building MENACEs for other games
MENACE at Manchester Science Festival
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The Mathematical Games of Martin Gardner

Comments

Comments in green were written by me. Comments in blue were not written by me.
 2019-07-06 
Can you make a version where the ai goes second?
Reply
Nukey
 2019-04-09 
@Sterling: To make a MENACE to play chess, you'd need to draw every possible position in chess. As I wrote in this blog post, if you had a matchbox for each possible chess position, your pile of matchboxes would be this big:



This number of positions is too big for this method of learning to work for chess, so you'd need a more advanced method...
Reply
Matthew
 2019-04-08 
After watching the video with yourself and Matt, you commented on teaching it chess are there any PDFs on how to teach, the machine chess. Ie what are the mechanics
Reply
Sterling
 2019-03-27 
@Alex: You can find the pdfs here. boxes0.pdf to boxes3.pdf are the states for a MENACE that goes first. second_boxes0.pdf to second_boxes2.pdf are the states for a MENACE that goes second.
Reply
Matthew
 2019-03-26 
I know this has been asked a lot but would it be possible if you could make a pdf of all possible states of the game on your website? If not please just send them to me directly. Thank you, alex
Reply
Alex
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 2015-03-15 
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 now 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!

Similar posts

Logical contradictions
Logic bot
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Comments

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 2014-11-26 
Last week, mathslogicbot started the long task of tweeting every tautology in propositional calculus. This post explains what this means and how I did it.

What is propositional calculus?

Propositional calculus is a form of mathematical logic, in which the formulae (the logical 'sentences') are made up of the following symbols:

Formulae

Formulae are defined recursively using the following rules:
For example, \((a\vee b)\), \(\neg f\) and \(((a\vee b)\rightarrow\neg f)\) are formulae.
Each of the variables is assigned a value of either "true" or "false", which leads to each formula being either true or false:

Tautologies

A tautology is a formula that is true for any assigment of truth values to the variables. For example:
\((a\vee \neg a)\) is a tautology because: if \(a\) is true then \(a\) or \(\neg a\) is true; and if \(a\) is false, then \(\neg a\) is true, so \(a\) or \(\neg a\) is true.
\((a\leftrightarrow a)\) is a tautology because: if \(a\) is true then \(a\) and \(a\) are both true; and if \(a\) is false then \(a\) and \(a\) are both false.
\((a\wedge b)\) is not a tautology because if \(a\) is true and \(b\) is false, then it is false.
The following are a few more tautologies. Can you explain why they are always true?

Python

If you want to play with the Logic Bot code, you can download it here.
In order to find all tautologies less than 140 characters long, one method is to first generate all formulae less than 140 characters then check to see if they are tautologies. (This is almost certainly not the fastest way to do this, but as long as it generates tautolgies faster than I want to tweet them, it doesn't matter how fast it runs.) I am doing this on a Raspberry Pi using Python in the following way.

All formulae

The following code is writing all the formulae that are less than 140 characters to a file called formulae.
 python 
from os.path import join
path = "/home/pi/logic"
First import any modules needed and set the path where the file will be saved.
 python 
def candidate(formula):
    global formulae

    if len(formula) <= 140 and formula not in formulae:
        formulae.append(formula)
        print formula
        f = open(join(path,"formulae"),"a")
        f.write(formula + "\n")
        f.close()
This function checks that a formula is not already in my list of formulae and shorter than 140 characters, then adds it to the list and writes it into the file.
 python 
variables = ["a""b""c""d""e""f""g""h""i""j",
             "k""l""m""n""o""p""q""r""s""t",
             "u""v""w""x""y""z""@""#""2""3",
             "4""5""6""7""8""9"]
This line says which characters are going to be used as variables. It is impossible to write a formula in less that 140 characters with more than 36 different variables so these will be sufficient. I haven't used 0 and 1 as these are used to represent false and true later.
 python 
= open(join(path,"formulae"))
formulae = f.readlines()
for i in range(0,len(formulae)):
    formulae[i] = formulae[i].strip("\n")
f.close()
These lines load the formulae already found from the file. This is needed if I have to stop the code then want to continue.
 python 
oldlen = 0
newlen = 26

while oldlen != newlen:
    for f in formulae + variables:
        candidate("-" + f)
    for f in formulae + variables:
        for g in formulae:
            for star in ["I""F""N""U"]:
                candidate("(" + f + star + g + ")")
    oldlen = newlen
    newlen = len(formulae)
The code inside the while loop goes through every formula already found and puts "-" in front of it, then takes every pair of formulae already found and puts "I", "F", "N" or "U" between them. These characters are used instead of the logical symbols as using the unicode characters leads to numerous python errors. The candidate function as defined above then adds them to the list (if they are suitable). This continues until the loop does not make the list of formulae longer as this will occur when all formulae are found.
 python 
= open(join(path,"formulae"),"a")
f.write("#FINISHED#")
f.close()
Once the loop has finished this will add the string "#FINISHED#" to the file. This will tell the truth-checking code when the it has checked all the formulae (opposed to having checked all those generated so far).

Tautologies

Now that the above code is finding all formulae, I need to test which of these are tautologies. This can be done by checking whether every assignment of truth values to the variables will lead to the statement being true.
 python 
from os.path import join
path = "/home/pi/logic"
First import any modules needed and set the path where the file will be saved.
 python 
def next(ar,i=0):
    global cont
    if i < len(ar):
        if ar[i] == "0":
            ar[i] = "1"
        else:
            ar[i] = "0"
            ar = next(ar, i + 1)
    else:
        cont = False
    return ar
Given an assignment of truth values, this function will return the next assignment, setting cont to False if all the assignments have been tried.
 python 
def solve(lo):
    lo = lo.replace("-0""1")
    lo = lo.replace("-1""0")

    lo = lo.replace("(0I0)""1")
    lo = lo.replace("(0I1)""1")
    lo = lo.replace("(1I0)""0")
    lo = lo.replace("(1I1)""1")

    lo = lo.replace("(0F0)""1")
    lo = lo.replace("(0F1)""0")
    lo = lo.replace("(1F0)""0")
    lo = lo.replace("(1F1)""1")

    lo = lo.replace("(0N0)""0")
    lo = lo.replace("(0N1)""0")
    lo = lo.replace("(1N0)""0")
    lo = lo.replace("(1N1)""1")

    lo = lo.replace("(0U0)""0")
    lo = lo.replace("(0U1)""1")
    lo = lo.replace("(1U0)""1")
    lo = lo.replace("(1U1)""1")

    return lo
This function will replace all instances of "NOT TRUE" with "FALSE" and so on. It will be called repeatedly until a formula is reduced to true or false.
 python 
= open(join(path,"formulae"))
formulae = f.readlines()
f.close()

= open(join(path,"donet"))
= int(f.read())
f.close()

variables = ["a""b""c""d""e""f""g""h""i""j",
             "k""l""m""n""o""p""q""r""s""t",
             "u""v""w""x""y""z""@""#""2""3",
             "4""5""6""7""8""9"]
These lines read the formulae from the file they are saved in and load how many have been checked if this script has been restarted. The the variables are set.
 python 
while formulae[-1] != "#FINISHED#" or i < len(formulae) - 1:
    if i < len(formulae):
        formula = formulae[i].strip("\n")
These lines will loop through all formulae until "#FINISHED#" is reached.
 python 
        insofar = True
        inA = []
        fail = False
        for a in variables:
            if a not in formula:
                insofar = False
            elif not insofar:
                fail = True
                break
            else:
                inA.append(a)
Here, the code checks that if a variable is in the formula, then all the previous variables are in the formula. This will prevent the Twitter bot from repeating many tautologies that are the same except for the variable a being replaced by b (although there will still be some repeats like this. Can you work out what these will be?).
 python 
        if not fail:
            valA = ["0"]*len(inA)
            cont = True
            taut = True
            while cont and taut:
                feval = formula
                for j in range(0,len(inA)):
                    feval = feval.replace(inA[j],valA[j])
                while feval not in ["0""1"]:
                    feval = solve(feval)
                if feval != "1":
                    taut = False
                valA = next(valA)
            if taut:
                f = open(join(path,"true"),"a")
                f.write(str(formula) + "\n")
                f.close()
    
        i += 1
        f = open(join(path,"donet"),"w")
        f.write(str(i))
        f.close()
Now, the formula is tested to see if it is true for every assignment of truth values. If it is, it is added to the file containing tautologies. Then the number of formulae that have been checked is written to a file (in case the script is stopped then resumed).
 python 
    else:
        f = open(join(path,"formulae"))
        formulae = f.readlines()
        f.close()
If the end of the formulae file is reached, then the file is re-loaded to include all the formulae found while this code was running.

Tweeting

Finally, I wrote a code that tweets the next item in the file full of tautologies every three hours (after replacing the characters with the correct unicode characters).

How long will it take?

Now that the bot is running, it is natural to ask how long it will take to tweet all the tautologies.
While it is possible to calculate the number of formulae with 140 characters or less, there is no way to predict how many of these will be tautologies without checking. However, the bot currently has over 13 years of tweets lines up. And all the tautologies so far are under 30 characters so there are a lot more to come...
Edit: Updated time left to tweet.

Similar posts

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

Comments

Comments in green were written by me. Comments in blue were not written by me.
 2017-07-05 
In part two you say a_{n+4} >= 2*a_n, and you have 13 years worth of tweets of length (say) 15-30. so there are 26 years worth length 19-34 characters, 13*2^n years worth of tweets of length between (15 + 4n) and (30 + 4n). In particular, setting n = 27, we have 13*2^{27} = 1744830464 years worth of tweets of length 123-138. I hope you have nice sturdy hardware!

Reply
Christian
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 2013-07-11 
On 19 June, the USB temperature sensor I ordered from Amazon arrived. This sensor is now hooked up to my Raspberry Pi, which is taking the temperature every 10 minutes, drawing graphs, then uploading them here. Here is a brief outline of how I set this up:

Reading the temperature

I found this code and adapted it to write the date, time and temperature to a text file. I then set cron to run this every 10 minutes. It writes the data to a text file (/var/www/temperature2) in this format:
2013 06 20 03 50,16.445019
2013 06 20 04 00,16.187843
2013 06 20 04 10,16.187843
2013 06 20 04 20,16.187843

Plotting the graphs

I found a guide somewhere on the internet about how to draw graphs with Python using Pylab/Matplotlib. If you have any idea where this could be, comment below and I'll put a link here.
In the end my code looked like this:
 python 
import time
import matplotlib as mpl
mpl.use("Agg")
import matplotlib.pylab as plt
import matplotlib.dates as mdates

ts = time.time()
import datetime
now = datetime.datetime.fromtimestamp(ts)
st = mdates.date2num(datetime.datetime(int(float(now.strftime("%Y"))),
                                       int(float(now.strftime("%m"))),
                                       int(float(now.strftime("%d"))),
                                       0, 0, 0))
weekst = st - int(float(datetime.datetime.fromtimestamp(ts).strftime("%w")))
= file("/var/www/temperature2","r")
= []
= []
tt = []
ss = []
= []
= []
= []
= []
= []
weekt = []
weektt = []
weeks = []
weekss = []
mini = 1000
maxi = 0
cur = -1
datC = 0

for line in f:
    fL = line.split(",")
    fL[0] = fL[0].split(" ")
    if cur == -1:
        cur = mdates.date2num(datetime.datetime(int(float(fL[0][0])),
                                                int(float(fL[0][1])),
                                                int(float(fL[0][2])),
                                                0,0,0))
        datC = mdates.date2num(datetime.datetime(int(float(fL[0][0])),
                                                 int(float(fL[0][1])),
                                                 int(float(fL[0][2])),
                                                 int(float(fL[0][3])),
                                                 int(float(fL[0][4])),
                                                 0))
    u.append(datC)
    v.append(fL[1])
    if datC >= st and datC <= st + 1:
        t.append(datC)
        s.append(fL[1])
    if datC >= st - 1 and datC <= st:
        tt.append(datC + 1)
        ss.append(fL[1])
    if datC >= weekst and datC <= weekst + 7:
        weekt.append(datC)
        weeks.append(fL[1])
    if datC >= weekst - 7 and datC <= weekst:
        weektt.append(datC + 7)
        weekss.append(fL[1])
    if datC > cur + 1:
        g.append(cur)
        h.append(mini)
        i.append(maxi)
        mini = 1000
        maxi = 0
        cur = mdates.date2num(datetime.datetime(int(float(fL[0][0])),
                                                int(float(fL[0][1])),
                                                int(float(fL[0][2])),
                                                0,0,0))
    mini = min(float(fL[1]),mini)
    maxi = max(float(fL[1]),maxi)
g.append(cur)
h.append(mini)
i.append(maxi)

plt.plot_date(x=t,y=s,fmt="r-")
plt.plot_date(x=tt,y=ss,fmt="g-")
plt.xlabel("Time")
plt.ylabel("Temperature (#^\circ#C)")
plt.title("Daily")
plt.legend(["today","yesterday"], loc="upper left",prop={"size":8})
plt.grid(True)
plt.gca().xaxis.set_major_formatter(mdates.DateFormatter("%H:%M"))
plt.gcf().subplots_adjust(bottom=0.15,right=0.99)
labels = plt.gca().get_xticklabels()
plt.setp(labels,rotation=90,fontsize=10)
plt.xlim(st,st + 1)
plt.savefig("/var/www/tempr/tg1p.png")

plt.clf()

plt.plot_date(x=weekt,y=weeks,fmt="r-")
plt.plot_date(x=weektt,y=weekss,fmt="g-")
plt.xlabel("Day")
plt.ylabel("Temperature (#^\circ#C)")
plt.title("Weekly")
plt.legend(["this week","last week"], loc="upper left",prop={"size":8})
plt.grid(True)
plt.gca().xaxis.set_major_formatter(mdates.DateFormatter("                     %A"))
plt.gcf().subplots_adjust(bottom=0.15,right=0.99)
labels = plt.gca().get_xticklabels()
plt.setp(labels,rotation=0,fontsize=10)
plt.xlim(weekst,weekst + 7)
plt.savefig("/var/www/tempr/tg2p.png")

plt.clf()

plt.plot_date(x=u,y=v,fmt="r-")
plt.xlabel("Date & Time")
plt.ylabel("Temperature (#^\circ#C)")
plt.title("Forever")
plt.grid(True)
plt.gca().xaxis.set_major_formatter(mdates.DateFormatter("%d/%m/%y %H:%M"))
plt.gcf().subplots_adjust(bottom=0.25,right=0.99)
labels = plt.gca().get_xticklabels()
plt.setp(labels,rotation=90,fontsize=8)
plt.savefig("/var/www/tempr/tg4p.png")


plt.clf()

plt.plot_date(x=g,y=h,fmt="b-")
plt.plot_date(x=g,y=i,fmt="r-")
plt.xlabel("Date")
plt.ylabel("Temperature (#^\circ#C)")
plt.title("Forever")
plt.legend(["minimum","maximum"], loc="upper left",prop={"size":8})
plt.grid(True)
plt.gca().xaxis.set_major_formatter(mdates.DateFormatter("%d %b"))
plt.gcf().subplots_adjust(bottom=0.15,right=0.99)
labels = plt.gca().get_xticklabels()
plt.setp(labels,rotation=90,fontsize=8)
plt.savefig("/var/www/tempr/tg3p.png")
If there's anything in there you don't understand, comment below and I'll try to fill in the gaps.

Uploading the graphs

Finally, I upload the graphs to mscroggs.co.uk/weather. To do this, I set up pre-shared keys on the Raspberry Pi and this server and added the following as a cron job:
 bash 
0 * * * * scp /var/www/tempr/tg*p.png username@mscroggs.co.uk:/path/to/folder
I hope this was vaguely interesting/useful. I'll try to add more details and updates over time. If you are building something similar, please let me know in the comments; I'd love to see what everyone else is up to.
Edit: Updated to reflect graphs now appearing on mscroggs.co.uk not catsindrag.co.uk.

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