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 2016-10-08 

Logical Contradictions

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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 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!

Similar Posts

Logical Contradictions
Logic Bot
How OEISbot Works
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 2014-11-26 

Logic Bot

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!

Christian
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