The Story of the Regexp and the Primes

If you are all sitting comfortably, then I shall begin...



Once upon a time, in a land an internet away, A python programmer was
browsing the web when he StumbledUpon a post with a curious regular
expression purporting to test numbers for primality. On further
investigation, the regexp was attributed in more than one place to
"Abigail" and took the form:

style="font-weight: bold; font-family: monospace;">^1?$|^(11+?)\1+$

Wanting to find out how it worked, the programmer was stumped, as all
the posts mentioning the regexp asked you to visit a page with a
supposedly excellent explanation of its inner workings, but the domain
name was no longer present.



The programmer decided to work it out for himself and came up with the
following...

A prime number is a
positive integer that is only divisible by itself and one. Zero and one
are not prime.



Let's say a number N, greater than one,  is not prime. Then
there exists two numbers, S and T, that when multiplied together equal
N. We have:

S x T = N

Lets us further assume
that T is greater than, or equal to S. (They can be swapped if
necessary to make this so). Then a small bit of manipulation gives:

(S x (T-1)) + S = N

(You take one less S in
the multiplication, then you need to add the S back).



Lets swap the terms and write this as equation (a):

style="font-weight: bold;">S + (S x (T-1)) = N  style="font-weight: bold;"> when S,T,N are integers greater
than one, and N is style="font-style: italic; font-weight: bold;">not style="font-weight: bold;"> prime

Abigail's regexp relies on representing integers as strings of that
number of ones:

Zero would be
    ''

One would be     '1'

two becomes     '11'

and so on...

The regexp gives a match if the string of ones does not represent a
prime.



To match zero or one occurrences of a 1 we use the first half of the
regexp:

style="font-family: monospace;">^1?$

Where ^anchors the following to a match from the beginning of the
string, ? will match zero or one occurrences of the preceding 1, and
the  trailing $ matches the end of the string, i.e. it matches
a string that contains only zero or one 1 characters.



Going back to equation (a), greater than one, is the same as two or
more.

So S will be represented by two or more 1's - we don't know how many,
but we do know that it is two or more. S, in a regexp, becomes
something like:

style="font-family: monospace;">11+

1 followed by one or more 1's. S is less than or equal to T,
which translates to using a less greedy form of the zero-or-one
matcher, which is +? giving a representation of S as:

style="font-family: monospace;">11+?

We want to find S followed by T-1 identical copies of S, so lets form a
group out of what matches S:

style="font-family: monospace;">(11+?)

Then by referring to this group, it is the first group in the regexp so
can be referred to as \1, we can search for extra copies of the group
by:

style="font-family: monospace;">\1+

S and its copies must cover the whole of the number so must start at ^
and finish at $:

style="font-weight: bold; font-family: monospace;">^(11+?)\1+$

Together with the earlier regexp fraction that covers the numbers zero
and one, they can be or'd together to cover all non primes as
the following:

style="font-family: monospace;">^1?$|^(11+?)\1+$

...Not knowing when to quit, the programmer opted to display some of
his Python code that defines a primality tester using the regexp:

>>> import re

>>> def isprime(n):

    return not re.match(r' style="font-weight: bold;">^1?$|^(11+?)\1+$', '1' * n)

>>> [i for i in range(40) if isprime(i)]

[2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37]

And another function that shows S and T  from the regexp match:

>>> def is style="font-weight: bold;">notprime(n):

    notprime = re.match(r'^(1?)$|^(11+?)(\2+)$', '1' * n)

    if notprime:

        if n <=1:

            return (len( notprime.groups()[0]), )

        else:

            S, SxT1 = notprime.groups()[1:]

            s = len(S)

            t = 1 + len(SxT1)//s

            return (s,t)

    else:

        return ()

>>> [(i, isnotprime(i)) for i in range(15)]

[(0, (0,)), (1, (1,)), (2, ()), (3, ()), (4, (2, 2)), (5, ()), (6, (2, 3)), (7, ()), (8, (2, 4)), (9, (3, 3)), (10, (2, 5)), (11, ()), (12, (2, 6)), (13, ()), (14, (2, 7))]

... And so to bed.



THE END.