Metadata-Version: 2.0
Name: python-category-equations
Version: 0.3.0
Summary: python-category-equations
Home-page: https://github.com/kummahiih/python-category-equations
Author: Pauli Rikula
Author-email: UNKNOWN
License: MIT
Platform: UNKNOWN
Classifier: License :: OSI Approved :: MIT License
Classifier: Programming Language :: Python :: 3.6
Requires-Python: ~=3.6

# python-category-equations


@copyright: 2010 - 2018 by Pauli Rikula <pauli.rikula@gmail.com>
@license: MIT <http://www.opensource.org/licenses/mit-license.php>


Create category like equations for the given operator in which
the underlaying '+' and '-' operations are basic set operations called union and discard.
The multiplication operator '*' connects sources to sinks. The equation system also has
a Identity 'I' and zerO 'O' terms. For futher details search 'category theory'
from the Wikipedia and do your own maths.




Here our connector operation is print function called 'debug' which
prints an arrow between two objects:

    >>> debug('a', 'b')
    a -> b

    >>> debug('b', 'a')
    b -> a

    >>> debug('a', 'a')
    a -> a

Get I and O singletons and class C, which use previously defined debug -function.
    >>> I, O, C = from_operator(debug)
    >>> I == I
    True
    >>> O == I
    False
    >>> C(1)
    C(1)

The items do have differing sinks and sources:

    >>> I.sinks
    {I}
    >>> I.sources
    {I}

    >>> O.sinks
    set()
    >>> O.sources
    set()

    >>> C(1).sinks
    {1}
    >>> C(1).sources
    {1}


You can write additions also with this notation

    >>> C(1,2) == C(1) + C(2)
    True


The multiplication connects sources to sinks like this:

    >>> (C(1,2) * C(3,4)).evaluate()
    1 -> 3
    1 -> 4
    2 -> 3
    2 -> 4

    >>> (C(3,4) * C(1,2)).sinks
    {3, 4}

    >>> (C(3,4) * C(1,2)).sources
    {1, 2}


By combining the two previous examples:

    >>> C(1,2) * C(3,4) == (C(1) + C(2)) * (C(3) + C(4))
    True

The order inside C(...) does not matter:

    >>> (C(1,2) * C(3,4)) == (C(2,1) * C(4,3))
    True

On the other hand you can not swap the terms like:

    >>> (C(1,2) * C(3,4)) == (C(3,4) * C(1,2))
    False

Because:

    >>> (C(3,4) * C(1,2)).evaluate()
    3 -> 1
    3 -> 2
    4 -> 1
    4 -> 2

The discard operation works like this:

    >>> (C(3,4) * C(1,2) - C(4) * C(1)).evaluate()
    3 -> 1
    3 -> 2
    4 -> 2

But

    >>> (C(3,4) * C(1,2) - C(4) * C(1)) == C(3) * C(1,2) + C(4) * C(2)
    False

Because sinks and sources differ:

    >>> (C(3,4) * C(1,2) - C(4) * C(1)).sinks
    {3}
    >>> (C(3) * C(1,2) + C(4) * C(2)).sinks
    {3, 4}

The right form would have been:

    >>> (C(3,4) * C(1,2) - C(4) * C(1)) == C(3) * C(1,2) + C(4) * C(2) - C(4) * O - O * C(1)
    True


The identity I and zero O work together like usual:

    >>> I * I == I
    True
    >>> O * I * O == O
    True


Identity 'I' works as a tool for equation simplifying.
For example:

    >>> C(1,2) * C(3,4) * C(5) + C(1,2) * C(5) == C(1,2) * ( C(3,4) + I ) * C(5)
    True

Because:

    >>> (C(1,2) * C(3,4) * C(5) + C(1,2) * C(5)).evaluate()
    1 -> 3
    1 -> 4
    1 -> 5
    2 -> 3
    2 -> 4
    2 -> 5
    3 -> 5
    4 -> 5

and

    >>> (C(1,2) * ( C(3,4) + I ) * C(5)).evaluate()
    1 -> 3
    1 -> 4
    1 -> 5
    2 -> 3
    2 -> 4
    2 -> 5
    3 -> 5
    4 -> 5

If two terms have the same middle part you can simplify equations via terminating loose sinks or sources with O:
For example:

    >>> (C(1) * C(2) * C(4) + C(3) * C(4)).evaluate()
    1 -> 2
    2 -> 4
    3 -> 4

    >>> (C(1) * C(2) * C(4) + O * C(3) * C(4)).evaluate()
    1 -> 2
    2 -> 4
    3 -> 4

    >>> (C(1) * ( C(2) + O * C(3) ) * C(4)).evaluate()
    1 -> 2
    2 -> 4
    3 -> 4

    >>> C(1) * C(2) * C(4) + O * C(3) * C(4) == C(1) * ( C(2) + O * C(3) ) * C(4)
    True


Note that the comparison wont work without the O -term because the sinks differ:

    >>> C(1) * C(2) * C(4) +  C(3) * C(4) == C(1) * ( C(2) + O * C(3) ) * C(4)
    False




