python-category-equations 0.3.7

python-category-equations

# python-category-equations

With the tools provided here you can create category like equations for the given operator. On the equations the underlaying ‘+’ and ‘-’ operations are basic set operations called union and discard and the multiplication operator ‘*’ connects sources to sinks. The equation system also has a Identity ‘I’ term and zerO -like termination term ‘O’. For futher details go https://en.wikipedia.org/wiki/Category_(mathematics)#Definition

## Usage

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