python-category-equations 0.8.0

python-category-equations

# python-category-equations

Category is way to represent and generate directed networks by using sinks, sources and connections from sources to sinks. With the tools provided here you can create and simplify category like equations for the given connector operator. On the equations the underlying + 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 term and zerO -like termination term O. For further 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}

Or

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

>>> (C(1) * C(2, I)).evaluate()
1 -> 2

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

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

And writing C(1,2) instead of C(1) + C(2) works with multiplication too:

>>> 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 won’t 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

## Equation solving and minimizations

The module contains also (quite inefficient) simplify -method, which can be used to expression minimization:

>>> I, O, C = from_operator(debug)
>>> m = EquationMap(I, O, C)
>>> a = C(1) + C(2)
>>> simplify(a, 300, m)
(C(1, 2), [C(1) + C(2), C(1, 2)])

>>> b = C(1) * C(3) + C(2) * C(3)
>>> simplified, path = simplify(b, 100, m)
>>> simplified
C(1, 2) * C(3)
>>> for p in path:
...    print(p)
C(1) * C(3) + C(2) * C(3)
(C(1) * I + C(2) * I) * C(3)
(C(1) + C(2) * I) * C(3)
(C(1) + C(2)) * C(3)
C(1, 2) * C(3)

For proofs use the get_route:

>>> I, O, C = from_operator(debug)
>>> m = EquationMap(I, O, C)
>>> a = C(1) * C(3) + C(2) * C(3)
>>> b = C(1, 2) * C(3)
>>> shortest, path = get_route(a,b, 100, m)
>>> for p in path:
...    print(p)
C(1) * C(3) + C(2) * C(3)
C(1) * C(3) + C(2) * I * C(3)
(C(1) * I + C(2) * I) * C(3)
(C(1) + C(2) * I) * C(3)
(C(1) + C(2)) * C(3)
C(1, 2) * C(3)