python-category-equations

## Project description

# 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 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}


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 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


## Equation solving and minimizations

The module contains also (quite unefficient) 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)


## Project details

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