piecewise_fold does not handle bound variables

[dniku]

Sympy 1.0-2 from Manjaro repository.

In [1]: import sympy as sp

In [2]: sp.init_session()
IPython console for SymPy 1.0 (Python 3.6.0-64-bit) (ground types: gmpy)

These commands were executed:
>>> from __future__ import division
>>> from sympy import *
>>> x, y, z, t = symbols('x y z t')
>>> k, m, n = symbols('k m n', integer=True)
>>> f, g, h = symbols('f g h', cls=Function)
>>> init_printing()

Documentation can be found at http://docs.sympy.org/1.0/

In [3]: k, i, n = sp.symbols('k, i, n', integer=True)

In [4]: p = sp.SeqFormula(sp.Piecewise((k, sp.Eq(i, 0)), (0, True)), (i, 0, 5))

In [5]: p
Out[5]: [k, 0, 0, 0, …]

In [6]: sp.piecewise_fold(p)
Out[6]: 
⎧[k, k, k, k, …]  for i = 0
⎨                          
⎩[0, 0, 0, 0, …]  otherwise

Needless to say, [5] and [6] should be identical.

[hanq08]

Maybe I'm missing something but I don't see any error here. Why [5] and [6] should be identical?

picewise_fold() takes an expression containing a piecewise function and returns the expression in piecewise form.

[6] seems to be a correct representation of the piecewise function p defined in [4].

[dniku]

@hanq08 in this case, i is a variable bound in the second argument of SeqFormula in [4]. It is an index that is fed into the function within SeqFormula. In [6], i becomes a free variable.

The object defined in [4] is printed in [5]. [6] is different from [5], hence it is not a correct representation of the object defined in [4].

[smichr]

Does this work with the new piecewise_fold function?

[dniku]

@smichr I haven't tested it and I probably won't for a long time.

[oscarbenjamin]

I get the same on current master.

[smichr]

I suspect that functions should have their own _eval_piecewise_fold that can override the assumed simplicity in the folding of a function (see inline notes of piecewise_fold).

[oscargus]

Maybe an easier to grasp example (found this when I was just about to post a new issue):

In [13]: Integral(Piecewise((1, x >= 1), (2, True)), (x, 0, 2))
Out[13]: 
2                 
⌠                 
⎮ ⎧1  for x ≥ 1   
⎮ ⎨             dx
⎮ ⎩2  otherwise   
⌡                 
0                 

In [14]: Integral(Piecewise((1, x >= 1), (2, True)), (x, 0, 2)).doit()
Out[14]: 3

In [15]: piecewise_fold(Integral(Piecewise((1, x >= 1), (2, True)), (x, 0, 2))).doit()
Out[15]: 
⎧2  for x ≥ 1
⎨            
⎩4  otherwise

It shouldn't be folding if the Piecewise is the argument of an integral or, ideally, do it in a different way

2        1     
⌠        ⌠     
⎮ 1 dx + ⎮ 2 dx
⌡        ⌡     
1        0     

which will be feasible for simple constraints.

For indefinite integrals it is more tricky though.

In [18]: (Integral(Piecewise((1, x >= 1), (2, True)), (x))).doit()
Out[18]: 
⎧ 2⋅x   for x ≤ 1
⎨                
⎩x + 1  otherwise

In [19]: piecewise_fold(Integral(Piecewise((1, x >= 1), (2, True)), (x))).doit()
Out[19]: 
⎧ x   for x ≥ 1
⎨              
⎩2⋅x  otherwise

In [20]: piecewise_fold(Integral(Piecewise((1, x >= 1), (2, True)), (x)).doit())
Out[20]: 
⎧ 2⋅x   for x ≤ 1
⎨                
⎩x + 1  otherwise