Structural zero detection with MX expressions

[Jonathan Frey]

Hi all,

I want to process a CasADi expression to detect linear and nonlinear parts, using structural zero detection "is_zero".

For SX variables, this works fine. I do something like in the following short example:

import casadi.*

x1    = SX.sym('x1');
x2    = SX.sym('x2');

x = vertcat(x1, x2);
x0 = rand(size(x));

f = [x1 + x2; 0 ];

% generate and eval f
f_fun = Function('f_fun', {x},{f});

f_new = SX(f)
f_new(1) = f_new(1) - 1 * x1
f_new(1) = f_new(1) - x2

f1 = f_new(1);
f1 = f1.simplify()

f1.is_zero()

This works fine and gives the final output "1"

Sadly for MX expressions this does not work.

import casadi.*

x1    = MX.sym('x1');
x2    = MX.sym('x2');

x = vertcat(x1, x2);
x0 = rand(size(x));

f = [x1 + x2; 0 ];

% generate and eval f
f_fun = Function('f_fun', {x},{f});

f_new = MX(f)
f_new(1) = f_new(1) - 1 * x1
f_new(1) = f_new(1) - x2

f1 = f_new(1);
f1 = f1.simplify()

f1.is_zero()

here the last output is "0".

The expression I want to analyze however contains a Spline call, such that using SX variables is not possible - see https://groups.google.com/forum/#!topic/casadi-users/-b8BL6H-tqw

Am I missing something when using MX here?

If not, I am quite interested in using the splines/lookup tables within SX CasADi expressions, which I read that Joris Gillis is working on, https://groups.google.com/d/msg/casadi-users/-b8BL6H-tqw/XDJZshLEAgAJ

Please let me know if this is ready to be tested!

Best regards,

Jonathan

[Joris Gillis]

Dear Jonathan,

Perhaps you are looking for 'which_depends'?

This function can classify expressions/variables..

Best,

  Joris

[Jonathan Frey]

Dear Joris,

unfortunately this does not do the trick.

The plan is to successively remove the linear parts and just keep the nonlinear part. This does not work with MX:

import casadi.*

% set up functions and variables
x1    = MX.sym('x1');

x2    = MX.sym('x2');

x = vertcat(x1, x2);

f = [x1 + x2;
        0];

% remove linear terms of f
f_new = MX(f);
f_new(1) = f_new(1) - x1;

f_new(1) = f_new(1) - x2;

f1 = f_new(1)
f1 = f1.simplify()

f1.is_zero() % should put out 1, but is 0

f1.which_depends(x) % should put out 0 0, but is 0 1

So do you think the way is to go with your "experimental build"?

Best,

Jonathan