Enforcing a rule in a symbolic expression

I have the following symbolic expression:

(3*s11)/2 + (3*s12)/2 + (3*s13)/2 + (3*s14)/2 + (3*s15)/2 + (s11*s12)/2 + (s11*s13)/2 + (s11*s14)/2 + (s12*s13)/2 + (s11*s15)/2 + (s12*s14)/2 + (s12*s15)/2 + (s13*s14)/2 + (s13*s15)/2 + (s14*s15)/2 + s11^2/4 + s12^2/4 + s13^2/4 + s14^2/4 + s15^2/4 + 9/4

It is stored as a symbolic expression variable. I would like to enforce the rule sij^2 = 1 i.e. the variables can be either -1 or +1. If I enforce the rule in the expression mentioned above, the expression will be as follows.

(3*s11)/2 + (3*s12)/2 + (3*s13)/2 + (3*s14)/2 + (3*s15)/2 + (s11*s12)/2 + (s11*s13)/2 + (s11*s14)/2 + (s12*s13)/2 + (s11*s15)/2 + (s12*s14)/2 + (s12*s15)/2 + (s13*s14)/2 + (s13*s15)/2 + (s14*s15)/2 + 1/4 + 1/4 + 1/4 + 1/4 + 1/4 + 9/4

How can I do this in Matlab?

Answer 1

[Michael Haderlein]

I suppose you want to solve the equation?

You can give additional arguments. In this case, I think, you must use strings:

solve('(3*s11)/2 + (3*s12)/2 + (3*s13)/2 + (3*s14)/2 + (3*s15)/2 + (s11*s12)/2 + (s11*s13)/2 + (s11*s14)/2 + (s12*s13)/2 + (s11*s15)/2 + (s12*s14)/2 + (s12*s15)/2 + (s13*s14)/2 + (s13*s15)/2 + (s14*s15)/2 + s11^2/4 + s12^2/4 + s13^2/4 + s14^2/4 + s15^2/4 + 9/4',...
's11^2=1','s12^2=1','s13^2=1','s14^2=1','s15^2=1' )

[Omar Shehab]

When I copy paste your code it give me following output, not the one I want.

Warning: 6 equations in 5 variables. 
> In C:\Program Files\MATLAB\R2014a\toolbox\symbolic\symbolic\symengine.p>symengine at 56
  In mupadengine.mupadengine>mupadengine.evalin at 97
  In mupadengine.mupadengine>mupadengine.feval at 150
  In solve at 170 

ans = 

      s11: [5x1 sym]
      s12: [5x1 sym]
      s13: [5x1 sym]
      s14: [5x1 sym]
      s15: [5x1 sym]

[Michael Haderlein]

That's strange, I don't get a warning. Directly copied from the command window:

>> solve('(3*s11)/2 + (3*s12)/2 + (3*s13)/2 + (3*s14)/2 + (3*s15)/2 + (s11*s12)/2 + (s11*s13)/2 + (s11*s14)/2 + (s12*s13)/2 + (s11*s15)/2 + (s12*s14)/2 + (s12*s15)/2 + (s13*s14)/2 + (s13*s15)/2 + (s14*s15)/2 + s11^2/4 + s12^2/4 + s13^2/4 + s14^2/4 + s15^2/4 + 9/4','s11^2=1','s12^2=1','s13^2=1','s14^2=1','s15^2=1')

ans = 

      s11: [5x1 sym]
      s12: [5x1 sym]
      s13: [5x1 sym]
      s14: [5x1 sym]
      s15: [5x1 sym]

Anyway, I read that solving the equation was not your target. Sorry for misunderstanding your question.

Answer 2

[Star Strider]

I’m not certain what you’re doing, so I can’t test your code with this, but you could use assume:

assume( s11 == -1 | s11 == 1 )
assume( s12 == -1 | s12 == 1 )
assume( s13 == -1 | s13 == 1 )
assume( s14 == -1 | s14 == 1 )
assume( s15 == -1 | s15 == 1 )

Expr = (3*s11)/2 + (3*s12)/2 + (3*s13)/2 + (3*s14)/2 + (3*s15)/2 + (s11*s12)/2 + (s11*s13)/2 + (s11*s14)/2 + (s12*s13)/2 + (s11*s15)/2 + (s12*s14)/2 + (s12*s15)/2 + (s13*s14)/2 + (s13*s15)/2 + (s14*s15)/2 + 1/4 + 1/4 + 1/4 + 1/4 + 1/4 + 9/4;

I can only claim that the Symbolic Math Toolbox accepts it. I have no idea how you would use it or evaluate it. (I created your statement as ‘Expr’ for my convenience.)

[Omar Shehab]

Shouldn't it affect all sij's? I want only the squared terms to be replaces by 1. When I tried to run your code I got following error.

Undefined function or variable 's11'.

[Star Strider]

I thought you already had s11 ... s15 in your workspace.

Put this loop above the assume statements to create them:

for k1 = 11:15
    eval(sprintf('s%d = sym(''s%d'')', k1, k1))
end

Or you could simply write them out in your syms statement, since there only five.

If sij is equal to either -1 or +1, sij^2=1.

[Omar Shehab]

Okay, I have created them like this.

syms s11 s12 s13 s14 s15

Then, I wrote the assumptions:

>>  assume(s11 == 1 | s11 == -1) 
>> assume(s12 == 1 | s12 == -1)
>> assume(s13 == 1 | s13 == -1)
>> assume(s14 == 1 | s14 == -1)
>> assume(s15 == 1 | s15 == -1)

Finally I expanded the expression:

expand(sym('(3*s11)/2 + (3*s12)/2 + (3*s13)/2 + (3*s14)/2 + (3*s15)/2 + (s11*s12)/2 + (s11*s13)/2 + (s11*s14)/2 + (s12*s13)/2 + (s11*s15)/2 + (s12*s14)/2 + (s12*s15)/2 + (s13*s14)/2 + (s13*s15)/2 + (s14*s15)/2 + s11^2/4 + s12^2/4 + s13^2/4 + s14^2/4 + s15^2/4 + 9/4'))

Still I am getting the same expression:

s11^2/4 + (s11*s12)/2 + (s11*s13)/2 + (s11*s14)/2 + (s11*s15)/2 + (3*s11)/2 + s12^2/4 + (s12*s13)/2 + (s12*s14)/2 + (s12*s15)/2 + (3*s12)/2 + s13^2/4 + (s13*s14)/2 + (s13*s15)/2 + (3*s13)/2 + s14^2/4 + (s14*s15)/2 + (3*s14)/2 + s15^2/4 + (3*s15)/2 + 9/4

:(

[Star Strider]

What do you want to do?

If you want to evaluate your expression for various values of s11 ... s15 that are each -1 or +1, it is best to do that numerically.

That is easiest if you use matlabFunction to create a function file or anonymous function from it. If you decide on an anonymous function, copy the anonymous function from the Command Window to your file, and evaluate it outside of the Symbolic Math Toolbox.

[Omar Shehab]

I don't want to evaluate my expression. I just want to enforce the rule sij^2 = 1. This will remove all the second order terms i.e. squared terms giving me a simpler expression. So, in the original expression, s11^2/4 will become 1/4 and so on.

[Star Strider]

The Symbolic Math Toolbox will simplify the sum of the fractions to (7/2), as it did when I used the subs function to evaluate only the squared terms as 1:

Expr = (3*s11)/2 + (3*s12)/2 + (3*s13)/2 + (3*s14)/2 + (3*s15)/2 + (s11*s12)/2 + (s11*s13)/2 + (s11*s14)/2 + (s12*s13)/2 + (s11*s15)/2 + (s12*s14)/2 + (s12*s15)/2 + (s13*s14)/2 + (s13*s15)/2 + (s14*s15)/2 + s11^2/4 + s12^2/4 + s13^2/4 + s14^2/4 + s15^2/4 + 9/4;

Expr1 = subs(Expr, {s11^2 s12^2  s13^2  s14^2  s15^2}, {1 1 1 1 1})

produces:

Expr1 =

(3*s11)/2 + (3*s12)/2 + (3*s13)/2 + (3*s14)/2 + (3*s15)/2 + (s11*s12)/2 + (s11*s13)/2 + (s11*s14)/2 + (s12*s13)/2 + (s11*s15)/2 + (s12*s14)/2 + (s12*s15)/2 + (s13*s14)/2 + (s13*s15)/2 + (s14*s15)/2 + 7/2

Does that do what you want?

[Omar Shehab]

Thanks. It worked.

[Omar Shehab]

I wanted to accept this answer but mistakenly clicked on the other one. Don't know how to fix it.

[Star Strider]

My pleasure!

I will get half-credit if you Vote for it. And since you wanted to Accept it, I’ll keep it up rather than deleting it (as I usually do with my Answers that aren’t accepted).

Unfortunately, it is impossible to ‘un-accept’ an Answer once accepted. That happens more often that you would imagine, and probably means that MathWorks needs to re-design MATLAB Answers with the help of a good human factors engineer.