Enforcing a rule in a symbolic expression
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Omar Shehab
on 18 Aug 2014
Commented: Michael Haderlein
on 19 Aug 2014
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?
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Accepted Answer
Michael Haderlein
on 18 Aug 2014
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' )
2 Comments
Michael Haderlein
on 19 Aug 2014
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.
More Answers (1)
Star Strider
on 18 Aug 2014
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.)
9 Comments
Star Strider
on 18 Aug 2014
Edited: Star Strider
on 18 Aug 2014
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.
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