# A MATLAB function with mixed scalar and vector as inputs fail

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Mohammad Shojaei Arani
on 7 Sep 2022

Commented: Walter Roberson
on 8 Sep 2022

Hello,

I have a question which I explain by a simple example first. Consider the matlab function

H=@(x,x0,par1,par2)[x+par1 x.^2+par2;1-x.^2-par2 x.^3+par1];

where here x and x0 are vectors but par1 and par2 are scalars. So, H should give us matrices as output (if x and x0 are also

scalars then such matrices are square, otherwise we get non-square matrices). Any of the combinations H([1; 2],[3; 4],1,2), H([1; 2],[3 4],1,2)

H([1 2],[3; 4],1,2) and H([1 2],[3 4],1,2) works and MATLAB does not through any error message.

Now, consider the attached matlab function H_LL(x,x0,par1,par2,par3,par4,par5) which is just longer and has 5 parameters, instead of 2. Unfortunately, this only works if x and x0 are only scalars (and I do not know why!!!)

It would be great if you have an idea what's wrong here?

Thanks

Babak

##### 2 Comments

dpb
on 7 Sep 2022

function h_LL = H_LL(x,x0,par1,par2,par3,par4,par5)

%H_LL

% H_LL = H_LL(X,X0,PAR1,PAR2,PAR3,PAR4,PAR5)

% This function was generated by the Symbolic Math Toolbox version 8.6.

% 07-Sep-2022 16:46:14

t2 = par4.^2;

t3 = x.^2;

t4 = x.^3;

t6 = 1.0./par2;

t8 = 1.0./par5.^2;

t9 = 1.0./par5.^3;

t10 = -x0;

t5 = t3.^2;

t7 = t6.^2;

t11 = t6.*x;

t12 = t2+t3;

t13 = t11-1.0;

t14 = 1.0./t12;

t15 = t14.^2;

t16 = t14.^3;

t17 = (par1.*t13.*x)./1.0e+2;

t18 = (par3.*t3.*t14)./1.0e+2;

t20 = (par1.*t4.*t7.*t8.*t13)./1.0e+2;

t21 = (par1.*t5.*t7.*t8.*t14)./1.0e+2;

t24 = (t4.*t8.*t13.*t14)./1.0e+2;

t19 = (par3.*par4.*t5.*t8.*t16)./5.0e+1;

t22 = (par1.*par3.*par4.*t5.*t7.*t8.*t15)./5.0e+1;

t25 = (par3.*par4.*t4.*t8.*t13.*t15)./5.0e+1;

t26 = -t24;

t27 = t10+t17+t18+x;

t23 = -t22;

t28 = t3.*t7.*t8.*t27;

t29 = par1.*t3.*t7.*t9.*t27.*2.0;

t31 = t9.*t13.*t27.*x.*2.0;

t32 = t3.*t9.*t14.*t27.*2.0;

t33 = par4.*t3.*t8.*t15.*t27.*2.0;

t34 = par3.*par4.*t3.*t9.*t15.*t27.*4.0;

t30 = -t29;

t35 = -t34;

t36 = t20+t28;

t37 = t19+t33;

h_LL = reshape([t3.*t8.*t13.^2.*(-1.0./1.0e+2),t36,t26,t25,t31,t36,par1.^2.*t5.*t7.^2.*t8.*(-1.0./1.0e+2)-par1.*t3.*t6.^3.*t8.*t27.*2.0,t21,t23,t30,t26,t21,t5.*t8.*t15.*(-1.0./1.0e+2),t37,t32,t25,t23,t37,par3.^2.*t2.*t5.*t8.*t15.^2.*(-1.0./2.5e+1)+par3.*t3.*t8.*t15.*t27.*2.0-par3.*t2.*t3.*t8.*t16.*t27.*8.0,t35,t31,t30,t32,t35,t8-t8.^2.*t27.^2.*3.0e+2],[5,5]);

What a nightmare!

But, all kinds of things have to be so for that to work -- specifically with respect to x, at a bare minimum without looking hard we find

...

t3 = x.^2;

...

t17 = (par1.*t13.*x)./1.0e+2;

t18 = (par3.*t3.*t14)./1.0e+2;

...

which will require x be same dimensions as par1 or a scalar that can multiply anything on an element-by-element basis.

Trying to make any sense of of this without what generated it is hopeless -- and trying to generalize it to anything other than that specific case is, as well....

### Accepted Answer

Walter Roberson
on 7 Sep 2022

h_LL = reshape([t3.*t8.*t13.^2.*(-1.0./1.0e+2),t36,t26,t25,t31,t36,par1.^2.*t5.*t7.^2.*t8.*(-1.0./1.0e+2)-par1.*t3.*t6.^3.*t8.*t27.*2.0,t21,t23,t30,t26,t21,t5.*t8.*t15.*(-1.0./1.0e+2),t37,t32,t25,t23,t37,par3.^2.*t2.*t5.*t8.*t15.^2.*(-1.0./2.5e+1)+par3.*t3.*t8.*t15.*t27.*2.0-par3.*t2.*t3.*t8.*t16.*t27.*8.0,t35,t31,t30,t32,t35,t8-t8.^2.*t27.^2.*3.0e+2],[5,5]);

You have 25 components in the [], each the same size as your x input. You ask to reshape() that to a 5 x 5 matrix. If x is scalar, that works out fine, a 1 x 25 reshaped to 5 x 5 fits perfectly. But if x were (for example) length 2, you would have a 1 x (25*2) array being reshaped to 5 x 5, and you cannot reshape 50 elements into 25 spaces.

You could edit the code to reshape the input x into the third dimension,

x = reshape(x, 1, 1, []);

and then in the assignment to h_LL you could change the reshape from [5,5] to 5, 5, [] . This would give you a 5 x 5 x numel(x) result... at the expense of having to manually edit the code.

You cannot convince the Symbolic Toolbox matlabFunction() to itself properly vectorize a function that returns an array (or a function that uses piecewise() or some other functions such as int()) .

##### 2 Comments

Walter Roberson
on 8 Sep 2022

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