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Asked by kevin
on 2 Apr 2012

i use the command

c=[1,2,3,4,5] perms(c)

to generate all permutation of c. and there are 2 equation which use the

j=sin(0*c)+sin(72*c)+sin(144*c)+sin(216*c)+sin(288*c)

k=cos(0*c)+cos(72*c)+cos(144*c)+cos(216*c)+cos(288*c)

and Z=j+k

how can i find what permutation for minimum z? the permutation must be the same for both j and k at the same time ie

j=sin(0*1)+sin(72*2)+sin(144*3)+sin(216*4)+sin(288*5)

k=cos(0*1)+cos(72*2)+cos(144*3)+cos(216*4)+cos(288*5)

i actually wrote the euqation wrong, sorry. but if

j=sin(0)*c+sin(72)*c+sin(144)*c+sin(216)*c+sin(288)*c

k=cos(0)*c+cos(72)*c+cos(144)*c+cos(216)*c+cos(288)*c

would that change anything?

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Answer by Thomas
on 2 Apr 2012

Accepted answer

Another way:

c=[1,2,3,4,5]; c=perms(c); z=[]; j=[]; k=[];

for i=1:length(c) j(i)=sin(0*c(i,1))*sin(72*c(i,2))+sin(144*c(i,3))+sin(216*c(i,4))+sin(288*c(i,5)); k(i)=cos(0*c(i,1))*cos(72*c(i,2))+cos(144*c(i,3))+cos(216*c(i,4))+cos(288*c(i,5)); z(i)=j(i)+k(i); end

[p,q,r]=find(z==min(z));

c(q,:) minz=min(z)

Show 3 older comments

Matt Tearle
on 2 Apr 2012

Actually, z = []; doesn't preallocate space. It only makes an empty matrix. A better approach would be

n = length(c);

j = zeros(n,1);

k = zeros(n,1);

for i = 1:n

...

Also, calculating z inside the loop isn't necessary. Just do z = j+k; at the end. Natural vectorized expressions like that are one of the main strengths of MATLAB. (Of course, I'd say that you don't need loops at all...)

Answer by Matt Tearle
on 2 Apr 2012

I think this is what you're after:

c=[1,2,3,4,5]; cp = perms(c); % 0*c1, 72*c2, 144*c3, 216*c4, 288*c5 allc = bsxfun(@times,[0 72 144 216 288],cp); sc = sin(allc); cc = cos(allc);

j = sc(:,1).*sc(:,2)+sum(sc(:,3:5),2); k = cc(:,1) + cc(:,2).*cc(:,3) + cc(:,4) + cc(:,5); z = j+k;

% which row of the c permutations corresponds to the minimum value of z? cp(z==min(z),:)

If `j` and `k` were all sums, this would be a bit neater, but I'm assuming the products there are deliberate.

**EDIT TO ADD**: From your comment in reply to Thomas, it seems like maybe these should all be sums (ie no products). In that case:

c=[1,2,3,4,5]; cp = perms(c);

% 0*c1, 72*c2, 144*c3, 216*c4, 288*c5 allc = bsxfun(@times,[0 72 144 216 288],cp); % sin(0*c1) + sin(72*c2) + ... j = sum(sin(allc),2); k = sum(cos(allc),2); z = j+k;

% which row of the c permutations corresponds to the minimum value of z? cp(z==min(z),:)

## 1 Comment

## Walter Roberson

Direct link to this comment:http://www.mathworks.com/matlabcentral/answers/34268#comment_71474

Is your earlier question http://www.mathworks.com/matlabcentral/answers/34267-ordering-a-list-of-number considered answered? If so please Accept the answer; otherwise there is the appearance that this is an extension of the previous question that should be merged with it.