Nested for loop help needed

Hi guys, can anyone help me out with this one:

x =

     A     B     C
     D     E     F
     G     H     I
     J     K     L

I need to produce a loop which for this matrix, would output 3 numbers defined as:

1st = DG + EF + FI

2nd = DJ + EK + FL

3rd = GJ + HK + IL

1st = the elements of the second row * the elements of the third

2nd = the elements of the second row * the elements of the fourth

3rd = the elements of the third row * the element of the fourth

So you see, the second row is held and multiplied with all rows below it. Then the loop moves down so the third row is held and multiplied with all rows below it.

Also it's required that the loop can accommodate a matrix with any number of columns, but always three rows. If anyone's interested, the number of permutations = ((N-1)(N-2)) / 2 where N = number of columns

Could anyone help me in figuring this out? I've been getting tangled up in loops for a while now, but I think the problem is probably quite simple for someone with experience of such things.

Kind regards,

Tom

[Cedric]

If there are always four rows, why a loop?

 x(2,:)

for example is the whole 2nd row of x. You probably know how to perform an element-wise multiplication to get a vector of products between x(2,:) and e.g. x(3,:). Then you can just compute a basic SUM of these terms..

[Tom]

Yes I see what you're saying, I'll give that a go. Thanks

[Tom]

I'm still struggling, Do I really not need to use any loops? I feel like I'm getting close with:

for j = 2:(n-1)

    for

    dotp(count1) = sum(x(j,:).*x(3,:))

    count1 = count1 + 1

    end
end

But the *x(3,:) term isn't right.

n = number of columns by the way.

Any thoughts?

[Cedric]

I answer assuming that this is for a homework; let me know if it's not the case. Look at the following

 >>  x = randi(10, 4, 6)
 x =
     9     7    10    10     5     7
    10     1    10     5    10     1
     2     3     2     9     8     9
    10     6    10     2    10    10

is a matrix of random numbers.. if you want to extract row 2:

 >> x(2,:)
 ans =
    10     1    10     5    10     1

As you can see, it's a vector.. that you could use in further computations.

Second hint: assume that you have two vectors:

 >> a = [3, 2, 4] ;
 >> b = [6, 2, 10] ;

you can easily compute the element-wise multiplication of these vectors:

 >> c = a .* b
 c =
    18     4    40

where you can check that each element of c is the product of corresponding (same column) elements in a and b.

Third hint: assume that you have a vector..

 >> y = [3, 7, 2, 5, 9, 1] ;

It is easy to get the sum of its elements using function SUM:

 >> sum(y)
 ans =
    27

[Tom]

It's not for homework per se, I am working on an algorithm which I need to use for my masters.

I have already implemented the element-wise multiplication of vectors as you can see in my above comment, thanks to your previous guidance. The problem I am having now is implementing (what I believe will ultimately be) a nested for loop to perform the sequence described in my original post. I am about to have another go at it, with hopefully more success than yesterday, but if you would know how to do it at a glance, I would be grateful if you could let me know.

Kind regards,

Tom

[Image Analyst]

For something this tiny, I wouldn't worry about loops if it gets the job done. Now if you had thousands of rows and columns, then you might start to worry. But for a few or a few dozen, it's no problem.

[Image Analyst]

Isn't this essentially the same as http://www.mathworks.com/matlabcentral/answers/104490-nested-for-loop-needed?

[Cedric]

The way I understand it is that this thread is about combinations and the one that you mention is about "all rows until the end" .. but I might have misunderstood.

Answer 1

[Cedric]

Following up on comments..

We usually try to avoid explicit loops when we can implement a matrix/vector approach. The latter is much more efficient than looping, especially looping over both rows and columns. My hints were showing how to get rid of the loop over columns:

 X = ...
 number1 = sum( X(2,:) .* X(3,:) ) ;
 number2 = sum( X(2,:) .* X(4,:) ) ;
 number3 = sum( X(3,:) .* X(4,:) ) ;

This is probably the best approach if you have only 4 rows in X, because it avoids the complication of looping over combinations of {2,3,4} taken 2 at a time. A loop for that would be more complex, probably less efficient, and involve more code.

If X had a variable number of rows, say nRow, and you needed to follow the same approach involving rows 2 to nRow, you could do it as follows:

 X       = randi(10, 6, 8) ;               % Example with 6 rows and 8 columns.
 combSet = nchoosek( 2:size(X,1), 2 ) ;    % Array of all comb. of 2 elements.
 results = zeros( size(combSet, 1), 1 ) ;  % Prealloc. for results.
 for cId = 1 : length( results )
     results(cId) = sum( X(combSet(cId,1),:) .* X(combSet(cId,2),:) ) ;
 end

With that, you get (you'll have different numbers due to RANDI)

 >> X
 X =
     9     3    10     8     7     8     7     8
    10     6     5    10     8     1     4     8
     2    10     9     7     8     3    10     2
    10    10     2     1     4     1     1     5
     7     2     5     9     7     1     5     5
     1    10    10    10     2     9     4     7

 >> combSet
 combSet =
     2     3
     2     4
     2     5
     2     6
     3     4
     3     5
     3     6
     4     5
     4     6
     5     6

 >> results
 results =
   318
   257
   314
   317
   200
   261
   359
   168
   196
   245

Check, taking e.g. combination 3

 >> sum( X(2,:) .* X(5,:) )
 ans =
   314

which is results(3), so it seems to be working.