Help me understand!

In the 3rd for loop , xloc(k) and yloc(k) are both vectors of length 6. And xp and yp are vectors of length 30 . If done alone the subtraction returns an error but here it doesn't. I wanna know what's so special here that it doesn't return an error?

if true
  % xloc = [1,7,8,17,22,27] ;
yloc = [28,18,16,2,10,8] ;
V = [3,7,4,5,2,6] ;
best_loc_x = 31 ;
best_loc_y = 31 ;
min_cost = 1e+6 ;
anotherx = [] ;
anothery = [] ;
C = [] ;
for xp = 1:30
  for yp = 1:30
      for k = 1:6
          d(k) = sqrt((xloc(k)-xp).^2+(yloc(k)-yp).^2) ;
          cost(k) = d(k)*V(k);
      end
      loc_cost(xp,yp) = sum(cost) ; 
      if loc_cost(xp,yp) < min_cost
          best_loc_x = xp;
          best_loc_y =  yp;
          min_cost = loc_cost(xp,yp);
      elseif (loc_cost(xp,yp)-min_cost) <=1
          anotherx = [anotherx,xp] ;
          anothery = [anothery,yp];
          C = [C,loc_cost(xp,yp)];
      end
  end
end
end

[dpb]

You didn't attach the relevant code snippets...

[Uzair Khan]

Check it now

[dpb]

Post the text we can read (formatted as code), not images...can't see all of that in one place.

[Uzair Khan]

Please don't mind my inefficiency, i am new here.

[dpb]

'Tis OK, but it makes more likely you'll get responses when make it easier for the folks to tryto help... :)

Do you see what's going on now w/ the Answer to help?

[Uzair Khan]

No haven't really spent much time. So what about this question?

Answer 1

[Walter Roberson]

for xp = 1:30          %so xp is a scalar
  for yp = 1:30        %so yp is a scalar
      for k = 1:6      %so k is a scalar
          d(k) = sqrt((xloc(k)-xp).^2+(yloc(k)-yp).^2) ;

vector indexed at a scalar gives a scalar, so xloc(k) is a scalar and yloc(k) is a scalar. You subtract the scalars xp and yp from them, giving scalars for the sub-expressions. You square, add, square-root, giving a scalar output, which you store in a scalar location. No problem.

          for xp = 1:30

does not make xp a vector of 30 elements: it tells MATLAB to iterate making xp hold each of the values 1:30 in turn, so at any one time xp is a scalar.

[Uzair Khan]

Thanks. U cleared the for loop too for me

[Uzair Khan]

Just one more question. Doesn't d(k) mean the first k number of calculations of d? If it does then we wouldn't find the distance for the rest of the 24 iterations, or would we? In other words please explain how d(k) works.

[Walter Roberson]

If you start from the beginning of a vector and count 1 for the first element, 2 for the second element, and so on, then d(k) refers to the content of the element for which the count is the same as the value k. For example d(7) is the seventh element. It is not the first seven elements together, just the one.

[Uzair Khan]

Ok thanks

Answer 2

[dpb]

The key is line 17...

[Uzair Khan]

Could you please elaborate