# Diskussion:Solution 8 Computational Intelligence Lab 2013

I think your proposed solution is wrong. I don't understand your first step. Also, at the end your conclusion is correct, but it cannot be made out of your last equation, the last line says, that the error is small when the large values of z are ignored,... totally the opposite of what we should get! I try to sketch my solution:

${\displaystyle \sigma ^{min}=\arg \min _{\sigma }||\mathbf {f} -\mathbf {f} _{\sigma }||_{2}^{2}=\arg \min _{\sigma }<(\mathbf {f} -\mathbf {f} _{\sigma }),(\mathbf {f} -\mathbf {f} _{\sigma })>}$
${\displaystyle =\arg \min _{\sigma }<\mathbf {f} ,\mathbf {f} >-2<\mathbf {f} ,\mathbf {f} _{\sigma }>+<\mathbf {f} _{\sigma },\mathbf {f} _{\sigma }>}$
the first term is constant and we can neglect it in our minimization therefore we only have
${\displaystyle \arg \min _{\sigma }<\mathbf {f} _{\sigma },\mathbf {f} _{\sigma }>-2<\mathbf {f} ,\mathbf {f} _{\sigma }>}$
${\displaystyle =\arg \min _{\sigma }\sum _{k=1}^{K'}(z_{\sigma (k)}^{2})-2<\sum _{k=1}^{K}z_{k}u_{k},\sum _{k=1}^{K'}z_{\sigma (k)}u_{\sigma (k)}>}$
${\displaystyle =\arg \min _{\sigma }\sum _{k=1}^{K'}z_{\sigma (k)}^{2}-2\sum _{i=1}^{K}\sum _{j=1}^{K'}(z_{i}z_{\sigma (j)})}$
since ${\displaystyle }$ is only 1 if ${\displaystyle i=\sigma (j)}$ (and 0 otherwise) and this can only be the case at most for one pair of i and j, we only need one sum and since K' < K we have to sum over K'.
${\displaystyle =\arg \min _{\sigma }\sum _{k=1}^{K'}z_{\sigma (k)}^{2}-2\sum _{k=1}^{K'}z_{\sigma (k)}^{2}}$
${\displaystyle =\arg \min _{\sigma }-\sum _{k=1}^{K'}z_{\sigma (k)}^{2}}$
${\displaystyle =\arg \max _{\sigma }\sum _{k=1}^{K'}z_{\sigma (k)}^{2}}$
obviously this is maximized, if we keep the largest ${\displaystyle z_{\sigma (k)}}$ values.
if anyone can agree with me, I'll take this to the solution page. --M@2 10:26, 18. Aug. 2012 (CEST)
i agree. --Ursedo 22:33, 19. Aug. 2012 (CEST)
I also agree. I updated the solution to 2013 as well. --Moeeeep (Diskussion) 14:38, 14. Aug. 2013 (CEST)