1.1 Binary file exercises/solutions/sol03.pdf has changed

     2.1 --- a/exercises/solutions/sol03.tex	Sat May 19 16:12:13 2012 +0200
     2.2 +++ b/exercises/solutions/sol03.tex	Sun May 20 23:03:17 2012 +0200
     2.3 @@ -91,7 +91,7 @@
     2.4  
     2.5  All projections of $R_S$ to a subset of $S$, so that $x_1$ is contained, basically reduces the network to the projection size, all properties remain unchanged, therefore the projection network of the projection of $R_S$ has only solutions which are in the projected relation (and vice versa).
     2.6  
     2.7 -All projections of $R_S$ to a subset of $S$ without $x_1$ mean a reduction to the constraint $2 > \sum_{i=1}^{m}x_i$ where m is the projection size. This has obviously also a binary representation with $R'_{x_i,x_j} = \{(0,0),(1,0),(0,1)\}$. Again it follows that the projection network of this projection has only solutions which are in the projected relation (and vice versa).
     2.8 +All projections of $R_S$ to a subset of $S$ without $x_1$ mean a reduction to the constraint $2 > \sum_{i=1}^{m}x_i$ where $m$ is the projection size. This has obviously also a binary representation with $R'_{x_i,x_j} = \{(0,0),(1,0),(0,1)\}$. Again it follows that the projection network of this projection has only solutions which are in the projected relation (and vice versa).
     2.9  
    2.10  Given that the relation $R_S$ itself and all its possible projections have binary representations, it holds that $R_S$ is binary decomposable. \qed\\\\
    2.11  (c) We will do part (c) before (b) because here we use the same constraint as in part (a). However, we will expand the domains with $D_i = \{0,1,2,3\}$. This gives us the relation $R_S = \{(x_1,(x_i)_{1<i\leq n}) \mid \sum_{i=2}^{n}x_i < x_1\}$.