Does the graph shown in problem $1$ have a $4-$clique? One way to answer this question is to

construct and solve an instance of $2$SAT$-o+.$ First, notice that vertices $7$ and $8$ could not be in

a $4-$clique because they each have degree Now, for $i=1,2,$dots,$6,$ let $x_i$ be the Boolean

variable which, if set to $1,$ means that vertex $i$ is in the $4-$clique. Now, notice that, if $(i,j)!$inE,

then i and $j$ cannot both be in a $4-$clique $($or any $k-$clique where $k2).$ Moreover, if a $4-$clique

does exist, then either $i$ or $j($but not both$)$ must be in the clique, since the remaining four

vertices can never form a $4-$clique $($of course this is not true for every graph, but is true for the

one shown in problem $1.$ verify!$).$ Therefore, we may form an instance of $2$SAT$-o+$ whose clause

set is

$2$SAT$-o+G(20pts)$

Does the graph shown in problem $1$ have a $4-$clique? One way to answer this question is to

construct and solve an instance of $2$ SAT$-o+.$ First, notice that vertices $7$ and $8$ could not be in

a $4-$clique because they each have degree Now, for $i=1,2,$dots,$6,$ let $x_i$ be the Boolean

variable which, if set to $1,$ means that vertex $i$ is in the $4-$clique. Now, notice that, if $(i,j)!$inE,

then i and $j$ cannot both be in a $4-$clique $($or any $k-$clique where $k2).$ Moreover, if a $4-$clique

does exist, then either $i$ or $j($but not both$)$ must be in the clique, since the remaining four

vertices can never form a $4-$clique $($of course this is not true for every graph, but is true for the

one shown in problem $1.$ verify!$).$ Therefore, we may form an instance of $2$SAT$-o+$ whose clause

set is

$2$SAT$-o+G(20pts)$