Can you reduce a precedence graph or do *all* relevant nodes need to be connected
Yes it is necessary.

According to the definition of precedence graph, a directed edge $T_i \longrightarrow T_j$  is created if one of the operations in $T_i$ appears in the schedule before some conflicting operation  in $T_j$.

It is clear from the definition that we have to consider every two transactions separately : $T_1$and $T_2$, $T_1$and $T_3$ and $T_2$ and $T_3$.

 T1  |  T2  |  T3
-----+------+-----
 w(x)|      |
     | w(x) |
     |      | w(x)


So an edge $T_1 \longrightarrow T_3$ is necessary because w(x) in $T_1$ and w(x) in $T_3$ are two conflicting¹ operations.

 T1  |  T2  |  T3
-----+------+-----
 w(x)|      |
 w(y)|      |
     | w(x) |
     | w(z) | 
     |      | w(y)
     |      | w(z)


For this schedule there will be three edges in the precedence graph:$T_1 \longrightarrow T_2$, $T_1 \longrightarrow T_3$ and $T_2 \longrightarrow T_3$.



1.Two operations are conflicting, if they are of different transactions, upon the same datum (data item), and at least one of them is write.