A graph is strongly connected if:
For all nodes u
for all nodes v not equal to u,
there is a path from u to v.
for the graph to not be strongly connected there must be one group of nodes that cannot reach another group of nodes. The most edges that you can have, is a complete graph (all possible edges between) of 5 nodes, and then all having edges to or from (but not both) the remaining node. This leads to 31 total edges.
consider six nodes with edges connecting them in order (as if around a hexagon); this is the smallest strongly connected graph. If there are fewer edges, then some node has no edge leaving it -- so the graph would not be strongly connected.
A strongly connected graph means every edge can eventaully get to every other edge. If the relation the graph represents is transitive then if you can ever get between nodes, you can get there in one step. Therefore, you can get between every pair of nodes in one step, meaning you have all possible edges. This is 36 edges.
St. Louis 28, Seattle 30 San Francisco 27, Arizona 34 Seattle 21, Arizona 10 St. Louis 24, Seattle 22 San Francisco 20, Seattle 14 San Francisco 20, St. Louis 13 St. Louis 16, Arizona 14 San Francisco 17, St. Louis 20
This should be directed; an edge exists from one team to another if this first team beats the other team.
wins and losses
yes, st. louis beat san fran, san fran beat seattle, st. lois beat seattle. (others exist).
it is not. it is not reflexive and there are also elements that are not symmetric (st. louis, seattle) and (seattle st. louis) are both in the graph.
Let A = {1,2,3,4,5}, and define the relation R as: {(a,b) | a <= b}
Let A = {1,2,3,4,5}, and define the relation R as: {(a,b) | a <= b} - { (4,5) }. so that the order between 4 and 5 is no longer defined.
You can do this, it turns out. define A = {1,2,3,4,s}
define R as "the usual" order on 1,2,3,4, then include also
(1,s) and (s,4).
(so that s has to come between 1 and 4). Then there are 3 consistent total orders:
1,s,2,3,4
1,2,s,3,4
1,2,3,s,4