Wiki User
∙ 14y agoSpanning Tree Election Criteria
Spanning Tree builds paths out from a central point along the fastest available links. It selects path according to the following criteria:
1. Lowest root bridge ID (BID)
2. Lowest path cost to the root
3. Lowest sender bridge ID
4. Lowest sender port ID (PID)
Therefore, the answer to your question is Lower Bridge ID
Wiki User
∙ 14y agoit becomes a Star polygon
First of all, if the rotation is 180 degrees then there is no difference clockwise and anti-clockwise so the inclusion of clockwise in the question is redundant. In terms of the coordinate plane, the signs of all coordinates are switched: from + to - and from - to +. So (2, 3) becomes (-2, -3), (-2, 3) becomes (2, -3), (2, -3) becomes (-2, 3) and (-2, -3) becomes (2, 3).
Generally none. But it can have four when it becomes a square.
Consumption becomes greater than supply
It is spilt and becomes part of the product molecules
lowest bridge ID
mass
Where it is produced
The mass of the host star.
The type of government determines how a government becomes centrally planned.
its original mass when it formed
If the president passes away, resigns, or is impeached.
A star becomes a star - "is born" - when the process of nuclear fusion begins in the core of the star.
The helper T cell count becomes very low
The helper T cell count becomes very low
Proving this is simple. First, you prove that G has a spanning tree, it is connected, which is pretty obvious - a spanning tree itself is already a connected graph on the vertex set V(G), thus G which contains it as a spanning sub graph is obviously also connected. Second, you prove that if G is connected, it has a spanning tree. If G is a tree itself, then it must "contain" a spanning tree. If G is connected and not a tree, then it must have at least one cycle. I don't know if you know this or not, but there is a theorem stating that an edge is a cut-edge if and only if it is on no cycle (a cut-edge is an edge such that if you take it out, the graph becomes disconnected). Thus, you can just keep taking out edges from cycles in G until all that is left are cut-gees. Since you did not take out any cut-edges, the graph is still connected; since all that is left are cut-edges, there are no cycles. A connected graph with no cycles is a tree. Thus, G contains a spanning tree. Therefore, a graph G is connected if and only if it has a spanning tree!
Proving this is simple. First, you prove that G has a spanning tree, it is connected, which is pretty obvious - a spanning tree itself is already a connected graph on the vertex set V(G), thus G which contains it as a spanning sub graph is obviously also connected. Second, you prove that if G is connected, it has a spanning tree. If G is a tree itself, then it must "contain" a spanning tree. If G is connected and not a tree, then it must have at least one cycle. I don't know if you know this or not, but there is a theorem stating that an edge is a cut-edge if and only if it is on no cycle (a cut-edge is an edge such that if you take it out, the graph becomes disconnected). Thus, you can just keep taking out edges from cycles in G until all that is left are cut-gees. Since you did not take out any cut-edges, the graph is still connected; since all that is left are cut-edges, there are no cycles. A connected graph with no cycles is a tree. Thus, G contains a spanning tree. Therefore, a graph G is connected if and only if it has a spanning tree!