ISIPhyNC - Network #3

JSON code of the network

{"nodes":["r","a","b","c","d","e","f","g","h","i","1","2","3","4","5"],"edges": [["r","a"],["a","b"],["b","1"],["b","c"],["c","d"],["d","2"],["a","e"],["e","c"],["e","3"],["r","f"],["f","g"],["g","4"],["g","h"],["h","d"],["f","i"],["i","h"],["i","5"]]}

Visualization of the network

r is the root and the leaves are labeled from 1 to 5. Arcs are oriented from the parent to the child.

Classes containing this network or not

Classes which contain this network (with direct proof)

• binary time-consistent: Here is a time-consistent labeling t:
  • t(r)=0
  • t(a)=t(f)=1
  • t(b)=t(c)=t(d)=t(e)=t(g)=t(h)=t(i)=2
  • t(1)=t(2)=t(3)=t(4)=t(5)=3
• binary nearly stable: Reticulation vertices c and d are the only ones which are not stable but in both cases, both their parents are.
• binary 2-nested: Easy to check.
• binary spread-2: Consider the order 1 2 3 4 5 of the leaves to obtain spread 2.
• binary level-3: Easy to check.

Classes which do not contain this network (with direct proof)

• binary tree-based: Suppose N is tree-based, and consider the tree T it is based on. If reticulation arc (c,d) is chosen in T, there would remain no leaf below h in T. A symmetric argument works for arc (h,d) and c, which leads to a contradiction.
• binary spread-1: Suppose there exists an order of the leaves which corresponds to spread 1. Vertices b, e, g, i are all above leaf 2 as well as above another leaf which is not below the other ones. So whatever two leaves, say 1 and 3, we choose to be around leaf 2 in the order, 4 and 5 will not be, so the leaves below g and h will not be an interval in this order.
• binary level-2: c, d and h are three reticulation vertices in the same blob.

All classes

In the inclusion diagram below, the names of classes containing this network are colored green and the name of the classes not containing it are colored red.