Solution to exercise 1 Given that V is a vertex and its degree

Solution to exercise 1
Given that V is a vertex and its degree is d ≥ ∆(T ). For each edge VE incident to V, use the
lengthiest path beginning with VE. Applying maximality, the vertex at the extreme of the path is
a leaf (Artur Czumaj, 2015). Applying the same to each of the d edges incident to V, we obtain d
paths beginning at V. They are disjoint except for V. the disjoint cuts off the cycle path and
makes it incomplete. Hence, every single path provides a dissimilar leaf, and we obtain d ≥ ∆(T )
leaves.
Solution to exercise 2
Every V of the graph on m vertices has0 ≤ d ≤ m−1. In instances where all the degrees (d)
differed in the graph G. The graph G must be composed of vertex of degree i for a i=0,1 ... . .
contrarily, the vertex of d=0 is not neighbouring to all other vertices of G and the vertex of
d=n−1 is neighbouring to another vertices of G which is unattainable (Edward R. Scheinerman,
2013). Therefore, only m−1 possible values for all d are attainable. In either case, m vertices and
m−1 possible values for d.
Solution to exercise 3
a)
Take H=(V , E) to be an d-uniform hypergraph with ¿ 2d −1 edges. Using two colours, randomly
colour V. For every edge e ∈ E take Ae to represent the event that the edge e is
unicolor/monochromatic (West, 2017).
It is open that
Pr ( Ae ) =21−d
Hence,
Pr ( ¿ e ∈ E Ae ) ≤ ∑
e∈ E
Pr ( Ae ) <1
And there exists a 2-coloring lacking unicolor/monochromatic edges.