Assignment On Matrix Of The Equation

Question 1
The graph can be divided as shown above into two graphs which share one edge.
6 vertices are available.
It has 7 edges . 2 are supposed to be deleted
=5 edges.
Both the graphs have 4 edges. Expected spanning tree 16
The number of deleted edges can be equal to 2.
Since they share the same one edge 16-1= 15.
The number of expected spanning trees will be equals to 15 spanning trees
As shown below. Two edges need to be deleted from above
the edges can be deleted as below in both the graphs

(a,e) (b,e) (d,e) (c,e) There are 15 spanning trees in the
above graph
(a,f) (b,f) (d,f) (c,f)
(a,h) (b,h) (d,h) (d,h)
(a,c) (b,c) (d,c)
Since the possible number of spanning trees present in the graph is the determinant of the matrix
of the values present in the matrix. the equation can be as below.
The matrix of the equation
0 1 2 3 4 5
1 -1 -1 -1 -1
2 0 -1 -1 -1 -1
3 -1 -1 -1 -1
4 -1 -1 -1 -1
5 -1 -1 -1 -1 -1
The spanning-trees=
{01,03,05,12,34,54(p-1)p,p(p+1),..,(2p-2)(2p-1),(2p-1)0}U{0P}

Question 2
G = (V, E) The equation of the circle
m = {u ∈ V : d(u, v) is even},
d(v, u) − d(v, w)| ≤ d(u, w) = 1
Since the graph can is reprsesnted by the equation k
mn
The total number of edges can be equal 2 m *n;
Now that the spanning trees will be equal to the total number of edges minus 1
Span= mn-1
The number of edges needed to be deleted is equal to D.
mn(m+n-1)
=(m-1)(n-1)
=Thus D is divisible by m-1
Which is(m−1)(n−1)
m−1
=n-1

b. Assume that the m>n
for the above, the number of edges is equal to 2 mn
∑ dgV
2
With the bove the degree of vertices is equal 2 ∑V
2
Since in the path, each vertice has a degree of 2
2 x+2 y +the degree of all the vertices ¿ the graph ¿
2
Where by x nd y are integers?
In this x=1 and y=1
2+ 2
2
=2
Thus the length is even.

Question 3
The number will be
The chromatic number can be
X(G) and Y(G)
g=1
Y(g)=3 √9
The chromatic number can be given by the bove equation
hence the chromatic value is 3