updated mat 419 test with corrections

Q1a.
                                                                   SOLUTION

from the format of the matrix we can get our 7×7 adjacency matrix using this format: 
│ a11 a12 a13 a14 a15 a16 a17│
│ a21 a22 a23 a24 a25 a26 a27│
│ a31 a32 a33 a34 a35 a36 a37│
│ a41 a42 a43 a44 a45 a46 a47│
│ a51 a52 a53 a54 a55 a56 a57│
│ a61 a62 a63 a64 a65 a66 a67│
│ a71 a72 a73 a74 a75 a76 a77│

Therefore:

            │ 2  0  1  0  1  0  1│
            │ 0  2  0  0  1  0  0│
            │ 1  0  2  0  1  0  1│
A(G)=       │ 0  0  0  2  0  0  0│
            │ 1  1  1  0  2  0  1│
            │ 0  0  0  0  0  2  0│
            │ 1  0  1  0   1 0  2│

Q1b. Then: C(v) = 3
then as you can see that we have two isolated vertices from our graph form by the above adjacency matrix which are V6 and V4 therefore V6 is a cut-vertex and V4 is also a cut-vertex while V1,V2,V3,V5 and V7 are vertex-cut.

Q2a.
ⅰ.

ⅱ. the dept of unary operation is 2.
because unary means SQRT(i.e 厂)  and if you can see of course it is in dept 2.

ⅲ. And the height is 6


Q3a.   K1+N4

There fore we can get the chromatic polynomial using the formula that will lead us to  Null graph
which is better and simple.

The formula is:
Π_k(G)= Π_k(Ge)- Π_k(G\e)
⇒

Q3b.if  K3 and C4 conjoined the result will be: 

ⅰ. And to find Z(L(G)) the inner bracket i.e L(G) which means line graph,need to be considered first then Z(L(G)) which means to find the center(i.e minimum eccentricity) of that L(G).

L(G)= 

And then Z(L(G))= {e_1,e_2,e₃,e₄,e₅,e_{6} 
because all the edges have the same radius which is 2,}and you know that Z means center of a graph and to get center of a graph you look for the set of vertices with minimum eccentricity.

ⅱ.And then G^{1 
which means complement of a the above graph as required.}

G¹⁼

and then to ascertain the validity of rad(G¹) ≤ diam(G¹) ≤ 2rad(G¹)
if we can see the
 rad(G¹)=2
diam(G¹)=4
2rad(G¹)= 2× 2 = 4

Hence 2 ≤ 4 ≤ 4

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