Buatlah sebuah digraph dengan order
9 dan out degree 3 dengan diameter 2 !
Penyelesaian :
v
≡ du
+ i ( mod
n )
dengan :
d = 3
u = vertex
i = d – 1 = 0, 1, dan 2
k = 2
n = dk = 32 =
9
vertices : 0, 1, 2, 3, 4, 5, 6, 7,
8
. vertex
0 ≡ 3 x 0 + 0 ( mod
9 ) = 0
vertex 0 ≡ 3 x 0 + 1 ( mod 9 ) = 1
vertex 0 ≡ 3 x 0 + 2 ( mod 9 ) = 2
. vertex
1 ≡ 3 x 1 + 0 ( mod
9 ) = 3
vertex 1 ≡ 3 x 1 + 1 ( mod 9 ) = 4
vertex 1 ≡ 3 x 1 + 2 ( mod 9 ) = 5
. vertex
2 ≡ 3 x 2 + 0 ( mod
9 ) = 6
vertex 2 ≡ 3 x 2 + 1 ( mod 9 ) = 7
vertex 2 ≡ 3 x 2 + 2 ( mod 9 ) = 8
. vertex
3 ≡ 3 x 3 + 0 ( mod
9 ) = 0
vertex 3 ≡
3 x 3 + 1 ( mod 9 ) = 1
vertex 3 ≡ 3 x 3 + 2 ( mod 9 ) = 2
. vertex
4 ≡ 3 x 4 + 0 ( mod
9 ) = 3
vertex 4 ≡ 3 x 4 + 1 ( mod 9 ) = 4
vertex 4 ≡ 3 x 4 + 2 ( mod 9 ) = 5
. vertex
5 ≡ 3 x 5 + 0 ( mod
9 ) = 6
vertex 5 ≡ 3 x 5 + 1 ( mod 9 ) = 7
vertex 5 ≡ 3 x 5 + 2 ( mod 9 ) = 8
. vertex
6 ≡ 3 x 6 + 0 ( mod
9 ) = 0
vertex 6 ≡ 3 x 6 + 1 ( mod 9 ) = 1
vertex 6 ≡ 3 x 6 + 2 ( mod 9 ) = 2
. vertex
7 ≡ 3 x 7 + 0 ( mod
9 ) = 3
vertex 7 ≡ 3 x 7 + 1 ( mod 9 ) = 4
vertex 7 ≡ 3 x 7 + 2 ( mod 9 ) = 5
. vertex
8 ≡ 3 x 8 + 0 ( mod
9 ) = 6
vertex 8 ≡ 3 x 8 + 1 ( mod 9 ) = 7
vertex 8 ≡ 3 x 8 + 2 ( mod 9 ) = 8
