COMP9020 Assignment 1

8.a.i f(a + b) = max{ f(a), f(b)}
Case 1: a > 0 and b > 0
max{ f(a), f(b)} = max{1, 1} = 1
f(a + b) = 1 (a > 0 and b > 0 → a + b > 0)
f(a + b) = max{ f(a), f(b)} = 1.

Case 2: a > 0 and b = 0
max{ f(a), f(b)} = max{1, 0} = 1
f(a + b) = 1 (a > 0 and b = 0 → a + b > 0)
f(a + b) = max{ f(a), f(b)} = 1.

Case 3: a = 0 and b > 0
Similarly:
max{ f(a), f(b)} = 1 = f(a + b) = 1

Case 4: a = 0 and b = 0
max{ f(a), f(b)} = max{0, 0} = 0
f(a + b) = 0 (a = 0 and b = 0 → a + b = 0)
f(a + b) = max{ f(a), f(b)} = 0.
From the above cases, we have shown that for all a, b ∈ N, f(a + b) = max{ f(a), f(b)}.

8.a.ii f(ab) = min{ f(a), f(b)}
Case 1: a > 0 and b > 0
min{ f(a), f(b)} = min{1, 1} = 1
f(ab) = 1 (a > 0 and b > 0 → ab > 0)
therefore f(ab) = min{ f(a), f(b)} = 1.

Case 2: a > 0 and b = 0
min{ f(a), f(b)} = min{1, 0} = 0
f(ab) = 0 (a > 0 and b = 0 → ab = 0)
therefore f(ab) = min{ f(a), f(b)} = 0.

Case 3: a = 0 and b > 0
min{ f(a), f(b)} = min{1, 0} = 0
f(ab) = 0 (a = 0 and b > 0 → ab = 0)
therefore f(ab) = min{ f(a), f(b)} = 0.

Case 4: a = 0 and b = 0
min{ f(a), f(b)} = min{0, 0} = 0
f(ab) = 0 (a = 0 and b = 0 → ab = 0)
therefore f(ab) = min{ f(a), f(b)} = 0.

From the above cases, we have shown that for all a, b ∈ N, f(ab) = min{ f(a), f(b)}.

8.b.i
Existence:
f(z’) = f(x’ + y’) = f(x’) + f(y’) = f(x) + f(y) = f(x + y) = f(z) → z’ = z. Thus, Z = [z] exists.

Uniqueness:
Since z’ = z’‘, it implies that [z’] = [z’']. Hence, the element Z = [z] satisfying (([x], [y]), Z) ∈ ⊞ is unique.

8.b.ii
Existence:
f(z’) = f(x’y’) = f(x’)f(y’) = f(x)f(y) = f(xy) = f(z), which means z’ = z. Thus, Z = [z] exists.

Uniqueness:
Since z’ = z’’ (both are equal to xy), it implies that [z’] = [z’']. Hence, the element Z = [z] satisfying (([x], [y]), Z) ∈ ⊡ is unique.

Therefore, we have shown that the relation ⊡ is a function.

8.c.i A ⊡ [1] = A

Let A = [a] for some a ∈ N.
A ⊡ [1] = [a] ⊡ [1] = [a · 1] = [a] = A.
Therefore, A ⊡ [1] = A holds for all A ∈ E.

8.c.ii A ⊞ B = B ⊞ A:

Let A = [a] and B = [b] for some a, b ∈ N.
A ⊞ B = [a] ⊞ [b] = [a + b] = [b + a] = [b] ⊞ [a] = B ⊞ A.
Therefore, A ⊞ B = B ⊞ A holds for all A, B ∈ E.

8.c.iii A ⊡ (B ⊞ C) = (A ⊡ B) ⊞ (A ⊡ C)

Let A = [a], B = [b], and C = [c] for some a, b, c ∈ N.
A ⊡ (B ⊞ C) = [a] ⊡ ([b] ⊞ [c]) = [a] ⊡ [b + c] = [a · (b + c)].

(A ⊡ B) ⊞ (A ⊡ C) = ([a] ⊡ [b]) ⊞ ([a] ⊡ [c]) = [a · b] ⊞ [a · c] = [(a · b) + (a · c)] = [a · (b + c)].
Therefore, A ⊡ (B ⊞ C) = (A ⊡ B) ⊞ (A ⊡ C) holds for all A, B, C ∈ E.

9.a.i
(i) Vertices:
The vertices of the graph represent the houses in the neighborhood. In this case, there are eight houses labeled as A, B, C, D, E, F, G, and H.

(ii) Edges:
The edges of the graph represent the interference between houses. Two houses will have an edge between them if they are either next to each other or directly opposite each other across the road.

9.a.ii
The graph representation can be visualized as follows:

The associated graph problem we need to solve is the Minimum Edge Coloring problem. In this problem, we aim to find the minimum number of different channels needed to color the vertices of the graph such that no two adjacent vertices have the same channels.

9.b
To find the minimum edge coloring,start with an empty set of colors and assign colors to the vertices one by one, making sure that no two adjacent vertices have the same color.

Assigning colors to the remaining vertices in the same order, get the following edge coloring:
Vertice color
A 1
B 2
C 1
D 2
E 2
F 1
G 2
H 1

The minimum number of Wi-Fi channels required for the neighborhood is the number of colors used, which in this case is 2. So, a minimum of 2 different channels are required.

9.c
The new graph representation of the neighborhood with the additional interference would be:

Use the Greedy Coloring algorithm to find a minimum edge coloring:

Assigning colors to the remaining vertices, get the following edge coloring:
Vertice color
A 1
B 2
C 1
D 2
E 3
F 4
G 3
H 4

The minimum number of Wi-Fi channels required for the neighborhood is the number of colors used, which in this case is 4. Therefore, a minimum of 4 different channels are required.