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please post the correct answer and provide the explanation A network topology sp

ID: 3907289 • Letter: P

Question

please post the correct answer and provide the explanation

A network topology specifies how computers, printers, and other devices are connected over a network. For example, ?-?-?-?-?-? is six nodes connected in a line. Given an adjacency matrix A[0…n-1; 0…n-1] representing the connection of the nodes, where n > 3, design a brute-force algorithm to indicate whether or not the graph has a line topology. Also, indicate its time efficiency class. Recall that in this case, the adjacency matrix will represent an undirected graph.

Explanation / Answer

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The adjacency matrix will look likw -

adjMat[n][n];

Initialization :

for i = 0 to n

for j = 0 to n

adjMat[i][j] = 0;

end

end

If there are six nodes in the network.Which are in linear order. Therefore, the edges will be -

(0,1) , (1,2) , (2,3) , (3,4) , (4,5) , (5,6)

We can mark these edges true in the graph-

Like : adjMat[i][j] = 1 (adjMat[0][1] ,adjMat[1][2] ,adjMat[2][3] ,adjMat[3][4] ,adjMat[4][5] ,adjMat[5][6] )

Adjacency matrix will look like :

0 1 0 0 0 0

0 0 1 0 0 0

0 0 0 1 0 0

0 0 0 0 1 0

0 0 0 0 0 1

0 0 0 0 0 0

Now, coming over to the second part :

Part 2) If we want to know whether the graph is linear or not.

Linear graph property :

1) There must be total of V-1 edges in it.

2) Every node can point to atmost one node [Why atmost? all nodes will point to one node except the last one].

How, can we know satisfy both the condition?

Proposed Algorithm

count = 0

for i = 0 : i < n ; i++

for j = 0 ; j < n ; j++

if arr[i][j] == 1

count++

end

end

if count != V-1

return 0; // Number of edges are more in the graph

// Now we know that the graph has V-1 edges

// We will check these vertices must be in this order -> (0,1) , (1,2) , (2,3) , (3,4) , (4,5) , (5,6)

for i = 0 : i < n : i++

if (adjMat[i][i+1] == 0)

return 0 // The following graph is not linear.

end

return 1

The complexity of the algorithm is O(n2) + O(n) == O(n2).

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