Someone who participates online for social reasons usually has a:

Which one of the following analog VOM ranges would be selected to measure a DC

<span>100 mA of course. You cannot use 10 mA because it will be over the range and the higher ranges are not accurate at 90 mA</span>

7 0

3 years ago

HELP FAST PLEASE!!! What is a glass tube containing electrodes used as a switch to produce on or off signals for the computer? F

Mekhanik [1.2K]

Answer:

The answer is... Vaccum Tubes...

This is one which was used in computers before the Phenomenal invention of Transisitors.

3 0

4 years ago

) A stub network has ______. (Points : 2) one backup route

jeyben [28]

Answer: One backup router

Explanation:

A stub network is the type of the packet network that basically describe the computer notwork. It typically capable for sending the complex data in the single network path when the network aware about its destination.

A stub network contain the one backup router as the stub routing is the typically designed for conserve the resources of the local router like the central processing unit (CPU) and the memory. It basically improve the stability of the network in the system.

6 0

3 years ago

There are n poor college students who are renting two houses. For every pair of students pi and pj , the function d(pi , pj ) ou

Nuetrik [128]

Answer:

Here the given problem is modeled as a Graph problem.

Explanation:

Input:- n, k and the function d(pi,pj) which outputs an integer between 1 and n2

Algorithm:-We model each student as a node. So, there would be n nodes. We make a foothold between two nodes u and v (where u and v denote the scholars pu and pv respectively) iff d(pu,pv) > k. Now, Let's call the graph G(V, E) where V is that the vertex set of the graph ( total vertices = n which is that the number of students), and E is that the edge set of the graph ( where two nodes have edges between them if and only the drama between them is bigger than k).

We now need to partition the nodes of the graph into two sets S1 and S2 such each node belongs to precisely one set and there's no edge between the nodes within the same set (if there's a foothold between any two nodes within the same set then meaning that the drama between them exceeds k which isn't allowed). S1 and S2 correspond to the partition of scholars into two buses.

The above formulation is akin to finding out if the graph G(V,E) is a bipartite graph. If the Graph G(V, E) is bipartite then we have a partition of the students into sets such that the total drama <= k else such a partition doesn't exist.

Now, finding whether a graph is bipartite or not is often found using BFS (Breadth First algorithm) in O(V+E) time. Since V = n and E = O(n2) , the worst-case time complexity of the BFS algorithm is O(n2). The pseudo-code is given as

PseudoCode:

// Input = n,k and a function d(pi,pj)

// Edges of a graph are represented as an adjacency list

1. Make V as a vertex set of n nodes.

2. for each vertex u ∈ V

$\rightarrow$ for each vertex v ∈ V

$\rightarrow\rightarrow$ if( d(pu, pj) > k )

$\rightarrow\rightarrow\rightarrow$ add vertex u to Adj[v] // Adj[v] represents adjacency list of v

$\rightarrow\rightarrow\rightarrow$ add vertex v to Adj[u] // Adj[u] represents adjacency list of u

3. bool visited[n] // visited[i] = true if the vertex i has been visited during BFS else false

4. for each vertex u ∈ V

$\rightarrow$ visited[u] = false

5. color[n] // color[i] is binary number used for 2-coloring the graph

6. for each vertex u ∈ V

$\rightarrow$ if ( visited[u] == false)

$\rightarrow\rightarrow$ color[u] = 0;

$\rightarrow\rightarrow$ isbipartite = BFS(G,u,color,visited) // if the vertices reachable from u form a bipartite graph, it returns true

$\rightarrow\rightarrow$ if (isbipartite == false)

$\rightarrow\rightarrow\rightarrow$ print " No solution exists "

$\rightarrow\rightarrow\rightarrow$ exit(0)

7. for each vertex u ∈V

$\rightarrow$ if (color[u] == 0 )

$\rightarrow\rightarrow$ print " Student u is assigned Bus 1"

$\rightarrow$ else

$\rightarrow\rightarrow$ print " Student v is assigned Bus 2"

BFS(G,s,color,visited)

1. color[s] = 0

2. visited[s] = true

3. Q = Ф // Q is a priority Queue

4. Q.push(s)

5. while Q != Ф {

$\rightarrow$ u = Q.pop()

$\rightarrow$ for each vertex v ∈ Adj[u]

$\rightarrow\rightarrow$ if (visited[v] == false)

$\rightarrow\rightarrow\rightarrow$ color[v] = (color[u] + 1) % 2

$\rightarrow\rightarrow\rightarrow$ visited[v] = true

$\rightarrow\rightarrow\rightarrow$ Q.push(v)

$\rightarrow\rightarrow$ else

$\rightarrow\rightarrow\rightarrow$ if (color[u] == color[v])

$\rightarrow\rightarrow\rightarrow\rightarrow$ return false // vertex u and v had been assigned the same color so the graph is not bipartite

}

6. return true

3 0

3 years ago

A window frame will expand to fill the entire desktop when you

natima [27]

A. Sometimes it is marked by outwards pointing arrows (<= =>) with the reverse (collapsing so as to not take up the whole screen) being in the same spot, marked by inwards arrows (=> <=)

8 0

3 years ago