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ESTRUCTURA DE DATOS

ESTRUCTURA DE DATOS

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Consider the following graph

What would the output be if we were to print each node in a Depth First Search starting from the node marked with value source? Assume we traverse starting from the leftmost neighbor (e.g. in node 5, we would push to the stack node 1, node 2, node 3 in that order)

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

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

5, 1, 2, 3, 4, 0, 1

Consider the following graph

What would the output be if we were to print each node in a Breadth-First Search starting from the node with value marked as source?

When traversing assume we go leftmost neighbor first.

3, 0, 4, 1, 5, 2

✅

3, 0, 4, 5, 1, 2

3, 0, 4, 5, 2, 1

0, 3, 5, 4, 2, 1

3, 4, 0, 5, 1, 2

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If G = (V, E) is an undirected graph, how much space is needed to represent G using an adjacency list? What about when G is a directed graph?

Cannot be determined from the premise

If G =(V, E) is a directed graph that has |V| = 20 , how many values would an adjacency matrix need to represent G ?

400

100%

200

Cannot be determined from the premise

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Consider the following directed graph G:

The Adjacency List representation of G is

and the Adjacency Matrix representation of G is

True

✅

False

❌

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Consider the following undirected graph G:

What would the adjacency matrix for G be like?

$M = \begin{bmatrix}0 & 1 & 0 & 0 & 1 \\1 & 0 & 1 & 1 & 1 \\0 & 1 & 0 & 1 & 0 \\0 & 1 & 1 & 0 & 1 \\1 & 1 & 0 & 1 & 0 \end{bmatrix}$ M = \begin{bmatrix}0 & 1 & 0 & 0 & 1 \\1 & 0 & 1 & 1 & 1 \\0 & 1 & 0 & 1 & 0 \\0 & 1 & 1 & 0 & 1 \\1 & 1 & 0 & 1 & 0 \end{bmatrix}

$M = \begin{bmatrix}0 & 1 & 0 & 1 & 0 \\0 & 0 & 1 & 1 & 1 \\0 & 1 & 0 & 1 & 0 \\0 & 1 & 1 & 0 & 1 \\1 & 1 & 0 & 1 & 0 \end{bmatrix}$ M = \begin{bmatrix}0 & 1 & 0 & 1 & 0 \\0 & 0 & 1 & 1 & 1 \\0 & 1 & 0 & 1 & 0 \\0 & 1 & 1 & 0 & 1 \\1 & 1 & 0 & 1 & 0 \end{bmatrix}

$M = \begin{bmatrix}0 & 0 & 0 & 1 & 0 \\0 & 0 & 1 & 1 & 1 \\0 & 1 & 0 & 1 & 0 \\0 & 1 & 0 & 0 & 1 \\1 & 1 & 0 & 1 & 0 \end{bmatrix}$ M = \begin{bmatrix}0 & 0 & 0 & 1 & 0 \\0 & 0 & 1 & 1 & 1 \\0 & 1 & 0 & 1 & 0 \\0 & 1 & 0 & 0 & 1 \\1 & 1 & 0 & 1 & 0 \end{bmatrix}

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To DownHeap (sifting down) we compare the parent node to its children to check it the heap property is upheld. If that is not the case, we swap the parent with one of its childrens and then recursively apply this until we reach a leaf.

This process is needed whenever we remove an entry from the Heap.

Please comment on YOUR OWN performance in the project. You should specifically address the following questions: