Functions are the building blocks of any application. This lesson covers how to declare and call functions.

extract all questions and give solutions

Here are the extracted questions and their solutions:

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1. The maximum number of non-zero entries in an adjacency matrix of a simple graph having $ n $ vertices can be

Options:

• $\boxed{n^2}$
• $\frac{n(n-1)}{2}$
• $\frac{n(n-1)}{4}$
• $n(n-1)$

Solution:

For a simple graph (no self-loops, undirected), the maximum number of edges is $\frac{n(n-1)}{2}$. But the adjacency matrix counts both (i, j) and (j, i) as nonzero for an edge between i and j. So the maximum number of nonzero entries is $n(n-1)$ (each edge counted twice in the adjacency matrix). Answer: $n(n-1)$

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2. We have a graph $G$ with 6 vertices. We write down the degrees of all vertices in $G$ in descending order. Which of the following is a possible listing of the degrees?

Options:

• (a) 6, 5, 4, 3, 2, 1
• (b) 5, 5, 2, 2, 1, 1
• (c) 5, 3, 3, 2, 2, 1
• (d) 2, 1, 1, 1, 1, 1

Solution:

• The sum of degrees in a graph with 6 vertices must be even (Handshaking Lemma).
• Maximum degree ≤ $n-1 = 5$.

(a) sum: 6+5+4+3+2+1=21 (odd, and 6>5 - invalid) (b) sum: 5+5+2+2+1+1 = 16 (valid) (c) sum: 5+3+3+2+2+1=16 (valid) (d) sum: 2+1+1+1+1+1=7 (odd - invalid)

But in (b), having 5 and 5 (i.e., two vertices connected to all except one each) is possible, with 2 and 2, and 1 and 1. So both (b) and (c) are possible degree sequences for a simple graph.

Answer:

• $5, 5, 2, 2, 1, 1$ (valid)
• $5, 3, 3, 2, 2, 1$ (valid)

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3. Data structure for keeping track of intersections in a maze

Question: What is a good data structure to keep track of the intersections we have visited to backtrack if we hit a dead end?

Options:

• List
• Stack
• Queue
• Array

Solution:

For maze solving with backtracking, a stack is appropriate (LIFO: last place visited is the first to return to).

Answer: Stack

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4. BFS tree for undirected graph (vertices 1 to 14)

Given the BFS tree rooted at node 1, which of the following cannot be an edge in the original graph?

Options:

• (8,11)
• (3,10)
• (4,5)
• (6,9)

Solution:

In a BFS tree, if vertices are on different levels with no parent-child relationship, the edge between them would have produced a shorter path in BFS.

(8,11):

• 8 is a descendant of 2 (from 6 -> 2 -> 8)
• 11 is child of 5
• Level of 8: 1→5→6→2→8 (level 4)
• Level of 11: 1→5→11 (level 3) No parent-child or sibling relationship.

But, in general, (8,11) connects two nodes where one is not ancestor/descendant of another. This means such an edge would produce a cross-edge, which is not possible in a BFS tree in an undirected graph. Answer: (8,11) cannot be an edge in the original graph.

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5. Which of the following graphs satisfies:

1. $|VC(G)| = 3$: minimum vertex cover is 3
2. $|PM(G)| = 3$: perfect matching of size 3
3. 3-colorable

Solution:

Let’s check each image:

First Graph:

• Looks like a hexagon with a triangle on top—does not have perfect matching of 3.

Second Graph:

• K3,3 (complete bipartite on 6: every U connects every V): perfect matching 3, vertex cover 3, 2-colorable (so 3-colorable).

Third Graph:

• Looks like a pentagon with chords, not obvious matching of 3.

Fourth Graph:

• Not perfect matching of 3.

Answer: The second graph (K3,3) satisfies all conditions.

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6. Which statements are true?

Options:

• (a) BFS can be used to identify the vertex which is at the farthest distance from $v$ in any graph, in terms of number of edges, where $v$ is the starting vertex.
• (b) BFS and DFS identifies all the vertices reachable from the starting vertex $v$.
• (c) BFS cannot be used to check for cycles in the graph while DFS can be used to check for cycles in the graph.
• (d) DFS can be used to identify the shortest distance from starting vertex $v$ to every other vertex in the graph, in terms of number of edges.

Solution:

• (a) True in connected unweighted graphs.
• (b) True for both BFS and DFS.
• (c) Incorrect. BFS can also identify cycles.
• (d) False. DFS cannot reliably find shortest path in unweighted graph; BFS does.

Answer: Statements (a) and (b) are true.

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7. Maximum Independent Set Cardinality (Numerical)

Adjacency matrix:

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

Solution:

Let the vertices be 1-6. Let’s try to find the largest independent set (set of vertices no two of which are adjacent):

• Vertex 1: adjacent to 2, 4.
• Vertex 2: adjacent to 1, 3, 4.
• Vertex 3: adjacent to 2, 5, 6.
• Vertex 4: adjacent to 1, 2.
• Vertex 5: adjacent to 3, 6.
• Vertex 6: adjacent to 3, 5.

Try picking 1, 5:

• 1 is connected to 2, 4.
• 5 is connected to 3, 6 (neither are 1 or 5). Try 1, 5, 4: 1 and 4 connected (invalid).

Try 1, 3, 6: 1 and 3 are not connected, 1 and 6 not connected, 3 and 6 are connected (invalid).

Try 1, 5, 3: 3 and 5 connected (invalid).

Try 4, 5, 6: 4 and 5 not connected, 4 and 6 not connected, 5 and 6 are connected (invalid).

Try 1, 5:

• 1 (not adjacent to 5).
• 5 (adjacent to 3 and 6).
• Add 4? 1 and 4 connected (invalid).
• Add 6? 5 and 6 connected (invalid).
• Add 2? 1 and 2 connected (invalid).

Try 1, 5 — that’s 2. Try 1, 6: 1 and 6 not connected.

• 1 (adjacent to 2, 4)
• 6 (adjacent to 3,5)
• Add 4? 1 and 4 connected (invalid).

Try 2, 5, 4:

• 2 and 5 not connected.
• 2 and 4 connected (invalid).

Try 4, 5:

• 4 and 5 not connected.

Try 4, 6:

• 4 and 6 not connected.

Try 4, 5, 2: 4 and 2 connected.

Try 1, 5, 6: 5 and 6 connected.

Try 1, 5.

Try 4, 6.

Try 2, 5.

Try 2 and 6.

No independent set of size 3.

But what about 3,4,5?

• 3 and 4 not connected.
• 3 and 5 connected.

So, maximum independent set is of size 2.

Answer: 2

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Summary Table

Qn	Answer/Explanation
1	$ n(n-1) $
2	(b) $ 5,5,2,2,1,1 $; (c) $ 5,3,3,2,2,1 $
3	Stack
4	(8,11)
5	Second graph
6	(a), (b)
7	2