AGT END 23-24 Solution

Question 2

Part (A) and (B)

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Question 3

Part (a) Prove: A graph $G$ is a tree $⟺$ $G$ is acyclic and has $n-1$ edges (where $n$ is the number of vertices).

We use induction on $n$ (the number of vertices).

Part 1: (Tree $⟹$ Acyclic and $m=n-1$)

• Base Case ($n=1$): A single vertex is a tree, is acyclic, and has $0=1-1$ edges. The statement holds.

• Inductive Hypothesis: Assume every tree with $k$ vertices is acyclic and has $k-1$ edges.

• Inductive Step ($n=k+1$): Let $G$ be a tree with $k+1$ vertices. A tree always has at least one leaf (a vertex with degree 1). Remove a leaf vertex and its connecting edge. This creates a new graph $G^{\prime }$.

  • $G^{\prime }$ is still a tree (removing a leaf can’t create a cycle or disconnect a connected graph).
  • $G^{\prime }$ has $k$ vertices.
  • By the inductive hypothesis, $G^{\prime }$ has $k-1$ edges.
  • $G$ had one more edge than $G^{\prime }$, so $G$ has $(k-1)+1=k$ edges. Since $k=(k+1)-1$, $G$ has $n-1$ edges.

Part 2: (Acyclic and $m=n-1$ $⟹$ Tree)

We need to show that if a graph $G$ is acyclic and has $n-1$ edges, it must be connected (and therefore a tree). We prove this by contradiction.

• Base Case ($n=1$): a single vertex is connect.

• Inductive Hypothesis: Assume every graph with $k$ vertices and is acyclic and has $k-1$ edges is connected.

• Inductive Step ($n=k+1$): Let $G$ acyclic with $k+1$ vertices. Suppose G is not connected.

  • Then, $G$ has multiple connected components: $C_1,C_2,...,C_l$ (where $l>1$, since it’s disconnected).
  • Each $C_i$ is acyclic (a subgraph of an acyclic graph is acyclic).
  • Let $n_i$ be the number of vertices in component $C_i$, and $m_i$ be the number of edges. Note, that ${\sum }_{i=1}^l{n}_i=k+1$.
  • Because each $C_i$ is acyclic and connected, by apply the part 1, $m_i=n_i-1$.
  • The total number of edges in $G$ is the sum of the edges in each component: $m={\sum }_{i=1}^l{m}_i={\sum }_{i=1}^l(n_i-1)=({\sum }_{i=1}^l{n}_i)-l=(k+1)-l$.
  • Since $l>1$ (disconnected), $(k+1)-l<k$. This means $m<k$.
  • But we started by assuming $G$ has $k$ edges ($m=n-1=k$). We have a contradiction. Our assumption that $G$ is disconnected is false.

• Therefore, $G$ must be connected. Since $G$ is acyclic and connected, it’s a tree.

Conclusion: We’ve shown both directions, so the statement is proven. A graph is a tree if and only if it is acyclic and has $n-1$ edges.

Part (B)

Question 5

Part (a)

Transclude of Eular,-Hamilton-Cycles;-Planer-Graph;-Coloring#5be995

Part (B)

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Question 6

Part (a)

AGT Complete, p.53

Part (b)

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Question 7

Part (a)

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Part (B)

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Question 8

Part (a)

Dijkstra’s Algorithm

Merits:

• Efficiency: For graphs with non-negative edge weights, Dijkstra’s algorithm is very efficient, typically having a time complexity of $O((V+E)\log V)$ using a priority queue, where $V$ is the number of vertices and $E$ the number of edges.
• Optimal Solution: It guarantees finding the shortest path from a source node to all other nodes in the graph if all edge weights are non-negative.

Demerits:

• Negative Edge Weights: Dijkstra’s algorithm fails to find the correct shortest paths when the graph contains negative edge weights. It may get stuck in an infinite loop or produce incorrect results.
• Single Source: The algorithm can only find the shortest path from the chosen source to every other node.

Example (Merit): Consider a graph where nodes represent cities and edges represent road distances (all positive). Dijkstra’s can efficiently find the shortest route from a starting city to all other cities.

Example (Demerit): Suppose there are nodes representing locations. There is an edge with a cost of -5. If we try to run Dijkstra’s, it may not output the most optimal path.

Bellman-Ford Algorithm

Merits:

• Negative Edge Weights: Bellman-Ford can handle graphs with negative edge weights, unlike Dijkstra’s algorithm.
• Negative Cycle Detection: It can detect the presence of negative-weight cycles, which are cycles where the sum of edge weights is negative.

Demerits:

• Efficiency: Bellman-Ford is generally less efficient than Dijkstra’s for graphs with non-negative weights. Its time complexity is $O(V\cdot E)$.
• Single Source: Similar to Dijkstra, the algorithm can only find the shortest path from a source to every other node.

Example (Merit): Consider a graph representing currency exchange rates. Some rates might represent a loss (negative edge). Bellman-Ford can find the most profitable exchange sequence or detect arbitrage opportunities (negative cycles). Example (Demerit): Given the city example again, for only non-negative edges, the Bellman Ford algorithm will run slower.