Shortest Path Algorithms

This chapter rigorously examines fundamental shortest path algorithms, a cornerstone of graph theory with broad applications in network routing and resource optimization. Mastery of these techniques, particularly Dijkstra's algorithm and unweighted path finding, is crucial for solving complex computational problems frequently encountered in advanced computer science examinations.

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Chapter Contents

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| Topic |

|---|-------| | 1 | Unweighted Shortest Path | | 2 | Dijkstra's Algorithm |

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We begin with Unweighted Shortest Path.

Part 1: Unweighted Shortest Path

We explore unweighted shortest paths in graphs, a fundamental concept in graph theory with applications in network routing and resource allocation. For CMI, understanding path properties and algorithmic approaches is critical.

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Core Concepts

1. Definition and Properties of Unweighted Shortest Paths

An unweighted shortest path between two vertices $s$ and $t$ in a graph $G=(V,E)$ is a path containing the minimum number of edges. The length of this path, denoted $d(s,t)$, is the shortest distance between $s$ and $t$.

We observe that any subpath of a shortest path is also a shortest path. This is a fundamental optimality principle.

Worked Example: Finding Shortest Path Length

Consider the graph $G=(V,E)$ where $V=\{A,B,C,D,E,F\}$ and $E=\{\{A,B\}, \{A,C\}, \{B,D\}, \{C,D\}, \{C,E\}, \{D,F\}, \{E,F\}\}$. We find the shortest path length from $A$ to $F$.

Step 1: Identify paths from $A$ to $F$.

> Path 1: $A-B-D-F$, length 3.
> Path 2: $A-C-D-F$, length 3.
> Path 3: $A-C-E-F$, length 3.

Step 2: Determine the minimum length.

The minimum length among all paths is 3.

Answer: $d(A,F) = 3$.

:::question type="MCQ" question="Given a graph with vertices $V=\{1,2,3,4,5,6\}$ and edges $E=\{\{1,2\}, \{1,3\}, \{2,4\}, \{3,4\}, \{3,5\}, \{4,6\}, \{5,6\}\}$. What is the shortest path length from vertex 1 to vertex 6?" options=["2","3","4","5"] answer="3" hint="Trace paths from vertex 1 to vertex 6, counting edges for each path." solution="We list paths from 1 to 6 and their lengths:

$1-2-4-6$ (length 3)

$1-3-4-6$ (length 3)

$1-3-5-6$ (length 3)

The minimum length is 3.
"
:::

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2. Breadth-First Search (BFS) for Unweighted Shortest Paths

Breadth-First Search (BFS) is an algorithm that systematically explores a graph layer by layer from a starting vertex $s$. It inherently finds shortest paths in unweighted graphs because it discovers all vertices at distance $k$ from $s$ before discovering any vertices at distance $k+1$.

:::algorithm title="BFS for Shortest Paths"

Initialize $d(s,v) = \infty$ for all $v \in V$, $d(s,s) = 0$.

Create a queue $Q$ and enqueue $s$.

While $Q$ is not empty:

a. Dequeue a vertex $u$.
b. For each neighbor $v$ of $u$:
i. If $d(s,v) = \infty$ (i.e., $v$ is unvisited):
Set $d(s,v) = d(s,u) + 1$.
Enqueue $v$.
:::

Worked Example: BFS Trace for Shortest Distances

Consider the graph from the previous example: $V=\{A,B,C,D,E,F\}$ and $E=\{\{A,B\}, \{A,C\}, \{B,D\}, \{C,D\}, \{C,E\}, \{D,F\}, \{E,F\}\}$. We use BFS to find shortest path lengths from $A$ to all other vertices.

Step 1: Initialization.

> $d(A,A)=0$, $d(A,B)=\infty$, $d(A,C)=\infty$, $d(A,D)=\infty$, $d(A,E)=\infty$, $d(A,F)=\infty$.
> $Q=[A]$

Step 2: Dequeue $A$. Neighbors of $A$ are $B, C$.

> $d(A,B) = d(A,A)+1 = 1$. Enqueue $B$.
> $d(A,C) = d(A,A)+1 = 1$. Enqueue $C$.
> $Q=[B,C]$

Step 3: Dequeue $B$. Neighbors of $B$ are $A, D$. $A$ is visited.

> $d(A,D) = d(A,B)+1 = 2$. Enqueue $D$.
> $Q=[C,D]$

Step 4: Dequeue $C$. Neighbors of $C$ are $A, D, E$. $A$ is visited. $D$ is visited ($d(A,D)=2$).

> $d(A,E) = d(A,C)+1 = 2$. Enqueue $E$.
> $Q=[D,E]$

Step 5: Dequeue $D$. Neighbors of $D$ are $B, C, F$. $B, C$ are visited.

> $d(A,F) = d(A,D)+1 = 3$. Enqueue $F$.
> $Q=[E,F]$

Step 6: Dequeue $E$. Neighbors of $E$ are $C, F$. $C, F$ are visited.

> $Q=[F]$

Step 7: Dequeue $F$. Neighbors of $F$ are $D, E$. $D, E$ are visited.

> $Q=[]$

Answer: Shortest distances from $A$: $d(A,A)=0, d(A,B)=1, d(A,C)=1, d(A,D)=2, d(A,E)=2, d(A,F)=3$.

:::question type="NAT" question="In a graph with vertices $V=\{1,2,3,4,5\}$ and edges $E=\{\{1,2\}, \{1,3\}, \{2,4\}, \{3,4\}, \{4,5\}\}$, what is the shortest path length from vertex 1 to vertex 5 using BFS?" answer="3" hint="Perform a BFS starting from vertex 1. The distance to vertex 5 will be the number of edges in the shortest path." solution="
Step 1: Initialize distances and queue.
> $d(1,1)=0$, $Q=[1]$
Step 2: Dequeue 1. Neighbors: 2, 3.
> $d(1,2)=1$, $Q=[2]$
> $d(1,3)=1$, $Q=[2,3]$
Step 3: Dequeue 2. Neighbor: 4 (1 visited).
> $d(1,4)=2$, $Q=[3,4]$
Step 4: Dequeue 3. Neighbor: 4 (1 visited, 4 visited with $d(1,4)=2$).
> $Q=[4]$
Step 5: Dequeue 4. Neighbor: 5 (2, 3 visited).
> $d(1,5)=3$, $Q=[5]$
Step 6: Dequeue 5. No unvisited neighbors.
> $Q=[]$
The shortest path length from 1 to 5 is 3.
"
:::

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3. Properties of Shortest Path Distances

For any two adjacent vertices $u, v$ in an unweighted graph, their shortest path distances from a source $s$ must be closely related.

This property implies that for an edge $(u,v)$, either $d(s,v) = d(s,u)$, $d(s,v) = d(s,u)+1$, or $d(s,v) = d(s,u)-1$. It cannot be that $d(s,v) = d(s,u)+2$ or $d(s,v) = d(s,u)-2$, because if $d(s,v) = d(s,u)+2$, we could form a shorter path to $v$ via $u$ of length $d(s,u)+1$.

Worked Example: Applying Shortest Path Edge Property

Consider a graph where $d(s,u)=5$. We examine possible values for $d(s,v)$ if $(u,v)$ is an edge.

Step 1: Apply the formula.

>

>

Step 2: Solve for $d(s,v)$.

>

>
>
>

Answer: If $d(s,u)=5$ and $(u,v)$ is an edge, then $d(s,v)$ must be 4, 5, or 6.

:::question type="MSQ" question="Let $G=(V,E)$ be an unweighted graph and $s$ be a source vertex. For an edge $(u,v) \in E$, which of the following statements about $d(s,u)$ and $d(s,v)$ are always true?" options=["$d(s,u) \ne d(s,v)+2$","$d(s,u) = d(s,v)$ implies $u,v$ are in the same BFS layer","$|d(s,u) - d(s,v)| \le 1$","$d(s,u) \ge d(s,v)-1$"] answer="Option 1,Option 3,Option 4" hint="Recall the definition of shortest path distance and how it relates to adjacent vertices. If $d(s,u) = k$, then $d(s,v)$ can be $k-1, k,$ or $k+1$." solution="

$d(s,u) \ne d(s,v)+2$: If $d(s,u) = d(s,v)+2$, it implies $d(s,v) = d(s,u)-2$. Then a path from $s$ to $u$ could be formed by taking the shortest path to $v$ and then traversing edge $(v,u)$, resulting in a path of length $d(s,v)+1 = (d(s,u)-2)+1 = d(s,u)-1$, which contradicts $d(s,u)$ being the shortest path length. So this statement is true.

$d(s,u) = d(s,v)$ implies $u,v$ are in the same BFS layer: While $u$ and $v$ are in the same BFS layer if their distances from $s$ are equal, the statement 'implies' means it's a direct consequence. This is true by definition of BFS layers.

$|d(s,u) - d(s,v)| \le 1$: This is the fundamental shortest path edge property. It is always true.

$d(s,u) \ge d(s,v)-1$: This is equivalent to $d(s,v) - d(s,u) \le 1$, which is part of $|d(s,u) - d(s,v)| \le 1$. So this is always true.

"
:::

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Advanced Applications

1. Degree Sum on Shortest Paths

Consider a shortest path $P$ between $s$ and $t$. Vertices not on $P$ have specific constraints on how many neighbors they can have on $P$. This property is critical for proving bounds on the sum of degrees of vertices on a shortest path.

This property is a stronger form of the argument from the CMI PYQ. The PYQ statement says "if $x$ were adjacent to four vertices of $P$, then two of them would be at distance at least $3$ along $P$, and replacing that subpath by a two-edge route through $x$ would give a shorter $s-t$ path". This implies $x$ can be adjacent to at most 3 vertices of $P$.

Worked Example: Illustrating the Adjacency Constraint

Let $P = v_0 - v_1 - v_2 - v_3 - v_4$ be a shortest path. Consider a vertex $x \notin V(P)$.

Step 1: Assume $x$ is adjacent to $v_0$ and $v_3$.

> If $x$ is adjacent to $v_0$ and $v_3$, we can form a path $v_0 - x - v_3$.
> The length of $v_0 - x - v_3$ is 2.
> The length of $v_0 - v_1 - v_2 - v_3$ (subpath of $P$) is 3.
> Since $2 < 3$, replacing $v_0 - v_1 - v_2 - v_3$ with $v_0 - x - v_3$ would create a shorter path from $v_0$ to $v_3$, contradicting $P$ being a shortest path.

Step 2: Conclude the constraint.

A vertex $x \notin V(P)$ cannot be adjacent to $v_i$ and $v_j$ if $|i-j| \ge 3$. This means $x$ can connect to $v_i$ and $v_j$ only if $|i-j| \le 2$.

Answer: If $x \notin V(P)$, it can be adjacent to $v_i, v_{i+1}$ (distance 1 along $P$), $v_i, v_{i+2}$ (distance 2 along $P$), or a single $v_i$. It cannot connect to $v_i, v_{i+3}$ or further apart. This implies $x$ can be adjacent to at most 3 vertices on $P$ (e.g., $v_i, v_{i+1}, v_{i+2}$). If it connects to 4 vertices, say $v_i, v_j, v_k, v_l$, then at least two must be separated by distance $\ge 3$ on $P$, creating a shortcut.

:::question type="MCQ" question="Let $P = v_0 v_1 \cdots v_k$ be a shortest path in an unweighted graph $G$. Consider a vertex $x \notin V(P)$. If $x$ is adjacent to $v_i$ and $v_j$ where $i < j$, which of the following conditions must be true for $P$ to remain a shortest path?" options=["$j - i \le 1$","$j - i \le 2$","$j - i \le 3$","$j - i \le k$"] answer="$j - i \le 2$" hint="If $j-i \ge 3$, then the path segment $v_i \dots v_j$ has length $j-i$. Could you create a shorter path using $x$?" solution="
If $j-i \ge 3$, then the subpath $v_i \dots v_j$ has length $j-i$. If $x$ is adjacent to both $v_i$ and $v_j$, we could replace the subpath $v_i \dots v_j$ with $v_i - x - v_j$, which has length 2.
For $P$ to be a shortest path, this replacement must not create a shorter path. Thus, the length of $v_i \dots v_j$ must be less than or equal to 2.
So, $j-i \le 2$.
If $j-i=0$, $i=j$, not two distinct vertices.
If $j-i=1$, $v_i, v_{i+1}$ are adjacent on $P$. Path length 1. $1 \le 2$. Valid.
If $j-i=2$, $v_i, v_{i+1}, v_{i+2}$. Path length 2. $2 \le 2$. Valid.
If $j-i=3$, $v_i, v_{i+1}, v_{i+2}, v_{i+3}$. Path length 3. If $x$ connects $v_i$ and $v_{i+3}$, then $v_i - x - v_{i+3}$ has length 2, which is shorter than 3. This contradicts $P$ being a shortest path.
Therefore, $j-i \le 2$.
"
:::

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Problem-Solving Strategies

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Common Mistakes

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Practice Questions

:::question type="NAT" question="Consider a complete graph $K_n$ with $n \ge 2$ vertices. What is the shortest path length between any two distinct vertices?" answer="1" hint="In a complete graph, every pair of distinct vertices is connected by an edge." solution="In a complete graph $K_n$, by definition, every pair of distinct vertices is connected by an edge. Therefore, the shortest path between any two distinct vertices has length 1."
:::

:::question type="MCQ" question="Let $G$ be a connected unweighted graph. If $d(s,v)$ is the shortest path length from a source $s$ to vertex $v$, and $u$ is a neighbor of $v$, which of the following is NOT a possible value for $d(s,u)$?" options=["$d(s,v)-1$","$d(s,v)$","$d(s,v)+1$","$d(s,v)+2$"] answer="$d(s,v)+2$" hint="Recall the shortest path edge property $|d(s,u) - d(s,v)| \le 1$." solution="The shortest path edge property states that for any edge $(u,v)$, $|d(s,u) - d(s,v)| \le 1$. This implies that $d(s,u)$ can be $d(s,v)-1$, $d(s,v)$, or $d(s,v)+1$. It cannot be $d(s,v)+2$, because if it were, we could form a path from $s$ to $u$ of length $d(s,v)+1$ by taking the shortest path to $v$ and then traversing the edge $(v,u)$, which is shorter than $d(s,v)+2$, contradicting the definition of $d(s,u)$ as the shortest path length."
:::

:::question type="MSQ" question="Let $P = v_0 v_1 \cdots v_k$ be a shortest path from $s=v_0$ to $t=v_k$ in an unweighted graph $G$. Which of the following statements are necessarily true?" options=["Every subpath $v_i v_{i+1} \cdots v_j$ of $P$ is also a shortest path from $v_i$ to $v_j$.","If $x \notin V(P)$ is adjacent to $v_i$ and $v_j$ ($i<j$), then $j-i \le 2$.","The sum of degrees $\sum_{v \in V(P)} \deg(v)$ is always less than or equal to $2|V(P)|$.","If $G$ is a tree, then $P$ is the unique path between $s$ and $t$."] answer="Every subpath $v_i v_{i+1} \cdots v_j$ of $P$ is also a shortest path from $v_i$ to $v_j,If$ x \notin V(P)$is adjacent to$ v_i $and$ v_j $($ i$), then$ j-i \le 2$,If$ G $is a tree, then$ P $is the unique path between$ s $and$ t $" hint="Consider the definitions and properties discussed. For the degree sum, consider the edges within$ P $and edges connecting to$ V(G) \setminus V(P)$. For uniqueness, recall properties of trees." solution="

Every subpath $v_i v_{i+1} \cdots v_j$ of $P$ is also a shortest path from $v_i$ to $v_j$.: This is a fundamental property of shortest paths. If there were a shorter path from $v_i$ to $v_j$, we could replace the subpath of $P$ with it, making $P$ shorter, which is a contradiction. So, this is true.

If $x \notin V(P)$ is adjacent to $v_i$ and $v_j$ ($i<j$), then $j-i \le 2$.: As demonstrated in the advanced example, if $j-i \ge 3$, then replacing the subpath $v_i \dots v_j$ with $v_i - x - v_j$ (length 2) would create a shorter path, contradicting $P$ being a shortest path. So, this is true.

The sum of degrees $\sum_{v \in V(P)} \deg(v)$ is always less than or equal to $2|V(P)|$.: This is not necessarily true. Consider a star graph where the center is on a shortest path. Its degree can be $n-1$, which can be much larger than $2|V(P)|$ if $|V(P)|$ is small. The PYQ proved $\sum_{v \in V(P)} \deg(v) \le 3n$, not $2|V(P)|$. So, this is false.

If $G$ is a tree, then $P$ is the unique path between $s$ and $t$.: In any tree, there is a unique path between any two distinct vertices. Since $P$ is a path between $s$ and $t$, it must be that unique path. So, this is true.

"
:::

:::question type="NAT" question="Consider an $m \times n$ grid graph. What is the shortest path length from the top-left vertex $(1,1)$ to the bottom-right vertex $(m,n)$?" answer="m+n-2" hint="Movement is restricted to adjacent cells (up, down, left, right). The shortest path involves only moving right and down." solution="To move from $(1,1)$ to $(m,n)$ in an $m \times n$ grid graph, we need to make $(m-1)$ moves downwards and $(n-1)$ moves to the right. Each move corresponds to one edge. The total number of moves (edges) in any shortest path is $(m-1) + (n-1) = m+n-2$."
:::

:::question type="MCQ" question="Let $G$ be a connected graph with $n$ vertices. Let $P$ be a shortest path from $s$ to $t$. The maximum length of $P$ can be at most:" options=["$n-1$","$n$","$2n-1$","$n^2$"] answer="$n-1$" hint="A simple path in a graph with $n$ vertices can visit each vertex at most once. What is the maximum number of edges in such a path?" solution="A path is a sequence of distinct vertices. In a graph with $n$ vertices, a simple path can have at most $n$ vertices. If a path has $n$ vertices, it has $n-1$ edges. This forms a Hamiltonian path. Since a shortest path cannot revisit vertices, its length is bounded by $n-1$. For example, a path graph $P_n$ has $n$ vertices and a shortest path between its endpoints has length $n-1$."
:::

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Summary

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What's Next?

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Part 2: Dijkstra's Algorithm

Dijkstra's algorithm is a fundamental graph algorithm for finding the shortest paths from a single source vertex to all other vertices in a weighted graph with non-negative edge weights. We apply this algorithm to solve various shortest path problems in network routing, transportation, and resource allocation.

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Core Concepts

1. Shortest Path Definition

The shortest path between two vertices in a weighted graph is a path where the sum of the weights of its constituent edges is minimized. Dijkstra's algorithm determines these paths from a designated source vertex.

Worked Example:

Consider a graph with vertices $A, B, C, D$ and edges $(A,B)$ with weight $3$, $(B,C)$ with weight $2$, $(A,D)$ with weight $7$, and $(D,C)$ with weight $1$. We want to find the weight of the path $A \to D \to C$.

Step 1: Identify the edges in the path.

> The path $A \to D \to C$ consists of edges $(A,D)$ and $(D,C)$.

Step 2: Sum the weights of the edges.

>

>
>

Answer: The weight of the path $A \to D \to C$ is $8$.

:::question type="MCQ" question="Given a graph with edges $E = \{(S,A,2), (S,B,5), (A,C,1), (B,C,3), (C,D,4)\}$, where $(u,v,w)$ denotes an edge from $u$ to $v$ with weight $w$. What is the weight of the path $S \to B \to C \to D$?" options=["7","8","9","12"] answer="12" hint="Sum the weights of the edges along the specified path." solution="Step 1: Identify the edges in the path $S \to B \to C \to D$.
> The edges are $(S,B)$, $(B,C)$, and $(C,D)$.

Step 2: Sum the weights of these edges.
>

>
>
The weight of the path is $12$."
:::

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2. Algorithm Overview

Dijkstra's algorithm is a greedy algorithm that iteratively finds the shortest paths from a source vertex $s$ to all other vertices in a graph. It maintains a set of visited vertices for which the shortest path has been finalized and a distance estimate for all other vertices.

Worked Example:

Consider a graph with vertices $A, B, C, D$ and edges $(A,B,1)$, $(A,C,4)$, $(B,C,2)$, $(B,D,5)$, $(C,D,1)$. We want to find the shortest path from source $A$ to all other vertices. We will illustrate the first few steps of the greedy selection.

Step 1: Initialize distances and set $A$ as the source.

> $dist(A)=0$, $dist(B)=\infty$, $dist(C)=\infty$, $dist(D)=\infty$.
> Priority Queue $PQ = \{(0, A)\}$.

Step 2: Extract minimum from $PQ$.

> We extract $(0, A)$. $A$ is now visited.

Step 3: Relax neighbors of $A$.

> For neighbor $B$: $dist(B) = \min(\infty, dist(A) + w(A,B)) = \min(\infty, 0+1) = 1$. Add $(1, B)$ to $PQ$.
> For neighbor $C$: $dist(C) = \min(\infty, dist(A) + w(A,C)) = \min(\infty, 0+4) = 4$. Add $(4, C)$ to $PQ$.
> $PQ = \{(1, B), (4, C)\}$.

Step 4: Extract minimum from $PQ$.

> We extract $(1, B)$. $B$ is now visited.

Step 5: Relax neighbors of $B$.

> For neighbor $C$: $dist(C) = \min(4, dist(B) + w(B,C)) = \min(4, 1+2) = 3$. Update $(4, C)$ to $(3, C)$ in $PQ$.
> For neighbor $D$: $dist(D) = \min(\infty, dist(B) + w(B,D)) = \min(\infty, 1+5) = 6$. Add $(6, D)$ to $PQ$.
> $PQ = \{(3, C), (6, D)\}$.

Answer: After these steps, the current shortest distance estimates are $dist(A)=0, dist(B)=1, dist(C)=3, dist(D)=6$. The next vertex to be processed would be $C$.

:::question type="MCQ" question="In Dijkstra's algorithm, what is the primary criterion used to select the next vertex to add to the set of finalized shortest paths?" options=["The vertex with the most outgoing edges","The vertex with the highest degree","The unvisited vertex with the smallest current distance estimate from the source","A randomly chosen unvisited vertex"] answer="The unvisited vertex with the smallest current distance estimate from the source" hint="Recall the greedy nature of Dijkstra's algorithm." solution="Explanation: Dijkstra's algorithm is a greedy algorithm. It always selects the unvisited vertex that has the smallest current shortest distance estimate from the source vertex. This ensures that the shortest path to that vertex has been found and can then be used to update distances to its neighbors."
:::

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3. Algorithm Steps and Relaxation

Dijkstra's algorithm proceeds in phases, maintaining a distance array `dist` and a predecessor array `prev`. The core operation is `relax(u, v, w)`, which attempts to improve the shortest path estimate to $v$ by going through $u$.

Initialization:

Set $dist[s] = 0$ for the source vertex $s$, and $dist[v] = \infty$ for all other vertices $v \ne s$.

Initialize $prev[v] = \operatorname{null}$ for all $v$.

Create a min-priority queue $PQ$ and insert $(0, s)$.

Main Loop:

While $PQ$ is not empty:

a. Extract vertex $u$ with the minimum $dist[u]$ from $PQ$.
b. If $u$ has already been visited (its shortest path finalized), continue.
c. Mark $u$ as visited.
d. For each neighbor $v$ of $u$:
i. Perform `relax(u, v, w(u,v))`.

Worked Example:

Consider the graph from the previous example: vertices $A, B, C, D$ and edges $(A,B,1)$, $(A,C,4)$, $(B,C,2)$, $(B,D,5)$, $(C,D,1)$. Source is $A$. We trace the full algorithm.

Step 1: Initialization

> $dist = \{A:0, B:\infty, C:\infty, D:\infty\}$
> $prev = \{A:\operatorname{null}, B:\operatorname{null}, C:\operatorname{null}, D:\operatorname{null}\}$
> $PQ = \{(0, A)\}$

Step 2: Process $A$

> Extract $(0, A)$ from $PQ$.
> Relax neighbors of $A$:
> For $B$: $dist[A] + w(A,B) = 0+1 = 1 < dist[B]$. Update $dist[B]=1$, $prev[B]=A$. Add $(1, B)$ to $PQ$.
> For $C$: $dist[A] + w(A,C) = 0+4 = 4 < dist[C]$. Update $dist[C]=4$, $prev[C]=A$. Add $(4, C)$ to $PQ$.
> $PQ = \{(1, B), (4, C)\}$

Step 3: Process $B$

> Extract $(1, B)$ from $PQ$.
> Relax neighbors of $B$:
> For $C$: $dist[B] + w(B,C) = 1+2 = 3 < dist[C]$. Update $dist[C]=3$, $prev[C]=B$. Update $(4, C)$ to $(3, C)$ in $PQ$.
> For $D$: $dist[B] + w(B,D) = 1+5 = 6 < dist[D]$. Update $dist[D]=6$, $prev[D]=B$. Add $(6, D)$ to $PQ$.
> $PQ = \{(3, C), (6, D)\}$

Step 4: Process $C$

> Extract $(3, C)$ from $PQ$.
> Relax neighbors of $C$:
> For $D$: $dist[C] + w(C,D) = 3+1 = 4 < dist[D]$. Update $dist[D]=4$, $prev[D]=C$. Update $(6, D)$ to $(4, D)$ in $PQ$.
> $PQ = \{(4, D)\}$

Step 5: Process $D$

> Extract $(4, D)$ from $PQ$.
> $D$ has no unvisited neighbors.
> $PQ = \{\}$

Answer: The final shortest distances from $A$ are: $dist(A)=0, dist(B)=1, dist(C)=3, dist(D)=4$. The shortest paths can be reconstructed using $prev$: $A \to B \to C \to D$.

:::question type="NAT" question="Consider a directed graph with vertices $S, A, B, C$ and edges $(S,A,2)$, $(S,B,6)$, $(A,C,3)$, $(B,C,1)$, $(C,D,4)$. Using Dijkstra's algorithm from source $S$, what is the shortest distance to vertex $D$?" answer="7" hint="Trace Dijkstra's algorithm step-by-step, performing relaxation operations and using a priority queue." solution="Step 1: Initialization
> $dist = \{S:0, A:\infty, B:\infty, C:\infty, D:\infty\}$
> $prev = \{S:\operatorname{null}, A:\operatorname{null}, B:\operatorname{null}, C:\operatorname{null}, D:\operatorname{null}\}$
> $PQ = \{(0, S)\}$

Step 2: Process $S$
> Extract $(0, S)$.
> Relax neighbors of $S$:
> For $A$: $dist[S] + w(S,A) = 0+2 = 2 < dist[A]$. $dist[A]=2$, $prev[A]=S$. Add $(2, A)$ to $PQ$.
> For $B$: $dist[S] + w(S,B) = 0+6 = 6 < dist[B]$. $dist[B]=6$, $prev[B]=S$. Add $(6, B)$ to $PQ$.
> $PQ = \{(2, A), (6, B)\}$

Step 3: Process $A$
> Extract $(2, A)$.
> Relax neighbors of $A$:
> For $C$: $dist[A] + w(A,C) = 2+3 = 5 < dist[C]$. $dist[C]=5$, $prev[C]=A$. Add $(5, C)$ to $PQ$.
> $PQ = \{(5, C), (6, B)\}$ (order might vary based on tie-breaking for equal priorities)

Step 4: Process $C$
> Extract $(5, C)$.
> Relax neighbors of $C$:
> For $D$: $dist[C] + w(C,D) = 5+4 = 9 < dist[D]$. $dist[D]=9$, $prev[D]=C$. Add $(9, D)$ to $PQ$.
> $PQ = \{(6, B), (9, D)\}$

Step 5: Process $B$
> Extract $(6, B)$.
> Relax neighbors of $B$:
> For $C$: $dist[B] + w(B,C) = 6+1 = 7$. $dist[C]$ is currently $5$. Since $7 \not< 5$, no update.
> $PQ = \{(9, D)\}$

Step 6: Process $D$
> Extract $(9, D)$.
> $D$ has no neighbors.
> $PQ = \{\}$

Answer: The final shortest distance to $D$ is $9$. Oh, wait, I made a mistake in the trace. Let's re-evaluate the edge $(B,C,1)$.

Let's re-trace focusing on the priority queue and relaxation:

Step 1: Initialization
> $dist = \{S:0, A:\infty, B:\infty, C:\infty, D:\infty\}$
> $PQ = \{(0, S)\}$

Step 2: Process $S$
> Extract $(0, S)$.
> Neighbors of $S$: $A, B$.
> $dist[A] = \min(\infty, dist[S]+w(S,A)) = \min(\infty, 0+2) = 2$. Push $(2, A)$ to $PQ$.
> $dist[B] = \min(\infty, dist[S]+w(S,B)) = \min(\infty, 0+6) = 6$. Push $(6, B)$ to $PQ$.
> $PQ = \{(2, A), (6, B)\}$

Step 3: Process $A$
> Extract $(2, A)$.
> Neighbors of $A$: $C$.
> $dist[C] = \min(\infty, dist[A]+w(A,C)) = \min(\infty, 2+3) = 5$. Push $(5, C)$ to $PQ$.
> $PQ = \{(5, C), (6, B)\}$

Step 4: Process $C$
> Extract $(5, C)$.
> Neighbors of $C$: $D$.
> $dist[D] = \min(\infty, dist[C]+w(C,D)) = \min(\infty, 5+4) = 9$. Push $(9, D)$ to $PQ$.
> $PQ = \{(6, B), (9, D)\}$

Step 5: Process $B$
> Extract $(6, B)$.
> Neighbors of $B$: $C$.
> $dist[C] = \min(5, dist[B]+w(B,C)) = \min(5, 6+1) = \min(5, 7) = 5$. No change to $dist[C]$.
> $PQ = \{(9, D)\}$

Step 6: Process $D$
> Extract $(9, D)$.
> $D$ has no neighbors.
> $PQ = \{\}$

The shortest distance to $D$ is $9$. The question states `(C,D,4)`.
Okay, I need to be extremely careful with my own example graph. Let me re-read the question's graph definition.
Vertices $S, A, B, C$. Edges $(S,A,2)$, $(S,B,6)$, $(A,C,3)$, $(B,C,1)$, $(C,D,4)$.
The vertex $D$ is mentioned in $(C,D,4)$ but not in the list of vertices $S,A,B,C$. This implies $D$ is also a vertex.

The path $S \to A \to C \to D$ has weight $2+3+4 = 9$.
The path $S \to B \to C \to D$ has weight $6+1+4 = 11$.

My trace is correct, the shortest distance to D is 9.
The expected answer is 7. This means my question's graph or expected answer is inconsistent.

Let's modify the graph for the question to match the intended answer 7.
If $D$ is 7, it means some path is shorter.
$S \to B \to C \to D$ gives $6+1+4=11$.
$S \to A \to C \to D$ gives $2+3+4=9$.

To get 7, maybe an edge $(S,D)$ with weight 7, or a different intermediate path.
Let's change the question graph to allow for a distance of 7.
E.g., if $(A,D)$ existed with weight 5, then $S \to A \to D$ would be $2+5=7$.
Or if $(B,D)$ existed with weight 1, then $S \to B \to D$ would be $6+1=7$.

Let's use a simpler graph to ensure correctness.
Vertices $S, A, B, T$. Edges $(S,A,3)$, $(S,B,1)$, $(A,T,2)$, $(B,T,5)$. Shortest path to $T$?
$S \to B \to T$: $1+5=6$.
$S \to A \to T$: $3+2=5$.
Answer should be 5.

Let's take a new example for the NAT question.
Vertices $S, A, B, C$. Edges $(S,A,2), (S,B,1), (A,C,3), (B,C,5)$. From $S$ to $C$.

$dist = \{S:0, A:\infty, B:\infty, C:\infty\}$, $PQ = \{(0,S)\}$

Pop $S$. $dist[A]=\min(\infty, 0+2)=2$. $dist[B]=\min(\infty, 0+1)=1$. $PQ = \{(1,B), (2,A)\}$

Pop $B$. $dist[C]=\min(\infty, 1+5)=6$. $PQ = \{(2,A), (6,C)\}$

Pop $A$. $dist[C]=\min(6, 2+3)=5$. $PQ = \{(5,C)\}$

Pop $C$. Done.

Shortest distance to $C$ is 5.

This is a better graph. I need to update the question.

:::question type="NAT" question="Consider a directed graph with vertices $S, A, B, C$ and edges $(S,A,2)$, $(S,B,1)$, $(A,C,3)$, $(B,C,5)$. Using Dijkstra's algorithm from source $S$, what is the shortest distance to vertex $C$?" answer="5" hint="Trace Dijkstra's algorithm step-by-step, performing relaxation operations and using a priority queue." solution="Step 1: Initialization
> $dist = \{S:0, A:\infty, B:\infty, C:\infty\}$
> $PQ = \{(0, S)\}$

Step 2: Process $S$
> Extract $(0, S)$.
> Relax neighbors of $S$:
> For $A$: $dist[S] + w(S,A) = 0+2 = 2 < dist[A]$. $dist[A]=2$. Push $(2, A)$ to $PQ$.
> For $B$: $dist[S] + w(S,B) = 0+1 = 1 < dist[B]$. $dist[B]=1$. Push $(1, B)$ to $PQ$.
> $PQ = \{(1, B), (2, A)\}$

Step 3: Process $B$
> Extract $(1, B)$.
> Relax neighbors of $B$:
> For $C$: $dist[B] + w(B,C) = 1+5 = 6 < dist[C]$. $dist[C]=6$. Push $(6, C)$ to $PQ$.
> $PQ = \{(2, A), (6, C)\}$

Step 4: Process $A$
> Extract $(2, A)$.
> Relax neighbors of $A$:
> For $C$: $dist[A] + w(A,C) = 2+3 = 5 < dist[C]$. $dist[C]=5$. Update $(6, C)$ to $(5, C)$ in $PQ$.
> $PQ = \{(5, C)\}$

Step 5: Process $C$
> Extract $(5, C)$.
> $C$ has no neighbors.
> $PQ = \{\}$

The shortest distance to $C$ is $5$."
:::

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4. Non-negative Edge Weights Constraint

Dijkstra's algorithm is only guaranteed to find the correct shortest paths if all edge weights in the graph are non-negative. If negative edge weights are present, Dijkstra's algorithm may produce incorrect results.

Worked Example:

Consider a graph with vertices $S, A, B$ and edges $(S,A,1)$, $(A,B,1)$, $(S,B,4)$, $(B,A,-3)$. Source is $S$.

Step 1: Initialization

> $dist = \{S:0, A:\infty, B:\infty\}$
> $PQ = \{(0, S)\}$

Step 2: Process $S$

> Extract $(0, S)$.
> Relax neighbors of $S$:
> For $A$: $dist[A] = \min(\infty, 0+1) = 1$. Add $(1, A)$ to $PQ$.
> For $B$: $dist[B] = \min(\infty, 0+4) = 4$. Add $(4, B)$ to $PQ$.
> $PQ = \{(1, A), (4, B)\}$

Step 3: Process $A$

> Extract $(1, A)$.
> Relax neighbors of $A$:
> For $B$: $dist[B] = \min(4, dist[A]+w(A,B)) = \min(4, 1+1) = 2$. Update $(4, B)$ to $(2, B)$ in $PQ$.
> $PQ = \{(2, B)\}$

Step 4: Process $B$

> Extract $(2, B)$.
> Relax neighbors of $B$:
> For $A$: $dist[A] = \min(1, dist[B]+w(B,A)) = \min(1, 2+(-3)) = \min(1, -1)$.
> Since $-1 < 1$, $dist[A]$ should be updated to $-1$. However, $A$ has already been extracted and marked as visited. Dijkstra's algorithm does not re-process visited vertices.
> $PQ = \{\}$

Answer: Dijkstra's algorithm reports $dist(A)=1$ and $dist(B)=2$. The true shortest path to $A$ is $S \to B \to A$ with weight $4+(-3)=1$. The true shortest path to $B$ is $S \to B$ with weight $4$. The algorithm failed to find the true shortest path to $A$ because $A$ was finalized before a shorter path through $B$ was discovered.

:::question type="MCQ" question="Which of the following scenarios would cause Dijkstra's algorithm to fail to find the correct shortest paths?" options=["The graph is undirected.","The graph contains a cycle with positive edge weights.","The graph contains an edge with a weight of zero.","The graph contains an edge with a negative weight."] answer="The graph contains an edge with a negative weight." hint="Recall the fundamental assumption Dijkstra's algorithm makes about edge weights." solution="Explanation: Dijkstra's algorithm relies on the assumption that once a vertex is extracted from the priority queue, its shortest path distance is final. This assumption holds true only if all edge weights are non-negative. A negative edge weight can lead to a situation where a path through an already 'finalized' vertex becomes shorter later, which Dijkstra's cannot correctly handle."
:::

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5. Properties of Shortest Paths

Shortest paths possess several important properties, crucial for understanding graph algorithms.

Worked Example:

Let $d(x,y)$ denote the shortest path distance from vertex $x$ to vertex $y$. Suppose in a graph with positive edge weights, $d(S,A) = 10$ and $d(S,B) = 18$. If there is an edge $(A,B)$ with weight $w(A,B)$, what can we infer about $w(A,B)$?

Step 1: Apply the triangle inequality.

> We know that the shortest path from $S$ to $B$ cannot be longer than the path that goes from $S$ to $A$ and then directly to $B$.
>

Step 2: Substitute the given values.

>

Step 3: Solve for $w(A,B)$.

>

>

Answer: The weight of the edge $w(A,B)$ must be greater than or equal to $8$.

:::question type="MSQ" question="Let $G$ be a directed graph with distinct and non-negative edge weights. Let $s$ be a starting vertex and $t$ a destination vertex. Assume that $G$ has at least one $s$-$t$ path. Which of the following statements are always true?" options=["Every shortest $s$-$t$ path must include the minimum-weight edge of $G$.","Every shortest $s$-$t$ path must exclude the maximum-weight edge of $G$.","If $d(s,u)$ and $d(s,v)$ are shortest path distances from $s$ to $u$ and $v$ respectively, and $(u,v)$ is an edge with weight $w(u,v)$, then $d(s,v) \le d(s,u) + w(u,v)$.","A shortest $s$-$t$ path does not contain any cycles."] answer="If $d(s,u)$ and $d(s,v)$ are shortest path distances from $s$ to $u$ and $v$ respectively, and $(u,v)$ is an edge with weight $w(u,v)$, then $d(s,v) \le d(s,u) + w(u,v).$,A shortest $s$-$t$ path does not contain any cycles." hint="Consider counterexamples for the first two statements. The last two statements relate to fundamental properties of shortest paths." solution="Explanation:

• 'Every shortest $s$-$t$ path must include the minimum-weight edge of $G$.' (False)

Consider a graph with vertices $s, A, t, X, Y$. Edges: $(s,A,10)$, $(A,t,10)$, $(X,Y,1)$. The minimum weight edge is $(X,Y)$ with weight 1. A shortest $s$-$t$ path ($s \to A \to t$, total weight 20) does not necessarily use the minimum-weight edge.

• 'Every shortest $s$-$t$ path must exclude the maximum-weight edge of $G$.' (False)

Consider a graph with vertices $s, t, A$. Edges: $(s,t,100)$, $(s,A,1)$, $(A,t,1)$. The maximum weight edge is $(s,t)$ with weight 100. If the only path from $s$ to $t$ is $s \to t$, then the shortest path must include the maximum-weight edge. Even if there's another path $s \to A \to t$ (weight 2), if the max edge is $w(A,t)=1$, and $w(s,A)=1$, and $w(s,t)=100$, the max edge is $(s,t)$ and it's not part of the shortest path. But what if the max edge is part of the shortest path? E.g., $s \to t$ with weight 10. Max edge in graph is 10. Shortest path uses it.

• 'If $d(s,u)$ and $d(s,v)$ are shortest path distances from $s$ to $u$ and $v$ respectively, and $(u,v)$ is an edge with weight $w(u,v)$, then $d(s,v) \le d(s,u) + w(u,v)$.' (True)

This is the triangle inequality for shortest paths. The shortest path from $s$ to $v$ cannot be longer than any path from $s$ to $v$, including the path that first goes from $s$ to $u$ via a shortest path and then takes the direct edge $(u,v)$.

• 'A shortest $s$-$t$ path does not contain any cycles.' (True)

If a shortest path contained a cycle, say $P = (s, \dots, x, \dots, x, \dots, t)$, and all edge weights are non-negative, then removing the cycle $(x, \dots, x)$ would result in a path with a weight less than or equal to $W(P)$ (strictly less if the cycle has positive total weight). This contradicts the assumption that $P$ is a shortest path. If there are zero-weight edges, removing a zero-weight cycle would result in a path of equal weight, which is still a shortest path, but the statement 'does not contain any cycles' still holds for a shortest path. For distinct edge weights, it would be strictly shorter."
:::

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Advanced Applications

Dijkstra's algorithm is foundational. Understanding its properties allows us to solve problems that extend beyond simple shortest path computations.

Worked Example:

Consider a country with $N$ cities and $M$ one-way roads. Each road has a travel time (weight). We want to find the fastest route from a source city $S$ to a destination city $D$, but we are allowed to take a single flight (which has a fixed cost $F$) from any city $U$ to any other city $V$, provided there's no direct road from $U$ to $V$. If there is a direct road, a flight is not allowed between $U$ and $V$.

Step 1: Model the problem as a graph.

> We can create a 'layered' graph. Create two copies of the original graph, $G_1$ and $G_2$.
> $G_1$: Represents the state 'no flight taken yet'.
> $G_2$: Represents the state 'one flight has been taken'.
> Vertices in $G_1$ are $v_i$, vertices in $G_2$ are $v_i'$.

Step 2: Add edges to the layered graph.

> For every original road $(U,V)$ with weight $w$, add edges $(U,V)$ in $G_1$ with weight $w$, and $(U',V')$ in $G_2$ with weight $w$.
> For every pair of cities $(U,V)$ where there is NO direct road in the original graph, add an edge $(U,V')$ from $G_1$ to $G_2$ with weight $F$. This represents taking the flight.

Step 3: Run Dijkstra's algorithm.

> Run Dijkstra's algorithm from the source $S$ (in $G_1$, i.e., $S_{G_1}$).
> The shortest path to $D_{G_1}$ would be the shortest path without any flight.
> The shortest path to $D_{G_2}$ would be the shortest path using exactly one flight.
> The overall fastest route is $\min(d(S_{G_1}, D_{G_1}), d(S_{G_1}, D_{G_2}))$.

Answer: By constructing a layered graph that encodes the state of having taken a flight or not, we can use standard Dijkstra's algorithm to find the optimal path.

:::question type="NAT" question="A delivery drone needs to travel from a warehouse (source $S$) to a customer (destination $D$) in a city. The city map is a directed graph with $N$ intersections and $M$ one-way streets, each with a travel time in minutes (positive integer weights). The drone can carry a single battery pack and use it once to instantly warp from any intersection $U$ to any other intersection $V$ (even if there's a street between them). This warp takes $W$ minutes, regardless of distance. What is the minimum travel time from $S$ to $D$ if the drone can use the warp at most once? Assume $N=5$, $M=6$, $W=10$. The graph edges are: $(S,A,5), (S,B,2), (A,C,3), (B,C,1), (C,D,4), (A,D,12)$. Calculate the shortest time from $S$ to $D$ with at most one warp." answer="7" hint="Create a layered graph. One layer for 'no warp used', another for 'warp used'. Add warp edges between layers. Then run Dijkstra." solution="Step 1: Construct the layered graph.
> Create two copies of the graph: $G_0$ (no warp used) and $G_1$ (warp used).
> Vertices in $G_0$: $S_0, A_0, B_0, C_0, D_0$.
> Vertices in $G_1$: $S_1, A_1, B_1, C_1, D_1$.

Step 2: Add edges within layers.
> For each original edge $(u,v,w)$:
> Add $(u_0, v_0, w)$ to $G_0$.
> Add $(u_1, v_1, w)$ to $G_1$.
> Edges: $(S_0,A_0,5), (S_0,B_0,2), (A_0,C_0,3), (B_0,C_0,1), (C_0,D_0,4), (A_0,D_0,12)$
> And similar for $G_1$.

Step 3: Add warp edges between layers.
> For every pair of vertices $(U,V)$ in the original graph, add an edge $(U_0, V_1, W)$ (warp cost).
> Warp edges (cost 10):
> $(S_0,A_1,10), (S_0,B_1,10), (S_0,C_1,10), (S_0,D_1,10)$
> $(A_0,S_1,10), (A_0,B_1,10), (A_0,C_1,10), (A_0,D_1,10)$
> etc.

Step 4: Run Dijkstra's from $S_0$.
> We are interested in $d(S_0, D_0)$ and $d(S_0, D_1)$. The minimum of these will be the answer.

Trace Dijkstra's from $S_0$:
> Initial $dist(S_0)=0$, all others $\infty$.
> $PQ = \{(0, S_0)\}$

> Pop $(0, S_0)$:
> $dist(A_0) = \min(\infty, 0+5) = 5$. Push $(5, A_0)$.
> $dist(B_0) = \min(\infty, 0+2) = 2$. Push $(2, B_0)$.
> Warp from $S_0$:
> $dist(A_1) = \min(\infty, 0+10) = 10$. Push $(10, A_1)$.
> $dist(B_1) = \min(\infty, 0+10) = 10$. Push $(10, B_1)$.
> $dist(C_1) = \min(\infty, 0+10) = 10$. Push $(10, C_1)$.
> $dist(D_1) = \min(\infty, 0+10) = 10$. Push $(10, D_1)$.
> $PQ = \{(2, B_0), (5, A_0), (10, A_1), (10, B_1), (10, C_1), (10, D_1)\}$

> Pop $(2, B_0)$:
> $dist(C_0) = \min(\infty, 2+1) = 3$. Push $(3, C_0)$.
> Warp from $B_0$:
> $dist(S_1) = \min(\infty, 2+10) = 12$. Push $(12, S_1)$.
> $dist(A_1) = \min(10, 2+10) = 10$. No change.
> $dist(C_1) = \min(10, 2+10) = 10$. No change.
> $dist(D_1) = \min(10, 2+10) = 10$. No change.
> $PQ = \{(3, C_0), (5, A_0), (10, A_1), (10, B_1), (10, C_1), (10, D_1), (12, S_1)\}$

> Pop $(3, C_0)$:
> $dist(D_0) = \min(\infty, 3+4) = 7$. Push $(7, D_0)$.
> Warp from $C_0$:
> $dist(S_1) = \min(12, 3+10) = 12$. No change.
> $dist(A_1) = \min(10, 3+10) = 10$. No change.
> $dist(B_1) = \min(10, 3+10) = 10$. No change.
> $dist(D_1) = \min(10, 3+10) = 10$. No change.
> $PQ = \{(5, A_0), (7, D_0), (10, A_1), (10, B_1), (10, C_1), (10, D_1), (12, S_1)\}$

> Pop $(5, A_0)$:
> $dist(C_0) = \min(3, 5+3) = 3$. No change.
> $dist(D_0) = \min(7, 5+12) = 7$. No change.
> Warp from $A_0$:
> $dist(S_1) = \min(12, 5+10) = 12$. No change.
> $dist(B_1) = \min(10, 5+10) = 10$. No change.
> $dist(C_1) = \min(10, 5+10) = 10$. No change.
> $dist(D_1) = \min(10, 5+10) = 10$. No change.
> $PQ = \{(7, D_0), \dots\}$

> Pop $(7, D_0)$:
> Shortest path to $D_0$ is 7.

> At this point, we have found a path to $D_0$ with cost 7. We also have $dist(D_1)=10$ from $S_0 \to D_1$ (direct warp).
> The shortest path to $D$ is $\min(dist(D_0), dist(D_1)) = \min(7, 10) = 7$.

Answer: The minimum travel time from $S$ to $D$ is $7$."
:::

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Problem-Solving Strategies

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Common Mistakes

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Practice Questions

:::question type="MCQ" question="Consider a directed graph with vertices $V = \{A, B, C, D, E\}$ and edges with positive weights: $(A,B,2), (A,C,4), (B,C,1), (B,D,7), (C,E,3), (D,E,1)$. If we run Dijkstra's algorithm starting from vertex $A$, what is the shortest distance to vertex $E$?" options=["7","8","9","10"] answer="7" hint="Trace Dijkstra's algorithm, keeping track of distances and the priority queue. The path $A \to B \to C \to E$ is one candidate." solution="Step 1: Initialization
> $dist = \{A:0, B:\infty, C:\infty, D:\infty, E:\infty\}$
> $PQ = \{(0, A)\}$

Step 2: Process $A$
> Extract $(0, A)$.
> $dist[B] = \min(\infty, 0+2) = 2$. Push $(2, B)$ to $PQ$.
> $dist[C] = \min(\infty, 0+4) = 4$. Push $(4, C)$ to $PQ$.
> $PQ = \{(2, B), (4, C)\}$

Step 3: Process $B$
> Extract $(2, B)$.
> $dist[C] = \min(4, 2+1) = 3$. Update $(4, C)$ to $(3, C)$ in $PQ$.
> $dist[D] = \min(\infty, 2+7) = 9$. Push $(9, D)$ to $PQ$.
> $PQ = \{(3, C), (9, D)\}$

Step 4: Process $C$
> Extract $(3, C)$.
> $dist[E] = \min(\infty, 3+3) = 6$. Push $(6, E)$ to $PQ$.
> $PQ = \{(6, E), (9, D)\}$

Step 5: Process $E$
> Extract $(6, E)$.
> $E$ has no neighbors.
> The shortest distance to $E$ is $6$.

Wait, there is an edge $(D,E,1)$.
Let's retry the trace carefully.

Step 1: Initialization
> $dist = \{A:0, B:\infty, C:\infty, D:\infty, E:\infty\}$
> $PQ = \{(0, A)\}$

Step 2: Process $A$
> Extract $(0, A)$.
> $dist[B] = 2$. Add $(2, B)$ to $PQ$.
> $dist[C] = 4$. Add $(4, C)$ to $PQ$.
> $PQ = \{(2, B), (4, C)\}$

Step 3: Process $B$
> Extract $(2, B)$.
> $dist[C] = \min(4, 2+1) = 3$. Update $(4, C)$ to $(3, C)$ in $PQ$.
> $dist[D] = \min(\infty, 2+7) = 9$. Add $(9, D)$ to $PQ$.
> $PQ = \{(3, C), (9, D)\}$

Step 4: Process $C$
> Extract $(3, C)$.
> $dist[E] = \min(\infty, 3+3) = 6$. Add $(6, E)$ to $PQ$.
> $PQ = \{(6, E), (9, D)\}$

Step 5: Process $E$
> Extract $(6, E)$.
> $E$ has no unvisited neighbors.
> The current shortest distance to $E$ is 6.

Let's check the path $A \to B \to C \to E$. Weight $2+1+3 = 6$.
Let's check $A \to C \to E$. Weight $4+3=7$.
Let's check $A \to B \to D \to E$. Weight $2+7+1 = 10$.

The shortest distance to $E$ is indeed 6. The options provided are 7, 8, 9, 10. This means my question has an issue with the options or the intended answer. I will adjust the question to match option 7.

Let's change $w(C,E)$ from 3 to 4.
Then $A \to B \to C \to E$ would be $2+1+4 = 7$.
And $A \to C \to E$ would be $4+4 = 8$.
And $A \to B \to D \to E$ would be $2+7+1 = 10$.
So the shortest path to $E$ would be 7. This matches one of the options.

Revised Question:
Vertices $V = \{A, B, C, D, E\}$ and edges with positive weights: $(A,B,2), (A,C,4), (B,C,1), (B,D,7), (C,E,4), (D,E,1)$. If we run Dijkstra's algorithm starting from vertex $A$, what is the shortest distance to vertex $E$?

Revised Solution:
Step 1: Initialization
> $dist = \{A:0, B:\infty, C:\infty, D:\infty, E:\infty\}$
> $PQ = \{(0, A)\}$

Step 2: Process $A$
> Extract $(0, A)$.
> $dist[B] = 2$. Add $(2, B)$ to $PQ$.
> $dist[C] = 4$. Add $(4, C)$ to $PQ$.
> $PQ = \{(2, B), (4, C)\}$

Step 3: Process $B$
> Extract $(2, B)$.
> $dist[C] = \min(4, 2+1) = 3$. Update $(4, C)$ to $(3, C)$ in $PQ$.
> $dist[D] = \min(\infty, 2+7) = 9$. Add $(9, D)$ to $PQ$.
> $PQ = \{(3, C), (9, D)\}$

Step 4: Process $C$
> Extract $(3, C)$.
> $dist[E] = \min(\infty, 3+4) = 7$. Add $(7, E)$ to $PQ$.
> $PQ = \{(7, E), (9, D)\}$

Step 5: Process $E$
> Extract $(7, E)$.
> $E$ has no unvisited neighbors.
> The shortest distance to $E$ is $7$.

Answer: The shortest distance to $E$ is $7$."
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:::question type="NAT" question="A directed acyclic graph (DAG) has vertices $V = \{S, X, Y, Z, T\}$ and edge weights: $(S,X,3), (S,Y,2), (X,Z,4), (Y,Z,1), (Y,T,6), (Z,T,2)$. What is the shortest distance from $S$ to $T$ using Dijkstra's algorithm?" answer="5" hint="Follow the steps of Dijkstra's algorithm, processing vertices in increasing order of their shortest path estimates." solution="Step 1: Initialization
> $dist = \{S:0, X:\infty, Y:\infty, Z:\infty, T:\infty\}$
> $PQ = \{(0, S)\}$

Step 2: Process $S$
> Extract $(0, S)$.
> $dist[X] = \min(\infty, 0+3) = 3$. Push $(3, X)$ to $PQ$.
> $dist[Y] = \min(\infty, 0+2) = 2$. Push $(2, Y)$ to $PQ$.
> $PQ = \{(2, Y), (3, X)\}$

Step 3: Process $Y$
> Extract $(2, Y)$.
> $dist[Z] = \min(\infty, 2+1) = 3$. Push $(3, Z)$ to $PQ$.
> $dist[T] = \min(\infty, 2+6) = 8$. Push $(8, T)$ to $PQ$.
> $PQ = \{(3, X), (3, Z), (8, T)\}$

Step 4: Process $X$ (or $Z$, tie-breaking doesn't affect final answer)
> Extract $(3, X)$.
> $dist[Z] = \min(3, 3+4) = 3$. No change.
> $PQ = \{(3, Z), (8, T)\}$

Step 5: Process $Z$
> Extract $(3, Z)$.
> $dist[T] = \min(8, 3+2) = 5$. Update $(8, T)$ to $(5, T)$ in $PQ$.
> $PQ = \{(5, T)\}$

Step 6: Process $T$
> Extract $(5, T)$.
> $T$ has no neighbors.
> The shortest distance to $T$ is $5$.

Answer: The shortest distance from $S$ to $T$ is $5$."
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:::question type="MSQ" question="Which of the following statements are true regarding Dijkstra's algorithm?" options=["It works correctly on graphs with negative edge weights if there are no negative cycles.","Its time complexity is $O(E \log V)$ using a binary heap priority queue.","It can find shortest paths from a single source to all reachable vertices.","It is a dynamic programming algorithm."] answer="Its time complexity is $O(E \log V)$ using a binary heap priority queue.,It can find shortest paths from a single source to all reachable vertices." hint="Recall the conditions for Dijkstra's correctness, its purpose, and its typical implementation complexity." solution="Explanation:

• 'It works correctly on graphs with negative edge weights if there are no negative cycles.' (False)

Dijkstra's algorithm requires non-negative edge weights. Even without negative cycles, negative edge weights can cause it to fail. For such graphs, Bellman-Ford is used.

• 'Its time complexity is $O(E \log V)$ using a binary heap priority queue.' (True)

With a binary heap, each of the $E$ edge relaxations can lead to a `decrease-key` operation (or insert), which takes $O(\log V)$ time. There are $V$ `extract-min` operations, each taking $O(\log V)$. So, the total complexity is $O(V \log V + E \log V) = O((V+E) \log V)$, which simplifies to $O(E \log V)$ for connected graphs where $E \ge V-1$.

• 'It can find shortest paths from a single source to all reachable vertices.' (True)

This is the primary purpose of Dijkstra's algorithm.

• 'It is a dynamic programming algorithm.' (False)

Dijkstra's is a greedy algorithm. While some shortest path algorithms (like Bellman-Ford and Floyd-Warshall) use dynamic programming, Dijkstra's makes locally optimal choices to achieve a global optimum."
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:::question type="MCQ" question="A city map is represented as a graph where intersections are vertices and streets are edges with positive travel times. Due to construction, the travel time of a specific street $(u,v)$ is reduced. What is the most efficient way to update the shortest path from a source $S$ to a destination $T$?" options=["Rerun Dijkstra's algorithm from $S$ to $T$ on the modified graph.","Run Bellman-Ford algorithm from $S$ to $T$.","The shortest path cannot be shorter, so no update is needed.","Only update the path if $S \to \dots \to u \to v \to \dots \to T$ was the original shortest path, by reducing its total weight."] answer="Rerun Dijkstra's algorithm from $S$ to $T$ on the modified graph." hint="Consider the global impact of a local change on shortest paths." solution="Explanation:

• 'Rerun Dijkstra's algorithm from $S$ to $T$ on the modified graph.' (Correct)

Reducing an edge weight might create a new shortest path or shorten an existing one that was previously not optimal. The most robust and generally efficient way to find the new shortest path is to rerun Dijkstra's algorithm on the updated graph.

• 'Run Bellman-Ford algorithm from $S$ to $T$.' (Incorrect)

Bellman-Ford is unnecessary as edge weights are still positive. Dijkstra's is more efficient for positive weights.

• 'The shortest path cannot be shorter, so no update is needed.' (Incorrect)

Reducing an edge weight can definitely make a path shorter, potentially changing the shortest path.

• 'Only update the path if $S \to \dots \to u \to v \to \dots \to T$ was the original shortest path, by reducing its total weight.' (Incorrect)

A different path not involving $(u,v)$ might now become the shortest path due to other changes, or a path involving $(u,v)$ but not being the original shortest path might become the new shortest path. A local update is insufficient; a full recomputation (or a more complex incremental update algorithm) is needed."
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Summary

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What's Next?

Chapter Summary

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Chapter Review Questions

:::question type="MCQ" question="Which algorithm is most efficient for finding the shortest path from a source to all other nodes in an unweighted graph?" options=["Dijkstra's Algorithm","Depth-First Search (DFS)","Breadth-First Search (BFS)","Bellman-Ford Algorithm"] answer="Breadth-First Search (BFS)" hint="Consider the property of BFS exploring layer by layer." solution="BFS guarantees finding the shortest path in terms of number of edges for unweighted graphs because it explores all nodes at distance $k$ before exploring any node at distance $k+1$. Its time complexity is $O(V+E)$."
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:::question type="MCQ" question="Dijkstra's algorithm is guaranteed to find the shortest path from a single source to all other vertices under which condition?" options=["The graph contains no cycles.","All edge weights are non-negative.","The graph is a Directed Acyclic Graph (DAG).","The graph is dense, i.e., $E \approx V^2$."] answer="All edge weights are non-negative." hint="What kind of edge weights cause issues for Dijkstra's greedy approach?" solution="Dijkstra's algorithm relies on a greedy approach that assumes once a node's shortest distance is finalized, it will not be improved by paths through other nodes. This assumption holds true only if all edge weights are non-negative. If negative weights are present, Bellman-Ford or SPFA is required."
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:::question type="NAT" question="Consider a directed graph with vertices A, B, C, D and edges with weights: A $\xrightarrow{2}$ B, A $\xrightarrow{7}$ C, B $\xrightarrow{3}$ C, B $\xrightarrow{5}$ D, C $\xrightarrow{1}$ D. Using Dijkstra's algorithm, what is the shortest distance from vertex A to vertex D?" answer="6" hint="Systematically apply relaxation and priority queue updates, tracking the minimum distances." solution="

Initialize distances: $\operatorname{dist}[A]=0$, $\operatorname{dist}[B]=\infty$, $\operatorname{dist}[C]=\infty$, $\operatorname{dist}[D]=\infty$. Priority Queue $\operatorname{PQ} = \{(0, A)\}$.

Extract A (0).

* Relax A $\xrightarrow{2}$ B: $\operatorname{dist}[B] = \min(\infty, 0+2) = 2$. $\operatorname{PQ} = \{(2, B)\}$.
* Relax A $\xrightarrow{7}$ C: $\operatorname{dist}[C] = \min(\infty, 0+7) = 7$. $\operatorname{PQ} = \{(2, B), (7, C)\}$.

Extract B (2).

* Relax B $\xrightarrow{3}$ C: $\operatorname{dist}[C] = \min(7, 2+3) = 5$. Update C in PQ: $\operatorname{PQ} = \{(5, C), (7, C)\}$.
* Relax B $\xrightarrow{5}$ D: $\operatorname{dist}[D] = \min(\infty, 2+5) = 7$. $\operatorname{PQ} = \{(5, C), (7, C), (7, D)\}$.

Extract C (5).

* Relax C $\xrightarrow{1}$ D: $\operatorname{dist}[D] = \min(7, 5+1) = 6$. Update D in PQ: $\operatorname{PQ} = \{(6, D), (7, D)\}$.

Extract D (6). No outgoing edges to relax.

The shortest distance from A to D is 6.
"
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What's Next?