Why Dijkstra's Algorithm Single Source Shortest Path in DSA C?

The Big Idea

Discover how computers find the quickest way home without getting lost!

The Scenario

Imagine you are in a new city and want to find the shortest way to reach every important place from your hotel. You try to remember every street and calculate distances by hand.

The Problem

Doing this manually is slow and confusing. You might forget some streets or make mistakes in adding distances. It is hard to keep track of all possible paths and find the shortest one quickly.

The Solution

Dijkstra's Algorithm helps by automatically checking all paths from your starting point and finding the shortest distance to every other place. It saves time and avoids errors by using a smart step-by-step method.

Before vs After

✗ Before

int distances[5] = {999, 999, 999, 999, 999};
distances[0] = 0;
// Manually update distances for each path and compare

✓ After

dijkstra(graph, 0, distances);
// Algorithm updates distances array with shortest paths automatically

What It Enables

It enables fast and reliable calculation of shortest paths from one point to all others in a network.

Real Life Example

GPS devices use this algorithm to find the quickest route from your current location to many destinations on the map.

Key Takeaways

Manual pathfinding is slow and error-prone.

Dijkstra's Algorithm finds shortest paths efficiently.

It is essential for navigation and network routing.

Practice

1. What is the main purpose of Dijkstra's Algorithm in graph theory?

easy

A. To sort the vertices of a graph topologically

B. To find the minimum spanning tree of a graph

C. To detect cycles in a graph

D. To find the shortest path from a single source to all other vertices

Solution

1. Step 1: Understand the goal of Dijkstra's Algorithm

   Dijkstra's Algorithm is designed to find the shortest distance from one starting point (source) to every other point in a graph with non-negative edge weights.

2. Step 2: Compare with other graph algorithms

   Minimum spanning tree algorithms find a tree connecting all vertices with minimum total edge weight, cycle detection finds loops, and topological sort orders vertices in directed acyclic graphs. These are different goals.

3. Final Answer:

   To find the shortest path from a single source to all other vertices -> Option D

4. Quick Check:

   Dijkstra's Algorithm = shortest paths from source [OK]

Hint: Dijkstra finds shortest paths from one start point [OK]

Common Mistakes:

• Confusing with minimum spanning tree algorithms
• Thinking it detects cycles
• Assuming it sorts vertices

2. Which of the following is the correct way to initialize the distance array in Dijkstra's Algorithm in C?

easy

A. int dist[V]; for(int i=0; i

B. int dist[V] = {INT_MAX};

C. int dist[V] = {0};

D. int dist[V] = NULL;

Solution

1. Step 1: Understand distance initialization

   Distances should start as infinity (or a very large number) except the source vertex, to represent unknown shortest paths.

2. Step 2: Check each option

   int dist[V] = {0}; initializes all distances to 0, which is incorrect. int dist[V]; for(int i=0; i

3. Final Answer:

   int dist[V]; for(int i=0; i -> Option A

4. Quick Check:

   Initialize distances to INT_MAX with loop [OK]

Hint: Use loop to set all distances to INT_MAX [OK]

Common Mistakes:

• Using {INT_MAX} initializes only first element
• Setting all distances to 0
• Assigning NULL to int array

3. What will be the output distances array after running Dijkstra's Algorithm from source vertex 0 on the following graph represented by adjacency matrix?

int graph[4][4] = {
  {0, 1, 4, 0},
  {1, 0, 2, 6},
  {4, 2, 0, 3},
  {0, 6, 3, 0}
};

medium

A. [0, 1, 4, 7]

B. [0, 1, 3, 5]

C. [0, 1, 3, 6]

D. [0, 2, 4, 6]

Solution

1. Step 1: Initialize distances and start from vertex 0

   Distances: dist[0]=0, dist[1]=∞, dist[2]=∞, dist[3]=∞. Update neighbors of 0: dist[1]=1, dist[2]=4.

2. Step 2: Pick vertex with smallest tentative distance and update neighbors

   Pick vertex 1 (dist=1). Update dist[2] = min(4, 1+2=3) = 3, dist[3] = min(∞, 1+6=7) = 7. Next pick vertex 2 (dist=3). Update dist[3] = min(7, 3+3=6) = 6. Finally pick vertex 3 (dist=6).

3. Final Answer:

   [0, 1, 3, 6] -> Option C

4. Quick Check:

   Distances after relaxation = [0,1,3,6] [OK]

Hint: Relax edges stepwise, pick smallest distance vertex next [OK]

Common Mistakes:

• Not updating distances after relaxation
• Choosing wrong next vertex
• Ignoring shorter paths through intermediate vertices

4. Identify the error in the following snippet of Dijkstra's Algorithm in C:

for (int count = 0; count < V-1; count++) {
  int u = minDistance(dist, sptSet);
  sptSet[u] = 1;
  for (int v = 0; v < V; v++) {
    if (!sptSet[v] && graph[u][v] && dist[u] != INT_MAX && dist[u] + graph[u][v] < dist[v]) {
      dist[v] = dist[u] + graph[u][v];
    }
  }
}

medium

A. The loop should run from 1 to V instead of 0 to V-1

B. The condition 'graph[u][v]' should be 'graph[u][v] != 0' to check edge existence

C. The sptSet array should be initialized with 1 instead of 0

D. The dist[u] check should be 'dist[u] == INT_MAX' instead of '!='

Solution

1. Step 1: Analyze edge existence condition

   The condition 'graph[u][v]' relies on implicit truthiness of non-zero values. In C, this works but is less clear. Using 'graph[u][v] != 0' explicitly checks for edge presence and improves readability.

2. Step 2: Check other conditions

   Loop range from 0 to V-1 is correct for V vertices. sptSet should start with 0 (false). The dist[u] check must be '!= INT_MAX' to ensure u is reachable before relaxing edges.

3. Final Answer:

   The condition 'graph[u][v]' should be 'graph[u][v] != 0' to check edge existence -> Option B

4. Quick Check:

   Explicit edge check improves clarity [OK]

Hint: Use explicit '!= 0' to check edges in adjacency matrix [OK]

Common Mistakes:

• Assuming loop range is wrong
• Initializing sptSet incorrectly
• Reversing dist[u] check logic

5. Given a weighted graph with vertices 0 to 5 and edges:

0-1: 7, 0-2: 9, 0-5: 14, 1-2: 10, 1-3: 15, 2-3: 11, 2-5: 2, 3-4: 6, 4-5: 9

Using Dijkstra's Algorithm from source 0, what is the shortest distance to vertex 4?

hard

A. 20

B. 26

C. 21

D. 19

Solution

1. Step 1: Initialize distances and start from vertex 0

   dist[0]=0, others = ∞. Update neighbors: dist[1]=7, dist[2]=9, dist[5]=14.

2. Step 2: Pick vertex with smallest dist and update neighbors

   Pick 1 (7): update dist[3] = min(∞, 7+15=22) = 22. Pick 2 (9): update dist[3] = min(22, 9+11=20) = 20, dist[5] = min(14, 9+2=11) = 11. Pick 5 (11): update dist[4] = min(∞, 11+9=20) = 20. Pick 3 (20): update dist[4] = min(20, 20+6=26) = 20. Pick 4 (20).

3. Final Answer:

   20 -> Option A

4. Quick Check:

   Shortest path to 4 = 20 [OK]

Hint: Relax edges carefully, update shortest distances stepwise [OK]

Common Mistakes:

• Not updating dist[5] to smaller value
• Ignoring shorter path through vertex 2
• Adding edges incorrectly