Bridges in Graph Tarjan's Algorithm in DSA C# - Execution Trace

Concept Flow - Bridges in Graph Tarjan's Algorithm

Start DFS from a node

↓

Assign discovery time & low value

↓

For each neighbor

↓

If not visited

→DFS on neighbor

↓

Update low of current node

↓

Check if low[neighbor

↓

If yes, mark edge as bridge

↓

If visited and neighbor != parent

→Update low of current node

↓

Repeat until all nodes visited

↓

Output all bridges found

The algorithm uses DFS to assign discovery and low times to nodes, updating low values based on neighbors. If a neighbor's low value is greater than the current node's discovery time, the edge is a bridge.

Execution Sample

DSA C#

void DFS(int u, int parent) {
  disc[u] = low[u] = time++;
  foreach (int v in graph[u]) {
    if (disc[v] == -1) {
      DFS(v, u);
      low[u] = Math.Min(low[u], low[v]);
      if (low[v] > disc[u]) bridges.Add((u, v));
    } else if (v != parent) {
      low[u] = Math.Min(low[u], disc[v]);
    }
  }
}

This DFS assigns discovery and low times, updates low values, and detects bridges by comparing low[v] and disc[u].

Execution Table

Step	Operation	Node Visited	Parent	Discovery Time (disc)	Low Value (low)	Bridges Found	Visual State
1	Start DFS	0	-1	disc[0]=0	low[0]=0	[]	Graph: 0--1--2--3--4 Visited: {0}
2	Visit neighbor	1	0	disc[1]=1	low[1]=1	[]	Visited: {0,1}
3	Visit neighbor	2	1	disc[2]=2	low[2]=2	[]	Visited: {0,1,2}
4	Visit neighbor	0	2	disc[0]=0	low[2]=0	[]	Back edge found, update low[2]=0
5	Visit neighbor	3	2	disc[3]=3	low[3]=3	[]	Visited: {0,1,2,3}
6	Visit neighbor	4	3	disc[4]=4	low[4]=4	[]	Visited: {0,1,2,3,4}
7	No unvisited neighbors	4	3	disc[4]=4	low[4]=4	[]	Return to 3
8	Update low[3]	3	2	disc[3]=3	low[3]=3	[(3,4)]	low[4] > disc[3], bridge found (3,4)
9	Return to 2	2	1	disc[2]=2	low[2]=0	[(3,4)]	low[3]=3 updated low[2]=0
10	Return to 1	1	0	disc[1]=1	low[1]=0	[(3,4)]	low[2]=0 updated low[1]=0
11	Return to 0	0	-1	disc[0]=0	low[0]=0	[(3,4)]	low[1]=0 updated low[0]=0
12	Visit neighbor	5	0	disc[5]=5	low[5]=5	[(3,4)]	Visited: {0,1,2,3,4,5}
13	No unvisited neighbors	5	0	disc[5]=5	low[5]=5	[(3,4)]	Return to 0
14	Update low[0]	0	-1	disc[0]=0	low[0]=0	[(3,4), (0,5)]	low[5] > disc[0], bridge found (0,5)
15	DFS complete	-	-	-	-	[(3,4), (0,5)]	All nodes visited, bridges found

💡 All nodes visited, DFS complete, bridges identified where low[v] > disc[u]

Variable Tracker

Variable	Start	After Step 1	After Step 2	After Step 3	After Step 4	After Step 5	After Step 6	After Step 7	After Step 8	After Step 9	After Step 10	After Step 11	After Step 12	After Step 13	After Step 14	Final
disc	[-1,-1,-1,-1,-1,-1]	[0,-1,-1,-1,-1,-1]	[0,1,-1,-1,-1,-1]	[0,1,2,-1,-1,-1]	[0,1,2,-1,-1,-1]	[0,1,2,3,-1,-1]	[0,1,2,3,4,-1]	[0,1,2,3,4,-1]	[0,1,2,3,4,-1]	[0,1,2,3,4,-1]	[0,1,2,3,4,-1]	[0,1,2,3,4,-1]	[0,1,2,3,4,5]	[0,1,2,3,4,5]	[0,1,2,3,4,5]	[0,1,2,3,4,5]
low	[-1,-1,-1,-1,-1,-1]	[0,-1,-1,-1,-1,-1]	[0,1,-1,-1,-1,-1]	[0,1,2,-1,-1,-1]	[0,1,0,-1,-1,-1]	[0,1,0,3,-1,-1]	[0,1,0,3,4,-1]	[0,1,0,3,4,-1]	[0,1,0,3,4,-1]	[0,1,0,3,4,-1]	[0,0,0,3,4,-1]	[0,0,0,3,4,-1]	[0,0,0,3,4,5]	[0,0,0,3,4,5]	[0,0,0,3,4,5]	[0,0,0,3,4,5]
bridges	[]	[]	[]	[]	[]	[]	[]	[]	[(3,4)]	[(3,4)]	[(3,4)]	[(3,4)]	[(3,4)]	[(3,4)]	[(3,4),(0,5)]	[(3,4),(0,5)]

Key Moments - 3 Insights

Why do we update low[u] with disc[v] when v is already visited and not the parent?

Because this means there is a back edge from u to an ancestor v, so low[u] should reflect the earliest discovery time reachable, as shown in step 4 where low[2] is updated to disc[0].

Why is an edge considered a bridge when low[v] > disc[u]?

Because low[v] > disc[u] means that v and its descendants cannot reach u or any ancestor of u without that edge, so removing it disconnects the graph. This is shown in step 8 where edge (3,4) is identified as a bridge.

What happens if the graph has multiple disconnected components?

The DFS should be started from each unvisited node to cover all components, ensuring all bridges are found. This is implied in the concept flow where DFS repeats until all nodes are visited.

Visual Quiz - 3 Questions

Look at the execution table at step 8, which edge is identified as a bridge?

A (3,4)

B (0,5)

C (2,0)

D (1,0)

💡 Hint

Check the 'Bridges Found' column at step 8 in the execution_table.

At which step does the low value of node 2 get updated due to a back edge?

A Step 3

B Step 4

C Step 5

D Step 6

💡 Hint

Look at the 'Low Value (low)' column for node 2 in the execution_table rows.

If the edge (0,5) was not present, how would the bridges list change at the end?

A It would contain (3,4) and (0,5)

B It would be empty

C It would still contain (3,4) only

D It would contain (0,5) only

💡 Hint

Refer to the 'Bridges Found' column at steps 14 and 15 in the execution_table.

Concept Snapshot

Tarjan's Algorithm finds bridges using DFS.
Assign discovery time and low value to each node.
Update low[u] with min of low[v] or disc[v] for neighbors.
If low[v] > disc[u], edge (u,v) is a bridge.
Repeat DFS for all nodes to cover disconnected graphs.

Full Transcript

Tarjan's algorithm for finding bridges in a graph uses a depth-first search (DFS) approach. Each node is assigned a discovery time and a low value, which represents the earliest visited vertex reachable from that node. During DFS, for each neighbor, if it is unvisited, DFS is called recursively. After returning, the low value of the current node is updated based on the neighbor's low value. If the neighbor's low value is greater than the current node's discovery time, the edge between them is a bridge. If the neighbor is already visited and is not the parent, the low value of the current node is updated with the neighbor's discovery time to account for back edges. This process continues until all nodes are visited, and all bridges are identified.

Practice

1. What is the main purpose of Tarjan's algorithm when finding bridges in a graph?

easy

A. To sort nodes in topological order

B. To find the shortest path between two nodes

C. To detect cycles in a graph

D. To find edges that, if removed, increase the number of connected components

Solution

1. Step 1: Understand what a bridge is in a graph

   A bridge is an edge that, when removed, increases the number of connected components in the graph.

2. Step 2: Identify Tarjan's algorithm goal

   Tarjan's algorithm finds such edges efficiently by using depth-first search and tracking discovery and low times.

3. Final Answer:

   To find edges that, if removed, increase the number of connected components -> Option D

4. Quick Check:

   Bridges = edges increasing components when removed [OK]

Hint: Bridges disconnect graph if removed [OK]

Common Mistakes:

• Confusing bridges with cycles
• Thinking it finds shortest paths
• Mixing bridges with articulation points

2. Which of the following is the correct way to declare the arrays used for discovery and low times in C# for Tarjan's algorithm?

easy

A. int disc = new int[n]; int low = new int[n];

B. int[] disc = new int[n]; int[] low = new int[n];

C. int disc[] = new int[n]; int low[] = new int[n];

D. int[] disc, low = new int[n];

Solution

1. Step 1: Recall C# array declaration syntax

   In C#, arrays are declared as type[] name = new type[size];

2. Step 2: Check each option

   int[] disc = new int[n]; int[] low = new int[n]; uses correct syntax for both arrays. int disc[] = new int[n]; int low[] = new int[n]; uses Java-style syntax, invalid in C#. int disc = new int[n]; int low = new int[n]; misses [] for arrays. int[] disc, low = new int[n]; declares only low as array, disc is just a variable.

3. Final Answer:

   int[] disc = new int[n]; int[] low = new int[n]; -> Option B

4. Quick Check:

   C# arrays = type[] name = new type[size] [OK]

Hint: Use type[] name = new type[size] in C# [OK]

Common Mistakes:

• Using Java array syntax in C#
• Forgetting [] in declaration
• Declaring multiple arrays incorrectly

3. Given the following graph edges and Tarjan's algorithm code snippet, what is the output list of bridges?

Edges: (0-1), (1-2), (2-0), (1-3)

// Assume graph is adjacency list
// Tarjan's algorithm finds bridges

Output bridges:

medium

A. [(0, 1)]

B. [(0, 1), (1, 3)]

C. [(1, 3)]

D. []

Solution

1. Step 1: Analyze the graph structure

   The edges (0-1), (1-2), and (2-0) form a cycle. The edge (1-3) connects node 3 to the cycle.

2. Step 2: Identify bridges

   Edges in a cycle are not bridges because removing them does not disconnect the graph. The edge (1-3) is a bridge because removing it disconnects node 3.

3. Final Answer:

   [(1, 3)] -> Option C

4. Quick Check:

   Only edge (1-3) disconnects graph [OK]

Hint: Edges in cycles are not bridges [OK]

Common Mistakes:

• Marking cycle edges as bridges
• Ignoring leaf edges as bridges
• Confusing nodes and edges

4. Identify the bug in this Tarjan's algorithm snippet for finding bridges in C#:

void DFS(int u, int parent) {
  disc[u] = time;
  low[u] = time;
  time++;
  foreach (int v in graph[u]) {
    if (disc[v] == -1) {
      DFS(v, u);
      low[u] = Math.Min(low[u], low[v]);
      if (low[v] > disc[u])
        bridges.Add((u, v));
    } else {
      low[u] = Math.Min(low[u], disc[v]);
    }
  }
}

medium

A. The check 'if (v == parent) continue;' is missing

B. The 'disc[u]' and 'low[u]' are not initialized before DFS

C. The 'time' variable is not incremented correctly

D. The condition 'if (low[v] > disc[u])' should be 'if (low[v] >= disc[u])'

Solution

1. Step 1: Examine the DFS traversal logic

   The code lacks 'if (v == parent) continue;' after foreach. This check prevents processing the parent edge as a back edge, which would incorrectly lower low[u] and miss bridges.

2. Step 2: Rule out other options

   The condition 'if (low[v] > disc[u])' should be 'if (low[v] >= disc[u])': 'low[v] > disc[u]' is correct for bridges (strict >). The 'disc[u]' and 'low[u]' are not initialized before DFS: disc[u] and low[u] are set inside DFS. The 'time' variable is not incremented correctly: time++ follows assignment correctly.

3. Final Answer:

   The check 'if (v == parent) continue;' is missing -> Option A

4. Quick Check:

   Parent check avoids revisiting parent [OK]

Hint: Always skip parent node in DFS to avoid false bridges [OK]

Common Mistakes:

• Forgetting to skip parent node
• Using wrong comparison operator for bridges
• Not initializing discovery times

5. You have a graph with nodes 0 to 5 and edges: (0-1), (1-2), (2-0), (1-3), (3-4), (4-5), (5-3). Using Tarjan's algorithm, which edges are bridges?

hard

A. [(1, 3)]

B. [(1, 3), (3, 4)]

C. [(1, 3), (4, 5)]

D. []

Solution

1. Step 1: Analyze graph cycles and connections

   Edges (0-1), (1-2), (2-0) form a cycle. Edges (3-4), (4-5), (5-3) form another cycle. Edge (1-3) connects these two cycles.

2. Step 2: Identify bridges

   Only edge (1-3) is a bridge; removing it disconnects {0,1,2} from {3,4,5}. Cycle edges like (3-4) are not bridges (path 3-5-4 remains), similarly for others.

3. Final Answer:

   [(1, 3)] -> Option A

4. Quick Check:

   Only edge (1-3) disconnects graph [OK]

Hint: Bridges connect cycles or leaf nodes [OK]

Common Mistakes:

• Missing bridges inside connected components
• Confusing cycle edges as bridges
• Ignoring multi-cycle connections