Bipartite Graph Check in DSA Typescript - Execution Trace

Concept Flow - Bipartite Graph Check

Start with uncolored graph

↓

Pick an unvisited node

↓

Assign color 0

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Use BFS/DFS to visit neighbors

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If neighbor uncolored, assign opposite color

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If neighbor colored same as current, stop: Not Bipartite

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Repeat for all nodes

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If no conflicts, graph is Bipartite

We color nodes using two colors while traversing. If a conflict arises, graph is not bipartite.

Execution Sample

DSA Typescript

function isBipartite(graph: number[][]): boolean {
  const color = new Array(graph.length).fill(-1);
  for (let i = 0; i < graph.length; i++) {
    if (color[i] === -1) {
      if (!bfsCheck(i, graph, color)) return false;
    }
  }
  return true;
}

function bfsCheck(start: number, graph: number[][], color: number[]): boolean {
  const queue: number[] = [start];
  color[start] = 0;
  while (queue.length > 0) {
    const node = queue.shift()!;
    for (const neighbor of graph[node]) {
      if (color[neighbor] === -1) {
        color[neighbor] = 1 - color[node];
        queue.push(neighbor);
      } else if (color[neighbor] === color[node]) {
        return false;
      }
    }
  }
  return true;
}

Checks if graph is bipartite by coloring nodes using BFS.

Execution Table

Step	Operation	Current Node	Neighbor	Color Array	Conflict Detected	Visual State
1	Start BFS from node 0	0	-	[-1, -1, -1, -1]	No	[0(-), 1(-), 2(-), 3(-)]
2	Color node 0 with 0	0	-	[0, -1, -1, -1]	No	[0(0), 1(-), 2(-), 3(-)]
3	Visit neighbor 1 of node 0	0	1	[0, -1, -1, -1]	No	[0(0), 1(-), 2(-), 3(-)]
4	Color neighbor 1 with 1	1	-	[0, 1, -1, -1]	No	[0(0), 1(1), 2(-), 3(-)]
5	Visit neighbor 3 of node 0	0	3	[0, 1, -1, -1]	No	[0(0), 1(1), 2(-), 3(-)]
6	Color neighbor 3 with 1	3	-	[0, 1, -1, 1]	No	[0(0), 1(1), 2(-), 3(1)]
7	Visit neighbor 0 of node 1	1	0	[0, 1, -1, 1]	No	[0(0), 1(1), 2(-), 3(1)]
8	Neighbor 0 already colored 0 (opposite), continue	1	0	[0, 1, -1, 1]	No	[0(0), 1(1), 2(-), 3(1)]
9	Visit neighbor 2 of node 1	1	2	[0, 1, -1, 1]	No	[0(0), 1(1), 2(-), 3(1)]
10	Color neighbor 2 with 0	2	-	[0, 1, 0, 1]	No	[0(0), 1(1), 2(0), 3(1)]
11	Visit neighbor 1 of node 2	2	1	[0, 1, 0, 1]	No	[0(0), 1(1), 2(0), 3(1)]
12	Neighbor 1 already colored 1 (opposite), continue	2	1	[0, 1, 0, 1]	No	[0(0), 1(1), 2(0), 3(1)]
13	Visit neighbor 3 of node 2	2	3	[0, 1, 0, 1]	No	[0(0), 1(1), 2(0), 3(1)]
14	Neighbor 3 already colored 1 (opposite), continue	2	3	[0, 1, 0, 1]	No	[0(0), 1(1), 2(0), 3(1)]
15	Visit neighbor 0 of node 3	3	0	[0, 1, 0, 1]	No	[0(0), 1(1), 2(0), 3(1)]
16	Neighbor 0 already colored 0 (opposite), continue	3	0	[0, 1, 0, 1]	No	[0(0), 1(1), 2(0), 3(1)]
17	Visit neighbor 2 of node 3	3	2	[0, 1, 0, 1]	No	[0(0), 1(1), 2(0), 3(1)]
18	Neighbor 2 already colored 0 (opposite), continue	3	2	[0, 1, 0, 1]	No	[0(0), 1(1), 2(0), 3(1)]
19	All nodes visited, no conflict	-	-	[0, 1, 0, 1]	No	[0(0), 1(1), 2(0), 3(1)]

💡 All nodes colored without conflict, graph is bipartite

Variable Tracker

Variable	Start	After Step 2	After Step 4	After Step 6	After Step 10	Final
color	[-1, -1, -1, -1]	[0, -1, -1, -1]	[0, 1, -1, -1]	[0, 1, -1, 1]	[0, 1, 0, 1]	[0, 1, 0, 1]
currentNode	-	0	1	3	2	-
conflictDetected	No	No	No	No	No	No

Key Moments - 3 Insights

Why do we assign colors only when a node is uncolored?

Because if a node is already colored, it means it was visited before and assigned a color. Reassigning could cause conflicts. See steps 7 and 11 where neighbors already have colors and no reassignment happens.

What happens if a neighbor has the same color as the current node?

This means the graph is not bipartite because two connected nodes share the same color. The algorithm stops immediately. In this example, no such conflict occurs, but if it did, the 'Conflict Detected' column would show 'Yes'.

Why do we start BFS from every unvisited node?

Because the graph can be disconnected. We must check each component separately. In this example, the graph is connected, so BFS starts only once at node 0.

Visual Quiz - 3 Questions

Look at the execution table at Step 4, what is the color assigned to node 1?

A 0

B 1

C -1 (uncolored)

D 2

💡 Hint

Check the 'Color Array' column at Step 4 in the execution table.

At which step does node 2 get its color assigned?

A Step 10

B Step 6

C Step 14

D Step 2

💡 Hint

Look for the step where 'Color neighbor 2 with 0' happens in the 'Operation' column.

If node 3 was colored 0 instead of 1 at Step 6, what would happen?

A No conflict, graph still bipartite

B Conflict detected later, graph not bipartite

C Algorithm would skip node 3

D Color array would remain unchanged

💡 Hint

Refer to the 'Conflict Detected' column and the rule that neighbors must have opposite colors.

Concept Snapshot

Bipartite Graph Check:
- Use two colors (e.g., 0 and 1) to color nodes.
- Traverse graph using BFS or DFS.
- Assign opposite color to neighbors.
- If a neighbor has same color, graph is not bipartite.
- Check all components for disconnected graphs.

Full Transcript

This visualization shows how to check if a graph is bipartite by coloring nodes with two colors during a BFS traversal. We start with all nodes uncolored (-1). We pick an unvisited node, color it 0, then color its neighbors 1, their neighbors 0, and so on. If at any point a neighbor has the same color as the current node, the graph is not bipartite. The execution table tracks each step, the current node, neighbors, color assignments, and conflicts. The variable tracker shows how the color array changes over time. Key moments clarify why colors are assigned only once, what conflicts mean, and why BFS starts at every unvisited node. The quiz tests understanding of color assignments and conflict detection. The snapshot summarizes the process in simple steps.