Floyd Warshall All Pairs Shortest Path in DSA Go - Cheat Sheet & Quick Revision

Recall & Review

beginner

What problem does the Floyd Warshall algorithm solve?

It finds the shortest paths between all pairs of nodes in a weighted graph, including graphs with negative edge weights but no negative cycles.

beginner

What is the main idea behind the Floyd Warshall algorithm?

It incrementally improves the shortest path between every pair of nodes by considering each node as an intermediate point in the path.

intermediate

What is the time complexity of the Floyd Warshall algorithm?

The time complexity is O(V³), where V is the number of vertices in the graph.

beginner

In Floyd Warshall, how do you update the distance between nodes i and j using an intermediate node k?

distance[i][j] = min(distance[i][j], distance[i][k] + distance[k][j])

intermediate

Can Floyd Warshall detect negative weight cycles? If yes, how?

Yes, by checking if distance[i][i] becomes negative for any node i after running the algorithm, indicating a negative cycle.

What type of graphs can Floyd Warshall handle?

A Graphs with negative cycles

B Only graphs with positive edges

C Only unweighted graphs

D Graphs with negative edges but no negative cycles

Floyd Warshall works with graphs that may have negative edges but must not have negative weight cycles.

What is the main data structure used to store distances in Floyd Warshall?

A Distance matrix

B Adjacency matrix

C Adjacency list

D Queue

Floyd Warshall uses a distance matrix to keep track of shortest distances between all pairs.

Which of the following best describes the Floyd Warshall algorithm approach?

A Greedy

B Divide and conquer

C Dynamic programming

D Backtracking

Floyd Warshall uses dynamic programming to build solutions from smaller subproblems.

What does a negative value on the diagonal of the distance matrix indicate after Floyd Warshall runs?

A Negative weight cycle detected

B Shortest path found

C No path exists

D Graph is disconnected

A negative diagonal value means a negative cycle exists involving that node.

What is the space complexity of Floyd Warshall algorithm?

A O(V)

B O(V²)

C O(V³)

D O(1)

The algorithm stores a V by V distance matrix, so space complexity is O(V²).

Explain how Floyd Warshall algorithm updates shortest paths using intermediate nodes.

Think about improving paths step by step by adding one node at a time.

Describe how Floyd Warshall can detect negative weight cycles in a graph.

Focus on what the diagonal values represent after the algorithm finishes.

Practice

1. What is the main purpose of the Floyd Warshall algorithm in graph theory?

easy

A. To perform a depth-first search traversal

B. To find the minimum spanning tree of a graph

C. To detect cycles in a graph

D. To find the shortest paths between all pairs of vertices in a weighted graph

Solution

1. Step 1: Understand Floyd Warshall's goal

   The Floyd Warshall algorithm is designed to find shortest paths between every pair of vertices in a weighted graph, including negative weights but no negative cycles.

2. Step 2: Compare with other graph algorithms

   Minimum spanning tree algorithms find a subset of edges connecting all vertices with minimum total weight, cycle detection finds cycles, and DFS is a traversal method. These are different from Floyd Warshall's all pairs shortest path goal.

3. Final Answer:

   To find the shortest paths between all pairs of vertices in a weighted graph -> Option D

4. Quick Check:

   Floyd Warshall = All pairs shortest paths [OK]

Hint: Floyd Warshall finds shortest paths between all vertex pairs [OK]

Common Mistakes:

• Confusing Floyd Warshall with minimum spanning tree algorithms
• Thinking it only finds shortest path from one source
• Assuming it detects cycles
• Mixing it with traversal algorithms like DFS

2. Which of the following is the correct way to initialize the distance matrix dist for Floyd Warshall in Go, given graph as adjacency matrix and V as number of vertices?

easy

A. dist = graph

B. for i := 0; i < V; i++ { dist[i] = graph[i] }

C. for i := 0; i < V; i++ { for j := 0; j < V; j++ { dist[i][j] = graph[i][j] } }

D. dist := make([]int, V*V)

Solution

1. Step 1: Understand matrix copying in Go

   In Go, to copy a 2D slice (matrix), you must copy each element individually. Assigning one slice to another copies the reference, not the data.

2. Step 2: Analyze options

   for i := 0; i < V; i++ { for j := 0; j < V; j++ { dist[i][j] = graph[i][j] } } copies each element explicitly, correctly initializing dist. for i := 0; i < V; i++ { dist[i] = graph[i] } copies slice references, not elements. dist = graph assigns reference, not copy. dist := make([]int, V*V) creates a 1D slice, not a 2D matrix.

3. Final Answer:

   for i := 0; i < V; i++ { for j := 0; j < V; j++ { dist[i][j] = graph[i][j] } } -> Option C

4. Quick Check:

   Copy element-wise for matrix copy [OK]

Hint: Copy 2D slices element-wise to avoid reference issues [OK]

Common Mistakes:

• Assigning slices directly causing shared references
• Using 1D slice instead of 2D for matrix
• Not initializing inner slices before assignment
• Assuming slice assignment copies data

3. What will be the output distance matrix after running Floyd Warshall on the following graph represented as adjacency matrix (9999 means no direct edge):

graph := [][]int{
  {0, 3, 9999, 7},
  {8, 0, 2, 9999},
  {5, 9999, 0, 1},
  {2, 9999, 9999, 0},
}

medium

A. [[0, 3, 5, 6], [7, 0, 2, 3], [3, 6, 0, 1], [2, 5, 7, 0]]

B. [[0, 3, 9999, 7], [8, 0, 2, 9999], [5, 9999, 0, 1], [2, 9999, 9999, 0]]

C. [[0, 3, 5, 7], [8, 0, 2, 3], [5, 8, 0, 1], [2, 5, 7, 0]]

D. [[0, 3, 4, 6], [7, 0, 2, 3], [3, 6, 0, 1], [2, 5, 7, 0]]

Solution

1. Step 1: Initialize distance matrix

   Start with given graph matrix where 9999 means no direct edge. Distances on diagonal are 0.

2. Step 2: Apply Floyd Warshall updates

   Check all triples (i, j, k) and update dist[i][j] = min(dist[i][j], dist[i][k] + dist[k][j]). After iterations, shortest paths are:
   dist[0][2] = min(9999, 3+2) = 5
   dist[0][3] = min(7, 3+2+1) = 6
   dist[1][0] = min(8, 2+1+2) = 7
   dist[1][3] = min(9999, 2+1) = 3
   dist[2][0] = min(5, 1+2) = 3
   dist[2][1] = min(9999, 1+2+3) = 6
   dist[3][1] = min(9999, 2+3) = 5
   dist[3][2] = min(9999, 2+3+2) = 7

3. Final Answer:

   [[0, 3, 5, 6], [7, 0, 2, 3], [3, 6, 0, 1], [2, 5, 7, 0]] -> Option A

4. Quick Check:

   Floyd Warshall updates shortest paths correctly [OK]

Hint: Update dist[i][j] using intermediate k for all triples [OK]

Common Mistakes:

• Not updating paths through intermediate vertices
• Confusing no edge (9999) with zero
• Forgetting to update all pairs in nested loops
• Incorrectly summing distances

4. Consider this Go code snippet for Floyd Warshall. What is the bug causing incorrect shortest path results?

for k := 0; k < V; k++ {
  for i := 0; i < V; i++ {
    for j := 0; j < V; j++ {
      if dist[i][k] + dist[k][j] < dist[i][j] {
        dist[i][j] = dist[i][k] + dist[k][j]
      }
    }
  }
}

medium

A. Missing check for dist[i][k] or dist[k][j] being infinity before addition

B. No bug; code is correct

C. Loop order should be i, j, k instead of k, i, j

D. Should use <= instead of < in if condition

Solution

1. Step 1: Understand the addition of distances

   When dist[i][k] or dist[k][j] is infinity (e.g., 9999), adding them can cause integer overflow or incorrect comparisons.

2. Step 2: Identify missing checks

   The code does not check if dist[i][k] or dist[k][j] is infinity before adding. This can cause wrong updates to dist[i][j]. Proper code should skip addition if either is infinity.

3. Final Answer:

   Missing check for dist[i][k] or dist[k][j] being infinity before addition -> Option A

4. Quick Check:

   Check infinity before adding distances [OK]

Hint: Always check for infinity before adding distances [OK]

Common Mistakes:

• Ignoring infinity checks causing wrong updates
• Changing loop order unnecessarily
• Using <= instead of < does not fix logic
• Assuming code is correct without boundary checks

5. Given a graph with vertices representing cities and edges representing travel times, how can Floyd Warshall be adapted to also reconstruct the shortest path between any two cities?

hard

A. Store all paths explicitly in a list during the algorithm

B. Maintain a separate predecessor matrix updated during relaxation to track paths

C. Run Dijkstra's algorithm for each city instead

D. Use BFS instead of Floyd Warshall for path reconstruction

Solution

1. Step 1: Understand Floyd Warshall's limitation

   Floyd Warshall computes shortest distances but does not store path details by default.

2. Step 2: Use predecessor matrix for path reconstruction

   By maintaining a predecessor matrix where predecessor[i][j] stores the vertex before j on the shortest path from i, we can reconstruct the path after the algorithm finishes.

3. Step 3: Other options analysis

   Storing all paths explicitly is inefficient. Running Dijkstra repeatedly is possible but less efficient for dense graphs. BFS is not suitable for weighted graphs.

4. Final Answer:

   Maintain a separate predecessor matrix updated during relaxation to track paths -> Option B

5. Quick Check:

   Predecessor matrix enables path reconstruction [OK]

Hint: Use predecessor matrix to rebuild shortest paths [OK]

Common Mistakes:

• Not tracking predecessors, so no path info
• Trying to store all paths explicitly causing inefficiency
• Confusing BFS with weighted shortest path algorithms
• Using Dijkstra repeatedly instead of Floyd Warshall