Floyd Warshall All Pairs Shortest Path in DSA Go - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Floyd Warshall on a simple graph

What is the output distance matrix after running Floyd Warshall on this graph?

DSA Go

package main

import "fmt"

func floydWarshall(graph [][]int) [][]int {
    n := len(graph)
    dist := make([][]int, n)
    for i := range dist {
        dist[i] = make([]int, n)
        copy(dist[i], graph[i])
    }

    for k := 0; k < n; k++ {
        for i := 0; i < n; i++ {
            for j := 0; j < n; j++ {
                if dist[i][k] != 9999 && dist[k][j] != 9999 && dist[i][k]+dist[k][j] < dist[i][j] {
                    dist[i][j] = dist[i][k] + dist[k][j]
                }
            }
        }
    }
    return dist
}

func main() {
    graph := [][]int{
        {0, 3, 9999, 7},
        {8, 0, 2, 9999},
        {5, 9999, 0, 1},
        {2, 9999, 9999, 0},
    }
    dist := floydWarshall(graph)
    fmt.Println(dist)
}

A [[0 3 9999 7] [8 0 2 9999] [5 9999 0 1] [2 9999 9999 0]]

B [[0 3 5 6] [7 0 2 3] [3 6 0 1] [2 5 7 0]]

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

D [[0 3 5 6] [8 0 2 3] [5 8 0 1] [2 5 7 0]]

💡 Hint

Look for the shortest paths updated through intermediate nodes.

The Floyd Warshall algorithm updates the distance matrix by checking if going through an intermediate node k gives a shorter path from i to j. The final matrix shows the shortest distances between all pairs.

🧠 Conceptual

intermediate

Understanding Floyd Warshall's time complexity

What is the time complexity of the Floyd Warshall algorithm for a graph with n vertices?

A O(n!)

B O(n^2)

C O(n log n)

D O(n^3)

💡 Hint

Consider the three nested loops over vertices.

Floyd Warshall uses three nested loops each running n times, resulting in O(n^3) time complexity.

🔧 Debug

advanced

Identify the error in this Floyd Warshall implementation

What error will this code produce when run?

DSA Go

func floydWarshall(graph [][]int) [][]int {
    n := len(graph)
    dist := make([][]int, n)
    for i := range dist {
        dist[i] = make([]int, n)
        copy(dist[i], graph[i])
    }

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

A Logic error: does not handle infinite distances properly

B No error, runs correctly

C Index out of range error

D Runtime error due to integer overflow

💡 Hint

Check how infinite distances are handled in the if condition.

The code does not check if dist[i][k] or dist[k][j] is infinite (e.g., 9999) before adding, which can cause incorrect updates.

🚀 Application

advanced

Using Floyd Warshall to detect negative cycles

How can Floyd Warshall algorithm be used to detect if a graph contains a negative weight cycle?

A Check if any element dist[i][j] is zero

B Check if the sum of all distances is negative

C Check if any diagonal element dist[i][i] becomes negative after running the algorithm

D Check if the graph has any edge with negative weight

💡 Hint

Think about what a negative cycle means for shortest path from a node to itself.

If dist[i][i] is negative, it means there is a cycle with negative total weight reachable from i.

❓ Predict Output

expert

Output of Floyd Warshall on a graph with unreachable nodes

What is the output distance matrix after running Floyd Warshall on this graph with unreachable nodes?

DSA Go

package main

import "fmt"

func floydWarshall(graph [][]int) [][]int {
    n := len(graph)
    dist := make([][]int, n)
    for i := range dist {
        dist[i] = make([]int, n)
        copy(dist[i], graph[i])
    }

    for k := 0; k < n; k++ {
        for i := 0; i < n; i++ {
            for j := 0; j < n; j++ {
                if dist[i][k] != 9999 && dist[k][j] != 9999 && dist[i][k]+dist[k][j] < dist[i][j] {
                    dist[i][j] = dist[i][k] + dist[k][j]
                }
            }
        }
    }
    return dist
}

func main() {
    graph := [][]int{
        {0, 9999, 3, 9999},
        {9999, 0, 9999, 7},
        {9999, 2, 0, 9999},
        {6, 9999, 9999, 0},
    }
    dist := floydWarshall(graph)
    fmt.Println(dist)
}

A [[0 5 3 12] [13 0 15 7] [8 2 0 9] [6 11 9 0]]

B [[0 9999 3 9999] [9999 0 9999 7] [9999 2 0 9999] [6 9999 9999 0]]

C [[0 5 3 7] [13 0 15 7] [8 2 0 9] [6 11 9 0]]

D [[0 5 3 14] [13 0 15 7] [8 2 0 9999] [6 11 9 0]]

💡 Hint

Check how unreachable nodes get updated through intermediate nodes.

The algorithm updates unreachable distances when a path exists through intermediate nodes, resulting in the matrix shown in option A.

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