Floyd Warshall All Pairs Shortest Path in DSA Go - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the distance matrix with the given graph.

DSA Go

dist := make([][]int, n)
for i := 0; i < n; i++ {
    dist[i] = make([]int, n)
    for j := 0; j < n; j++ {
        if i == j {
            dist[i][j] = [1]
        } else if graph[i][j] != 0 {
            dist[i][j] = graph[i][j]
        } else {
            dist[i][j] = 1 << 30
        }
    }
}

Drag options to blanks, or click blank then click option'

A1

B-1

C100

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Setting dist[i][i] to -1 or a large number instead of 0.

Not initializing the diagonal elements properly.

The distance from any node to itself is zero, so dist[i][i] should be 0.

2fill in blank

medium

Complete the code to update the distance matrix using the Floyd Warshall algorithm.

DSA Go

for k := 0; k < n; k++ {
    for i := 0; i < n; i++ {
        for j := 0; j < n; j++ {
            if dist[i][k] != 1<<30 && dist[k][j] != 1<<30 && dist[i][k] + dist[k][j] [1] dist[i][j] {
                dist[i][j] = dist[i][k] + dist[k][j]
            }
        }
    }
}

Drag options to blanks, or click blank then click option'

A>

B==

C<

D!=

Attempts:

3 left

💡 Hint

Common Mistakes

Using > instead of < causes no updates or wrong updates.

Using == or != does not correctly find shorter paths.

We update dist[i][j] only if the new path through k is shorter, so we check if dist[i][k] + dist[k][j] < dist[i][j].

3fill in blank

hard

Fix the error in the code that prints the shortest distance matrix after Floyd Warshall completes.

DSA Go

for i := 0; i < n; i++ {
    for j := 0; j < n; j++ {
        if dist[i][j] == [1] {
            fmt.Print("INF ")
        } else {
            fmt.Print(dist[i][j], " ")
        }
    }
    fmt.Println()
}

Drag options to blanks, or click blank then click option'

A1 << 30

B-1

C100000

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Checking for 0 or -1 instead of the large infinity value.

Printing wrong values for unreachable nodes.

We used 1 << 30 as infinity, so distances equal to that mean no path exists and should print INF.

4fill in blank

hard

Fill both blanks to complete the code that initializes the graph and runs Floyd Warshall.

DSA Go

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

n := len(graph)
// Initialize dist and run Floyd Warshall algorithm

Drag options to blanks, or click blank then click option'

A0

C100000

DINF

Attempts:

3 left

💡 Hint

Common Mistakes

Using 0 for missing edges causes confusion with actual zero-weight edges.

Using 'INF' as a string instead of a number causes errors.

We use a large number (100000) to represent no direct edge between nodes in the graph.

5fill in blank

hard

Fill all three blanks to complete the code that updates the distance matrix and prints the final shortest paths.

DSA Go

for k := 0; k < n; k++ {
    for i := 0; i < n; i++ {
        for j := 0; j < n; j++ {
            if dist[i][k] != [1] && dist[k][j] != [1] && dist[i][k] + dist[k][j] [2] dist[i][j] {
                dist[i][j] = dist[i][k] + dist[k][j]
            }
        }
    }
}

for i := 0; i < n; i++ {
    for j := 0; j < n; j++ {
        if dist[i][j] == [3] {
            fmt.Print("INF ")
        } else {
            fmt.Print(dist[i][j], " ")
        }
    }
    fmt.Println()
}

Drag options to blanks, or click blank then click option'

A0

B1 << 30

C<

D>

Attempts:

3 left

💡 Hint

Common Mistakes

Using 0 instead of infinity for checks causes wrong updates.

Using > instead of < causes no updates.

Printing wrong values for unreachable nodes.

We check that dist[i][k] and dist[k][j] are not infinity (1 << 30), update if new path is shorter (<), and print INF for distances equal to infinity.

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