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Floyd Warshall All Pairs Shortest Path in DSA Python - Build from Scratch

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Floyd Warshall All Pairs Shortest Path
📖 Scenario: You are working as a city planner. You have a map of cities connected by roads with distances. You want to find the shortest distance between every pair of cities to plan the best routes.
🎯 Goal: Build a program that uses the Floyd Warshall algorithm to find the shortest distances between all pairs of cities in the map.
📋 What You'll Learn
Create a 2D list called dist representing the distance matrix between cities
Create a variable n for the number of cities
Use three nested for loops with variables k, i, and j to update shortest paths
Print the final shortest distance matrix dist after applying the algorithm
💡 Why This Matters
🌍 Real World
City planners and GPS systems use all pairs shortest path algorithms to find the best routes between locations.
💼 Career
Understanding Floyd Warshall helps in roles involving network optimization, route planning, and graph algorithms.
Progress0 / 4 steps
1
Create the distance matrix
Create a 2D list called dist with these exact values representing distances between 4 cities. Use float('inf') for no direct path:
dist = [
[0, 5, float('inf'), 10],
[float('inf'), 0, 3, float('inf')],
[float('inf'), float('inf'), 0, 1],
[float('inf'), float('inf'), float('inf'), 0]
]
DSA Python
Hint

Use a list of lists to represent the matrix. Use float('inf') for no direct road.

2
Set the number of cities
Create a variable called n and set it to the number of cities in the dist matrix.
DSA Python
Hint

Use the len() function on dist to get the number of cities.

3
Apply the Floyd Warshall algorithm
Use three nested for loops with variables k, i, and j from 0 to n to update dist[i][j] with the minimum of dist[i][j] and dist[i][k] + dist[k][j].
DSA Python
Hint

Check if going through city k gives a shorter path from i to j. If yes, update dist[i][j].

4
Print the shortest distance matrix
Print the dist matrix after applying the Floyd Warshall algorithm. Use a for loop with variable row to print each row.
DSA Python
Hint

Use a for loop to print each row of the matrix on a new line.

Practice

(1/5)
1. What is the main purpose of the Floyd Warshall algorithm in graph theory?
easy
A. To find the shortest paths between all pairs of vertices in a weighted graph
B. To find the minimum spanning tree of a graph
C. To perform a depth-first search traversal
D. To detect cycles in a graph

Solution

  1. Step 1: Understand Floyd Warshall's goal

    The Floyd Warshall algorithm calculates shortest paths between every pair of vertices in a weighted graph.
  2. Step 2: Compare with other graph algorithms

    Minimum spanning tree, DFS, and cycle detection serve different purposes than Floyd Warshall.
  3. Final Answer:

    To find the shortest paths between all pairs of vertices in a weighted graph -> Option A
  4. Quick Check:

    Floyd Warshall = all pairs shortest paths [OK]
Hint: Floyd Warshall finds shortest paths between all vertex pairs [OK]
Common Mistakes:
  • Confusing with minimum spanning tree algorithms
  • Thinking it finds shortest path from one source only
  • Mixing with traversal algorithms like DFS or BFS
2. Which of the following is the correct way to initialize the distance matrix dist for Floyd Warshall in Python for a graph with n vertices?
easy
A. dist = [[0] * n] * n
B. dist = [[float('inf')] * n for _ in range(n)]
C. dist = [float('inf')] * n * n
D. dist = [0 for _ in range(n)]

Solution

  1. Step 1: Understand matrix initialization

    We need a 2D list with size n x n, each element set to infinity initially.
  2. Step 2: Check each option

    dist = [[float('inf')] * n for _ in range(n)] correctly creates a new list for each row with float('inf'). dist = [float('inf')] * n * n creates a 1D list, not 2D. dist = [[0] * n] * n creates shallow copies causing aliasing. dist = [0 for _ in range(n)] creates 1D list.
  3. Final Answer:

    dist = [[float('inf')] * n for _ in range(n)] -> Option B
  4. Quick Check:

    2D list with inf values = dist = [[float('inf')] * n for _ in range(n)] [OK]
Hint: Use list comprehension for 2D matrix initialization [OK]
Common Mistakes:
  • Using multiplication for 2D lists causing shared references
  • Creating 1D list instead of 2D
  • Initializing with zeros instead of infinity
3. What will be the output distance matrix after running Floyd Warshall on this graph represented by adjacency matrix?
graph = [
  [0, 3, float('inf'), 7],
  [8, 0, 2, float('inf')],
  [5, float('inf'), 0, 1],
  [2, float('inf'), float('inf'), 0]
]
medium
A. [[0, 3, 5, 6], [5, 0, 2, 3], [3, 6, 0, 1], [2, 5, 7, 0]]
B. [[0, 3, 10, 7], [8, 0, 2, 3], [5, 8, 0, 1], [2, 5, 7, 0]]
C. [[0, 3, 7, 7], [8, 0, 2, 3], [5, 8, 0, 1], [2, 5, 7, 0]]
D. [[0, 3, 5, 7], [8, 0, 2, 3], [5, 8, 0, 1], [2, 5, 7, 0]]

Solution

  1. Step 1: Initialize dist matrix same as graph

    Start with dist = graph matrix, where float('inf') means no direct edge.
  2. Step 2: Update distances using Floyd Warshall logic

    Iterate k from 0 to 3, update dist[i][j] = min(dist[i][j], dist[i][k] + dist[k][j]) for all i,j. After all iterations, shortest paths are computed.
  3. Final Answer:

    [[0, 3, 5, 6], [5, 0, 2, 3], [3, 6, 0, 1], [2, 5, 7, 0]] -> Option A
  4. Quick Check:

    All pairs shortest paths matrix = [[0, 3, 5, 6], [5, 0, 2, 3], [3, 6, 0, 1], [2, 5, 7, 0]] [OK]
Hint: Update dist with min of current and via intermediate vertex [OK]
Common Mistakes:
  • Not updating distances through intermediate vertices
  • Confusing indices in nested loops
  • Leaving float('inf') where shorter paths exist
4. Consider this snippet of Floyd Warshall implementation in Python:
for k in range(n):
    for i in range(n):
        for j in range(n):
            if dist[i][k] + dist[k][j] < dist[i][j]:
                dist[i][j] = dist[i][k] + dist[k][j]

What is the main bug if dist[i][k] or dist[k][j] is float('inf')?
medium
A. It causes incorrect updates because adding infinity is not handled
B. It causes incorrect addition leading to overflow errors
C. No bug, the code works fine with infinity values
D. It causes syntax errors due to float('inf') usage

Solution

  1. Step 1: Understand addition with infinity in Python

    float('inf') + any finite number = float('inf'), and float('inf') + float('inf') = float('inf').
  2. Step 2: Check the condition

    The condition if dist[i][k] + dist[k][j] < dist[i][j] will be False whenever the sum is inf, because inf is not less than any finite or inf value. Thus, no incorrect updates occur; invalid paths are correctly skipped.
  3. Final Answer:

    No bug, the code works fine with infinity values -> Option C
  4. Quick Check:

    inf + x = inf, inf < y always False for path updates [OK]
Hint: Python's float('inf') handles additions and comparisons correctly for FW [OK]
Common Mistakes:
  • Believing inf causes incorrect updates
  • Confusing with integer overflow issues
  • Thinking syntax error with inf
5. Given a graph with negative edge weights but no negative cycles, how does Floyd Warshall algorithm handle these edges to find shortest paths?
hard
A. It requires a separate algorithm to preprocess negative edges before running
B. It fails because it cannot handle negative weights at all
C. It ignores negative edges and only uses positive weights
D. It updates distances by considering intermediate vertices, correctly handling negative edges without cycles

Solution

  1. Step 1: Understand Floyd Warshall with negative edges

    The algorithm can handle negative edges by updating distances through intermediate vertices, as long as no negative cycles exist.
  2. Step 2: Confirm no negative cycle condition

    If negative cycles exist, shortest paths are undefined. Without them, Floyd Warshall converges to correct shortest paths.
  3. Final Answer:

    It updates distances by considering intermediate vertices, correctly handling negative edges without cycles -> Option D
  4. Quick Check:

    Negative edges allowed if no negative cycles [OK]
Hint: Floyd Warshall handles negative edges if no negative cycles exist [OK]
Common Mistakes:
  • Thinking Floyd Warshall cannot handle any negative edges
  • Confusing negative edges with negative cycles
  • Assuming preprocessing is always needed
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