Fixed Length Eulerian Trail Queries

Problem Overview

Learning Goals

After solving this problem, you will be able to:

• Understand Eulerian trail conditions for directed and undirected graphs
• Apply Hierholzer’s algorithm for finding Eulerian paths
• Use matrix exponentiation to count paths of fixed length
• Combine graph theory conditions with efficient path counting

Problem Statement

Problem: Given a directed graph with n nodes and q queries, determine if there exists an Eulerian trail of exactly k edges from node a to node b.

Input:

• Line 1: n (nodes) and q (queries)
• Next n lines: n x n adjacency matrix (1 = edge exists, 0 = no edge)
• Next q lines: a b k (start node, end node, path length)

Output:

• For each query: 1 if such a trail exists, 0 otherwise

Constraints:

• 1 <= n <= 100
• 1 <= q <= 10^5
• 1 <= k <= 10^9
• 1 <= a, b <= n

Example

Input:
3 2
0 1 1
1 0 1
1 1 0
1 3 5
2 1 4

Output:
1
0

Explanation:

• Query 1: Path 1->2->3->1->2->3 has length 5 from node 1 to 3. Answer: 1
• Query 2: No valid path of length 4 exists from node 2 to 1. Answer: 0

Intuition: How to Think About This Problem

Pattern Recognition

Key Question: This combines two classic problems: (1) Does an Eulerian path exist? (2) How many paths of length k exist between two nodes?

The trick is recognizing that an Eulerian trail visits every edge exactly once. For a trail to exist:

• Undirected graphs: At most 2 vertices have odd degree
• Directed graphs: At most one vertex has out-degree - in-degree = 1, and at most one has in-degree - out-degree = 1

Breaking Down the Problem

1. What are we looking for? A path of exactly k edges that could be an Eulerian trail
2. What information do we have? Graph structure, start/end nodes, required length
3. What’s the relationship? k must equal total edges for a true Eulerian trail

Analogies

Think of this like planning a mail delivery route - you want to visit every street exactly once and return (or end at a specific location). The length of your route is fixed by the number of streets.

Solution 1: Brute Force DFS

Idea

Try all possible paths of length k using DFS, tracking which edges we have used.

Algorithm

1. Start DFS from node a
2. Track used edges in a set
3. If we reach node b with exactly k edges, return true
4. Backtrack and try other edges

Code

def brute_force(n, adj, a, b, k):
  """
  Check if Eulerian trail of length k exists from a to b.

  Time: O(n^k) - exponential, only for small k
  Space: O(k) - recursion depth
  """
  def dfs(node, remaining, used_edges):
    if remaining == 0:
      return node == b

    for next_node in range(n):
      edge = (node, next_node)
      if adj[node][next_node] and edge not in used_edges:
        used_edges.add(edge)
        if dfs(next_node, remaining - 1, used_edges):
          return True
        used_edges.remove(edge)
    return False

  return 1 if dfs(a - 1, k, set()) else 0

Complexity

Why This Works (But Is Slow)

Guarantees correctness by exhaustive search, but k up to 10^9 makes this impossible.

Solution 2: Matrix Exponentiation (Optimal)

Key Insight

The Trick: The number of paths of length k from node i to j equals (A^k)[i][j] where A is the adjacency matrix. We can compute A^k in O(n^3 log k) using matrix exponentiation.

Algorithm Overview

1. Precompute graph properties: Check if Eulerian trail is possible
2. Matrix exponentiation: Compute A^k to check if any path of length k exists
3. Answer queries: Combine both conditions

Matrix Exponentiation

If A is the adjacency matrix:

• A[i][j] = 1 means there is an edge from i to j
• A^2[i][j] = number of paths of length 2 from i to j
• A^k[i][j] = number of paths of length k from i to j

Code (Python)

def matrix_mult(A, B, mod=None):
  """Multiply two matrices."""
  n = len(A)
  C = [[0] * n for _ in range(n)]
  for i in range(n):
    for j in range(n):
      for k in range(n):
        C[i][j] += A[i][k] * B[k][j]
        if mod:
          C[i][j] %= mod
  return C

def matrix_pow(M, k, mod=None):
  """Compute M^k using binary exponentiation."""
  n = len(M)
  # Identity matrix
  result = [[1 if i == j else 0 for j in range(n)] for i in range(n)]

  while k > 0:
    if k & 1:
      result = matrix_mult(result, M, mod)
    M = matrix_mult(M, M, mod)
    k >>= 1

  return result

def solve(n, adj, queries):
  """
  Answer Eulerian trail queries using matrix exponentiation.

  Time: O(n^3 log k) per unique k value
  Space: O(n^2)
  """
  results = []
  cache = {}  # Cache matrix powers

  for a, b, k in queries:
    if k not in cache:
      cache[k] = matrix_pow(adj, k)

    # Check if path of length k exists (non-zero entry)
    if cache[k][a-1][b-1] > 0:
      results.append(1)
    else:
      results.append(0)

  return results

# Main execution
def main():
  import sys
  input_data = sys.stdin.read().split()
  idx = 0

  n, q = int(input_data[idx]), int(input_data[idx+1])
  idx += 2

  adj = []
  for i in range(n):
    row = [int(input_data[idx+j]) for j in range(n)]
    adj.append(row)
    idx += n

  queries = []
  for _ in range(q):
    a, b, k = int(input_data[idx]), int(input_data[idx+1]), int(input_data[idx+2])
    queries.append((a, b, k))
    idx += 3

  results = solve(n, adj, queries)
  print('\n'.join(map(str, results)))

Complexity

Dry Run Example

Let’s trace through with n=3, adjacency matrix, and query (1, 3, 2):

Adjacency Matrix A:
    1  2  3
1 [ 0, 1, 1 ]
2 [ 1, 0, 1 ]
3 [ 1, 1, 0 ]

Computing A^2:

A^2[0][0] = A[0][0]*A[0][0] + A[0][1]*A[1][0] + A[0][2]*A[2][0]
         = 0*0 + 1*1 + 1*1 = 2

A^2[0][2] = A[0][0]*A[0][2] + A[0][1]*A[1][2] + A[0][2]*A[2][2]
         = 0*1 + 1*1 + 1*0 = 1

Result A^2:
    1  2  3
1 [ 2, 1, 1 ]
2 [ 1, 2, 1 ]
3 [ 1, 1, 2 ]

Query (1, 3, 2): A^2[0][2] = 1 > 0, so answer is 1
Paths of length 2 from 1 to 3: 1->2->3

Common Mistakes

Mistake 1: Integer Overflow

Problem: k up to 10^9 causes overflow in path counts. Fix: Use modular arithmetic or just check if value > 0.

Mistake 2: Not Caching Matrix Powers

# WRONG - recomputes for every query
for a, b, k in queries:
  result = matrix_pow(adj, k)  # Expensive!

Problem: Same k value may appear multiple times. Fix: Cache computed matrix powers.

Mistake 3: Confusing Trail vs Path vs Walk

Walk:    Can repeat vertices and edges
Path:    No repeated vertices
Trail:   No repeated edges (vertices can repeat)
Circuit: Trail that starts and ends at same vertex

Problem: Using wrong definition. Fix: Trail allows vertex repetition but not edge repetition.

Edge Cases

When to Use This Pattern

Use Matrix Exponentiation When:

• You need to count/check paths of length k
• k is very large (up to 10^9)
• Graph is small enough (n <= 100)
• Multiple queries with same or similar k values

Don’t Use When:

• k is small - use BFS/DFS instead
• n is very large - O(n^3) per multiplication is too slow
• You need the actual path, not just existence/count

Pattern Recognition Checklist:

• Need paths of exact length k? -> Matrix exponentiation
• Large k value? -> Binary exponentiation is essential
• Small n? -> Matrix approach is feasible

Hierholzer’s Algorithm (Bonus)

For actually finding an Eulerian path (not just checking existence):

def find_eulerian_path(adj, n, start):
  """
  Find Eulerian path using Hierholzer's algorithm.

  Time: O(E) where E is number of edges
  Space: O(E) for storing the path
  """
  # Make a copy of adjacency list
  graph = [list(row) for row in adj]
  out_degree = [sum(row) for row in graph]

  stack = [start]
  path = []

  while stack:
    v = stack[-1]
    if out_degree[v] == 0:
      path.append(v)
      stack.pop()
    else:
      # Find next edge
      for u in range(n):
        if graph[v][u] > 0:
          graph[v][u] -= 1
          out_degree[v] -= 1
          stack.append(u)
          break

  return path[::-1]  # Reverse to get correct order

Related Problems

Easier (Do These First)

Similar Difficulty

Harder (Do These After)

Key Takeaways

1. The Core Idea: Matrix exponentiation transforms path counting into matrix multiplication
2. Time Optimization: From O(n^k) brute force to O(n^3 log k) using binary exponentiation
3. Space Trade-off: O(n^2) space for matrices enables efficient computation
4. Pattern: This is a classic application of linear algebra in graph theory

Practice Checklist

Before moving on, make sure you can:

• State Eulerian path/circuit conditions for directed and undirected graphs
• Implement matrix exponentiation from scratch
• Explain why A^k[i][j] counts paths of length k
• Recognize when matrix exponentiation applies to a problem

Additional Resources

• CP-Algorithms: Matrix Exponentiation
• CP-Algorithms: Eulerian Path
• CSES Mail Delivery - Eulerian circuit problem