Transitive_Clo_Alg_4
You are working with a Graph class implemented using an adjacency map, where each vertex maps to a dictionary of its neighbors. The graph is directed, and the methods get_edge(u, v) and insert_edge(u, v) are used to check for and add edges.
The function TRANSITIVE_CLOSURE(graph) shown below modifies a deep copy of the input graph by potentially adding new edges based on the structure of the graph:

function TRANSITIVE_CLOSURE(graph): closure ← deep copy of input graph
vertices ← list of vertices in closure n ← number of vertices
for k from 0 to n-1: for i from 0 to n-1: if i ≠ k and edge (i, k) exists in closure: for j from 0 to n-1: if i ≠ j ≠ k and edge (k, j) exists in closure: if edge (i, j) does not exist in closure: insert edge (i, j) into closure
return closure

What is the time complexity of this algorithm, assuming get_edge and insert_edge each take constant time?

Question

Transitive_Clo_Alg_4
You are working with a Graph class implemented using an adjacency map, where each vertex maps to a dictionary of its neighbors. The graph is directed, and the methods get_edge(u, v) and insert_edge(u, v) are used to check for and add edges.
The function TRANSITIVE_CLOSURE(graph) shown below modifies a deep copy of the input graph by potentially adding new edges based on the structure of the graph:
 
function TRANSITIVE_CLOSURE(graph):    closure ← deep copy of input graph
    vertices ← list of vertices in closure    n ← number of vertices
    for k from 0 to n-1:        for i from 0 to n-1:            if i ≠ k and edge (i, k) exists in closure:                for j from 0 to n-1:                    if i ≠ j ≠ k and edge (k, j) exists in closure:                        if edge (i, j) does not exist in closure:                            insert edge (i, j) into closure
    return closure

What is the time complexity of this algorithm, assuming get_edge and insert_edge each take constant time?

BlackTom AI · Accepted Answer

We are given a directed graph represented by an adjacency map, and a TRANSITIVE_CLOSURE procedure that iterates through vertices to potentially add edges. The time complexity hinges on how many times the inner operations run.

Option: O(n²)
This would imply only two nested levels of iteration with constant work inside. However, the algorithm uses three nested loops: for k from 0 to n-1, for i from 0 to n-1, and for j from 0 to n-1, with additional conditional checks inside. Even if each edge check/insert is constant time, the sheer number of (k, i, j) triplets is on the order of n³, so O(n²) underestimates the work.

Option: O(n³)
Within the code, there are three explicit loops each ranging over n: k, i, and j. Inside the innermost loop, there are constant-time operations (edge existence checks and possibly insertion) when conditions are met. Therefore the total number of iterations scales as n × n × n = n³, making O(n³) the correct asymptotic bound.

Option: O(n log n)
There is no logarithmic behavior or data structure operation that would typically yield a log n factor here. The loops dominate, so this option mismatches the actual cubic growth.

Option: O(n² log n)
Similarly, there is no log factor multiplying a quadratic term; the triple-nested loop drives cubic complexity rather than any combination of n² with a log n factor. This option is not consistent with the implementation.

In summary, because the algorithm performs three independent passes over the n vertices (k, i, and j) with constant-time work inside, the overall time complexity scales as n³.

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