Efficient Counting of 3-Term Arithmetic Progressions Using On Log N Algorithm

Counting the number of 3-term arithmetic progressions in a subset of [1... n] can be achieved efficiently using an algorithm that runs in On log n time. This method is particularly useful for large datasets where traditional quadratic approaches would be too slow. Let's explore this algorithm in detail.

Algorithm Overview

To count the number of 3-term arithmetic progressions in a subset of [1... n] efficiently, the algorithm follows a structured approach that includes sorting, hash table lookups, and pair iteration. The objective is to maintain an efficient time complexity of On log n through these steps.

Steps of the Algorithm

1. Input the Subset: Let S be the subset of [1... n]. The first step is to sort the subset S.

2. Use a Hash Set for Fast Lookups: Create a hash set or a dictionary to store the elements of S for faster lookup operations. This provides an average-time complexity of O1 for lookups.

3. Iterate Through Pairs: For each pair ai, aj in the sorted subset S where i Calculate the potential middle term: m (ai aj) / 2 Check if m is an Integer: If the sum ai aj is even, proceeded to check if m is an integer. Check If m Exists in the Set: If m is an integer and exists in the hash set, then ai, m, aj forms a 3-term arithmetic progression. Increment the counter by one.

Pseudocode

def count_arithmetic_progressions(S):
    # Step 1: Sort the subset
    ()
    S_set  set(S)
    count  0
    # Step 3: Iterate through pairs
    n  len(S)
    for i in range(n):
        for j in range(i   1, n):
            a_i  S[i]
            a_j  S[j]
            # Calculate the middle term
            if (a_i   a_j) % 2  0: # Check if the sum is even
                m  (a_i   a_j) // 2
                # Check if m exists in the set
                if m in S_set:
                    count   1
    return count

Complexity Analysis

The overall complexity of the algorithm is dominated by the sorting step, which takes On log n time. The nested loops iterate through pairs, leading to a worst-case time complexity of On^2, but the operations inside the loop, due to hash table lookups, average to O1.

Conclusion

This algorithm provides an efficient way to count 3-term arithmetic progressions in a subset of [1... n] by leveraging sorting and hash sets for fast lookups. The On log n time complexity ensures that it can handle large datasets without excessive computational overhead.