What is the smallest possible number of iterations of the while-loop in Gale-Shapley for instances with n employers?

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The provided data for this question is incomplete: there are no answer options given, and the 'answer' field is empty. Because multiple-choice questions require selecting from a set of options, we cannot identify which choice is to be evaluated.

Nevertheless, we can discuss the general behavior of the Gale-Shapley algorithm (the stable marriage/assignment algorithm) with respect to the number of iterations of its outer loop when run with n employers (and n workers, typically men proposing to women, or vice versa).

- Best-case scenario (smallest possible number of iterations): In the most favorable arrangement of preferences, every proposer could be accepted on the first round. Each free proposer proposes to their top-choice employer, and if all top choices are distinct and all are tentatively accepted (with no rejections or re-proposals needed), all matches are formed in a single iteration. Hence, the smallest possible number of iterations can be 1.

- Worst-case scenario (largest possible number of iterations): In less favorable configurations, many proposers must advance through several choices, and rejections lead to further proposals in subsequent iterations. In the classic analysis, the total number of proposals is at most O(n^2), since each of the n proposers can propose to each of the n employers at most once. The number of iterations (rounds) is bounded above by n in the standard formulation: in each iteration, each currently free proposer makes one proposal, and after some proposers get accepted and others get rejected, the next iteration proceeds with the remaining free proposers. In the worst case, it can take up to n−1 iterations to settle all matches.

- Important distinctions: The exact number of iterations can depend on the precise definitions used for an 'iteration' (some sources count a full pass of proposals and rejections as one iteration; others may partition work differently). The core takeaway is that the minimum number of iterations is 1 (best case), while the maximum grows with n, up to on the order of n in common analyses.

Without the available answer options, we cannot map these insights to a specific choice. If options were present, you would identify the choice corresponding to the best-case scenario (typically stating that all matches occur in the first pass) as the one describing the smallest possible number of iterations, and then you would contrast it with other options that describe higher iteration counts or mischaracterize the process.

CPSC_V 320 201/202/203 2024W2 Reading Quiz #1 (Stable Matching Problem)

What is the smallest possible number of iterations of the while-loop in Gale-Shapley for instances with n employers?

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