Chapter 3 - Section 3.1 - Algorithms - Exercises - Page 204: 62

$Proof$ $by$ $contradiction$ Let $n$ be the number of men (note that n is then also equal to the number of women) Let us assume, for the sake of contradiction, that the algorithm never terminates. This is only possible when there exists an integer $i \in\{1,2, \ldots, n\}$ such that the proposal list $P_{i}$ of women $w_{i}$ remains empty in every iteration (thus the list needs to be empty in every iteration of the algorithm, because once the list contains a proposal, there will be e a remaining proposal in the list in every iteration of the algorithm). There are an equal number of women and men, which means that there needs to exist an integer $j_{k} \in\{1,2, \ldots, n\}$ with $i \neq j_{k}$ such that $P_{j}$ contains at least two proposals in the $k$th iteration of the algorithm. Next, all proposals, except the best proposal, will be rejected and thus at least one proposal will be rejected in the $k$th iteration. Since the algorithm never terminates, that implies that there are an infinite amount of irritations and the thus are also an infinite amount of rejection proposals (as at least one proposal is rejected in each iteration). However, this is impossible as there are at most $n^{2}$ rejected proposals (when every man would propose to every woman and the woman would reject all proposals) and thus we have derived a contradiction. This then implies that our assumption “the algorithm never terminates” is incorrect and thus the deferred acceptance algorithm terminates.

$Proof$ $by$ $contradiction$ Let $n$ be the number of men (note that n is then also equal to the number of women) Let us assume, for the sake of contradiction, that the algorithm never terminates. This is only possible when there exists an integer $i \in\{1,2, \ldots, n\}$ such that the proposal list $P_{i}$ of women $w_{i}$ remains empty in every iteration (thus the list needs to be empty in every iteration of the algorithm, because once the list contains a proposal, there will be e a remaining proposal in the list in every iteration of the algorithm). There are an equal number of women and men, which means that there needs to exist an integer $j_{k} \in\{1,2, \ldots, n\}$ with $i \neq j_{k}$ such that $P_{j}$ contains at least two proposals in the $k$th iteration of the algorithm. Next, all proposals, except the best proposal, will be rejected and thus at least one proposal will be rejected in the $k$th iteration. Since the algorithm never terminates, that implies that there are an infinite amount of irritations and the thus are also an infinite amount of rejection proposals (as at least one proposal is rejected in each iteration). However, this is impossible as there are at most $n^{2}$ rejected proposals (when every man would propose to every woman and the woman would reject all proposals) and thus we have derived a contradiction. This then implies that our assumption “the algorithm never terminates” is incorrect and thus the deferred acceptance algorithm terminates.