Marriage problem says {b_1,b_2,…,b_n} and {g_1,g_2,…,g_n} are n boys and n girls respectively. Let G_i=set of girls b_i likes.

Hayley Mcclain

Hayley Mcclain

Answered question

2022-11-07

Marriage problem says { b 1 , b 2 , , b n } and { g 1 , g 2 , , g n } are n boys and n girls respectively. Let G i =set of girls b i likes. Then a necessary and sufficient condition for a monogamous marriage to happen such that each b i gets married to a girl in G i (a compatible matching) is that for any 1 k n, and any 1 i 1 < i 2 < < i k n, cardinality of G i 1 G i k should be bigger than k that is, take any k boys then the total number of girls that these k boys together like should be at least k.
There is apparently a version of marriage problem in which one considers both the sets- the set of girls whom each boy likes and set of the boys whom each girl likes. In that case the condition reduces to considering any k boys out of n / 2 instead of n above.

Answer & Explanation

embutiridsl

embutiridsl

Beginner2022-11-08Added 26 answers


If B and G are the sets of boys and girls, respectively, for each b B let G ¯ ( b ) be the set of girls whom 𝑏 does not like, and for each g G let B ¯ ( g ) be the set of boys whom g does not like. (It’s more convenient here to use these rather than G ( b ) and B ( g ), the sets of boys or girls liked by an individual, but the two are obviously interchangeable.) Then the boys and girls can be married off iff there do not exist B 0 B and G 0 G such that | B 0 | + | G 0 | > n, G 0 b B 0 G ¯ ( b ), and B 0 g G 0 B ¯ ( g ). Clearly if such B 0 and G 0 did exist, one of them would have to have cardinality at most n / 2 .
Suppose that for each B 1 B and G 1 G of cardinality at most n / 2 , | b B 1 G ¯ ( b ) | < n | B 1 | and | g G 1 B ¯ ( b ) | < n | G 1 | . Suppose, to get a contradiction, that B 0 B and G 0 G are such that | B 0 | + | G 0 | > n, G 0 b B 0 G ¯ ( b ), and B 0 g G 0 B ¯ ( g ); without loss of generality | B 0 n / 2 . But then | b B 0 G ¯ ( b ) | < n | B 0 | < | G 0 | , so G 0 b B 0 G ¯ ( b ).
It only remains to check that the nonexistence of sets B 0 B and G 0 G such that, G 0 b B 0 G ¯ ( b ), and B 0 g G 0 B ¯ ( g ) is equivalent to the usual formulation of the marriage condition, but this is pretty straightforward if one thinks in terms of the adjacency matrix of the obvious bipartite graph.

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