The marriage problem requires us to match n girls with the set of n boys. Each girl (after a long and no doubt exhausting deliberation) submits a list of boys she likes. We also make an assumption
that being of noble character no boy will break a heart of a girl who likes him by turning her down. So, although, girls appear to seize the initiative by advertising their preferences, the situation is quite symmetric and is best represented by a zero-one matrix. An element aij in row i and column j is 1 iff the marriage between the girl #i and the boy number #j is feasible. aij is 0, otherwise. Sometimes all the girls can be given away, sometimes no complete match is possible.
The marriage condition can be formulated in several equivalent ways:
Every set of r girls, 1rn likes at least r boys.
Pick up any s columns. Concentrate on rows that have at least one 1 in the selected columns.
The number of such rows must not be less than s.
Every set of s boys, 1sn likes at least s girls.
Pick up any r rows. Concentrate on columns that have at least one 1 in the selected rows.
The number of such columns must not be less than r.
No zero r by s submatrix may satisfy r + s > n.
If such a matrix exists then some r girls can marry only (n - s) boys outside the submatrix. Since r > n - s, there are just too few boys to satisfy all r girls.
The marriage condition and the Marriage theorem are due to the English
mathematician Philip Hall (1935.)
Marriage Theorem
Hall's condition is both sufficient and necessary for a complete match.
where vertical bars denote the cardinality (the number of elements) of a set.
Proof #2
Assuming that the marriage condition holds for the sets A1, A2, ..., An,
let the sets be depleted until a family F = {B1, B2, ..., Bn} is reached
such that removal of 1 more element from any of the Bi would
cause the marriage condition to be violated. We assert that each
member of F consists of a single element; because these elements
are distinct (by the marriage condition), F itself is the required SDR.
Suppose, on the contrary, that, say B1 has 2 elements,
Let's say, B1 contains at least two elements, x and y.
By the minimality of the collection, removal of either x or y would violate the matching condition. Therefore, there exist two sets of indices P and Q such that
X = (B1-{x})({Bi: iP}) and Y = (B1-{y})({Bi: iQ})
with |X||P| and |Y||Q|.
Consequently, by the Inclusion-Exclusion priciple,