Suppose we have a job allocation problem as before, but the number of jobs is smaller than the number of people, so that not everyone will get a job. Suppose that there are some enthusiastic people who really want a job, and some other people who do not care so much. Can we arrange a job allocation in which every enthusiastic person gets a job? Obviously this will be easier if every job has many enthusiastic candidates. We can give a precise result as follows.
Consider a job allocation problem with a set of people and a set of jobs such that . Suppose we are also given a subset of enthusiastic people. Suppose that for each subset we have
(so is plausible, as in Definition 11.2)
.
Then there is a matching which allocates all the jobs, in such a way that all the enthusiastic people get a job.
Imagine a set of additional fake jobs that can only be done by unenthusiastic people (watching TV, lying on the beach and so on). The number of additional jobs should be , so that the set has . We declare that all the unenthusiastic people are qualified for all of the jobs in , and that none of the enthusiastic people are qualified. This gives a new job allocation problem, with candidate sets for say. We claim that this extended problem is still plausible, or in other words that for all . Indeed, we can write as , where is a set of real jobs and is a set of fake jobs. If is empty then and the candidate set is the same as before, i.e. , so by condition (a) in the statement of the theorem. Suppose instead that , so contains at least one fake job, so all unenthusiastic people are candidates for , as well as the real candidates for all the real jobs in . In symbols, we have . We claim that . To see this, consider the following Venn diagram:
The shaded region is We have and and and ; the claim is clear from this. We also have by our original assumptions, so
On the other hand, we have with and so so . Putting this together, we get as required. This proves that our new allocation problem is plausible, so Hall’s Theorem tells us that there is a solution, say . This allocates all the real jobs and all the fake jobs. As , we see that everyone gets a job (either real or fake) for which they are qualified. The enthusiastic people are not qualified for the fake jobs, so they must all be allocated a real job. Thus, if we just ignore the fake jobs, we have a solution to the original problem in which every enthusiastic person is employed. ∎
For later use, it will be convenient to reformulate condition (b) of the theorem slightly.
In the context of Theorem 12.1, the condition is equivalent to .
is the disjoint union of the sets and , so . Also, it is clear that . Thus, the condition is equivalent to , and this can be rearranged as . ∎
After this reformulation, it becomes clear that condition (b) is necessary as well as sufficient. Indeed, the people in need to be given jobs (because they are enthusiasts) but they cannot be given jobs from (because they are unqualified) so they need to be given jobs from . This will be impossible unless .
In the above results, we have repeatedly used the column/candidate sets:
Recall that we also defined the row sets:
Suppose that there is a constant such that
For every job we have , so every job has at least candidates.
For every person we have , so no person is qualified for more than jobs.
Then the matching problem is plausible, and so has a solution by Hall’s Theorem.
Consider a subset ; we must show that . We will consider a set , which we can describe in three different ways:
A moment’s thought should convince you that these are three ways of saying the same thing. From the third description of , we get
On the other hand, the second description gives
If then is not qualified for any of the jobs in so , so the corresponding term is zero. On the other hand, if then we have . We therefore have
By putting these inequalities together, we get . As , we can divide by to get , as required. ∎
Suppose that there is a constant such that
For every job we have , so every job has precisely candidates.
For every person we have , so no person is qualified for more than jobs.
We will say that a person is talented if . Then there is a job allocation in which every talented person has a job.
Let be the set of talented people. Proposition 12.4 is still applicable, so we know that every subset is plausible, i.e. . By Hall’s Theorem, this implies that the job allocation problem is solvable. However, it does not guarantee that the allocation can be arranged so that everyone in gets a job. For that, we need to check the additional criterion in Theorem 12.1, which is . Now , so the required inequality is equivalent to or . For this, we let be the set of pairs such that is talented and qualified for , but is not qualified for any job in . This ensures that cannot be in , so . Once we have chosen , we can try to choose ; this must in particular be an element of , and , so there are at most choices for . This analysis gives . Alternatively, we can start by choosing . This can be any element of , and then can be any of the jobs for which is qualified. As is talented, there are precisely choices for . This analysis gives . Putting these together, we get . As this gives as required. ∎
We now give another theorem that is mathematically equivalent to Hall’s Theorem, but thinly disguised.
Suppose we have a list of finite sets . A transversal is a list of elements such that for all , and the elements are all different. The list is plausible if for every sequence of indices , we have
The phrase distinct set of representatives means the same as transversal.
Consider the following list:
The following choices give a transversal:
There exists a transversal iff the list is plausible.
Put and . Define
this gives a matching problem with candidate sets . Thus, if with , we have . This makes it clear that the list is plausible (according to Definition 12.6) iff the matching problem is plausible (according to Definition 11.2). Also, Hall’s Theorem (together with Lemma 11.5) tells us that the matching problem is plausible iff there exists a full matching. If is a full matching, then it assigns each to some element , in such a way that are all different, so we have a transversal. This construction gives a bijection between full matchings and transversals, so in particular, a transversal exists iff a full matching exists. Putting all this together, we see that a transversal exists iff the list of sets is plausible. ∎
Consider the following list:
We note that
Here we have taken of the sets, and their union only has size , which violates the plausibility condition. Thus, there is no transversal.
We now consider a version of the job allocation problem in which jobs can require a team of several workers.
A team allocation problem consists of a set of people, a set of jobs, a set of pairs where person is qualified for job , and numbers for each . The problem is to choose a team for each job , with , such that each person in is qualified to do job , and the sets are disjoint (so that no one has to do more than one job).
For any subset , we define (so is the total number of people required for all the jobs in ). We say that is plausible if . We say that the whole team allocation problem is plausible if every subset is plausible.
Suppose we have solved the team allocation problem, by choosing a team for each job . Then contains , and the sets are disjoint and have size , so , so . Thus, if the team allocation is solvable, then it is plausible. By the contrapositive, if the problem is not plausible, then there is no solution.
It is not hard to analyse the team allocation problem by converting it into an ordinary allocation problem of the type that we have considered already. We just imagine making a set of badges, labelled by elements of the set
For example, if we need bakers, we make badges marked Baker 1 up to Baker 5. We then put
corresponding to the idea that is qualified to wear badge iff is qualified to do job . Now the problem of choosing teams is equivalent to the problem of assigning badges. Using this, we can prove the following result.
If a team allocation problem is plausible, then it is solvable.
Consider a plausible team allocation problem as before, and the corresponding badge allocation problem . As we have explained, it will be enough to prove that the badge allocation problem is solvable. By Hall’s theorem, it will be enough to show that the badge allocation problem is plausible. Consider a subset ; we must show that . Put
In other words, is a set of badges, and is the set of job titles that appear on at least one of those badges. Now put
In other words, is the set of badges that share a job title with one of the badges in , so , so . It is clear that , so the relation becomes . The team allocation problem is assumed to be plausible, which implies that . On the other hand, a person is qualified for one of the badges in iff they are qualified for one of the jobs in , so . Putting this together, we get as required. ∎