- The match making problem:
- We have 2 sets of items
(e.g., jobs and workers)
- We must find a match up that satisfies a given condition
- We have 2 sets of items
(e.g., jobs and workers)
- Example 1:
assigning tasks to
skilled workers
- 4 tasks are being assigned to
4 workers:
- Worker 1 is qualified to perform tasks A and B
- Worker 2 is qualified to perform tasks A, B, C and D
- Worker 3 is qualified to perform tasks A and B
- Worker 4 is qualified to perform task B
- Graphical representation:
- Question:
- Can all 4 workers be assigned to different tasks for which they are qualified ???
- 4 tasks are being assigned to
4 workers:
- Example 2: the marriage problem
- 4 men are getting married to
4 women:
- Man 1 knows women A and B
- Man 2 knows women A, B, C and D
- Man 3 knows women A and B
- Man 4 knows woman B
- Graphical representation:
- Question:
- Can all 4 men be married to different women for which they know the woman that he marries ???
- 4 men are getting married to
4 women:
- Definition: Bi-partite graph
- Bi-partite graph =
a graph where the vertex set
can be divide into
2 disjoint sets
X and Y such that:
- For every edge e =
(a, b):
-
a ∈ X
b ∈ Y
- For every edge e =
(a, b):
- Example:
Notice that every edge starts in a node of one set and ends in a node of the other set
- Bi-partite graph =
a graph where the vertex set
can be divide into
2 disjoint sets
X and Y such that:
- Definition:
Matching
- Matching =
a pairing of
some elements from
set X to
set Y where:
- Two different elements from set X are paired with different elements in set Y
Example:
- These are not matchings
Explanation:
- In the first example, the same element 1 ∈ X is paired with 2
- Matching =
a pairing of
some elements from
set X to
set Y where:
- Marriage Theorem:
- Given:
- n men
- x women (note: x ≥ n)
- There is a matching
of n men (to
n women)
if and only if:
- For every subset of m men (m ≤ n), these m men know at least m women
- Given:
- Proof of the then part
- We will first prove that:
- There is a matching
of n men (to
n women)
then:
- For every subset of m men (m ≤ n), the m men know ≥ m women
We will show this claim by showing that the contraposition statement is true.
I.e., we show that:
- If there is a subset of m men (m ≤ n), and these m men know < m women, then there is no matching of n men (to n women)
- There is a matching
of n men (to
n women)
then:
- Consider a subset A of
m men
(m ≤ n), and
these m men
know < m women
(subset B):
Fact:
- Only women ∈ B
can be used to pair with a
man ∈ A
- Only one women ∈ B can be used to pair with one man ∈ A
- Only women ∈ B
can be used to pair with a
man ∈ A
- Since
|B|
< |A|,
we conclude:
- At least one man ∈ A will not be able to be paired with any woman ∈ B.
Hence: there is no matching possible.
- We will first prove that:
- Proof of the "if" part:
- We will next prove that:
- There is a matching
of n men (to
n women)
if:
- For every subset of m men (m ≤ n), the m men know at least m women
Re-written:
- If
every subset of
m men
(m ≤ n),
these m men
know at least m women,
then:
- There is a matching of n men (to n women)
We show this by construction (it's actually an induction proof)
- There is a matching
of n men (to
n women)
if:
- Base case of the induction proof:
- We can find a matching for 1 (m = 1) man
Proof:
- Since every subset of 1 men the 1 men knows at least 1 women, we just match that man with any one of the woman that he knows....
- Suppose that we have found a
matching for
m men and
m < n:
We will prove that:
- There is a matching for m+1 men
- Since m < n,
there is at least one man (p) that
is unmatched:
There are 2 possibilities:
- p knows
some q ∉ B
In this case, it is trivial to prove that there is a matching for m+1 men:
- p
only knows
women q ∈ B
The reason that p cannot find a match is:
- The women that
p
can be matched with
are
all taken by men
in the set A
(Because otherwise, p would find someone outside B to match with !!!)
- The women that
p
can be matched with
are
all taken by men
in the set A
We will now show how to construct a match for this difficulyt case.
- p knows
some q ∉ B
- We construct a
m+1 matching
using:
- p and
- All the elements in the set A
as follows:
- Start with man p
- Given: for every subset of
m = 1 man
the 1 man (p)
know at least 1 woman
Do this:
- Match p up with someone he knows.
- Note: we know that she is some element ∈ B
- Given: for every subset of
m = 1 man
the 1 man (p)
know at least 1 woman
-
Now, 2 men
(p and q) that
are matched up with the
same woman:
Because we have:
- For every subset of m = 2 men the 2 men know at least 2 woman
There are 2 possibilities:
Case 1: the second woman ∉ B Case 2 the second woman ∈ B 
- In the first case
(when the
other woman ∉ B)
we have found a m+1 matching:
- In the second case
we repeat the argument
using
3 elements:
Because we are given that:
- For every subset of m = 3 men the 3 men know at least 3 woman
There are again 2 possibilities:
Case 1: the second woman ∉ B Case 2: the second woman ∈ B 
- And so on...
- Eventually, we will reach
m+1
Then we will have found a m+1 matching
I.e.,: m+1 men can be matched with a different woman that they know
- Therefore, we have shown that:
- If there is a matching of m men and m < n, we can always find a larger matching of m+1 men
Therefore:
- There is a matching of n men
- We will next prove that: