Stable Matchings P vs NP - Carnegie Mellon School of ...aada/courses/15251s16/www/slides/15251-s16-lecture...The Traditional Marriage Algorithm For each day that some boy gets a “No”

15-251: Great Theoretical Ideas in Computer Science

Stable Matchings +

P vs NP

Lecture 13

Question: How do we pair them?

Dating Scenario

There are n boys and n girls

Each girl has her own ranked preference list of all the boys

Each boy has his own ranked preference list of the girls

The lists have no ties

3,5,2,1,4

1

5,2,1,4,3

2

4,3,5,1,2

3

1,2,3,4,5

4

2,3,4,1,5

5

1

3,2,5,1,4

2

1,2,5,3,4

3

4,3,2,1,5

4

1,3,4,2,5

5

1,2,4,5,3

Rogue Couples

Suppose we match all the boys and girls

Now suppose that some boy and some girl prefer each other to the people to whom they are paired

They will be called a rogue couple

A pairing of boys and girls is called stable if it contains no rogue couples

Stable Pairings

3,2,1

1

2,1,3

2

3,1,2

3

1

3,2,1

2

1,2,3

3

3,2,1

The Traditional Marriage Algorithm

For each day that some boy gets a “No” do:

Morning

• Each girl stands on her balcony

• Each boy proposes to the best girl whom he has not yet crossed off

Afternoon (for girls with at least one suitor)

• To today’s best: “Maybe, return tomorrow”

• To any others: “No, I will never marry you”

Evening

• Any rejected boy crosses the girl off his list

If no boys get a “No”, each girl marries the boy to whom she just said “maybe”

3,5,2,1,

4

1

5,2,1,4,3

4,3,5,1,2

3

1,2,3,4,5

4

2,3,4,1,5

5

1

3,2,5,1,4

2

1,2,5,3,4

3

4,3,2,1,5

4

1,3,4,2,5

5

1,2,4,5,3

2

Traditional Marriage Algorithm Example

3,5,2,1,

4

1

5,2,1,4,3

4,3,5,1,2

3

1,2,3,4,5

4

2,3,4,1,5

5

1

3,2,5,1,4

2

1,2,5,3,4

3

4,3,2,1,5

4

1,3,4,2,5

5

1,2,4,5,3

2


3,5,2,1,

4

1

5,2,1,4,3

4,3,5,1,2

3

1,2,3,4,5

4

2,3,4,1,5

5

1

3,2,5,1,4

2

2,5,3,4

3

4,3,2,1,5

4

3,4,2,5

5

1,2,4,5,3

2


3,5,2,1,

4

1

5,2,1,4,3

4,3,5,1,2

3

1,2,3,4,5

4

2,3,4,1,5

5

1

3,2,5,1,4

2

2,5,3,4

3

4,3,2,1,5

4

3,4,2,5

5

1,2,4,5,3

2


3,5,2,1,

4

1

5,2,1,4,3

4,3,5,1,2

3

1,2,3,4,5

4

2,3,4,1,5

5

1

3,2,5,1,4

2

2,5,3,4

3

4,3,2,1,5

4

3,4,2,5

5

1,2,4,5,3

2


3,5,2,1,

4

1

5,2,1,4,3

4,3,5,1,2

3

1,2,3,4,5

4

2,3,4,1,5

5

1

2,5,1,4

2

2,5,3,4

3

4,3,2,1,5

4

3,4,2,5

5

1,2,4,5,3

2


3,5,2,1,

4

1

5,2,1,4,3

4,3,5,1,2

3

1,2,3,4,5

4

2,3,4,1,5

5

1

2,5,1,4

2

2,5,3,4

3

4,3,2,1,5

4

3,4,2,5

5

1,2,4,5,3

2


3,5,2,1,

4

1

5,2,1,4,3

4,3,5,1,2

3

1,2,3,4,5

4

2,3,4,1,5

5

1

2,5,1,4

2

2,5,3,4

3

4,3,2,1,5

4

3,4,2,5

5

1,2,4,5,3

2


3,5,2,1,

4

1

5,2,1,4,3

4,3,5,1,2

3

1,2,3,4,5

4

2,3,4,1,5

5

1

5,1,4

2

2,5,3,4

3

4,3,2,1,5

4

3,4,2,5

5

1,2,4,5,3

2


3,5,2,1,

4

1

5,2,1,4,3

4,3,5,1,2

3

1,2,3,4,5

4

2,3,4,1,5

5

1

5,1,4

2

2,5,3,4

3

4,3,2,1,5

4

3,4,2,5

5

1,2,4,5,3

2


Does TMA Always Terminate?

It might encounter a situation where the algorithm does not specify what to do next.

It might keep on going for an infinite number of days

Does TMA Always Terminate?

It might encounter a situation where the algorithm does not specify what to do next

It might keep on going for an infinite number of days

cannot happen, every day some name is crossed off

this could happen if some boy crosses all names off his list!

we’ll show it doesn’t…

Improvement Lemma: If a girl has a boy on a string, then she will always have someone at least as good on a string (or for a husband)

She would only let go of him in order to “maybe” someone better

She would only let go of that guy for someone even better

She would only let go of that guy for someone even better

AND SO ON…

Corollary: Each girl will marry her absolute favorite of the boys who visit her during the TMA

Contradiction

Lemma: No boy can be rejected by all the girls

Proof (by contradiction):

Suppose boy b is rejected by all the girls

At that point:

Each girl must have a suitor other than b

(By Improvement Lemma, once a girl has a suitor she will always have at least one)

The n girls have n suitors, and b is not among them. Thus, there are at least n+1 boys

Theorem: The TMA always terminates in at most n2 days

A “master list” of all n of the boys lists starts with a total of n x n = n2 girls on it

Each day that at least one boy gets a “No”, so at least one girl gets crossed off the master list

Therefore, the number of days is bounded by the original size of the master list

Great! We know that TMA terminates and produces a matching

But is it stable?

g b

g*

I rejected you when you came to my balcony.

Now I’ve got someone better

Theorem: The matching T produced by TMA is stable.

Let b and g be any couple in T, and b prefers g* to g. We claim that g* prefers her husband to b.

During TMA, b proposed to g* before he proposed to g. Hence, at some point g* rejected b for someone

she preferred to b. By the Improvement lemma, the person she married

was also preferable to b. Thus, every boy will be rejected by any girl

he prefers to his wife. T is stable.

Opinion Poll

Forget TMA for a Moment…

How should we define “the optimal girl for boy b”?

Flawed Attempt: “The girl at the top of b’s list”

, ,

, ,

She is the best girl he can conceivably get in a stable world. Presumably, she might

be better than the girl he gets in the stable pairing output by TMA

The Optimal Girl

A boy’s optimal girl is the highest ranked girl for whom there is some stable pairing

in which the boy gets her

A boy’s pessimal girl is the lowest ranked girl for whom there is some stable pairing

in which the boy gets her

The Pessimal Girl

She is the worst girl he can conceivably get in a stable world

, , , ,

, , , ,

, , , ,

Dating Heaven and Hell A pairing is male-optimal if every boy gets

his optimal mate. This is the best of all possible worlds for every boy

simultaneously

A pairing is male-pessimal if every boy gets his pessimal mate. This is the worst of

all possible worlds for every boy simultaneously

BTW, is a male-optimal pairing stable? How about a male-pessimal one?

Dating Heaven and Hell A pairing is female-optimal if every girl gets

her optimal mate. This is the best of all possible stable worlds for every girl

simultaneously

A pairing is female-pessimal if every girl gets her pessimal mate. This is the worst of

all possible stable worlds for every girl simultaneously

not clear if either exists!

we’ll show that both exist…

The Naked Mathematical Truth!

The Traditional Marriage Algorithm always produces

a male-optimal, female-pessimal pairing

Theorem: TMA produces a male-optimal pairing

Suppose, for a contradiction, that some boy gets rejected by his optimal girl during TMA

At time t, boy b got rejected by his optimal girl g because she said “maybe” to a preferred b*

Therefore, b* likes g at least as much as his optimal

Let t be the earliest time at which this happened

By the definition of t, b* had not yet been rejected by his optimal girl

Some boy b got rejected by his optimal girl g because she said “maybe” to a preferred b*. b* likes g at least as much as his optimal girl

Also, there must exist a stable pairing S in which b and g are married (by def. of “optimal”)

b* wants g more than his wife in S:

g wants b* more than her husband in S:

Contradiction

g is at least as good as his optimal and he is not matched to g in stable pairing S

b is her husband in S and she rejects him for b* in TMA

g

Suppose TMA

is not male-optimal

b

b*

g b

b* g’

likes b*

more than b

likes g*

at least

as much as

his optimal

(not been

rejected by

optimal yet)

consider the

first moment

in TMA when

some boy

is rejected by

his optimal girl

since g is b’s optimal,

there is a stable matching S

where g and b are matched

g’ at most

as good as

his optimal

likes b*

more than b

contradicts

stability of S!!!

Theorem: The TMA pairing, T, is female-pessimal

We know it is male-optimal. Suppose there is a stable pairing S where some girl g does worse than in T

Let b be her mate in T

Let b* be her mate in S

By assumption, g likes b better than her mate b* in S

b likes g better than his mate in S (we already know that g is his optimal girl)

Therefore, S is not stable Contradiction

Theorem: The TMA pairing, T, is female-pessimal

suppose not

T

g

…and a stable matching S

that matches g-b*,

where b* is worse for g

b g

b*

b

g is his

optimal girl

prefers b

to b*

contradicts

stability of S!!!

(boy-optimal)

there exists

a girl g…

TMA = Gale Shapley Alg.

David Gale Not Shapley

Gale Shapley and Roth

David Gale Alvin E. Roth

Gale and Roth won the 2012 Nobel Price in Economics for applying stable matchings to assigning medical students to hospitals

not Lloyd Shapley

Theorem: The Gale Shapley Algorithm terminates after O(n2) iterations and produces a stable matching which is male-optimal and female-pessimal.

P vs. NP

$1,000,000

— the prize for solving any

of the Millennium Prize Problems

Millennium Prize Problems

1. Birch and Swinnerton-Dyer Conjecture

2. Hodge Conjecture

3. Existence & smoothness for Navier–Stokes

4. Poincaré Conjecture

5. P vs. NP

6. Riemann Hypothesis

7. Yang–Mills existence and mass gap



2. Hodge Conjecture



5. P vs. NP



The only one with

philosophical &

metamathematical

implications.



2. Hodge Conjecture



5. P vs. NP



Solved in 2003

by Grisha Perelman.

What is the P vs. NP problem?

Sudoku

Sudoku 3×3 × 3×3

Sudoku 4×4 × 4×4

Sudoku 4×4 × 4×4

No-Promises Sudoku

5

This one has no solution.

4×4 × 4×4

No-Promises Sudoku

5

This one has multiple solutions.

4×4 × 4×4

No-Promises Sudoku n×n × n×n

Given a partially filled n×n×n×n Sudoku grid,

output YES or NO: can it be validly completed?

Naive decision algorithm:

For each empty cell (≤ n4), try each possible digit.

Check if that’s a valid solution. Overall time ≈ nn4.

Smart decision algorithm: ???

Verifying a proposed solution: Time O(n4).


Naive decision algorithm: Time ≈ nn4.

Verifying a proposed solution: Time O(n4).

For n = 100 (meaning 10,000×10,000 grids):

Verifying a solution: ≈ 100M steps.

Your cell phone can do this in 1 second.

Naive algorithm: a number with ≈ 200M digits.

Insanely larger than # of quarks in the universe.


Question:

Is there a fixed constant c and an algorithm A

such that A solves the decision problem in

time O(nc)?

This is equivalent to

the P vs. NP problem!

Is this famous $1,000,000 problem

really about Sudoku?? Yes and no.

Here’s how P vs. NP is usually (informally) stated:

Let L be an algorithmic task.

Suppose there is an efficient algorithm

for verifying solutions to L.

Is there always also an efficient algorithm

for finding solutions to L?

“L∈NP”

“L∈P”

Isn’t Sudoku just one particular instance

of this question?

Let L be an algorithmic task.

Suppose there is an efficient algorithm

for verifying solutions to L.

Is there always also an efficient algorithm

for finding solutions to L?

“L∈NP”

“L∈P”

We’ll see: It’s true for all problems

if and only if it’s true for Sudoku!

Let’s develop these notions formally…

We’ll start by describing some

sample algorithmic problems.

3-Coloring

Input: A graph

Task: Decide if there is a 3-coloring.

If so, find one.

Circuit-Sat

Input: A boolean circuit C

Task: Decide if there is a 0/1

setting to the input

wires which “satisfies” C

(makes output wire 1).

If so, find such a setting.

AND OR

NOT AND

OR

x1 x2 x3

1 0 0

Hamiltonian Cycle

Input: A graph

Task: Decide if there is a Hamiltonian Cycle

in it, meaning a cycle that visits each

vertex exactly once. If so, find one.

Bipartite Perfect Matching

Input: A bipartite graph

Task: Decide if there is a perfect matching.

If so, find one.

Input: A partially filled

n2×n2 Sudoku grid

Task: Decide if there is a valid Sudoko

completion. If so, find one.

(No-Promises) Sudoku n×n × n×n

Decision vs. Search

Each of these problems was of the form,

“Does a solution exist? If so, find one.”

Decision problem Search problem

For simplicity, we focus on decision problems.

(Given a decision algorithm, it’s usually

easy to use it to solve the search problem.)

Reducing search to decision

AND OR

NOT AND

OR

x1 x2 x3

Example: Circuit-Sat

Suppose you have a good

decision alg. for Circuit-Sat.

How can you get a good alg.

for solving the search problem?

Hint:

Try fixing x1 to 0, fixing x1 to 1,

and running the decision alg. in both cases.

Decision problems as languages

Given G, does it have

a Hamiltonian cycle?

Given G, does

it have a perfect matching?

Given circuit C, does it have

a “satisfying” input string?

Given graph G, is it

3-colorable?

Given partially filled Sudoku

grid S, can it be completed?

Decision problems Languages in {0,1}*

HAM-CYCLE = {⟨G⟩ : G contains a

Hamiltonian cycle}

PMATCH = {⟨G⟩ : G has a

perfect matching}

CIRCUIT-SAT = {⟨C⟩ : C has a

satisfying input}

3-COL = {⟨G⟩ : G is 3-colorable}

SUDOKU = {⟨S⟩ : S can be validly

completed}

Given TM M and input x,

does M(x) halt? HALTS = {⟨M,x⟩ : M(x) halts}

Decision problems as languages

Languages in {0,1}*

PMATCH = {⟨G⟩ : G has a

perfect matching}

CIRCUIT-SAT = {⟨C⟩ : C has a

satisfying input}

3-COL = {⟨G⟩ : G is 3-colorable}

SUDOKU = {⟨S⟩ : S can be validly

completed}

HALTS = {⟨M,x⟩ : M(x) halts}

No Turing Machine

(or Java algorithm)

can decide this one.

Is there a TM

(or Java algorithm)

which decides

the others?

Of course!

HAM-CYCLE = {⟨G⟩ : G contains a

Hamiltonian cycle}

Efficiency

HAM, PMATCH, CIRCUIT-SAT, 3-COL, SUDOKU

can all be decided by “trying all possibilities”.

E.g., there is a naive algorithm for deciding

3-COL which runs in ≈ 3|V|

time.

We care about more than just

“Is there an algorithm?”

We care about

“Is there a reasonably ‘efficient’ algorithm?”

Polynomial time

50 years of computer science experience

shows it’s a very compelling definition:

• It’s independent of the “machine model”:

poly-time on a TM = poly-time on a RAM

= poly-time in Java = poly-time in Python

• It’s “robust”: plug a poly-time subroutine

into a poly-time algorithm: still poly-time.

• Empirically, it seems that most natural

problems with poly-time algorithms also

have efficient-in-practice algorithms.

Polynomial time

P

The set of all languages L such that

there is a fixed constant c

and an algorithm (TM) A

such that A decides L and

A runs in time O(|x|c) on all inputs x.

=

Definitions:

Decision/search problems

P

Algorithms and analysis:

Gale-Shapley Stable Matching Algorithm and Analysis

Study Guide

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Stable Matchings P vs NP - Carnegie Mellon School of ...aada/courses/15251s16/www/slides/15251-s16-lecture...The Traditional Marriage Algorithm For each day that some boy gets a “No”