1. Blah blah blah

2. Zoe’s shark

3. Thanks to the conference artist
Phoebe

4. A Constraint Programming Approach to The Stable Marriage Problem

6. What is the Stable Marriage Problem? (SM)
We have n men
and n women
Each man ranks the n women
and each woman ranks the men
Men
Women 1 2 3 4 : : : : 1 2 3 4 : : : :
Each man has to marry a woman.
Bigamy is not allowed.
Marriages must be stable
done

7. No Bigamy?
There is a bijection from men to women
man mi marries woman wj  woman wj marries man mi
No, not bigamy … bijection 

8. Stable?
An example 1 2 3 4 : : : : : : : :
man 2 marries woman <2,4>
man 3 marries woman <3,1>
man 2 prefers woman 1 to woman 4
woman 1 prefers man 2 to man 3
man 2 and woman 1 will elope

9. There are algorithms for this
The Gale Shapely algorithm, O(n2)
problem has n variables, each with n partners
and is of size O(n2)
The complexity of the algorithm is linear in the
size of the input

10. The Extended Gale Shapley Algorithm (man optimal)
1 L := list of free men
2. While L  
2.1 m := selectAndRemoveFrom(L) // note, m is removed from L
2.2 w := first(preferenceList(m)) // m’s most preferred partner
2.3 if isMarried(w) // unconditional remarriage!
L := add(L,partner(w)) // w’s partner is free … again!
delete(preferenceList(w),partner(w))
delete(preferenceList(partner(w)),w)
2.4 marry(m,w) // m and w get married. 
2.5 for each m’ in preferenceList(w)
if w prefers m to m’ // remove unacceptable pairs
delete(preferenceList(w),m’)
delete(preferenceList(m’),w)
Now you’ve seen it

11. The Extended Gale Shapley Algorithm (man optimal)
“… reduces the preference lists by eliminating pairs that can readily be identified as not belonging to any stable matching” Gusfield and Irving, 1989

12. male-oriented extended Gale-Shapely algorithm
An example of the
male-oriented extended
Gale-Shapely algorithm
Male-oriented EGS

13. Free: 1,2,3,4
m = 1 (first free man)
w = 3 (man 1’s first choice)
engaged(m,w)
delete pairs : <3,3>,<4,3>,<2,3>
Men 1: 2: 3: 4:
Women 1: 2: 3: 4:
Men 1: 2: 3: 4:
Women 1: 2: 3: 4:
Men 1: 2: 3: 4:
Women 1: 2: 3: 4:
Men 1:
2: 2 1 4 3: 4:
Women 1: 2:
3: 1 4:

14. Free: 2,3,4
m = 2 (first free man)
w = 2 (man 2’s first choice)
engaged(m,w)
delete pairs : <1,2>,<4,2>
Men 1:
2: 2 1 4 3: 4:
Women 1: 2:
3: 1 4:
Men 1:
2: 2 1 4 3: 4:
Women 1: 2:
3: 1 4:
Men 1:
2: 2 1 4 3: 4:
Women 1:
2: 3 2
3: 1 4:
Men 1:
2: 2 1 4 3: 4:
Women 1: 2:
3: 1 4:

15. Free: 3,4
m = 3 (first free man)
w = 1 (man 3’s first choice)
engaged(m,w)
delete pairs : <4,1>,<1,1>
Men 1:
2: 2 1 4 3: 4:
Women 1:
2: 3 2
3: 1 4:
Men 1:
2: 2 1 4 3: 4:
Women 1:
2: 3 2
3: 1 4:
Men 1:
2: 2 1 4 3: 4:
Women 1:
2: 3 2
3: 1 4:
Men 1:
2: 2 1 4 3: 4:
Women 1:
2: 3 2
3: 1 4:
Men 1:
2: 2 1 4 3:
4: 4
Women
1: 2 3
2: 3 2
3: 1 4:

16. Free: 4
m = 4 (first free man)
w = 4 (man 4’s first choice)
engaged(m,w)
delete pairs : <1,4>,<2,4>,<3,4>
Men 1:
2: 2 1 4 3:
4: 4
Women
1: 2 3
2: 3 2
3: 1 4:
Men 1:
2: 2 1 4 3:
4: 4
Women
1: 2 3
2: 3 2
3: 1 4:
Men
1: 3
2: 2 1 3:
4: 4
Women
1: 2 3
2: 3 2
3: 1
Men 1:
2: 2 1 4 3:
4: 4
Women
1: 2 3
2: 3 2
3: 1 4:

17. Male-oriented EGS Free:  Men 1: 3 1 2 4 2: 2 1 4 3 3: 1 3 2 44:
Women 1: 2: 3: 4:
before
Men
1: 3
2: 2 1 3:
4: 4
Women
1: 2 3
2: 3 2
3: 1
after
done

18. The Extended Gale Shapley Algorithm (man optimal)
On completion we have man optimal GS-lists
if men marry their 1st partners we get
man optimal stable marriages
female pessimal stable marriages
We can also apply female oriented EGS
The intersection of the man and woman lists
gives us GS-lists

19. A Constraint Encoding (a 1st stab)2 3 4 : : : : : : : :
2n variables, each with a domain (1 .. n)
n2 channeling constraints
vi = j  vj = i
n2 stability constraints
2 allDiff’s (one for men, one for women)
done

20. What’s a stability constraint?
n2 stability constraints 1 2 3 4 : : : : : : : :
An example, between man m4 and woman w3
(1) locate position of w3 in m4’s preference list
(2) locate position of m4 in w3’s preference list
(3) compute the Cartesian product beyond these points
{(1,2),(2,2)}
these are nogoods
done

21. Arc consistency does not compute the GS-lists We require search
Encoding is O(n4)
allDiff’s are redundant
No good news then?

22. Hey! It was a 1st stab!

23. Combining the Constraints
The channeling constraints guarantee a bijection 1 2 3 4 : : : : : : : :
m4 = 3  w3 = 4
Nogoods: m4 marries w3 and w3 marries some one else!
{(3,1),(3,3),(3,2)}
Nogoods: w3 marries m4 and m4 marries some one else!
{(4,4),(1,4),(2,4)}

24. Combining the Constraints1 2 3 4 : : : : : : : :
(m4 = 3  w3 = 4)  stable(m4,w3)
Combined nogoods:
{(3,1),(3,3),(3,2),(4,4),(1,4),(2,4),(1,2),(2,2)}
Do this for all man/woman pairs

25. Combining the Constraints
Arc consistency now computes the GS-lists
we have a proof
We have failure free enumeration
with a value ordering heuristic
The encoding trivially extends to SMTI
(an NP-complete problem)
The encoding is O(n4)
So, some good news then?

26. The constraint has structure1 2 3 4 : : : : : : : :
Combined nogoods:
{(3,1),(3,3),(3,2),(4,4),(1,4),(2,4),(1,2),(2,2)} m4 w3 m4 w3
Arrange in preference order
Can’t see it?

27. The constraint has structure1 2 3 4 : : : : : : : :
Combined nogoods:
{(3,1),(3,3),(3,2),(4,4),(1,4),(2,4),(1,2),(2,2)} m4 w3
I is illegal, bigamy
B is blocking, unstable
A is allowed
More generally

28. Revision of constraints
Revising only certain shapes of constraints has an effect mi wk mi wk
wk is top of mi’s list
mi is top of wk’s list
Revise(i,k)
Revise(i,k)
Done

29. Nice, interesting, … but still O(n4) 

30. An O(n2) encoding (a 1st stab)
In a man’s world
The intuition, part 1a
Man to woman: I love you. Marry me. Please say yes!
Woman to man: that’s the best offer I’ve had. Okay.
(and by the way, all you other guys, beat it!)
The intuition, part 1b
Man to woman: I love you. Marry me. Please say yes!
Woman to man: Who do you think you are! I’ve married someone
better than you. Beat it!
Man to man: Och well. There’s always someone else.

31. An O(n2) encoding (a 1st stab)
The intuition, part 1.1a
Assume man mi has proposed to his jth choice, woman wk
assume wk is not engaged to someone she prefers to mi
wk should reject any proposals from men less preferred than mi
The intuition, part 1.1b
Assume man mi has proposed to his jth choice, woman wk
assume wk is engaged to someone she prefers to mi
mi should then propose to his j+1th choice
The following description is male-oriented
and has a symmetric female orientation

32. An O(n2) encoding (a 1st stab)
mi,j = 1  mi has proposed to his jth choice or worse
wk,h = 0  wk has accepted a proposal from better
than her hth choice
wk,h = 0  wk,h+1 = 0
If wk has accepted a proposal from her hth choice or better
then wk has accepted a proposal from her h+1th choice or better

33. An O(n2) encoding (a 1st stab)
m3 proposes to 2nd choice
w6 has been proposed to by better than her 5th

34. An O(n2) encoding, the stability constraints
preference
person
Man m3 proposed to his 2nd choice, w6
Suppose man mi has a partner no better than jth choice, i.e. mi,j = 1 and jth choice is woman wk
Suppose woman wk has a partner better than hth choice, i.e. wk,h = 0 and hth choice is man mi
preference
person
Woman w6 has a partner better than her 7th choice, m3
Therefore mi is rejected by wk, and mi has a partner no better
than his j+1th choice, i.e. mi,j+1 = 1

35. An O(n2) encoding, the stability constraints
 mi is rejected by wk, and mi has a partner no better
than his j+1th choice, i.e. mi,j+1 = 1
wk is mi’s jth choice
mi is wk’s hth choice
the constraint is
This increments the pointer to the head of mi’s preference list

36. An O(n2) encoding, the stability constraints
Otherwise mi becomes engaged to wk, and wk now has a partner better
than her h+1th choice, i.e. wk,h+1 = 0
wk is mi’s jth choice
mi is wk’s hth choice
This deletes the tail of wk’s preference list

37. An O(n2) encoding
Obviously, symmetrically we have the female orientation

38. An O(n2) encoding
On the completion of (male-oriented) propagation
the preference list for mi
the maximum value of j such that mij = 1 gives the head of the list
the preference list for wi
minimum value of j such that wi,j+1 = 0 gives the tail of the list
when we do male and female propagation
we get the XGS-list, a superset of the GS-list
= (6,5,3)
= (5,6)

39. An O(n2) encoding … the complexity
2n2 variables
domains are {0,1}
6n(n-1) constraints
arc consistency complexity is O(edr)
where e is number of constraints
d is domain size
r is arity of constraints
for our encoding e = 6n(n-1), d = 2, r  3
O(n2)
This is same complexity as the GS algorithm, and is linear in the size of the input. It is optimal

40. A 2nd stab at the 0/1 encoding
Generalises to incomplete list (SMI)
we prove that the AC on the 0/1 model gives XGS-lists
we prove that we get failure free enumeration

41. Conclusions
A naïve encoding as a csp
However, combining the constraints
AC gives us GS-lists and failure free enumeration
easily extend to SMTI
O(n4) space
0/1 encoding
gives us XGS-lists and failure free enumeration for SMI
O(n2), optimal!
Non-trivial
SM constraint may be part of a bigger problem
something like allDiff
Question
“If there is a special algorithm, such as GS, is it worth encoding as a csp?”
In this case “Yes.”

42. Questions?