The%20Stable%20Marriage%20Problem%20and%20Constraint%20Programming

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The Stable Marriage Problem and Constraint
Programming
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What is a SM?
MEN
WOMEN
1 2 3 6 1 4 5 1 6 5 1 3 2 4 2 4 6 1 2 5
3 2 2 4 6 1 3 5 3 1 4 5 3 6 2 3 4 3
6 2 5 1 4 6 5 3 4 2 1 4 1 3 5 4 2 6 5 2 3
1 4 5 6 5 3 2 6 1 4 5 6 3 1 2 5 6 4
6 5 1 3 4 6 2
What do you mean by stable? - assume mx is
married to wx, and mx prefers wy to wx
and wx
prefers my to mx - this is okay so long as wy
prefers her partner to mx
and my prefers his partner to wx
Harrison Ford is not going to leave his wife and
take Andrea away from me
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A Blocking Pair
MEN
WOMEN
1 2 3 6 1 4 5 1 6 5 1 3 2 4 2 4 6 1 2 5
3 2 2 4 6 1 3 5 3 1 4 5 3 6 2 3 4 3
6 2 5 1 4 6 5 3 4 2 1 4 1 3 5 4 2 6 5 2 3
1 4 5 6 5 3 2 6 1 4 5 6 3 1 2 5 6 4
6 5 1 3 4 6 2
M6 marries W6, his 5th choice W5 marries M1, her
4th choice M6 prefers W5 to W6 W5 prefers M6 to
M1 This is a blocking pair, and is unstable
Romeo marries Liz, Richard marries Juliet is
this stable?
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A Solution
MEN
WOMEN
1 2 3 6 1 4 5 1 6 5 1 3 2 4 2 4 6 1 2 5
3 2 2 4 6 1 3 5 3 1 4 5 3 6 2 3 4 3
6 2 5 1 4 6 5 3 4 2 1 4 1 3 5 4 2 6 5 2 3
1 4 5 6 5 3 2 6 1 4 5 6 3 1 2 5 6 4
6 5 1 3 4 6 2
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The Basic Gale Shapley (GS) Algorithm
1. L list of free men 2. While
isNotEmpty(L) 2.1 m selectFrom(L) // but
do not delete m from L 2.2 w
first(preferenceList(m)) // ms most preferred
partner 2.3 if isUnmarried(w) 2.3.1
marry(m,w) 2.3.2 L remove(L,m) 2.4 else //
w is already married 2.4.1 if w prefers m to
partner(w) 2.4.1.1 L add(L,partner(w))
// free, at last! 2.4.1.2 delete(preferenceList(
w),partner(w)) // divorced ( 2.4.1.3
delete(preferenceList(partner(w)),w) // divorced
( 2.4.1.4 marry(m,w) // w gets a better
partner ) 2.4.2 else 2.4.2.1
delete(preferenceList(w),m) // w will not
remarry for m 2.4.2.1 delete(preferenceList(m),w)
// and m stays in L
Note 1. Man optimal, men propose 2.
The order of L is immaterial 3. The
algorithm terminates with a man optimal (woman
pessimal) stable matching always
(there is always a stable matching) 4.
O(n2), and uses data structures initialised in
O(n2)
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The Extended Gale Shapley Algorithm (man optimal)
1. L list of free men 2. While
isNotEmpty(L) 2.1 m selectAndRemoveFrom(L)
// note, removed from L 2.2 w
first(preferenceList(m)) 2.3 if isMarried(w)
// unconditional remarriage! 2.3.1 L
add(L,partner(w)) 2.3.2 delete(preferenceList(w)
,partner(w)) 2.3.3 delete(preferenceList(partner(
w)),w) 2.4 marry(m,w) 2.5 for each m in
preferenceList(w) 2.5.1 if w prefers m to m //
remove unacceptable pairs 2.5.1.1
delete(preferenceList(w),m) 2.5.1.2
delete(preferenceList(m),w)
reduces the preference lists by eliminating
pairs that can readily be identified as not
belonging to any stable matching Gusfield and
Irving, 1.2.4, page 15, 1989
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The Extended Gale Shapley Algorithm

• We can apply the algorithm for men, and then for
  women
• The reduced preference lists are called the
  GS-lists
• The GS lists contain all matchings
• We can generate a man optimal matching by
  marrying each man
• to his first preference

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A (naïve) CP encoding of SM
MEN
WOMEN
1 2 3 6 1 4 5 1 6 5 1 3 2 4 2 4 6 1 2 5
3 2 2 4 6 1 3 5 3 1 4 5 3 6 2 3 4 3
6 2 5 1 4 6 5 3 4 2 1 4 1 3 5 4 2 6 5 2 3
1 4 5 6 5 3 2 6 1 4 5 6 3 1 2 5 6 4
6 5 1 3 4 6 2

• a permutation problem with n variables, each
  with n values
• n man variables and n woman variables, domain
  size of n
• a marriage constraint between each (man,woman)
  pair
• if manX marries womanY then womanY marries manX
• a stability constraint between each (man,woman)
  pair
• list all blocking pairs between a man and woman
• an allDifferent constraint

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The Stability Constraint
MEN
WOMEN
1 2 3 6 1 4 5 1 6 5 1 3 2 4 2 4 6 1 2 5
3 2 2 4 6 1 3 5 3 1 4 5 3 6 2 3 4 3
6 2 5 1 4 6 5 3 4 2 1 4 1 3 5 4 2 6 5 2 3
1 4 5 6 5 3 2 6 1 4 5 6 3 1 2 5 6 4
6 5 1 3 4 6 2

• by example, consider man 6 and woman 5
• the following marriages for (m6,w5) are
  unstable
• (6,1) m6 prefers w5 to w6 and w5 prefers m6 to
  m1
• (6,4) m6 prefers w5 to w6 and w5 prefers m6 to
  m4
• (6,5) m6 w5 w6 w5
  m6 m5
• (4,1) m6 prefers w5 to w4 and w5 prefers m6 to
  m1
• (4,4) w4
  m4
• (4,5) w4
  m5
• easily generated, as cross product of tails of
  preference lists
• a binary constraint, extensional representation,
  list of nogoods

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The Marriage Constraint
MEN
WOMEN
1 2 3 6 1 4 5 1 6 5 1 3 2 4 2 4 6 1 2 5
3 2 2 4 6 1 3 5 3 1 4 5 3 6 2 3 4 3
6 2 5 1 4 6 5 3 4 2 1 4 1 3 5 4 2 6 5 2 3
1 4 5 6 5 3 2 6 1 4 5 6 3 1 2 5 6 4
6 5 1 3 4 6 2

• a binary constraint between (man,woman) pairs
• manX Y -gt womanY X
• this is called a channeling constraint
• the allDiff constraint is now redundant

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Improving Propagation
MEN
WOMEN
1 1 3 6 2 4 5 1 1 5 6 3 2 4 2 4 6 1 2 5
3 2 2 4 6 1 3 5 3 1 4 5 3 6 2 3 4 3
6 2 5 1 4 6 5 3 4 2 1 4 1 3 5 4 2 6 5 2 3
1 4 5 6 5 3 2 6 1 4 5 6 3 1 2 5 6 4
6 5 1 3 4 6 2

• The problem has been slightly changed
• it should be obvious that in all stable
  matchings
• man 1 marries woman 1 (anything else is
  unstable)
• nogoods stability (3,5),(3,6), (3,4), ,
  (5,5), ,(5,4)
• achieving arc consistency does NOT detect this
• marriage (channeling) constraint and stability
  constraint operate
• independently
• however we can COMBINE marriage and stability
  constraints
• nogoods marriage (2,1),(3,1),,(6,1),(1,2),(1,3)
  ,,(1,6)
• we now have one (not two) tighter constraint
  capturing all the nogoods
• (2,1),(3,1),,(6,1), (1,2),(1,3),,(1,6),(3,5),(
  3,6),,(3,4),,(5,5),,(5,4)

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So?

• AC now detects inevitability of marriage of man
• 1 to woman 1
• AC reduces the domains more than the initial
• encoding
• However O(n4) encoding in space and time
• there are n(n-1) constraints
• each constraint is of size O(n2)
• this is also true of the naïve encoding

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AC and EGS, the main results

• the Arc Consistent domains are (extended) Gale
  Shapley lists
• Enumerating all solutions using MAC can result
  in failure driven backtracking
• we have examples of this
• the domains are not GAC
• they may contain pairs that are not in any
  stable matching
• SAC would achieve GAC
• Using value ordering and MAC results in failure
  free enumeration
• AC on SM is O(n4) whereas GS is O(n2)
• There is an O(n2) algorithm for achieving GAC

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A brief history on how we got there
1. Naïve encoding of Davids exam question in
choco (494) 2. Ian and Toby investigate
constrainedness of SM (489,500) 3. Barbara
encodes SM in Solver (497) it would be nice
if it could be solved polynomially in
constraint programming. and it might be
possible to show that AC by itself is sufficient
to remove all values which cannot participate
in a solution 4. David, Rob, Ian and Patrick
improve propagation (495) 5. Barbara shows that
enumerating solutions is not failure free
(498) 6. Proof that AC domains are GS-lists
(502) 7. Barbara finds a small counter example
for (5) above From 1 to 7 in 3 weeks, with 6
people
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Still to do
1. Prove that value ordering and MAC allows
failure free enumeration 2. Explore the
connection between rotations and MAC
enumeration 3. Investigate how to efficiently
encode AC and GAC for SM into a CP
language 4. Investigate AC and SMI