A branch and bound method to compute a median permutation

Irène Charon, Olivier Hudry

Télécom ParisTech

olivier.hudry@telecom-paristech.fr

A permutation problem in voting theory

Given a profile Π = (σ1, σ2, …, σm) of m permutations

(i.e. linear orders) σi (1 ≤ i ≤ m) on a set X of n = |X|

elements, how to aggregate them into a unique permutation which summarizes Π as accurately as possible?

In voting theory (Condorcet, 1784): we want to rank n candidates from the rankings provided by m voters.

Example

X = {a, b, c, d, e, f}, m = 5

voter 1: σ1 = a > b > c > f > d > e

voter 2: σ2 = a > c > f > b > d > e

voter 3: σ3 = e > d > a > f > b > c

voter 4: σ4 = b > c > d > e > f > a

voter 5: σ5 = c > f > b > e > a > d.

A combinatorial optimization problem

Symmetric difference distance d between R and R′ : d(R, R′ ) = |{(x, y) �∈ X2 with [xRy and not xR′ y]

or [not xRy and xR′ y]}|.

Let Σ be the set of all the permutations defined on X. Then, for Π = (σ1, σ2, …, σm):

Minimize ρΠ(σ) = for σ �∈ Σ

(cf. J.-P. Barthélemy, B. Monjardet, 1981)

∑=

σσm

iid

1),(

d(R, R′ ) measures the number of disagreements between R and R′.

ρΠ(σ) (= remoteness of σ from Π) measures the total number of disagreements between σ and Π.

σ* minimizing ρΠ over Σ is called a median permutation (or a median linear order) of Π.

Theorem (J.J. Bartholdi III et alii, 1989; O. Hudry, 1989; C. Dwork et alii, 2001):

The computation of σ* is NP-hard.

σ = (σxy)(x, y)∈�X2 with σxy = 1 if σ ranks x better than y (x >σ y) and σxy = 0 otherwise.

mxy = m – 2|{i: 1 ≤ i ≤ m and x >σi y}| = –myx

Then: ρΠ(σ) = C +

with :

∀ x �∈ X, σxx = 1 (reflexivity)

∀ (x, y) �∈ X2, x ≠ y, σxy + σyx = 1 (antisymmetry)

∀ (x, y, z) ∈ �X3, σxy + σyz – σxz ≤ 1 (transitivity)

∀ (x, y) �∈ X2, σxy �∈ {0, 1} (binarity)

A 0-1 linear programming problem

∑∈

σ2),( Xyx

xyxym

Lagrangean relaxation

Relaxation of the transitivity constraints:

∀ (x, y, z) �∈ X3, σxy + σyz – σxz ≤ 1

Lagrangean function L for σ = (σxy)(x, y)∈X2 with σxy ∈ {0, 1}, σxx = 1, σxy + σyx = 1, and Λ = (λxyz)(x, y, z)∈X3 with λxyz ≥ 0:

L(σ, Λ) = ρΠ(σ) +

= C +

with ∑∈

λ−λ+λ+=ΛXz

xzyzxyxyzxyxy ma )()(

( )∑∈

−σ−σ+σλ3),,(

1Xzyx

xzyzxyxyz

∑∑∈∈

λ−σΛ32 ),,(),(

)(Xzyxxyz

Xyxxyxya

Lagrangean relaxation (end)

Dual function for Λ = (λxyz)(x, y, z)∈X3 with λxyz ≥ 0:

D(Λ) = min{L(σ, Λ) with σ ∈ A}

with A = {reflexive and antisymmetric relations defined on X}.

Dual problem: maximize D(Λ) for Λ ≥ 0.

The maximum of D gives a lower bound of the minimum of ρΠ.

Computation of D(Λ) for a given Λ:

if axy ≥ 0, set σxy = 0, and σxy = 1 otherwise.

Resolution of the dual problem by subgradient methods.

The components of the BB algorithm

Initial bound: found by a metaheuristic (a self-tuned noising method; I. Charon and O. Hudry, 1993, 2009)

Evaluation function: provided by the Lagrangean relaxation.

Branching rule (J.-P. Barthélemy, A. Guénoche, O. Hudry, 1989; I. Charon, A. Guénoche, O. Hudry, F. Woirgard, 1996):

The root of the BB-tree contains all the permutations defined on X.A node of the BB-tree contains the permutations sharing a given beginning

section S (i.e. a permutation of a subset of X): S(xj1, xj2, …, xjp) = xj1 >σ xj2 >σ … >σ xjp.

The branching principle consists in expanding this beginning section:S(xj1, xj2, …, xjp, x) = xj1 >σ xj2 >σ … xjp >σ x

with x ∉{xj1, xj2, …, xjp}.

Shape of the BB-tree

x 1x 2

x n – 1x nx j 1

x j 1x j 2

x j p – 1> >> …

x j 1x j 2

x j p – 1> >> … x j p

>

… x j 1x j p – 1

>> x j p> x h n – p

>

……

… …

x j 1x j 2

x j p – 1>> x j p

> x h 1>… >

> …x j 2

… … … … … …

… …

Other components to prune the BB-tree

Hamiltonian permutations.

* We may summarize a profile Π of permutations by a tournament T (weighted by –mxy > 0): there is an arc (x, y) if a majority of voters prefer x to y (we assume that there is no tie).

* We say that a permutation σ is Hamiltonian if it induces a Hamiltonian path in T.

* Theorem (R. Remage and W.A. Thompson, 1966): a median permutation is Hamiltonian.

→ xj1 >σ xj2 >σ … >σ xjp is expanded into xj1 >σ … >σ xjp >σ x only if a majority of voters prefer xjp to x.

Example X = {a, b, c, d, e, f}

σ1 = a > b > c > f > d > eσ2 = a > c > f > b > d > eσ3 = e > d > a > f > b > cσ4 = b > c > d > e > f > aσ5 = c > f > b > e > a > d.

Here, a > c > f > b > d > eis a median permutation and induces a Hamiltonian path.

a b

c

d

f

e

a b

c

d

f

e

1

1

1

1

1

3

3

11

33

3

1

1

1

We compute the variation of ρΠ when, from a permutation σ beginning with S = xj1 >σ xj2 >σ … >σ xjp, we take an interval xjh >σ … >σ xjp (1 ≤ h ≤ p) and we shift it at the end of σ, after the elements of X – S (= OS = « out of section »):

σ = xj1 >σ xj2 >σ … xjh–1 >σ xjh >σ … >σ xjp >σ (OS)

becomes

σ′ = xj1 >σ′ xj2 >σ′ … xjh–1 >σ′ (OS) >σ′ xjh >σ′ … >σ′ xjp.

If ρΠ decreases, we do not keep the node associated with S.

OSmoves will count this kind of cuts.

Other components to prune the BB-tree

When we deal with a new beginning section

S = xj1 >σ xj2 >σ … xjh–1 >σ xjh >σ … >σ xjp >σ x,

we consider the beginning sections that we can get by moving, inside S, an “interval” of S including x, i.e., the beginning sections with the following shape:

xjh >σ′ … >σ′ xjp >σ′ x >σ′ xj1 >σ′ xj2 >σ′ … xjh–1.

If ρΠ decreases, we do not keep the node associated with S.

Smoves will count this kind of cuts.

Other components to prune the BB-tree (end)

An experimental result on the efficiency of the branch and bound components

Numbers of cuts for an instance on 39 candidates

0

1 0 0 0 0

2 0 0 0 0

3 0 0 0 0

4 0 0 0 0

5 0 0 0 0

6 0 0 0 0

7 0 0 0 0

8 0 0 0 0

h a m : 2 8 8 , 3 8 8

O S m o v e s : 2 0 7 , 0 2 2

S m o v e s : 9 3 , 2 9 6

l e x : 2 , 3 7 7

B e g S e c : 3 , 2 1 2

r e l a x : 9 2 , 9 3 5

n u m b e r o f s e p a r a t e d n o d e s : 2 1 , 3 2 3n u m b e r o f c u t n o d e s : 6 8 7 , 2 3 0

1 2 3 4 5 6 7 8 9 1 0 1 1 1 2 1 3 1 4 . . . 2 8

d e p t h - l e v e l

n u m b e r o f n o d e s

CPU times for m ∈ {3, 4, 100, 101}

CPU times in seconds (Rk: order = n).

0 , 1

1

1 0

1 0 0

1 0 0 0

1 5 2 0 2 5 3 0 3 5 4 0 4 5 5 0 5 5

3 : 0 . 8 0

4 : 0 . 6 0

1 0 0 : 0 . 6 7

1 0 1 : 0 . 6 8

a v e r a g e C . P . U . t i m e

o r d e r

Number of median permutations versus number of Hamiltonian permutations

Let M(n) and H(n) denote respectively the maximum number of median permutations or of Hamiltonian permutations for instances on n candidates.

If n is even with n ≥ 2: M(n) = n!

If n is odd: M(n) ≤ H(n).

Theorem (N. Alon, 1990): H(n) ≤ (c × n1.5 × n!)/2n where c is a constant.

Theorem (I. Charon, O. Hudry, 2000): for n = 3k,

30.75(n – 1)/n2 ≤ M(n).

Thank you for your attention!