Discrete Convexity in Joint Winner Property

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by  Yuni Iwamasa

Discrete Convexity in Joint Winner Property
Yuni Iwamasa∗ Kazuo Murota† Stanislav Zˇivny´‡
January 27, 2017
Abstract
In this paper, we reveal a relation between joint winner property (JWP) in the field of
7
valued constraint satisfaction problems (VCSPs) and M♮-convexity in the field of discrete
1
convex analysis (DCA). We introduce the M♮-convex completion problem, and show that
0
2 a function f satisfying JWP is Z-free if and only if a certain function f associated with f
is M♮-convex completable. This means that if a function is Z-free, then the function can
n
a be minimized in polynomial time via M♮-convex intersection algorithms. Furthermore we
J propose a new algorithm for Z-free function minimization, which is faster than previous

6 algorithms for some parameter values.
2

] 1 Introduction
M
D
A valued constraint satisfaction problem (VCSP) is a general and important framework for dis-
s. crete optimization (see [14] for details). Informally, the VCSP framework deals with the mini-
c mization problem of a function represented as the sum of “small” arity functions. An important
[
line of VCSP research is to investigate what instances are NP-hard and what instances are solv-

1 able in polynomial time. Cooper–Zˇivny´ [1] showed that if a function represented as the sum of
v
5 unary or binary functions satisfies the joint winner property (JWP), then the function can be
4 minimized in polynomial time.
6
Discrete convex analysis (DCA) [7] is a theory of convex functions on the integer lattice. It
7
0 is well known that a variety of polynomially solvable problems in discrete optimization can be
1. regarded as being related to discrete convexity. M♮-convex functions play an important role in
0 DCA, appearing in many areas such as operations research, economics, and game theory (see
7
e.g., [7, 8, 9]).
1
: The results of this paper are summarized as follows:
v
i • WerevealarelationbetweenJWPandM♮-convexity. Thatis,wegiveaDCAinterpretation
X
of polynomial-time solvability of JWP.
r
a
• To describe the connection of JWP and M♮-convexity, we introduce the M♮-convex comple-
tion problem, and give a characterization of M♮-convex completability.
• By utilizing a DCA interpretation of JWP, we propose a new algorithm for Z-free function
minimization, which is faster than previous algorithms for some parameter values.
Thisstudywillhopefullybethefirststeptowards fruitfulinteractions betweenVCSPsandDCA.
∗DepartmentofMathematicalInformatics,GraduateSchoolofInformationScienceandTechnology,University
of Tokyo,Tokyo, 113-8656, Japan.
Email: yuni [email protected]
†School of Business Administration, TokyoMetropolitan University,Tokyo, 192-0397, Japan.
Email: [email protected]
‡Department of Computer Science, Universityof Oxford, Oxford,OX1 3QD,United Kingdom.
Email: [email protected]
1
Notations. Let R and R denote the sets of reals and nonnegative reals, respectively. In this
+
paper, functions can take the infinite value +∞, where a < +∞, a+∞ = +∞ for a ∈ R, and
0·(+∞) = 0. Let R := R∪{+∞} and R := R ∪{+∞}. For a function f : {0,1}n → R,
+ +
the effective domain is denoted as dom f := {x ∈ {0,1}n | f(x) < +∞}. For a positive integer
k, we define [k] := {1,2,...,k}. For x = (x ,x ,...,x ) ∈ Rn, we define supp+(x) := {i ∈ [n] |
1 2 n
x >0}.
i
2 Preliminaries
Joint Winner Property. Let d ≥ 2 be a positive integer and D := [d ] for i ∈ [r]. We
i i i
consider a function f : D ×D ×···×D → R represented as the sum of unary or binary
1 2 r +
functions as
f(x ,x ,...,x ) = c (x )+ c (x ,x ), (1)
1 2 r i i ij i j
i∈[r] 1≤i<j≤r
X X
where c : D → R is a unary function for i ∈ [r] and c : D ×D → R is a binary function
i i + ij i j +
for 1 ≤ i < j ≤ r. Furthermore we assume c = c for distinct i,j ∈ [r]. A function f of the
ij ji
form (1) is said [1] to satisfy the joint winner property (JWP) if it holds that
c (a,b) ≥ min{c (b,c),c (a,c)} (2)
ij jk ik
for all distinct i,j,k ∈ [r] and all a ∈ D ,b ∈ D ,c ∈D . A function f of the form (1) satisfying
i j k
JWP is said to be Z-free if it satisfies that
|argmin{c (a,c),c (a,d),c (b,c),c (b,d)}| ≥ 2 (3)
ij ij ij ij
for any i,j ∈[r] (i 6= j), {a,b} ⊆ D (a 6= b), and {c,d} ⊆ D (c 6= d).
i j
Cooper–Zˇivny´ [1] showed that if f of the form (1) satisfies JWP, then f can be minimized in
polynomial time. In fact, they showed that if f satisfies JWP, then f can be transformed into a
certain Z-free function f′ in polynomial time such that a minimizer of f′ is also a minimizer of
f. Moreover they showed that a Z-free function can be minimized in polynomial time.
M♮-Convexity. Afunctionf :{0,1}n → Rissaid[7,8]tobeM♮-convexifforallx,y ∈ {0,1}n
and all i ∈ supp+(x−y) there exists j ∈ supp+(y−x)∪{0} such that
f(x)+f(y)≥ f(x−χ +χ )+f(y+χ −χ ), (4)
i j i j
where χ is the ith unit vector and χ is the zero vector. A function f : {0,1}n → R is said [7]
i 0
to be M♮-convex if f can be represented as the sum of two M♮-convex functions. It is well known
2
that M♮-convex functions can beminimized in polynomial time. Furthermore if we are given two
M♮-convex functions g and h, we can minimize an M♮-convex function f = g+h in polynomial
2
time by solving the so-called “M♮-convex intersection problem.”
Theorem 1 ([11, Theorem 10][13, Theorem 6.5]; see also [9, Theorem 3.3]). A function f :
{0,1}n → R with the zero vector in dom f is M♮-convex if and only if f satisfies the following
two conditions:
Condition 1: For all distinct i,j,k ∈ [n] and all z ∈ {0,1}n with supp+(z) ⊆ [n]\{i,j,k}, it
holds that
f(z+χ +χ )+f(z+χ ) ≥ min{f(z+χ +χ )+f(z+χ ),f(z+χ +χ )+f(z+χ )}.
i j k j k i i k j
(5)
2
Condition 2: For all distinct i,j ∈ [n] and all z ∈ {0,1}n with supp+(z) ⊆ [n]\{i,j}, it holds
that
f(z+χ +χ )+f(z) ≥ f(z+χ )+f(z+χ ).
i j i j
We pay special attention to quadratic M♮-convex functions. Using Theorem 1, we provide a
necessary and sufficient condition for the M♮-convexity of a function f : {0,1}n → R of the form
f(x ,x ,...,x ):= h x + h x x ((x ,x ,...,x ) ∈ {0,1}n), (6)
1 2 n i i ij i j 1 2 n
i∈[n] 1≤i<j≤n
X X
where we assume h = h and h < +∞ for i,j ∈[n].
ij ji i
Lemma 2. A function f of the form (6) is M♮-convex if and only if it satisfies the following:
• h ≥ min{h ,h } (i,j,k : distinct).
ij ik jk
• h ≥ 0 (i,j : distinct).
ij
In Lemma 2, h can take the infinite value +∞, whereas all h ’s are assumed to be finite in
ij ij
thecharacterization in[5]and[8]. Inparticular,werefertothefirstconditionh ≥ min{h ,h }
ij ik jk
(i,j,k : distinct) as the anti-ultrametric property. Note that no conditions are imposed on h .
i
The proof of Lemma 2 is in Section 5
By Lemma 2, we know that M♮-convexity of a function of the form (6) depends only on
quadraticcoefficients (h ) . We saythatafunctionf oftheform(6)isdefinedby(h )
ij i,j∈[n] ij i,j∈[n]
if the quadratic coefficients of f is equal to (h ) .
ij i,j∈[n]
3 M♮-Convexity in Joint Winner Property
M♮-Convex Completion Problem. We introduce the M♮-convex completion problem, and
give a characterization of an M♮-convex completable function on {0,1}n defined by (h ) .
ij i,j∈[n]
The M♮-convex completion problem is the following:
Given: (h ) such that h ∈ R or h is undefined for every distinct i,j ∈ [n].
ij i,j∈[n] ij ij
Question: By assigning appropriate values in R to “undefined” elements of (h ) , can we
ij i,j∈[n]
construct an M♮-convex function f : {0,1}n → R of the form (6)?
It should be clear that a defined element can be equal to +∞ and the infinite value (+∞)
may be assigned to undefined elements. If there is an appropriate assignment of (h ) ,
ij i,j∈[n]
then (h ) is said to be M♮-convex completable. If h < 0 or h < min{h ,h } holds
ij i,j∈[n] ij ij jk ik
for some defined elements h ,h ,h , then we obviously know that (h ) is not M♮-convex
ij jk ik ij i,j∈[n]
completable. Hence in considering the M♮-convex completion problem, we assume that
h ≥ 0, (7)
ij
h ≥ min{h ,h } (8)
ij jk ik
for all defined elements h ,h ,h .
ij jk ik
For quadratic coefficients H := (h ) containing undefined elements, we define the as-
ij i,j∈[n]
signmentgraphofH asagraphG = ([n],E ;w),whereE := {{i,j} |i 6= j and h is defined}
H H H ij
and w : E → R is defined by w({i,j}) := h for {i,j} ∈ E . Then the following theorem
H + ij H
holds.
3
Theorem 3. H := (h ) is M♮-convex completable if and only if |argmin w(e)| ≥ 2
ij i,j∈[n] e∈C
holds for every chordless cycle C of G .
H
The proof of Theorem 3 is in Section 5.
Remark 4. Farach–Kannan–Warnow [3] introduced the matrix sandwich problem for ultramet-
ric property, which contains the M♮-convex completion problem as a special case. They also
constructed an O(m+nlogn)-time algorithm for the matrix sandwich problem for ultrametric
property, where m is the number of defined elements. By using this algorithm, we can obtain an
appropriate M♮-convex completion in O(m+nlogn) time if one exists .
In this paper, we present a graphic characterization of M♮-convex completability. With this
characterization, we provide a DCA interpretation of polynomial-time solvability of JWP.
Transformation into a Function over {0,1}. To connect JWP and M♮-convexity, we intro-
duce a transformation of a function f : D ×D ×···×D → R into a function fˆ:{0,1}U → R,
1 2 r
where U is the set of all assignments to variables, that is,
U := {(1,1),(1,2),...,(1,d ),(2,1),(2,2),...,(2,d ),...,(r,1),(r,2),...,(r,d )}.
1 2 r
We consider the following correspondence between x = (x ,x ,...,x )∈ D ×D ×···×D and
1 2 r 1 2 r
xˆ = (xˆ ,...,xˆ ,xˆ ,...,xˆ ,...,xˆ ,...,xˆ ) ∈ {0,1}U:
(1,1) (1,d1) (2,1) (2,d2) (r,1) (r,dr)
(1,x1) (2,x2) (r,xr)
(x ,x ,...,x )7→ (0,...,0, ˇ1 ,0,...,0,0,...,0, ˇ1 ,0,...,0,...,0,...,0, ˇ1 ,0,...,0).
1 2 r
d1 d2 dr
(9)
| {z } | {z } | {z }
That is, xˆ = 1 means that we assign a to x , and xˆ = 0 means that we do not. In view
(i,a) i (i,a)
of (9), define a function fˆby
f(x) if there exists x satisfying (9),
fˆ(xˆ) := (xˆ ∈ {0,1}U).
(+∞ otherwise
Note that minimizing f is equivalent to minimizing fˆ.
Now we consider the transformation of f of the form (1) into fˆ, where f is given in terms of
c for i ∈ [r] and c for i,j ∈[r]. We define f :{0,1}U → R by
i ij
f(xˆ) := c (a)xˆ + h xˆ xˆ (xˆ ∈ {0,1}U), (10)
i (i,a) (i,a),(j,b) (i,a) (j,b)
(i,a)∈U (i,a),(j,b)∈U, (i,a)6=(j,b)
X X
where
c (a,b) if i6= j,
ij
h := (11)
(i,a),(j,b)
(undefined if i= j.
We also define δ : {0,1}U → R by
U
0 if there exists x satisfying (9),
δ (xˆ) := (xˆ ∈ {0,1}U),
U
(+∞ otherwise
which is the indicator function for the feasible assignments. Then we have
fˆ(xˆ)= f(xˆ)+δ (xˆ) (xˆ ∈ {0,1}U).
U
4
It is clear that δ is M♮-convex (dom δ is the base family of a partition matroid, which is
U U
a direct sum of matroids of rank 1). Moreover, the values assigned to the undefined elements
in (11) do not affect the value of fˆ. Hence if (h ) has an M♮-convex completion
(i,a),(j,b) (i,a),(j,b)∈U
(h˜ ) , then f defined by (h˜ ) is M♮-convex and fˆ = f + δ is
(i,a),(j,b) (i,a),(j,b)∈U (i,a),(j,b) (i,a),(j,b)∈U U
M♮-convex. This means that fˆ can be minimized in polynomial time. We need the values of
2
(h˜ ) in a minimization algorithm of M♮-convex functions.
(i,a),(j,b) (i,a),(j,b)∈U 2
A function of the form (1) satisfies JWP if and only if it holds that h ≥ 0 and
(i,a),(j,b)
h ≥ min{h ,h } for defined elements h ,h ,h given
(i,a),(j,b) (j,b),(k,c) (i,a),(k,c) (i,a),(j,b) (j,b),(k,c) (i,a),(k,c)
in (10). This means that (h ) satisfies the assumptions (7) and (8) for the M♮-
(i,a),(j,b) (i,a),(j,b)∈U
convex completion problem. Theorem 3 implies the following theorem (the proof is in Section 5).
Theorem 5. For a function f of the form (1), let (h ) be defined by (11). Then
(i,a),(j,b) (i,a),(j,b)∈U
(h ) is M♮-convex completable if and only if f is Z-free.
(i,a),(j,b) (i,a),(j,b)∈U
4 Algorithm
By using a general algorithm for the M♮-convex intersection (minimization of M♮-convex func-
2
tions), we can minimize Z-free functions of the form (1) in polynomial time. Suppose that we
are given c : D → R for i ∈ [r], c : D ×D → R for 1 ≤ i < j ≤ r, and a Z-free function
i i + ij i j +
f defined as (1). We can minimize f by minimizing fˆ= f +δ with an M♮-convex intersection
U
algorithm.
Here we take advantage of the fact that all the vectors in dom δ have a constant component
U
sum, i.e., xˆ = r for all xˆ ∈ dom δ . This implies that δ is an M-convex function [7]
(i,a)∈U (i,a) U U
and we can use an M-convex intersection algorithm, which is slightly simpler to describe. Since
P
the functions are definedon {0,1}n, the proposed algorithm is actually a variant of valuated ma-
′
troid intersection algorithms [6]. Specifically, let f denote the restriction of f to the hyperplane
containing dom δ , i.e.,
U
f(xˆ) if xˆ =r,
(i,a)
′
f (xˆ) :=  (i,Xa)∈U
+∞ otherwise.
Then minimizing f + δ is equivalentto minimizing f′ + δ , where f′ and δ are M-convex
U U U
functions.
The proposed algorithm consists of three steps.
Step 1: Onthebasisof Theorem5, weconstructanM♮-convex functionf :{0,1}U → Rin(10)
through an M♮-convex completion of (h ) in (11).
(i,a),(j,b) (i,a),(j,b)∈U
′
Step 2: We find a minimizer of f , to be used as an initial solution in Step 3.
′
Step 3: Wefindaminimizer off +δ by thesuccessive shortestpathalgorithm withpotentials
U
for the M-convex intersection [7] (see also [6, Section 5.2]).
′
In Step 3 of the algorithm, we usethe auxiliary graph G =(V,E ) definedfor xˆ ∈ dom f
xˆ,yˆ xˆ,yˆ
5
and yˆ∈ dom δ by
U
V := {s,t}∪U, (12)
′
E := {((i,a),(j,b)) |(i,a),(j,b) ∈ U, xˆ+χ −χ ∈ dom f }, (13)
xˆ (j,b) (i,a)
E := {((i,a),(j,b)) |(i,a),(j,b) ∈ U, yˆ+χ −χ ∈ dom δ }, (14)
yˆ (i,a) (j,b) U
E+ := {(s,(i,a)) |(i,a) ∈ supp+(xˆ−yˆ)}, (15)
E− := {((j,b),t) | (j,b) ∈ supp+(yˆ−xˆ)}, (16)
E := E ∪E ∪E+ ∪E− (17)
xˆ,yˆ xˆ yˆ
with the arc length function ℓ =ℓ : E → R given by
xˆ,yˆ xˆ,yˆ
′ ′
f (xˆ+χ −χ )−f (xˆ) if (u,v) ∈ E ,
v u xˆ
ℓ(u,v) := (18)
(0 otherwise.
Note that, by the definition of δ , we can also describe E as E = {((i,a),(i,b)) |i ∈[r], a,b ∈
U yˆ yˆ
D , (i,a) 6∈ supp+(yˆ), (i,b) ∈ supp+(yˆ)}.
i
Algorithm for Z-free function minimization.
Step 1: Find an M♮-convex completion (h˜ ) of (h ) , and define
(i,a),(j,b) (i,a),(j,b)∈U (i,a),(j,b) (i,a),(j,b)∈U
f :{0,1}U → R by
f(xˆ) = c (a)xˆ + h˜ xˆ xˆ (xˆ ∈ {0,1}U).
i (i,a) (i,a),(j,b) (i,a) (j,b)
(i,a)∈U (i,a),(j,b)∈U, (i,a)6=(j,b)
X X
Step 2: Let xˆ∗ ∈ {0,1}U be the zero vector. While xˆ∗ < r, do the following:
(i,a)∈U (i,a)
P
Step 2-1: Obtain (i,a)∗ ∈ argmin{f(xˆ∗+χ ) |(i,a) ∈ U \supp+(xˆ∗)}.
(i,a)
Step 2-2: xˆ∗ ← xˆ∗+χ .
(i,a)∗
Step 3: Let p : V → R be a potential defined by p(v) := 0 for v ∈ {s,t} ∪ U. Take any
yˆ∗ ∈ dom δ . While xˆ∗ 6= yˆ∗, do the following:
U
Step 3-1: MaketheauxiliarygraphGxˆ∗,yˆ∗. Definethemodifiedarclengthℓp : Exˆ∗,yˆ∗ → R
by ℓ (u,v) := ℓ(u,v)+p(u)−p(v) for (u,v) ∈ E
p xˆ∗,yˆ∗
Step 3-2: For each v ∈ V, compute the length ∆p(v) of an s-v shortest path in G
xˆ∗,yˆ∗
with respect to the modified arc length ℓ . Let P be an s-t shortest path having the
p
smallest number of arcs in G with respect to the modified arc length ℓ .
xˆ∗,yˆ∗ p
Step 3-3: For (i,a) ∈ U,
xˆ∗ −1 if ((i,a),(j,b)) ∈ P ∩E ,
(i,a) xˆ∗
xˆ∗(i,a) ← xˆ∗(i,a) +1 if ((j,b),(i,a)) ∈ P ∩Exˆ∗,

xˆ∗ otherwise,
(i,a)
yˆ(∗i,a)+1 if ((i,a),(j,b)) ∈ P ∩Eyˆ∗,
yˆ(∗i,a) ← yˆ(∗i,a)−1 if ((j,b),(i,a)) ∈ P ∩Eyˆ∗,

yˆ∗ otherwise.
(i,a)
For v ∈ V, p(v) ← p(v)+∆p(v). ✷
6
The time complexity of this algorithm is as follows, where n := |U| = d (the proof is
i∈[r] i
in Section 5).
P
Theorem 6. The proposed algorithm runs in O(nr3+nrlogn+n2) time.
By improving the algorithm of running time O(n3) given in [1], Cooper–Zˇivny´ [2] gave an
O(n2lognlogr)-time algorithm for minimizing Z-free functions of the form (1). Our proposed
algorithm is faster than Cooper–Zˇivny´’s for some r (e.g., r = O(n1/3)).
′
Remark 7. At the end of Step 2, we obtain a minimizer of f . The validity of Step 2 is given
in [10] in Japanese. For reader’s convenience we reproduce the proof of the validity here. For
′
ρ= 0,1,...,r define f by
ρ
f(xˆ) if xˆ = ρ,
(i,a)
′
fρ(xˆ) :=  (i,Xa)∈U
+∞ otherwise.
Then we have f′ = f′. It suffices toshow that if xˆ∗ ∈ argminf′, then there exists (i,a) ∈
r ρ
U \supp+(xˆ∗) such that xˆ∗+χ ∈argminf′ .
(i,a) ρ+1
Let xˆ be the vector in argminf′ with kxˆ − xˆ∗k minimum. Since xˆ >
ρ+1 1 (i,a)∈U (i,a)
xˆ∗ , supp+(xˆ − xˆ∗) is nonempty. Take (i,a) ∈ supp+(xˆ − xˆ∗). By the definition
(i,a)∈U (i,a) P
of M♮-convexity (4), there exists v ∈ supp+(xˆ∗−xˆ)∪{0} such that
P
f(xˆ)+f(xˆ∗)≥ f(xˆ−χ +χ )+f(xˆ∗+χ −χ ). (19)
(i,a) v (i,a) v
If v ∈supp+(xˆ∗−xˆ), then xˆ−χ +χ ∈ dom (f′ ) and xˆ∗+χ −χ ∈ dom (f′), which
(i,a) v ρ+1 (i,a) v ρ
imply
f(xˆ−χ +χ ) ≥ f(xˆ), f(xˆ∗+χ −χ ) ≥ f(xˆ∗). (20)
(i,a) v (i,a) v
′
By (19)and(20), weobtain xˆ−χ +χ ∈ argminf . Thisisacontradiction tothechoice of
(i,a) v ρ+1
xˆ, sincek(xˆ−χ +χ )−xˆ∗k = kxˆ−xˆ∗k −2. Hencewehavev = 0. Thenf(xˆ∗+χ ) = f(xˆ)
(i,a) v 1 1 (i,a)
and therefore, xˆ∗+χ ∈ argminf′ .
(i,a) ρ+1
5 Proofs
In this section, we give the proofs of Lemma 2, Theorem 3, Theorem 5, and Theorem 6.
Proof of Lemma 2. (only-if part). Suppose that there exist distinct i,j,k ∈ [n] such that
h < min{h ,h }. Note that h < +∞ holds. Then
ij jk ik ij
f(χ +χ )+f(χ )= h +h +h +h < h +h +h +h =f(χ +χ )+f(χ ),
i j k i j k ij i j k jk j k i
f(χ +χ )+f(χ )= h +h +h +h < h +h +h +h = f(χ +χ )+f(χ )
i j k i j k ij i j k ik i k j
hold since h ,h ,h ,h < +∞. By Condition 1 of Theorem 1, f is not M♮-convex.
i j k ij
Suppose that there exist distinct i,j ∈ [n] such that h < 0(< +∞). Then
ij
f(χ +χ )+f(χ ) = h +h +h < h +h = f(χ )+f(χ )
i j 0 i j ij i j i j
holds since h ,h < +∞. By Condition 2 of Theorem 1, f is not M♮-convex.
i j
7
(if part). Take arbitrary distinct i,j,k ∈ [n] and z ∈ {0,1}n with supp+(z) ⊆ [n]\{i,j,k}.
If f(z +χ +χ ) = +∞ or f(z +χ ) = +∞ holds, then Condition 1 of Theorem 1 obviously
i j k
holds. We assume f(z+χ +χ )< +∞ and f(z+χ )< +∞.
i j k
It holds that
f(z+χ +χ ) = f(z)+h +h + h + h +h , (21)
i j i j ip jp ij
p∈supp+(z) p∈supp+(z)
X X
f(z+χ ) = f(z)+h + h . (22)
k k kp
p∈suXpp+(z)
Note that all terms appearing in (21) and (22) have finite values since f(z+χ +χ ) < +∞ and
i j
f(z+χ )< +∞ hold. Then we have
k
f(z+χ +χ )+f(z+χ )≥ f(z+χ +χ )+f(z+χ )
i j k j k i
⇔ 2f(z)+h +h +h + h + h + h +h
i j k ip jp kp ij
p∈suXpp+(z) p∈suXpp+(z) p∈suXpp+(z)
≥ 2f(z)+h +h +h + h + h + h +h
j k i jp kp ip jk
p∈supp+(z) p∈supp+(z) p∈supp+(z)
X X X
⇔ h ≥ h .
ij jk
Also we have
f(z+χ +χ )+f(z+χ ) ≥ f(z+χ +χ )+f(z+χ )
i j k j k i
⇔ h ≥ h .
ij ik
By the assumption, it holds that h ≥ min{h ,h }. Hence we obtain
ij jk ik
f(z+χ +χ )+f(z+χ )≥ min{f(z+χ +χ )+f(z+χ ),f(z+χ +χ )+f(z+χ )}.
i j k j k i i k j
By the assumption of h ≥ 0, we also obtain
ij
f(z+χ +χ )+f(z) ≥ f(z+χ )+f(z+χ )
i j i j
for all distinct i,j ∈ [n].
Proof of Theorem 3. First we give a graphical interpretation for the anti-ultrametric prop-
erty. For H := (h ) and α∈ R, let us define Eα and Vα by
ij i,j∈[n] H H
Eα := {{i,j} ∈ E |h ≥α}, (23)
H H ij
Vα := {i | ∃e∈ Eα such that i∈ e}. (24)
H H
Let Gα := (Vα,Eα). Then the following lemma holds:
H H H
Lemma 8. H := (h ) satisfies the anti-ultrametric property if and only if each connected
ij i,j∈[n]
component of Gα is a complete graph for every α ∈ R.
H
Proof. (only-if part). We show the contraposition. Suppose that for some α ∈ R there exists a
non-complete graph among the connected components of Gα. Then there exist distinct i,j,k ∈
H
[n] with {i,j},{j,k} ∈ Eα 6∋ {i,k}. By the definition of Eα, it holds that min{h ,h } ≥ α >
H H ij jk
h . This means that {h ,h ,h } does not satisfy the anti-ultrametric property.
ik ij jk ik
8
(if part). Suppose that each connected component of Gα is a complete graph for all α ∈ R.
H
To show the anti-ultrametric property of (h ) , it suffices to prove h = h for all distinct
ij i,j∈[n] jk ik
i,j,k satisfying h > h . If h ≥ h , then there exists a non-complete graph among the
ij jk ik ij
connected components of Gα for α = h , which is a contradiction. If h > h > h , then
H ij ij ik jk
there exists a non-complete graph among the connected components of Gα for α = h , which
H ik
is a contradiction. If h > h , then there exists a non-complete graph among the connected
jk ik
components of Gα for α = h , which is a contradiction. Therefore we must have h = h . ✷
H jk jk ik
We are now ready to prove Theorem 3.
Proof of Theorem 3. (only-if part). Suppose to the contrary that H := (h ) is M♮-convex
ij i,j∈[n]
completableandthatthereexistsachordlesscycleC ofG with|argmin w(e)| = 1. LetC =
H e∈C
{{i ,i },{i ,i },...,{i ,i }}, and consider the corresponding entries {h ,h ,...,h } of
1 2 2 3 m 1 i1i2 i2i3 imi1
H. Note that h is undefined for p,q ∈ [m] with |p − q| =6 1 mod m. We may assume
ipiq
α := h = min{h ,h ,...,h }. By the assumption of |argmin w(e)| = 1, we have
i1i2 i1i2 i2i3 imi1 e∈C
min{h ,...,h } > α. Since {h ,h ,h } should satisfy the anti-ultrametric prop-
i2i3 imi1 i1i2 i2i3 i1i3
erty, we have to assign α to h to obtain an M♮-convex completion. Since {h ,h ,h }
i1i3 i1i3 i3i4 i1i4
should satisfy the anti-ultrametric property, we have to assign α to h to obtain an M♮-
i1i3
convex completion. By repeating this procedure, we arrive at h = α. This is a contra-
i1im−1
diction, since h < min{h ,h } and hence the anti-ultrametric property fails for
i1im−1 im−1im i1im
{h ,h ,h }.
i1im−1 im−1im imi1
(if part). For α ∈ R , define Sα by
+ H
Sα := {{i,j} 6∈ E | i,j ∈ Vα, i and j are connected in G [Eα]},
H H H H H
where G [Eα] is the subgraph of G induced by Eα. Let G˜α := (Vα,Eα ∪Sα). Recall that
H H H H H H H H
Eα and Vα are defined in (23) and (24).
H H
Firstwe showthatif each connected componentof G˜α is acomplete graphfor every α ∈ R ,
H +
then H is M♮-convex completable. Let α > α > ··· > α be the distinct values of defined
1 2 m
elements of (h ) (α can be the infinite value). Note that for each undefined element h ,
ij i,j∈[n] 1 ij
{i,j} is in S for some α ∈ {α ,α ,...,α }. We assign α to each undefined element h such
α 1 2 m 1 ij
that {i,j} ∈ Sα1, and α to each h such that {i,j} ∈ Sαk \Sαk−1. Then we obtain a certain
H k ij H H
completion H˜ := (h˜ ) of (h ) . It is clear that each connected component of Gα is a
ij i,j∈[n] ij i,j∈[n] H˜
complete graph for every α ∈ R. By Lemma 8, H˜ satisfies the anti-ultrametric property. This
means that H is M♮-convex completable.
Next we show that if |argmin w(e)| ≥ 2 holds for every chordless cycle C of G , then
e∈C H
each connected component of G˜α is a complete graph for every α ∈ R . Take arbitrary α ∈ R
H + +
and i and j which are connected in G˜α (Note that vertex sets of connected components of G˜α
H H
are the same as those of G [Eα]). It suffices to prove that {i,j} ∈ Eα or {i,j} ∈ Sα holds.
H H H H
Suppose to the contrary that there exist i and j such that {i,j} 6∈ Eα and {i,j} 6∈ Sα hold. Let
H H
I be the set of such {i,j}. Let {i ,j } ∈ I be a pair of vertices such that the number of edges
0 0
of a shortest i -j path on G [Eα] is minimum in I. Since i and j are connected in G [Eα]
0 0 H H 0 0 H H
and {i ,j } 6∈ Sα, we have {i ,j } ∈ E . Moreover since {i ,j } 6∈ Eα, h < α holds. Take
0 0 H 0 0 H 0 0 H i0j0
a i -j shortest path P . Then P ∪{i ,j } is a chordless cycle of G . Indeed, if P ∪{i ,j }
0 0 0 0 0 0 H 0 0 0
has a chord in G , there exist i′ and j′ satisfying {i′,j′} =6 {i ,j } in P ∪{i ,j } such that
H 0 0 0 0 0
Eα 6∋ {i′,j′} ∈ E by the minimality of |I|. Then {i′,j′} ∈ I and the number of edges of a
H H
shortest i′-j′ path is smaller that those of P . However this is a contradiction to the minimality
0
of {i ,j }. Hence P ∪{i ,j } isa chordless cycle of G . Itholdsthat h ≥ αfor {i,j} ∈P and
0 0 0 0 0 H ij 0
h < α. Therefore we obtain |argmin w(e)| = 1. This contracts the assumption of
i0j0 e∈P0∪{i0,j0}
|argmin w(e)| ≥ 2. Hence we have {i,j} ∈ Eα or {i,j} ∈ Sα. ✷
e∈C H H
9
Proof of Theorem 5. Let H := (h ) , where the entries h with i = j
(i,a),(j,b) (i,a),(j,b)∈U (i,a),(j,b)
are undefined. Recall that G = (U,E ;w) is the assignment graph of H. By the definition
H H
of E and h in (11), we have E = {{(i,a),(j,b)} | i 6= j, a ∈ D , b ∈ D }. By
H (i,a),(j,b) H i j
Theorem 3, H is M♮-convex completable if and only if every chordless cycle C satisfies the
condition |argmin w(e)| ≥ 2.
e∈C
First we show that chordless cycles in G have length 3 or 4. Take any chordless cycle
H
C = {{(i ,a ),(i ,a )},{(i ,a ),(i ,a )},...,{(i ,a ),(i ,a )}} of G . SinceC is chordless, we
1 1 2 2 2 2 3 3 k k 1 1 H
have i = i for 3 ≤ p ≤ k−1 and i = i for 4 ≤ q ≤ k. This implies k ≤ 4, since otherwise we
1 p 2 q
obtain i = i = i , contradicting the existence of an edge between (i ,a ) and (i ,a ).
1 4 2 1 1 2 2
Fora(chordless)cycleoflength3,say,C = {{(i ,a ),(i ,a )},{(i ,a ),(i ,a )},{(i ,a ),(i ,a )}}
1 1 2 2 2 2 3 3 3 3 1 1
with i 6= i 6= i 6= i , the condition |argmin w(e)| ≥ 2 is equivalent to (2) for JWP. For a
1 2 3 1 e∈C
chordlesscycleoflength4,say,C = {{(i ,a ),(i ,a )},{(i ,a ),(i ,a )},{(i ,a ),(i ,a )},{(i ,a ),(i ,a )}}
1 1 2 2 2 2 3 3 3 3 4 4 4 4 1 1
we have i 6= i , i 6= i , i = i , i = i , a 6= a , a 6= a , and then the condition
1 2 3 4 1 3 2 4 1 3 2 4
|argmin w(e)| ≥ 2 is equivalent to (3) for Z-freeness.
e∈C
Proof of Theorem 6. We investigate each step in turn.
(Step 1). Since the number of defined elements of (h ) is O(n2− r d2) =
(i,a),(j,b) (i,a),(j,b)∈U i=1 i
O(n2), we can find an M♮-convex completion in O(n2+nlogn) time (recall Remark 4).
P
(Step 2). If we have the value of f(xˆ∗), we can compute the value of f(xˆ∗ +χ ) in O(r)
(i,a)
time since f(xˆ∗+χ )= f(xˆ∗)+c (a)+ h˜ . Hence the time complexity
(i,a) i (j,b)∈supp+(xˆ∗) (i,a),(j,b)
of Step 3 is O(nr2) time.
P
(Step 3). Recall the definition of G in (12)–(18). We have |E | = O(r(n−r)) = O(nr),
xˆ∗,yˆ∗ xˆ∗
|E | = O(n), |E+| = O(r), and |E−| = O(r). Hence |E | = O(nr). Furthermore we need to
yˆ∗ xˆ∗,yˆ∗
compute ℓ only on E , since ℓ is equal to zero on other arcs. If we have the value of f(xˆ∗) at
xˆ∗
hand, we can compute the value of f(xˆ∗+χ −χ ) in O(r) time since
(j,b) (i,a)
f(xˆ∗+χ −χ )
(j,b) (i,a)
= f(xˆ∗)− c (a)+ h˜ + c (b)+ h˜ .
i (i,a),(k,c) j (j,b),(k,c)
   
(k,c)∈sXupp+(xˆ∗) (k,c)∈suppX+(xˆ∗−χ(i,a))
   
Therefore we can construct the auxiliary graph G in O(nr2) time.
xˆ∗,yˆ∗
The modified arc length ℓ is nonnegative [6, Section 5.2]. Hence we can compute ∆p(v) for
p
v ∈ V and a shortest path P in Step 3-2 in O(nr+nlogn) time by using Dijkstra’s algorithm
with Fibonacci heaps [4] (see also [12, Section 7.4]). We can update xˆ∗, yˆ∗, and p in Step 3-3 in
O(nr) time. By one iteration of Step 3, the value of kxˆ∗ −yˆ∗k is decreased by two. Hence the
1
number of iterations of Step 3 is bounded by O(r). Therefore the time complexity of Step 3 is
O(nr3+nrlogn).
By the above discussion, we see that the proposed algorithm runs in O(nr3+nrlogn+n2)
time.
Acknowledgments
We thank Kazutoshi Ando and Takanori Maehara for information on the paper [3] in Remark 4.
ThisresearchwasinitiatedattheTrimesterProgram“CombinatorialOptimization”atHausdorff
Institute of Mathematics, 2015. The first author’s research was supported by JSPS Research
Fellowship for Young Scientists. The second author’s research was supported by The Mitsubishi
Foundation, CREST, JST, and JSPS KAKENHI Grant Number 26280004. The last author’s
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