Simple fair division of a square

Journal of Mathematical Economics 86 (2020) 35–40

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Simple fair division of a square Jerzy Legut Faculty of Pure and Applied Mathematics, Wrocław University of Science and Technology, 50-370 Wrocław, Poland

article

info

Article history: Received 31 May 2019 Received in revised form 28 August 2019 Accepted 5 November 2019 Available online 15 November 2019 Keywords: Equitable division Land division Cake-cutting

a b s t r a c t Suppose we are given a cake represented by the unit interval to be divided among agents evaluating the pieces of the cake by nonatomic probability measures. It is known that we can divide the unit interval into contiguous and connected pieces and assign them to the agents in such a way that the values of the pieces are equal according to the individual agents measures. Such division is said to be equitable and simple. In this paper we show that an equitable and simple division also exists in the case of dividing two-dimensional cake represented by the unit square. In this case, by simple division we mean dividing the unit square firstly by horizontal cuts, and then partition the resulting rectangles by vertical cuts. We give a method of obtaining a proportional and simple division of this cake. Furthermore, we prove the existence of proportional, equitable and simple division. © 2019 The Author(s). Published by Elsevier B.V. This is an open access article under the CC BY-NC-ND license (http://creativecommons.org/licenses/by-nc-nd/4.0/).

1. Introduction 1.1. Preliminaries Let (X , BX ) be a measurable space, where X represents an object (e.g. a cake) to be divided among n agents (players) and BX is a set of all admissible parts of a partition. Assume that nonatomic probability measures µ1 , . . . , µn defined on BX are used by the agents to evaluate the size of sets A ∈ BX . Denote by I := {1, . . . , n} the set of numbered agents. For the purpose of this paper we introduce the following definition of a division. Definition 1. By an ordered division (partition) P = {Ai }ni=1 of the cake X among the players i ∈ I with assigned to them probability measures µi , i ∈ I, is meant a collection of BX -measurable pairwise disjoint subsets {A1 , . . . , An } satisfying

( ) µi ∪nl=1 Al = 1 for all i ∈ I . In classical theory of fair division it is assumed that ∪ni=1 Ai = X . In order to simplify the topology of the sets Ai it is convenient in this paper to consider divisions in the sense of Definition 1. The problem of fair division of the cake is to divide X among the agents in a way that would be ‘‘fair’’ according to some fairness notion accepted by all players. There exist several notions of fair divisions:

• • • •

proportional division: µi (Ai ) ≥ 1/n for all i ∈ I; envy-free division: µi (Ai ) ≥ µi (Aj ) for all i, j ∈ I; exact division: µi (Aj ) = 1/n for all i, j ∈ I; equitable division: µi (Ai ) = µj (Aj ) for all i, j ∈ I.

E-mail address: [email protected].

A simple and well-known method for realizing the proportional division for two players is ‘‘for one to cut, the other to choose’’. Steinhaus in 1944 asked whether a fair procedure could be found for dividing a cake among n agents for n > 2. He found a solution for n = 3 and Banach and Knaster (cf. Knaster, 1946; Steinhaus, 1948, 1949) showed that the solution for n = 2 can be extended to arbitrary n. 1.2. One-dimensional case The most interesting results in fair division (cake-cutting) problems were obtained for cutting the unit interval X = (0, 1) into n contiguous subintervals {(xk−1 , xk )}nk=1 , where 0 = x0 < x1 < · · · < xn−1 < xn = 1. Such partitions are called simple divisions. Stromquist (1980) and independently Woodall (1980) proved under some weak assumptions the existence of envy-free simple divisions of a cake represented by the unit interval (0, 1). Cechlárová et al. (2013) used the notion of the generalized inverse of a function to prove the existence of simple equitable divisions. The same result was obtained by Aumann and Dombb (2015) which used a compactness argument. Legut (2017) considered an equitable optimal partition of the unit interval (0, 1) for measures with piecewise linear density functions. He showed that under some assumptions it is possible to obtain simple equitable optimal division P = {Ai }ni=1 maximizing the number µi (Ai ). An interesting and very short proof of the existence of an equitable and simple division of the unit interval (0, 1) relied on the classical Borsuk–Ulam theorem and was presented by Chéze (2017). Assume that the probability measures µ1 , . . . , µn are defined by integrable probability densities fi , i ∈ I, that is:

µi (A) =

∫

fi dx,

A ∈ B(0,1) .

A

https://doi.org/10.1016/j.jmateco.2019.11.001 0304-4068/© 2019 The Author(s). Published by Elsevier B.V. This is an open access article under the CC BY-NC-ND license (http://creativecommons.org/licenses/bync-nd/4.0/).

36

J. Legut / Journal of Mathematical Economics 86 (2020) 35–40

A partition {(xk−1 , xk )}nk=1 , where the interval (xk−1 , xk ) belongs to the kth agent is an equitable and simple division if the following system of equations is satisfied:

∫

x1

∫

x2

f1 dx = 0

∫

1

fn dx.

f2 dx = · · · = x1

• equitable; • proportional; • equitable and proportional.

(1)

2. The model

xn−1

Denote by S m , m ∈ N, the sphere: S m = {e = (e1 , . . . , em+1 ) ∈ Rm+1 :

Denote by Cn the following set of finite sequences m+1 ∑

e2i = 1}.

(2)

Cn := {c = (c1 , . . . , cr ) : cj ∈ N, j = 1, . . . , r , r ∑

i=1

We recall the following version of the Borsuk–Ulam theorem stating that for every continuous antipodal mapping F : S m → Rm (i.e. F (−x) = −F (x) for all x ∈ S m ) there exists a point x∗ ∈ S m satisfying F (x∗ ) = 0. For the purpose of this paper we repeat here the Chéze’s (2017) proof of the following theorem: Theorem 1. For all density functions fi , i ∈ I, the system of Eqs. (1) has a solution. Proof. Define a function F : S n−1 → Rn−1 by:

∫ Fi (e) = sgn(ei+1 )

e21 +···+e2i +e2i+1 e21 +···+e2i

e21

∫ fi+1 dx − sgn(e1 )

f1 dx,

(3)

0

cj = n, 2 ≤ r ≤ n}.

j=1

For a given c = (c1 , . . . , cr ) ∈ Cn consider the following procedure of fair division: 1. Divide the square (0, 1)2 using horizontal cuts starting at heights {h1 , . . . , hr −1 }. 2. Next, each of the resulting r rectangles (0, 1) × (hj−1 , hj ), j = 1, . . . , r, where hr = 1 and h0 = 0, we leave unpartitioned (if cj = 1) or we divide it by cuts parallel to the Oy axis into cj > 1 parcels. Using this procedure we obtain n rectangles to be assigned to the n agents. We introduce a two-dimensional version of the one-dimensional definition of simple divisions.

where F (e) = (F1 (e), . . . , Fn−1 (e)). It is easy to check that F is continuous and antipodal, then by the Borsuk–Ulam theorem there exists e∗ ∈ S n−1 such that F (e∗ ) = 0. If we set x0 = 0 and xi = xi−1 + (e∗i )2 , we get the solution of the equation system (1).

Definition 2. A partition Pc = {Ai }ni=1 is called a simple division of (0, 1)2 for c = (c1 , . . . , cr ) ∈ Cn , if there exist numbers {hj }rj=0

1.3. Two-dimensional case

0 = h0 < h1 < · · · < hr −1 < hr = 1,

Consider a fair division problem of a land represented by the open unit square X = (0, 1)2 where ith agent evaluates the Borel subsets of (0, 1)2 using nonatomic probability measure νi . We assume throughout the paper that the measures νi are absolutely continuous with respect to the Lebesgue measure λ defined on Borel subsets of (0, 1)2 and that there exist density functions ui (x, y) : (0, 1)2 → R such that

and

ui (x, y) > 0

νi (A) =

∫∫ A

for all

(x, y) ∈ (0, 1)2

ui (x, y)dxdy,

for all

and

A ∈ B(0,1)2 .

(4) (5)

Partition of land is regarded by economists as one of the most important applications of fair division theory in practice. In the land division literature, there are presented many models considering procedures and algorithms for constructing fair partitions satisfying given criteria (e.g. Berliant et al., 1992; Legut et al., 1994; Dall’ Aglio and Maccheroni, 2009; Nicoló et al., 2012). Hill (1983) showed the existence of a division of a disputed territory among n neighbouring countries such that each country would get a connected piece adjacent to its border and worth at least 1/n of the total value of the territory. Segal-Halevi et al. (2017) studied a land division problem with some geometric constraints imposed to the parcels. They found other interesting techniques for fair cutting of a square into pieces having nice shapes like squares and so-called fat rectangles. Woodall (1980) noticed that the problem of fair division of the square can be reduced to a problem of cutting one dimensional interval by simple projecting (0, 1)2 to the interval (0, 1). Unfortunately, if the number of agents is large we obtain thin rectangular parcels that are useless in practical applications. In the next section we introduce a definition of two-dimensional simple divisions and we show that in special cases, it is possible to divide the square into more useful shaped rectangles and obtain resulting simple partitions being:

(j) cj

and {xk }k=0 satisfying (6)

(j)

(j)

0 = x0 < x1 < · · · < x(j) cj = 1 ,

(7)

such that each set Ai is given by: (j)

(j)

Ai := (xk−1 , xk ) × (hj−1 , hj ) ⇔ i =

j−1 ∑

cm + k,

(8)

m=0

where c0 = 0. It follows from the assumption νi ≪ λ, i ∈ I, that the boundaries of sets Ai have measure 0 according to νi , i ∈ I. Hence

) ( νi ∪nl=1 Al = 1,

i ∈ I,

and the partition Pc = {Ai }ni=1 defined above is a division in the sense of Definition 1. The division scheme described in Definition 2 is illustrated by an example presented in Fig. 1. As it was mentioned in the Introduction the problem of dividing the square can be reduced to a problem of dividing the unit interval. Let fi : (0, 1) → R+ be density functions defined by: 1

∫

ui (x, y) dx,

fi (y) :=

i ∈ I.

(9)

0

It is easy to see that if {(yi−1 , yi )}ni=1 is an equitable division of (0, 1) for the densities (9), where y0 = 0, yn = 1, then the partition {Ai }ni=1 defined by Ai = (0, 1)×(yi−1 , yi ), i ∈ I, is a simple equitable division of the square (0, 1)2 for c = (1, 1, . . . , 1) ∈ Cn . 3. The main results Now we prove the two-dimensional version of Theorem 1. Theorem 2. For any probability nonatomic measures νi ≪ λ, i ∈ I, defined by (5), satisfying (4) and for any c = (c1 , . . . , cr ) ∈ Cn

J. Legut / Journal of Mathematical Economics 86 (2020) 35–40

37

Fig. 3. An example illustrating Lemma 2, where n = 4 and c = (2, 2). It is easy (1) (2) (1) (2) (2) (1) to see that for simple divisions Pc and Pc we have A3 ⊆ A3 and A1 ⊆ A1 .

Fig. 1. An example of the division scheme of a square (0, 1)2 according to Definition 2 for n = 8 and c = (3, 2, 3).

(1)

(1)

(2)

(2)

Fig. 2. An example illustrating Lemma 1, where t = 3 and (z1 , z2 ) ⊆ (z1 , z2 ).

(j) cj

there exist numbers {hj }rj=0 and {xk }k=0 fulfilling (6), (7) such that the following equalities hold

νi (Ai ) = νm (Am ) for all i, m ∈ I ,

Fig. 4. Function gk is an integral of wk on a certain vertical segment.

where Pc = {Ai }ni=1 is a simple division defined by (8). Moreover, the division Pc = {Ai }ni=1 is unique. For proving Theorem 2 we need first to show three lemmas.

It follows from Lemma 1 that there exist 1 ≤ j0 ≤ r such that (2) (2) (1) (1) (hj −1 , hj ) ⊆ (hj −1 , hj ) and 1 ≤ k0 ≤ cj0 satisfying 0

0

(m)

(m)

(m)

Lemma 1. Let 0 = z0 < z1 < · · · < zt = 1, t ∈ N, m = 1, 2, be two increasing sequences of real numbers. Then there exists (2) (1) (1) (2) 1 ≤ k0 ≤ t such that (zk −1 , zk ) ⊆ (zk −1 , zk ). 0

0

0

0

(1)

≤ z1(2) then obviously (z0(1) , z1(1) ) ⊆ (z0(2) , z1(2) ). Suppose that > z1(2) . If z2(1) ≤ z2(2) then (z1(1) , z2(1) ) ⊆ (z1(2) , z2(2) ) (1) (2) (cf. Fig. 2). Since zt = zt = 1, it is easy to check that there exists 1 ≤ k0 ≤ t such that:

Proof.

(1) zk −1 0

If z1

>

(1) z1

(2) zk −1 0

(1) zk 0

and

≤

(2) zk 0

(1)

(m)

(1)

(2)

(2)

{ }n = A(m) , m = 1, 2, be two simple divisions i i=1

⊆

(2)(j ) (xk −01 0

(2)(j0 )

, xk0

).

∑j

−1

(1)

(2)

0 It is easy to see that for i0 = l=0 cl + k0 , we have Ai0 ⊆ Ai0 (cf. Fig. 3) which completes the proof.

Let wk (x, y) : (0, 1)2 → R+ , k = 1, . . . , t, denote positive density functions. For 0 ≤ y1 < y2 ≤ 1 define functions gk (y1 , y2 , ·) : (0, 1) → (0, 1), k = 1, . . . , t, by (see Fig. 4):

∫

y2

wk (x, y)dy.

(10)

The proof of Theorem 1 does not require that fi are density functions (integrable to one), hence it is also true for arbitrary integrable positive functions. Then, for y1 and y2 , there exists a division {(zk−1 , zk )}tk=1 such that: z1

∫ 0

g1 (y1 , y2 , x)dx =

∫

z2

g2 (y1 , y2 , x)dx

z1

∫

0

(2)

Ai .

1

gt (y1 , y2 , x)dx

= ··· =

0

(11)

z t −1

Proof. Let (m) Ai

0

0

y1

obtained according to the scheme presented in Definition 2 for a (1) given c = (c1 , . . . , cr ) ∈ Cn . Then there exists i0 such that Ai ⊆

{

,

(1)(j ) xk 0 ) 0

gk (y1 , y2 , x) :=

,

which implies that (zk −1 , zk ) ⊆ (zk −1 , zk ), and the proof is 0 0 0 0 complete.

Lemma 2. Let Pc

(1)(j ) (xk −01 0

}n i=1

{

(m)(j) (xk−1

= {

,

(m)(j) xk )

×

(m) (hj−1

,

r (m) cj hj ) k=1 j=1

}

}

, m = 1, 2.

It follows from a result of Cechlárová et al. (2013) that if functions gk are positive then the equitable division {(zk−1 , zk )}tk=1 is unique. Hence for a rectangle (0, 1) × (y1 , y2 ) obtained by horizontal cuts

38

J. Legut / Journal of Mathematical Economics 86 (2020) 35–40

starting at points 0 ≤ y1 < y2 ≤ 1 we can define a function α (y1 , y2 ) by: z1

∫

α (y1 , y2 ) :=

g1 (y1 , y2 , x)dx = · · · =

0

1

−

(1)

(1) zk −1 1

1

∫

(1)

zk

∫

gt (y1 , y2 , x)dx. (12)

∫

(2)

zk

1

≤

z t −1

gk1 (y1 , y2 , x)dx

(1)

(2)

gk1 (y2 , y2 , x)dx =

(2)

∫

(2)

(m)

α

(m) z1

(m) (y1 y2 )

,

=

(m) g1 (y1 y2

,

, x)dx = · · · =

∫

1

(m) gt (y1 y2 (m) zt −1

0

,

, x)dx,

(1)

where y1 < y2 . It follows from Lemma 1 that there exists (m) (1) (1) (2) (2) 1 ≤ k0 ≤ t depending on y2 such that (zk −1 , zk ) ⊆ (zk −1 , zk ). 0

0

0

0

Since wk0 (x, y) > 0 for all x, y ∈ (0, 1) we conclude from (10) that

(2) gk0 (y1 y2

,

α

, x) >

∫

(2) (y1 y2 )

,

(1) gk0 (y1 y2

,

, x). Then we have

(2)

zk

0

= (2)

(2) gk0 (y1 y2

∫

,

, x)dx >

zk −1 0

0

(1)

(1) gk0 (y1 y2

,

, x)dx

z k −1 0

⏐ ⏐∫ ∫ ′ ⏐ ⏐ 1 y ⏐ ⏐ ′ w (x, y)dxdy ⏐ . β (y, y ) = ⏐ ⏐ ⏐ 0 y

1

(2)

(1) gk1 (y1 y2

,

, x)dx −

(1)

(1) gk1 (y1 y2

,

, x)dx ≤ 0,

and, by (10), (1)

(1)

(2)

gk1 (y1 , y2 , x) = gk1 (y1 , y2 , x) + gk1 (y2 , y2 , x), we have (1) α (y1 , y(2) 2 ) − α (y1 , y2 ) =

(1) zk 1

−

(1) zk −1 1 (2) zk 1

∫ =

∫

(2)

zk

1

(2)

(2) zk −1 1

gk1 (y1 , y2 , x)dx

(1) (1) (2) |α (y1 , y(2) 2 ) − α (y1 , y2 )| < β (y2 , y2 ). (1)

∫

1

(2) (1) (2) |α (y(1) 1 , y2 ) − α (y1 , y2 )| < β (y1 , y1 ).

Using the triangle inequality we obtain

+|α (y1 , y(2) 2 ) − α (y1 , y2 )|. Thus

(2)

1

(2)

z k −1 1

(2)

(14) (2)

Since β (y1 , y1 ) → 0 and β (y2 , y2 ) → 0 when y1

→ y1

(2) y2

and → y2 , then (14) implies the continuity of the function α (y1 , y2 ). Proof of Theorem 2. For 0 ≤ y1 < y2 ≤ 1 and i ∈ I define positive functions: y2

∫

ui (x, y)dy.

(15)

Let 0 = h0 < h1 < · · · < hr −1 < hr = 1 and denote for

i=

j−1 ∑

cm + k,

m=0

where j = 1, . . . , r , k = 1, . . . , cj , c0 = 0. It follows from (j) Theorem 1 applied to functions fk (hj−1 , hj , x) (with respect to (j) variable x) that there exist equitable simple divisions {(xk−1 , (j)

cj

xk )}k=1 of the unit interval (0, 1). For j = 1, . . . , r define functions:

αj (hj−1 , hj ) :=

(j)

x1

∫

(j)

f(c 0

(1)

j−1 +1)

(hj−1 , hj , x)dx = · · ·

1 (j) xc − 1 j

gk1 (y1 , y2 , x)dx

fc(j) (hj−1 , hj , x)dx. j

Analogously to the proof of Theorem 1, we define function F : S r −1 → Rr −1 by:

(1)

gk1 (y1 , y2 , x)dx

(2)

zk

(2)

In a similar way one can prove that for any 0 < y1 < y1 ≤ y2 ≤ 1

Fj (e) = sgn(ej+1 )αj+1 (e21 + · · · + e2j , e21 + · · · + e2j + e2j+1 )

−sgn(e1 )α1 (0, e21 ),

(1) zk −1 1

=

(2) w(x, y)dxdy = β (y(1) 2 , y2 ).

(1) y2

It follows from the monotonicity of α (y1 , y2 ) and (13) that

=

(1)

zk

−

(2)

y2

∫

(13)

z k −1 1

∫

0

∫

(1) (2) [gk1 (y1 , y(1) 2 , x) + gk1 (y2 , y2 , x)]dx

(2)

(1) y2

1

∫

fk (hj−1 , hj , x) := fi (hj−1 , hj , x)

zk −1 1

∫

wk1 (x, y)dydx ≤

(j)

(1) zk 1

∫

zk −1 1

(2)

(2) zk −1 1

(2)

y2

y1

1

1

∫

fi (y1 , y2 , x) :=

Lemma 1 implies that there exists 1 ≤ k1 ≤ t such that (1) (1) (2) (1) (2) (zk −1 , zk ) ⊆ (zk −1 , zk ). Since for y1 < y2 (2) zk 1

1

(2) (2) (2) |α (y(2) 1 , y2 ) − α (y1 , y2 )| ≤ β (y1 , y1 ) + β (y2 , y2 ).

Hence the function α (y1 , y2 ) is strictly increasing with respect to variable y2 . In a similar way one can prove that α (y1 , y2 ) is strictly decreasing with respect to variable y1 . Denote by w (x, y) = max{wk (x, y) : k = 1, . . . , t } and for any y, y′ ∈ (0, 1) define function β by

1

wk1 (x, y)dydx.

y2

(2) (2) (2) (2) |α (y(2) 1 , y2 ) − α (y1 , y2 )| ≤ |α (y1 , y2 ) − α (y1 , y2 )|

= α (y1 , y(1) 2 ).

∫

(2)

zk

(1)

zk

∫

(1)

(2)

(m)

satisfying 0 < y2 < y2 ≤ 1. Denote by {(zk−1 , zk )}tk=1 , m = 1, 2, two equitable partitions of (0, 1) such that:

∫

(2)

y2

∫

By definition of function w (x, y) w obtain

Proof. First we show the monotonicity of α (y1 , y2 ) with respect (m) to the variable y2 . Let y2 , m = 1, 2, be arbitrary numbers (1)

1

zk −1 1

zk −1 1

Lemma 3. For every set of functions gk , k = 1, . . . , t, given by (10) the function α (y1 , y2 ) defined by (12) is continuous.

(2)

zk

(1)

gk1 (y1 , y2 , x)dx +

∫

(2)

zk

1

(2)

z k −1 1

(1)

(2)

gk1 (y2 , y2 , x)dx

where j = 1, . . . , r − 1. It follows from Lemma 3 that the function F satisfies the assumptions of the Borsuk–Ulam theorem and then there exists e∗ ∈ S r −1 such that F (e∗ ) = 0. We set h∗0 = 0 and h∗j = h∗j−1 + (e∗j )2 , j = 1, . . . , r. Denote by

{

c

(j) ∗(j) j {(x∗k− 1 , xk )}k=1

}r

j=1

equitable divisions of (0, 1) corresponding

J. Legut / Journal of Mathematical Economics 86 (2020) 35–40 (j)

to functions fk (h∗j−1 , h∗j , x). It is easy to check that {A∗i }ni=1 defined by: ∗(j)

∗(j)

A∗i = (xk−1 , xk ) × (h∗j−1 , h∗j )

for

j−1 ∑

i=

cm + k,

where j = 1, . . . , r , k = 1, . . . , cj , is an equitable simple division of the square. Now we prove that the equitable simple division is unique. Suppose that there exist two different equitable simple divisions (1) (2) Pc and Pc where Pc(m)

=

{

(m) Ai

}n

{

(m)(j) (xk−1

= {

i=1

,

(m)(j) xk )

×

(m) (hj−1

,

r (m) cj hj ) k=1 j=1

}

}

,

m = 1, 2. It follows from Lemma 2 that there exists i0 and i1 such that (1)

Ai

0

Thus

ν

(2) i0 (Ai0 )

(2) i1 (Ai1 )

≤ν

=ν

(1)

(1)

0

1

≤ν

(1) i1 (Ai1 )

ν

.

.

(1)(j ) xk 0 ) 0

,

×

(16)

(2)(j0 ) 0 −1

(xk

(2)(j0 )

, xk0

,

(1) hj ) 0

̸=

(2)(j ) (xk −01 0

,

(2)(j ) xk 0 ) 0

×

(1)

(1) (hj −1 1

,

(1) hj ) 1

⊂

(1) (hj −1 0

(2) (hj −1 0

(1) hj ) 0

0 −1

.

(1)(j0 )

) =

(2) (hj −1 1

,

(2) hj ) 1

,

, h(1) j0 ) is unique.

(2) hj ). 0

, ̸= Then there exists j1 ≤ j0 such that

Then we must have

<

,

). As it was mentioned in the proof of Lemma 3, the

(2) hj − 1 . 0

Suppose that

ν

=α

(1) (hj −1 1

,

.

(2)

(2)

, hj 0 0 −1

(1) hj ) 1

<α

(2) (hj −1 1

,

(2) hj ) 1

(2) i0 (Ai0 )

=ν

0

0

, and the proof is complete.

Now we present a method of obtaining proportional simple division of (0, 1)2 . Theorem 3. For any probability nonatomic measures νi ≪ λ, i ∈ I, defined by (5), satisfying (4) and for any c = (c1 , . . . , cr ) ∈ Cn , (j) cj there exist numbers {hj }rj=0 and {xk }k=0 fulfilling (6), (7) and a permutation σ : I → I such that the following inequalities hold

νσ (i) (Ai ) ≥

1 n

for all

i ∈ I,

(17)

where Pc = {Ai }ni=1 is a simple proportional division defined by (8). Proof. At first we divide the unit square by horizontal cuts using the following recursive procedure starting from j = 1 up to j = r − 1. Step 1 (j = 1). (1) (1) Since νi ≪ λ, i ∈ I, then there exist numbers y1 , . . . , yn satisfying

νi ((0, 1) × (0,

(1) yi ))

νi ((0, 1) × (h1 , 1)) ≥ 1 −

c1

i ∈ I1

i ∈ I \ I1 .

for all

n

and

Step 2 (j = 2). (2) Next we proceed in a similar way. Denote by yi , i ∈ I \ I1 , the numbers satisfying c2 n

,

i ∈ I \ I1 .

(2) (2) for all i ∈ I2 and yi ≥ yi ≤ y(1) i2 2 for all i ∈ I \ (I1 ∪ I2 ). (2)

Denote h2 = yi . Then we have

=

c1 n

,

c2

for all

n

i ∈ I2

and

c2 − for all i ∈ I \ (I1 ∪ I2 ). n n Applying recursively this procedure up to j = r − 1, we finally obtain a partition {(0, 1) × (hj−1 , hj )}rj=1 of the square, where h0 = 0 and hr = 1, satisfying

νi ((0, 1) × (h2 , 1)) ≥ 1 −

c1

cj n

for all

i ∈ Ij , j = 1, . . . , r ,

where Ir = I \ ∪rj=−11 Ij .

which contradicts the equality (16). In a similar way we con(1) (0) sider remaining cases of the relationships between hj −1 , hj and hj

for all

n

νi ((0, 1) × (hj−1 , hj )) ≥

Hence (1) i0 (Ai0 )

νi ((0, 1) × (0, h1 )) ≥

νi ((0, 1) × (h1 , h2 )) ≥

(2) hj ) 0

(2) (2) 0 , h(1) j0 ) = (hj0 −1 , hj0 ) then (xk0 −1 , xk0

equitable division of the rectangle (0, 1) × (hj (1) hj −1 0

(2) (hj −1 0 (1)(j )

0 −1

1

c1

2

(1) (hj −1 0

(1)

Note that if (hj

(1)

(2)

Suppose that for 1 ≤ j0 ≤ r and 1 ≤ k0 ≤ cj0 we have (1)(j ) (xk −01 0

(1) (1) ≤ y(1) for all i ∈ I1 and yi ≥ yi for all i ∈ I \ I1 . i1 1

For simpler notation denote h1 = yi . It easy to see that

yi

(2) i0 (Ai0 )

=ν

(1)

There exists i2 ∈ I \ I1 and a coalition I2 ⊂ I \ I1 of c2 agents such that

Since νi0 (Ai ) = νi1 (Ai ) we get (1) i0 (Ai0 )

yi

νi ((0, 1) × (h1 , y(2) i )) =

(2) (1) ⊆ A(2) and Ai ⊆ Ai . i0 1 1

(1) i0 (Ai0 )

(1)

In practice finding the numbers y1 , . . . , yn for arbitrary measures νi requires solving complex equations and we can obtain only approximate solutions. There exists i1 ∈ I and a coalition I1 ⊂ I of c1 agents such that (1)

m=0

39

i ∈ I.

Now we divide each of the rectangles (0, 1) × (hj−1 , hj ) among the agents i ∈ Ij , j = 1, . . . , r by vertical cuts using a classical procedure of proportional division (cf. Knaster, 1946; Steinhaus, 1948, 1949). The assignment of rectangles to each agent obtained in the above procedure determines a permutation σ : I → I satisfying (17) and the proof is complete. Now we prove the existence of a proportional and equitable division of the square (0, 1)2 . Theorem 4. For any probability nonatomic measures νi ≪ λ, i ∈ I, defined by (5), satisfying (4) and for any c = (c1 , . . . , cr ) ∈ Cn , there exists a permutation σ : I → I and an equitable simple and proportional division {Ai }ni=1 defined by (8) such that:

νσ (i) (Ai ) ≥

1

for all

n

i ∈ I.

(18)

Proof. It follows from Theorem { 3 that }n there exist a permutation (p) σ and a proportional division Ai such that: i=1

νσ (i) (A(p) i ) ≥

1 n

for all

i ∈ I.

(19)

Theorem 2 implies that { for permutation σ satisfying (19) there }n (e) exists equitable division Ai satisfying i=1

(e) σ (i) (Ai )

ν

=ν

(e) σ (m) (Am )

for all

i, m ∈ I .

40

J. Legut / Journal of Mathematical Economics 86 (2020) 35–40

equitable divisions of the parts N and S: (y)

(y)

PN = {Ai }8i=5 ,

(y)

PS

4 = {A(y) i }i=1 .

Define two functions: (y) αN (y) := ν5 (A(y) 5 ) = · · · = ν8 (A8 )

and (y) αS (y) := ν1 (A(y) 1 ) = · · · = ν4 (A4 ).

It can be verified that the function αN is strictly decreasing, and the function αS is strictly increasing with respect to variable y ∈ (0, 1). In a similar way to the proof of Lemma 3 one can show that functions αN and αS are continuous. Then there exists y∗ such that αS (y∗ ) = αN (y∗ ) which means that the partition ∗)

P (y

Fig. 5. An example of the division scheme obtained by mixed horizontal and vertical cuts.

(p)

(e)

It follows from Lemma 2 that there exist i0 such that Ai ⊆ Ai . 0 0 Then we have: (e) σ (i0 ) (Ai0 )

ν

≥ν

(p) σ (i0 ) (Ai0 )

≥

1

n which completes the proof.

,

is equitable. Analogously to the proofs of Theorems 3 and 4 one can prove the existence of equitable and proportional division according to the scheme shown in Fig. 2. Based on the idea of the above example, it can be proved that Theorem 4 can be generalized to arbitrary division schemes by applying interchangeably horizontal and vertical cuts. 4.2. Open question The open question is whether for any c ∈ Cn there exists an envy-free division of the square (0, 1)2 being simple in the sense of Definition 2.

4. Remarks

References

4.1. Extensions of the main results Presented results for simple divisions described by Definition 2 can be extended for more complex partitions. Suppose n = 8. We start the division scheme with one horizontal cut, then each of the two resulting rectangles we divide by two vertical cuts. We finish with four horizontal cuts of the remaining four rectangles obtaining in the end 8 parcels. This procedure is illustrated in Fig. 5. Let y denote a number for which (0, 1) × {y} represents the first cut in the above procedure separating the square (0, 1)2 into two parts: the northern : N = (0, 1) × (y, 1) = ∪8i=5 Ai

∗

) 8 = {A(y }i=1 i

and

the southern : S = (0, 1) × (0, y) = ∪4i=1 Ai . It is easy to see that each of the parts N and S is partitioned according to the scheme given in Definition 2. In the proof of Theorem 2 in the definition (15) of the functions fi , i = 1, . . . , n we can also consider arbitrary integrable positive functions ui (x, y) (not necessarily to be probability density functions). Then Theorem 2 is also true for such functions and can be applied to N and S instead to (0, 1)2 . Thus, for all y ∈ (0, 1) there exist

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