1
Cutting a Pie Is Not a Piece of Cake
Julius B. Barbanel and Steven J. Brams
May 2007
1. INTRODUCTION. The general problem of fair division and the specific problem of
cutting a cake fairly have received much attention in recent years (for overviews, see [1],
[2], [4], [5], [8]). Cutting a pie into wedge-shaped sectors, by contrast, has received far
less attention, though it would seem that the connection between cake-cutting and piecutting is close ([3], [11]). Roughly speaking, if a cake is represented by a line segment,
then it becomes a pie when its endpoints are connected to form a circle.
Suppose each of n players attributes values to pieces of pie. We ask whether it is always
possible to divide the pie into n sectors and assign one sector to each player in a way that
causes no envy. If so, is there always a “best” possible solution? Some mathematical
formalism makes these questions precise. The particular goal of this paper is to provide a
partial answer, in the pie-cutting context, to a question posed by David Gale ([7]) about
whether a certain kind of envy-free division is possible.
Let’s deem a pie mathematically equivalent to the circle S1 = R / Z, with representative
points in the unit interval [0,1) . A pair of real numbers α,β∈[0,1) with α<β can be used
to specify either the sector of pie given by the subinterval [α,β) or its complement
[0,1)\[α,β), which we shall denote by [β,α). Cutting along a diameter, for example,
divides the pie into two sectors [α,α+0.5) and [α+0.5,α) for some α∈[0,1). By the
“length” of a sector [α,β) or [β,α) for α<β, we mean the value β-α or 1-(β-α),
respectively. Diameter cuts, for example, produce sectors of length 0.5. (One comment:
Our notation does not allow for the possibility that one player receives a sector of length
zero. We do allow for this possibility, but we will not need notation for it.)
In order to assess values of pieces, let’s assume that each player uses an integrable
probability density function p:[0,1]→R satisfying p(0) = p(1) (and, of course,
∫
p(t) dt =1 ). This player then assigns the value (“measure”)
∫
p(t) dt to the sector
[α , β )
[0,1]
[α,β). We say that a player prefers one sector over another if her measure of the first
sector is strictly greater than her measure of the second.
Players’ measures of sectors are continuous in the following sense: If (α n ) and (β n ) are
sequences in S 1 converging to α∈S1 and β∈S1, respectively, with respect to the absolute
value function induced from R, then
∫
[α n , β n )
p(t) dt →
∫
p(t) dt . This corresponds to the
[α , β )
intuitive notion that, as one endpoint of some sector moves through the interval [0,1),
each player views the value of that sector as changing continuously. It is understood that
different players may operate with different probability density functions and, therefore,
different measures.
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It will sometimes be convenient to assume that players’ measures are absolutely
continuous with respect to one another, meaning that if one player assigns a value zero to
a particular sector, then all players do. In this case, by contracting to a point each sector
to which players simultaneously assign the value zero, we can assume without loss of
generality that no player assigns a value zero to a sector of positive length.
To state Gale’s question, let’s call an allocation of sectors among a number of players:
•
•
•
envy-free if no player prefers another sector to her own;
undominated if, based on each player’s measure, every other possible
allocation of sectors either gives the same values or else decreases the value
assigned to at least one player; and
equitable if all players assign exactly the same value to the sectors they
receive (and so no player envies another’s “degree of happiness”).
We emphasize that “undominated” here means “undominated with respect to other
allocations into sectors, one sector per player,” the only type of allocation we consider in
this paper. It is certainly possible that an allocation can be dominated by an allocation
that assigns some players more than one disjoint sector, for instance.
Gale’s query ([7]) is simple: Does there always exist an envy-free and undominated
allocation of a pie? We have made some progress towards answering this question
For two players, the answer to Gale’s question with respect to pie is affirmative. In fact,
if players’ measures are absolutely continuous with respect to one another, an allocation
that is simultaneously envy-free, undominated, and equitable is sure to exist. We show
this in section 2.
For the case of four or more players, the answer to Gale’s question is negative (at least if
we do not insist that the measures be absolutely continuous with respect to one another):
We exhibit in section 3 a specific example showing this to be the case. Surprisingly,
resolution of the three-player situation remains open.
Finding procedures, sometimes only approximate ([10]), for producing certain desirable
allocations—as opposed to merely demonstrating that such allocations exist—is a central
concern in the fair-division literature. In section 4 we present two “moving-knife”
procedures.
2. TWO-PLAYER ALLOCATIONS. Consider two players A and B who assign to the
sectors [α,β) and [β,α) the values A(α,β) and B(β,α), respectively. The pair
(A(α,β),B(β,α)) is a point in the closed unit square [0,1] x [0,1]. Let S denote the set of
points that arise this way. The points (0,1) and (1,0), for instance, are always points of S.
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Any point of S lying on the main diagonal of the square corresponds to an equitable
allocation, and any point in the upper-right quarter [0.5,1] × [0.5,1] to an envy-free
allocation. To determine whether or not a point P of S corresponds to an undominated
allocation, draw horizontal and vertical axes centered at P and label the four (closed)
quadrants of the plane defined by them I P , II P , III P , and IVP , as illustrated is Figure 1.
If the only point of S contained in quadrant I P is P, then P corresponds to an
undominated allocation. Our goal is to determine whether or not points with some, or all,
of these properties exist.
Figure 1. Quadrants used to determine whether an allocation is undominated
The continuity of players’ measures shows that S possesses four key features:
i)
S is closed.
If P ∈[0,1] × [0,1] is a limit point of S, then there exist points Pn =
(A(αn,βn),B(βn,αn)) of S such that Pn → P (with respect to the standard metric on
R2) as n grows. As S 1 is compact, (αn) possesses a convergent subsequence.
Working with this subsequence, we may as well assume that (αn) converges to
α∈S1. The corresponding sequence (βn) possesses a convergent subsequence.
Again, without loss of generality, we can assume that this sequence converges to
a value β∈S1. Continuity now shows that P = (A(α,β),B(β,α)) ∈ S .
ii)
S is path connected.
For instance, a path connecting P = (A(α,β),B(β,α)) to Q = (A(γ,δ),B(δ,γ)) is given
by (A ((1-t)α + tγ , (1-t)β + tδ) , B ((1-t)β + tδ , (1-t)α + tγ) for 0 ≤ t ≤ 1
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iii)
(
)
S is symmetric with respect to the point 0.5,0.5 .
This follows from the fact that (A(α,β),B(β,α)) = (1-A(β,α),1-B(α,β)).
iv)
If P and Q are two points of S with the same x-coordinate or the same ycoordinate, then line segment PQ lies in S.
Without loss of generality, suppose P and Q have the same x-coordinate a. Let
(a,b) be any point on the segment PQ and define U and V as follows:
U = {α∈[0,1] : for some β∈[0,1], A(α,β) = a and B(β,α)≥b}
V = {α∈[0,1] : for some β∈[0,1], A(α,β) = a and B(β, α)≤b}
We claim that U and V are both closed. Suppose that (αn) is a sequence in U that
converges to α. Then there is a sequence (βn) in [0,1] such that for each n,
A(αn,βn) = a and B(βn,αn)≥b. Since (βn) has a convergent subsequence, we lose
no generality in assuming that (βn) converges to some β. It follows from the
continuity of A and B that A(α,β) = a, B(β,α)≥b, and that hence α∈U. This
establishes that U is closed. The proof for V is similar.
Next, we claim that U  V = [0,1]. Clearly U  V ⊆ [0,1]. For any α∈[0,1],
A(α,β) = a for some β∈[0,1], and thus either α∈U or α∈V. It follows that U  V
= [0,1].
Since P and Q are both in S, we know that U≠∅ and V≠∅. The interval [0,1]
cannot be expressed as the union of two disjoint closed sets and hence U  V≠∅.
Choose any α∈U  V. Then, for some β1,β2∈[0,1], A(α,β1) = A(α,β2) = a and
B(β1,α)≤b≤B(β2,α). The continuity of B and the intermediate value theorem
imply that for some β between β1 and β2, B(β,α)=b. Since β is between β1 and β2,
we must have A(α,β) = a. Thus, (a,b)∈S. This establishes that line segment PQ
lies in S.
Properties iii) and iv) establish that (0.5,0.5) is always a point of S. It is also not difficult
to see that (1,0) and (0,1) are the only points of S that belong to the boundary of the unit
square if, and only if, the measures of the two players are absolutely continuous with
respect to one another. In this case, each horizontal and each vertical cross-section of S
away from (1,0) and (0,1) has endpoints in the interior of the unit square.
Figure 2 illustrates four possibilities for the set S. If the players’ measures are equal, then
S consists only of the diagonal between (1,0) and (0,1), as shown in Figure 2a. The set S
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in Figures 2a and 2b corresponds to measures that are absolutely continuous with respect
to one another, while the set S in Figures 2c and 2d corresponds to measures that are not.
Figure 2. Some possibilities for the set S
Theorem 2.1. For two players, each with an integrable probability density function, an
envy-free, equitable allocation of sectors exists and an envy-free, undominated allocation
of sectors also exists. If players’ measures are also absolutely continuous with respect to
one another, then an envy-free, undominated, and equitable allocation of sectors exists.
Proof: We have already shown that (0.5,0.5) is in S. This point corresponds to an
allocation that is envy-free and equitable.
Let Sur denote the intersection of S with the closed upper-right quarter of the unit square
(i.e., [0.5,1]x[0.5,1]). Then Sur is a closed set. The function F : (x, y)  x + y is a
continuous function from Sur to R. By the extreme-value theorem, there exists a point P
in Sur for which F attains a maximal value. Any allocation that corresponds to this point
is envy-free and undominated.
Assume now that players’ measures are absolutely continuous with respect to one
another, and let P be the right-most point of S on the main diagonal of the unit square.
(Restrict the function G : (x, y)  x to the intersection of the main diagonal and S—itself
a closed set—and apply the extreme-value theorem to show that such a point exists.)
Clearly, any allocation that corresponds to P is envy-free and equitable. Our goal is to
show that any such allocation is also undominated.
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Suppose, to the contrary, there is a point Q of S different from P in quadrant I P . This
point does not lie on the main diagonal of the unit square. Assume, without loss of
generality, that it lies above the main diagonal.
Now Q = (A(α,β),B(β,α)) for some sector [α,β). Using continuity, and the fact that the
measures are absolutely continuous with respect to one another, we can adjust the values
of α and β to produce a second point lying in the interior of I P with x-coordinate larger
than the x-coordinate of Q (and y-coordinate necessarily smaller). Thus, without loss of
generality, we can assume Q has x-coordinate larger than P’s.
Since S is path connected, there is a path γ from (0.5,0.5) to (1,0) consisting of points
from S. As P is the right-most point of S on the diagonal, γ does not cross the main
diagonal in I P . Consequently there is a point R on γ with the same x-coordinate as Q
lying below the main diagonal. That QR ⊂ S provides a contradiction on the choice of
P.

The final two paragraphs of this proof can be readily modified to show that when the
measures are absolutely continuous measures with respect to one another, the rightmost
point along any horizontal cross-section of S corresponds to an undominated allocation of
sectors, as does the topmost point of any vertical cross-section. Loosely speaking, every
point along the upper boundary of S is undominated.
The assumption of absolute continuity is necessary for the above proof. Consider, for
example, players A and B working with respective probability density functions:
⎧2 for t ∈[0,0.25] ∪ [0.5,0.75]
p(t) = ⎨
elsewhere
⎩0
q(t) = 1 for all t ∈[0,1]
This produces the set S shown in Figure 2d with no point that corresponds to an envyfree, equitable, and undominated allocation.
Comment: Suppose we insist that the two players A and B are allowed to cut only along
a diameter of the pie. Construct the subset S of the closed unit square given as the set of
all points of the form (A(α,α+0.5),B(α+0.5,α)) for α∈S1. Then S is a (possibly selfintersecting) loop within the closed unit square, symmetric about the center of the square.
As before, one is guaranteed allocations that are envy-free and equitable, or envy-free
and undominated, but not all three properties. Consider, for example, players A and B
working with respective probability density functions:
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⎧1.5 for t ∈[0,0.5]
p(t) = ⎨
⎩0.5 for t ∈[0.5,1]
q(t) = 1 for all t ∈[0,1]
In this case, S is the horizontal line segment between (.25,.5) and (.75,.5), and no point of
S corresponds to an allocation that is simultaneously envy-free, equitable, and
undominated. Note too that these players’ measures are absolutely continuous with
respect to one another.
3. FOUR-PLAYER ALLOCATIONS. We exhibit an example that shows it is possible
for four players to possess measures for which no envy-free and undominated allocation
of sectors can exist among those players. The measures in this example are not absolutely
continuous with respect to one another.
Let players A and B each operate with the density function p given by:
⎧2 for x ∈[0,0.25] ∪ [0.5,0.75]
p(x) = ⎨
otherwise
⎩0
and players C and D each with density function q given by:
⎧2.4 for x ∈[0.25,0.5]
⎪
q(x) = ⎨1.6 for x ∈[0.75, 1]
⎪0
otherwise
⎩
We shall refer to the four portions of pie determined by the points 0, 0.25, 0.5, and 0.75
as quadrants. We make two observations.
Observation 1. In any undominated allocation, if player A’s piece and player C’s piece
are adjacent, then they must meet on one of the four boundaries between quadrants. The
same statement holds for players A and D, for players B and C, and for players B and D.
Reason. If players A and C have adjacent pieces, and these pieces meet at some point
other than a quadrant boundary, then the meeting point must be in the interior of a
quadrant. But since each quadrant has positive value throughout for one of these two
players and value zero for the other, the boundary can be moved so as to increase one
player’s value and not change the other’s value. This violates our assumption that the
original allocation is undominated. The same argument works for players A and D, for
players B and C, and for players B and D.
Observation 2. No envy-free and undominated allocation can give player A and player B
adjacent pieces or can give player C and player D adjacent pieces.
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Reason. Suppose, by way of contradiction, that some allocation that is envy-free and
undominated gives player A and player B adjacent pieces. Observation 1 implies that the
union of player A’s piece and player B’s piece is precisely one quadrant, precisely two
quadrants, or precisely three quadrants of the pie. We will show that each of these leads
to a contradiction.
Case 1. The union of player A’s piece and player B’s piece is precisely one quadrant.
Clearly, this must be either the [0,0.25] quadrant or the [0.5,0.75] quadrant. Assume,
without loss of generality, that it is the [0.5,0.75] quadrant. In order that neither player A
nor player B envy the other, these players must split this quadrant (which each values at
one half the pie) into pieces that each values at one quarter of the pie. Then, in order that
neither player C nor player D envies the other, the boundary between their two pieces
must be inside the [0.25,0.5] quadrant. But then players A and B will each envy the
player that gets all of the [0,0.25] quadrant, since each values this quadrant at half the pie.
Case 2. The union of player A’s piece and player B’s piece is precisely two quadrants.
Since these quadrants are adjacent, one of them must be the [0.25,0.5] quadrant or else
one of them must be the [0.75,1] quadrant. These quadrants have value zero to player A
and to player B, and this violates our assumption that the allocation is undominated. This
follows from that fact that moving the boundary that is an endpoint of this quadrant so as
to shrink player A’s piece or player B’s piece would make no difference to player A or
player B, but it would add to the value of player C’s piece or to player D’s piece.
Case 3. The union of player A’s piece and player B’s piece is precisely three quadrants.
Then the union of player C’s piece and player D’s piece is precisely one quadrant. The
argument then proceeds in a manner similar to that of Case 1, with the roles of players A
and B on the one hand, and players C and D on the other hand, reversed.
Since players A and B cannot have adjacent pieces, it follows immediately that players C
and D cannot have adjacent pieces.
Our two observations together imply that each player receives one of the quadrants.
Then player C or player D did not receive the [0.25,0.5] quadrant, and this player will
envy the player that did receive this quadrant. This establishes that there is no envy-free
and undominated allocation.
In our example, the players' measures are not absolutely continuous with respect to one
another. We do not know whether there is, for every four-player example, an envy-free
and undominated allocation when the players' measures are absolutely continuous with
respect to one another.
It is straightforward to extend this construction to show that undominated and envy-free
allocations need not exist for more than four players. To create a five-player example, for
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instance, simply add a new sector that player E views as 100% of the value of the pie and
the other four players interpret as valueless.
4. PROCEDURAL RESULTS. To establish the existence of an allocation with certain
desired properties is not the same as giving a procedure to produce it. In this section we
consider so-called moving-knife procedures for finding desirable allocations of a pie.
These procedures are to be distinguished from discrete procedures, in which cuts are
made at specified points that do not depend on continuously moving knives (e.g., around
the circumference of a pie) or calling “stop” when a moving knife reaches a certain point
(e.g., that creates a piece tied for largest with some other piece).
Moving-knife procedures were first applied to dividing a cake among n players using n –
1 parallel, vertical cuts (which is the minimal number). It is not difficult to show that if a
division is envy-free, it will be undominated ([7]). Two minimal-cut envy-free movingknife procedures for three players have been found ([1],[9]), but no four-player envy-free
procedure is known. (An envy-free procedure for four players that requires up to five
cuts—two more than the minimal number of three— is known, but it may necessitate that
players receive disconnected pieces [1].)
For pie, as we have seen, there may be no envy-free, undominated allocation using n
radial cuts—the minimal number—if there are four or more players. Unlike cake, we do
not know whether there is an envy-free, undominated allocation of a pie for three players,
but we do know from the analysis in section 2 that there is one for two players. In fact,
for two players, there is an allocation that is envy-free, undominated, and equitable if the
players’ measures are absolutely continuous with respect to one another. How to find
allocations satisfying these desirable properties is the question we address next.
We present two procedures, one for two players and one for three players. The two-player
procedure will produce an allocation that is envy-free and undominated but need not be
equitable, and the three-player procedure will produce an allocation that is envy-free but
need be neither undominated nor equitable.
Two-player procedure
This procedure produces an envy-free and undominated allocation for two players whose
measures are absolutely continuous with respect to one another. Call the two players
player A and player B, whom we shall refer to as “she” and “he,” respectively. Player A
holds two radial knives above the pie in such a way that, in her view, the two pieces of
pie determined by these knives each have value 1/2. She then rotates these knives
continuously all the way around the pie, maintaining this 1/2 value of the pieces, until the
knives return to their original positions. After observing this process, player B identifies
the position that, in his view, gives the maximum value to one of the two pieces so
determined. (Ties can be broken randomly. The continuity of the players’ measures,
along with the extreme-value theorem, guarantees that there is such a maximum-value
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piece.) Player B takes this piece, and player A receives the other piece. Call this
allocation P.
We claim that P is envy-free and undominated. To see that P is envy-free, we first
observe that player A certainly believes that her piece has size exactly 1/2, and so she
will not envy player B. Player B does not envy player A since, if he does, then he must
have picked the smaller, rather than the larger piece, at his chosen position. Thus, P is
envy-free.
Suppose, by way of contradiction, that some allocation Q dominates P. Then, both
players receive at least as much pie in allocation Q as in allocation P (in each’s own
view), and at least one player receives strictly more. The absolute continuity of the
measures with respect to one another allows us to alter Q, if necessary, so as to give
player A less pie and player B more pie, and in this way to obtain an allocation R such
that player A receives the same value of pie (in her view) in allocation R as in allocation
P (i.e., value 1/2), and player B receives strictly more value of the pie (in his view) in
allocation R than in allocation P. Then, the piece that player B obtains in allocation R is
one of the pieces that he would have seen as player A rotated the knives around the pie.
This contradicts the fact that player B chose the largest piece that he saw.
This establishes that the allocation is envy-free and undominated. It will be equitable if
and only if any piece of pie that player A considers to be half of the pie is also considered
to be half of the pie by player B. In general, of course, this will not be the case.
What happens if one or both players are not truthful? For example, player A could rotate
the knives in such a way as to maintain a 1/3-2/3 balance in her view and, if player B then
chooses the 1/3 piece, player A ends up with what she thinks is 2/3 of the pie instead of
1/2. But, of course, player A could also end up with 1/3 of the pie instead of 1/2. Thus,
we see that players can do better, or worse, by not being truthful. What is true, however,
is that by being truthful, each player is guaranteed a piece of size at least 1/2 (in each’s
own view) and, hence, will not envy the other player, regardless of whether or not the
other player is truthful. Players who are risk-averse will presumably like this procedure,
because it maximizes the minimum-value piece that A and B can ensure for themselves.
To tie this procedure with the set S in section 2, note that the procedure moves along the
line segment that is the intersection of the line x = 1/2 and the set S. In picking his largest
piece, player B is identifying an allocation that corresponds to the point on the line
segment with greatest y coordinate. (There is a largest such point, because S is a closed
set.)
We do not know whether or not there is a moving-knife procedure to produce an
allocation that is envy-free, undominated, and equitable. We do know, however, that if
the players’ measures fail to be absolutely continuous with respect to one another, then
there may not exist such an allocation.
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Finally, we note that if the measures are not absolutely continuous with respect to one
another, then this procedure might not be well defined. For example, if player A were to
assign value zero to some sector that has positive value to player B, then, for a fixed
position of one of player A’s knives, there would be multiple positions for the other knife
which yield value 1/2 to player A for each of the pieces so determined. We cannot
simply have player A pick one of these positions arbitrarily, since these different
positions yield different values for player B.
Three-Player Procedure
This procedure produces an envy-free allocation for three players. We call the three
players player A, player B, and player C and refer to them as “she,” “he,” and “it,”
respectively. We assume that the three players’ measures are absolutely continuous with
respect to one another. Player A rotates three radial knives continuously around the pie,
maintaining what she believes to be 1/3-1/3-1/3 portions. Player B calls “stop” when he
thinks two of the pieces are tied for largest, which must occur for at least one set of
positions in the rotation (see below). The players then choose pieces in the order C first,
B second, and A third.
We must show that at some point player B will think that two of the pieces are tied for
largest, and that the allocation produced by this procedure is envy-free.
To show that there must be at least one set of knife positions in the rotation at which
player B thinks there are two pieces that tie for most-valued, let us call the three pieces
determined by the beginning positions of the knives piece i, piece ii, and piece iii. (These
pieces will change as player A rotates the knives.) Let player B specify his most-valued
piece at the start of the rotation. If there is a tie, then we are done. If not, then player A
begins rotating the three radial knives. We assume, without loss of generality, that player
B’s most-valued piece at the start of the rotation is piece i, and that player A rotates the
three knives in such a way that piece i moves toward the original position of piece ii.
Because, in player A’s view, each of the three pieces is 1/3 of the pie, piece i will
eventually occupy the position of the original piece ii, which we make a requirement of
the procedure. At this point, piece iii occupies the original position of piece i, which we
also make a requirement, and hence player B must think that this new piece iii is the
largest piece. Because, in player B’s view, piece i starts out largest and another piece
becomes largest as the rotation proceeds, it follows from the continuity of the players’
measures and the intermediate-value theorem that there must be a position in the rotation
when player B views two pieces as tied for largest.
To see that the procedure gives an envy-free allocation, note that the first player to
choose, player C, can take a most-valued piece, so it will not be envious. If player C
takes one of player B’s tied-for-most-valued pieces, player B can take the other one;
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otherwise, player B can choose either of his two tied-for-most-valued pieces. Because
player A values all three pieces equally, it does not matter which piece she gets.
Finally, we make two observations. First, this procedure may fail to give an allocation
that is either undominated or equitable, just as the two-player procedure may not give an
allocation that is equitable. Also, like the two-player procedure, the three-player
procedure does not rely on the players’ being truthful. In other words, if any player
misrepresents her or his or its valuations of pieces of pie (for example, if player A moves
the knives in such a way that the pieces are not maintained at size 1/3-1/3-1/3 in her view,
or if player B calls “stop” at some time other than when there is a tie for largest piece in
his view), it is still the case that any player that is truthful will not envy any other player’s
portion in the resulting allocation, regardless of the truthfulness of the other players.
5. CONCLUSIONS. In summary, we have shown that for two players,
•
•
•
an envy-free, equitable allocation exists;
an envy-free, undominated allocation exists; and
if the players’ measures are absolutely continuous with respect to one another,
then an envy-free, undominated, and equitable allocation exists.
For four players, we established that there need not be an allocation that is envy-free and
undominated if the players’ measures are not absolutely continuous with respect to one
another.
We gave a procedure for two players (working with measures that are absolutely
continuous with respect to one another) that produces an allocation that is envy-free and
undominated, and a procedure for three players that produces an allocation that is envyfree. The three-player procedure establishes that an envy-free allocation for three players
always exists.
We close by restating three questions we were not able to answer.
Open Question 1. For two players using measures that are absolutely continuous with
respect to one another, does there exist a moving-knife procedure that produces an
allocation that is envy-free, undominated, and equitable? (If by a “procedure” one allows
the players to submit their measures to a referee, the referee can determine the cuts that
give such an allocation, as shown in [3].)
Open Question 2. For three players, does there always exist an allocation that is envyfree and undominated (with or without the assumption that the measures are absolutely
continuous with respect to one another)?
Open Question 3. For four or more players, does there always exist an allocation that is
envy-free and undominated if the players’ measures are absolutely continuous with
respect to one another?
13
ACKNOWLEDGMENTS. The authors would like to thank David Gale, Michael Jones,
Christian Klamler, Peter Landweber, James Tanton, William Thomson, Daniel Velleman,
and an anonymous referee for valuable comments.
14
REFERENCES
1. J. Barbanel and S. Brams, Cake Division with Minimal Cuts: Envy-Free Procedures
for 3 Persons, 4 Persons, and Beyond, Mathematical Social Sciences 48 (2004) 251-270.
2. S. Brams, M. Jones, and C. Klamler, Better Ways to Cut a Cake, Notices of the AMS
35 (2006) 1314-1321.
3. S. Brams, M. Jones, and C. Klamler, Proportional Pie-Cutting, International Journal of
Game Theory (forthcoming).
4. S. Brams, A. Taylor, Fair Division: From Cake-Cutting to Dispute Resolution,
Cambridge University Press, New York, 1996.
5. S. Brams, A. Taylor, and W. Zwicker, Old and New Moving-Knife Schemes,
Mathematical Intelligencer 17, (1995) 30-35.
6. S. Brams, A. Taylor, and W. Zwicker, A Moving-Knife Solution to the Four-Person
Envy-Free Cake Division Problem, Proceedings of the American Mathematical Society
125 (1997) 547-554.
7. D. Gale, Mathematical Entertainments, Mathematical Intelligencer 15 (1993) 48-52.
8. J. Robertson and W. Webb, Cake-Cutting Algorithms: Be Fair If You Can, A K Peters,
Natick, MA, 1998
9. W. Stromquist, How to Cut a Cake Fairly, American Mathematical Monthly 87 (1980)
640-644.
10. F. E. Su, Rental Harmony: Sperner’s Lemma in Fair Division, American
Mathematical Monthly 106 (1999) 922-934.
11. W. Thomson, Children Crying at Birthday Parties. Why?, Economic Theory 31
(2007) 501-521.
Department of Mathematics, Union College, Schenectady, NY 12308,
[email protected]
Department of Politics, New York University, New York, NY 10003,
[email protected]