Transcript A Mathematical View of Our World 1 ed.

A Mathematical View
of Our World
1st ed.
Parks, Musser, Trimpe,
Maurer, and Maurer
Chapter 4
Fair Division
Section 4.1
Divide and Choose Methods
• Goals
• Study fair-division problems
• Continuous fair division
• Discrete fair division
• Mixed fair division
• Study fair-division procedures
• Divide-and-choose method for 2 players
• Divide-and-choose method for 3 players
• Last-diminisher method for 3 or more players
4.1 Initial Problem
• The brothers Drewvan, Oswald, and
Granger are to share their family’s 3600acre estate.
• Drewvan:
• Values vineyards three times as much as fields.
• Values woodlands twice as much as fields.
• Oswald:
• Values vineyards twice as much as fields.
• Values woodlands three times as much as fields.
4.1 Initial Problem, cont’d
• Granger:
• values vineyards twice as much as fields.
• Values fields three times as much as woodlands.
• How can the brothers fairly divide the
estate?
• The solution will be given at the end of the
section.
Fair-Division Problems
• Fair-division problems involve fairly dividing
something between two or more people,
without the aid of an outside arbitrator.
• The people who will share the object are
called players.
• The solution to a problem is called a fairdivision procedure or a fair-division scheme.
Types of Fair-Division Problems
• Continuous fair-division problems:
• The object(s) can be divided into pieces of
any size with no loss of value.
• An example is dividing a cake or an amount
of money among two or more people.
Types of Fair-Division, cont’d
• Discrete fair-division problems:
• The object(s) will lose value if divided.
• We assume the players do not want to sell
everything and divide the proceeds.
• However, sometimes money must be used
when no other fair division is possible
• An example is dividing a car, a house, and a
boat among two or more people.
Types of Fair-Division, cont’d
• Mixed fair-division problems:
• Some objects to be shared can be divided
and some cannot.
• This type is a combination of continuous
and discrete fair division.
• An example is dividing an estate consisting of
money, a house, and a car among two or
more people.
Question:
Three cousins will share an
inheritance. The estate includes a
house, a car, and cash. What type of
fair-division problem is this?
a. continuous
b. discrete
c. mixed
Types of Fair-Division, cont’d
• This section will consider only
continuous fair-division problems.
• We make the assumption that the value
of a player’s share is determined by his
or her values.
• Different players may value the same
share differently.
Value of a Share
• In a fair-division problem with n players,
a player has received a fair share if that
player considers his or her share to be
worth at least 1/n of the total value
being shared.
• A division that results in every player
receiving a fair share is called
proportional.
Value of a Share, cont’d
• We assume that a player’s values in a
fair-division problem cannot change
based on the results of the division.
• We also assume that no player has any
knowledge of any other player’s values.
Fair Division for Two Players
• The standard procedure for a
continuous fair-division problem with
two players is called the divide-andchoose method.
• This method is described as dividing a
cake, but it can be used to fairly divide
any continuous object.
Divide-And-Choose Method
•
Two players, X and Y, are to divide a cake.
1) Player X divides the cake into 2 pieces that he
or she considers to be of equal value.
• Player X is called the divider.
2)
Player Y picks the piece he or she considers
to be of greater value.
• Player Y is called the chooser.
3) Player X gets the piece that player Y did not
choose.
Divide-And-Choose Method, cont’d
• This method produces a proportional
division.
• The divider thinks both pieces are equal,
so the divider gets a fair share.
• The chooser will find at least one of the
pieces to be a fair share or more than a
fair share. The chooser selects that
piece, and gets a fair share.
Example 1
• Margo and Steven will share a $4
pizza that is half pepperoni and half
Hawaiian.
• Margo likes both kinds of pizza equally.
• Steven likes pepperoni 4 times as much
as Hawaiian.
Example 1, cont’d
•
Margo cuts the
pizza into 6 pieces
and arranges them
as shown.
a) What monetary
value would Margo
and Steven each
place on the
original two halves
of the pizza?
Example 1, cont’d
• Solution: The whole pizza is worth $4.
• Margo values both kinds of pizza
equally. To her each half is worth half of
the total value, or $2.
• Steven values pepperoni 4 times as
much as Hawaiian. To him the
pepperoni half is worth 4/5 of the total
value, or $3.20. The Hawaiian half is
worth 1/5 of the total, or $0.80.
Example 1, cont’d
b) What value
would each
person place on
each of the two
plates of pizza?
c) What plate will
Steven choose?
Example 1, cont’d
b) Solution: The whole pizza is worth $4.
• Margo values both kinds of pizza
equally. To her each plate of pizza is
worth half of the total value, or $2.
Example 1, cont’d
b) Solution, cont’d:
• Steven values pepperoni 4 times as
much as Hawaiian.
• To him each pepperoni slice is worth
$3.20/3 = $1.067 and each Hawaiian slice
is worth $0.80/3 = $0.267.
• The first plate is worth 2($0.267) +
1($1.067) = $1.60.
• The second plate is worth 1($0.267) +
2($1.067) = $2.40
Example 1, cont’d
c) Solution:
• Steven will choose the second plate,
with one slice of Hawaiian and two slices
of pepperoni.
• Margo gets a plate of pizza that she
feels is worth half the value.
• Steven gets a plate of pizza that he feels
is worth more than half the value.
Example 2
•
Caleb and Diego will drive 6 hours during
the day and 4 hours at night.
•
Caleb prefers night to day driving 2 to 1.
•
Diego prefers them equally, or 1 to 1.
•
How should they divide the driving into 2
shifts if Caleb is the divider and Diego is
the chooser?
Example 2, cont’d
• Solution:
• Caleb can assign 2 points to each hour
of night driving and 1 point to each hour
of day driving.
• Caleb values the entire drive at 1(6) + 2(4)
= 14 points.
• To Caleb a fair share will be worth half the
total value, or 7 points.
Example 2, cont’d
• Solution, cont’d:
• A possible fair division for Caleb is to
create shifts of:
• 6 hours of daytime driving and 0.5 hours of
nighttime driving.
• 3.5 hours of nighttime driving.
• Both shifts are worth 7 points to Caleb.
Example 2, cont’d
• Solution, cont’d:
• Diego can assign 1 point to each hour of
night driving and 1 point to each hour of
day driving.
• Diego values the entire drive at 1(6) + 1(4)
= 10 points.
• To Diego a fair share will be worth half the
total value, or 5 points.
Example 2, cont’d
• Solution, cont’d:
• Diego values the first shift at 1(6) +
1(0.5) = 6.5 points.
• Diego values the second shift at 1(3.5) =
3.5 points.
• Diego will choose the first shift, because
it is worth more to him.
Two Players, cont’d
•
•
Notice that in both of the previous
examples:
•
The divider got a share he or she felt was
equal to exactly half of the total value.
•
The chooser got a share he or she felt was
equal to more than half of the total value.
It is often advantageous to be the chooser,
so the roles should be randomly chosen.
Fair Division for Three Players
• In a continuous fair-division problem
with 3 players, it is still possible to
have one player divide the object and
the other players choose.
• This method is also called the lonedivider method.
Divide-And-Choose Method
•
Three players, X, Y, and Z are to divide a cake.
1) Player X (the divider) divides the cake into 3
pieces that he/she considers to be of equal
value.
2) Players Y and Z (the choosers) each decide
which pieces are worth at least 1/3 of the total
value.
• These pieces are said to be acceptable.
3) The choosers announce their acceptable
pieces.
Divide-And-Choose Method, cont’d
3) There are 2 possibilities:
a) If at least 1 piece is unacceptable to both Y
and Z, Player X gets that piece.
• If Y and Z can each choose acceptable
pieces, they do so.
• If Y and Z cannot each choose acceptable
pieces, they put the remaining pieces
back together and use the two player
method to re-divide.
Divide-And-Choose Method, cont’d
3) Cont’d:
b) If every piece is acceptable to both Y
and Z, they each take an acceptable
piece. Player X gets the leftover piece.
•
Note: The divide-and-choose method can
be extended to more than 3 players. The
more players, the more complicated the
process becomes.
Question:
The divide-and-choose method for 3 players is being
used to divide a pizza. Player A has cut a pizza into what
she views as 3 equal shares. Player B thinks that only
shares 2 and 3 are acceptable. Player C thinks that only
share 2 is acceptable. What is the fair division?
a. Player A gets share 3, Player B gets share 1, and
Player C gets share 2.
b. Player A gets share 1, Player B gets share 3, and
Player C gets share 2.
c. Player A gets share 2, Player B gets share 3, and
Player C gets share 1.
d. Player A gets share 1, Player B gets share 2, and
Player C gets share 3.
Example 3
•
•
Emma, Fay, and Grace will divide 24 ounces of
ice cream, which is made up of equal amounts of
vanilla, chocolate, and strawberry.
•
Emma likes the 3 flavors equally well.
•
Fay prefers chocolate 2 to 1 over either other
flavor and prefers vanilla and strawberry
equally well.
•
Grace prefers vanilla to chocolate to
strawberry in the ratio 1 to 2 to 3.
If Emma is the divider, what are the results of the
divide-and-choose method for 3 players?
Example 3, cont’d
•
Solution: Suppose Emma divides the ice cream
into 3 equal parts, each consisting of one of the
flavors.
Example 3, cont’d
•
Solution, cont’d: Fay is one of the choosers.
•
Faye finds portions 1 and 3 unacceptable.
Example 3, cont’d
•
Solution, cont’d: Grace is the other chooser.
•
She finds portion 1 unacceptable
Example 3, cont’d
•
Solution, cont’d: All of the players’ values are
summarized in the table below.
Example 3, cont’d
•
•
Solution, cont’d:
•
Portion 1 is unacceptable to both Fay and
Grace. As the divider, Emma will receive
portion 1.
•
Only portion 2 is acceptable to Faye.
•
Portions 2 and 3 are acceptable to Grace.
The division is Emma: portion 1; Fay:
portion 2; Grace: portion 3.
Last-Diminisher Method
•
A method for continuous fair-division
problems with 3 or more players is called
the last-diminisher method.
•
Suppose any number of players X, Y, …
are dividing a cake.
1) Player X cuts a piece of cake that he or she
considers to be a fair share.
Last-Diminisher Method, cont’d
2) Each player, in turn, judges the fairness
of the piece.
a) If a player considers the piece fair or less
than fair, it passes to the next player.
b) If a player considers the piece more than
fair, the player trims the piece to make it fair,
returning the trimming to the undivided
portion and passing the trimmed piece to
the next player.
Last-Diminisher Method, cont’d
3) The last player to trim the piece, gets the
piece as his or her share.
• If no player trimmed the piece, player X gets
the piece.
4) After one player gets a piece of cake,
the process begins again without that
player and that piece.
• When only 2 players remain, they use the
divide-and-choose method.
Example 4
•
Hector, Isaac, and James will divide 24
ounces of ice cream, which is equal parts
vanilla, chocolate, and strawberry.
• Hector values vanilla to chocolate to
strawberry 1 to 2 to 3.
• Isaac likes the 3 flavors equally.
• James values vanilla to chocolate to
strawberry 1 to 2 to 1.
Example 4, cont’d
•
Using the last-diminisher method with
Hector as the first divider and Isaac as the
first judge, find the results of the division.
•
Solution:
•
Hector assigns 1 point to each ounce of
vanilla, 2 points to each ounce of chocolate,
and 3 points to each ounce of strawberry.
Example 4, cont’d
•
Solution, cont’d: A fair share of ice cream,
to Hector, is worth 48/3 = 16 points.
Example 4, cont’d
• Solution, cont’d:
• One possible fair share for Hector would
be all 8 ounces of vanilla plus 4 ounces
of chocolate.
• This share is worth 1(8) + 2(4) = 16
points to Hector, so he would be happy
with this share.
• Next, Isaac must decide whether the
share is fair, according to his values.
Example 4, cont’d
• Solution, cont’d:
• Isaac assigns 1 point to each ounce
of vanilla, 1 point to each ounce of
chocolate, and 1 point to each
ounce of strawberry.
• Isaac values all of the ice cream at
1(8) + 1(8) + 1(8) = 24 points.
• A fair share to Isaac is 8 points.
Example 4, cont’d
•
Solution, cont’d:
• Isaac’s value for Hector’s serving is 1(8)
+ 1(4) = 12 points.
• Isaac thinks it is more than a fair share.
• Isaac trims off 4 points worth of ice
cream.
• Suppose he trims off the 4 ounces of
chocolate.
Example 4, cont’d
•
Solution, cont’d:
• Next, James must judge the share.
• James assigns 1 point to each ounce of
vanilla, 2 points to each ounce of
chocolate, and 1 point to each ounce of
strawberry.
• James values all of the ice cream at 1(8)
+ 2(8) + 1(8) = 32 points.
• A fair share to James is worth 32/3 points.
Example 4, cont’d
• Solution, cont’d:
• The existing share is now just 8
ounce of vanilla.
• To James, the share is worth 1(8) =
8 points.
• James thinks this is less than a fair
share.
• James will not trim the share.
Example 4, cont’d
•
Solution, cont’d:
• Isaac was the last-diminisher, and gets
the share of ice cream.
• Hector and James will divide the
remaining ice cream using the divideand-choose method.
•
Note: This is only one of many different
possible solutions.
4.1 Initial Problem Solution
• The brothers Drewvan, Oswald, and Granger are
to share their family’s estate, which is 1200 acres
each of vineyards, woodlands, and fields.
• Drewvan prefers vineyards to woodlands to fields
3 to 2 to 1.
• Oswald prefers vineyards to woodlands to fields
2 to 3 to 1.
• Granger prefers vineyards to woodlands to fields
2 to 1 to 3.
Initial Problem Solution, cont’d
• Use the divide-and-choose method for 3
players.
• Let Drewvan be the divider.
Initial Problem Solution, cont’d
• Drewvan values the entire estate at 7200 points.
Initial Problem Solution, cont’d
• To Drewvan, a fair share is worth 7200/3 =
2400 points.
• One possible fair division is shown below.
Initial Problem Solution, cont’d
• Next, the two choosers will consider this
division.
• Granger and Oswald both value the
entire estate at 7200 points also.
• To Oswald, a fair share is worth 2400
points.
• To Granger, a fair share is worth 2400
points.
Initial Problem Solution, cont’d
• Oswald considers piece 1 to be unacceptable.
Initial Problem Solution, cont’d
• Granger considers pieces 1 and 2 to be
unacceptable.
Initial Problem Solution, cont’d
• Both choosers think piece 1 is
unacceptable, so Drewvan gets piece 1.
• Granger thinks only piece 3 is
acceptable, so he gets that piece.
• Oswald thinks pieces 2 and 3 are
acceptable, so Oswald gets piece 2.
Section 4.2
Discrete and Mixed
Division Problems
• Goals
• Study discrete fair-division problems
• The method of sealed bids
• The method of points
• Study mixed fair-division problems
• The adjusted winner procedure
4.2 Initial Problem
• When twins Zack and Zeke turned 16 they
received:
• A pickup truck
• A horse
• A cow
• How can they share these three things?
• The solution will be given at the end of the
section.
Discrete Fair Division
• Recall that discrete fair division
problems involve sharing objects that
cannot be divided without losing value.
• Two methods for solving discrete fairdivision problems are:
• The method of sealed bids.
• The method of points.
Method of Sealed Bids
•
Any number of players, N, are to share any
number of items.
•
If necessary, money will be used to insure
fairness.
1) All players submit sealed bids, stating
monetary values for each item.
2) Each item goes to the highest bidder.
•
The highest bidder places the dollar amount of
his or her bid into a compensation fund.
Method of Sealed Bids, cont’d
3) From the compensation fund, each
player receives 1/N of his or her bid
on each item.
4) Any money leftover in the fund is
distributed equally to all players.
• Note: This method is also called the
Knaster Inheritance Procedure.
Example 1
•
Three sisters Maura, Nessa, and Odelia
will share a house and a cottage.
•
Apply the method of sealed bids to divide
the property, using the bids shown below.
Example 1, cont’d
•
Solution: Each piece of property goes to
the highest bidder.
•
Odelia gets the family home and places
$301,000 into the compensation fund.
•
Nessa gets the cottage and places $203,000
into the compensation fund.
Example 1, cont’d
• Solution, cont’d: The compensation
fund now contains a total of $203,000
+ $301,000 = $504,000.
Example 1, cont’d
•
Solution, cont’d: Each sister receives 1/3
of her total bids from the compensation
fund.
•
Maura receives 289000  188000  $159, 000
3
•
286000  203000
 $163, 000
Nessa receives
3
•
Odelia receives 301000  182000  $161, 000
3
Example 1, cont’d
• Solution, cont’d: After the
distributions, there is $504,000 –
($159,000 + $160,000 + $161,000)
= $21,000 left in the fund.
• The leftover money is distributed
equally to the three sisters in the
amount of $7000 each.
Example 1, cont’d
• Solution, cont’d: The final shares
are:
• Maura receives $166,000 and no
property.
• Nessa receives the cottage, for which
she paid a net amount of $33,000.
• Odelia receives the family home, for
which she paid a net amount of
$133,000.
Example 1, cont’d
•
Solution, cont’d: Note that the division is
proportional because each sister receives what
she considers to be a fair share.
Method of Points
• Three players will share three items.
1) Each player assigns points to each
item, so that the points for each
player total to 100.
2) List all 6 possible arrangements of
players and items, along with the
point values.
Method of Points, cont’d
3) Note the smallest number of points
assigned to an item in each
arrangement.
• Choose the arrangement with the
largest value of the smallest number.
• If there is not only one such
arrangement, keep all the
arrangements with that smallest point
value and go to step 4.
Method of Points, cont’d
4) For each arrangement kept in step
3, note the middle point value.
• Choose the arrangement with the
largest value of the middle number.
• If there is not only one such
arrangement, keep all the
arrangements with that middle point
value and go to step 5.
Method of Points, cont’d
5) For each arrangement kept in step
4, note the largest point value.
• Choose any arrangement with the
largest value of the largest number.
Example 2
•
Three couples, A, B, and C, need to decide
who gets which room in a hotel.
•
The couples assign points to each room as
shown below.
Example 2, cont’d
•
Solution:
Example 2, cont’d
•
Solution, cont’d: The largest minimum
point value is 40, in row 4.
Example 2, cont’d
• Solution, cont’d: The arrangement
selected is:
• Couple A, Room 2
• Couple B, Room 3
• Couple C, Room 1
• Two couples got their first choice
and one got their second choice.
Example 3
•
Three couples, A, B, and C, need to decide
who gets which room in a cabin.
•
The couples assign points to each room as
shown below.
Example 3, cont’d
Example 3, cont’d
• Table from
previous slide
•
Solution, cont’d: The largest minimum point value
is 12, which occurs in 4 arrangements.
Example 3, cont’d
•
Solution, cont’d: Keep those 4 arrangements
and look at the middle point values.
Example 3, cont’d
•
Solution, cont’d: The largest middle point value
is 20, which occurs in 3 arrangements.
Example 3, cont’d
•
Solution, cont’d: Keep those 3 arrangements
and look at the largest point values.
•
The maximum largest number is 71.
Example 3, cont’d
• Solution, cont’d: The arrangement
selected is:
• Couple A, Room 2
• Couple B, Room 1
• Couple C, Room 3
Question:
The assignments were Couple A, Room 2;
Couple B, Room 1; and Couple C, Room 3.
Considering the point values each couple
assigned to the three rooms, is this
division proportional?
a. yes
b. no
Example 3, cont’d
• Solution, cont’d: Sometimes the
method of points produces a division
that is not proportional. However it
is still the best division that can be
made.
Mixed Fair Division
• Recall that mixed fair division problems
involve sharing a mixture of discrete
and continuous objects.
• A method for solving mixed fair-division
problems is the adjusted winner
procedure.
Adjusted Winner Procedure
• Two players are to fairly divide any
number of items.
• The items may be discrete and/or
continuous.
• Ownership of some items may be
shared.
1) Each player assigns points to each
item, for a total of 100 points.
Adjusted Winner Procedure, cont’d
2) Each player tentatively receives
items to which he or she assigned
the most points.
• The points are added to the player’s
total.
• If 2 players tie for an item, the item
goes to the player with the lowest point
total so far.
Adjusted Winner Procedure, cont’d
3) Look at the players’ points.
• If the point totals are equal, the
process is done.
• If Player X has more points than Player
Y, then give Player Y the item for which
the ratio of the number of points
assigned by X to the number of points
assigned by Y is the smallest.
Adjusted Winner Procedure, cont’d
4) Re-examine the players’ points.
a) If the point totals are equal, the
process is done.
b) If Player X still has more points than
Player Y, repeat step 3.
c) If Player Y now has more points than
Player X, move a fraction of the last
item moved back to X.
Adjusted Winner Procedure, cont’d
4) Cont’d: The formula for case C is as
follows, where q = fraction of item to be
moved, TX = Player X’s point total without
this item, TY = Player Y’s point total without
this item, PX = number of points Player X
assigned to this item, PY = number of points
Player Y assigned to this item.
TY  TX  PY
q
PX  PY
Question:
Three players will divide a car, a boat, and an
RV. Their point assignments are shown in the
table below. Complete step 2 in the adjusted
winner procedure.
Player 1 Player 2 Player 3
car
15
30
35
boat
35
20
20
RV
50
50
45
a. Player 1: boat; Player 2: RV; Player 3: car
b. Player 1: boat, RV; Player 2: nothing;
Player 3: car
c. Player 1: boat; Player 2: car, RV; Player 3:
nothing
Example 4
•
Two players, A and B, need to divide a house,
a boat, a cabin, and a condominium.
•
Both have assigned points as shown below.
Example 4, cont’d
• Solution: The tentative assignment
of items, along with the current point
totals, is shown in the table below.
Example 4, cont’d
•
Solution, cont’d:
Player A has more
points than Player B.
•
Determine which item
to move from A to B
by considering the
ratios of both items
currently assigned to
Player A.
Example 4, cont’d
• Solution, cont’d: Move the house,
which has the smaller ratio, from
Player A to Player B.
Example 4, cont’d
• Solution, cont’d: Now Player B has
more points than Player A.
• A fraction of the house must be given
back to Player A.
• The values for the formula are:
• TX = TA = 30
• TY = TB = 45
• PX = PA = 45
• PY = PB = 35
Example 4, cont’d
• Solution, cont’d: The calculation is:
TB  TA  PB 45  30  35 50 5
q



PA  PB
45  35
80 8
• Player B keeps 3/8 of the house and
5/8 of the house goes back to Player
A.
Example 4, cont’d
• Solution, cont’d: Re-examine the
points totals.
• Player A has 30 + 45(5/8) = 58.125
points.
• Player B has 25 + 20 + 35(3/8) =
58.125 points.
• The division is now proportional.
4.2 Initial Problem Solution
• Zack and Zeke need to share a pickup truck, a
horse, and a cow.
• Solution: They could use the adjusted winner
procedure to share the items. Suppose they
assign points as shown.
Initial Problem Solution, cont’d
• Tentatively, the assignments are:
• Zack: truck and cow, 73 points
• Zeke: horse, 35 points.
• Zack has more points, so something
must be given to Zeke.
• Check the ratios for the truck and the cow.
Initial Problem Solution, cont’d
• The ratios are
shown in the
table.
• The ratio for the
truck is smaller.
• Move the truck
from Zack to
Zeke.
Initial Problem Solution, cont’d
• Now:
• Zack has 40 points.
• Zeke has 65 points.
• Since the points are not equal, a fraction of
the truck must be given back to Zack.
• The values for the formula are:
• TX = TZack = 40; PX = PZack = 33
• TY = TZeke = 35; PY = PZeke = 30
Initial Problem Solution, cont’d
• Solution, cont’d: The calculation is:
TZeke  TZack  PZeke 35  40  30 25
q


 39.7%
PZack  PZeke
33  30
63
• Zeke keeps 60.3% of the truck and
39.7% of the truck goes back to
Zack.
Initial Problem Solution, cont’d
• Solution, cont’d: Re-examine the
points totals.
• Zack has 40 + 33(0.397) = 53.095
points.
• Zeke has 35 + 30(0.603) = 53.095
points.
• For a proportional division to be
created, the truck must be shared.
Section 4.3
Envy-Free Division
• Goals
• Study envy-free division
• Continuous envy-free division for 3 players
4.3 Initial Problem
• Dylan, Emery, and Fordel will share a
cake that is half chocolate and half yellow.
• Dylan likes chocolate cake twice as much as
yellow.
• Emery likes chocolate and yellow cake
equally well.
• Fordel likes yellow cake twice as much as
chocolate.
• How should they divide the cake?
• The solution will be given at the end of the section.
Envy-Free Division
• A division, among n players, is considered
envy-free if each player feels that:
• He or she has received at least 1/n of the total
value
• No other player’s share is more valuable than his
or her own.
• Note: A proportional division is not always
envy-free.
Question:
Ice cream was divided among three
players. The players’ values of the shares
are shown in the table below.
The division was Emma, Portion 1; Fay,
Portion 2; Grace, Portion 3.
Is the division envy free?
a. yes b. no
Continuous Envy-Free Division
• This section covers envy-free divisions
for continuous fair-division problems
involving three players.
• The procedure is split into two parts:
• Part 1 distributes the majority of the item.
• Part 2 distributes the excess.
Envy-Free Division Part 1
• Players A, B, and C are to share a
cake (or some other item).
1) Player A (the divider) divides the cake
into 3 pieces he or she considers to be
of equal value.
2) Player B chooses the one most
valuable piece of these 3 pieces.
Envy-Free Division Part 1, cont’d
3) Player B (the trimmer) trims the most
valuable piece so it is equal to the
second most valuable piece.
• The excess is set aside.
4) Player C (the chooser) chooses the
piece he or she considers to have the
greatest value.
Envy-Free Division Part 1, cont’d
5) Player B chooses next.
• Player B gets the trimmed piece if it is
available.
• If the trimmed piece is gone, B chooses
the most valuable piece from the 2
remaining pieces.
6) Player A gets the 1 remaining piece.
Example 1
• Gabi, Holly, and Izzy will share a cake
that is ¼ chocolate, ¼ white, ¼ yellow,
and ¼ spice cake.
Example 1, cont’d
• The girls’ preference ratios are given
below:
Example 1, cont’d
• Solution: Let Gabi be the divider, Holly the
trimmer, and Izzy the chooser.
• Gabi assigns point values to slices of cake.
Example 1, cont’d
• Solution, cont’d: Gabi could divide the
cake as shown below into 3 pieces of
equal value to her.
Example 1, cont’d
• Solution, cont’d: Holly is the trimmer.
Example 1, cont’d
• Solution, cont’d: Holly trims piece 3 to
make it equal in value to pieces 1 and 2.
Example 1, cont’d
Example 1, cont’d
• Solution, cont’d:
• Izzy chooses piece 2.
• Holly gets the trimmed piece, piece 3.
• Gabi gets the last piece, piece 1.
Example 1, cont’d
Envy-Free Division Part 2
•
Players A, B, and C are to share a cake (or
some other item) and have completed Part
1.
1) Of Players B and C, the player who
received the trimmed piece becomes the
second chooser.
•
The other player becomes the second divider.
2) The second divider divides the excess into
3 pieces of equal value.
Envy-Free Division Part 2
3) The second chooser takes the piece
he or she considers to be of the
greatest value.
4) Player A chooses the remaining piece
that he or she considers to be of the
greatest value.
5) The second divider gets the last
remaining piece.
Question:
In Part 1 of an envy-free division, the
results were as follows:
Player A (the divider) got share 2;
Player B (the trimmer) got share 1;
and Player C (the chooser) got share
3, which had been trimmed.
Who is the second divider for Part 2?
a. Player A
b. Player B
c. Player C
Example 2
• Gabi, Holly, and Izzy will complete the
division of the cake.
• The excess portion is shown below.
Example 2, cont’d
• Solution: Recall that Gabi was the first
divider and Holly received the trimmed piece.
• Holly will be the second chooser and Izzy will
be the second divider.
Example 2, cont’d
• Solution, cont’d: Izzy divides the excess
into 3 pieces.
Example 2, cont’d
Example 2, cont’d
• Solution, cont’d:
• Since the pieces are all of equal value to
Holly, she arbitrarily chooses a piece, say
piece 3.
• Gabi, the first divider, chooses either
equally valuable piece, say piece 1.
• Izzy receives the last piece, piece 2.
Example 2, cont’d
Example 2, cont’d
Example 2, cont’d
Example 2, cont’d
• Solution, cont’d: The division is envy-free because
each player feels she received a share worth as
much or more as every other player’s share.
Example 3
• Jenny, Kara, and Lindsey need to divide 10
yards each of beige linen, red silk, and
yellow gingham.
• Each assigns points per yard as shown below.
Example 3, cont’d
• Solution: Jenny divides the fabric.
Example 3, cont’d
Example 3, cont’d
Example 3, cont’d
Example 3, cont’d
• Solution, cont’d: Lindsey took the
trimmed share, so Kara cannot have it.
• Kara chooses the most valuable remaining
share, share 2.
• The remaining piece, share 1, goes to
Jenny.
• Note: This completes Part 1.
Example 3, cont’d
• Solution, cont’d: The excess is divided
into 3 equal pieces.
• Since the excess pieces are all the same,
there is no real choosing to do in Part 2.
• Each sister gets 2/3 yard of silk as her
share of the excess.
Example 3, cont’d
4.3 Initial Problem Solution
• Dylan, Emery, and Fordel will share a
cake that is half chocolate and half yellow.
• Dylan likes chocolate cake twice as much as
yellow.
• Emery likes chocolate and yellow cake
equally well.
• Fordel likes yellow cake twice as much as
chocolate.
• How should they divide the cake?
Initial Problem Solution, cont’d
• Let Dylan be the divider, Emery the
trimmer, and Fordel the chooser.
Initial Problem Solution, cont’d
• Part 1: Dylan evaluates the cake:
Initial Problem Solution, cont’d
• Dylan creates 3 equal shares:
Initial Problem Solution, cont’d
Initial Problem Solution, cont’d
• Emery trims piece 3 so that it is equal in
value to pieces 1 and 2:
Initial Problem Solution, cont’d
Initial Problem Solution, cont’d
• Fordel chooses piece 3, the trimmed piece.
• The remaining pieces are identical, so
Emery and Dylan each take one.
• Part 2: The excess is all yellow cake, so it
can merely be divided into 3 equal-sized
pieces and shared among the players.
• The final division is shown on the next
slides.
Initial Problem Solution, cont’d
Initial Problem Solution, cont’d
Initial Problem Solution, cont’d