Title: A Mathematical View of Our World
1A Mathematical View of Our World
- 1st ed.
- Parks, Musser, Trimpe, Maurer, and Maurer
2Chapter 4
3Section 4.1Divide 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
44.1 Initial Problem
- The brothers Drewvan, Oswald, and Granger are to
share their familys 3600-acre 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.
54.1 Initial Problem, contd
- 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.
6Fair-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
fair-division procedure or a fair-division
scheme.
7Types 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.
8Types of Fair-Division, contd
- 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.
9Types of Fair-Division, contd
- 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.
10Question
- 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
11Types of Fair-Division, contd
- This section will consider only continuous
fair-division problems. - We make the assumption that the value of a
players share is determined by his or her
values. - Different players may value the same share
differently.
12Value 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.
13Value of a Share, contd
- We assume that a players 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 players values.
14Fair Division for Two Players
- The standard procedure for a continuous
fair-division problem with two players is called
the divide-and-choose method. - This method is described as dividing a cake, but
it can be used to fairly divide any continuous
object.
15Divide-And-Choose Method
- Two players, X and Y, are to divide a cake.
- Player X divides the cake into 2 pieces that he
or she considers to be of equal value. - Player X is called the divider.
- Player Y picks the piece he or she considers to
be of greater value. - Player Y is called the chooser.
- Player X gets the piece that player Y did not
choose.
16Divide-And-Choose Method, contd
- 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.
17Example 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.
18Example 1, contd
- Margo cuts the pizza into 6 pieces and arranges
them as shown. - What monetary value would Margo and Steven each
place on the original two halves of the pizza?
19Example 1, contd
- 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.
20Example 1, contd
- What value would each person place on each of the
two plates of pizza? - What plate will Steven choose?
21Example 1, contd
- 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.
22Example 1, contd
- Solution, contd
- 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
23Example 1, contd
- 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.
24Example 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?
25Example 2, contd
- 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.
26Example 2, contd
- Solution, contd
- 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.
27Example 2, contd
- Solution, contd
- 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.
28Example 2, contd
- Solution, contd
- 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.
29Two Players, contd
- 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.
30Fair 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 lone-divider
method.
31Divide-And-Choose Method
- Three players, X, Y, and Z are to divide a cake.
- Player X (the divider) divides the cake into 3
pieces that he/she considers to be of equal
value. - 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.
- The choosers announce their acceptable pieces.
32Divide-And-Choose Method, contd
- There are 2 possibilities
- 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.
33Divide-And-Choose Method, contd
- Contd
- 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.
34Question
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.
35Example 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?
36Example 3, contd
- Solution Suppose Emma divides the ice cream into
3 equal parts, each consisting of one of the
flavors.
37Example 3, contd
- Solution, contd Fay is one of the choosers.
- Faye finds portions 1 and 3 unacceptable.
38Example 3, contd
- Solution, contd Grace is the other chooser.
- She finds portion 1 unacceptable
39Example 3, contd
- Solution, contd All of the players values are
summarized in the table below.
40Example 3, contd
- Solution, contd
- 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.
41Last-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. - Player X cuts a piece of cake that he or she
considers to be a fair share.
42Last-Diminisher Method, contd
- Each player, in turn, judges the fairness of the
piece. - If a player considers the piece fair or less than
fair, it passes to the next player. - 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.
43Last-Diminisher Method, contd
- 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. - 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.
44Example 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.
45Example 4, contd
- 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.
46Example 4, contd
- Solution, contd A fair share of ice cream, to
Hector, is worth 48/3 16 points.
47Example 4, contd
- Solution, contd
- 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.
48Example 4, contd
- Solution, contd
- 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.
49Example 4, contd
- Solution, contd
- Isaacs value for Hectors 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.
50Example 4, contd
- Solution, contd
- 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.
51Example 4, contd
- Solution, contd
- 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.
52Example 4, contd
- Solution, contd
- Isaac was the last-diminisher, and gets the share
of ice cream. - Hector and James will divide the remaining ice
cream using the divide-and-choose method. - Note This is only one of many different possible
solutions.
534.1 Initial Problem Solution
- The brothers Drewvan, Oswald, and Granger are to
share their familys 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.
54Initial Problem Solution, contd
- Use the divide-and-choose method for 3 players.
- Let Drewvan be the divider.
55Initial Problem Solution, contd
- Drewvan values the entire estate at 7200 points.
56Initial Problem Solution, contd
- To Drewvan, a fair share is worth 7200/3 2400
points. - One possible fair division is shown below.
57Initial Problem Solution, contd
- 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.
58Initial Problem Solution, contd
- Oswald considers piece 1 to be unacceptable.
59Initial Problem Solution, contd
- Granger considers pieces 1 and 2 to be
unacceptable.
60Initial Problem Solution, contd
- 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.
61Section 4.2Discrete 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
624.2 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.
63Discrete Fair Division
- Recall that discrete fair division problems
involve sharing objects that cannot be divided
without losing value. - Two methods for solving discrete fair-division
problems are - The method of sealed bids.
- The method of points.
64Method of Sealed Bids
- Any number of players, N, are to share any number
of items. - If necessary, money will be used to insure
fairness. - All players submit sealed bids, stating monetary
values for each item. - Each item goes to the highest bidder.
- The highest bidder places the dollar amount of
his or her bid into a compensation fund.
65Method of Sealed Bids, contd
- From the compensation fund, each player receives
1/N of his or her bid on each item. - Any money leftover in the fund is distributed
equally to all players. - Note This method is also called the Knaster
Inheritance Procedure.
66Example 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.
67Example 1, contd
- 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.
68Example 1, contd
- Solution, contd The compensation fund now
contains a total of 203,000 301,000
504,000.
69Example 1, contd
- Solution, contd Each sister receives 1/3 of her
total bids from the compensation fund. - Maura receives
- Nessa receives
- Odelia receives
70Example 1, contd
- Solution, contd 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.
71Example 1, contd
- Solution, contd 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.
72Example 1, contd
- Solution, contd Note that the division is
proportional because each sister receives what
she considers to be a fair share.
73Method of Points
- Three players will share three items.
- Each player assigns points to each item, so that
the points for each player total to 100. - List all 6 possible arrangements of players and
items, along with the point values.
74Method of Points, contd
- 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.
75Method of Points, contd
- 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.
76Method of Points, contd
- For each arrangement kept in step 4, note the
largest point value. - Choose any arrangement with the largest value of
the largest number.
77Example 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.
78Example 2, contd
79Example 2, contd
- Solution, contd The largest minimum point value
is 40, in row 4.
80Example 2, contd
- Solution, contd 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.
81Example 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.
82Example 3, contd
83Example 3, contd
- Table from previous slide
- Solution, contd The largest minimum point value
is 12, which occurs in 4 arrangements.
84Example 3, contd
- Solution, contd Keep those 4 arrangements and
look at the middle point values.
85Example 3, contd
- Solution, contd The largest middle point value
is 20, which occurs in 3 arrangements.
86Example 3, contd
- Solution, contd Keep those 3 arrangements and
look at the largest point values. - The maximum largest number is 71.
87Example 3, contd
- Solution, contd The arrangement selected is
- Couple A, Room 2
- Couple B, Room 1
- Couple C, Room 3
88Question
- 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
89Example 3, contd
- Solution, contd Sometimes the method of points
produces a division that is not proportional.
However it is still the best division that can be
made.
90Mixed 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.
91Adjusted 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.
- Each player assigns points to each item, for a
total of 100 points.
92Adjusted Winner Procedure, contd
- Each player tentatively receives items to which
he or she assigned the most points. - The points are added to the players total.
- If 2 players tie for an item, the item goes to
the player with the lowest point total so far.
93Adjusted Winner Procedure, contd
- 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.
94Adjusted Winner Procedure, contd
- Re-examine the players points.
- If the point totals are equal, the process is
done. - If Player X still has more points than Player Y,
repeat step 3. - If Player Y now has more points than Player X,
move a fraction of the last item moved back to X.
95Adjusted Winner Procedure, contd
- Contd The formula for case C is as follows,
where q fraction of item to be moved, TX
Player Xs point total without this item, TY
Player Ys 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.
96Question
- 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. - 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
Player 1 Player 2 Player 3
car 15 30 35
boat 35 20 20
RV 50 50 45
97Example 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.
98Example 4, contd
- Solution The tentative assignment of items,
along with the current point totals, is shown in
the table below.
99Example 4, contd
- Solution, contd 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.
100Example 4, contd
- Solution, contd Move the house, which has the
smaller ratio, from Player A to Player B.
101Example 4, contd
- Solution, contd 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
102Example 4, contd
- Solution, contd The calculation is
- Player B keeps 3/8 of the house and 5/8 of the
house goes back to Player A.
103Example 4, contd
- Solution, contd 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.
1044.2 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.
105Initial Problem Solution, contd
- 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.
106Initial Problem Solution, contd
- The ratios are shown in the table.
- The ratio for the truck is smaller.
- Move the truck from Zack to Zeke.
107Initial Problem Solution, contd
- 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
108Initial Problem Solution, contd
- Solution, contd The calculation is
- Zeke keeps 60.3 of the truck and 39.7 of the
truck goes back to Zack.
109Initial Problem Solution, contd
- Solution, contd 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.
110Section 4.3Envy-Free Division
- Goals
- Study envy-free division
- Continuous envy-free division for 3 players
1114.3 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.
112Envy-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 players share is more valuable than his
or her own. - Note A proportional division is not always
envy-free.
113Question
- 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
114Continuous 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.
115Envy-Free Division Part 1
- Players A, B, and C are to share a cake (or some
other item). - Player A (the divider) divides the cake into 3
pieces he or she considers to be of equal value. - Player B chooses the one most valuable piece of
these 3 pieces.
116Envy-Free Division Part 1, contd
- Player B (the trimmer) trims the most valuable
piece so it is equal to the second most valuable
piece. - The excess is set aside.
- Player C (the chooser) chooses the piece he or
she considers to have the greatest value.
117Envy-Free Division Part 1, contd
- 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. - Player A gets the 1 remaining piece.
118Example 1
- Gabi, Holly, and Izzy will share a cake that is ¼
chocolate, ¼ white, ¼ yellow, and ¼ spice cake.
119Example 1, contd
- The girls preference ratios are given below
120Example 1, contd
- Solution Let Gabi be the divider, Holly the
trimmer, and Izzy the chooser. - Gabi assigns point values to slices of cake.
121Example 1, contd
- Solution, contd Gabi could divide the cake as
shown below into 3 pieces of equal value to her.
122Example 1, contd
- Solution, contd Holly is the trimmer.
123Example 1, contd
- Solution, contd Holly trims piece 3 to make it
equal in value to pieces 1 and 2.
124Example 1, contd
125Example 1, contd
- Solution, contd
- Izzy chooses piece 2.
- Holly gets the trimmed piece, piece 3.
- Gabi gets the last piece, piece 1.
126Example 1, contd
127Envy-Free Division Part 2
- Players A, B, and C are to share a cake (or some
other item) and have completed Part 1. - Of Players B and C, the player who received the
trimmed piece becomes the second chooser. - The other player becomes the second divider.
- The second divider divides the excess into 3
pieces of equal value.
128Envy-Free Division Part 2
- The second chooser takes the piece he or she
considers to be of the greatest value. - Player A chooses the remaining piece that he or
she considers to be of the greatest value. - The second divider gets the last remaining piece.
129Question
- 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
130Example 2
- Gabi, Holly, and Izzy will complete the division
of the cake. - The excess portion is shown below.
131Example 2, contd
- 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.
132Example 2, contd
- Solution, contd Izzy divides the excess into 3
pieces.
133Example 2, contd
134Example 2, contd
- Solution, contd
- 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.
135Example 2, contd
136Example 2, contd
137Example 2, contd
138Example 2, contd
- Solution, contd The division is envy-free
because each player feels she received a share
worth as much or more as every other players
share.
139Example 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.
140Example 3, contd
- Solution Jenny divides the fabric.
141Example 3, contd
142Example 3, contd
143Example 3, contd
144Example 3, contd
- Solution, contd 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.
145Example 3, contd
- Solution, contd 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.
146Example 3, contd
1474.3 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?
148Initial Problem Solution, contd
- Let Dylan be the divider, Emery the trimmer, and
Fordel the chooser.
149Initial Problem Solution, contd
- Part 1 Dylan evaluates the cake
150Initial Problem Solution, contd
- Dylan creates 3 equal shares
151Initial Problem Solution, contd
152Initial Problem Solution, contd
- Emery trims piece 3 so that it is equal in value
to pieces 1 and 2
153Initial Problem Solution, contd
154Initial Problem Solution, contd
- 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.
155Initial Problem Solution, contd
156Initial Problem Solution, contd
157Initial Problem Solution, contd