Dividing a Cake Fairly

1
Dividing a Cake Fairly
among n players
Thomas Yeo Advisor Prof. B. Prabhakar
Network Group, Electrical Engineering
Department, Stanford University
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Topics Outline

• The Different Definitions of Fairness
• Simple Fairness, Case n 2
• Simple Fairness, Case n 3, Finite Algorithm
• Simple Fairness, Case n 3, Infinite Algorithm
• Minimizing the Number of Cuts Required
• Other Criteria An Envy-Free Algorithm Example
• Why Disagreeing might be Better
• Conclusion Applicability to Real Life

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Different Definitions of Fairness

• Problem Statement Given a cake, devise a
  procedure to divide it fairly among n players
• Aim In this presentation, we shall try to
  provide some intuitive understanding of cake
  sharing algorithms. For detailed proofs, we
  recommend the book by Robertson and Webb
• Examples of definitions of fairness
• 1) Simple Fair Each gets at least 1/n
• 2) Strong Fair Each gets more than 1/n
• 3) Envy-Free No one thinks another has a
  strictly larger piece
• 4) Super Envy-Free Each thinks everyone else
  gets less than 1/n
• 5) Exact Everyone thinks everyone gets exactly
  1/n.
• Some criteria are subsets of others. For example,
  Simple Fair is the weakest criterion, implied by
  all the other four criteria

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Simple Fairness, case n 2

• A simple solution is
• 1) Player A cuts the cake into what he considers
  to be two equal halves
• 2) Player B picks one of the 2 pieces
• 3) Player A gets the remaining piece
• Why is this simple fair?
• Although both A and B are guaranteed a piece
  which they consider to be at least half, would
  you rather be player A or B?
• Note that this algorithm is also envy-free. Why?
  But, is it strong fair, super envy-free, or exact?

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Simple Fairness, n 3

• n 3 is harder than n 2, but since we know how
  to do it for the latter, can we extend the
  previous method inductively to n 3?
• Perform the previous algorithm for 2 players.
  Now, players A and B each has at least 1/2 of the
  cake.
• Players A and B now divide their shares into 3
  equal parts.
• Player C will choose 1 part from player A and 1
  part from player B
• Why is player C ensured at least 1/3 of the cake?
• It is fairly straight-forward to extend this
  algorithm to the general n-players case.
• However, this algorithm requires many cuts and
  therefore generates many small pieces of cake for
  everyone. Is it possible to use lesser number of
  cuts?

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Accomplishing n 3 With 2 (or n-1) Cuts

• This can be accomplished by the so-called Moving
  Knife Algorithm
• A knife is moved continuously over the cake from
  left to right. When a player thinks that the cake
  on the left side of the knife is 1/3 of the whole
  cake, he calls out, Stop, and gets that piece
• This is repeated for the remaining 2 players. The
  player who did not call out this time, will get
  the piece on the right
• However, this algorithm is not finite, because
  the players have to make a decision at every
  instance of time
• In fact, it is not possible to accomplish the
  task with n-1 cuts (except for the case of n
  2), unless an infinite algorithm is used
• What then is the minimum number of cuts that can
  be attained with a finite algorithm?

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Min. Cuts for Finite Simple Fair Algorithms

• The inspiration comes from a problem commonly
  encountered in programming inserting an element
  into a sorted list of length L. We know that an
  efficient algorithm is to recursively reduce the
  size of the problem by half at each step of the
  algorithm
• Using this divide and conquer strategy, we can
  develop a cutting algorithm with O(n) ? n x log2n
  
• In general, supposed the problem has been reduced
  to k (assume even) players
• Then all but one of the k players are told to cut
  the cake into the ratio k/2. So there are in
  total k-1 cuts in this iteration
• The non-cutter now goes to the middle cut and
  picks the side which he thinks is bigger. He then
  joins the group of players ((k-2)/2 of them)
  whose cuts landed on the side he picks
• We now have a k/2 problem. If k is odd, only a
  slight modification is required

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Other Criteria An Envy-Free Algorithm

• Because the simple fair criterion is the subset
  of the other four criteria mentioned before, it
  is therefore the easiest to be accomplished.
• To give everyone a taste of how much harder the
  other criteria are, an envy-free algorithm for 3
  players is shown here (attributed to Selfridge)
• Player A cuts cake into 3 equal pieces. player B
  ranks pieces ?, ?, ?, from largest to smallest.
• Player B trims ? from ? so that ? ? - ? ?
• Leaving ? out, the three pieces are split among
  players C, B, A in that order. Player B has to
  pick ? if available. Therefore B or C must have
  picked ?. Call this person P1 and the other P2.
• P2 now cuts ? into 3 equal pieces, and P1, player
  A, P2 pick the pieces in that order.
• Why does this work intuitively? Why is everyone
  happy with step 3? Why is everyone happy with
  step 4?

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Other Criteria Why Harder?

• As can be seen, the envy-free algorithm for 3
  players is much more complicated than that of the
  simple fair algorithm.
• In fact, a finite bounded algorithm for the
  general case has yet to be found.
• Why are the other criteria so much harder to
  accomplish?
• Intuitively, this is because unlike the simple
  fair criterion, each player not only cares about
  what he gets, but also cares about what the other
  players get. For example, in exact division, each
  player thinks every other player has the same
  piece as him.
• Therefore, each player must be somehow involved
  in the decision making process throughout the
  algorithm. So a recursive algorithm (such as that
  in divide and conquer) which reduces the problem
  to a smaller size at each step by eliminating
  some players will not work.

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Why Disagreeing might be Better Part 1

• Interestingly, players who have different
  measures of cake size can both benefit from the
  disagreement.
• For example, from the simple fair algorithm
  mentioned for n 2

Player A divides Cake into X1 X2 , in his
opinion
B might think X1 60 and X2 40
B takes X1 leaving A with X2
B divides X1 into 12 pieces and A takes one piece
B now has 55 and A now has 54 of the cake
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Why disagreeing might be Better Part 2

• Notice that we have just achieved super envy-free
  division for n 2
• In fact super envy-free division is possible for
  n players if and only if the n measures are
  independent.
• The following concluding example once again
  illustrates how disagreement can make everyone
  happier.
• Consider three players A, B and C splitting a
  family inheritance a car, boat and painting.
• The players are told to submit their individual
  valuation of the three properties, so we will get
  a matrix like this

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Why disagreeing might be Better Part 3
3. Each of the three items is given to the
players who value them the most and the players
pay the difference between their valuation of the
items they receive and their fair share of the
inheritance to a common pool. Therefore

• Because 13333 8000 3667 9000, there is an
  excess of 9000 and each player gets an extra
  3000 beyond his valuation of a fair share.
• We have achieve strong fairness.
• Why does this work? Intuitively, since the person
  who values an item the most gets the item, the
  other players will think that he is paying more
  than he should to the pool. For example, if
  player B gets the boat instead of the painting,
  he would only pay 23000 14000 9000 to the
  pool instead of 13333.

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Conclusion Applicability to Real Life

• While the previous example seems promising for
  solving real life sharing
• dispute, what might be some of the problems that
  arise?
• As the sharing procedure continues, people might
  change their measures of the shared resource
  during the procedure.
• People might also disagree on what the
  appropriate share should be, for example in a
  territorial dispute. Player A might want 70 of
  the land, while player B might want 50 of it.
  This is much more difficult than the unequal
  sharing problem which actually has a finite
  bounded solution.

Source Cake-Cutting Algorithms by Jack Robertson
and William Webb