Bakery%20Algorithm%20-%20Proof

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Bakery Algorithm - Proof

• Operating Systems
• Spring 2003

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Bakery algorithm for n processes
R
T
D
T-D
C
E
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Structure of the algorithm

• R code prior to using Bakery algorithm
• T creating of a ticket and awaiting for
  permission to enter critical section
• D creation of a number (first part of a ticket)
• T-D awaiting for the permission to enter
  critical section
• C critical section itself
• E code executed upon exit from critical section

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Basic Lemma
Lemma 1
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Lemma 1 - proof

• Denote s time point where lemmas conditions
  hold
• At time s, is in C (critical section)
• numberi has been selected and still exists
• numberj has been selected and still exists
• Exist time point t1lts, where performs a
  check (choosingj0) and passes it
• Exists time point t2, t1ltt2lts, where
  performs a check
• (numberj!0) and (numberi,i)lt(numberj,j)

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Lemma 1 proof (cont)

• Since at time t1 exists (choosingj0) one of
  the following has to be true
• CASE A numberj was chosen after time t1 and
  before time s
• CASE B numberj was chosen before t1

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Lemma 1 proof CASE A

• Since at time t1 already checks for
  permission to enter critical section, computation
  of numberi was computed before that and
  persists until s
• Thus, at the time starts to compute its
  numberj, it has to take into account of max
  value of numberi. So it creates a value which
  is greater then numberi at least by 1, and it
  persists until time s
• That is (numberi,i)lt(numberj,j) at time s

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Lemma 1 proof CASE B

• Both numberi and numberj were computed before
  t1, thus also before time t2 and persisted
  until s
• At time t2 performed check (numberj!0)
  (numberj,j)lt(numberi,i), which failed,
  since is in C at time s
• numberj was chosen before t2 and persisted,
  thus first part of the check could not fail, also
  i and j are different, so
• (numberi,i)lt(numberj,j) at time s

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Lemmata 2,3,4

• Lemma 2 mutual exclusion
• Bakery Algorithm satisfies mutual exclusion
  property
• Lemma 3 progress
• Bakery Algorithm guarantees progress
• Lemma 4 fairness
• Bakery Algorithm guarantees lockout-freedom

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Lemma 2 Proof

• Assume in contradiction that there are two
  different processes that have entered critical
  section
• Then conditions of Lemma 1 are true for both
  processes simmetrically that is
  (numberi,i)lt(numberj,j) and
  (numberj,j)lt(numberi,i) contradiction
• We conclude that mutual exclusion is satisfied

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Lemma 3 Proof

• Suppose progress is not guaranteed
• Then eventually a point is reached after which
  all processes are in T or R
• By the code, all the processes in T eventually
  complete D and reach T-D
• Then the process with the lowest (number,ID) pair
  is not blocked from reaching C, that is enters
  critical section
• We conclude that Bakery algorithm satisfies
  Progress

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Lemma 4 Proof

• Consider a particular process in T and
  suppose it never reaches C
• The process eventually completes D and reaches
  T-D
• After that any new process that enters D
  perceives numberi and chooses a higher number
• Thus, since does not reach C, none of these
  new processes reach C either
• But by Lemma 3 there must be continuing progress,
  that is infinitely many entries to C
• Contradiction blocks the entry to C

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Remark on Fairness

• A process that obtained a ticket (numberk,k)
  will wait at most for (n-1) turns, when other
  processes will enter the critical section
• For example if all the processes obtained their
  tickets at the same time they will look like
  (q,1),(q,2)(q,n)
• In which case process will wait for processes
• to complete the critical section

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Bibliography

• Nancy Ann Lynch, Distributed Algorithms, JMC
  243.9 LY53
• Leslie Lamport
• A new solution of Dijkstras concurrent
  programming problem
• Communications of the ACM
• 17(8)453-455, 1974