Sorting an Intransitive Total Ordered Set

1
Sorting an Intransitive Total Ordered Set

• Sequential and Parallel Implementation
• for COMP 5704 Parallel Algorithms
• by
• Johannes Singler, Carleton University

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Presentation Outline

• Introduction to the Problem
• Prerequisites
• Sequential Algorithm
• Parallelizing for a PRAM
• Questions?!

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Introduction

• Definition of Transitivity
• Traditional sorting relies upon (Transitive)
  Total Ordered Set (e. g. natural numbers and ).
• New Problem Sorting of Intransitive Total
  Ordered Set (ITOS)
• Relying on intransitive relation

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Intransitive Sorted Sequence

• An intransitive sorted sequence has to satisfy
• Theorem There exists always a solution (proof
  later if time remaining).
• The solution is not unique in general.
• Real-world example for an ITOS tournament in
  sports Every player wins or loses against each
  other.
• Therefore ITOS also referred to as a Tournament.

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Representation as a Graph

• Tournament can be represented as a complete
  directed graph.
• Every two nodes are connected by an edge,
  direction denotes relation (who wins).
• A intransitive sorted sequence then corresponds
  to a Hamiltonian Path.
• Definition of Hamiltonian PathA path in a graph
  that visits every node exactly once.

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Sorting Prerequisites

• Sorting using Semi-Heap according to Wu 2
• Definition of
• On 3 elements.
• Set in general (not unique any more).
• If one player beats both the others, he is the
  only maximum element, otherwise all three are
  maximal.
• Definition of a Semi-Heap
• Binary tree (complete in the beginning).
• Similar to a heap.
• Semi-Heap condition
• 4 valid possibilities (and 4 invalid)

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Actual Sorting

• Similar to Heap-Sort
• Build up heap
• Iteratively remove root, replace by rightmost
  leaf of deepest level before pushing this down
  the heap.
• Intransitive Sorting using Semi-Heap
• Build up semi-heap
• Iteratively remove root, update semi-heap
  top-down.
• Tree may not be complete any more after first
  removal.

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Pseudo-Code

• Pseudo-Code
• semi-heap-sortbuild-semi-heapWhile nodes
  left Remove root and append to sorted
  sequence Recursively replace by winning child
  top-down
• build-semi-heap (Q(n))
• for iheapsize/2 downto 1 semi-heapify(i)
• semi-heapify(i) if(i not max) exchange with
  max child semi-heapify(former child position)
  recursively

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Sequential Complexity

• Demo
• Sequential Complexity O(n log n)

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Parallelizing the Algorithm for a PRAM

• Parallelizing for fine-grained multiprocessing
  using Pipelining.
• One processor per two consecutive tree levels
  (overlapping) Q(log n) processors needed.
• Two phases Either processors with even numbers
  or processor with odd numbers are active.
• While updating the tree top-down, the work is
  handed from one processor to the next.
• Demo

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Complexity Issues

• Parallel Complexity
• Q(n) on an EREW PRAM using Q(log n) processors,
  therefore cost-optimal.
• Building the initial semi-heap takes Q(n)
  sequentially and thus needs no parallelization.
• No concurrent access needed because of phases.
• Can be modified to use less processors according
  to Brents Law.

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Outlook

• Extension
• A Hamiltonian Cycle exists in every strong
  connected Tournament.
• Definition of strong connectedEvery node is
  reachable from each other
• Hamiltonian Cycle Every node is visited exactly
  once and there is an edge from the last node to
  the first one

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Questions?

• Do you have any questions left?

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Questions!

• Which of the following subtrees satisfy the
  semi-heap condition?
• What does Pipelining mean?
• How many possible tournaments of size n are there?

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Conclusion

• Presentation and demo program available at my
  website http//www.jsingler.de/COMP5704/
• References1 Danny Soroker. Fast parallel
  algorithms for finding Hamiltonian paths and
  cycles in a tournament. Journal of Algorithms,
  9(2)276286, June 1988.2 J. Wu. On sorting
  an intransitive total ordered set using
  semi-heap. In International Parallel and
  Distributed Processing Symposium, pages 257262,
  2000.

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Appendix Proof of Theorem

• By induction according to Soroker 1
• Result clear for n2
• n ? n1
• Let v be vertex of V(T), then V(T)-v has
  Hamiltonian Path v1,,vn
• If v ? v1, then v,v1,,vn is a valid result
• Otherwise let I be the largest index such that vi
  ? v, then v1,,vi,v,vi1,,vn or v1,,vn,v
  respectively (in) is a valid result.