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More Quiz Questions ITCS 4145/5145 Parallel Programming Parallel Strategies

More Quiz Questions ITCS 4145/5145 Parallel Programming Parallel Strategies. What is meant by the term “Embarrassingly Parallel” computation?. All parallel computations are embarrassingly parallel computations

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More Quiz Questions ITCS 4145/5145 Parallel Programming Parallel Strategies

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  1. More Quiz QuestionsITCS 4145/5145Parallel ProgrammingParallel Strategies ITCS 4/5145 Parallel Programming, UNC-Charlotte, B. Wilkinson, 2013, QuizQuestions3.ppt Jan 28, 2013

  2. What is meant by the term “Embarrassingly Parallel” computation? • All parallel computations are embarrassingly parallel computations • A computation that can be divided into parallel parts in an obvious fashion without the parts generally needing to communicate with each other. • One that is simple. • A computation that cannot be divided into parallel parts. • A P2P (peer-to-peer) computation • None of the other answers

  3. What is the sequential time complexity of bucket sort if there are N numbers and M buckets?Assume the values of the numbers are uniformly distributed and a sorting algorithm such as Quicksort to sort the buckets. • O ( N log N/M ) • O ( log N ) • O ( N/M log N/M ) • O (N log N ) • O (N2 ) • None or the other answers. • From N steps to put numbers into separate buckets plus N/M log N/M steps to sort each bucket assuming N/M numbers and M buckets done one after the other, gives O ( N + N log N/M ) = O ( N log N/M). • Note if N = kM, get O (N) which is better than compare and exchange sorting algorithms but assumes numbers are uniformly distributed.

  4. What is the parallel time complexity of bucket sort using the “simple” approach in notes, with same assumptions as before and each parallel process operates in synchronism at the same time (P processes, P = M)? • O ( N log N/M ) • O ( log N ) • O ( N + N/M log N/M ) • O (N + N/M log N ) • O (N2 ) • None or the other answers. (c) Each process: N steps to put N/M numbers into its bucket plus N/M log N/M steps to sort its bucket gives O ( N + N/M log N/M ) = O ( N/M log N/M ) as eventually log N/M will be greater than M for fixed M.

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