Tight Comparison Bounds On the Complexity of Parallel Sorting

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Tight Comparison Bounds On the Complexity of Parallel Sorting
Yassi Azar
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To see this, simply simulate the AKS network by slowing it down to work in k rounds.
For k = \, c(\, n) = ^(n'^-n). This is since any sorting algorithm which works in one round must perform all comparisons. Otherwise, suppose that a dispensed comparison is between two successive elements in the sorted order; the algorithm will clearly fail to distinguish their order. On the other hand, observe that performing all comparisons simultaneously yields an one round algorithm in the parallel compariso
...n model that matches exactly this lower bound, i. E. , ui l, /i. ) = %(«"-«).
So, it remains to consider the situation for 1 k( - n) for an V A:, A? > 1, where ^ is the base of the natural logarithm.
e Corollaries of the main result: Suppose we have p processors with the interpretation that each processor can perform at most one comparison at each round. Observed that kp^c{k, n) or p^cik, n)/k . Therefore, Corollary 1.


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