By Henri Casanova, Arnaud Legrand, Yves Robert
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Extra resources for Parallel Algorithms (manuscript)
Example text
Applying a split network to a bitonic sequence yields two equal-length bitonic sequences. 9). Let us prove the above result more formally. Assume that there are more 1’s than 0’s in the initial bitonic sequence b. • If b is of the form 1i 0j 1k . – If k – If i n 2 n 2 n 2 then the output sequence is 1i 0j 1k . n n then the output sequence is 1i− 2 0j 1k 1 2 . n – If i < and k < n2 (note that we then necessarily have i + k 2 since there are more 1’s than 0’s) then the output sequence is of n n n n the form 0 2 −k 1 2 −j 0 2 −i 1 2 .
In both cases, it moves to the right from step 2 and for all following steps until it is at position n. Before step 2, the last 1 is at least at position 2 and thus it always has enough time to arrive at position n in n − 1 steps. Let us now follow the moves of the next-to-last 1. 6). As the last 1 moves to the right from step 2 on (at least), the next-to-last 1 is never blocked by the last 1 when moving to the right. Therefore, from step 3 on, the next-to-last 1 moves to the right until it reaches position n − 1.
Hence, there are at most 2(hX − 1) + 1 + 2(hY + 1) + 1 = 2r + 2 elements of X |Y between e1 , e2 , . . , er . Case 2 (er ∈ Y ): Let us add an element e0 ∈ Y preceding e1 and an element er+1 ∈ X following er . Then, elements from X and Y lying between e1 , e2 , . . er come from elements for X lying between hX + 1 elements of X and elements of Y lying between hY + 1 elements of Y . Therefore, we have (2hX + 1) + (2hY + 1) = 2r + 2 elements of X |Y . Coming back to the proof of the lemma, let us define Z = REDUCE(X|Y ) and Z = REDUCE(X |Y ).