Consider the following technique for performing the compare-split operation. Let x1 , x2 , ..., xk be the elements stored at process Pi in increasing order, and let y1 , y2 , ..., yk be the elements stored at process Pj in decreasing order. Process Pi sends x1 to Pj . Process Pj compares x1 with y1 and then sends the larger element back to process Pi and keeps the smaller element for itself. The same procedure is repeated for pairs (x 2 , y2 ), (x3 , y3 ), ..., (xk , yk ). If for any pair (xl , yl ) for 1≤ l≤ k , xl yl , then no more exchanges are needed. Finally, each process sorts its elements.

Required:
a. Show that this method correctly performs a compare-split operation.
b. Analyze its run time, and compare the relative merits of this method to those of the method presented in the text.
c. Is this method better suited for MIMD or SIMD parallel computers?

Cómo pongo un tomo de llamada sin pagar?

Match the function to its purpose. fast worth 50pts.

The idea that, for each pair of devices v and w, there’s a strict dichotomy between being “in range” or “out of range” is a simplified abstraction. more accurately, there’s a power decay function f (·) that specifies, for a pair of devices at distance δ, the signal strength f(δ) that they’ll be able to achieve on their wireless connection. (we’ll assume that f (δ) decreases with increasing δ.) we might want to build this into our notion of back-up sets as follows: among the k devices in the back-up set of v, there should be at least one that can be reached with very high signal strength, at least one other that can be reached with moderately high signal strength, and so forth. more concretely, we have values p1 ≥ p2 ≥ . . ≥ pk, so that if the back-up set for v consists of devices at distances d1≤d2≤≤dk,thenweshouldhavef(dj)≥pj foreachj. give an algorithm that determines whether it is possible to choose a back-up set for each device subject to this more detailed condition, still requiring that no device should appear in the back-up set of more than b other devices. again, the algorithm should output the back-up sets themselves, provided they can be found.\