By Russ Miller, Laurence Boxer

With multi-core processors exchanging conventional processors and the circulate to multiprocessor workstations and servers, parallel computing has moved from a area of expertise quarter to the center of computing device technology. which will offer effective and reasonably-priced strategies to difficulties, algorithms has to be designed for multiprocessor platforms. Algorithms Sequential and Parallel: A Unified strategy 2/E offers a state of the art method of an algorithms path. The booklet considers algorithms, paradigms, and the research of options to severe difficulties for sequential and parallel types of computation in a unified model. this offers practising engineers and scientists, undergraduates, and starting graduate scholars a history in algorithms for sequential and parallel algorithms inside of one textual content. necessities comprise basics of information buildings, discrete arithmetic, and calculus.

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**Additional resources for Algorithms Sequential & Parallel: A Unified Approach (Electrical and Computer Engineering Series)**

**Sample text**

Therefore, we have shown that for all integers n, d n x = nx n1 . dx Recursion A subprogram that calls upon itself (either directly or indirectly) is called recursive. To the beginner unfamiliar with this notion, it may sound like a recipe for an infinite loop, as indeed it can be if not used with care. In fact, recursion is often used as a form of looping. However, recursion should be used so that a recursive subprogram’s self-reference is made only with “simpler” data. That is, each time a program calls itself, it does so with a smaller/simpler instance of the problem.

2 k for some integer k v 4 . Based on this, we want to show that ( k + 1)! > 2 k +1 . Now, ( k + 1)! ) , which (by the inductive hypothesis and the assumption that k v 4 ) is an expression at least as large as 5(2k) > 2(2k) = 2k+1, as desired. This completes the proof. EXAMPLE d n x = nx n1 , for all integers n. dx Proof: Even though this statement is about all integers, we can use mathematical induction to give the proof for n, an arbitrary positive integer, and then use fundamental rules of calculus to handle other values of n.

Based on this analysis, we can now attempt to produce an asymptotically tight bound on the function g(n). First, we consider a lower bound on g(n). Our analysis shows that µ n 1 n 1 1 dx f ¨ . x k =1 k 18 Chapter 1 Asymptotic Analysis Because µ n 1 n 1 dx = ln x ¼¾ 1 = ln n ln 1 = ln n, x we know that g(n) is bounded below by ln n. Next, we consider an upper bound on g(n). Notice that if we apply the result of our analysis for a nonincreasing function blindly, we obtain n 1 ¨k f µ k =1 n 1 dx = ln x ¼¾ 0 = h.