Algorithms: Design Techniques and Analysis (Lecture Notes by M. H. Alsuwaiyel

By M. H. Alsuwaiyel

Challenge fixing is a necessary a part of each medical self-discipline. It has elements: (1) challenge identity and formula, and (2) resolution of the formulated challenge. it is easy to resolve an issue by itself utilizing advert hoc innovations or stick with these thoughts that experience produced effective ideas to related difficulties. This calls for the knowledge of assorted set of rules layout options, how and while to exploit them to formulate recommendations and the context acceptable for every of them. This e-book advocates the learn of set of rules layout thoughts by means of offering lots of the worthwhile set of rules layout recommendations and illustrating them via quite a few examples.

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1 ~ shows ~approx~mate ~ running 2 times ~of algorithms . with 24 Basic Concepts in Algorithmic Analysis time complexities log n, n, nlogn, n2,n3 and 2n, for n = 23, 2*, . . ,PoM one million, assuming that each operation takes one nanosecond. Note the explosive running time (measured in centuries) when it is of the order 2n. 1 Running times for different sizes of input. “nsec” stands for nanoseconds, 1‘ I1 ‘ p IS one microsecond and “cent” stands for centuries. 1 We denote by an “elementary operation” any computational step whose cost is always upperbounded by a constant amount of time regardless of the input data or the algorithm used.

Let Algorithm MODBINARYSEARCH be some implementation of this binary search technique. Thus, MODBINARYSEARCH({2,3,6,8,9},7) = 4. The modified sorting algorithm is given in Algorithm MODINSERTIONSORT. SORT. n] of n elements. Output: A ( l . n ] sorted in nondecreasing order. 1 . for i t 2 to n 2. z+- A[i] 3. Z - 1],2) for j+- i - 1 downto k 4. 5. A[j + 11 Abl 6. end for 7. A(k]+-z 8. 1), it follows that the total MODBINARYSEARCH. + How to Estimate the Running Time of an Algorithm 41 number of comparisons done by Algorithm MODINSERTIONSORT is at most n n-1 n- 1 z ( [ l o g ( i - l ) J + l ) = n - l + ~ j l o g i J< n - l + ~ l o g i = Q ( n l o g n ) .

We have that n! Since lim n-oo we conclude that n! 12 ~ (n n! = lim 1 + I)! It follows that n! ) Consider the series xy=llogj. Clearly, n n That is, n C l o g j = O(nl0gn). Also, Thus, n C l o g j = f2(nlogn). nlogn). 13 We want to find an exact bound for the function f ( n ) = logn!. First, note that logn! = logj. It follows that logn! = 0(nIogn). 14 Since logn! )but n! is not 0(2n). Similarly, since logZna = n2 > n l o g n , and logn! 13), it follows that n! ). l Tame Complexity 31 However, this upper bound is not useful since it is not tight.

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