Data Structures and Algorithm Analysis - Computer Science at ...

Sec. 9.1 Searching Unsorted and Sorted Arrays 303n−1∑T(n) = np n + (i + 1)p i .What happens to the equation if we assume all the p i ’s are equal (except p 0 )?i=0n−1∑T(n) = p n n + (i + 1)p= p n n + pi=0n∑ii=1n(n + 1)= p n n + p2= p n n + 1 − p nnn(n + 1)2= n + 1 + p n(n − 1)2Depending on the value of p n , n+12≤ T(n) ≤ n.For large collections of records that are searched repeatedly, sequential searchis unacceptably slow. One way to reduce search time is to preprocess the recordsby sorting them. Given a sorted array, an obvious improvement over simple linearsearch is to test if the current element in L is greater than K. If it is, then we knowthat K cannot appear later in the array, and we can quit the search early. But thisstill does not improve the worst-case cost of the algorithm.We can also observe that if we look first at position 1 in sorted array L and findthat K is bigger, then we rule out position 0 as well as position 1. Because moreis often better, what if we look at position 2 in L and find that K is bigger yet?This rules out positions 0, 1, and 2 with one comparison. What if we carry this tothe extreme and look first at the last position in L and find that K is bigger? Thenwe know in one comparison that K is not in L. This is very useful to know, butwhat is wrong with the conclusion that we should always start by looking at the lastposition? The problem is that, while we learn a lot sometimes (in one comparisonwe might learn that K is not in the list), usually we learn only a little bit (that thelast element is not K).The question then becomes: What is the right amount to jump? This leads usto an algorithm known as Jump Search. For some value j, we check every j’thelement in L, that is, we check elements L[j], L[2j], and so on. So long as K isgreater than the values we are checking, we continue on. But when we reach a