Section II.6: Analysis of Iterative Searches and Sorts

Transcription

1 Sectio II.6: Aalysis of Iterative Searches ad Sorts Biary Search Biary Search is a efficiet algorithm that searches for a item X i a sorted array. The item X searched for is compared with the value of the midpoit of the list. If X is foud, the the positio i the list is give. If X is less tha this value, we search the left half of the list (excludig the midpoit). If X is greater tha this value, we search the right half of the list (excludig the midpoit). This process is cotiued util the legth of the sublist is 1. If X is foud, the positio i the list is retured. If X is ot foud, the the positio 0 is retured. Iterative pseudo code for Biary Search, foud i Epp, p. 51 is give below. Algorithm Biary Search Iput: positive iteger, array A [1], A [],..., A[] i ascedig order, item X of the same data type as the elemets of the array. Output: 0 (zero) if X is ot foud, otherwise the positio i the array where X is foud. Algorithm Body: idex := 0; bot := 1; top := ; while (bot top ad idex = 0) bot + top mid := if X = A[mid] the idex := mid else if X < A[mid] the top := mid - 1 else [X>A[mid]] bot := mid + 1; ed while retur idex ed Algorithm Biary Search Example II.6.1: (a) Trace the actio of Biary Search o the list 1, 8, 10, 1, 17, 19, 3 if we are searchig for item 10. (b) Trace the actio of Biary Search o the list 1, 8, 10, 1, 17, 19, 3 if we are searchig for item 15. Solutio: (a) Iteratio Number Variable mid Names X? A[mid] -- < > = bot top idex The item is foud i positio 3. 1

2 Solutio: (b) Iteratio Number Variable mid Names X? A[mid] -- > < < bot top idex The item is ot foud, idex 0 is retured. Tree Structure for Biary Search Defiitios (Iformal): A tree is a data structure such that each ode has at most oe predecessor (paret) ode. A ode may have may successor (childre) odes. Trees are formally covered later but are iformally used from ow o for illustrative purposes. There is oe ode without ay paret (the root), ad the leaves are odes that have o childre. The level of a ode is the legth of the path from the root to that ode. The root has level 0. A Biary Tree is oe where each ode has at most two childre. The biary tree for Biary Search is costructed as follows. At each level the midpoit of that sublist is the paret ad the midpoits of the left ad right sublists are the childre. The root of the tree is the midpoit of the origial list. The leaves are the odes for which the left ad right sublists are empty. Example II.6. (a): Draw the tree for the list i Example The solutio is show above. The first umber i each ode is the positio of the item i the list, ad the secod is the item itself. Example II.6. (b): Draw the tree for list 1, 4, 7, 8, 10, 1, 15, 17, 19,,

3 The solutio is show above. The first umber i each ode is the positio of the item i the list, ad the secod is the item itself. Aalysis of Iterative Biary Search We wish to fid (a) the ru time T() ad (b) a fuctio U() with oe term, ad leadig coefficiet 1, where T( = Θ(U()). Solutio: (a) k Case 1: If the size of the array is = 1 for some positive iteger k, the the tree structure is full, i. e., each level has the maximum umber of odes. (Example 1 is such a case for k = 3). The maximum umber of subdivisios of the origial array is the maximum umber of odes from the root to the leaf level of the tree ad this value is k. Sice k = + 1, we have k = log ( + 1) = log ( +1) sice log ( + 1) is a iteger. Therefore the ru time T() = log ( +1). Case : k < k for some positive iteger k. The tree for biary search requires at least oe ode at the ext level of the tree. So the maximal legth of a path through the tree is log ( +1). We must use the ceilig fuctio i this case sice log ( + 1) is ot a iteger. Therefore the ru time T() = log ( +1). I either case, the, T() = log ( +1). Solutio: (b) We show T() = Θ(log ). To show T()= O(log ), ote that log ( + 1) log + 1 log for all itegers. I the defiitio of big O, we ca take C =, X 0 =. I the above iequality we use 1 log ad fact that the slope of the graph of log ( + 1) is less tha the slope of the graph of log + 1. We also use the fact that log ( + 1) = log + 1for = 1. To show log = O( log ( +1) ), ote that log < log ( + 1) log ( +1). So C = 1 ad X 0 = 1 from the defiitio of big O ca be used Isertio Sort Isertio Sort is a sort that creates from the iput array A of size, a positive iteger, a sorted subarray of size i for each i, 1 i. Oce a sorted array of size i-1 is obtaied, for i, we assig A[i] to v. v is compared to A[i-1]. If v < A[i-1] the oe copies A[i-1] ito A[i]. The v is compared to A[i-]. If v < A[i-] the the elemets A[i-] is moved up. This process cotiues util v A[j] for some j, 0 j i. If the coditio v < A[1] is reached, the we must compare v to the setiel A[0]. Oe obtais v > A[0], ad this causes the while loop to termiate. This process is repeated for i =, i =3,... up to i =. Fially we get a sorted array of size. 3

4 The pseudo code for Isertio Sort is as follows: Algorithm Isertio Sort Iput: Positive iteger, array A of items. Output: Sorted array i ascedig order. Body of algorithm: A[0] := setiel [ put a value less tha all others i the list ito A[0]] for i := to v := A[i]; j := i; while A[j-1] > v A[j] := A[j-1]; j := j 1; ed while; A[j] := v; ext i; Ed Algorithm Isertio Sort Aalysis of Isertio Sort Worst Case: The list is i descedig order. For i betwee ad, there are i comparisos whe A[i] is smaller tha all previous values. The umber of comparisos, icludig a compariso agaist the setiel, is T() = i = i = ( +1) - 1 = + = Θ( ). Best Case: The list is already sorted. Whe A[i+1] is compared with A[i], it is always larger. So oly oe compariso is made for each i. The umber of comparisos is T() = -+1 = 1 = Θ(). Average Case: Assume that i ay subarray of A of legth i, the elemets are uiformly distributed, i.e., the maximum elemet is equally likely to be i ay positio. Let C be the total umber of comparisos eeded, ad C i be the umber of comparisos eeded whe we isert the elemet at idex i i its proper positio i the subarray A[1],, A[i] where A[1],, A[i-1] is already sorted. The T() = E[C] = E[C + + C ] = E[C i ] i = by the liearity of expectatio (Theorem I..1). We ow fid E[C i ]. Let P i (k) be the probability that max {A[1],...,A[i]} is i the k-th positio, 1 k i. The P i (k) = i 1. 4

5 Let C i (k) be the umber of comparisos eeded if the elemet at idex i is to be put i idex k, where the subarray A[1],, A[i-1] is already sorted. The C i (k) = i k + 1. We the have i i i (i +1) i i +1 E[C i ] = P i (k) C i (k) = (i k + 1) = ( j k =1 k =1 i i ) = = j =1 i Therefore +1 i + 1 i T () = E[C i ] = = = 1 ( +1)( + ) 1 - (1 + ) i = i = = = = Θ ( ). 4 4 Note that the average ru time is about half that of the worst case. i = 3 Example II.6.3: Give a array A that is 4, 3, 1, 9, 6 (a) Give the arragemet of the array A at the ed of each iteratio of the for loop i the Isertio Sort algorithm. (b) Compute the total umber of comparisos eeded. (c) Compute the umber of comparisos eeded i the best case, the worst case, ad the average case. Solutio: (a) Iitially: value of i array A (b) value of i Number of Comparisos 3 1 Total = 8. Note that if A[i] < A[1], the A[i] must be compared to the setiel A[0]. Therefore, we have comparisos for i = ad 3 comparisos for i = 3. (c) Plug i 5 for i above formulas for the best case, the worst case, ad the average case. Best case: T() = 5 1 = Worst Case: T() = = Average case: T() = = =

6 Exercises: (1) Let A be the sorted array 1, 7, 9, 1, 17, 1, 30. Use the format of Example II.6.1, parts (a) ad (b). (a) Fill i the trace table for Biary Search whe X = 9 is searched for. (b) Fill i the trace table for Biary Search whe X = 31 is searched for. (c) Draw a tree structure for this list (see Example II.6 ). () Let A be the sorted array 1, 3, 4, 5, 9,, 4, 31, 35, 36, 39, 41. Use the format of Example II.6.1, parts (a) ad (b). (a) Fill i the trace table for Biary Search whe X = 39 is searched for. (b) Fill i the trace table for Biary Search whe X= 8 is searched for. (c) Draw a tree structure for this list (see Example II.6.). (3) Compute the best case, the average case, ad the worst case for the ru time T() of isertio sort for a array of 16 elemets. (4) Let A be a array of elemets 5, 8,, 13, 4, 1, 15, 9. (a) Usig the method of example II.6.3, give the arragemet of array A at the ed of each iteratio of the for loop i the Isertio Sort algorithm. (b) Compute the actual umber of comparisos of A [i] to previous elemets for each i betwee ad 8. (c) Get the total umber of comparisos eeded. 6