09/28/2015. Problem Rearrange the elements in an array so that they appear in reverse order.

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1 Unit 4 The array is a powerful that is widely used in computing. Arrays provide a special way of sorting or organizing data in a computer s memory. The power of the array is largely derived from the fact that it provides us with a very simple and efficient way of referring to and performing computations on collections of data that share come common attributes. All the elements in the array share a common name and has a unique address which distinguishes it from all other elements in the list. Arrays play an integral part in many computer algorithms. They simplify the implementation of algorithms that must perform the same computations on collections of data. Employments of arrays often leads to implementations that are more efficient then they would otherwise be. Rearrange the elements in an array so that they appear in reverse order. development The problem of reversing the order of an array of numbers appears to be completely straightforward. We start the design of this algorithm by careful examination of the array before and after it has been reversed; for example, First element ends up in the last position. Carrying this process through we get the following set of exchanges. In terms of suffixes the exchanges are : What we observe from our diagram is that the 1

2 Examining the effects of these exchanges we discover that after step 3 the array is completely reversed. In setting up our algorithm we need a pair of suffixes that model this increasing-decreasing behavior. Our increasing suffix can be the variable i which is incremented by1. For our decreasing suffix we might try n-i since this decreases by 1 with each increase in i by 1. This means that when i=1 we find that n-i is equal to n-1 rather than n. We can correct this by adding 1 So the suffix n-i-1 can be used at other end rather than using n-i. This gives our following algorithm. Arrayreversal.txt Vector and Matrix Processing Given a set of n student s examination marks (in the range 0 to 100 ) make a count of the number of students that obtained each possible mark. What we are required to do in this case is obtain the distribution of a set of marks. This is a typical frequency counting problem. One approach is to declare 101 variables C0,C1,C2,C3,.,C100 each corresponding to a particular mark. The counting strategy we could then employ might be as follows: 2

3 The difficulty with this approach is that we need to make 101 test (only one of which is successful) just to update the count for one particular mark. A better solution would be to examine each mark, and depending upon its value we place a star (*) in corresponding marks slot. At this point we need to recognize that an array can be useful in the solution of the problem. We can easily set up an array with 101 locations, each location corresponding to a particular mark value. The value of a particular mark leads us directly to the particular slot that must be updated. If we store in each array location the count of the number of students that obtained that mark we will have the required solution to the problem. The solution can be summarized as follows histogram.txt Statistical analysis Find the maximum number in a set of n numbers. 3

4 Development Before we begin to work on the algorithm for finding the maximum we need to have a clear idea of the definition of a maximum. A maximum is a number which is greater than or equal to all other numbers in the set. To start in the algorithm development for this problem let us examine a particular set of number. For example After studying this example we can conclude that all numbers need to be examined to establish the maximum. A second conclusion is that comparison of the relative magnitude of numbers must be made. The simplest and most systematic way to examine every item in a list is to start at the beginning of the list and work through, number by number until the end of the list. Here we need to examine and compare the each and every number. With a task of finding largest among n numbers, we start with the first number. As there is no other number to compare with we make this number as a temporary candidate against whom we start comparing. Then start taking the second number in the set. Three situations are possible: The second number can be less than our temporary candidate. The second number can be equal to our temporary candidate The second number can be greater than our temporary candidate. If situation 1or 2 arrives, no change is done (temporary candidate is itself maximum number). However in 3 rd case the second number is made the temporary candidate(i.e maximum number). We then move on and compare the third number and the process is continued. The whole process is continued until all the elements in the set have been examined. Maximum.txt Plotting Scaling And Sorting 4

5 Remove all duplicates from an ordered array and contract the array accordingly. Development As a starting point for this design let us focus on specific example so that we have a clear idea of exactly what is required. After a brief examination of the array we will be able to produce the contracted array below Comparing the two arrays we see that all elements apart from 1 have shifted their position in the array. In other words each unique element has been moved so far to the left of the array. A duplicate pair is identified when two adjacent elements are equal in value. With each comparison only two situations are possible: A pair of duplicates has been encountered; The two elements are different. To try to understand, consider what happens when the four 23 s are encountered. We have When 15 is compared with 23(step 4) the 23 is the most recent unique element encountered and so it should be moved as far to the left as possible. At next step, two 23 s are compared. Since the 23 has already been accounted for in the previous step, the only action we must take is to move to the next pair. Once again two 23 s are compared and so again all we can do is move to the next step. At step (8) 26 is compared with 23. Here the 26 is the most recently encountered unique element. It must therefore be appropriately relocated in the array. Summarizing the basic steps so far in our mechanism we have: While all adjacent pairs of elements have not been compared do If they are not equal, shift the rightmost element in the next pair to the array position determined by the current unique element count. 5

6 removedup.txt Application Data Compression and text processing problems Given a randomly ordered array of n elements, partition the elements into two subsets such that elements x are in one subset and elements > x are in the other subset. Development Given a random data set below, we are asked to partition it into two subsets, one containing elements <= 17 and the other containing elements >17. of the array. One straightforward way of making this transfer is to sort the array into ascending order. When we do this we get the configuration below. Clearly we need to be able to separate two subsets. To do this, we could put those elements > 17 at the top end of the array and <= 17 at the bottom After sorting we can easily find the location that partitions the elements into the two subsets we require. With this solution to the problem, we have actually ordered the two subsets in addition to separating them. In original statement of the problem, it was not required that the elements to be ordered. For example Sorting of data is costly operation. Therefore the less costly solution to the problem is to rearrange the elements without sorting. Comparing these two data sets, we see that elements at the left-hand end >17 must be moved to the right hand end of the portioning point p. 6

7 Elements <=17 that are to left of p need not be moved because they are already in their proper partition. When we are initially presented with the random set we do not know how many elements are <=17. One way to overcome this would be to make a pass through that array counting all values <=17 Once we know the value of p we can then make another pass through the array. This time when we encounter a value >17 on the left side of p we must move it to the right of p. For example, the 28 could be placed where the 12 is and the 12 could be placed in the 28 s original position. This idea can be extended and so we end up with the steps illustrated below. We might therefore propose the following basic partitioning mechanism: while the two partitions have not met do Extend the left and right partitions inwards exchanging any wrongly placed pairs in the process. Then increase and decrease the left and right partition and repeat the process a till all the elements are placed in it proper position. Paritition.txt Sorting Statistical classification : Given a randomly ordered array of n elements determine the k th smallest element in the set. 7

8 Development The current problem is a generalization of this problem. In partitioning problem, we knew in advance the value x about which the array was to be partitioned but we did not know how many values were to be partitioned on either side of x. The current problem represents the complementary situation where we are given how the array is to be partitioned but we do not know in advance the value of x. The partitioning algorithm can only partition data into two subsets when it has in advance the value about which the two subsets are to be partitioned. Since we do not have any idea in advance what the kth smallest value is, we might therefore be tempted to choose a value x at random from the array and partition the array about x. The variables l and u are initially assigned to the bounds of the array. For example The value of x will cause the original data set to be divided into two smaller subsets. The kth smallest value will have to be in one of the subsets A or B. If for example, the kth smallest value is in subset B. Then we can completely disregard subset A and start trying to find the kth smallest value in subset B. The easiest way to do this is to replace l by l and start searching for the kth smallest value again in the smaller subset B. If we repeatedly apply this partition process to smaller and smaller subsets that contain the kth smallest value we will eventually obtain the desired result. 8

9 Smallestelement.txt Finding the median and percentile Given a set of n distinct numbers, find the length of the longest monotone increasing subsequence. Studying random sequences File comparision 9