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Friday, March 21, 2014
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Binary search applied to one-dimensional array

8:30 AMFriday, March 21, 2014

Binary search applied to one-dimensional array

Before reading this section find out the meaning of queue in any English dictionary. You will not search the word looking page by page. No need of page by page search because you know that the words are arranged alphabetical order. Then have you reached to the word directly? I don’t think so. You skip few pages and might have reached to a page of dictionary containing the words starting with the letter ‘k’. Then you further skip the next few pages not previous ones because you know the letter ‘q’ comes after ‘k’. Now if you reach to the page of letter ‘r’, you skip few previous pages, then finally reach to the page containing word ‘queue’.
      As we have already seen that binary search can be applied to only one-dimensional sorted (arrangement of elements in ascending or descending order) or ordered array. In order to apply binary search to one dimensional array we follow a method of exact division of array in two halves. So the name binary search. Divide the array into two halves to get the middle element. compare the key to be searched with the middle element, if both are equal, the search is successful. Otherwise repeat the application of process either to the left half or to the right half depending on the possible placement of element in the array.
          To find the middle element of the array for the first time, we can use the first index and the last index of the array. find out the middle element’s index by finding the sum of first and last index and dividing the sum by 2. Compare the key with the middle element. If both are equal search is successful in the first comparison otherwise the key may be greater than or less than the middle element. 
If key is greater than the middle element, change the first index as middle index plus 1 and keeping the last index as it is in case of first step repeat the process. 
If key is less than middle element, change the last index as middle index minus 1 keeping the first index as it is in case of first step repeat the process.
The process of finding out the middle index and comparison of key with the middle element must be stopped if the first index is greater than the last index and in that case the search is unsuccessful.
        So, when binary search is applied for one-dimensional array we consider the lower bound as first index and upper bound as last index. Usually the lower bound of an array is 1 and the upper bound of array is N, the size of the array. In case of C language the lower bound of array is 0 and the upper bound of the array is N-1, where N is the size of the array. so take two variables FIRST and LAST and initialize FIRST variable with lower bound value and LAST variable with upper bound value. Take another variable MID and initialize that variable with integer part of [(FIRST+LAST)/2].Compare the given key with MID index element of the array.

 Three possibilities arise from the comparison, MID indexed element may be:

=                   report search successful 
 <                 go for the left half of the array by changing the value of variable LAST as MID-1.Keep the                            value of variable first unchanged. Find out the MID value again and repeat the process of                              comparison.
<                  go for the right half of the array by changing the value of variable FIRST as MID+1. Keep the                       value of variable LAST unchanged. Find out the MID value again and repeat the process of                           comparison.
               The process of comparison terminates when the search is successful or the value of variable FIRST is greater than that of LAST (in that case obviously the search is unsuccessful).
              The above case of binary search is based on the assumption of array elements stored in the ascending order. If the array elements stored in descending order then, the change in the values of variables LAST and FIRST is reversed. 

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