Demonstrate interpolation search

Demonstrate interpolation search, Data Structures, Binary search

1. Demonstrate Interpolation Search Motapalukula Manoj 20951a0585

2. Introduction Interpolation search is an improved variant of binary search. This search algorithm works on the probing position of the required value. For this algorithm to work properly, the data collection should be in a sorted form and equally distributed. Binary search has a huge advantage of time complexity over linear search. Linear search has worst-case complexity of Ο(n) whereas binary search has Ο(log n). There are cases where the location of target data may be known in advance. For example, in case of a telephone directory, if we want to search the telephone number of Morpheus. Here, linear search and even binary search will seem slow as we can directly jump to memory space where the names start from 'M' are stored.

3. Positioning in Binary Search In binary search, if the desired data is not found then the rest of the list is divided in two parts, lower and higher. The search is carried out in either of them.

4. Even when the data is sorted, binary search does not take advantage to probe the position of the desired data. Position Probing in Interpolation Search Interpolation search finds a particular item by computing the probe position. Initially, the probe position is the position of the middle most item of the collection.

5. If a match occurs, then the index of the item is returned. To split the list into two parts, we use the following method − mid = Lo + ((Hi - Lo) / (A[Hi] - A[Lo])) * (X - A[Lo]) where − A = list Lo = Lowest index of the list Hi = Highest index of the list A[n] = Value stored at index n in the list If the middle item is greater than the item, then the probe position is again calculated in the sub-array to the right of the middle item. Otherwise, the item is searched in the subarray to the left of the middle item. This process continues on the sub-array as well until the size of subarray reduces to zero. Runtime complexity of interpolation search algorithm is  Ο(log (log n))  as compared to  Ο(log n)  of BST in favorable situations.

6. Interpolation Search Given a sorted array of n uniformly distributed values array[], write a function to search for a particular element x in the array.  Linear Search finds the element in O(n) time,  Jump search  takes O(√ n) time and Binary search take O(Log n) time. 

7. The Interpolation Search is an improvement over  Binary Search  for instances, where the values in a sorted array are uniformly distributed. Binary Search always goes to the middle element to check. On the other hand, interpolation search may go to different locations according to the value of the key being searched. For example, if the value of the key is closer to the last element, interpolation search is likely to start search toward the end side.

8. // The idea of formula is to return higher value of pos // when element to be searched is closer to array[hi] . And // smaller value when closer to array[lo] pos = lo + [ (x-array[lo])*(hi-lo) / (array[hi]-array[Lo]) ] array[] ==> Array where elements need to be searched x ==> Element to be searched lo ==> Starting index in array[] hi ==> Ending index in array[] To find the position to be searched, it uses following formula. 

9. Let's assume that the elements of the array are linearly distributed. General equation of line : y = m*x + c. y is the value in the array and x is its index. Now putting value of lo , hi and x in the equation array[hi] = m * hi + c ----(1) Array[lo] = m * lo + c ----(2) x = m*pos + c ----(3) m = (array[hi] - array[lo] )/ (hi - lo) subtracting equation (2) from (3) x - array[lo] = m * (pos - lo) lo + (x - array[lo])/m = pos pos = lo + (x - array[lo]) *(hi - lo)/(array[hi] - array[lo])

10. Algorithm Rest of the Interpolation algorithm is the same except the above partition logic.  Step1:  In a loop, calculate the value of “pos” using the probe position formula.  Step2:  If it is a match, return the index of the item, and exit.  Step3:  If the item is less than array[pos], calculate the probe position of the left sub-array. Otherwise calculate the same in the right sub-array.  Step4:  Repeat until a match is found or the sub-array reduces to zero. Below is the implementation of algorithm. 

11. code 

12. # Python3 program to implement # interpolation search # with recursion # If x is present in array[0..n-1], then # returns index of it, else returns -1. def interpolationSearch(array, lo, hi, x):        # Since array is sorted, an element present      # in array must be in range defined by corner      if (lo < = hi and x > = array[lo] and x < = array[hi]):            # Probing the position with keeping          # uniform distribution in mind.          pos = lo + ((hi - lo) // (array[hi] - array[lo]) * (x - array[lo])) # Condition of target found          if array[pos] == x:              return pos          # If x is larger, x is in right subarray          if array[pos] < x:              return interpolationSearch(array, pos + 1 ,hi, x)          # If x is smaller, x is in left subarray          if array[pos] > x:              return interpolationSearch(array, lo, pos - 1 , x)      return - 1

13. # Driver code   # Array of items in which # search will be conducted array = [ 10 , 12 , 13 , 16 , 18 , 19 , 20 ,         21 , 22 , 23 , 24 , 33 , 35 , 42 , 47 ] n = len (array)   # Element to be searched x = 18 index = interpolationSearch(array, , n - 1 , x)   if index ! = - 1 :      print ( "Element found at index" , index) else :      print ( "Element not found" ) Output Element found at index 4