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Military Technical Academy Data Structures and Algorithms

Military Technical Academy Data Structures and Algorithms. Lecturer: Dr. Nguyen Nam Hong Tel: 04 8781 437 Mob: 0912 312 816 Email: nguyennamhong2003@yahoo.com.au Lecture 8. Searching Algorithms. Lecture 8. Searching Algorithms (1/2). References:

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Military Technical Academy Data Structures and Algorithms

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  1. Military Technical Academy Data Structures and Algorithms Lecturer: Dr. Nguyen Nam Hong Tel: 04 8781 437 Mob: 0912 312 816 Email: nguyennamhong2003@yahoo.com.au Lecture 8. Searching Algorithms. Dr. Nguyen Nam Hong, Le Quy Don Technical University

  2. Lecture 8. Searching Algorithms (1/2) References: • Data structures and Algorithms Searching.htm • Kyle Loudon Mastering Algorithms Chapter 12 Sorting and Searching • Lecture 19 Sequential and Binary Search.htm • Sedgewick Algorithms 14. Elementary Searching Methods Dr. Nguyen Nam Hong, Le Quy Don Technical University

  3. Lecture 8. Searching Algorithms (2/2) Contents: 8.1. Searching Concept (3) 8.2. Linear Search (7) 8.3. Binary Search (8) 8.4. Interpolation Search (7) Dr. Nguyen Nam Hong, Le Quy Don Technical University

  4. 8.1. Searching Concepts (1/3) • The problem of locating an element in a list (ordered or not) occurs in many contexts. • For instance, a program that checks the spelling of words searches for them in a dictionary, which is just an ordered list of words. • Problems of this kind are called searching problems. Dr. Nguyen Nam Hong, Le Quy Don Technical University

  5. 8.1. Searching Concepts (2/3) • There are many searching algorithms. • The natural searching method is linear search (or sequential search, or exhaustive search), which is very simple but takes a long time when applying with large lists. Dr. Nguyen Nam Hong, Le Quy Don Technical University

  6. 8.1. Searching Concepts (3/3) • A binary search repeatedly subdivides the list to locate an item and for larger lists it is much faster than linear search. • Like a binary search, an interpolation search repeatedly subdivides the list to locate an item. • Interpolation search is much faster than binary search because it makes a reasonable guess about where the target item should lie. Dr. Nguyen Nam Hong, Le Quy Don Technical University

  7. 8.1. Linear Search (1/8) • This is a very simple algorithm. • It uses a loop to sequentially step through an array, starting with the first element. • It compares each element with the value being searched for and stops when that value is found or the end of the array is reached. Dr. Nguyen Nam Hong, Le Quy Don Technical University

  8. 8.1. Linear Search (2/8) Sub LinearSearch(x:int, a[]: Int, loc: Int) i:=1 While (i<=n) And (x<>a[i]) i:=i+1 End While If i<=n Then loc = i Else loc = 0 End Sub Dr. Nguyen Nam Hong, Le Quy Don Technical University

  9. 8.1. Linear Search (3/8) • Array numlist contains • Searching for the the value 11, linear search examines 17, 23, 5, and 11 -> Found (loc = 4) • Searching for the the value 7, linear search examines 17, 23, 5, 11, 2, 29, and 3 -> Not Found (loc = 0) Dr. Nguyen Nam Hong, Le Quy Don Technical University

  10. 8.1. Linear Search (4/8) • The advantage is its simplicity. • It is easy to understand • Easy to implement • Does not require the array to be in order • The disadvantage is its inefficiency • If there are 20,000 items in the array and what you are looking for is in the 19,999th element, you need to search through the entire list. Dr. Nguyen Nam Hong, Le Quy Don Technical University

  11. 8.1. Linear Search (5/8) • Whenever the number of entries doubles, so does the running time, roughly. • If a machine does 1 million comparisons per second, it takes about 30 minutes for 4 billion comparisons. Dr. Nguyen Nam Hong, Le Quy Don Technical University

  12. 8.1. Linear Search (6/8) Dr. Nguyen Nam Hong, Le Quy Don Technical University

  13. 8.1. Linear Search (7/8) Use a Sentinel to Improve the Performance Sub LinearSearch2(x:int, a[]: Int, loc: Int) a[n+1] = x: n = n + 1: i = 1 While (x<>a[i]) i = i+1 End While If i<=n Then loc = i Else loc = 0 End Sub Dr. Nguyen Nam Hong, Le Quy Don Technical University

  14. 8.1. Linear Search (8/8) Apply Linear Search to Sorted Lists Sub LinearSearch3(x:int, a[]: Int, loc: Int) i = 1 While (x > a[i]) i = i+1 End While If a[i] = x Then loc = i Else loc = 0 End Sub Dr. Nguyen Nam Hong, Le Quy Don Technical University

  15. 8.2. Binary Search (1/9) Can We Search More Efficiently? • Yes, provided the list is in some kind of order, for example alphabetical order with respect to the names. • If this is the case, we use a “divide and conquer” strategy to find an item quickly. • This strategy is what one would use in a “number guessing game”, for example. Dr. Nguyen Nam Hong, Le Quy Don Technical University

  16. 8.2. Binary Search (2/9) I’m Thinking of A Number… • … between 1 and 1000. Guess it! • Is it 500? Nope, too low. • Is it 750? Nope, too high. • Is it 625? … etc… This strategy guarantees a correct guess in no more than ten guesses! Dr. Nguyen Nam Hong, Le Quy Don Technical University

  17. 8.2. Binary Search (3/9) Apply This Strategy to Searching • The resulting algorithm is called the “Binary Search” algorithm. • We check the middle key in our list. • If it is beyond what we are looking for (too high), we look only at the top half of the list. • If it’s not far enough in (too low), we look at the bottom half. • Then iterate! Dr. Nguyen Nam Hong, Le Quy Don Technical University

  18. 8.2. Binary Search (4/9) • Divide a sorted array into three sections. • middle element • elements on one side of the middle element • elements on the other side of the middle element • If the middle element is the correct value, done. Otherwise, go to step 1, using only the half of the array that may contain the correct value. Dr. Nguyen Nam Hong, Le Quy Don Technical University

  19. 8.2. Binary Search (5/9) • Continue steps 1 and 2 until either the value is found or there are no more elements to examine. Dr. Nguyen Nam Hong, Le Quy Don Technical University

  20. 8.2. Binary Search (6/9) Binary Search Example • Array numlist2 contains • Searching for the the value 11, binary search examines 11 and stops. Found. • Searching for the the value 7, binary search examines 11,3,5,and stops. Not Found. Dr. Nguyen Nam Hong, Le Quy Don Technical University

  21. 8.2. Binary Search (7/9) Algorithm for Binary search Sub BinarySearch(x:int, a[]: int, loc: Int) i =1: j =n while i<j begin m =(i + j) \ 2 if x > a[m] then i=m+1 else j=m end if x=a[i] then loc=i else loc=0 End Sub Dr. Nguyen Nam Hong, Le Quy Don Technical University

  22. 8.2. Binary Search (8/9) • The worst case number of comparisons grows by only 1 comparison every time list size is doubled. • Only 32 comparisons would be needed on a list of 4 billion using Binary Search. (Sequential Search would need 4 billion comparisons and would take 30 minutes!) Dr. Nguyen Nam Hong, Le Quy Don Technical University

  23. 8.2. Binary Search (9/9) • Benefit • Much more efficient than linear search. • For array of N elements, performs at most log2N comparisons. • Disadvantage • Requires that array elements be sorted. Dr. Nguyen Nam Hong, Le Quy Don Technical University

  24. 8.3. Interpolation Search (1/9) • Binary search is a great improvement over linear search because it eliminates large portion of the list without actually examing all the eliminated values. • If we know that the values are fairly evenly distributed, we can use interpolation to eliminate even more values at each step. Dr. Nguyen Nam Hong, Le Quy Don Technical University

  25. 8.3. Interpolation Search (2/9) • Interpolation is the process of using known values to gess where an unknown value lies. • We use the indexes of known values in the list to gess what index the target value should have. • Interpolation search selects the dividing point by interpolation using the following code m = l + (x – a[l])*(r-l)/(a[r]-a[l]) Dr. Nguyen Nam Hong, Le Quy Don Technical University

  26. 8.3. Interpolation Search (3/9) • Compare x to a[m] • If x = a[m]: Found. • If x<a[m]: set r = m-1 • If x > a[m]: set l = m + 1 • If searching is still not finish, continue searching with new l and r. • Stop searching when Found or x<a[l] or x>a[r]. Dr. Nguyen Nam Hong, Le Quy Don Technical University

  27. 8.3. Interpolation Search (4/9) Example: Find the key x = 32 in the list 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 1 4 7 9 9 12 13 17 19 21 24 32 36 44 45 54 55 63 66 70 1: l=1, r=20 -> m=1+(32-1)*(20-1)/(70-1) = 10 a[10]=21<32=x -> l=11 2: l=11, r=20 -> m=11+(30-24)*(20-11)/(70-24) = 12 a[12]=32=x -> Found at m = 12 Dr. Nguyen Nam Hong, Le Quy Don Technical University

  28. 8.3. Interpolation Search (5/9) Example: Find the key x = 30 in the list 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 1 4 7 9 9 12 13 17 19 21 24 32 36 44 45 54 55 63 66 70 1: l=1, r=20 -> m=1+(30-1)*(20-1)/(70-1) = 9 a[9]=19<30=x -> l=10 2: l=10, r=20 -> m=10+(30-21)*(20-10)/(70-21) = 12 a[12]=32>30=x -> r = 11 3: l=10, r=11 -> m=10+(30-24)*(11-10)/(24-21) = 12 m=12>11=r: Not Found Dr. Nguyen Nam Hong, Le Quy Don Technical University

  29. 8.3. Interpolation Search (6/9) Private Sub Interpolation(a[]: Int, x: Int, n: Int, Found: Boolean) l = 1: r = n Do While (r > l) m = l + ((x – a[l]) / (a[r] – a[l])) * (r - l) ‘Verify and Decise What to do next Loop End Sub Dr. Nguyen Nam Hong, Le Quy Don Technical University

  30. 8.3. Interpolation Search (7/9) ‘Verify and Decide what to do next If (a[m] = x) Or (m < l) Or (m > r) Then Found = iif(a[m] = x, True, False) Exit Do ElseIf (a[m] < x) Then l = m + 1 ElseIf (a[m] > x) Then r = m – 1 End If Dr. Nguyen Nam Hong, Le Quy Don Technical University

  31. 8.3. Interpolation Search (8/9) • Binary search is very fast (O(logn)), but interpolation search is much faster (O(loglogn)). • For n = 2^32 (four billion items) • Binary search took 32 steps of verification • Interpolation search took only 5 steps of verification. Dr. Nguyen Nam Hong, Le Quy Don Technical University

  32. 8.3. Interpolation Search (9/9) • Interpolation search performance time is nearly constant for a large range of n. • Interpolation is still more usefull if the data had been stored on a hard disk or other relatively slow device. Dr. Nguyen Nam Hong, Le Quy Don Technical University

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