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Searching
Tiziana Ligorio
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Today’s Plan
Searching algorithms and their analysis
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Searching Tiziana Ligorio 1 Todays Plan Searching algorithms and - - PowerPoint PPT Presentation
Searching Tiziana Ligorio 1 Todays Plan Searching algorithms and - - PowerPoint PPT Presentation
Searching Tiziana Ligorio 1 Todays Plan Searching algorithms and their analysis 2 Searching Looking for something! In this discussion we will assume searching for an element in an array 3 Linear search Most intuitive
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SLIDE 3
Searching
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Looking for something! In this discussion we will assume searching for an element in an array
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Linear search
Most intuitive Start at first position and keep looking until you find it
int linearSearch(int a[], int size, int value) { for (int i = 0; i < size; i++) { if (a[i] == value) { return i; } } return-1; }
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How long does linear search take?
If you assume value is in the array and probability of finding it at any location is uniform, on average n/2 If value is not in the array (worst case) n Either way it’s O(n)
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What if you know array is sorted? Can you do better than linear search?
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Lecture Activity
You are given a sorted array of integers. How would you search for 115? ( try to do it in fewer than n steps: don’t search sequentially) You can write pseudocode or succinctly explain your algorithm
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We have done this before! When?
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Look in ?
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Binary Search
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14 43 76 100 108 158 195 200 274 523 543 599 3
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Binary Search
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14 43 76 100 108 158 195 200 274 523 543 599 3
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Binary Search
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14 43 76 100 108 158 195 200 274 523 543 599 3
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Binary Search
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14 43 76 100 108 158 195 200 274 523 543 599 3
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Binary Search
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14 43 76 100 108 158 195 200 274 523 543 599 3
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Binary Search
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14 43 76 100 108 158 195 200 274 523 543 599 3
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Binary Search
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14 43 76 100 108 158 195 200 274 523 543 599 3
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Binary Search
What is happening here?
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Binary Search
What is happening here? Size of search is cut in half at each step
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Binary Search
What is happening here? Size of search is cut in half at each step Let T(n) be the running time and assume n = 2k T(n) = T(n/2) + 1
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Simplification: assume n is a power of 2 so it can be evenly divided in two parts The running time One comparison Search lower OR upper half
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Binary Search
What is happening here? Size of search is cut in half at each step Let T(n) be the running time and assume n = 2k T(n) = T(n/2) + 1 T(n/2) = T(n/4) +1
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One comparison Search lower OR upper half of n/2
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Binary Search
What is happening here? Size of search is cut in half at each step Let T(n) be the running time and assume n = 2k T(n) = T(n/2) + 1 T(n/2) = T(n/4) +1 T(n) = T(n/4) + 1 + 1
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Binary Search
What is happening here? Size of search is cut in half at each step Let T(n) be the running time and assume n = 2k T(n) = T(n/2) + 1 T(n) = T(n/4) + 2 . . .
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22 2 21 1
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Binary Search
What is happening here? Size of search is cut in half at each step Let T(n) be the running time and assume n = 2k T(n) = T(n/2) + 1 T(n) = T(n/4) + 2 . . . T(n) = T(n/2k) + k
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Binary Search
What is happening here? Size of search is cut in half at each step Let T(n) be the running time and assume n = 2k T(n) = T(n/2) + 1 T(n) = T(n/4) + 2 . . . T(n) = T(n/2k) + k T(n) = T(1) + log2(n)
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n/n = 1 The number to which I need to raise 2 to get n And we said n = 2k
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Binary Search
What is happening here? Size of search is cut in half at each step Let T(n) be the running time and assume n = 2k T(n) = T(n/2) + 1 T(n) = T(n/4) + 2 . . . T(n) = T(n/2k) + k T(n) = T(1) + log2(n)
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