1. COMPUTER 2430 Object Oriented Programming and Data Structures I

2. Search
Linear Search: O(N) Binary Search: O(log n) Any better search algorithm? Is O(1) possible? Hashing!
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Program due! 2 2 2

3. Hashing Function
A function from an object to an integer (as array index) O(1)
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Program due! 3 3 3

4. obj Hashing function h O(1) Where to store the object? index = h(obj)
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Program due!
Where is the object?
index = h(obj) 4 4 4

5. Hashing Function I
Employee [] employees = new Employee[100]; // at most 100 employees // IDs are unique // all IDs are between 1 and 100 Hashing function: index = h(id) = id - 1 Perfect hashing function! Not feasible in many cases.
Quiz is coming?
Program due! 5 5 5

6. Hashing Function II (Division)
Employee [] employees = new Employee[MAX_SIZE]; // MAX_SIZE is a large prime number Hashing function: index = h(SSN) = SSN % MAX_SIZE Collision! MAX_SIZE:
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Program due! 6 6 6

7. Integer Division No Calculators on Quiz6, Test3, and Final!
20 % % % % 17 3 16 13 1

8. Hashing Function III
Employee [] employees = new Employee[MAX_SIZE]; // MAX_SIZE is a large prime number // The key field is a string Hashing function: index = h(key) sum = 0 for each char ch in key sum = sum + int(ch) sum = sum * ANOTHER_PRIME sum = sum % MAX_SIZE Still Collision!
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Program due! 8 8 8

9. Handling Collision Linear Probe (Open Addressing):
Go to the next available slot (circular)
Double Hashing:
Use a second hashing function if collision occurs (circular)
Buckets
Each array slot points to an array (2-D array)
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Program due! 9 9 9

10. Example
We will assume the division method of hashing: key is an integer table_size: 11 (prime number) Index = h(key) = key % table_size = key % 11
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Program due! 10 10 10 10

11. Linear Probe Index = key % 11 Index ++ when occupied (circular)8
Key: 23; index: 23 % 11 = 1 23 8
Key: 32; index: 32 % 11 = 10
Key value 8 23 32 45 55 75 86
Hash value 8 1 10 9 23 8 32
Key: 45; index: 45 % 11 = 1
 2 23 45 8 32
Key: 55; index: 55 % 11 = 0
Quiz is coming?
Program due! 55 23 45 8 32
Key: 75; index: 75 % 11 = 9 55 23 45 8 75 32
Key: 86; index: 86 % 11 = 9
 10
 0
 1
 2
 3 55 23 45 86 8 75 32 11

12. Double Hashing
Friday

13. Buckets 1 2 3 4 5 6 7 8 9 10 55
Index = key % 11 Hash value 8: 8%11  8 23: 23%11  1 32: 32%11  10 45: 45%11  1 55: 55%11  0 75: 75%11  9 86: 86%11  9 23 45
Quiz is coming?
Program due! 8 75 86 32 13 13 13

14. Quiz 5