1. 1 What is it? A side order for your eggs? A form of narcotic intake? A combination of the two?

2. 2 Searching O(n) O(lg n) O(n) O(lg n) balanced O(n) - skewed

3. 3 Can we do better? n What if we have students in a class that have been assigned id numbers in the range 0-50? n Just store each student in his/her student number position in the array: n What is the order of insert, remove, find for such a list? n Would this be a reasonable way to store student records at a university? JaneStevePaulMary... [0][1][2][3]...

4. 4 n Can we use the previous method if we want to use SSN as the key field? n How may positions would our array need? n How many of those would we actually use? n What we need is a way to map the range of possible keys (10 9 for SSNs) onto the available array positions. n For SSNs we could just use the last 4 digits of the SSN to determine where to store the element.

5. 5 n This process of mapping a range of keys onto an address space is known as hashing. n The function (usually more complicated than just key mod 10,000 ) used to find the hash value for a key is called the hashing function. n What is the order of insertion and searching using hashing? n What is the order of remove? n Is there a problem with this technique?

6. 6 Collisions n When two keys have the same hash value (they map to the same location) we have what is called a collision. n When using a hash table we must have some collision resolution method. n Some collision resolution methods include: linear probing, quadratic probing, separate chaining, double hashing

7. 7 Linear Probing n With linear probing, when a collision occurs, the element is placed in the next available position 0 2 1 3 4 5 6 7 9 8 Hash( 89) = 9 Hash( 18) = 8 Hash( 49) = 9 Hash( 58) = 8 Hash( 9) = 9 89 18 49 58 9 Problem: primary clustering

8. 8 Quadratic Probing n If the element hashes to location H, the next location we try is H+1 2, then H+2 2, H+3 2, … H+i 2 0 2 1 3 4 5 6 7 9 8 Hash( 89) = 9 Hash( 18) = 8 Hash( 49) = 9 Hash( 58) = 8 Hash( 9) = 9 89 18 49 58 9

9. 9 Double Hashing Use two hash functions If M is prime, eventually will examine every position in the table double_hash_insert(K) if(table is full) error probe = h1(K) offset = h2(K) while (table[probe] occupied) probe = (probe + offset) mod M table[probe] = K Many of same (dis)advantages as linear probing Distributes keys more uniformly than linear probing

10. 10 Double Hashing Example h1(K) = K mod 13 h2(K) = 8 - K mod 8 we want h2 to be an offset to add 0 2 1 3 4 5 6 7 9 8 41 22 73 44 10 11 12 18 59 32 31

11. 11 Separate Chaining n Maintain an array of linked lists. The hash function tells us which list to insert an element X, and where to find an element X. 0 2 1 3 4 5 6 7 9 8 Hash( 89) = 9 Hash( 18) = 8 Hash( 49) = 9 Hash( 58) = 8 Hash( 9) = 9 891849958

12. 12 Pseudo-code (chaining implementation) Has 3 basic methods, and the constructor which: creates the table of M lists insert(K) index = h(K) insert into table[index] find(K) index = h(K) walk down list at table[index], looking for a match return what was found (or error) remove(K) index = h(K) walk down list at table[index], looking for a match remove what was found (or error)

13. 13 Hash Functions Need to choose a good hash function –quick to compute –distributes keys uniformly throughout the table How to deal with hashing non-integer keys: –find some way of turning the keys into integers –for example, if the keys are phone numbers, remove the hyphen in 863-7639 to get 8637639! –for a String, add up the ASCII values of the characters of your string –then use a standard hash function on the integers

14. 14 Hash Functions Use the remainder –h(K) = K mod M –K is the key, M the size of the table Need to choose M M = b e (bad) –if M is a power of two, h(K) gives the e least significant bits of K –all keys with the same ending go to the same place M prime (good) –helps ensure uniform distribution –take a number theory class to understand why

15. 15 Hash Functions (cont.) Mid-Square – h(K) = middle digits of K 2 –I.E. Table size power of 10 h(4150130) = 21526 4436 17100 h(415013034) = 526447 3522 151420 h(1150130) = 13454 2361 7100 –I.E. Table power is power of 2 h(1001) = 10 100 01 h(1011) = 11 110 01 h(1101) = 101 010 01

16. 16 Theoretical Results Let  the load factor: average number of keys per array index Analysis is probabilistic, rather than worst-case

17. 17

18. 18 Sources: Some information in these slides was taken from the slides that accompany Data Structures and Algorithms in Java by Goodrich & Tamassia, Wiley 2014.