CSC 172 DATA STRUCTURES

1. CSC 172 DATA STRUCTURES

2. SETS and HASHING Unadvertised in-store special: SETS! in JAVA, see Weiss 4.8 Simple Idea: Characteristic Vector HASHING...The main event.

3. Representation of Sets List Simple O(n) dictionary operations Binary Search Trees O(log n) average time Range queries, sorting Characteristic Vector O(1) dictionary ops, but limited to small sets Hash Table O(1) average for dictionary ops Tricky to expand, no range queries

4. Characteristic Vectors Boolean Strings whose position corresponds to the members of some fixed “universal” set A “1” in a location means that the element is in the set A “0” means that it is not

5. MUSIC THEORY A chord is a set of notes played at the same time. Represented by a 12 bit vector called a “pitch class” {B,A#,A,G#,G,F#,F,E,D#,D,C#,C} 000010010001 represents C major 000010001001 represents C minor Rotation is “transposition” Bit reversal is “inversion”

6. UNIX file privileges {user, group, others} x {read, write, execute} 9 possible privileges Type “ls –l” on UNIX total 142 -rw-rw-r-- 1 pawlicki none 76 Jun 20 2000 PKG416.desc -rw-rw-r-- 1 pawlicki none 28906 Jun 20 2000 PKG416.pdf -rw-rw-r-- 1 pawlicki none 1849 Jun 20 2000 let.1 -rw-rw-r-- 1 pawlicki none 0 Apr 2 13:03 out -rw-rw-r-- 1 pawlicki none 39891 Jun 20 2000 stapp.uu

7. UNIX files The order is rwx for each of user (owner), group, and others So, a protection mode of 110100000 means that the owner may read and write (but not execute), the group can read only and others cannot even read

8. GAMBLING A deck has 52 cards {2C,2H,2S,2D,3C, .... KD,AC,AH,AS,AD} Represent a “hand” as a vector of 52 bits 0000000000000000000000000000000000000000000000000101 is a pair of aces In “Texas Hold'em” everyone gets two “hole” cards and 5 “board” cards We can use bitwise & to find “hands”

9. CV advantages If the universal set is small, sets can be represented by bits packed 32 to a word Insert, delete, and lookup are O(1) on the proper bit Union, intersection, difference are implemented on a word-by-word basis O(m) where m is the size of the set Small constant factor (1/32) Fast, machine operations

10. Hashing A cool way to get from an element x to the place where x can be found An array [0..B-1] of buckets Bucket contains a list of set elements B = number of buckets A hash function that takes potential set elements and quickly produces a “random” integer [0..B-1]

11. Example If the set elements are integers then the simplest/best hash function is usually h(x) = x % B or h(x) = x - (x%B), (never 0). Suppose B = 6 and we wish to store the integers {70, 53, 99, 94, 83, 76, 64, 30} They belong in the buckets 4, 5, 3, 4, 5, 4, 4, and 0 Note: If B = 7 0,4,1,3,6,6,1,2

12. Pitfalls of Hash Function Selection We want to get a uniform distribution of elements into buckets Beware of data patterns that cause non-uniform distribution

13. Example If integers were all even, then B = 6 would cause only buckets 0,2, and 4 to fill If we hashed words in the UNIX dictionary into 10 buckets by length of word then 20% go into bucket 7

14. Dictionary Operations Lookup Go to head of bucket h(x) Search for bucket list. If x is in the bucket Insertion: append if not found Delete – list deletion from bucket list

15. Analysis If we pick B to be new N, the number of elements in the set, then the average list is O(1) long Thus, dictionary ops take O(1) time Worst case: all elements go into one bucket O(n)

16. Managing Hash Table Size If n gets as high as 2B, create a new hash table with 2B buckets “Rehash” every element into the new table O(n) time total There were at least n inserts since the last “rehash” All these inserts took time O(n) Thus, we “amortize” the cost of rehashing over the inserts since the last rehash Constant factor, at worst So, even with rehashing we get O(1) time ops

17. Collisions A collision occurs when two values in the set hash to the same value There are several ways to deal with this Chaining (using a linked list or some secondary structure) Open Addressing Double hashing Linear Probing

18. Chaining 0 70  1 99 64  2 30  94  3 4 53  5 6 83 76  Very efficient Time Wise Other approaches Use less space 

19. Open Addressing When a collision occurs, if the table is not full find an available space Linear Probing Quadratic Probing Double Hashing

20. Linear Probing If the current location is occupied, try the next table location LinearProbingInsert(K) { if (table is full) error; probe = h(K); while (table[probe] is occupied) probe = ++probe % M; table[probe] = K; } Walk along table until an empty spot is found Uses less memory than chaining (no links) Takes more time than chaining (long walks) Deleting is a pain (mark a slot as having been deleted)

21. Linear Probing 18 0 1 2 3 4 5 6 7 8 9 10 11 12 h(K) = K % 13 Insert: 18, 41, 22, 59, 32, 31, 73 h(K) : 5,

22. Linear Probing 41 18 0 1 2 3 4 5 6 7 8 9 10 11 12 h(K) = K % 13 Insert: 18, 41, 22, 59, 32, 31, 73 h(K) : 5, 2,

23. Linear Probing 41 18 22 0 1 2 3 4 5 6 7 8 9 10 11 12 h(K) = K % 13 Insert: 18, 41, 22, 59, 32, 31, 73 h(K) : 5, 2, 9,

24. Linear Probing 41 18 59 22 0 1 2 3 4 5 6 7 8 9 10 11 12 h(K) = K % 13 Insert: 18, 41, 22, 59, 32, 31, 73 h(K) : 5, 2, 9, 7,

25. Linear Probing 41 18 32 59 22 0 1 2 3 4 5 6 7 8 9 10 11 12 h(K) = K % 13 Insert: 18, 41, 22, 59, 32, 31, 73 h(K) : 5, 2, 9, 7, 6,

26. Linear Probing 41 18 32 59 22 0 1 2 3 4 5 6 7 8 9 10 11 12 h(K) = K % 13 Insert: 18, 41, 22, 59, 32, 31, 73 h(K) : 5, 2, 9, 7, 6, 5,

27. Linear Probing 41 18 32 59 22 0 1 2 3 4 5 6 7 8 9 10 11 12 h(K) = K % 13 Insert: 18, 41, 22, 59, 32, 31, 73 h(K) : 5, 2, 9, 7, 6, 5,

28. Linear Probing 41 18 32 59 22 0 1 2 3 4 5 6 7 8 9 10 11 12 h(K) = K % 13 Insert: 18, 41, 22, 59, 32, 31, 73 h(K) : 5, 2, 9, 7, 6, 5,

29. Linear Probing 41 18 32 59 31 22 0 1 2 3 4 5 6 7 8 9 10 11 12 h(K) = K % 13 Insert: 18, 41, 22, 59, 32, 31, 73 h(K) : 5, 2, 9, 7, 6, 5,

30. Linear Probing 41 18 32 59 31 22 0 1 2 3 4 5 6 7 8 9 10 11 12 h(K) = K % 13 Insert: 18, 41, 22, 59, 32, 31, 73 h(K) : 5, 2, 9, 7, 6, 5, 8

31. Linear Probing 41 18 32 59 31 22 0 1 2 3 4 5 6 7 8 9 10 11 12 h(K) = K % 13 Insert: 18, 41, 22, 59, 32, 31, 73 h(K) : 5, 2, 9, 7, 6, 5, 8 73

32. Double Hashing If the current location is occupied, try another table location Use two hash functions If M is prime, eventually will examine every location DoubleHashInsert(K) { if (table is full) error; probe = h1(K); offset = h2(K); while (table[probe] is occupied) probe = (probe+offset) % M; table[probe] = K; } Many of the same (dis)advantages as linear probing Distributes keys more evenly than linear probing

33. Quadratic Probing Don't step by 1 each time. Add i2 to the h(x) hashed location (mod B of course) for i = 1,2,...

34. Double Hashing 0 1 2 3 4 5 6 7 8 9 10 11 12 h1(K) = K % 13 h1(K) = 8 - K % 8 Insert: 18, 41, 22, 59, 32, 31, 73 h1(K) : 5, 2, 9, 7, 6, 5, 8 h2(K) : 6, 7, 2, 5, 8, 1, 7

35. Double Hashing 41 18 32 59 22 0 1 2 3 4 5 6 7 8 9 10 11 12 h1(K) = K % 13 h1(K) = 8 - K % 8 Insert: 18, 41, 22, 59, 32, 31, 73 h1(K) : 5, 2, 9, 7, 6, 5, 8 h2(K) : 6, 7, 2, 5, 8, 1, 7 31

36. Double Hashing 41 18 32 59 22 0 1 2 3 4 5 6 7 8 9 10 11 12 h1(K) = K % 13 h1(K) = 8 - K % 8 Insert: 18, 41, 22, 59, 32, 31, 73 h1(K) : 5, 2, 9, 7, 6, 5, 8 h2(K) : 6, 7, 2, 5, 8, 1, 7 73 31

37. Theoretical Results Not Found Found Chaining Linear Probing Double Hashing

38. Expected Probes Linear Probing Double Hashing Chaining 1.0 0.5 1.0