Array-based Hashtables

Transcription

1 Array-based Hashtables For simplicity, we will assume that we only insert numeric keys into the hashtable hash(x) = x % B; where B is the number of 5

2 Implementation class Hashtable { int [B]; bool occupied[b]; Insert(5) h=hash(5) = 5 % 9 =

3 Insert class Hashtable { int [B]; bool occupied[b]; Insert(5) h=hash(5) = 5 % 9 = Insert() h=hash() = % 9 =

4 Insert class Hashtable { int [B]; bool occupied[b]; Insert(5) 0 2 h=hash(5) = 5 % 9 = 8 Insert() h=hash() = % 9 =

5 Insert class Hashtable { int [B]; bool occupied[b]; 0 2 insert(x) { h = hash(x); [h] = x; occupied[h] = true;

6 Insert class Hashtable { int [B]; bool occupied[b]; 0 2 Insert() h=hash() = % 9 = Collision

7 Collision Resolution If h(x) is occupied, we try other We try h 0 x, h x, h 2 x, until an empty bucket is found h i (x) = hash(x) + f(i) where f is called the collision resolution function 2

8 Linear Probing f i = i Insert() h=hash() = % 9 = 4 h 0 =4 + 0 = 4 (Collision) h =4 + =

9 Linear Probing f i = i Insert() h=hash() = % 9 = 4 h 0 =4 + 0 = 4 (Collision) h =4 + =

10 Linear Probing f i = i Insert(20) h=hash(20) = 20 % 9 = 2 h 0 =2 + 0 =

11 Linear Probing f i = i Insert(20) h=hash(20) = 20 % 9 = 2 h 0 =2 + 0 =

12 Linear Probing f i = i Insert(26) h=hash(26) = 26 % 9 = 8 h 0 =8 + 0 = 8 (Collision) h =8 + =

13 Linear Probing f i = i Insert(26) h=hash(26) = 26 % 9 = 8 h 0 =8 + 0 = 8 (Collision) h =8 + =

14 Linear Probing f i = i Insert(40) h=hash(40) = 40 % 9 = 4 h 0 =4 + 0 = 4 (Collision) h =4 + = 5 (Collision) h 2 =4 + 2 =

15 Linear Probing f i = i Insert(40) h=hash(40) = 40 % 9 = 4 h 0 =4 + 0 = 4 (Collision) h =4 + = 5 (Collision) h 2 =4 + 2 =

16 Find Find(40) h=hash(40) = 40 % 9 = 4 h 0 =4 + 0 = ?

17 Find Find(40) h=hash(40) = 40 % 9 = 4 h 0 =4 + 0 = 4 (No match) h =4 + = 5 (No match) ?

18 Find Find(40) h=hash(40) = 40 % 9 = 4 h 0 =4 + 0 = 4 (No match) h =4 + = 5 (No match) h 2 =4 + 2 = 6 (Match)

19 Find Find(22) h=hash(22) = 22 % 9 = 4 h 0 =4 + 0 = ?

20 Find Find(22) h=hash(22) = 22 % 9 = 4 h 0 =4 + 0 = 4 (No match) h =4 + = 5 (No match) ?

21 Find Find(22) h=hash(22) = 22 % 9 = 4 h 0 =4 + 0 = 4 (No match) h =4 + = 5 (No match) h 2 =4 + 2 = 6 (No match) ?

22 Find Find(22) h=hash(22) = 22 % 9 = 4 h 0 =4 + 0 = 4 (No match) h =4 + = 5 (No match) h 2 =4 + 2 = 6 (No match) h =4 + = (Empty) ? 8 5 6

23 Find Find(22) h=hash(22) = 22 % 9 = 4 h 0 =4 + 0 = 4 (No match) h =4 + = 5 (No match) h 2 =4 + 2 = 6 (No match) h =4 + = (Empty) Return false (not found) ? 8 5

24 Insert & Find void insert(x) { h = hash(x); while (occupied[h]) { h++; [h] = x; occupied[h] = true; Do you spot an error? bool find(x) { h = hash(x); while (occupied[h]) { if ([h] == x) return true; h++; return false; 8

25 Insert & Find void insert(x) { h = hash(x); while (occupied[h]) { h = (h+) % B; [h] = x; occupied[h] = true; bool find(x) { h = hash(x); while (occupied[h]) { if ([h] == x) return true; h = (h+) % B; return false; 9

26 Deletion bool delete(x) { h = hash(x); while (occupied[h]) { if ([h] == x) { occupied[h] = false; return true; h = (h+) % B; return false; 40

27 Delete Delete() h=hash() = % 9 = 4 h 0 =4 + 0 = 4 (Match) 0 26 Is this correct??

28 Delete Find(40) h=hash(40) = 40 % 9 = 4 h 0 =4 + 0 = 4 (No match) return false!!! ?

29 Revisit the Design class Hashtable { int [B]; enum {E, O, D status[b]; The status of each bucket is either Empty, Occupied, or Deleted Initially all are empty 4

30 Status Delete Delete() h=hash() = % 9 = 4 h 0 =4 + 0 = 4 (Match) 0 O 26 E 2 O 20 E? 4 O 5 O 6 O 40 E 8 O 5 44

31 Status Delete Delete() h=hash() = % 9 = 4 h 0 =4 + 0 = 4 (Match) Find(40) h=hash(40) = 40 % 9 = 4 h 0 =4 + 0 = 4 (No match) h =4 + = 5 (No match) h 2 =4 + 2 = 6 (Match) return true? 0 O 26 E 2 O 20 E 4 D 5 O 6 O 40 E 8 O 5 45

32 Insert, Find & Delete void insert(x) { h = hash(x); while (status[h]== O ) { h = (h+) % B; [h] = x; status[h] = O ; bool delete(x) { h = hash(x); while (status[h]!= E ) { if ([h] == x) { status[h] = D ; return true; h = (h+) % B; return false; bool find(x) { h = hash(x); while (status[h]!= E ) { if (status[h] == O && [h] == x) return true; h = (h+) % B; return false; 46

33 Primary Clustering A cluster (a block) of are all occupied Any key that hits the cluster will have to linearly probe until the end of the cluster 0 O 26 E 2 O 20 E 4 O Primary clustering 5 O 6 O 40 E 8 O 5 4

34 Status Quadratic Probing f i = i 2 Insert(5) 0 E E 2 E E 4 E 5 E 6 E E 8 E 48

35 Status Quadratic Probing f i = i 2 Insert(5) Insert(2) 0 E E 2 E E 4 E 5 E 6 O 5 E 8 E 49

36 Status Quadratic Probing f i = i 2 Insert(5) Insert(2) Insert(4) 0 E E 2 E E 4 E 5 O 2 6 O 5 E 8 E 50

37 Status Quadratic Probing f i = i 2 Insert(5) Insert(2) Insert(4) h 0 =5+0 (Occupied) h =5+ (Occupied) h 2 =5+4=0 (Empty) 0 E E 2 E E 4 E 5 O 2 6 O 5 E 8 E 5

38 Status Quadratic Probing f i = i 2 Insert(5) Insert(2) Insert(4) h 0 =5+0 (Occupied) h =5+ (Occupied) h 2 =5+4=0 (Empty) Insert() 0 O 4 E 2 E E 4 E 5 O 2 6 O 5 E 8 E 52

39 Status Quadratic Probing f i = i 2 Insert(5) Insert(2) Insert(4) h 0 =5+0 (Occupied) h =5+ (Occupied) h 2 =5+4=0 (Empty) Insert() h 0 =6+0 (Occupied) h =6+ (Empty) 0 O 4 E 2 E E 4 E 5 O 2 6 O 5 E 8 E 5

40 Status Quadratic Probing f i = i 2 Insert(4) h 0 =5+0 (Occupied) h =5+ (Occupied) h 2 =5+4=0 (Occupied) h =5+9=5 (Occupied) h 4 =5+6= (Empty) 0 O 4 E 2 E O 4 4 E 5 O 2 6 O 5 O 8 E 54

41 Status Secondary Clustering For two distinct keys x and x 2, if h(x ) = h(x 2 ), then h i x = h i x 2, i In other words, if two keys start at the same position, they will probe the same sequence of 0 O 4 E 2 E O 4 4 E 5 O 2 6 O 5 O 8 E 55

42 Double Hashing f i = i hash 2 x Where hash 2 is an alternative hash function Double hashing can eliminate both primary and secondary clustering 56