CSE 2331 Table Doubling 11.1 CSE 2331 Dynamic Arrays Allocate an array of size 10. What if you try to insert 11 elements in the array? Need to reallocate the array. Reallocate to size 11. Insert element. Reallocate to size 12. Insert element. etc. How much time does it take to insert 100 elements? How much time does it take to insert n elements? 11.2 CSE 2331 Open Address Hashing Hash table of size m = 20. For Θ(1) expected running time, need: # elements n ≤ 20/2 = 10. After inserting 10 elements, need to increase hash table size m. 11.3 CSE 2331 Array Doubling Start with array of size 2. After inserting 2 elements, replace with array of size 4. After inserting 4 elements, replace with array of size 8. After inserting 8 elements, replace with array of size 16. etc. 11.4 CSE 2331 Array Doubling function Insert(x, A, m, n) /* A is an array of size m. */ /* n = # elements in A. */ 1 if (n = m) then 2 Create new array A2 with size 2m; 3 for i ← 1 to m do 4 A2[i] ← A[i]; 5 end 6 Replace array A with A2; 7 m ← 2m; 8 end 9 n ← n + 1; 10 A[n] ← x; 11.5 CSE 2331 Example A : [ , ] Insert: 21 A : [21, ] Insert: 22 A : [21, 22] Insert: 41 A : [21, 22, 41, ] (Array size m = 4.) Insert: 42 A : [21, 22, 41, 42] Insert: 81 A : [21, 22, 41, 42, 81, , , ] (Array size m = 8.) Insert: 82 Insert: 83 Insert: 84 A : [21, 22, 41, 42, 81, 82, 83, 84] Insert: 61 A : [21, 22, 41, 42, 81, 82, 83, 84, 61, , , , , , , ] (m = 16.) 11.6 CSE 2331 Running Time Analysis function Insert(x, A, m, n) 1 if (n = m) then 2 Create new array A2 with size 2m; 3 for i ← 1 to m do 4 A2[i] ← A[i]; 5 end 6 Replace array A with A2; 7 m ← 2m; 8 end 9 n ← n + 1; 10 A[n] ← x; What is the running time of Insert when m 6= n? What is the running time of Insert when m = n? 11.7 CSE 2331 Running Times # elements Array size m Insert time 0 2 c 1 2 c 2 2 c +4c 3 4 c 4 4 c +8c 5 8 c 6 8 c 7 8 c 8 8 c + 16c 9 16 c 10 16 c . 15 16 c 16 16 c + 32c 17 32 c 11.8 CSE 2331 Running Time Analysis Total running time for n insertions: T (n)= cn + (Cost of doubling) = cn + (4c + 8c + 16c + 32c + ... + nc/2+ nc + 2nc) = cn + (2nc + nc + nc/2+ ... + 32c + 16c + 8c + 4c) = cn + 2nc(1+1/2 + 1/4+ ... + 2/n) ≤ cn + 4cn = 5cn. T (n)= cn + (Cost of doubling) ≥ cn. Since cn ≤ T (n) ≤ 5cn, T (n) ∈ Θ(n). 11.9 CSE 2331 Amortized Analysis Cost of one single operation may vary greatly. Average cost is much lower than the highest cost. Example: Array doubling. Insert cost: c or c + 2cn. Total cost of n Insert’s: 5cn. Average Insert cost: 5cn/n = 5c ∈ Θ(1). 11.10 CSE 2331 Hash Table Doubling function Dict.Insert(K, D) 1 m ← HashTable.size; 2 if (HashTable.NumElements ≥ m/2) then 3 Create new hash table HashTable2 with size 2m; 4 for i ← 1 to HashTable.size do 5 if (HashTable[i] is not empty) then 6 K ← HashTable[i].key; 7 HashTable2.Insert(K, HashTable[i].data); 8 end 9 end 10 Replace HashTable with HashTable2; 11 end 12 HashTable.Insert(K,D); 11.11 CSE 2331 Deletions from Dynamic Tables 11.12 CSE 2331 Pop function Pop(x, A, m, n) /* A is an array of size m. */ /* n = # elements in A. */ 1 if (n = 0) then error “Empty array.”; 2 x ← A[n]; 3 n ← n − 1; 4 return (x); 11.13 CSE 2331 Dynamic Table Deletions m = size of array A. n = number of elements in A. Want to shrink the array if n is a lot less than m. Proposal: Create new array of size m/2 if n ≤ m/2. What’s the problem? 11.14 CSE 2331 Insert/Delete problem Operation # elements Array size m Time 32 32 Insert c + 64c 33 64 Delete c + 32c 32 32 Insert c + 64c 33 64 Delete c + 32c 32 32 Insert c + 64c 33 64 Delete c + 32c 32 32 11.15 CSE 2331 Dynamic Table Deletions m = size of array A. n = number of elements in A. Want to shrink the array if n is a lot less than m. Solution: Create new array of size m/2 if n ≤ m/4. 11.16 CSE 2331 Dynamic Table Pop function DynamicPop(x, A, m, n) /* A is an array of size m. */ /* n = # elements in A. */ 1 if (n = 0) then error “Empty array.”; 2 x ← A[n]; 3 n ← n − 1; 4 if ((n ≤ m/4) and (m ≥ 4)) then 5 Create new array A2 with size m/2; 6 for i ← 1 to n do 7 A2[i] ← A[i]; 8 end 9 Replace array A with array A2; 10 m ← m/2; 11 end 12 return (x); 11.17 CSE 2331 Example A : [31, 32, 33, 34, 35, 36, , , , , , , , , , ] (Array has size m = 16.) DynamicPop A : [31, 32, 33, 34, 35, , , , , , , , , , , ] DynamicPop A : [31, 32, 33, 34, , , , ]. (Array has size m = 8.) DynamicPop A : [31, 32, 33, , , , , ]. DynamicPop A : [31, 32, , ] (Array has size m = 4.) 11.18 CSE 2331 Dynamic Table Pop function DynamicPop(x, A, m, n) 1 if (n = 0) then error “Empty array.”; 2 x ← A[n]; 3 n ← n − 1; 4 if ((n ≤ m/4) and (m ≥ 4)) then 5 Create new array A2 with size m/2; 6 for i ← 1 to n do 7 A2[i] ← A[i]; 8 end 9 Replace array A with array A2; 10 m ← m/2; 11 end 12 return (x); What is the running time of DynamicPop when n 6= m/4? What is the running time of DynamicPop when n = m/4? 11.19 CSE 2331 Running Times # elements Array size m DynamicPop time 18 64 c 17 64 c + 32c 16 32 c . 10 32 c 9 32 c + 16c 8 16 c 7 16 c 6 16 c 5 16 c +8c 4 8 c 3 8 c +4c 2 4 c +2c 1 2 c 11.20 CSE 2331 Running Time Analysis Assume number of elements n = array size m = 2k. Total running time for n deletes: T (n)= cn + (Cost of table halving) = cn +(cn + cn/2+ cn/4+ cn/8+ ... + 2c) = cn + cn(1+1/2 + 1/4 + 1/8+ ... + 2/n) ≤ cn + 2cn = 3cn. T (n)= cn + (Cost of table halving) ≥ cn. Since cn ≤ T (n) ≤ 3cn, T (n) ∈ Θ(n). 11.21 CSE 2331 Insert & Delete Running Time n inserts. n deletes. All inserts do not necessarily precede all deletes. Total running time of 2n operations is still Θ(n). 11.22 CSE 2331 Data Structures for Disjoint Sets 11.23 CSE 2331 Union-Find Data Structure Represent disjoint sets: Each element is in exactly one set. Operations: MakeSet(x) - Create a new set containing element x. Union(x, y) - Union of the sets containing x and y. FindSet(x) - Return a reference to a representative element of the set containing x. 11.24 CSE 2331 Connected Component v12 v1 KK 0 KK 0 KK 0 KK 0 v11 KK0 v2 ZZZZZZZ Z Kc0K0cccccc << ZZZZZZcZccccccc 0KK ppp < c ccccccc ZZ ZZZZZZ0 pKpK ccc<c<ccc pZp0ZZZKKZZZ v ccccccc < ppp 0 KK ZZZZZZ v 10 << pp 00 KK 3 < pp 0 KK < pp 0 KK << ppp 0 KK << ppp 00 v9 [[[[[ pp< 0 v4 [[[[ [[[ pp << 0 [pp[p[[[[[[< 0 pp [<[[ [[[[[ 0 pp << [[[[[[[[ v8 < [[ v5 << << < v7 v6 Are vi and vj in the same connected component? 11.25 CSE 2331 Connected Component /* Create a set for each vertex */ 1 foreach vertex vi do MakeSet(vi); 2 while (Not Done) do ... /* Add edge (vi, vj ) */ 3 if (FindSet(vi) 6= FindSet(vj )) then 4 Union (vi, vj ); 5 end ... /* Report if vx and vy are in the same component */ 6 if (FindSet(vx) = FindSet(vy )) then 7 Print “vx and vy are in the same component.” 8 end ... 9 end 11.26 CSE 2331 Linked List Based Union-Find . x1 o / x2 o / x3 o / x4 o / x5 . x6 o / x7 o / x8 o / x9 o / x10 o / x11 11.27 CSE 2331 Linked List Based Union-Find . x1 o / x2 o / x3 o / x4 o / x5 . x6 o / x7 o / x8 o / x9 o / x10 o / x11 11.28 CSE 2331 Linked List Based Union-Find rwyx . x1 / x2 / x3 / x4 / x5 9 tail {rwyx . x6 / x7 / x8 / x9 / x10 / x11 7 tail 11.29 CSE 2331 Linked List Based Union-Find rwyx . x1 / x2 / x3 / x4 / x5 9 tail {rwyx . x6 / x7 / x8 / x9 / x10 / x11 7 tail procedure FindSet(x) 1 return (x.head); 11.30 CSE 2331 Linked List Based Union-Find }} sw / x1 / x2 / x3 / x4 / x5 procedure Union(x, y) : ′ 1 x ← FindSet (x); tail ′ 2 y ← FindSet (y); ′ ′ 3 x .tail.next ← y ; ′ } 4 w ← y ; sw / x6 / x7 / x8 / x9 5 while (w 6= NULL) do = ′ 6 w.head ← x ; tail 7 w ← w.next; 8 end ′ ′ 9 x .tail ← y .tail; 11.31 CSE 2331 Linked List Based Union-Find procedure Union(x, y) ′ procedure MakeSet(x) 1 x ← FindSet (x); ′ 1 x.head ← x; 2 y ← FindSet (y); ′ ′ 2 x.tail ← x; 3 x .tail.next ← y ; ′ 3 x.next ← NULL; 4 w ← y ; 5 while (w 6= NULL) do ′ 6 w.head ← x ; procedure FindSet(x) 7 w ← w.next; 1 return (x.head); 8 end ′ ′ 9 x .tail ← y .tail; 11.32 CSE 2331 Linked List: Weighted Union procedure WeightedUnion(x, y) ′ 1 x ← FindSet (x); ′ 2 y ← FindSet (y); ′ ′ 3 if (x = y ) then return; ′ ′ 4 if (x .length ≥ y .length) then ′ ′ 5 x .tail.next ← y ; ′ 6 w ← y ; 7 while (w 6= NULL) do ′ 8 w.head ← x ; 9 w ← w.next; 10 end ′ ′ ′ 11 x .length = x .length + y .length; ′ ′ 12 x .tail ← y .tail; 13 else 14 WeightedUnion(y, x); 15 end 11.33 CSE 2331 Weighted Union: Analysis For a given node xi, how many times does xi.head change? Min size of set containing xi Cost of changing xi.head 1 c 2 c 4 c 8 c .