Union-Find Problem. Using Arrays And Chains. A Set As A Tree. Result Of A Find Operation

Transcription

1 Union-Find Problem Given set {,,, n} of n elements. Initilly ech element is in different set. ƒ {}, {},, {n} An intermixed sequence of union nd find opertions is performed. A union opertion combines two sets into one. ƒ Ech of the n elements is in exctly one set t ny time. A find opertion identifies the set tht contins prticulr element. Using Arrys And Chins See Section. for pplictions s well s for solutions tht use rrys nd chins. Best time complexity obtined in Section. is O(n + u log u + f), where u nd f re, respectively, the number of union nd find opertions tht re done. Using tree (not binry tree) to represent set, the time complexity becomes lmost O(n + f) (ssuming t lest n/ union opertions). A Set As A Tree Result Of A Find Opertion S = {,,,,,, } Some possible tree representtions: find(i) is to identify the set tht contins element i. In most pplictions of the union-find problem, the user does not provide set identifiers. The requirement is tht find(i) nd find(j) return the sme vlue iff elements i nd j re in the sme set. find(i) will return the element tht is in the tree root.

2 Strtegy For find(i) Trees With Prent Pointers Strt t the node tht represents element i nd climb up the tree until the root is reched. Return the element in the root. To climb the tree, ech node must hve prent pointer. 0 6 Possible Node Structure Exmple Use nodes tht hve two fields: element nd prent. ƒ Use n rry tble[] such tht tble[i] is pointer to the node whose element is i. ƒ To do find(i) opertion, strt t the node given by tble[i] nd follow prent fields until node whose prent field is null is reched. ƒ Return element in this root node. tble[] 0 0 (Only some tble entries re shown.)

3 Better Representtion Use n integer rry prent[] such tht prent[i] is the element tht is the prent of element i. union(i,j) Union Opertion ƒ i nd j re the roots of two different trees, i!= j. To unite the trees, mke one tree subtree of the other. ƒ prent[j] = i prent[] Union Exmple The Find Method union(,) public int find(int theelement) { while (prent[theelement]!= 0) theelement = prent[theelement]; // move up return theelement; }

4 The Union Method Time Complexity Of union() O() public void union(int roota, int rootb) {prent[rootb] = roota;} Time Complexity of find() Tree height my equl number of elements in tree. ƒ union(,), union(3,), union(,3), union(,) 3 u Unions nd f Find Opertions O(u + uf) = O(uf) Time to initilize prent[i] = 0 for ll i is O(n). Totl time is O(n + uf). Worse thn solution of Section.! Bck to the drwing bord. So complexity is O(u).

5 Smrt Union Strtegies union(,) Which tree should become subtree of the other? Height Rule Mke tree with smller height subtree of the other tree. Brek ties rbitrrily. union(,) Weight Rule Mke tree with fewer number of elements subtree of the other tree. Brek ties rbitrrily. union(,) Implementtion Root of ech tree must record either its height or the number of elements in the tree. When union is done using the height rule, the height increses only when two trees of equl height re united. When the weight rule is used, the weight of the new tree is the sum of the weights of the trees tht re united.

6 Height Of A Tree Suppose we strt with single element trees nd perform unions using either the height or the weight rule. The height of tree with p elements is t most floor (log p) +. Proof is by induction on p. See text. b Sprucing Up The Find Method e d c f g find() Do dditionl work to mke future finds esier. 0 6, b, c, d, e, f, nd g re subtrees 0 Pth Compction Mke ll nodes on find pth point to tree root. find() Pth Splitting Nodes on find pth point to former grndprent. find() f g 0 e d 0 6, b, c, d, e, f, nd g re subtrees b c Mkes two psses up the tree. f g 0 e 0 6 d, b, c, d, e, f, nd g re subtrees b c Mkes only one pss up the tree.

7 Pth Hlving Prent pointer in every other node on find pth is chnged to former grndprent. find() b d c e f g , b, c, d, e, f, nd g re subtrees Chnges hlf s mny pointers. 0 Ackermnn s function. Time Complexity ƒ A(i,j) = j, i = nd j >= ƒ A(i,j) = A(i-,), i >= nd j = ƒ A(i,j) = A(i-,A(i,j-)), i, j >= Inverse of Ackermnn s function. ƒ lph(p,q) = min{z>= A(z, p/q) > log q}, p >= q >= Time Complexity Ackermnn s function grows very rpidly s i nd j re incresed. ƒ A(,) = 6,36 The inverse function grows very slowly. ƒ lph(p,q) < until q = A(,) ƒ A(,) = A(,6) >>>> A(,) In the nlysis of the union-find problem, q is the number, n, of elements; p = n + f; nd u >= n/. For ll prcticl purposes, lph(p,q) <. Time Complexity Theorem. [Trjn nd Vn Leeuwen] Let T(f,u) be the mximum time required to process ny intermixed sequence of f finds nd u unions. Assume tht u >= n/. *(n + f*lph(f+n, n)) <= T(f,u) <= b*(n + f*lph(f+n, n)) where nd b re constnts. These bounds pply when we strt with singleton sets nd use either the weight or height rule for unions nd ny one of the pth compression methods for find.