Connected-components. Summary of lecture 9. Algorithms and Data Structures Disjoint sets. Example: connected components in graphs
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1 Prm University, Mth. Deprtment Summry of lecture 9 Algorithms nd Dt Structures Disjoint sets Summry of this lecture: (CLR.1-3) Dt Structures for Disjoint sets: Union opertion Find opertion Mrco Pellegrini Spring 00, Lecture 9 4//03 1 4//03 Disjoint set opertions. We hve collection of elements, which we identify with numers 1,,3, etc. A Collection of sets S 1, S,..S k. The property we ssume is tht n element is t most in one set. Tht is the sets re disjoint. We shll see the following opertions: Mke-set(x): tke element x not in ny set nd genertes the set contining x, tht is: {x}. Union(x,y): mkes the union of the set contining x nd the set contining y. Find-set(x): returns pointer to the representtive of the set contining x. Ech set hs unique representtive. 4//03 3 Exmple: connected components in grphs V={1,,3,4,5,6,7,8,9, 10,11,1,13,14} E = {(1,3)(,3)(,11)(4,8) (7,5)(5,10)(10,9)(6,1) (1,13)(1,14)} 4//03 4 Connected-components. We wnt to find the components of the grph. A component is group of nodes tht we cn rech y following edges strting from node. Connected-component(V,E) FOR ech vertex v in V DO Mke-set(v) END FOR FOR ech edge (u,v) in E DO IF Find-set(u) ž Find-set(v) THEN Union(u,v) END FOR Sme-component(u,v) RETURN [Find-set(u) = Find-set(v)] Exercise Run the procedure Connected-component on the grph (tht is V nd E) given in the exmple. How mny sets (tht is components) do you hve t the end. Hint: nturlly I hve not told you yet HOW mke-set, union nd find-set work ut you know WHAT they re supposed to do. 4//03 5 4//03 6 1
2 Representing disjoint sets y lists. We cn use circulr singly linked lists nd use the first element of the list s representtive next rep 4//03 7 Opertions Mke-set nd find-set cn e done in O(1) on circulr list with ck-pointer to the hed. Union(x,y) is trickier. Appending two circulr lists cn e done in O(1), ut to updte the pointer to the representtive we must scn one of the two lists. If n is the numer of elements, m the numer of opertions, in the worst cse we might use time O(m ) to perform the m opertions. This hppens when we grow list y one element t time nd we lwys scn the longest list. 4//03 8 Size heuristic. Improvement: when we mke the union scn the shortest of the two list. To know which is the shortest we crry field SIZE linked to the representtive which we updte when we perform the union. This simple rule decreses the time complexity to O(m + n log n) for n elements, m opertions. Proof. By chrging rgument. Fix one element x, initilly it is in set of size 1. Then t every union the size of the set contining it t lest doules, tht is,4,8...,n. This cn hppen only log n times. At every union we updte repr(x). In totl n log n updtes. All other costs re O(m). 4//03 9 Homework HW.3 (CLR.-1) Write pseudocode for mke-set,find-set, union using singly linked lists nd the weighted union rule. Ech oject x, hs: ) field repr[x] pointing to the representtive of the set contining x, ) field lst[x] pointing to the lst ojectin the list contining x, c) field size[x] giving the size of the listcontining x. Size[x] nd Lst[x] re correct only when x is representtive. 4//03 10 Forest-of-trees implementtion. Exmples Cn we do etter thn the linked lists DS? Yes, uy it is hrd to prove it! Ide: represent set y tree, we need only pointer from node to the prent. We could keep trck of sizes, ut for proof purposes we keep trck of the rnk of the trees. z c e the root points to itself The rnk is the length of the longest pth. g Rnk = 3 4// //03 1
3 Union y rnk The rnk of the tree is the length of the longest pth from root to lef (without pth compression) Union y rnk is: ttch the lowest rnk tree under the root of the highest rnk tree. 4//03 13 Union y rnk If we merge two trees A,B of different rnk the rnk of the resulting tree is mx{rnk(a), rnk(b)}. If they hve the sme rnk, thn the rnk increses y 1. Mke-set(x) prent[x] := x rnk[x] := 0 Union(x,y) := find-set(x), := findset(y) IF rnk[] > rnk[] THEN prent[] := ELSE prent[] := IF rnk[] = rnk[] THEN rnk[] := rnk[]+1 4//03 14 Find-set with pth compression. g h d d h g find-set(g) 4//03 15 Find set + pth compression. To implement find-set(x) we follow pointers from x to the root nd then we go ck mking every node we visit on the pth child of the root. Find-set(x) IF x ž p[x] THEN prent[x] := Find-set(prent[x]) ENDIF RETURN prent[x] Exercise: simulte this code on the exmple of the previous slide find-set(g). Show for ech cll the node in input nd the one in output nd the chnge to the prent field. 4//03 16 Progrmming ssignment Implement in C the Disjoins-sets DS using lists with weighted union rule. [DS1] Implement in C the Disjoint-set DS using trees with the rnk rule nd the pth-compression. [DS] Generte rndom Grph G with n nodes nd n edges. Implement the connected components lgorithm twice using DS1 the first time nd DS the second time.[a1,a] Run the A1 nd A on the grph G (check tht the output (tht is the components) is the sme. Plot the running times of A1 nd A for n=100,00, , mximum numer efore the progrm crshes. 4//03 17 Performnce of Union-Find DS n elements, m opertions (union nd find), f find opertions. UF= Union find, PC= pth compression, WH= weight heuristic, RH= rnk heuristic. UF+WH is O(m + n log n). UF+RH is O(m log n) UF+PC is O(n+f log n) if f<n, O(f log (1+f/n) n) if f>n. UF+PC+RH is O(m log* n) Where log*n= min{i log (i) n ˆ1}. And log (0) n = log n, log (i) n= log (log (i-1) n). 4//
4 ... F(i)= F(1) = 1 F() = 4 F(3) = 16 F(4) = 65,536 i times Log*(F(x))=x. F(5) = > 10 4//03 19 Anlysis of UF+PC+RH The function G(n)=log* n is the pseudo-inverse of F(n). For ll prcticl inputs log* n < 5. The min ide is tht to estimte the numer of nodes visited y find opertions, we cn insted count the numer of find opertions visiting ny node. 4//03 0 Lemm 1. For node x tht is not root, rnk[x] < rnk[prent[x]]. Proof: y construction. Lemm. For node x, rnk[x] is n incresing function of time (non-decresing). Constnt when x ceses to e root. Proof: y construction. For root x, size(x) is the numer of nodes in the tree rooted t x. Lemm 3. Size(x) rnk(x). Proof: y induction. Initilly ech node is root nd ech node hs rnk 0, so the lemm is true. Inductive step: suppose rnk(x)<rnk(y). After the union we hve size t the root size(x)+size(y) > size(y) rnk(y). The rnk of the root fter the union is rnk(y). Suppose rnk(x)=rnk(y). After the union we hve sixe t the root size(x)+size(y) rnk(y) + rnk(y) rnk(y)+1. The rnk of the root fter the union is rnk(y)+1. 4//03 1 4//03 Lemm 4. For ech integer r, there re t most n/ r nodes of rnk r t ny given time. Proof. Since the rnk is n incresing function over ech pth from lef to the root, no two nodes of sme rnk r re one the ncestor of the other. So for x nd y oth of rnk r the sutrees rooted t x nd y re disjoint. Ech sutree hs t let r nodes, so there re t most n/ r nodes of rnk r. Lemm 5. No vertex hs rnk greter thn log n. Chrging scheme. We split the possile rnks r=[0,, log n ] into groups. Rnk r goes into group G(r ). Equivlently group j contins rnks [F(j-1)+1,..,F(j)]. Cse 1. If find visits x nd group(x) ž group(prent(x)), or x is the root, chrge 1 to the find opertion. Cse. If find visits x nd group(x) = group(prent(x)) chrge 1 to the node x. ( is not the root). 4//03 3 4//03 4 4
5 Costs Cse 1. A single find opertion incur in cse (1) when crossing the oundries etween two groups, so t most G(log n) = G(n) -1 times. So n find opertions re chrged O(nG(n)). Cse. While x nd prent(x) re in the sme group g, since the rnk is incresing, we cn chrge x t most F(g)- F(g-1) times < F(g). How mny nodes do we hve in group g? F ( g ) r= F ( g 1) + 1 r F n ( n ) ( g 1) + 1 i= 0 i ( n F ( g 1) ) = n F( g) So the totl chrge to group g is n. The numer of groups is G(n). The totl chrge y cse is O(nG(n)). Theorem 6. Union Find with Pth Compression nd Rnk Heuristic, n elements nd opertions tkes time O(n log* n). 4//03 5 4//03 6 Conclusions We hve seen two DS for Disjoint-sets, sometimes clled (Union-find ADS). The first sed on lists, the second sed on trees with union-y-rnk nd pth-compression. The nlysis tells us the second is symptoticlly fster. Your progrmming ssignment should test whether this hppens lso in experimentl tests. Actully n even more complex nlysis cn prove tht the right ound is O(m α(n,m)) where α(n,m) is function tht is even slower thn log*(n). 4//03 7 5
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