2. CHAPTER 8 THE DISJOINT SET ADT §1 Equivalence Relations 【 Definition 】 A relation R is defined on a set S if for every pair of elements (a, b), a, b  S, a R b is either true or false. If a R b is true, then we say that a is related to b. 【 Definition 】 A relation, ~, over a set, S, is said to be an equivalence relation over S iff it is symmetric, reflexive, and transitive over S. 【 Definition 】 Two members x and y of a set S are said to be in the same equivalence class iff x ~ y. 1/12

3. §2 The Dynamic Equivalence Problem Given an equivalence relation ~, decide for any a and b if a ~ b. 〖 Example 〗 Given S = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 } and 9 relations: 12  4, 3  1, 6  10, 8  9, 7  4, 6  8, 3  5, 2  11, 11  12. The equivalence classes are { 2, 4, 7, 11, 12 }, { 1, 3, 5 }, { 6, 8, 9, 10 } Algorithm: { /* step 1: read the relations in */ Initialize N disjoint sets; while ( read in a ~ b ) { if ( ! (Find(a) == Find(b)) ) Union the two sets; } /* end-while */ /* step 2: decide if a ~ b */ while ( read in a and b ) if ( Find(a) == Find(b) ) output( true ); else output( false ); } (Union / Find) Dynamic (on-line) 2/12

4. §2 The Dynamic Equivalence Problem  Elements of the sets: 1, 2, 3,..., N  Sets : S 1, S 2,...... and S i  S j =  ( if i  j ) —— disjoint 〖 Example 〗 S 1 = { 6, 7, 8, 10 }, S 2 = { 1, 4, 9 }, S 3 = { 2, 3, 5 } 10 68 7 4 19 2 35 A possible forest representation of these sets Note: Pointers are from children to parents  Operations : (1) Union( i, j ) ::= Replace S i and S j by S = S i  S j (2) Find( i ) ::= Find the set S k which contains the element i. 3/12

5. §3 Basic Data Structure  Union ( i, j ) Idea: Make S i a subtree of S j, or vice versa. That is, we can set the parent pointer of one of the roots to the other root. 10 68 7 4 19 4 19 68 7 S1  S2S1  S2 S2  S1S2  S1 Implementation 1: S1 S2 S3    name[ ] 10 68 7 4 19 2 35 S2  S1S2  S1  S 2 4/12

6. Here we use the fact that the elements are numbered from 1 to N. Hence they can be used as indices of an array. §3 Basic Data Structure Implementation 2: S [ element ] = the element’s parent. Note: S [ root ] = 0 and set name = root index. 10 68 7 4 19 2 35 〖 Example 〗 The array representation of the three sets is ( S 1  S 2  S 1 )  S [ 4 ] = 10  Find ( i ) Implementation 1: name[k] S  j i... find ( i ) =‘S’ Implementation 2: SetType Find ( ElementType X, DisjSet S ) { for ( ; S[X] > 0; X = S[X] ) ; return X ; } void SetUnion ( DisjSet S, SetType Rt1, SetType Rt2 ) { S [ Rt2 ] = Rt1 ; } [1] 4 [2] 0 [10] 0 [9] 4 [8] 10 [7] 10 [6] 10 [5] 2 [4] 0 [3] 2 S 10 5/12

7. §3 Basic Data Structure  Analysis Practically speaking, union and find are always paired. Thus we consider the performance of a sequence of union- find operations. 〖 Example 〗 Given S = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 } and 9 relations: 12  4, 3  1, 6  10, 8  9, 7  4, 6  8, 3  5, 2  11, 11  12. We have 3 equivalence classes { 2, 4, 7, 11, 12 }, { 1, 3, 5 }, and { 6, 8, 9, 10 } Algorithm using union-find operations { Initialize S i = { i } for i = 1,..., 12 ; for ( k = 1; k <= 9; k++ ) { /* for each pair i  j */ if ( Find( i ) != Find( j ) ) SetUnion( Find( i ), Find( j ) ); } Can you think of a worst case example? Sure. Try this one: union(2, 1), find(1); union(3, 2), find(1);...,... ; union(N, N  1), find(1). N N1N1  1 T =  ( N 2 ) ! That’s not good. 6/12

8. §4 Smart Union Algorithms  Union-by-Size -- Always change the smaller tree S [ Root ] = – size; /* initialized to be –1 */ Now T = O( N ) for the worst case example I gave. 1 2 N  【 Lemma 】 Let T be a tree created by union-by-size with N nodes, then Proof: By induction. (Each element can have its set name changed at most log 2 N times.) Time complexity of N Union and M Find operations is now O( N + M log 2 N ).  Union-by-Height -- Always change the shallow tree Please read Figure 8.13 on p.273 for detailed implementation. 7/12

9. §5 Path Compression SetType Find ( ElementType X, DisjSet S ) { if ( S[ X ] <= 0 ) return X; else return S[ X ] = Find( S[ X ], S ); } SetType Find ( ElementType X, DisjSet S ) { ElementType root, trail, lead; for ( root = X; S[ root ] > 0; root = S[ root ] ) ; /* find the root */ for ( trail = X; trail != root; trail = lead ) { lead = S[ trail ] ; S[ trail ] = root ; } /* collapsing */ return root ; } Note: Not compatible with union-by- height since it changes the heights. Just take “height” as an estimated rank. Slower for a single find, but faster for a sequence of find operations. 8/12

10. §6 Worst Case for Union-by-Rank and Path Compression 【 Lemma (Tarjan) 】 Let T( M, N ) be the maximum time required to process an intermixed sequence of M  N finds and N  1 unions. Then: k 1 M  ( M, N )  T( M, N )  k 2 M  ( M, N ) for some positive constants k 1 and k 2.  Ackermann’s Function and  ( M, N )  O( log * N )  4 http://mathworld.wolfram.com/AckermannFunction.html log * N (inverse Ackermann function) = # of times the logarithm is applied to N until the result  1. log * 2 65536 = 5 since logloglogloglog ( 2 65536 )= 1 9/12

11. Home work: p.282 8.7 A formatted version We have a network of computers and a list of bi-directional connections. Each of these connections allows a file transfer from one computer to another. Is it possible to send a file from any computer on the network to any other? Input: Input consists of several test cases. For each test case, the first line contains N (<=10,000), the total number of computers in a network. Each computer in the network is then represented by a positive integer between 1 and N. Then in the following lines, the input is given in the format: I c1 c2 where I stands for inputting a connection between c1 and c2; or C c1 c2 where C stands for checking if it is possible to transfer files between c1 and c2; or S where S stands for stopping this case. 10/12

12. Output: For each C case, print in one line the word yes or no if it is possible or impossible to transfer files between c1 and c2, respectively. At the end of each case, print in one line The network is connected. if there is a path between any pair of computers; or There are k components. where k is the number of connected components in this network. Print a blank line between test cases. Sample Input: 3 C 1 2 I 1 2 C 1 2 S 3 I 3 1 I 2 3 C 1 2 S Sample Output: no yes There are 2 components. yes The network is connected. Must use union-by-size and path compression to obtain a worst-case running time of O( M  ( M, N ) ). 11/12

13. Bonus Problem 2 Attack of Panda (2 points) Due: Friday, November 13th, 2009 at 10:00pm Detailed requirements can be downloaded from http://acm.zju.edu.cn/dsaa/ 12/12