Today’s Outline

• Announcements – Assignment #4 coming soon . Disjoint Sets and Dynamic Equivalence Relations • Today’s Topics: – Leftist & Skew Heaps – Disjoint Sets & Dynamic Equivalence CSE 373 Data Structures and Algorithms

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Motivation Disjoint Sets

Some kinds of data analysis require keeping track of • Two sets S 1 and S 2 are disjoint if and only if they transitive relations. have no elements in common . Equivalence relations are one family of transitive relations. ∩ ∅ •S1 and S 2 are disjoint iff S1 S2 = Grouping pixels of an image into colored regions is (the intersection of the two sets is the empty set) one form of data analysis that uses “dynamic equivalence relations”. Creating mazes without cycles is another application. For example {a, b, c} and {d, e} are disjoint. Later we’ll learn about “minimum spanning trees” for networks, and how the dynamic equivalence But {x, y, z} and {t, u, x} are not disjoint. relations help out in computing spanning trees.

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Equivalence Relations Induced Equivalence Relations

• A binary relation R on a set S is an equivalence • Let S be a set, and let P be a partition of S.

relation provided it is reflexive, symmetric, and P = { S 1, S 2, . . . , Sk } transitive: P being a partition of S means that: ≠ → ∩ ∅ • Reflexive - R(a,a) for all a in S. i j Si Sj = and S ∪ S ∪ . . . ∪ S = S • Symmetric - R(a,b) → R(b,a) 1 2 k • Transitive - R(a,b) ∧ R(b,c) → R(a,c) • P induces an equivalence relation R on S: R(a,b) provided a and b are in the same subset (same element of P). Is ≤ an equivalence relation on integers? Is “is connected by roads ” an equivalence relation on So given any partition P of a set S, there is a corresponding cities? equivalence relation R on S.

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1 Example Introducing the UNION-FIND ADT • S = {a, b, c, d, e} • Also known as the Disjoint Sets ADT or the P = { S , S , S } 1 2 3 Dynamic Equivalence ADT. S = {a, b, c}, S = {d}, S = {e} 1 2 3 • There will be a set S of elements that does not P being a partition of S means that: ≠ → ∩ ∅ change. i j Si Sj = and ∪ ∪ ∪ S1 S2 . . . Sk = S • We will start with a partition P 0, but we will • P induces an equivalence relation R on S: modify it over time by combining sets. R = { (a,a), (b,b), (c,c), (a,b), (b,a), (a,c), (c,a), • The combining operation is called “UNION” (b,c), (c,b), • Determining which set (of the current partition) an (d,d), element of S belongs to is called the “FIND” (e,e) } operation.

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Example Union

• Maintain a set of pairwise disjoint * sets. • Union(x,y) – take the union of two sets named x – {3,5,7}, {4,2,8}, {9}, {1,6} and y • Each set has a unique name: one of its members – {3, 5,7} , {4,2, 8}, { 9}, { 1,6} – {3, 5,7}, {4,2, 8}, { 9}, { 1,6} – Union(5,1) {3, 5,7,1,6}, {4,2, 8}, { 9},

To perform the union operation, we replace sets x and y by (x ∪ y) *Pairwise Disjoint: For any two sets you pick, their intersection will be empty)

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Find Application: Building Mazes

• Find(x) – return the name of the set containing x. • Build a random maze by erasing edges. – {3, 5,7,1,6}, {4,2, 8}, { 9}, – Find(1) = 5 – Find(4) = 8

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2 Building Mazes (2) Building Mazes (3)

• Pick Start and End • Repeatedly pick random edges to delete.

Start Start

End End

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Desired Properties A Cycle

• None of the boundary is deleted • Every cell is reachable from every other cell. • Only one path from any one cell to another (There are no cycles – no cell can reach itself by a path Start unless it retraces some part of the path.)

End

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A Good Solution A Hidden Tree

Start Start

End End

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3 Number the Cells Basic Algorithm •P = set of sets of connected cells •E = set of edges We have disjoint sets P ={ {1}, {2}, {3}, {4},… {36} } each cell is unto itself. We have all possible edges E ={ (1,2), (1,7), (2,8), (2,3), … } 60 edges total. • Maze = set of maze edges (initially empty) While there is more than one set in P { pick a random edge (x,y) and remove from E Start 1 2 3 4 5 6 u := Find(x); v := Find(y); 7 8 9 10 11 12 if u ≠ v then // removing edge (x,y) connects previously non- 13 14 15 16 17 18 // connected cells x and y - leave this edge removed! Union(u,v) 19 20 21 22 23 24 else // cells x and y were already connected, add this // edge to set of edges that will make up final maze. 25 26 27 28 29 30 add (x,y) to Maze } 31 32 33 34 35 36 End All remaining members of E together with Maze form the maze 10/29/2010 19 10/29/2010 20

Example Step Example P P Pick (8,14) P {1,2, 7,8,9,13,19} {1,2, 7,8,9,13,19,14,20 26,27} {1,2, 7,8,9,13,19} {3} Find(8) = 7 {3} {3} {4} Find(14) = 20 {4} Start 1 2 3 4 5 6 {4} {5} {5} {5} {6} {6} 7 8 9 10 11 12 {6} {10 } Union(7,20) {10 } {10 } {11, 17 } 13 14 15 16 17 18 {11, 17 } {11, 17 } {12 } {12 } {12 } {14, 20 ,26,27} 19 20 21 22 23 24 {15, 16 ,21} {14, 20 ,26,27} {15, 16 ,21} . 25 26 27 28 29 30 {15, 16 ,21} . . . . End {22,23,24,29,39,32 31 32 33 34 35 36 . {22,23,24,29,39,32 33, 34 ,35,36} {22,23,24,29,30,32 33, 34 ,35,36} 10/29/2010 33, 34 ,35,36} 21 10/29/2010 22

Example Example at the End

Pick (19,20) P P {1,2, 7,8,9,13,19 {1,2,3,4,5,6, 7,… 36} 14,20,26,27} Start 1 2 3 4 5 6 {3} Start 1 2 3 4 5 6 {4} E 7 8 9 10 11 12 {5} 7 8 9 10 11 12 Maze {6} 13 14 15 16 17 18 13 14 15 16 17 18 {10 } 19 20 21 22 23 24 {11, 17 } 19 20 21 22 23 24 {12 } 25 26 27 28 29 30 {15, 16 ,21} 25 26 27 28 29 30 . 31 32 33 34 35 36 End . 31 32 33 34 35 36 End {22,23,24,29,39,32 10/29/2010 33, 34 ,35,36} 23 10/29/2010 24

4 Implementing the Disjoint Sets ADT Up-Tree for Disjoint Union/Find • n elements , Total Cost of: m finds, ≤ n-1 unions Initial state: 1 2 3 4 5 6 7

• Target complexity: O(m+n) i.e. O(1) amortized After several 1 3 7 Unions: • O(1) worst-case for find as well as union would be 2 5 4 great, but… Known result : both find and union cannot be done Roots are the names of each set. 6 in worst-case O(1) time

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Find Operation Union Operation

Find(x) - follow x to the root and return the root Union(x,y) - assuming x and y are roots, point y to x.

1 3 7 Union(1,7) 1 3 7 2 5 4 2 5 4 Find(6) = 7 6 6

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Simple Implementation Implementation • Array of indices int Find(int x) { void Union(int x, int y) { while(up[x] != 0) { up[y] = x; Up[x] = 0 means x = up[x]; } 1 2 3 4 5 6 7 x is a root. } up 0 1 0 7 7 5 0 return x; runtime for Union(): } 1 3 7 runtime for Find():

2 5 4 runtime for m Finds and n-1 Unions: 6 10/29/2010 29 10/29/2010 30

5 Find Solutions Now this doesn’t look good
Can we do better? Yes!

Recursive Find(up[] : integer array, x : integer) : integer { 1. Improve union so that find only takes Θ(log n) //precondition: x is in the range 1 to size// • Union-by-size if up[x] = 0 then return x else return Find(up,up[x]); • Reduces complexity to Θ(m log n + n) } Iterative 2. Improve find so that it becomes even better! Find(up[] : integer array, x : integer) : integer { //precondition: x is in the range 1 to size// • Path compression while up[x] ≠ 0 do • Reduces complexity to almost Θ(m + n) x := up[x]; return x; }

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A Bad Case Weighted Union

• Weighted Union 1 2 3 … n Union(2,1) – Always point the smaller (total # of nodes) tree to the root of the larger tree 2 3 … n

1 Union(3,2) W-Union(1,7) 3 … n : : 1 3 7 2 2 1 4 n Union(n,n-1) 1 2 5 4 3

2 Find(1) n steps!! 6 1 10/29/2010 33 10/29/2010 34

Example Again Analysis of Weighted Union

With weighted union an up-tree of height h has weight at h 1 2 3 … n least 2 . W-Union(2,1) … 2 3 n • Proof by induction 0 1 W-Union(3,2) – Basis : h = 0. The up-tree has one node, 2 = 1 2 … n : – Inductive step : Assume true for all h’ < h. : 1 3 W-Union(n,2) h-1 T W(T 1) > W(T 2) > 2 2 Minimum weight Weighted Induction h-1 1 3 … n up-tree of height h union hypothesis Find(1) constant time T1 T formed by 2 W( T) > 2h-1 + 2 h-1 = 2 h weighted unions 10/29/2010 35 10/29/2010 36

6 Worst Case for Weighted Union Analysis of Weighted Union (cont) Let T be an up-tree of weight n formed by weighted union. Let h be its height. n/2 Weighted Unions n > 2h

log 2 n > h n/4 Weighted Unions • Find(x) in tree T takes O(log n) time. – Can we do better?

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Example of Worst Case (cont’) Array Implementation

After n/2 + n/4 + …+ 1 Weighted Unions: 1 3 7 2 1 4

2 5 4

log 2n 6 1 2 3 4 5 6 7 Find up -1 1 -1 7 7 5 -1 If there are n = 2 k nodes then the longest weight 2 1 4 path from leaf to root has length k.

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Weighted Union Nifty Storage Trick

W-Union(i,j : index){ • Use the same array representation as before //i and j are roots new runtime for Union(): wi := weight[i]; –1 wj := weight[j]; • Instead of storing for the root, if wi < wj then simply store –size up[i] := j; new runtime for Find(): weight[j] := wi + wj; else up[j] :=i; [Read section 8.4, page 299] weight[i] := wi +wj; } runtime for m finds and n-1 unions =

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7 How about Union-by-height ? Path Compression • Can still guarantee O(log n) worst case depth • On a Find operation point all the nodes on the search path directly to the root. Left as an exercise!

1 7 • Problem: Union-by-height doesn’t combine very well with the new find optimization technique we’ll see next 2 5 4 PC-Find(3)

6 8 9

3 10

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Student Activity Draw the result of Find(e): Self-Adjustment Works

c g

a f h PC-Find(x)

b d x

i e

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Path Compression Find Interlude: A Really Slow Function

PC-Find(i : index) { Ackermann’s function is a really big function A( x, y) with r := i; inverse α(x, y) which is really small while up[r] ≠ -1 do //find root r := up[r]; How fast does α(x, y) grow? // Assert: r= the root, up[r] = -1 α ≠ (x, y) = 4 for x far larger than the number of atoms in the if i r then // if i was not a root universe (2 300 ) temp := up[i]; while temp ≠ r do // compress path α shows up in: up[i] := r; – Computation Geometry (surface complexity) i := temp; – Combinatorics of sequences temp := up[temp]

return(r) (New?) runtime for Find: } 10/29/2010 49 10/29/2010 50

8 A More Comprehensible Slow Function Complex Complexity of Union-by-Size + Path Compression log* x = number of times you need to compute log to bring value down to at most 1 Tarjan proved that, with these optimizations, p union and find operations on a set of n elements have worst case complexity of O( p ⋅ α(p, n)) E.g. log* 2 = 1 log* 4 = log* 2 2 = 2 log* 16 = log* 2 22 = 3 (log log log 16 = 1) For all practical purposes this is amortized constant time: 2 log* 65536 = log* 2 22 = 4 (log log log log 65536 = 1) O( p ⋅ 4) for p operations! log* 2 65536 = …………… = 5 • Very complex analysis – worse than splay tree analysis etc. that Take this: α(m,n) grows even slower than log* n !! we skipped!

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Disjoint Union / Find Amortized Complexity with Weighted Union and PC • Worst case time complexity for a W-Union is O(1) • For disjoint union / find with weighted union and and for a PC-Find is O(log n). path compression. • Time complexity for m ≥ n operations on n – average time per operation is essentially a constant. elements is O(m log* n) where log* n is a very – worst case time for a PC-Find is O(log n). slow growing function. • An individual operation can be costly, but over – Log * n < 7 for all reasonable n. Essentially constant time the average cost per operation is not. time per operation!

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