CSE 326: Data Structures Part 7: The Dynamic (Equivalence) Duo: Weighted Union & Path CompressionToday’s OutlineWhat’s a Good Maze?Slide 4The Maze Construction ProblemThe Middle of the MazeMaze Construction AlgorithmEquivalence RelationsSlide 9Disjoint Set Union/Find ADTExampleExample, First StepExample, ContinuedUp-Tree IntuitionUp-Tree Union-Find Data StructureFindUnionFor Your Reading Pleasure...The Whole Example (1/11)The Whole Example (2/11)The Whole Example (3/11)The Whole Example (4/11)The Whole Example (5/11)The Whole Example (6/11)The Whole Example (7/11)The Whole Example (8/11)The Whole Example (9/11)The Whole Example (10/11)The Whole Example (11/11)Nifty storage trickImplementationRoom for Improvement: Weighted UnionWeighted Union CodeWeighted Union Find AnalysisAlternatives to Weighted UnionRoom for Improvement: Path CompressionPath Compression ExamplePath Compression CodeDigression: Inverse Ackermann’sComplex Complexity of Weighted Union + Path CompressionTo DoCSE 326: Data StructuresPart 7: The Dynamic (Equivalence) Duo:Weighted Union & Path CompressionHenry KautzAutumn Quarter 2002Whack!!ZINGPOWBAM!Today’s Outline•Making a “good” maze•Disjoint Set Union/Find ADT•Up-trees•Weighted Unions•Path CompressionWhat’s a Good Maze?What’s a Good Maze?1. Connected2. Just one path between any two rooms3. RandomThe Maze Construction Problem•Given: –collection of rooms: V–connections between rooms (initially all closed): E•Construct a maze:–collection of rooms: V = V–designated rooms in, iV, and out, oV–collection of connections to knock down: E  Esuch that one unique path connects every two roomsThe Middle of the Maze•So far, a number of walls have been knocked down while others remain.•Now, we consider the wall between A and B.•Should we knock it down?When should we not knock it?ABMaze Construction AlgorithmWhile edges remain in E Remove a random edge e = (u, v) from EHow can we do this efficiently?If u and v have not yet been connected-add e to E-mark u and v as connectedHow to check connectedness efficiently?Equivalence RelationsAn equivalence relation R must have three properties–reflexive:–symmetric:–transitive:Connection between rooms is an equivalence relation–Why?Equivalence RelationsAn equivalence relation R must have three properties–reflexive: for any x, xRx is true–symmetric: for any x and y, xRy implies yRx–transitive: for any x, y, and z, xRy and yRz implies xRzConnection between rooms is an equivalence relation–any room is connected to itself–if room a is connected to room b, then room b is connected to room a–if room a is connected to room b and room b is connected to room c, then room a is connected to room cDisjoint Set Union/Find ADT•Union/Find operations–create–destroy–union–find•Disjoint set partition property: every element of a DS U/F structure belongs to exactly one set with a unique name•Dynamic equivalence property: Union(a, b) creates a new set which is the union of the sets containing a and b{1,4,8}{7}{6}{5,9,10}{2,3}find(4)8union(3,6){2,3,6}name of setExampleConstruct the maze on the rightInitial (the name of each set is underlined):{a}{b}{c}{d}{e}{f}{g}{h}{i}Randomly select edge 1Order of edges in blueadbecfg h i32411101796812 5Example, First Step{a}{b}{c}{d}{e}{f}{g}{h}{i}find(b)  bfind(e)  efind(b)  find(e) so:add 1 to Eunion(b, e){a}{b,e}{c}{d}{f}{g}{h}{i}adbecfg h iOrder of edges in blue32411101796812 5Example, Continued{a}{b,e}{c}{d}{f}{g}{h}{i}Order of edges in blueadbecfg h i3241110796812 5Up-Tree IntuitionFinding the representative member of a set is somewhat like the opposite of finding whether a given key exists in a set.So, instead of using trees with pointers from each node to its children; let’s use trees with a pointer from each node to its parent.Up-Tree Union-Find Data Structure•Each subset is an up-tree with its root as its representative member•All members of a given set are nodes in that set’s up-tree•Hash table maps input data to the node associated with that dataa c g hd beUp-trees are not necessarily binary!f iFinda c g hd bef ifind(f)find(e)adbecfg h i111079812Just traverse to the root!runtime:Uniona c g hd bef iunion(a,c)adbecfg h i11109812Just hang one root from the other!runtime:For Your Reading Pleasure...The Whole Example (1/11)ef g ha b c d iunion(b,e)e f g ha b c d iadbecfg h i32411101796812 5The Whole Example (2/11)union(a,d)adbecfg h i3241110796812 5ef g ha b c d if g ha b c id eThe Whole Example (3/11)union(a,b)adbecfg h i341110796812 5f g ha b c id ef g habc ideThe Whole Example (4/11)find(d) = find(e)No union!adbecfg h i41110796812 5f g habc ideWhile we’re finding e, could we do anything else?The Whole Example (5/11)union(h,i)adbecfg h i1110796812 5f g habc idef g habcideThe Whole Example (6/11)union(c,f)adbecfg h i1110796812f g habcidefg habcideThe Whole Example (7/11)find(e)find(f)union(a,c)adbecfg h i111079812fg habcidefg habcideCould we do a better job on this union?The Whole Example (8/11)adbecfg h i11109812fghabcidefg habcidefind(f)find(i)union(c,h)The Whole Example (9/11)find(e) = find(h) and find(b) = find(c)So, no unions for either of these.adbecfg h i1110912fghabcideThe Whole Example (10/11)find(d)find(g)union(c, g)adbecfg h i1112fghabcidefghabcideThe Whole Example (11/11)find(g) = find(h) So, no union.And, we’re done!adbecfg h i12fghabcideadbecfg h iOoh… scary!Such a hard maze!fg habcide0 -1 0 1 2 -1 -1 7-10 (a) 1 (b) 2 (c) 3 (d) 4 (e) 5 (f) 6 (g) 7 (h) 8 (i)Nifty storage trickA forest of up-trees can easily be stored in an array.Also, if the node names are integers or characters, we can use a very simple, perfect hash.up-index:ImplementationID find(Object x) { assert(HashTable.contains(x)); ID xID = HashTable[x]; while(up[xID] != -1) { xID = up[xID]; } return xID;}ID union(Object x, Object y) { ID rootx = find(x); ID rooty = find(y); assert(rootx != rooty); up[y] = x;}typedef ID int;ID up[10000];runtime: O(depth) or … runtime: O(1)Room for Improvement:Weighted Union•Always makes the root of the larger tree the new root•Often cuts down on height of the new up-treefg habcidefg habcideCould we do a better job on this union?Weighted union!fg habcideWeighted Union CodeID union(Object x, Object y) { rx = Find(x); ry = Find(y); assert(rx != ry); if (weight[rx] > weight[ry]) { up[ry] = rx; weight[rx] += weight[ry]; } else { up[rx] = ry;