Announcements. HW4 due Friday. Summer 2016 CSE373: Data Structures & Algorithms 1
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1 Announcements HW4 due Friday Summer 0
2 Minimum Spanning Trees Hunter Zahn Summer 0 Summer 0
3 Spanning Trees A simple problem: Given a connected undirected graph G=(V,E), find a minimal subset of edges such that G is spll connected A graph G=(V,E) such that G is connected and removing any edge from E makes G disconnected Summer 0 3
4 ObservaPons. Any solupon to this problem is a tree Recall a tree does not need a root; just means acyclic For any cycle, could remove an edge and spll be connected. SoluPon not unique unless original graph was already a tree 3. Problem ill- defined if original graph not connected So E >= V - 4. A tree with V nodes has V - edges So every solupon to the spanning tree problem has V - edges Summer 0 4
5 MoPvaPon A spanning tree connects all the nodes with as few edges as possible Example: A phone tree so everybody gets the message and no unnecessary calls get made Bad example since would prefer a balanced tree In most compelling uses, we have a weighted undirected graph and we want a tree of least total cost Example: Electrical wiring for a house or clock wires on a chip Example: A road network if you cared about asphalt cost rather than travel Pme This is the minimum spanning tree problem Will do that next, a`er intuipon from the simpler case Summer 0 5
6 Two Approaches Different algorithmic approaches to the spanning- tree problem:. Do a graph traversal (e.g., depth- first search, but any traversal will do), keeping track of edges that form a tree. Iterate through edges; add to output any edge that does not create a cycle Summer 0
7 Spanning tree via DFS spanning_tree(graph G) { for each node i: i.marked = false for some node i: f(i) } f(node i) { i.marked = true for each j adjacent to i: if(!j.marked) { add(i,j) to output f(j) // DFS } } Correctness: DFS reaches each node. We add one edge to connect it to the already visited nodes. Order affects result, not correctness. Time: O( E ) Summer 0 7
8 Stack f() Output: Summer 0 8
9 Stack (boeom) 3 f() 7 f() 4 5 Output: (,) Summer 0 9
10 Stack (boeom) 3 f() f() 7 f(7) 4 5 Output: (,), (,7) Summer 0 0
11 Stack (boeom) 3 f() f() 7 f(7) 4 f(5) 5 Output: (,), (,7), (7,5) Summer 0
12 Stack (boeom) f() f() f(7) f(5) f(4) Output: (,), (,7), (7,5), (5,4) Summer 0
13 Stack (boeom) f() f() f(7) f(5) f(4) f(3) Output: (,), (,7), (7,5), (5,4),(4,3) Summer 0 3
14 Stack (boeom) f() f() f(7) f(5) f(4) f() f(3) Example 7 5 Output: (,), (,7), (7,5), (5,4), (4,3), (5,) 3 4 Summer 0 4
15 Stack (boeom) f() f() f(7) f(5) f(4) f() f(3) Example 7 5 Output: (,), (,7), (7,5), (5,4), (4,3), (5,) 3 4 Summer 0 5
16 Second Approach Iterate through edges; output any edge that does not create a cycle Correctness (hand- wavy): Goal is to build an acyclic connected graph When we add an edge, it adds a vertex to the tree Else it would have created a cycle The graph is connected, so we reach all verpces Efficiency: Depends on how quickly you can detect cycles Reconsider a`er the example Summer 0
17 Edges in some arbitrary order: (,), (3,4), (5,), (5,7),(,5), (,), (,7), (,3), (4,5), (4,7) Output: Summer 0 7
18 Edges in some arbitrary order: (,), (3,4), (5,), (5,7),(,5), (,), (,7), (,3), (4,5), (4,7) Output: (,) 3 Summer 0 8
19 Edges in some arbitrary order: (,), (3,4), (5,), (5,7),(,5), (,), (,7), (,3), (4,5), (4,7) Output: (,), (3,4) 3 Summer 0 9
20 Edges in some arbitrary order: (,), (3,4), (5,), (5,7),(,5), (,), (,7), (,3), (4,5), (4,7) Output: (,), (3,4), (5,), 3 Summer 0 0
21 Edges in some arbitrary order: (,), (3,4), (5,), (5,7),(,5), (,), (,7), (,3), (4,5), (4,7) Output: (,), (3,4), (5,), (5,7) Summer 0
22 Edges in some arbitrary order: (,), (3,4), (5,), (5,7), (,5), (,), (,7), (,3), (4,5), (4,7) Output: (,), (3,4), (5,), (5,7), (,5) Summer 0
23 Edges in some arbitrary order: (,), (3,4), (5,), (5,7), (,5), (,), (,7), (,3), (4,5), (4,7) Output: (,), (3,4), (5,), (5,7), (,5) 3 Summer 0 3
24 Edges in some arbitrary order: (,), (3,4), (5,), (5,7), (,5), (,), (,7), (,3), (4,5), (4,7) Output: (,), (3,4), (5,), (5,7), (,5) 3 Summer 0 4
25 Edges in some arbitrary order: (,), (3,4), (5,), (5,7), (,5), (,), (,7), (,3), (4,5), (4,7) Output: (,), (3,4), (5,), (5,7), (,5), (,3) 5 Can stop once we have V - edges Summer 0 5
26 Cycle DetecPon To decide if an edge could form a cycle is O( V ) because we may need to traverse all edges already in the output So overall algorithm would be O( V E ) But there is a faster way we know: use union- find! IniPally, each item is in its own - element set Union sets when we add an edge that connects them Stop when we have one set Summer 0
27 Using Disjoint- Set Can use a disjoint- set implementapon in our spanning- tree algorithm to detect cycles: Invariant: u and v are connected in output- so- far iff u and v in the same set IniPally, each node is in its own set When processing edge (u,v): If find(u) equals find(v), then do not add the edge Else add the edge and union(find(u),find(v)) O( E ) operapons that are almost O() amorpzed Summer 0 7
28 Summary So Far The spanning- tree problem Add nodes to parpal tree approach is O( E ) Add acyclic edges approach is almost O( E ) Using union- find as a black box But really want to solve the minimum- spanning- tree problem Given a weighted undirected graph, give a spanning tree of minimum weight Same two approaches will work with minor modificapons Both will be O( E log V ) Summer 0 8
29 Gelng to the Point Algorithm # Shortest- path is to Dijkstra s Algorithm as Minimum Spanning Tree is to Prim s Algorithm (Both based on expanding cloud of known verpces, basically using a priority queue instead of a DFS stack) Algorithm # Kruskal s Algorithm for Minimum Spanning Tree is Exactly our nd approach to spanning tree but process edges in cost order Summer 0 9
30 Prim s Algorithm Idea Idea: Grow a tree by adding an edge from the known verpces to the unknown verpces. Pick the edge with the smallest weight that connects known to unknown. Recall Dijkstra picked edge with closest known distance to source That is not what we want here Otherwise idenpcal (!) Summer 0 30
31 The Algorithm. For each node v, set v.cost = and v.known = false. Choose any node v a) Mark v as known b) For each edge (v,u) with weight w, set u.cost=w and u.prev=v 3. While there are unknown nodes in the graph a) Select the unknown node v with lowest cost b) Mark v as known and add (v, v.prev) to output c) For each edge (v,u) with weight w, if(w < u.cost) { u.cost = w; u.prev = v; } Summer 0 3
32 A C F B 5 D 0 E 5 3 G vertex known? cost prev A?? B?? C?? D?? E?? F?? G?? Summer 0 3
33 0 A B 5 C D 0 F E 5 3 G vertex known? cost prev A Y 0 B A C A D A E?? F?? G?? Summer 0 33
34 0 A B 5 C D 0 F E G vertex known? cost prev A Y 0 B A C D D Y A E D F D G 5 D Summer 0 34
35 0 A B 5 C D 0 F E G vertex known? cost prev A Y 0 B A C Y D D Y A E D F C G 5 D Summer 0 35
36 0 A B 5 C D 0 F E G vertex known? cost prev A Y 0 B E C Y D D Y A E Y D F C G 3 E Summer 0 3
37 0 A B 5 C D 0 F E G vertex known? cost prev A Y 0 B Y E C Y D D Y A E Y D F C G 3 E Summer 0 37
38 0 A B 5 C D 0 F E G vertex known? cost prev A Y 0 B Y E C Y D D Y A E Y D F Y C G 3 E Summer 0 38
39 0 A B 5 C D 0 F E G vertex known? cost prev A Y 0 B Y E C Y D D Y A E Y D F Y C G Y 3 E Summer 0 39
40 Analysis Correctness?? A bit tricky IntuiPvely similar to Dijkstra Run- Pme Same as Dijkstra O( E log V ) using a priority queue Costs/prioriPes are just edge- costs, not path- costs Summer 0 40
41 Kruskal s Algorithm Idea: Grow a forest out of edges that do not grow a cycle, just like for the spanning tree problem. But now consider the edges in order by weight So: Sort edges: O( E log E ) (next course topic) Iterate through edges using union- find for cycle detecpon almost O( E ) Somewhat beeer: Floyd s algorithm to build min- heap with edges O( E ) Iterate through edges using union- find for cycle detecpon and deletemin to get next edge O( E log E ) Not beeer worst- case asymptopcally, but o`en stop long before considering all edges Summer 0 4
42 Pseudocode. Sort edges by weight (beeer: put in min- heap). Each node in its own set 3. While output size < V - Consider next smallest edge (u,v) if find(u,v) indicates u and v are in different sets output (u,v) union(find(u),find(v)) Recall invariant: u and v in same set if and only if connected in output- so- far Summer 0 4
43 C A F D B 5 0 E 5 3 G Edges in sorted order: : (A,D), (C,D), (B,E), (D,E) : (A,B), (C,F), (A,C) 3: (E,G) 5: (D,G), (B,D) : (D,F) 0: (F,G) Output: Note: At each step, the union/find sets are the trees in the forest Summer 0 43
44 C A F D B 5 0 E 5 3 G Edges in sorted order: : (A,D), (C,D), (B,E), (D,E) : (A,B), (C,F), (A,C) 3: (E,G) 5: (D,G), (B,D) : (D,F) 0: (F,G) Output: (A,D) Note: At each step, the union/find sets are the trees in the forest Summer 0 44
45 C A F D B 5 0 E 5 3 G Edges in sorted order: : (A,D), (C,D), (B,E), (D,E) : (A,B), (C,F), (A,C) 3: (E,G) 5: (D,G), (B,D) : (D,F) 0: (F,G) Output: (A,D), (C,D) Note: At each step, the union/find sets are the trees in the forest Summer 0 45
46 C A F D B 5 0 E 5 3 G Edges in sorted order: : (A,D), (C,D), (B,E), (D,E) : (A,B), (C,F), (A,C) 3: (E,G) 5: (D,G), (B,D) : (D,F) 0: (F,G) Output: (A,D), (C,D), (B,E) Note: At each step, the union/find sets are the trees in the forest Summer 0 4
47 C A F D B 5 0 E 5 3 G Edges in sorted order: : (A,D), (C,D), (B,E), (D,E) : (A,B), (C,F), (A,C) 3: (E,G) 5: (D,G), (B,D) : (D,F) 0: (F,G) Output: (A,D), (C,D), (B,E), (D,E) Note: At each step, the union/find sets are the trees in the forest Summer 0 47
48 C A F D B 5 0 E 5 3 G Edges in sorted order: : (A,D), (C,D), (B,E), (D,E) : (A,B), (C,F), (A,C) 3: (E,G) 5: (D,G), (B,D) : (D,F) 0: (F,G) Output: (A,D), (C,D), (B,E), (D,E) Note: At each step, the union/find sets are the trees in the forest Summer 0 48
49 C A F D B 5 0 E 5 3 G Edges in sorted order: : (A,D), (C,D), (B,E), (D,E) : (A,B), (C,F), (A,C) 3: (E,G) 5: (D,G), (B,D) : (D,F) 0: (F,G) Output: (A,D), (C,D), (B,E), (D,E), (C,F) Note: At each step, the union/find sets are the trees in the forest Summer 0 49
50 C A F D B 5 0 E 5 3 G Edges in sorted order: : (A,D), (C,D), (B,E), (D,E) : (A,B), (C,F), (A,C) 3: (E,G) 5: (D,G), (B,D) : (D,F) 0: (F,G) Output: (A,D), (C,D), (B,E), (D,E), (C,F) Note: At each step, the union/find sets are the trees in the forest Summer 0 50
51 C A F D B 5 0 E 5 3 G Edges in sorted order: : (A,D), (C,D), (B,E), (D,E) : (A,B), (C,F), (A,C) 3: (E,G) 5: (D,G), (B,D) : (D,F) 0: (F,G) Output: (A,D), (C,D), (B,E), (D,E), (C,F), (E,G) Note: At each step, the union/find sets are the trees in the forest Summer 0 5
52 Kruskal s Algorithm: Correctness It clearly generates a spanning tree. Call it T K. Suppose T K is not minimum: Pick another spanning tree T min with lower cost than T K Pick the smallest edge e =(u,v) in T K that is not in T min T min already has a path p in T min from u to v Adding e to T min will create a cycle in T min Pick an edge e in p that Kruskal s algorithm considered a;er adding e (must exist: u and v unconnected when e considered) cost(e ) cost(e ) can replace e with e in T min without increasing cost! Keep doing this unpl T min is idenpcal to T K T K must also be minimal contradicpon! Summer 0 5
53 MST ApplicaPon: Clustering Given a collecpon of points in an r- dimensional space, and an integer K, divide the points into K sets that are closest together Summer 0 53
54 Distance clustering Divide the data set into K subsets to maximize the distance between any pair of sets dist (S, S ) = min {dist(x, y) x in S, y in S } Summer 0 54
55 Divide into clusters Summer 0 55
56 Divide into 3 clusters Summer 0 5
57 Divide into 4 clusters Summer 0 57
58 Distance Clustering Algorithm Let C = {{v }, {v },..., {v n }}; T = { } while C > K Let e = (u, v) with u in C i and v in C j be the minimum cost edge joining distinct sets in C Replace C i and C j by C i U C j Summer 0 58
59 K- clustering Summer 0 59
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