CSE 332: Data Structures & Parallelism Lecture 22: Minimum Spanning Trees. Ruth Anderson Winter 2018

Transcription

1 SE 33: Data Structures & Parallelism Lecture : Minimum Spanning Trees Ruth nderson Winter 08

2 Minimum Spanning Trees iven an undirected graph =(V,E), find a graph =(V, E ) such that: E is a subset of E E = V - is connected is a minimum spanning tree. c uv is minimal ( u, v) E' pplications: Example: Electrical wiring for a house or clock wires on a chip Example: road network if you cared about asphalt cost rather than travel time /8/08

3 Student ctivity j 4 7 ind the MST k 9 m 5 n B 7 6 H D 4 E /8/08 3

4 Two Different pproaches Prim s lgorithm lmost identical to Dijkstra s Kruskals s lgorithm ompletely different! /8/08 4

5 Two Different pproaches Prim s lgorithm lmost identical to Dijkstra s One node, grow greedily Kruskals s lgorithm ompletely different! orest of MSTs, Union them together. (Need a new data structure for this) /8/08 5

6 Prim s algorithm Idea: row a tree by picking a vertex from the unknown set that has the smallest cost. Here cost = cost of the edge that connects that vertex to the known set. Pick the vertex with the smallest cost that connects known to unknown. node-based greedy algorithm Builds MST by greedily adding nodes v known /8/08 6

7 Prim s lgorithm vs. Dijkstra s Recall: Dijkstra picked the unknown vertex with smallest cost where cost = distance to the source. Prim s pick the unknown vertex with smallest cost where cost = distance from this vertex to the known set (in other words, the cost of the smallest edge connecting this vertex to the known set) Otherwise identical ompare to slides in Dijkstra lecture! /8/08 7

8 Prim s lgorithm for MST. or each node v, set v.cost = and v.known = false. hoose any node v. (this is like your start vertex in Dijkstra) a) Mark v as known b) or 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 (the MST) c) or each edge (v,u) with weight w, if(w < u.cost) { } u.cost = w; u.prev = v; /8/08 8

9 Example: ind MST using Prim s B 5 E D vertex known? cost prev 0 B D Order added to known set: E /8/08 9

10 Example: ind MST using Prim s 0 B 6 5 D 0 E 5 3 Order added to known set: vertex known? cost prev Y 0 B D E?????? /8/08 0

11 Example: ind MST using Prim s 0 B D 0 E 5 3 Order added to known set:, D 5 vertex known? cost prev Y 0 B D D Y E D 6 D 5 D /8/08

12 Example: ind MST using Prim s 0 B 6 5 D 0 E 5 3 Order added to known set:, D, 5 vertex known? cost prev Y 0 B Y D D Y E D 5 D /8/08

13 Example: ind MST using Prim s 0 B 6 5 D 0 E 5 3 Order added to known set:, D,, E 3 vertex known? cost prev Y 0 B E Y D D Y E Y D 3 E /8/08 3

14 Example: ind MST using Prim s 0 B 6 5 D 0 E 5 3 Order added to known set:, D,, E, B 3 vertex known? cost prev Y 0 B Y E Y D D Y E Y D 3 E /8/08 4

15 Example: ind MST using Prim s 0 B 6 5 D 0 E 5 3 Order added to known set:, D,, E, B, 3 vertex known? cost prev Y 0 B Y E Y D D Y E Y D Y 3 E /8/08 5

16 Example: ind MST using Prim s 0 B 6 5 D 0 E 5 3 Order added to known set:, D,, E, B,, 3 vertex known? cost prev Y 0 B Y E Y D D Y E Y D Y Y 3 E /8/08 6

17 Student ctivity Start with V ind MST using Prim s v 4 4 v v v 3 7 v 5 v V Kwn Distance path 5 8 v 4 6 v v3 v4 v5 v6 v7 V Order Declared Known: Total ost: v 7 /8/08 7

18 Prim s nalysis orrectness?? bit tricky Intuitively similar to Dijkstra Might return to this time permitting (unlikely) Run-time Same as Dijkstra O( E log V ) using a priority queue /8/08 8

19 Kruskal s MST lgorithm Idea: row a forest out of edges that do not create a cycle. Pick an edge with the smallest weight. =(V,E) v /8/08 9

20 Kruskal s lgorithm for MST n edge-based greedy algorithm Builds MST by greedily adding edges. Initialize with empty MST all vertices marked unconnected all edges unmarked. While all vertices are not connected a. Pick the lowest cost edge (u,v) and mark it b. If u and v are not already connected, add (u,v) to the MST and mark u and v as connected to each other /8/08 0

21 side: Union-ind aka Disjoint Set DT Union(x,y) take the union of two sets named x and y iven sets: {3,5,7}, {4,,8}, {9}, {,6} Union(5,) Result: {3,5,7,,6}, {4,,8}, {9}, To perform the union operation, we replace sets x and y by (x y) ind(x) return the name of the set containing x. iven sets: {3,5,7,,6}, {4,,8}, {9}, ind() returns 5 ind(4) returns 8 We can do Union in constant time. We can get ind to be amortized constant time (worst case O(log n) for an individual ind operation). /8/08

22 Kruskal s pseudo code void raph::kruskal(){ int edgesccepted = 0; DisjSet s(num_verties); } while (edgesccepted < NUM_VERTIES ){ e = smallest weight edge not deleted yet; // edge e = (u, v) uset = s.find(u); vset = s.find(v); if (uset!= vset){ edgesccepted++; s.unionsets(uset, vset); } } /8/08

23 Kruskal s pseudo code void raph::kruskal(){ int edgesccepted = 0; DisjSet s(num_verties); while (edgesccepted < NUM_VERTIES ){ e = smallest weight edge not deleted yet; // edge e = (u, v) uset = s.find(u); vset = s.find(v); if (uset!= vset){ edgesccepted++; s.unionsets(uset, vset); } } } E finds V unions E heap ops /8/08 3

24 Kruskal s pseudo code void raph::kruskal(){ } int edgesccepted = 0; DisjSet s(num_verties); while (edgesccepted < NUM_VERTIES ){ } e = smallest weight edge not deleted yet; // edge e = (u, v) uset = s.find(u); vset = s.find(v); if (uset!= vset){ } edgesccepted++; s.unionsets(uset, vset); E log E + E log V + V O( E log E ) = O( E log V ) b/c log E < log V = log V E finds V unions On heap of edges Deletemin = log E E heap ops One for each vertex in the edge ind = log V Union = O() /8/08 4

25 Example: ind MST using Kruskal s D B E 5 3 Edges in sorted order: : (,D), (,D), (B,E), (D,E) : (,B), (,), (,) 3: (E,) 5: (D,), (B,D) 6: (D,) 0: (,) Output: Note: t each step, the union/find sets are the trees in the forest /8/08 5

26 Example: ind MST using Kruskal s D B E 5 3 Edges in sorted order: : (,D), (,D), (B,E), (D,E) : (,B), (,), (,) 3: (E,) 5: (D,), (B,D) 6: (D,) 0: (,) Output: (,D) Note: t each step, the union/find sets are the trees in the forest /8/08 6

27 Example: ind MST using Kruskal s D B E 5 3 Edges in sorted order: : (,D), (,D), (B,E), (D,E) : (,B), (,), (,) 3: (E,) 5: (D,), (B,D) 6: (D,) 0: (,) Output: (,D), (,D) Note: t each step, the union/find sets are the trees in the forest /8/08 7

28 Example: ind MST using Kruskal s D B E 5 3 Edges in sorted order: : (,D), (,D), (B,E), (D,E) : (,B), (,), (,) 3: (E,) 5: (D,), (B,D) 6: (D,) 0: (,) Output: (,D), (,D), (B,E) Note: t each step, the union/find sets are the trees in the forest /8/08 8

29 Example: ind MST using Kruskal s D B E 5 3 Edges in sorted order: : (,D), (,D), (B,E), (D,E) : (,B), (,), (,) 3: (E,) 5: (D,), (B,D) 6: (D,) 0: (,) Output: (,D), (,D), (B,E), (D,E) Note: t each step, the union/find sets are the trees in the forest /8/08 9

30 Example: ind MST using Kruskal s D B E 5 3 Edges in sorted order: : (,D), (,D), (B,E), (D,E) : (,B), (,), (,) 3: (E,) 5: (D,), (B,D) 6: (D,) 0: (,) Output: (,D), (,D), (B,E), (D,E) Note: t each step, the union/find sets are the trees in the forest /8/08 30

31 Example: ind MST using Kruskal s D B E 5 3 Edges in sorted order: : (,D), (,D), (B,E), (D,E) : (,B), (,), (,) 3: (E,) 5: (D,), (B,D) 6: (D,) 0: (,) Output: (,D), (,D), (B,E), (D,E), (,) Note: t each step, the union/find sets are the trees in the forest /8/08 3

32 Example: ind MST using Kruskal s D B E 5 3 Edges in sorted order: : (,D), (,D), (B,E), (D,E) : (,B), (,), (,) 3: (E,) 5: (D,), (B,D) 6: (D,) 0: (,) Output: (,D), (,D), (B,E), (D,E), (,) Note: t each step, the union/find sets are the trees in the forest /8/08 3

33 Example: ind MST using Kruskal s D B E 5 3 Edges in sorted order: : (,D), (,D), (B,E), (D,E) : (,B), (,), (,) 3: (E,) 5: (D,), (B,D) 6: (D,) 0: (,) Output: (,D), (,D), (B,E), (D,E), (,), (E,) Note: t each step, the union/find sets are the trees in the forest /8/08 33

34 Student ctivity ind MST using Kruskal s 3 B H 4 D E 4 Total ost: Now find the MST using Prim s method. Under what conditions will these methods give the same result? /8/08 34

35 D B E 4 H Draw the UpTree Nodes B D E H Parent Size /8/08 35

36 orrectness Kruskal s algorithm is clever, simple, and efficient But does it generate a minimum spanning tree? How can we prove it? irst: it generates a spanning tree Intuition: raph started connected and we added every edge that did not create a cycle Proof by contradiction: Suppose u and v are disconnected in Kruskal s result. Then there s a path from u to v in the initial graph with an edge we could add without creating a cycle. But Kruskal would have added that edge. ontradiction. Second: There is no spanning tree with lower total cost /8/08 36

37 The inductive proof set-up Let (stands for forest ) be the set of edges Kruskal has added at some point during its execution. laim: is a subset of one or more MSTs for the graph (Therefore, once = V -, we have an MST.) Proof: By induction on Base case: =0: The empty set is a subset of all MSTs Inductive case: =k+: By induction, before adding the (k+) th edge (call it e), there was some MST T such that -{e} T /8/08 37

38 Staying a subset of some MST laim: is a subset of one or more MSTs for the graph So far: -{e} T: Two disjoint cases: If {e} T: Then T and we re done Else e forms a cycle with some simple path (call it p) in T Must be since T is a spanning tree /8/08 38

39 Staying a subset of some MST laim: is a subset of one or more MSTs for the graph So far: -{e} T and e forms a cycle with p T e There must be an edge e on p such that e is not in Else Kruskal would not have added e laim: e.weight == e.weight /8/08 39

40 Staying a subset of some MST laim: is a subset of one or more MSTs for the graph So far: -{e} T e forms a cycle with p T e on p is not in e e laim: e.weight == e.weight If e.weight > e.weight, then T is not an MST because T-{e}+{e} is a spanning tree with lower cost: contradiction If e.weight < e.weight, then Kruskal would have already considered e. It would have added it since T has no cycles and -{e} T. But e is not in : contradiction /8/08 40

41 Staying a subset of some MST laim: is a subset of one or more MSTs for the graph So far: -{e} T e forms a cycle with p T e on p is not in e.weight == e.weight e e laim: T-{e}+{e} is an MST It s a spanning tree because p-{e}+{e} connects the same nodes as p It s minimal because its cost equals cost of T, an MST Since T-{e}+{e}, is a subset of one or more MSTs Done. /8/08 4