Minimum Trees. The problem. connected graph G = (V,E) each edge uv has a positive weight w(uv)

Transcription

1 /16/015 Minimum Trees Setting The problem connected graph G = (V,E) each edge uv has a positive weight w(uv) Find a spanning graph T of G having minimum total weight T is a tree w T = w uv uv E(T) is minimal T is known as a Minimum Spanning Tree 1

2 /16/015 MSTs are not SPTs SPT MST SPT Generic greedy algorithm Let A be a subset of edges of a minimum spanning tree An edge uv is safe for A if A {uv} is also a subset of a minimum spanning tree A := while (V,A) does not form a spanning tree find an edge uv that is safe for A A := A {uv}

3 /16/015 Adding safe edges Definitions Cut is partition of the node set into two sets An edge crosses a cut if its nodes are in different sets of a partition A cut respects a set of edges if none crosses the cut Proposition A is a subset of edges of a minimum spanning tree (S, V - S) is any cut that respects A uv is a lightest edge crossing (S, V - S) uv is a safe edge for A Proof x y set of nodes S minimum spanning tree T set of edges A lightest edge crossing (S, V S) u v uv is a lightest edge crossing (S, V S) xy is the edge in the unique tree path between u to v that crosses (S, V S) T = T {uv} {xy} is also a spanning tree and contains uv definition of T implies w(t ) w(t) w(uv) w(xy) implies w(t ) w(t) w(t ) = w(t) 3

4 /16/015 Prim s algorithm Prim(G, c) set A is {{v, parent[v]}: v V Q {s}} for each v key[v] := ; parent[v] := NIL key[s] := 0 Q := V while Q select a node u of Q for which key[u] is smallest Q := Q {u} for each uv if v Q and c(uv) < key[v] key[v] := c(u,v) parent[v] := u complexity O(m log(n)) Prim s algorithm: example I

5 /16/015 Prim s algorithm: example II Kruskal s algorithm Kruskal(G, c) A := for each v V MakeSet(v) sort the edges E into nondecreasing order by weight for each uv E, taken in nondecreasing order of weight a := FindSet(u); b := FindSet(v) if a b A := A {uv} Union(a,b) 5

6 /16/015 Kruskal s algorithm: example I Kruskal s algorithm: example II

7 /16/015 Data structure for disjoint sets v ranks are indicated as superscripts y 0 x 1 z 1 u 0 w 0 The nodes of a disjoint set (connected component) form an inbranching with a self-loop at the root Each node has a rank that corresponds to length of the longest path to it in the in-branching Operations on disjoint sets MakeSet(v) parent[v] := v rank[v] := 0 FindSet(v) while parent[v] v v := parent[v] return v Union(a, b) if a b if rank[a] > rank[b] parent[b] := a else parent[a] := b if rank[a] = rank[b] rank[b] := rank[b] +1

8 /16/015 Example - I MakeSet(u), MakeSet(u),, MakeSet(w) u 0 v 0 x 0 y 0 z 0 w 0 Union(y, u), Union(z, v), Union(w,x) u 1 v 1 x 1 y 0 z 0 w 0 Example - II u x 1 Union(v, u) y 0 v 1 w 0 z 0 Union(x, u) u y 0 v 1 x 1 z 0 w 0

9 /16/015 Properties Any node of rank k has at least k descendants (proof by induction) There can be at most n / k nodes of rank k, where n is the number of nodes in the graph The maximum rank is log n Complexity of Kruskal s algorithm O(m log m) = O(m log n) for ordering the weights O(n) for MakeSet O(m) FindSet operations; each FindSet operation is O(log n) n-1 Union operations Complexity is O(m log n)

10 /16/015 Path compression FindSet(v) if parent[v] v parent[v] := FindSet(parent[v]) return parent[v] u 3 FindSet(s) u 3 y 0 v x 1 y 0 v x 1 z 1 s 0 z 1 w 0 w 0 t 0 s 0 t 0 Clustering Classify a set objects into clusters of similarity Distance function on pairs of objects represents similarity Partition of the objects such that objects in the same cluster are similar and objects in different clusters are dissimilar in some sense A whole area of knowledge: we just touch the surface

11 /16/015 The problem A set U of n objects, p 1, p,, p n A distance function d(p i, p j ) d(p i, p i ) = 0 d(p i, p j ) = d(p j, p i ) d(p i, p j ) > 0 for i j Spacing of a clustering is the minimum distance between pairs of objects lying in different classes Seek the clustering with k clusters of maximum spacing Example Distances are euclidean

12 /16/015 Algorithm for a k-clustering Run Kruskal s algorithm and stop just before it would start adding the k 1 last edges Run a minimum spanning tree algorithm and delete from the tree the k 1 heavier edges Proof of correctness Let C = (C 1, C,, C k ) be the clustering found by the algorithm; the spacing d(c) of C is the weight of the (k 1) st weightier edge of the minimum spanning tree Let C = (C 1, C,, C k ) be the any other clustering with k clusters; we have to show that its spacing d(c ) is at most d(c) There must exist a cluster C r that is not contained in any C k There are p and q belonging to the same C r, but belonging to different C s and C t There is a unique path within C r joining p to q; there is an edge p q where p belongs to C s and q belongs to C t The weight of p q is at most the spacing of C (d(c) d(p,q )) The spacing of C is at most the weight of p q (d(p,q ) d(c )) 1

13 /16/015 All pairs maximum capacity tree Each edge in the graph is has positive capacity Is there a spanning tree such that all paths in the tree are of maximum capacity? Yes Use Kruskal s algorithm with edges ordered from higher capacity to lower capacity How would you argue that Kruskal s algorithm as described solves the all pairs maximum capacity tree? Example u 5 v x y 15 0 w 15 z uxyv is a path of maximum capacity (15) between u to v wx is a path of maximum capacity (15) between w and x zyv is a path of maximum capacity (0) from z to v 13