Graph Algorithms

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Graph Algorithms
3/7/2007
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Minimum Spanning Trees (MST) Union - Find Dana
Shapira
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Spanning tree
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????? ?? ?? ??????? (??? ?? ????? ??????? ????).
A spanning tree of G is a subset T ? E of edges,
such that the sub-graph G'(V,T) is connected and
acyclic.
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Minimum Spanning Tree
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?????? ??? ??? ??????? (MST)
Given a graph G (V, E) and an assignment of
weights w(e) to the edges of G, a minimum
spanning tree T of G is a spanning tree with
minimum total edge weight
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How To Build A Minimum Spanning Tree
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???? ???.

• General strategy
• Maintain a set of edges A such that (V, A) is a
  spanning forest of G and such that there exists a
  MST (V, F) of G such that A?F.
• As long as (V, A) is not a tree, find an edge
  that can be added to A while maintaining the
  above property.
• Generic-MST(G(V,E))
• A ?
• while (A is not a spanning tree of G) do
• choose a safe edge e(u,v)?E
• AA?e
• return A

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Cuts
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A cut (X, Y) of a graph G (V, E) is a partition
of the vertex set V into two sets X and Y V \
X. An edge (v, w) is said to cross the cut (X, Y)
if v ? X and w ?Y.
A cut (X, Y) respects a set A of edges if no edge
in A crosses the cut.
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A Cut Theorem
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A Cut Theorem
Theorem Let A be a subset of the edges of some
minimum spanning tree of G let (X, Y) be a cut
that respects A and let e be a minimum weight
edge that crosses (X, Y). Then A ?e is also a
subset of the edges of a minimum spanning tree of
G edge e is safe.
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??? ????? T. T ???? ??? ????? ????????.
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A Cut Theorem
Theorem Let A be a subset of the edges of some
minimum spanning tree of G let (X, Y) be a cut
that respects A and let e be a minimum weight
edge that crosses (X, Y). Then A ?e is also a
subset of the edges of a minimum spanning tree of
G edge e is safe.
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A Cut Theorem
u
e
e
v
f
T
w(e) w(f)
w(e) w(f) w(T') w(T)
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Proof

• Let T be a MST such that A?T.
• If e (u,v) ? T, add e to T.
• The edge e (u,v) forms a cycle with edges on
  the path p from u to v in T. Since u and v are on
  opposite sides of the cut, there is at least one
  edge f (x,y) in T on the path p that also
  crosses the cut.
• f ?A since the cut respects A. Since f is on the
  unique path from u to v in T, removing it breaks
  T into two components.
• w(e) w(f) (why?)
• Let T ' T f ?e ?w(T ') w(T).

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A Cut Theorem
Corollary Let G(V,E) be a connected undirected
graph and A a subset of E included in a minimum
spanning tree T for G, and let C(VC,EC) be a
tree in the forest GA(V,A). If e is a light edge
connecting C to some other component in GA, then
e is safe for A.

• Proof The cut (VC, VVC) respects A, and e is a
  light edge for this cut. Therefore, e is safe.

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Kruskals Algorithm
Kruskal(G) 1 A ? Ø 2 for every edge e (v, w) of
G, sorted by weight 3 do if v and w belong to
different connected components of (V, A) 4 then
add edge e to A
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A. ?????? ???? ??????? e.
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a
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(a, d)1 (h, i)1 (c, e)1 (f, h)2 (g, h)2 (b,
c)3 (b, f)3 (b, e)4 (c, d)5 (f, g)5 (e,
i)6 (d, g)8 (a, b)9 (c, f)12
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Correctness Proof
Sorted edge sequence e1, e2, e3, e4, e5, e6, ,
ei, ei 1, ei 2, ei 3, , en
Every edge ej that cross the cut have a weight
w(ej) w(ei).
Hence, edge ei is safe.
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Union-Find Data Structures
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???? ????? ??????? union-find

• Given a set S of n elements, maintain a partition
  of S into subsets S1, S2, , Sk
• Support the following operations
• Union(x, y) Replace sets Si and Sj such that x
  ? Si and y ? Sj with Si ? Sj in the current
  partition
• Find(x) Returns a member r(Si) of the set Si
  that contains x
• In particular, Find(x) and Find(y) return the
  same element if and only if x and y belong to the
  same set.
• It is possible to create a data structure that
  supports the above operations in O(a(n))
  amortized time, where a is the inverse Ackermann
  function.

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Kruskals Algorithm Using Union-Find Data
Structure
Kruskal(G,w) A ? ? for each vertex v?V
do Make-Set(v) sort the edges in E in
non-decreasing weight order w for each edge
(u,v)?E do if Find-Set(u) ? Find-Set(v)
then A ? A ? (u,v)
Union(u,v) return A
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Kruskals Algorithm Using Union-Find Data
Structure
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• Analysis
• O(E log E) time for everything except the
  operations on S
• Cost of operations on S
• O(a(E,V)) amortized time per operation on S
• V 1 Union operations
• E Find operations
• Total O((V E)a(E,V)) running time
• Total running time O(E lg E).

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Union/Find
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• Assumptions
• The Sets are disjoint.
• Each set is identified by a representative of the
  set.
• Initial state
• A union/find structure begins with n elements,
  each considered to be a one element set.
• Functions
• Make-Set(x) Creates a new set with element x in
  it.
• Union(x,y) Make one set out of the sets
  containing x and y.
• Find-Set(x) Returns a pointer to the
  representative of the set containing x.

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Basic Notation

• The elements in the structure will be numbered 0
  to n-1
• Each set will be referred to by the number of one
  of the element it contains
• Initially we have sets S0,S1,,Sn-1
• If we were to call Union(S2,S4), these sets would
  be removed from the list, and the new set would
  now be called either S2 or S4
• Notations
• n Make-Set operations
• m total operations
• n?m

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First Attempt
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• Represent the Union/Find structure as an array
  arr of n elements
• arri contains the set number of element i
• Initially, arrii (Make-Set(i))
• Find-Set(i) just returns the value of arri
• To perform Union(Si,Sj)
• For every k such that arrkj, set arrki

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????? ???- O(1)
Union- O(n) ???? ??????? ?? ?? ?????
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Analysis
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• The worst-case analysis
• Find(i) takes O(1) time
• Union(Si,Sj) takes ?(n) time
• A sequence of n Unions will take ?(n2) time

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Second Attempt
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• Represent the Union/Find structure using linked
  lists.
• Each element points to another element of the
  set.
• The representative is the first element of the
  set.
• Each element points to the representative.
• How do we perform Union(Si,Sj)?

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Analysis
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• The worst-case analysis
• Find(i) takes O(1) time
• Make-Set(i) takes O(1) time
• Union(Si,Sj) takes ?(n) time (Why?)
• A sequence of n Unions-Find will take ?(n2) time
  (Example?)

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????? n ??????. Make-set ????? O(n) ??????
??????? ??"? T(n2) ?? union-find. ???? ?? ??????
?????? ?? ???? union(x2,x1)

union(x3,x2) n-1
?????? union(xn,xn-1) ??
? ????? ????????
nO(1) (n-1) T(n) ( O(n) O(n2) ) / 2
T(n2)
???? ???' ???????
union
Make-set
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Up-Trees
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• A simple data structure for implementing disjoint
  sets is the up-tree.
• We visualize each element as a node
• A set will be visualized as a directed tree
• Arrows will point from child to parent
• The set will be referred to by its root

H
X
F
A
W
B
R
H, A and W belong to the same set. H is the
representative
X, B, R and F are in the same set. X is the
representative
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Operations in Up-Trees
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• Follow pointer to representative element.
• find(x)
• if (x?p(x)) // not the representative
• then p(x)?find(p(x))
• return p(x)

P(x)- ??? ?? x
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Union

• Union is more complicated.
• Make one representative element point to the
  other, but which way?Does it matter?

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Union(H, X)
?????
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?????? ??????. ?? ????? ???? ????? ?? find.
H
X
F
X points to H B, R and F are now deeper
A
W
B
R
H
X
F
H points to X A and W are now deeper
A
W
B
R
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A worst case for Union

• Union can be done in O(1), but may cause find to
  become O(n)

A
B
C
D
E
Consider the result of the following sequence of
operations Union (A, B) Union (C, A) Union
(D, C) Union (E, D)
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E
D
C
B
A
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Array Representation of Up-tree
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• Assume each element is associated with an integer
  i0n-1. From now on, we deal only with i.
• Create an integer array, An
• An array entry is the elements parent
• A -1 entry signifies that element i is the
  representative element.

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Array Representation of Up-tree
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• Now the union algorithm might be
• Union(x,y)
• Ay x // attaches y to x
• The find algorithm would be
• find(x)
• if (Ax lt 0)
• return(x)
• else
• return(find(Ax))
• Performance ???

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Analysis
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• Worst case
• Union(Si,Sj) take O(1) time
• Find(i) takes O(n) time
• Can we do better in an amortized analysis?
• What is the maximum amount of time n operations
  could take us?
• Suppose we perform n/2 unions followed by n/2
  finds
• The n/2 unions could give us one tree of height
  n/2-1
• Thus the total time would be n/2 (n/2)(n/2)
  O(n2)
• This strategy doesnt really help

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