Min Spanning

8/12/2019 Min Spanning

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Subhash Suri UC Santa Barbara

Minimum Spanning Trees

Given an undirected graph G= (V, E),with edge costs cij.

A spanning tree T of G is a cycle-freesubgraph that spans all the nodes.

The cost of T is the sum of the costs ofthe edges in T.

MST is the smallest cost spanning tree.

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Graph G MST


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Applications of MST

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Graph G MST

Direct applications: interconnection ofentities.

1. electrical devices (circuit boards)2. utilities (gas, oil)3. computers or communication devices by

high speed lines.4. cable service customers

Indirect applications.

1. Optimal message passing.2. Data storage.3. Cluster analysis


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Optimality Conditions

Greedy incremental flavor: add one edgeat a time.

Each step colors an edge of G blue(accept) orred (reject).

ColorInvariant (CI): MST containingall blue edges, and no red edges.

Recall that a cut in G= (V, E) is apartition of it vertices (X, V X).


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Blue and Red Rules

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Cut Rule (Blue) Red Rule (Cycle)

Blue (Cut) Rule: Select a cut not crossed byany blue edge. Among the uncolored edgescrossing the cut, make the minimum cost

edge blue.

Red (Cycle) Rule: Select a simple cycle with

no red edges. Among all uncolored edges of the cycle, make the maximum cost one red.


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Generic MST Algorithm

Theorem: Apply red and blue rules inarbitrary order until neither rule applies.

The resulting set of blue edges forms a MST.

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Graph G MST

Proof has two parts:

1. The Color Invariant (CI) holds.

2. All edges are ultimately colored.


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Correctness of Blue Rule

Let T be the MST guaranteed by the CIbefore the last coloring step.

ee

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MST T*

MST satisfyingcolor invariant

Cut

Suppose the last step was to color e blue.Consider the cut (X, V X) to which bluerule applied. Some edge e of T mustcross this cut.

The graph T + e contains a cycle

containing both e and e, andcost(e)cost(e). (Why?)

So T + e e is also a MST.


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Correctness of Red Rule

MST T*

MST satisfyingcolor invariant

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Cycle

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Suppose edge e colored red. If eT, thenT still satisfies CI. Otherwise, consider

T e. It has two components.

The cycle used in coloring e has some edgee with one end in each of thesecomponents.

By choice, cost(e

)cost(e). Thus, the treeT e + e is also an MST.


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Proof of Completion

Blue edges form a forest.

Suppose edge e left uncolored, and neitherrule applies.

If both endpoints ofe in same blue tree B,

then red rule applies to the cycle B+ e.

If endpoints in different blue trees B1, B2,then blue rule applies across the cutseparating B1 and B2.

Thus, the generic MST algorithm iscorrect.


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Prims Algorithm

Start at any node s, and set T ={s}.

repeat n 1 times

Let T be the current tree.Choose a minimum cost uncolored

(u, v) with uT and vT.Color (u, v) blue, and add v to T.

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Graph GAfter 3 steps.


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Prims Algorithm

1. Input: Graph G= (V, E). For each vertexv, A(v) is list of its neighbors.

2. key(w) is the cost of the cheapest edge(v, w) with vT. blue(v, w) is the identityof this edge.

3. Initialize:

key(v) =, for all vV. H=makeHeap(). v=s.

4. while v=N IL do

for wA(v) ifcost(v, w)< key(w) key(w) =cost(v, w) blue(w) = (v, w) if wH insert(w,H,key(w)) else DecreaseKey(w,H,key(w)) end for v=ExtractMin(H).


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Illustration

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After 3 steps

After 4 steps Final MST


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Illustration

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After 3 steps

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Blue edges and keys after 3 steps

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Kruskals Algorithm

Initially make each node of V a singletontree.

Scan edges ofE in non-decreasing order ofcost.

scan edge e:

If both endpoints of e in the sametree, color it red. Otherwise color eblue, and merge the two trees.


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Illustration

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8, 10 colored red. 12 colored blue

After 4 edges scanned

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Final MST


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Complexity Results

Heap implementation of Prim: O(m log2 n).

d-Heap: O(nd logd n + m logd n).

Fibonacci Heap: O(m + n log n).

Kruskal with Union-Find:O(m log n + m(n)).

Round-Robin: O(m log log n).

Latest theoretical bound: O(m(n)).


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Balancing MST and SP

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Graph G MST SPT

MST minimizes total interconnectioncost.

SPT minimizes individual path lengths,from a root source.

How does each do with the other costmetric?

In Fig. 2, dmst(a, c) = 20, but dspt(a, c) = 15.

In Fig. 3, cost(spt) = 35, but cost(mst) = 30.


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A Pathological Example

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(i) (ii)

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Fig. (i) shows an example wheredmst(a, z) =n, while dspt= 2.

As n , this ratio is unbounded.

Fig. (ii) shows an example where

cost(spt) =nM, while cost(mst) =n 1 + M.

As M , this ratio is unbounded.


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Balanced Tree Theorem

Theorem: Pick any constant >1, and let= 1 + 2

1. Given G= (V, E) and a node s,there always exists a spanning tree T rooted

at s such that

dT(s, v) dspt(s, v), for any vV.

cost(T) cost(M ST).

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Which T for these pathological cases?


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Pre-Order Numbering

Pre-Order Lemma: Let T be a spanning treewith root s. Let z0, z1, . . . , zk be any k+ 1nodes listed in their pre-order sequence.

Then,k

i=1dT(zi1, zi) 2cost(T).

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Draw T in the plane. Let W be thedoubling walk around T (pre-order).

Each edge visted twice, cost(W) = 2cost(T).


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Proof

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dT(u, v) is tree path length.

Due to pre-ordering, firstoccurrence of

zi1 is beforethat of zi.

Mark off portions of W that join firstoccurence of zi1 to that of zi.

No edge of W marked more than once.

Thus,ki=1

dT(zi1, zi) cost(W) 2cost(T).


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Algorithm

= 1 + 21. E.g. = 2, = 3.

Compute MST and its pre-order numbers,starting with s.

Compute dspt(s, v), for all v.

Initialize H=M ST.

for each node v in pre-order doif dH(s, v)> dspt(s, v) thenadd to H all edges of path Pspt(s, v).

Output SPT of H, rooted at s.

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Graph G MST

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Illustration

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MST (1,5) added

(1,6) added Final tree T output

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Analysis

T satisfies dT(s, v) dspt(s, v); algorithmadds SP whenever needed.

Let z0, z1, . . . , zk be the pre-order sequenceof vertices that caused SP edges to beadded to H.

When zi is examined, H contains theshortest path to zi1. So, we must have

dspt(s, zi1) + dmst(zi1, zi) > dspt(s, zi).

Sum these over all i:

dspt(s, z1) dspt(s, z0) < dmst(z0, z1)

dspt(s, z2) dspt(s, z1) < dmst(z1, z2)

...

dspt(s, zk) dspt(s, zk1) < dmst(zk1, zk)

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i=1( 1)dspt(s, zi)0,

Ev =ki=1

w(ei) < 2w(M ST)

t 1 .


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Spanner Results

Althofer et al. result gives spanners with

O(n1+ 2t1) edges.

Peleg-Schaffer shows that there aren-vertex graphs, whose all t-spannerscontain (n1+

1t ) edges.

Much better results know if distances areEuclidean.

In Euclidean k-space (constant k)spanners with O(n) edges exist for anyconstant stretch factors t.

Such spanners possible with O(M ST)weight, O(1) degree, O(log n) diameter.

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