Discrete Wiskunde II, 2009 Lecture 5: Shortest Paths ...uetzm/dw/DWII_Lecture5.pdfShortest Path...

Discrete Wiskunde II, 2009Lecture 5: Shortest Paths & Spanning Trees

Marc UetzUniversity of Twente

[email protected]

wwwhome.math.utwente.nl/~uetzm/dw/

Marc Uetz Discrete Wiskunde II

mailto:[email protected]

http://wwwhome.math.utwente.nl/~uetzm/dw/

Shortest Path Problem

Given

directed graph (digraph) G = (V ,E ), |V | = n and |E | = m,arc lengths c : E → R, two vertices s, t ∈ V

want: shortest path from s to t (from s to all nodes v ∈ V )

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Assumption: all vertices are reachable from s


Three main cases

Arbitrary digraphs

includes negative cycle detection

Label correcting algorithms

O( nm ) computation time

Nonnegative edge lengths (cvw ≥ 0 for all (v ,w) ∈ E )

Dijkstra’s Algorithm

O( n2 ) computation time

Acyclic graphs (6 ∃ directed cycle)

Dynamic programming

O( n + m ) computation time


Shortest Path Optimality

Let d(v), v ∈ V be arbitrary node labels such thatd(v) ≥ shortest path length from s to v , v ∈ V

Theorem

Labels d(v) are shortest (s, v)-path lengths if and only if the4-inequality holds for all edges (v ,w) ∈ E

d(w) ≤ d(v) + cvw

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Proof

Necessity

trivial, for otherwise there is a shortcut to w via v

Sufficiency

Let `(w) be the (true) shortest path length to w , `(w) ≤ d(w).pick shortest (s,w)-path P = (s, . . . u, v ,w), length `(w)use 4-inequality along this path

d(w) ≤ d(v) + cvw

≤ d(u) + cuv + cvw

...

≤ d(s) + length(P) = length(P)

= `(w)


Dijkstra’s Algorithm (1959)

all arc lengths cvw are nonnegative

Idea

Start with distance labels d(v) ≥ shortest path lengths, thencorrect violated 4-inequalities in a clever way

Algorithm

d(s) = 0, d(v) =∞ and pred(v) = ∅ ∀ v ∈ V \ s, S = ∅while (S 6= V )

pick v ∈ S̄ := V \ S with smallest label d(v)

S = S ∪ v /*d(v) will remain unchanged!*/

For all w with (v ,w) ∈ E , if (d(w) > d(v) + cvw )

d(w) = d(v) + cvw

pred(w) = v

pred(v) = predecessor of v on a shortest (s, v)-path


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Correctness Dijkstra I

We prove by induction on # iterations

Claim 1: d(u) ≤ d(w) for all u ∈ S , w 6∈ S

Iteration 1: S = s, d(s) = 0, other d(w) ≥ 0√

Iteration where S = S ∪ v :First argue about v : d(v) ≤ d(w) ∀w 6∈ S , by our choice of v

Next, argue about any other u ∈ S :If d(w), w 6∈ S , was not updated, d(u) ≤ d(w) for all u ∈ S byinduction hypothesis

If d(w), w 6∈ S , has been updated,d(w) = d(v) + cvw ≥ d(v), as cvw ≥ 0and d(u) ≤ d(v) ∀u ∈ S by induction hypothesisso d(u) ≤ d(w) ∀ u ∈ S also after this update

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Correctness Dijkstra II

We prove by induction on # iterations

Claim 2: labels within S are the correct shortest path lengths

Iteration 1: S = s, d(s) = 0√

Iteration where S = S ∪ v :By induction hypothesis, only need to check that d(v) is correct.Theorem ⇒ only need to verify 4-inequality for edges into v

Case 1: consider any (u, v), u ∈ S :From iteration S = S ∪ u (which was before), d(v) ≤ d(u) + cuv

Case 2: consider any (w , v), w 6∈ S :In iteration S = S ∪ w (which is later), our claim 1 yieldsd(v) ≤ d(w) ≤ d(w) + cwu (d(v) remains unchanged)

√

Upon termination, S = V , so Dijkstra works correctly


Computation Time

Simple

Initialization O( n )

n iterations of while-loop, in each need to find smallest labelin V \ S , which is doable in O( n ) O( n2 )

in total, at most m updates of d(v), O( 1 ) each O( m )

Which gives O( n ) + n · O( n ) + O( m ) ∈ O( n2 ) time

Less Simple

An O( m log n ) implementation, which uses priority queue (alsocalled heap) to manage finding minimal d(v) in O( 1 ) time.(log n comes from overhead in organization of the heap)


BFS - breadth-first-search

Let all arc lengths cvw = 1. What about Dijkstra’s algorithm?

distance labels d(v) = minimal # arcs to reach v from s(shortest path tree = BFS tree)

d(v) =∞ for non-reachable v (“reachability algorithm”)

Computation time?

once assigned a finite value, no node needs relabeling(otherwise contradiction to choice of v = argmin{d(v)})hence organize such nodes as a FIFO list Q (queue):when d(w) is assigned a (finite) value, w → end of Q

then minimal d(v) is found in O( 1 ) time, at top of Q

Total time O( n + m )


Minimum Spanning Trees (MST)

MST

Given an edge-weighted, connected, undirected graphG = (V ,E , c), with |V | = n and |E | = m, find a MST.(cheapest connected acyclic subgraph)

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Spanning Trees and Cuts

Definitions

For subset of nodes W ⊆ V , δ(W ) denotes the cut inducedby W : all edges “leaving” W [δ(W ) = δ(V \W )]

Given a spanning tree T of G , let C (e) be the cut induced bydeleting edge e from T

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The Cut Condition

Theorem

Given a graph G = (V ,E ) with edge costs ce , e ∈ E , then aspanning tree T ⊆ E is an MST if and only if

ce ≤ cf for all edges e ∈ T and all edges f ∈ C (e) .

(i.e., T has the cheapest connection that crosses the cut)

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The Path Condition

In a tree T , denote by PT (v ,w) the (unique) path from v to w

Theorem

Given a graph G = (V ,E ) with edge costs ce , e ∈ E , then aspanning tree T ⊆ E is an MST if and only if

∀ edges f = {v ,w} ∈ E \ T : cf ≥ ce ∀ edges e ∈ PT (v ,w)

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Theorem

Kruskals algorithm solves MST problem in time O( m log m + n2 ).


Kruskal’s Algorithm: Correctness

Proof 2nd Invariant (Path Condition)

any edge f = {v ,w} not added to T creates cycle, namely{v ,w} ∪ PT (v ,w),in particular, all edges e ∈ PT (v ,w) are already in T

by the sorting of the edges ⇒ cf ≥ ce ∀e ∈ PT (v ,w)

T connected

for any cut δ(W ) of the graph (V ,E ), the cheapest (so atleast one!) edge e ∈ δ(W ) must be in T

thus (V ,T ) must be connected(connected :⇔ graph contains at least one edge from any cutδ(W ), W ⊆ V )


Kruskal’s Algorithm: Computation Time

To start with

O( m log m ) for sorting the ce values (MergeSort)

need to do m times: Is T ∪ e acyclic?, and if so, add e to T

We need a clever data structure!

Store & update to which component any node belongs:

Initialize c(v) = v ∀ v ∈ V [n components] O( n )

m times we do for an edge {v ,w}:Check (T ∪ {v ,w} acyclic) ⇔ c(v) 6= c(w) O( 1 )

If so, add {v ,w} to T [merge 2 components, n − 1 times]:For all u ∈ V : If c(u) == c(w) let c(u) = c(v) O( n )

O( m log m ) + O( n ) + mO( 1 ) + (n− 1)O( n ) ∈ O( m log m + n2 )


More about MSTs

With a little more care (data structure & analysis), Kruskal’salgorithm can be implemented to run in O( m log m ) time

Prim’s algorithm (1959)

Start with W = {v0}, any v0 ∈ V , T = ∅For (i = 1, . . . , n − 1)

Let edge ei = {v ,w} be the cheapest edge in δ(W )(where v ∈W and w 6∈W )Let T = T ∪ ei and W = W ∪ {w}

Correctness follows directly from Cut Condition

Computation time can be shown in O( n2 )


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