adm i | exercise session 9 · v c (v; ) for all (v;w) 2a y a free for all a 2a a solution is called...

ADM I — Exercise Session 9

TU Berlin

December 11, 2013

TU Berlin Shortest Path 1 / 19

Shortest Path Problem

Given: digraph D = (V ,A), node r ∈ V , arc costs ca, a ∈ A.Task: for each v ∈ V , find dipath from r to v of least cost

Notes:

I single-source shortest path problem

I there may not be a path (unconnected pair of nodes)

I there may not be a shortest path (cycle with negative length)


Shortest Path – LP

LP formulation

Idea for the LP formulation:

I yv is the shortest distance from s to v

I triangle inequality: yv + c(v ,w) ≤ yw for all (v ,w) ∈ A

Linear Program, shortest r -s path:

max ys − yrs.t. yw − yv ≤ c(v ,w) for all (v ,w) ∈ Aya free for all a ∈ A

a solution is called a feasible potential

Observation:ya, a ∈ A is a feas. pot. ⇒ y ′a = {ya + c}, c ∈ R is a feas. pot. assume yr = 0


Shortest Path – dual LP

Primal and Dual LP FormulationDual and primal linear program for the shortest path problem

Primal LP:max ys − yrs.t. yw−yv ≤ c(v ,w)

∀(v ,w) ∈ A

Dual LP:min cT · xs.t.

∑a∈δ−(v) xa −

∑a∈δ+(v) xa = bv ,

∀v ∈ Vxa ≥ 0 for all a ∈ A

interpretation:

I ya is bound on distance to r ,

I xa is the ”number“ of time the edge a is used


Shortest Path – dual LP

Primal and Dual LP FormulationDual and primal linear program for the shortest path problem

Primal LP:max yss.t. yw−yv ≤ c(v ,w)

∀(v ,w) ∈ Ayr = 0


∑a∈δ−(v) xa −

∑a∈δ+(v) xa = bv ,

∀v ∈ Vxa ≥ 0 for all a ∈ A

interpretation:

I ya is bound on distance to r ,

I xa is the ”number“ of time the edge a is used


Example

r

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v1

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11 1

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I directed graph D

I shortest r -v path

I feasible solutionfor the primal LP

I set yr = 0 andobeying triangleinequality

dual solution:set of arcs

I arcs r → 1, 2→ 3 ,3→ 5, 5→ v used once

I xa = 1 for a ∈ {(r , 2), (2, 3), (3, 5), (5, v)}, all other xa = 0


Example

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v1

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-3.5

-2.5

-1.5

-0.5

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16

I directed graph D








Example

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I directed graph D








Shortest Path – Dual LP

dual LP:min cT · xs.t.

∑a∈δ−(v) xa −

∑a∈δ+(v) xa = bv ,

∀v ∈ Vxa ≥ 0 for all a ∈ Awith br = −1, bv = 1, bv = 0 for all v ∈ V \ {r , v}

interpretation:

I r has one outgoing edge

I v has one incoming edge

I v ∈ V \ {r , v} has samenumber of incoming andoutgoing edges

I ⇒ set {a|xa = 1} contains ar -v path

r

4

2

3

5

v6

7


Shortest Path – Dual LP

dual LP:min cT · xs.t.

∑a∈δ−(v) xa −

∑a∈δ+(v) xa = bv ,

∀v ∈ Vxa∈ {0, 1} for all a ∈ Awith br = −1, bv = 1, bv = 0 for all v ∈ V \ {r , v}

interpretation:I r has one outgoing edge

I v has one incoming edge

I v ∈ V \ {r , v} has samenumber of incoming andoutgoing edges

I ⇒ set {a|xa = 1} contains ar -v path

r

4

2

3

5

v6

7


Shortest Path – integrality of the dual LP

integrality dual LP:min cT · xs.t.

∑a∈δ−(v) xa −

∑a∈δ+(v) xa = bv ,

∀v ∈ Vxa ∈ {0, 1} for all a ∈ A

fractional solution:

I if contains cycle with cost 0

I decrease values alongcycle, remove at least oneedge

I degree constraints still valid

I optimal solution withoutcycles




∑a∈δ−(v) xa −

∑a∈δ+(v) xa = bv ,

∀v ∈ Vxa ∈ {0, 1} for all a ∈ A


r

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v6

7

0.4

0.3

0.2








∑a∈δ−(v) xa −

∑a∈δ+(v) xa = bv ,

∀v ∈ Vxa ∈ {0, 1} for all a ∈ A


r

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v6

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0.2

0.1








∑a∈δ−(v) xa −

∑a∈δ+(v) xa = bv ,

∀v ∈ Vxa ∈ {0, 1} for all a ∈ A


r

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v6

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0.3

0.2

0.7

0.3

0.4

I if contains parallel paths

I decrease values along onepath, increase along others

I at least one edge vanishes

I optimality all parallelpaths have the same length

I optimal solution withoutparallel paths

⇒ we constructed integral solution




∑a∈δ−(v) xa −

∑a∈δ+(v) xa = bv ,

∀v ∈ Vxa ∈ {0, 1} for all a ∈ A


r

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3

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v6

7

0.6

0.5

0.4 0.1

I if contains parallel paths

I decrease values along onepath, increase along others

I at least one edge vanishes

I optimality all parallelpaths have the same length

I optimal solution withoutparallel paths

⇒ we constructed integral solution


Shortest Path – integality of the dual LP

simplex algorithm find integral solutions:

I Lemma 5.16: basis is a spanning tree

I r -v using edges a with xa > 0 is unique

I exactly one path with xa = 1

alternative proof via solution of the dual, using proof ofFord’s Algorithm


Shortest Path – LP for single source shortest path

LP model for all end nodes, also integral solution

Primal LP:

max∑

ys − yrs.t. yw−yv ≤ c(v ,w)

∀(v ,w) ∈ Ayr = 0


∑a∈δ−(v) xa −

∑a∈δ+(v) xa = bv ,

∀v ∈ Vbr = −n + 1, bv = 1 for all v ∈ V \ {r}xa ≥ 0 for all a ∈ A

I integrality ensured due to the simplex algorithm


Shortest Path – Ford’s Algorithm

Algorithm 1 Ford’s Algorithm

1: yr := 0, p(r) := r2: yv :=∞ for all v ∈ V3: pv :=null for all v ∈ V4: while ∃(v ,w) ∈ A with yw > yv + c(v ,w) do5: yw := yv + c(v ,w)

6: p(w) := v7: end while

Running time: O(∑

a∈A ca · n2) not polynomial

I sequence of labeling edges is important

I running time should not depend on the weight of the edges


Shortest Path – Ford’s Algorithm

Hint: exercise 45b), tight example for the bound of the runningtime

I use parallel edges

I store used edge for each vertex v ,instead of previous node (p[v ])

I parallel edges can be removed using an intermediate vertex

I but handling parallel edges is easier

I number of changes in distance label yv is exponential

1 2 1 2

3

4

5


Shortest Path – Bellmanford

Algorithm 2 Bellman–Ford Algorithm

1: yr := 0, p(r) := r2: yv :=∞ for all v ∈ V3: pv :=null for all v ∈ V4: for i = 1 to n − 1 do5: for all (v ,w) ∈ A do6: if yw > yv + c(v ,w) then7: yw := yv + c(v ,w)

8: p(w) := v9: end if

10: end for11: end for

Running time: O(n ·m) polynomial!

I labeling sequence: n − 1 times all edges in a fixed order


Shortest Paths – Bellman–Ford vs. Ford

Example – Ford’s Algorithm: Sequence 1

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Ford’s Algorithm:

I initialization

I checking triangleinequality




r

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v1

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0

∞

∞

∞

∞

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Ford’s Algorithm:

I initialization





r

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∞

∞

∞

∞

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Ford’s Algorithm:

I initialization


edge no.1: (r , 2)




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0 ∞

∞

∞

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Ford’s Algorithm:

I initialization


edge no.1: (r , 2)




r

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0 ∞

∞

∞

∞

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Ford’s Algorithm:

I initialization


edge no.2: (1, 3)




r

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0

∞

∞

∞

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2Ford’s Algorithm:

I initialization


edge no.2: (1, 3)




r

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I initialization


edge no.3: (3, 5)




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Ford’s Algorithm:

I initialization


edge no.3: (3, 5)




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∞

∞

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Ford’s Algorithm:

I initialization


edge no.4: (5, v)




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∞

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Ford’s Algorithm:

I initialization


edge no.4: (5, v)




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∞

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Ford’s Algorithm:

I initialization


edge no.5: (r , 4)




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Ford’s Algorithm:

I initialization


edge no.5: (r , 4)




r

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Ford’s Algorithm:

I initialization


5 labeling steps




r

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Ford’s Algorithm:

I initialization


5 labeling steps

all labeled edges in shortest paths tree, perfect sequence




r

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Ford’s Algorithm:

I initialization





r

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∞

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Ford’s Algorithm:

I initialization





r

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∞

∞

∞

∞

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Ford’s Algorithm:

I initialization


edge 1: (r , 3) + 7 arcs




r

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∞

∞

∞

∞


I initialization


edge 1: (r , 3) + 7 arcs




r

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∞

∞

∞

∞


I initialization


edge 2: (3, v) + 6 arcs




r

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v1

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∞

∞∞

3

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Ford’s Algorithm:

I initialization


edge 2: (3, v) + 6 arcs




r

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∞

∞∞

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Ford’s Algorithm:

I initialization


edge 3: (r , 4) + 6 arcs




r

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∞

∞

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Ford’s Algorithm:

I initialization


edge 3: (r , 4) + 6 arcs




r

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∞

∞

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Ford’s Algorithm:

I initialization


edge 4: (4, 5) + 6 arcs




r

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∞

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Ford’s Algorithm:

I initialization


edge 4: (4, 5) + 6 arcs




r

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∞

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Ford’s Algorithm:

I initialization


edge 5: (5, v) + 6 arcs




r

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∞

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Ford’s Algorithm:

I initialization


edge 5: (5, v) + 6 arcs




r

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∞

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Ford’s Algorithm:

I initialization


edge 6: (r , 2) + 8 arcs




r

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Ford’s Algorithm:

I initialization


edge 6: (r , 2) + 8 arcs




r

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Ford’s Algorithm:

I initialization


edge 7: (2, 3) + 7 arcs




r

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I initialization


edge 7: (2, 3) + 7 arcs




r

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61


I initialization


edge 8: (2, v) + 7 arcs




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2 5

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Ford’s Algorithm:

I initialization


edge 8: (2, v) + 7 arcs




r

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2 5

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Ford’s Algorithm:

I initialization


edge 9: (3, 5) + 9 arcs




r

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Ford’s Algorithm:

I initialization


edge 9: (3, 5) + 9 arcs




r

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Ford’s Algorithm:

I initialization


edge 10: (5, v) + 9 arcs




r

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Ford’s Algorithm:

I initialization


edge 10: (5, v) + 9 arcs




r

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Ford’s Algorithm:

I initialization


10 labeling steps

81 edges checked, bound is (2C + 1) · n2 = (2 · 4 + 1) · 62 = 324



Example – Bellman Ford Algorithm:

r

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4Bellman Ford Algorithm:

I initialization

I sequence:(r , 2), (r , 3), (r , 4), (2, 3), (2, v)(4, 3), (3, v), (3, 5), (4, 5), (5, v)

I first run




r

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Bellman Ford Algorithm:

I initialization


I first run




r

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0 ∞

∞

∞

∞


I initialization


I first run




r

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I initialization


I first run




r

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∞

∞1

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I initialization


I first run




r

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∞

∞1

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I initialization


I first run




r

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∞

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I initialization


I first run




r

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I initialization


I first run




r

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I initialization


I first run




r

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I initialization


I first run

no changes after first run, bound is (n − 1) ·m = 5 · 9 = 45


Bellman-Ford Speed-up

differences to algorithms presented in CoMa:

I single source shortest path, instead all pairs shortest path

I running time n3 instead of n4

I improvement via faster matrix multiplication ...

I ... not possible, idea does not work

I good order to label probably better, not using old distances


Shortest Path – Floyd Warshall

Idea: calculate shortest distances between all pairs of jobsuse yu,v for the distance from u to v

Algorithm 3 Floyd Warshall Algorithm

1: yu,v := c(u,v) for all u, v ∈ V2: for i = 1 to n do3: for all v ,w ∈ V do4: yv ,w := min(yv ,w , yv ,i + yi ,w )5: end for6: end for

Running time: O(n3) as fast as Bellman Ford algorithm for a single source

Correctness: ...






Running time:

O(n3) as fast as Bellman Ford algorithm for a single source

Correctness: ...






Running time: O(n3) as fast as Bellman Ford algorithm for a single source

Correctness: ...



Correctness – brief sketch

I important formula yv ,w := min(yv ,w , yv ,i + yi ,w )

I considers node i as part of the path in loop i

I after i-th loop, yv ,w ≤ shortest distance between v ,wusing inner nodes from set {1, . . . , i}

I at the end: yv ,w shortest v ,w distance with all possible innernodes

u v

iyu,i yi ,v

yu,v


adm i | exercise session 9 · v c (v; ) for all (v;w) 2a y a free for all a 2a a solution is called...

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