design and analysis of algorithms (daa 2018)...master’s programme in computer science juha...

Master’s Programme in Computer Science

Juha Kärkkäinen

Based on slides by Veli Mäkinen

DESIGN AND ANALYSIS OF ALGORITHMS (DAA 2018)

17/09/2018 1 Design and Analysis of Algorithms 2018 week 3

Shortest paths and network flows

Week 3

Shortest paths

Some dynamic programming you might already have seen


RECURRENCES WITH REPEATING SUBSTRUCTURES

• Example: Fibonacci numbers

• Defined by recurrence: F(i) = F(i-2) + F(i-1) and F(0)=F(1)=1

• Recursive computation is slow (O(F(n)) time): F(i) = F(i-2)+F(i-1) = F(i-4)+F(i-3)+F(i-3)+F(i-2) = … because same values computed repeatedly.

• Memoization (storing computed values) is faster (O(n) time). Proper evaluation order helps F(0)=F(1)=1, F(2)=F(0)+F(1), F(3)=F(1)+F(2), …

• In general, a proper evaluation order is a topological order in a directed acyclic graph (DAG) denoting the recurrence dependencies:

17/09/2018 Design and Analysis of Algorithms 2018 week 3 4

1 1 2 3 5 …


SHORTEST S-T PATH IN A DAG Find shortest path from s to t in DAG G=(V,E)

• Base case: d(s) = 0

• Recurrence: d(v) = min{d(u)+c(u,v) : u is an in-neighbor of v}

• Output: d(t)

• Topological sort, compute each d(v) in |N-(v)| time, total O(|V|+|E|) time, where V and E are the set of vertices and edges of the DAG, respectively.

• Traceback to output an optimal path


s t

v u

…

N-(v)

… …


SHORTEST PATH IN A DAG (EXAMPLE)


Topological sort

5

10

12

5

10

21

10

5 4 5

12

3

11

1. Topological sort




Topological sort

5

10

12

5

10

21

10

5 4 5

12

3

11

0 10 15 19 24 26

1. Topological sort

2. Compute distances

(from left to right)




Topological sort

5

10

12

5

10

21

10

5 4 5

12

3

11

0 10 15 19 24 26

1. Topological sort

2. Compute distances

3. Traceback

(from right to left)




Topological sort

5

10

12

5

10

21

10

5 4 5

12

3

11

26


SHORTEST S-T PATH IN A GENERAL GRAPH

• General directed graph

• May have cycles

• May have negative edge weights

• But no negative cycles (why?)

• Cannot use Dijkstra’s algorithms because of negative weights

• The standard algorithm is Bellman-Ford (Section 24.1 in book)

• Recurrence: d(v) = min{d(u,v) + c(u,v) : u in N-(v)}

• No trivial topological order because of cycles



DERIVING BELLMAN-FORD 1

• Make n=|V| copies of the graph and redirect all edges from copy (i-1) to copy i for i in [1..n-1], like in example below:

• In the resulting DAG find shortest path from any s to t


s t

0 1 2

s t

3

s s


DERIVING BELLMAN-FORD 2

Recurrence for Bellman-Ford

• Base case: d(s,k)=0, for all k, d(v,0)=∞ when v≠s

• Recurrence: d(v,k) = min{d(u,k-1) + c(u,v) : u in N-(v)}

• k = the number of the graph copy

• k = maximum number of edges allowed on the path from s to v

• Output: d(t) = d(t,|V|-1)

• Topological order by copy number

• Order within copy does not matter

• O(|V||E|) time


Network flows

An application of shortest path algorithms and a

fundamental problem to reduce other problems to


FLOWS

• Recall Kirchhoff’s laws from high-school physics

• Flow f in a directed graph (flow network):

• unique source s and unique target t

• flow f(e) defined on each edge e=(u,v), s.t. flow conservation property holds on each vertex v except s and t

• value of flow is imbalance at source and sink:


v

(flow conservation property)

)( )()( )(

),(),(),(),(tNu tNwsNw sNu

wtftufsufwsff

)( )(

),(),(vNu vNw

wvfvuf


FLOW NETWORK • Flow network is a directed graph

• No self loops or antiparallel edges: if (u,v) is an edge, (v,u) is not

• Capacities c(u,v) ≥ 0 for each edge (u,v)

• A flow is feasible if it satisfies capacity constraints: 0 ≤ f(u,v) ≤ c(u,v) for every edge (u,v)


Unconstrained flow Flow with capacity constraint


A FEASIBLE FLOW (EXAMPLE)


11/14

1/4

7/7

12/12

capacity flow

Flow value =11+8=15+4=19


MAXIMUM FLOW PROBLEM

• Given a flow network, find a feasible flow f maximizing |f|.


?/14

?/4

?/7

?/12

Flow value = max


FORD-FULKERSON METHOD (FIRST CONCEPTS)

• Idea: Given a feasible flow f, find another flow f’ s.t. augmenting f with f’ increases the flow value.

(f↑f’)(u,v) = f(u,v) + f’(u,v) - f’(v,u)

• Flow f’ is a feasible flow in a residual network Gf with capacities

• cf(u,v) = c(u,v) - f(u,v)

• cf(v,u) = f(u,v) (antiparallel edge)

• f↑ f’ is always a feasible flow and |f↑f’| = |f| + |f’| (see Lemma 26.1 in book)



RESIDUAL NETWORK (EXAMPLE)


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?/1 ?/11

?/8

?/15

Residual transform

Zero capacity edges

can be omitted


FORD-FULKERSON METHOD (SINGLE STEP)

• Let P be a simple s-t path in the residual network Gf.

• No edges with zero residual capacity

• P is called augmenting path

• Let cf(P) = min{cf(u,v) : (u,v) in P}

• Bottleneck capacity of P

• Let fP(u,v) = cf(P) if (u,v) is in P and fP(u,v)=0 otherwise

• Maximum augmenting flow along P

LEMMA (Lemma 26.3 in book)

• f↑fP is a feasible flow and |f↑fP| = |f| + cf(P)



FORD-FULKERSON METHOD (SINGLE STEP)

Proof sketch.

• Local changes in flow along P:

• → v → : inflow and outflow increase by cf(P)

• ← v → : outflow increases by cf(P) - cf(P) = 0

• → v ← : inflow increases by cf(P) - cf(P) = 0

• ← v ← : inflow and outflow decrease by cf(P)

• s →: ouflow increases by cf(P)

• s ←: inflow decreases by cf(P)

• → t: inflow increases by cf(P)

• ← t: ouflow decreases by cf(P)

• u → v: flow increases by cf(P) ≤ cf(u,v) = c(u,v)-f(u,v)

• u ← v: flow decreases by cf(P) ≤ cf(u,v) = f(u,v)


Flow

conservation

maintained

Flow value

increases

by cf(P)

Capacity

constraints

maintained


AUGMENTING PATH (EXAMPLE)


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?/7

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?/11

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?/1 ?/11

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?/15

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12/12


FORD-FULKERSON METHOD (ALL STEPS)

• Section 26.2 in book

• Start with a zero flow f.

• Repeat until flow f does not improve anymore

• Find an augmenting s-t path P in the residual graph and augment f with the bottleneck flow induced by path P.

• Running time is O(|f*||E|), where f* is the maximum flow in a graph with integral capacities.

• Worst case: always choose augmenting paths with tight bottleneck


1


FORD-FULKERSON METHOD (EXAMPLE)


11/14

1/4

7/7

12/12

?/7

?/12

?/5

?/4

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?/1 ?/11

?/8

?/15

11/14

1/4

7/7

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?/7

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?/11

?/3

?/1 ?/11

?/1

?/19

No augmenting path to increase the flow.


EDMONDS-KARP

• Augmenting path = shortest s-t path in the residual graph

• Shortest = fewest edges

• Otherwise identical to Ford-Fulkerson

• O(|V||E|2) running time!

• Lecture blackboard or Theorem 26.8 in book



APPLICATIONS OF MAX FLOW

• Vertex-disjoint paths

• Find the maximum number of vertex disjoint paths from a set of source vertices to a set of sink vertices

• Maximum bipartite matching

• Bipartite graph: vertices partitioned into two distinct subsets, no edges within subset

• Matching: subset of edges with no common end vertices

• Find maximum matching in a bipartite graph

• Both problems have a reduction to max-flow

• See lecture blackboard and separate writeup


design and analysis of algorithms (daa 2018)...master’s programme in computer science juha...

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