CS 401 Max Flow Applications Xiaorui Sun 1 Survey Design Survey - - PowerPoint PPT Presentation

CS 401

Max Flow Applications

Xiaorui Sun

1

Survey Design

Survey Design

Survey design.

Design survey asking n1 consumers about n2 products. Can only survey consumer i about a product j if they own it. Ask consumer i between ci and ci' questions. Ask between pj and pj' consumers about product j.

• Goal. Design a survey that meets these specs, if possible.

1 3 1' 3' 2 4 2' 4' [c1, c1’] = [2, 4] [2, 4] [2, 4] [2, 4] [p1, p1’] = [2, 4] [2, 4] [2, 4] [2, 4]

consumers products

Survey Design

Survey design.

Design survey asking n1 consumers about n2 products. Can only survey consumer i about a product j if they own it. Ask consumer i between ci and ci' questions. Ask between pj and pj' consumers about product j.

• Goal. Design a survey that meets these specs, if possible.

Bipartite perfect matching. Special case when ci = ci' = pi = pi' = 1.

Survey Design

• Algorithm. Formulate as a circulation problem with lower

bounds.

Include an edge (i, j) if customer own product i. Demands = 0 for all vertices Integer circulation Û feasible survey design.

s

1 3 5 1' 3' 5'

t

2 4 2' 4' [c1, c1'] [0, 1]

consumers

[p1, p1'] [0, ¥]

products

Image Segmentation

Image Segmentation

Given an image we want to separate foreground from background

• Important problem in image processing.
• Divide image into coherent regions.

7

Foreground / background segmentation

Label each pixel as foreground/background. ! = set of pixels, " = pairs of neighboring pixels. #$ is the original image. #$ ≫ 0 means we prefer to label ' in foreground. ($,* ≥ 0 is separation penalty for labeling one of ' and j as foreground, and the other as background. Goals: Find partition (-, -) that minimizes: − 0

$∈2

#$ +

$,* ∈4 $∈2,*∈2

($,* where - is the foreground.

8

Min cut Formulation

!′ = (%′, '′). Add s to correspond to foreground; Add t to correspond to background; Use two anti-parallel edges instead of undirected edge.

9

pij pij pij

s t

i j

pij *+ if *+ > 0

!′

−*+ if *+ < 0

Min cut Formulation (cont’d)

Consider min cut (", ") in G’. (" = foreground.)

10

%&' ", " = )

*∈,

−&* 1/012 + )

*∈,

&* 1/042 + )

*,5 ∈6 *∈,,5∈,

'*,5 = − ∑*∈, &* + ∑*∈, &* 1/042 + ∑*∈, &* 1/042 + ∑⋯'*,5 = − ∑*∈, &* + ∑* &* + ∑⋯'*,5 = − ∑*∈, &* + ∑⋯'*,5 + constant Precisely, what we want to minimize.

s t

i j

pij

?′ "

&5 if &5 > 0 −&5 if &5 < 0

Reality

The main difficulty is to come up with a good model. Segmentation may be real-valued instead of {0,1}. There are many more than 1 objects. May need labeling. Augmenting path is not great for GPU.

11

Summary

Min s-t cut problem. Find an s-t cut of minimum capacity. Max flow problem. Find s-t flow of maximum value. Ford-Fulkerson Algorithm. Augment flow along augmenting path

• O(nm) for unit capacity graph
• O(m2 log C) for integer capacity graph

Max flow applications

• Bipartite matching, disjoint paths, circulation, image segmentation
• Techniques: add source/sink, reduce to circulation, turn negative

values to (positive) capacities

12

Longest Common Subsequence

Sequence alignment Common subsequence length = # matches

13

a g c d e f

• f

b c d e a

• Common subsequence = af

2 matches, 8 gaps

a

• g

c d e f f b c d e

• a

Common subsequence = acde 4 matches, 2 gaps, 1 mismatch

Longest Common Subsequence

14

LCS: alignment with largest number of matches Let !"#(%, ') be the number of matches for the best alignment

)*, … , ), and -*, … , -. Case 1: OPT matches ),, -.

• !"#(% − 1, ' − 1)+1 if ), = -., otherwise !"#(% − 1, ' − 1)

Case 2: OPT leaves ), unmatched

• !"# % − 1, '

Case 3: OPT leaves -. unmatched

• !"#(%, ' − 1)