Rolf Harren- Approximating the Orthogonal Knapsack Problem for Hypercubes

8/3/2019 Rolf Harren- Approximating the Orthogonal Knapsack Problem for Hypercubes

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Introduction Square Packing Hypercube Packing Summary

Approximating the Orthogonal Knapsack

Problem for Hypercubes

Rolf Harren

University of Dortmund

2007.09.07

Rolf Harren Approx. the Orthogonal Knapsack Problem for Hypercubes
http://find/http://goback/


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Outline

1 IntroductionProblem

Preparation

2 Square Packing

Separation into Large, Medium and Small ItemsPacking the Large Items Optimally

Adding the Small Items

Putting Everything Together

3 Hypercube PackingGeneralization

4 Summary



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Introduction Square Packing Hypercube Packing Summary Problem Preparation

Outline

1 IntroductionProblem

Preparation

2 Square Packing




3 Hypercube PackingGeneralization

4 Summary



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Multidimensional Orthogonal Packing Problems

Input:

d-dimensional cuboid items a1, . . . , an

. . . . . .

Objective:

an orthogonal, non-rotational and non-overlapping packing intoa given space such that...



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Multidimensional Orthogonal Packing Problems

BIN PACKING...the number of bins

is minimized . . .k = 1 2 3

STRIP PACKING

...the total height

is minimizedh

KNAPSACK PACKING

...the profit of the packed

selection of items

is maximized . . .

. . .

selection



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Orthogonal Knapsack Packing for Hypercubes

Considered problemAll items are squares, cubes or hypercubes 0 < a1, . . . , an 1Items have profits piBin has unit size

. . . . . .

selection

1


I d i S P ki H b P ki S P bl P i


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Results

BIN PACKING STRIP PACKING

2- dim 1.525.. Bansal, Correa, AFPTAS Kenyon, Rmila

APX-complete Sviridenko

3- dim 3.382.. 1.691.. Bansal, Han, Iwama

Sviridenko, Zhang

d- dim open open

Hypercube BIN PACKING Hypercube STRIP PACKING

2- dim APTAS Bansal, Correa, AFPTAS

Kenyon, Sviridenko

d- dim APTAS Bansal, Correa, APTAS

Kenyon, Sviridenko


I t d ti S P ki H b P ki S P bl P ti


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Results

BIN PACKING STRIP PACKING

2- dim 1.525.. Bansal, Correa, AFPTAS Kenyon, Rmila

APX-complete Sviridenko

3- dim 3.382.. 1.691.. Bansal, Han, Iwama

Sviridenko, Zhang

d- dim open open

Hypercube BIN PACKING Hypercube STRIP PACKING

2- dim APTAS Bansal, Correa, AFPTAS

Kenyon, Sviridenko

d- dim APTAS Bansal, Correa, APTAS

Kenyon, Sviridenko




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Results

KNAPSACK PACKINGGeneral Hypercube

2- dim 2 + Jansen, Zhang 54 +

3- dim 7 + Diedrich, H., Jansen 9

8

+ APX-complete Thle, Thomas

d- dim open 2d+12d

+

Further results exist onPACKING WITH LARGE RESOURCES,MAXIMIZING THE VOLUME,MAXIMIZING THE NUMBER, ...




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Results

KNAPSACK PACKINGGeneral Hypercube

2- dim 2 + Jansen, Zhang 54 +

3- dim 7 + Diedrich, H., Jansen 9

8

+ APX-complete Thle, Thomas

d- dim open 2d+12d

+

Further results exist onPACKING WITH LARGE RESOURCES,MAXIMIZING THE VOLUME,MAXIMIZING THE NUMBER, ...




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Open Question

NP-Completeness

It is unknown for all previous packing problems whether therestriction to Hypercube packing is NP-hard for d 3.




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Applications

Cutting Problems

All packing problems can be seen as cutting problems, e.g.,

cutting textile or wood

Transportation Industry

Arranging container on a ship

Arranging items inside a container




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Applications

Cutting Problems

All packing problems can be seen as cutting problems, e.g.,

cutting textile or wood

Transportation Industry

Arranging container on a ship

Arranging items inside a container




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q g yp g y p

Applications

Advertisement Placement

Arranging ads in a newspaper

Arranging ads on a flash page

Scheduling

Bounded running time on a computer with a grid layout for

the processors

Tasks need a fixed running time on a rectangular grid ofprocessors




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q g yp g y p

Applications

Advertisement Placement

Arranging ads in a newspaper

Arranging ads on a flash page

Scheduling

Bounded running time on a computer with a grid layout for

the processors

Tasks need a fixed running time on a rectangular grid of

processors




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Outline

1 Introduction

Problem

Preparation

2 Square Packing




3

Hypercube PackingGeneralization

4 Summary




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NFDH

NEXT-FIT-DECREASING-HEIGHT (NFDH)

is a very efficient layer based packing

algorithm for small items

l1

l2

l3

l4

Lemma

For small items ai , the unfilled volume is bounded byd




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Gaps in a Packing

Lemma

Given a packing P of m squares we can partition the free spaceinto at most3m rectangles

At least one item in P has to be aligned to the bottom of the bin




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Shifting Technique

w = 1h . . .S

1 S3 S4 Sl

hL

L

. . .SiS2

. . .

dispose

. . .

Lemma

For small items ai it is possible to free a given line L byshifting the items into a gap losing not more than O()p(I) ofthe profit.




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Shifting Technique

w = 1h . . .S

1 S3 S4 Sl

h

L

L

. . .SiS2

. . .

dispose

. . .

Lemma

For small items ai it is possible to free a given line L byshifting the items into a gap losing not more than O()p(I) ofthe profit.




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Rectangle Packing with Large Resources

If the bin is much bigger than the items we can derive a goodapproximation ratio

Lemma

There is an approximation algorithm for RECTANGLE PACKING

into a bin B = (a, b) where a= 1 and b 14 with approximation

ratio(1 + )

Idea:

Bin has strip-like shapePack a selection of items with the AFPTAS for STRIP PACKING and

apply a shifting technique to the overhang




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Rectangle Packing with Large Resources

If the bin is much bigger than the items we can derive a goodapproximation ratio

Lemma

There is an approximation algorithm for RECTANGLE PACKING

into a bin B = (a, b) where a= 1 and b 14 with approximation

ratio(1 + )

Idea:

Bin has strip-like shapePack a selection of items with the AFPTAS for STRIP PACKING and

apply a shifting technique to the overhang


Introduction Square Packing Hypercube Packing Summary Separation Large Items Small Items Everything


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Outline

1 Introduction

Problem

Preparation

2 Square Packing

Separation into Large, Medium and Small Items

Packing the Large Items Optimally



3 Hypercube Packing

Generalization

4 Summary



O li


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Outline

1 Introduction

Problem

Preparation

2 Square Packing





3 Hypercube Packing

Generalization

4 Summary



S ti


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Separation

A little bit technical...

Separation

Let r = 1

and 0 = , i+1 = 4i

Divide (unknown) optimal solution Iopt

Mi = {s Iopt : s [i+1, i[} for 1 i r i {1, . . . , r} with p(Mi) p(Iopt)

Large, medium and small items

Lopt := {s Iopt : s i

}M := Mi

S := {s I : s < i+1}



S ti


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Separation


Separation

Let r = 1

and 0 = , i+1 = 4i





}M := Mi

S := {s I : s < i+1}



S ti


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Separation


Separation

Let r = 1

and 0 = , i+1 = 4i





}M := Mi

S := {s I : s < i+1}



Separation


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Separation


Separation

Let r = 1

and 0 = , i+1 = 4i





}M := Mi

S := {s I : s < i+1}



Separation


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Separation


Separation

Let r = 1

and 0 = , i+1 = 4i





}M := Mi

S := {s I : s < i+1}



Separation


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Separation


Separation

Let r = 1

and 0 = , i+1 = 4i





}M := Mi

S := {s I : s < i+1}



Separation


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Separation

In a nutshell

Large items are large (i.e., i)

Small items are small (i.e., < i+1)

Medium items are unimportant (i.e., p(M) OPT(I)).



Separation


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Separation

In a nutshell






Separation


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Separation

In a nutshell






Outline


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Outline

1 Introduction

Problem

Preparation

2 Square Packing





3 Hypercube Packing

Generalization

4 Summary



Enumeration


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Enumeration

Enumeration

All relevant values are constant:

r, i and i {1, . . . , r}

at most1

2i large items fit into the bin (volume argument)

try all values for i and all possible selection of 1

2i

large

items

we assume i

and an optimal packing of the large items areknown



Enumeration


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Enumeration

Enumeration


r, i and i {1, . . . , r}

at most1



2i

large

items

we assume i




Enumeration


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Enumeration

Enumeration


r, i and i {1, . . . , r}

at most1



2i

large

items

we assume i




Enumeration


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Enumeration

Enumeration


r, i and i {1, . . . , r}

at most1



2i

large

items

we assume i




Unfilled Volume


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U ed o u e

Gaps in packing

Small items

NFDH

Packing ofLopt

At most 3m gapsUnfilled volume per gap 2



Unfilled Volume


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Gaps in packing

Small items

NFDH

Packing ofLopt

At most 3m gapsUnfilled volume per gap 2



Unfilled Volume


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Adding small items with NEXT-FIT-DECREASING-HEIGHT

Unfilled volume

number of gaps

3m

unfilled volume per gap

2

3 1

2i

2i+1

= 3 1

2i

24i

2i



Outline


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1 Introduction

ProblemPreparation

2 Square Packing


Packing the Large Items OptimallyAdding the Small Items


3 Hypercube Packing

Generalization

4 Summary



3 Methods


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Enough remaining space

Several large items

Only one very large item



Enough Remaining Space Vol(Lopt) 1 i


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( p )

S S

constantpacking

. . . . . .

NFDH

Gaps in packingGuess Lopt

3|Lopt| gapspartition free space in

comple

te

enume

ration

separated by i

Input I

selection with

FracKnap+(S, V 22i)

Packing of Lopt

p(S) V22

i

V OPT(S) (1 22

i

i)OPT(S)

p(Lopt S) (1 O())OPT(I)





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p

S S

constantpacking

. . . . . .

NFDH



comple

te

enume

ration

separated by i

Input I

selection with

FracKnap+(S, V 22i)

Packing of Lopt

p(S) V22

i

V OPT(S) (1 22

i

i)OPT(S)






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S S

constantpacking

. . . . . .

NFDH



comple

te

enume

ration

separated by i

Input I

selection with

FracKnap+(S, V 22i)

Packing of Lopt

p(S) V22

i

V OPT(S) (1 22

i

i)OPT(S)




Several Large Items


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S S

constantpacking

. . . . . .

NFDH

Lopt \ {ak}

Knapsack(S, 1 Vol(Lopt), )

P1

P2

constantpacking

complete

enumeration

selection with

separated by i

partition free space in

Input I

Guess Lopt

Same packing without ak Gaps in packing

k 1 items

k items

Packing of Lopt

3|Lopt|gaps



Several Large Items


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Profit P1 k pkProfit P2 p(Lopt S

) pk

0 0.25 0.5 0.75 1

0.5

1

P1 P2

max(P1, P2) (k

k + 1

O())OPT(I)



Several Large Items


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Profit P1 k pkProfit P2 p(Lopt S

) pk

0 0.25 0.5 0.75 1

0.5

1

P1 P2

max(P1, P2) (k

k + 1

O())OPT(I)



Only One Very Large Item amax 1 4


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S3

S2

S1

g1

g2

amaxg1

g2

amax

1 amax

S5

1 + amax

1 amax

1 amax

S4

amax

Use RECTANGLE PACKING WITH LARGE RESOURCES for the free

space




Only One Very Large Item amax 1 4


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S3

S2

S1

g1

g2

amaxg1

g2

amax

1 amax

S5

1 + amax

1 amax

1 amax

S4

amax

Use RECTANGLE PACKING WITH LARGE RESOURCES for the free

space




Outline


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1 Introduction

ProblemPreparation

2 Square Packing




3 Hypercube Packing

Generalization

4 Summary





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We derived methods for

Case 1 Enough remaining space (1 O())OPT(I)

Case 2 Several large items ( kk+1 O())OPT(I)Case 3 Only one very large item (1 O())OPT(I)

We show that k < 4 can be reduced to Case 1 or Case 3



Main Idea


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Three similarly large squares cannot fill a square bin almostcompletely


Introduction Square Packing Hypercube Packing Summary Generalization

Outline


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1 Introduction

ProblemPreparation

2 Square Packing




3 Hypercube Packing

Generalization

4 Summary



Methods for Hypercube Packing


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Direct adoption of 2-dim methodsWith suitable separation parameters

Case 1 Enough remaining space

Case 2 Several large items

work as well

More work needed

For Case 3 we need an approximation algorithm for

ORTHOGONAL KNAPSACK PACKING WITH LARGE

RESOURCES FOR HYPERCUBESwith ratio (1 + ) if the bin is big enough



Methods for Hypercube Packing


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Direct adoption of 2-dim methodsWith suitable separation parameters

Case 1 Enough remaining space

Case 2 Several large items

work as well

More work needed

For Case 3 we need an approximation algorithm for

ORTHOGONAL KNAPSACK PACKING WITH LARGE

RESOURCES FOR HYPERCUBESwith ratio (1 + ) if the bin is big enough



Wellstructured Packing


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amax big enough

amax

x1

x2

x3

1

free of items

space for



Construction of a Wellstructured Packing


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amax big enough

amax

x1

x2

x3

C

H1 H1

H2

H2



Packing the Small Items


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Applying the algorithm for ORTHOGONAL KNAPSACK PACKING

WITH LARGE RESOURCES FOR HYPERCUBES

amax

x1

x2

x3space for

S3

S1S2



Improving Approximation Ratio


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Seven similarly large cubes cannot fill a cube bin almost

completely

In generalFor d-dim Hypercube Packing, we can reduce Case 2 with

k < 2d to Case 1 or Case 3



Improving Approximation Ratio


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Seven similarly large cubes cannot fill a cube bin almost

completely

In generalFor d-dim Hypercube Packing, we can reduce Case 2 with

k < 2d to Case 1 or Case 3



Outline


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1 Introduction

ProblemPreparation

2 Square Packing




3 Hypercube Packing

Generalization

4 Summary



Summary


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ResultWe developed an approximation algorithm with approximation

ratio 1 + 12d

+ for d-dimensional ORTHOGONAL KNAPSACKPACKING FOR HYPERCUBES

Main StepsSeparation of large, medium and small items

Packing the large items

Adding the small items

Three similarly large squares cannot fill a square binalmost completely



Summary


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Result

We developed an approximation algorithm with approximation

ratio 1 + 12d

+ for d-dimensional ORTHOGONAL KNAPSACKPACKING FOR HYPERCUBES

Main StepsSeparation of large, medium and small items

Packing the large items

Adding the small items

Three similarly large squares cannot fill a square binalmost completely



Additional notes


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Practical application of this algorithm

The running time is dominated by huge enumerations, making

the algorithm practically unusable.

Asymptotic behavior

The structure of the problem does not allow asymptotic

algorithms. Neither in the size of the input, nor in the value of

an optimal solution.



Additional notes


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Practical application of this algorithm

The running time is dominated by huge enumerations, making

the algorithm practically unusable.

Asymptotic behavior

The structure of the problem does not allow asymptotic

algorithms. Neither in the size of the input, nor in the value of

an optimal solution.



The End


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Thanks for your attention


Rolf Harren- Approximating the Orthogonal Knapsack Problem for Hypercubes

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