[PPT] - Advanced Algorithms COMS31900 Approximation algorithms part two PowerPoint Presentation

SLIDE 1

Advanced Algorithms – COMS31900 Approximation algorithms part two more constant factor approximations

Rapha¨ el Clifford Slides by Benjamin Sach

SLIDE 2

Approximation Algorithms Recap

An algorithm A is an α-approximation algorithm for problem P if,

A runs in polynomial time
A always outputs a solution with value s

Here P is an optimisation problem with optimal solution of value Opt

If P is a maximisation problem, Opt

α

s Opt

within an α factor of Opt

If P is a minimisation problem, Opt s α · Opt

We have seen a 3/2-approximation algorithm for Bin Packing (and a faster 2-approximation)

SLIDE 3

Scheduling Jobs on Parallel Machines

1 2 3 4 5

m identical

machines

n jobs

time taken Goal: minimise the (wall-clock) time taken to process all jobs

SLIDE 4

Scheduling Jobs on Parallel Machines

1 2 3 4 5

m identical

machines

n jobs

time taken Goal: minimise the (wall-clock) time taken to process all jobs (it’s NP-hard)

SLIDE 5

Scheduling Jobs on Parallel Machines

1 2 3 4 5 Goal: minimise the (wall-clock) time taken to process all jobs

wall-clock time (also called makespan)

SLIDE 6

Scheduling Jobs on Parallel Machines

1 2 3 4 5 Goal: minimise the (wall-clock) time taken to process all jobs

wall-clock time (also called makespan)

Job j takes tj time units

SLIDE 7

Scheduling Jobs on Parallel Machines

1 2 3 4 5 Goal: minimise the (wall-clock) time taken to process all jobs

wall-clock time (also called makespan)

We say that j ∈ J(i) iff job j is assigned to machine i
Job j takes tj time units

SLIDE 8

Scheduling Jobs on Parallel Machines

1 2 3 4 5 Goal: minimise the (wall-clock) time taken to process all jobs

wall-clock time (also called makespan)

We say that j ∈ J(i) iff job j is assigned to machine i
The load of machine i is Li =

j∈J(i) tj

Job j takes tj time units

SLIDE 9

Scheduling Jobs on Parallel Machines

1 2 3 4 5 Goal: minimise the (wall-clock) time taken to process all jobs

wall-clock time (also called makespan)

We say that j ∈ J(i) iff job j is assigned to machine i
The load of machine i is Li =

j∈J(i) tj

Job j takes tj time units

L2 L3 L1

SLIDE 10

Scheduling Jobs on Parallel Machines

1 2 3 4 5 Goal: minimise the (wall-clock) time taken to process all jobs

wall-clock time (also called makespan)

We say that j ∈ J(i) iff job j is assigned to machine i
The load of machine i is Li =

j∈J(i) tj

Job j takes tj time units
So the wall-clock time is maxi Li (which we want to minimise)

L2 L3 L1

SLIDE 11

Scheduling Jobs on Parallel Machines

1 2 3 4 5 Goal: minimise the (wall-clock) time taken to process all jobs

wall-clock time (also called makespan)

We say that j ∈ J(i) iff job j is assigned to machine i
The load of machine i is Li =

j∈J(i) tj

Job j takes tj time units
So the wall-clock time is maxi Li (which we want to minimise)

L2 L3 L1 L1

SLIDE 12

Scheduling Jobs on Parallel Machines

1 2 3 4 5

SLIDE 13

Scheduling Jobs on Parallel Machines

1 2 3 4 5

Algorithm: Put job j on the machine i with smallest (current) load

SLIDE 14

Scheduling Jobs on Parallel Machines

1 2 3 4 5

Algorithm: Put job j on the machine i with smallest (current) load

SLIDE 15

Scheduling Jobs on Parallel Machines

1 2 3 4 5

Algorithm: Put job j on the machine i with smallest (current) load

SLIDE 16

Scheduling Jobs on Parallel Machines

1 2 3 4 5

Algorithm: Put job j on the machine i with smallest (current) load

SLIDE 17

Scheduling Jobs on Parallel Machines

1 2 3 4 5

Algorithm: Put job j on the machine i with smallest (current) load

SLIDE 18

Scheduling Jobs on Parallel Machines

1 2 3 4 5

Algorithm: Put job j on the machine i with smallest (current) load

SLIDE 19

Scheduling Jobs on Parallel Machines

1 2 3 4 5

Algorithm: Put job j on the machine i with smallest (current) load

SLIDE 20

Scheduling Jobs on Parallel Machines

1 2 3 4 5

Algorithm: Put job j on the machine i with smallest (current) load

SLIDE 21

Scheduling Jobs on Parallel Machines

1 2 3 4 5

Algorithm: Put job j on the machine i with smallest (current) load

SLIDE 22

Scheduling Jobs on Parallel Machines

1 2 3 4 5

Algorithm: Put job j on the machine i with smallest (current) load

SLIDE 23

Scheduling Jobs on Parallel Machines

1 2 3 4 5

Algorithm: Put job j on the machine i with smallest (current) load

SLIDE 24

Scheduling Jobs on Parallel Machines

1 2 3 4 5

Algorithm: Put job j on the machine i with smallest (current) load

SLIDE 25

Scheduling Jobs on Parallel Machines

1 2 3 4 5

Algorithm: Put job j on the machine i with smallest (current) load

SLIDE 26

Scheduling Jobs on Parallel Machines

1 2 3 4 5

Algorithm: Put job j on the machine i with smallest (current) load

SLIDE 27

Scheduling Jobs on Parallel Machines

1 2 3 4 5

Algorithm: Put job j on the machine i with smallest (current) load

SLIDE 28

Scheduling Jobs on Parallel Machines

1 2 3 4 5

Algorithm: Put job j on the machine i with smallest (current) load

SLIDE 29

Scheduling Jobs on Parallel Machines

1 2 3 4 5

Algorithm: Put job j on the machine i with smallest (current) load How long does it take to compute this schedule?

SLIDE 30

Scheduling Jobs on Parallel Machines

1 2 3 4 5

Algorithm: Put job j on the machine i with smallest (current) load How long does it take to compute this schedule?

m machines n jobs

SLIDE 31

Scheduling Jobs on Parallel Machines

1 2 3 4 5

Algorithm: Put job j on the machine i with smallest (current) load How long does it take to compute this schedule?

O(nm) time naively, O(n log m) time using a priority queue m machines n jobs

SLIDE 32

Scheduling Jobs on Parallel Machines

1 2 3 4 5

Algorithm: Put job j on the machine i with smallest (current) load How long does it take to compute this schedule?

O(nm) time naively, O(n log m) time using a priority queue

(it’s also an online solution)

m machines n jobs

SLIDE 33

Scheduling Jobs on Parallel Machines

1 2 3 4 5

Algorithm: Put job j on the machine i with smallest (current) load How long does it take to compute this schedule?

O(nm) time naively, O(n log m) time using a priority queue

How good is it? (it’s also an online solution)

m machines n jobs

SLIDE 34

The greedy approximation

Theorem The greedy algorithm given is a 2-approximation algorithm Let Opt denote the time taken by the optimal scheduling of jobs Let Tg denote the time taken by the greedy schedule Job j takes tj time units

Li is the load of

machine i

m machines n jobs

SLIDE 35

The greedy approximation

Before we prove this, we prove two useful facts,

Theorem The greedy algorithm given is a 2-approximation algorithm Let Opt denote the time taken by the optimal scheduling of jobs Let Tg denote the time taken by the greedy schedule Job j takes tj time units

Li is the load of

machine i

m machines n jobs

SLIDE 36

The greedy approximation

Before we prove this, we prove two useful facts,

Fact Opt maxj tj Theorem The greedy algorithm given is a 2-approximation algorithm Let Opt denote the time taken by the optimal scheduling of jobs Let Tg denote the time taken by the greedy schedule Job j takes tj time units

Li is the load of

machine i

m machines n jobs

SLIDE 37

The greedy approximation

Before we prove this, we prove two useful facts,

Fact Opt maxj tj

Some machine must process the largest job

Theorem The greedy algorithm given is a 2-approximation algorithm Let Opt denote the time taken by the optimal scheduling of jobs Let Tg denote the time taken by the greedy schedule Job j takes tj time units

Li is the load of

machine i

m machines n jobs

SLIDE 38

The greedy approximation

Before we prove this, we prove two useful facts,

Fact Opt maxj tj Fact Opt

j tj

m

Some machine must process the largest job

Theorem The greedy algorithm given is a 2-approximation algorithm Let Opt denote the time taken by the optimal scheduling of jobs Let Tg denote the time taken by the greedy schedule Job j takes tj time units

Li is the load of

machine i

m machines n jobs

SLIDE 39

The greedy approximation

Before we prove this, we prove two useful facts,

Fact Opt maxj tj Fact Opt

j tj

m

Some machine must process the largest job
There is a total of

j tj time units of work to be done

Theorem The greedy algorithm given is a 2-approximation algorithm Let Opt denote the time taken by the optimal scheduling of jobs Let Tg denote the time taken by the greedy schedule Job j takes tj time units

Li is the load of

machine i

m machines n jobs

SLIDE 40

The greedy approximation

Before we prove this, we prove two useful facts,

Fact Opt maxj tj Fact Opt

j tj

m

Some machine must process the largest job
There is a total of

j tj time units of work to be done

Some machine i must have load Li at least
j tj

m

Theorem The greedy algorithm given is a 2-approximation algorithm Let Opt denote the time taken by the optimal scheduling of jobs Let Tg denote the time taken by the greedy schedule Job j takes tj time units

Li is the load of

machine i

m machines n jobs

SLIDE 41

The greedy approximation

Before we prove this, we prove two useful facts,

Fact Opt maxj tj Fact Opt

j tj

m

Some machine must process the largest job
There is a total of

j tj time units of work to be done

Some machine i must have load Li at least
j tj

m

(the m machines can’t all have below average load) Theorem The greedy algorithm given is a 2-approximation algorithm Let Opt denote the time taken by the optimal scheduling of jobs Let Tg denote the time taken by the greedy schedule Job j takes tj time units

Li is the load of

machine i

m machines n jobs

SLIDE 42

The greedy approximation

Theorem The greedy algorithm given is a 2-approximation algorithm Let Opt denote the time taken by the optimal scheduling of jobs Let Tg denote the time taken by the greedy schedule

Li is the load of

machine i Job j takes tj time units

m machines n jobs

SLIDE 43

The greedy approximation

Theorem The greedy algorithm given is a 2-approximation algorithm Proof Consider the machine i with largest load Tg = Li Let Opt denote the time taken by the optimal scheduling of jobs Let Tg denote the time taken by the greedy schedule

Li is the load of

machine i Job j takes tj time units

m machines n jobs

SLIDE 44

The greedy approximation

Theorem The greedy algorithm given is a 2-approximation algorithm Proof Consider the machine i with largest load Tg = Li Let Opt denote the time taken by the optimal scheduling of jobs Let Tg denote the time taken by the greedy schedule

Li is the load of

machine i Job j takes tj time units

1 2 3 4 5

Li = Tg

m machines n jobs

SLIDE 45

The greedy approximation

Theorem The greedy algorithm given is a 2-approximation algorithm

Let j denote the last job machine i completes

Proof Consider the machine i with largest load Tg = Li Let Opt denote the time taken by the optimal scheduling of jobs Let Tg denote the time taken by the greedy schedule

Li is the load of

machine i Job j takes tj time units

1 2 3 4 5

Li = Tg

m machines n jobs

SLIDE 46

The greedy approximation

Theorem The greedy algorithm given is a 2-approximation algorithm

Let j denote the last job machine i completes

Proof Consider the machine i with largest load Tg = Li Let Opt denote the time taken by the optimal scheduling of jobs Let Tg denote the time taken by the greedy schedule

Li is the load of

machine i Job j takes tj time units

1 2 3 4 5

Li = Tg

m machines n jobs

job j

SLIDE 47

The greedy approximation

Theorem The greedy algorithm given is a 2-approximation algorithm

Let j denote the last job machine i completes

Proof Consider the machine i with largest load Tg = Li

When job j was assigned, machine i had the smallest load, Li − tj

Let Opt denote the time taken by the optimal scheduling of jobs Let Tg denote the time taken by the greedy schedule

Li is the load of

machine i Job j takes tj time units

1 2 3 4 5

Li = Tg

m machines n jobs

job j

SLIDE 48

The greedy approximation

Theorem The greedy algorithm given is a 2-approximation algorithm

Let j denote the last job machine i completes

Proof Consider the machine i with largest load Tg = Li

When job j was assigned, machine i had the smallest load, Li − tj

Let Opt denote the time taken by the optimal scheduling of jobs Let Tg denote the time taken by the greedy schedule

Li is the load of

machine i Job j takes tj time units

1 2 3 4 5

Li = Tg

m machines n jobs

SLIDE 49

The greedy approximation

Theorem The greedy algorithm given is a 2-approximation algorithm

Let j denote the last job machine i completes

Proof Consider the machine i with largest load Tg = Li

When job j was assigned, machine i had the smallest load, Li − tj

Let Opt denote the time taken by the optimal scheduling of jobs Let Tg denote the time taken by the greedy schedule

Li is the load of

machine i Job j takes tj time units

1 2 3 4 5

Li = Tg

m machines n jobs

job j

SLIDE 50

The greedy approximation

Theorem The greedy algorithm given is a 2-approximation algorithm

Let j denote the last job machine i completes

Proof Consider the machine i with largest load Tg = Li

When job j was assigned, machine i had the smallest load, Li − tj
So. . .

Let Opt denote the time taken by the optimal scheduling of jobs Let Tg denote the time taken by the greedy schedule

Li is the load of

machine i Job j takes tj time units

1 2 3 4 5

Li = Tg

m machines n jobs

job j

SLIDE 51

The greedy approximation

Theorem The greedy algorithm given is a 2-approximation algorithm

Let j denote the last job machine i completes

Proof Consider the machine i with largest load Tg = Li

When job j was assigned, machine i had the smallest load, Li − tj

Li − tj Lk for all 1 k m,

So. . .

Let Opt denote the time taken by the optimal scheduling of jobs Let Tg denote the time taken by the greedy schedule

Li is the load of

machine i Job j takes tj time units

1 2 3 4 5

Li = Tg

m machines n jobs

job j

SLIDE 52

The greedy approximation

Theorem The greedy algorithm given is a 2-approximation algorithm

Let j denote the last job machine i completes

Proof Consider the machine i with largest load Tg = Li

When job j was assigned, machine i had the smallest load, Li − tj

Li − tj Lk for all 1 k m,

So. . .
If we then sum over all k,

Let Opt denote the time taken by the optimal scheduling of jobs Let Tg denote the time taken by the greedy schedule

Li is the load of

machine i Job j takes tj time units

1 2 3 4 5

Li = Tg

m machines n jobs

job j

SLIDE 53

The greedy approximation

Theorem The greedy algorithm given is a 2-approximation algorithm

Let j denote the last job machine i completes

Proof Consider the machine i with largest load Tg = Li

When job j was assigned, machine i had the smallest load, Li − tj

Li − tj Lk for all 1 k m,

So. . .
If we then sum over all k,

m(Li − tj)

m

k=1

Lk

Let Opt denote the time taken by the optimal scheduling of jobs Let Tg denote the time taken by the greedy schedule

Li is the load of

machine i Job j takes tj time units

1 2 3 4 5

Li = Tg

m machines n jobs

job j

SLIDE 54

The greedy approximation

Theorem The greedy algorithm given is a 2-approximation algorithm

Let j denote the last job machine i completes

Proof Consider the machine i with largest load Tg = Li

When job j was assigned, machine i had the smallest load, Li − tj

Li − tj Lk for all 1 k m,

So. . .
If we then sum over all k,

m(Li − tj)

m

k=1

Lk

so (Li − tj)

m

k=1 Lk

m

Opt (by the second fact)

Let Opt denote the time taken by the optimal scheduling of jobs Let Tg denote the time taken by the greedy schedule

Li is the load of

machine i Job j takes tj time units

1 2 3 4 5

Li = Tg

m machines n jobs

job j

SLIDE 55

The greedy approximation

Theorem The greedy algorithm given is a 2-approximation algorithm

Let j denote the last job machine i completes

Proof Consider the machine i with largest load Tg = Li

When job j was assigned, machine i had the smallest load, Li − tj

Li − tj Lk for all 1 k m,

So. . .
If we then sum over all k,

m(Li − tj)

m

k=1

Lk

so (Li − tj)

m

k=1 Lk

m

Opt (by the second fact)

Let Opt denote the time taken by the optimal scheduling of jobs Let Tg denote the time taken by the greedy schedule

Li is the load of

machine i Job j takes tj time units

1 2 3 4 5

Li = Tg

m machines n jobs

job j

Fact Opt

j tj

m

SLIDE 56

The greedy approximation

Theorem The greedy algorithm given is a 2-approximation algorithm

Let j denote the last job machine i completes

Proof Consider the machine i with largest load Tg = Li

When job j was assigned, machine i had the smallest load, Li − tj

Li − tj Lk for all 1 k m,

So. . .
If we then sum over all k,

m(Li − tj)

m

k=1

Lk

so (Li − tj)

m

k=1 Lk

m

Opt (by the second fact)

Let Opt denote the time taken by the optimal scheduling of jobs Let Tg denote the time taken by the greedy schedule

Li is the load of

machine i Job j takes tj time units

1 2 3 4 5

Li = Tg

m machines n jobs

job j

SLIDE 57

The greedy approximation

Theorem The greedy algorithm given is a 2-approximation algorithm

Let j denote the last job machine i completes

Proof Consider the machine i with largest load Tg = Li

When job j was assigned, machine i had the smallest load, Li − tj

Li − tj Lk for all 1 k m,

So. . .
If we then sum over all k,

m(Li − tj)

m

k=1

Lk

so (Li − tj)

m

k=1 Lk

m

Opt (by the second fact)

also tj Opt (by the first fact) Let Opt denote the time taken by the optimal scheduling of jobs Let Tg denote the time taken by the greedy schedule

Li is the load of

machine i Job j takes tj time units

1 2 3 4 5

Li = Tg

m machines n jobs

job j

SLIDE 58

The greedy approximation

Theorem The greedy algorithm given is a 2-approximation algorithm

Let j denote the last job machine i completes

Proof Consider the machine i with largest load Tg = Li

When job j was assigned, machine i had the smallest load, Li − tj

Li − tj Lk for all 1 k m,

So. . .
If we then sum over all k,

m(Li − tj)

m

k=1

Lk

so (Li − tj)

m

k=1 Lk

m

Opt (by the second fact)

also tj Opt (by the first fact) Let Opt denote the time taken by the optimal scheduling of jobs Let Tg denote the time taken by the greedy schedule

Li is the load of

machine i Job j takes tj time units

1 2 3 4 5

Li = Tg

m machines n jobs

job j

Fact Opt maxj tj

SLIDE 59

The greedy approximation

Theorem The greedy algorithm given is a 2-approximation algorithm

Let j denote the last job machine i completes

Proof Consider the machine i with largest load Tg = Li

When job j was assigned, machine i had the smallest load, Li − tj

Li − tj Lk for all 1 k m,

So. . .
If we then sum over all k,

m(Li − tj)

m

k=1

Lk

so (Li − tj)

m

k=1 Lk

m

Opt (by the second fact)

also tj Opt (by the first fact) Let Opt denote the time taken by the optimal scheduling of jobs Let Tg denote the time taken by the greedy schedule

Li is the load of

machine i Job j takes tj time units

1 2 3 4 5

Li = Tg

m machines n jobs

job j

SLIDE 60

The greedy approximation

Theorem The greedy algorithm given is a 2-approximation algorithm

Let j denote the last job machine i completes

Proof Consider the machine i with largest load Tg = Li

When job j was assigned, machine i had the smallest load, Li − tj

Li − tj Lk for all 1 k m,

So. . .
If we then sum over all k,

m(Li − tj)

m

k=1

Lk

so (Li − tj)

m

k=1 Lk

m

Opt (by the second fact)

also tj Opt (by the first fact) Therefore, Tg = Li = (Li − tj) + tj Opt + Opt = 2Opt Let Opt denote the time taken by the optimal scheduling of jobs Let Tg denote the time taken by the greedy schedule

Li is the load of

machine i Job j takes tj time units

1 2 3 4 5

Li = Tg

m machines n jobs

job j

SLIDE 61

Longest Processing Time (LPT)

1 2 3 4 5

Step 1: Sort the jobs into non-increasing order (job 1 is now largest)

SLIDE 62

Longest Processing Time (LPT)

1 2 3 4 5

Step 1: Sort the jobs into non-increasing order (job 1 is now largest)

SLIDE 63

Longest Processing Time (LPT)

1 2 3 4 5

Step 1: Sort the jobs into non-increasing order (job 1 is now largest) Step 2: Put job j on the machine i with smallest (current) load

SLIDE 64

Longest Processing Time (LPT)

1 2 3 4 5

Step 1: Sort the jobs into non-increasing order (job 1 is now largest) Step 2: Put job j on the machine i with smallest (current) load

SLIDE 65

Longest Processing Time (LPT)

1 2 3 4 5

Step 1: Sort the jobs into non-increasing order (job 1 is now largest) Step 2: Put job j on the machine i with smallest (current) load

SLIDE 66

Longest Processing Time (LPT)

1 2 3 4 5

Step 1: Sort the jobs into non-increasing order (job 1 is now largest) Step 2: Put job j on the machine i with smallest (current) load

SLIDE 67

Longest Processing Time (LPT)

1 2 3 4 5

Step 1: Sort the jobs into non-increasing order (job 1 is now largest) Step 2: Put job j on the machine i with smallest (current) load

SLIDE 68

Longest Processing Time (LPT)

1 2 3 4 5

Step 1: Sort the jobs into non-increasing order (job 1 is now largest) Step 2: Put job j on the machine i with smallest (current) load

SLIDE 69

Longest Processing Time (LPT)

1 2 3 4 5

Step 1: Sort the jobs into non-increasing order (job 1 is now largest) Step 2: Put job j on the machine i with smallest (current) load

SLIDE 70

Longest Processing Time (LPT)

1 2 3 4 5

Step 1: Sort the jobs into non-increasing order (job 1 is now largest) Step 2: Put job j on the machine i with smallest (current) load

SLIDE 71

Longest Processing Time (LPT)

1 2 3 4 5

Step 1: Sort the jobs into non-increasing order (job 1 is now largest) Step 2: Put job j on the machine i with smallest (current) load

SLIDE 72

Longest Processing Time (LPT)

1 2 3 4 5

Step 1: Sort the jobs into non-increasing order (job 1 is now largest) Step 2: Put job j on the machine i with smallest (current) load

SLIDE 73

Longest Processing Time (LPT)

1 2 3 4 5

Step 1: Sort the jobs into non-increasing order (job 1 is now largest) Step 2: Put job j on the machine i with smallest (current) load

SLIDE 74

Longest Processing Time (LPT)

1 2 3 4 5

Step 1: Sort the jobs into non-increasing order (job 1 is now largest) Step 2: Put job j on the machine i with smallest (current) load

SLIDE 75

Longest Processing Time (LPT)

1 2 3 4 5

Step 1: Sort the jobs into non-increasing order (job 1 is now largest) Step 2: Put job j on the machine i with smallest (current) load

SLIDE 76

Longest Processing Time (LPT)

1 2 3 4 5

Step 1: Sort the jobs into non-increasing order (job 1 is now largest) Step 2: Put job j on the machine i with smallest (current) load

SLIDE 77

Longest Processing Time (LPT)

1 2 3 4 5

Step 1: Sort the jobs into non-increasing order (job 1 is now largest) Step 2: Put job j on the machine i with smallest (current) load

SLIDE 78

Longest Processing Time (LPT)

1 2 3 4 5

Step 1: Sort the jobs into non-increasing order (job 1 is now largest) Step 2: Put job j on the machine i with smallest (current) load

SLIDE 79

Longest Processing Time (LPT)

1 2 3 4 5

Step 1: Sort the jobs into non-increasing order (job 1 is now largest) Step 2: Put job j on the machine i with smallest (current) load

greedy solution

SLIDE 80

Longest Processing Time (LPT)

1 2 3 4 5

Step 1: Sort the jobs into non-increasing order (job 1 is now largest) Step 2: Put job j on the machine i with smallest (current) load

greedy solution

How long does it take to compute this schedule?

SLIDE 81

Longest Processing Time (LPT)

1 2 3 4 5

Step 1: Sort the jobs into non-increasing order (job 1 is now largest) Step 2: Put job j on the machine i with smallest (current) load

greedy solution

How long does it take to compute this schedule?

O(n log n) time (to sort the jobs)

SLIDE 82

Longest Processing Time (LPT)

1 2 3 4 5

Step 1: Sort the jobs into non-increasing order (job 1 is now largest) Step 2: Put job j on the machine i with smallest (current) load

greedy solution

How long does it take to compute this schedule?

O(n log n) time (to sort the jobs)

How good is it?

SLIDE 83

The LPT approximation

Theorem The LPT algorithm is a 3/2-approximation algorithm

Let Tl denote the time taken by the LPT schedule

Li is the load of

machine i

m machines n jobs

Job j takes tj time units

SLIDE 84

The LPT approximation

Theorem The LPT algorithm is a 3/2-approximation algorithm

Before we prove this, we prove another useful fact and a Lemma
Let Tl denote the time taken by the LPT schedule

Li is the load of

machine i

m machines n jobs

Job j takes tj time units

SLIDE 85

The LPT approximation

Theorem The LPT algorithm is a 3/2-approximation algorithm

Before we prove this, we prove another useful fact and a Lemma

Fact If there are at most m jobs (n m) then LPT is optimal

Let Tl denote the time taken by the LPT schedule

Li is the load of

machine i

m machines n jobs

Job j takes tj time units

SLIDE 86

The LPT approximation

Theorem The LPT algorithm is a 3/2-approximation algorithm

Before we prove this, we prove another useful fact and a Lemma

Fact If there are at most m jobs (n m) then LPT is optimal LPT gives each job its own machine so maxi Li maxj tj Opt

Let Tl denote the time taken by the LPT schedule

Li is the load of

machine i

m machines n jobs

Job j takes tj time units If there are at most m jobs then

SLIDE 87

The LPT approximation

Theorem The LPT algorithm is a 3/2-approximation algorithm

Before we prove this, we prove another useful fact and a Lemma

Fact If there are at most m jobs (n m) then LPT is optimal Lemma If n > m then Opt 2t(m+1) (after sorting) LPT gives each job its own machine so maxi Li maxj tj Opt

Let Tl denote the time taken by the LPT schedule

Li is the load of

machine i

m machines n jobs

Job j takes tj time units If there are at most m jobs then

SLIDE 88

The LPT approximation

Theorem The LPT algorithm is a 3/2-approximation algorithm

Before we prove this, we prove another useful fact and a Lemma

Fact If there are at most m jobs (n m) then LPT is optimal Lemma If n > m then Opt 2t(m+1) (after sorting) LPT gives each job its own machine so maxi Li maxj tj Opt

Let Tl denote the time taken by the LPT schedule

Proof

Li is the load of

machine i

m machines n jobs

Job j takes tj time units If there are at most m jobs then

SLIDE 89

The LPT approximation

Theorem The LPT algorithm is a 3/2-approximation algorithm

Before we prove this, we prove another useful fact and a Lemma

Fact If there are at most m jobs (n m) then LPT is optimal Lemma If n > m then Opt 2t(m+1) (after sorting) LPT gives each job its own machine so maxi Li maxj tj Opt

Note that t1 t2 t3 . . . tm t(m+1)
Let Tl denote the time taken by the LPT schedule

Proof

Li is the load of

machine i

m machines n jobs

Job j takes tj time units If there are at most m jobs then

SLIDE 90

The LPT approximation

Theorem The LPT algorithm is a 3/2-approximation algorithm

Before we prove this, we prove another useful fact and a Lemma

Fact If there are at most m jobs (n m) then LPT is optimal Lemma If n > m then Opt 2t(m+1) (after sorting) LPT gives each job its own machine so maxi Li maxj tj Opt

Note that t1 t2 t3 . . . tm t(m+1)
One of the m machines must be assigned
Let Tl denote the time taken by the LPT schedule

Proof (at least) two of these m + 1 jobs under any schedule

Li is the load of

machine i

m machines n jobs

Job j takes tj time units If there are at most m jobs then

SLIDE 91

The LPT approximation

Theorem The LPT algorithm is a 3/2-approximation algorithm

Before we prove this, we prove another useful fact and a Lemma

Fact If there are at most m jobs (n m) then LPT is optimal Lemma If n > m then Opt 2t(m+1) (after sorting) LPT gives each job its own machine so maxi Li maxj tj Opt

Note that t1 t2 t3 . . . tm t(m+1)
One of the m machines must be assigned
Let Tl denote the time taken by the LPT schedule

Proof (at least) two of these m + 1 jobs under any schedule

So we have that any schedule takes at least 2t(m+1) time

Li is the load of

machine i

m machines n jobs

Job j takes tj time units If there are at most m jobs then

SLIDE 92

The LPT approximation

Theorem The LPT algorithm is a 3/2-approximation algorithm

Before we prove this, we prove another useful fact and a Lemma

Fact If there are at most m jobs (n m) then LPT is optimal Lemma If n > m then Opt 2t(m+1) (after sorting) LPT gives each job its own machine so maxi Li maxj tj Opt

Note that t1 t2 t3 . . . tm t(m+1)
One of the m machines must be assigned
Let Tl denote the time taken by the LPT schedule

Proof (at least) two of these m + 1 jobs under any schedule

So we have that any schedule takes at least 2t(m+1) time

Li is the load of

machine i

m machines n jobs

in particular Opt 2t(m+1) Job j takes tj time units If there are at most m jobs then

SLIDE 93

The LPT approximation

Theorem The LPT algorithm is a 3/2-approximation algorithm

Li is the load of

machine i

m machines n jobs

Job j takes tj time units

SLIDE 94

The LPT approximation

Theorem The LPT algorithm is a 3/2-approximation algorithm Proof Consider the machine i with largest load Tl = Li

Li is the load of

machine i

m machines n jobs

Job j takes tj time units

SLIDE 95

The LPT approximation

Theorem The LPT algorithm is a 3/2-approximation algorithm

Let j denote the last job machine i completes

Proof Consider the machine i with largest load Tl = Li

Li is the load of

machine i

m machines n jobs

Job j takes tj time units

SLIDE 96

The LPT approximation

Theorem The LPT algorithm is a 3/2-approximation algorithm

Let j denote the last job machine i completes

Proof Consider the machine i with largest load Tl = Li

Li is the load of

machine i

m machines n jobs

Job j takes tj time units

job j 1 2 3 4 5

SLIDE 97

The LPT approximation

Theorem The LPT algorithm is a 3/2-approximation algorithm

Let j denote the last job machine i completes

Proof Consider the machine i with largest load Tl = Li

Using the same argument as before, we have that,

Li is the load of

machine i

m machines n jobs

Job j takes tj time units

job j 1 2 3 4 5

SLIDE 98

The LPT approximation

Theorem The LPT algorithm is a 3/2-approximation algorithm

Let j denote the last job machine i completes

Proof Consider the machine i with largest load Tl = Li

Using the same argument as before, we have that,

(Li − tj) Opt Li is the load of

machine i

m machines n jobs

Job j takes tj time units

job j 1 2 3 4 5

SLIDE 99

The LPT approximation

Theorem The LPT algorithm is a 3/2-approximation algorithm

Let j denote the last job machine i completes

Proof Consider the machine i with largest load Tl = Li

Using the same argument as before, we have that,

(Li − tj) Opt

If n m then we are done so assume n > m

Li is the load of

machine i

m machines n jobs

Job j takes tj time units

job j 1 2 3 4 5

SLIDE 100

The LPT approximation

Theorem The LPT algorithm is a 3/2-approximation algorithm

Let j denote the last job machine i completes

Proof Consider the machine i with largest load Tl = Li

Using the same argument as before, we have that,

(Li − tj) Opt

If n m then we are done so assume n > m

Li is the load of

machine i

m machines n jobs

Job j takes tj time units

job j 1 2 3 4 5 because LPT is optimal in this case

SLIDE 101

The LPT approximation

Theorem The LPT algorithm is a 3/2-approximation algorithm

Let j denote the last job machine i completes

Proof Consider the machine i with largest load Tl = Li

Using the same argument as before, we have that,

(Li − tj) Opt

If n m then we are done so assume n > m

Li is the load of

machine i

m machines n jobs

Job j takes tj time units

job j 1 2 3 4 5

SLIDE 102

The LPT approximation

Theorem The LPT algorithm is a 3/2-approximation algorithm

Let j denote the last job machine i completes

Proof Consider the machine i with largest load Tl = Li

Using the same argument as before, we have that,

(Li − tj) Opt

If n m then we are done so assume n > m
Further if (Li − tj) = 0 then Tl = Li = tj Opt

Li is the load of

machine i

m machines n jobs

Job j takes tj time units

job j 1 2 3 4 5

SLIDE 103

The LPT approximation

Theorem The LPT algorithm is a 3/2-approximation algorithm

Let j denote the last job machine i completes

Proof Consider the machine i with largest load Tl = Li

Using the same argument as before, we have that,

(Li − tj) Opt

If n m then we are done so assume n > m
Further if (Li − tj) = 0 then Tl = Li = tj Opt

Li is the load of

machine i

m machines n jobs

Job j takes tj time units

job j 1 2 3 4 5

Fact Opt maxj tj

SLIDE 104

The LPT approximation

Theorem The LPT algorithm is a 3/2-approximation algorithm

Let j denote the last job machine i completes

Proof Consider the machine i with largest load Tl = Li

Using the same argument as before, we have that,

(Li − tj) Opt

If n m then we are done so assume n > m
Further if (Li − tj) = 0 then Tl = Li = tj Opt

Li is the load of

machine i

m machines n jobs

Job j takes tj time units

job j 1 2 3 4 5

SLIDE 105

The LPT approximation

Theorem The LPT algorithm is a 3/2-approximation algorithm

Let j denote the last job machine i completes

Proof Consider the machine i with largest load Tl = Li

Using the same argument as before, we have that,

(Li − tj) Opt

If n m then we are done so assume n > m
Further if (Li − tj) = 0 then Tl = Li = tj Opt

so assume that (Li − tj) > 0

Li is the load of

machine i

m machines n jobs

Job j takes tj time units

job j 1 2 3 4 5

SLIDE 106

The LPT approximation

Theorem The LPT algorithm is a 3/2-approximation algorithm

Let j denote the last job machine i completes

Proof Consider the machine i with largest load Tl = Li

Using the same argument as before, we have that,

(Li − tj) Opt

If n m then we are done so assume n > m
Further if (Li − tj) = 0 then Tl = Li = tj Opt

so assume that (Li − tj) > 0

Therefore machine i was assigned at least two jobs

Li is the load of

machine i

m machines n jobs

Job j takes tj time units

job j 1 2 3 4 5

SLIDE 107

The LPT approximation

Theorem The LPT algorithm is a 3/2-approximation algorithm

Let j denote the last job machine i completes

Proof Consider the machine i with largest load Tl = Li

Using the same argument as before, we have that,

(Li − tj) Opt

If n m then we are done so assume n > m
Further if (Li − tj) = 0 then Tl = Li = tj Opt

so assume that (Li − tj) > 0

Therefore machine i was assigned at least two jobs

By the algorithm description, we have that j m + 1

Li is the load of

machine i

m machines n jobs

Job j takes tj time units

job j 1 2 3 4 5

SLIDE 108

The LPT approximation

Theorem The LPT algorithm is a 3/2-approximation algorithm

Let j denote the last job machine i completes

Proof Consider the machine i with largest load Tl = Li

Using the same argument as before, we have that,

(Li − tj) Opt

If n m then we are done so assume n > m
Further if (Li − tj) = 0 then Tl = Li = tj Opt

so assume that (Li − tj) > 0

Therefore machine i was assigned at least two jobs

By the algorithm description, we have that j m + 1

Li is the load of

machine i

m machines n jobs

Job j takes tj time units

job j 1 2 3 4 5

it doesn’t assign a second job to any machine until

every machine has at least one job

SLIDE 109

The LPT approximation

Theorem The LPT algorithm is a 3/2-approximation algorithm

Let j denote the last job machine i completes

Proof Consider the machine i with largest load Tl = Li

Using the same argument as before, we have that,

(Li − tj) Opt

If n m then we are done so assume n > m
Further if (Li − tj) = 0 then Tl = Li = tj Opt

so assume that (Li − tj) > 0

Therefore machine i was assigned at least two jobs

By the algorithm description, we have that j m + 1

Li is the load of

machine i

m machines n jobs

Job j takes tj time units

job j 1 2 3 4 5

SLIDE 110

The LPT approximation

Theorem The LPT algorithm is a 3/2-approximation algorithm

Let j denote the last job machine i completes

Proof Consider the machine i with largest load Tl = Li

Using the same argument as before, we have that,

(Li − tj) Opt

If n m then we are done so assume n > m
Further if (Li − tj) = 0 then Tl = Li = tj Opt

so assume that (Li − tj) > 0

Therefore machine i was assigned at least two jobs

By the algorithm description, we have that j m + 1

tj tm+1 Opt/2 (by the Lemma) Li is the load of

machine i

m machines n jobs

Job j takes tj time units

job j 1 2 3 4 5

SLIDE 111

The LPT approximation

Theorem The LPT algorithm is a 3/2-approximation algorithm

Let j denote the last job machine i completes

Proof Consider the machine i with largest load Tl = Li

Using the same argument as before, we have that,

(Li − tj) Opt

If n m then we are done so assume n > m
Further if (Li − tj) = 0 then Tl = Li = tj Opt

so assume that (Li − tj) > 0

Therefore machine i was assigned at least two jobs

By the algorithm description, we have that j m + 1

tj tm+1 Opt/2 (by the Lemma) Li is the load of

machine i

m machines n jobs

Job j takes tj time units

job j 1 2 3 4 5

Lemma If n > m then Opt 2t(m+1) (after sorting)

SLIDE 112

The LPT approximation

Theorem The LPT algorithm is a 3/2-approximation algorithm

Let j denote the last job machine i completes

Proof Consider the machine i with largest load Tl = Li

Using the same argument as before, we have that,

(Li − tj) Opt

If n m then we are done so assume n > m
Further if (Li − tj) = 0 then Tl = Li = tj Opt

so assume that (Li − tj) > 0

Therefore machine i was assigned at least two jobs

By the algorithm description, we have that j m + 1

tj tm+1 Opt/2 (by the Lemma) Li is the load of

machine i

m machines n jobs

Job j takes tj time units

job j 1 2 3 4 5

SLIDE 113

The LPT approximation

Theorem The LPT algorithm is a 3/2-approximation algorithm

Let j denote the last job machine i completes

Proof Consider the machine i with largest load Tl = Li

Using the same argument as before, we have that,

(Li − tj) Opt

Therefore, Tl = Li = (Li − tj) + tj Opt + Opt/2 = (3/2) · Opt

If n m then we are done so assume n > m
Further if (Li − tj) = 0 then Tl = Li = tj Opt

so assume that (Li − tj) > 0

Therefore machine i was assigned at least two jobs

By the algorithm description, we have that j m + 1

tj tm+1 Opt/2 (by the Lemma) Li is the load of

machine i

m machines n jobs

Job j takes tj time units

job j 1 2 3 4 5

SLIDE 114

Scheduling conclusions

Theorem The LPT algorithm is a 3/2-approximation algorithm which runs in O(n log n) time Theorem The greedy algorithm is a 2-approximation algorithm which runs in O(n log m) time and it’s online In fact, LPT is a 4/3-approximation algorithm (using better analysis)

m machines n jobs

SLIDE 115

k-centers

SLIDE 116

k-centers

SLIDE 117

k-centers

n points (sites) in 2D space

SLIDE 118

k-centers

n points (sites) in 2D space

The distance between points si, sj is

(xi − xj)2 + (yi − yj)2

SLIDE 119

k-centers

n points (sites) in 2D space

The distance between points si, sj is

(xi − xj)2 + (yi − yj)2

(i.e. ‘normal’ euclidean distance)

SLIDE 120

k-centers

n points (sites) in 2D space

The distance between points si, sj is

(xi − xj)2 + (yi − yj)2

Select k sites to be centers

(i.e. ‘normal’ euclidean distance)

SLIDE 121

k-centers

n points (sites) in 2D space

The distance between points si, sj is

(xi − xj)2 + (yi − yj)2

Select k sites to be centers

(i.e. ‘normal’ euclidean distance)

SLIDE 122

k-centers

n points (sites) in 2D space

The distance between points si, sj is

(xi − xj)2 + (yi − yj)2

Select k sites to be centers

SLIDE 123

k-centers

n points (sites) in 2D space

The distance between points si, sj is

(xi − xj)2 + (yi − yj)2

Goal Minimise the largest distance from any site to the closest center Select k sites to be centers

SLIDE 124

k-centers

n points (sites) in 2D space

The distance between points si, sj is

(xi − xj)2 + (yi − yj)2

Goal Minimise the largest distance from any site to the closest center Select k sites to be centers

SLIDE 125

k-centers

n points (sites) in 2D space

The distance between points si, sj is

(xi − xj)2 + (yi − yj)2

Goal Minimise the largest distance from any site to the closest center Select k sites to be centers

r r

SLIDE 126

k-centers

n points (sites) in 2D space

The distance between points si, sj is

(xi − xj)2 + (yi − yj)2

Goal Minimise the largest distance from any site to the closest center Select k sites to be centers

SLIDE 127

k-centers

n points (sites) in 2D space

The distance between points si, sj is

(xi − xj)2 + (yi − yj)2

Goal Minimise the largest distance from any site to the closest center Select k sites to be centers

SLIDE 128

k-centers

n points (sites) in 2D space

The distance between points si, sj is

(xi − xj)2 + (yi − yj)2

Goal Minimise the largest distance from any site to the closest center Select k sites to be centers

r

SLIDE 129

k-centers

n points (sites) in 2D space

The distance between points si, sj is

(xi − xj)2 + (yi − yj)2

Goal Minimise the largest distance from any site to the closest center (in general it’s NP-hard) Select k sites to be centers

r

SLIDE 130

A Greedy approximation

SLIDE 131

Start by picking any point to be a center

A Greedy approximation

SLIDE 132

Start by picking any point to be a center

A Greedy approximation

SLIDE 133

Start by picking any point to be a center Repeatedly pick the site which is furthest from any existing center

A Greedy approximation

SLIDE 134

Start by picking any point to be a center Repeatedly pick the site which is furthest from any existing center

A Greedy approximation

SLIDE 135

Start by picking any point to be a center Repeatedly pick the site which is furthest from any existing center

A Greedy approximation

SLIDE 136

Start by picking any point to be a center Repeatedly pick the site which is furthest from any existing center

A Greedy approximation

SLIDE 137

Start by picking any point to be a center Repeatedly pick the site which is furthest from any existing center

A Greedy approximation

SLIDE 138

Start by picking any point to be a center Repeatedly pick the site which is furthest from any existing center

A Greedy approximation

SLIDE 139

Start by picking any point to be a center Repeatedly pick the site which is furthest from any existing center

A Greedy approximation

SLIDE 140

Start by picking any point to be a center Repeatedly pick the site which is furthest from any existing center

A Greedy approximation

SLIDE 141

Start by picking any point to be a center Repeatedly pick the site which is furthest from any existing center

A Greedy approximation

SLIDE 142

Start by picking any point to be a center Repeatedly pick the site which is furthest from any existing center

A Greedy approximation

SLIDE 143

Start by picking any point to be a center Repeatedly pick the site which is furthest from any existing center

A Greedy approximation

SLIDE 144

Start by picking any point to be a center Repeatedly pick the site which is furthest from any existing center

A Greedy approximation

SLIDE 145

Start by picking any point to be a center Repeatedly pick the site which is furthest from any existing center

A Greedy approximation

SLIDE 146

Start by picking any point to be a center Repeatedly pick the site which is furthest from any existing center

A Greedy approximation

SLIDE 147

Start by picking any point to be a center Repeatedly pick the site which is furthest from any existing center

A Greedy approximation r

SLIDE 148

Start by picking any point to be a center Repeatedly pick the site which is furthest from any existing center This takes O(nk) time

A Greedy approximation

SLIDE 149

Start by picking any point to be a center Repeatedly pick the site which is furthest from any existing center This takes O(nk) time

A Greedy approximation

but is it any good?

SLIDE 150

The Greedy approximation

Theorem The Greedy algorithm for k-center is a 2-approximation algorithm Let Cg (resp. COpt) denote the set of centers selected by Greedy (resp. Optimal) Proof Let rg (resp. Opt) denote largest site-center distance using Greedy (resp. Optimal)

SLIDE 151

The Greedy approximation

Theorem The Greedy algorithm for k-center is a 2-approximation algorithm Let Cg (resp. COpt) denote the set of centers selected by Greedy (resp. Optimal) Proof Let rg (resp. Opt) denote largest site-center distance using Greedy (resp. Optimal) Case 1: No si, si′ ∈ Cg are closest to the same sj ∈ COpt

SLIDE 152

The Greedy approximation

Theorem The Greedy algorithm for k-center is a 2-approximation algorithm Let Cg (resp. COpt) denote the set of centers selected by Greedy (resp. Optimal) Proof Let rg (resp. Opt) denote largest site-center distance using Greedy (resp. Optimal) Case 1: No si, si′ ∈ Cg are closest to the same sj ∈ COpt Optimal centers

SLIDE 153

The Greedy approximation

Theorem The Greedy algorithm for k-center is a 2-approximation algorithm Let Cg (resp. COpt) denote the set of centers selected by Greedy (resp. Optimal) Proof Let rg (resp. Opt) denote largest site-center distance using Greedy (resp. Optimal) Case 1: No si, si′ ∈ Cg are closest to the same sj ∈ COpt Optimal centers

Opt

SLIDE 154

The Greedy approximation

Theorem The Greedy algorithm for k-center is a 2-approximation algorithm Let Cg (resp. COpt) denote the set of centers selected by Greedy (resp. Optimal) Proof Let rg (resp. Opt) denote largest site-center distance using Greedy (resp. Optimal) Case 1: No si, si′ ∈ Cg are closest to the same sj ∈ COpt Optimal centers

Opt

Sites

SLIDE 155

The Greedy approximation

Theorem The Greedy algorithm for k-center is a 2-approximation algorithm Let Cg (resp. COpt) denote the set of centers selected by Greedy (resp. Optimal) Proof Let rg (resp. Opt) denote largest site-center distance using Greedy (resp. Optimal) Case 1: No si, si′ ∈ Cg are closest to the same sj ∈ COpt Optimal centers

Opt

Sites

purposes only Disclaimer: for illustrative

SLIDE 156

The Greedy approximation

Theorem The Greedy algorithm for k-center is a 2-approximation algorithm Let Cg (resp. COpt) denote the set of centers selected by Greedy (resp. Optimal) Proof Let rg (resp. Opt) denote largest site-center distance using Greedy (resp. Optimal) Case 1: No si, si′ ∈ Cg are closest to the same sj ∈ COpt Optimal centers

Opt

Sites

purposes only Disclaimer: for illustrative

Greedy centers

SLIDE 157

The Greedy approximation

Theorem The Greedy algorithm for k-center is a 2-approximation algorithm Let Cg (resp. COpt) denote the set of centers selected by Greedy (resp. Optimal) Proof Let rg (resp. Opt) denote largest site-center distance using Greedy (resp. Optimal) Case 1: No si, si′ ∈ Cg are closest to the same sj ∈ COpt Optimal centers

Opt

Sites

purposes only Disclaimer: for illustrative

Greedy centers

SLIDE 158

The Greedy approximation

Theorem The Greedy algorithm for k-center is a 2-approximation algorithm Let Cg (resp. COpt) denote the set of centers selected by Greedy (resp. Optimal) Proof Let rg (resp. Opt) denote largest site-center distance using Greedy (resp. Optimal) Case 1: No si, si′ ∈ Cg are closest to the same sj ∈ COpt Optimal centers

Opt

Sites

purposes only Disclaimer: for illustrative

Greedy centers Distance at most 2Opt

SLIDE 159

The Greedy approximation

Theorem The Greedy algorithm for k-center is a 2-approximation algorithm Let Cg (resp. COpt) denote the set of centers selected by Greedy (resp. Optimal) Proof Let rg (resp. Opt) denote largest site-center distance using Greedy (resp. Optimal) Case 1: No si, si′ ∈ Cg are closest to the same sj ∈ COpt Optimal centers

Opt

Sites

purposes only Disclaimer: for illustrative

Greedy centers Distance at most 2Opt so rg 2Opt

SLIDE 160

The Greedy approximation

Theorem The Greedy algorithm for k-center is a 2-approximation algorithm Let Cg (resp. COpt) denote the set of centers selected by Greedy (resp. Optimal) Proof Let rg (resp. Opt) denote largest site-center distance using Greedy (resp. Optimal) Case 2: Some si, si′ ∈ Cg are closest to the same sj ∈ COpt

SLIDE 161

The Greedy approximation

Theorem The Greedy algorithm for k-center is a 2-approximation algorithm Let Cg (resp. COpt) denote the set of centers selected by Greedy (resp. Optimal) Proof Let rg (resp. Opt) denote largest site-center distance using Greedy (resp. Optimal) Case 2: Some si, si′ ∈ Cg are closest to the same sj ∈ COpt

si si′

SLIDE 162

The Greedy approximation

Theorem The Greedy algorithm for k-center is a 2-approximation algorithm Let Cg (resp. COpt) denote the set of centers selected by Greedy (resp. Optimal) Proof Let rg (resp. Opt) denote largest site-center distance using Greedy (resp. Optimal) Case 2: Some si, si′ ∈ Cg are closest to the same sj ∈ COpt Assume wlog. that Greedy made si a center after si′

si si′

SLIDE 163

The Greedy approximation

Theorem The Greedy algorithm for k-center is a 2-approximation algorithm Let Cg (resp. COpt) denote the set of centers selected by Greedy (resp. Optimal) Proof Let rg (resp. Opt) denote largest site-center distance using Greedy (resp. Optimal) Case 2: Some si, si′ ∈ Cg are closest to the same sj ∈ COpt Assume wlog. that Greedy made si a center after si′

si si′ si was added as a center because it was

SLIDE 164

The Greedy approximation

Theorem The Greedy algorithm for k-center is a 2-approximation algorithm Let Cg (resp. COpt) denote the set of centers selected by Greedy (resp. Optimal) Proof Let rg (resp. Opt) denote largest site-center distance using Greedy (resp. Optimal) Case 2: Some si, si′ ∈ Cg are closest to the same sj ∈ COpt Assume wlog. that Greedy made si a center after si′

si si′ si was added as a center because it was

the furthest from any existing Greedy center

SLIDE 165

The Greedy approximation

Theorem The Greedy algorithm for k-center is a 2-approximation algorithm Let Cg (resp. COpt) denote the set of centers selected by Greedy (resp. Optimal) Proof Let rg (resp. Opt) denote largest site-center distance using Greedy (resp. Optimal) Case 2: Some si, si′ ∈ Cg are closest to the same sj ∈ COpt Assume wlog. that Greedy made si a center after si′

si si′ si was added as a center because it was

the furthest from any existing Greedy center However, si is at most 2Opt away from si′

SLIDE 166

The Greedy approximation

Theorem The Greedy algorithm for k-center is a 2-approximation algorithm Let Cg (resp. COpt) denote the set of centers selected by Greedy (resp. Optimal) Proof Let rg (resp. Opt) denote largest site-center distance using Greedy (resp. Optimal) Case 2: Some si, si′ ∈ Cg are closest to the same sj ∈ COpt Assume wlog. that Greedy made si a center after si′

si si′ si was added as a center because it was

the furthest from any existing Greedy center However, si is at most 2Opt away from si′

SLIDE 167

The Greedy approximation

Theorem The Greedy algorithm for k-center is a 2-approximation algorithm Let Cg (resp. COpt) denote the set of centers selected by Greedy (resp. Optimal) Proof Let rg (resp. Opt) denote largest site-center distance using Greedy (resp. Optimal) Case 2: Some si, si′ ∈ Cg are closest to the same sj ∈ COpt Assume wlog. that Greedy made si a center after si′

si si′ si was added as a center because it was

the furthest from any existing Greedy center However, si is at most 2Opt away from si′ So even before adding si as a center, all sites were 2Opt away from a Greedy center

SLIDE 168

The Greedy approximation

Theorem The Greedy algorithm for k-center is a 2-approximation algorithm Let Cg (resp. COpt) denote the set of centers selected by Greedy (resp. Optimal) Proof Let rg (resp. Opt) denote largest site-center distance using Greedy (resp. Optimal) Case 2: Some si, si′ ∈ Cg are closest to the same sj ∈ COpt Assume wlog. that Greedy made si a center after si′

si si′ si was added as a center because it was

the furthest from any existing Greedy center However, si is at most 2Opt away from si′ So even before adding si as a center, all sites were 2Opt away from a Greedy center Therefore, rg 2Opt

SLIDE 169

k-center Conclusions

Theorem The Greedy algorithm for k-center is a 2-approximation algorithm which runs in O(nk) time

SLIDE 170

k-center Conclusions

Theorem The Greedy algorithm for k-center is a 2-approximation algorithm which runs in O(nk) time

The approximation works for any (metric) distance function,

SLIDE 171

k-center Conclusions

Theorem The Greedy algorithm for k-center is a 2-approximation algorithm which runs in O(nk) time

The approximation works for any (metric) distance function,

d(si, sj) = L1 or L∞ for example

SLIDE 172

k-center Conclusions

Theorem The Greedy algorithm for k-center is a 2-approximation algorithm which runs in O(nk) time

The approximation works for any (metric) distance function,

d(si, sj) = L1 or L∞ for example d(x, y) = d(y, x), d(x, y) 0

Distance function d is a metric iff

(d(x, y) = 0 iff x = y) and d(x, z) d(x, y) + d(y, z)

SLIDE 173

k-center Conclusions

Theorem The Greedy algorithm for k-center is a 2-approximation algorithm which runs in O(nk) time

The approximation works for any (metric) distance function,

d(si, sj) = L1 or L∞ for example

For a general (metric) d, the problem is not α-approximable with α < 2

d(x, y) = d(y, x), d(x, y) 0

Distance function d is a metric iff

(d(x, y) = 0 iff x = y) and d(x, z) d(x, y) + d(y, z)

SLIDE 174

k-center Conclusions

Theorem The Greedy algorithm for k-center is a 2-approximation algorithm which runs in O(nk) time

The approximation works for any (metric) distance function,

d(si, sj) = L1 or L∞ for example

For a general (metric) d, the problem is not α-approximable with α < 2

d(x, y) = d(y, x), d(x, y) 0

For d = L2 , the problem is not α-approximable with α <

√ 3 ≈ 1.73

Distance function d is a metric iff

(d(x, y) = 0 iff x = y) and d(x, z) d(x, y) + d(y, z)

SLIDE 175

k-center Conclusions

Theorem The Greedy algorithm for k-center is a 2-approximation algorithm which runs in O(nk) time

The approximation works for any (metric) distance function,

d(si, sj) = L1 or L∞ for example

For a general (metric) d, the problem is not α-approximable with α < 2

d(x, y) = d(y, x), d(x, y) 0

For d = L2 , the problem is not α-approximable with α <

√ 3 ≈ 1.73

For d = L1 or d = L∞ , the problem is not α-approximable with α < 2
Distance function d is a metric iff

(d(x, y) = 0 iff x = y) and d(x, z) d(x, y) + d(y, z)