Poscat seminar 6

POSCAT Seminar 6 :Greedy Approach

yougatup @ POSCAT

1

Topic

Topic today

− Greedy Approach

• Basic Concept

• Interval Scheduling

• Interval Partitioning

• Fractional Knapsack

• Huffman Encoding

• Other Problems

POSCAT Seminar 1-24 July 2014

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Greedy Approach

Problem Solving Paradigm

− Follows the problem solving heuristic of making the locally

optimal choose at each stage

− You should prove that greedy approach guarantees the global

optimal

Very difficult normally

− Finding a solution is very difficult, but

− Code is very simple because it choose locally optimal usually


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Greedy Approach

Starting at A, our greedy algorithm should find ‘M’, not ‘m’.

Therefore, we must prove that greedy algorithm always find ‘M’, although we always choose locally optimal path


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Greedy Approach

Problem

Find a path which has largest sum


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3

7

9 2 1 2

6

Greedy Approach

Problem


Greedy approach : Choose what appears to be the optimal


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3

7

9 2 1 2

6

Greedy Approach

Problem




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3

7

9 2 1 2

6

Greedy Approach

Problem



Is it our solution ?


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3

7

9 2 1 2

6

Greedy Approach

Problem



We must prove that our choice guarantees global optimal value


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3

7

9 2 1 2

6

Greedy Approach

Problem



We must prove that our choice guarantees global optimal value


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3

7

9 2 1 2

6

Completely Wrong !

Interval Scheduling

Problem

− Job 𝑗 starts at 𝑠𝑗 and finishes at 𝑓𝑗

− Two jobs compatible if they don’t overlap

− Find maximum subset of mutually compatible jobs


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Time

Interval Scheduling

Problem

− Job 𝑗 starts at 𝑠𝑗 and finishes at 𝑓𝑗

− Two jobs compatible if they don’t overlap

− Find maximum subset of mutually compatible jobs


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TimeThey don’t overlap each other.

Interval Scheduling

Problem

Idea : What have to be the first interval ?


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Time

Interval Scheduling

Problem



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Time

Which one is better ? Why ?

Interval Scheduling

Problem



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Time

Which one is better ? Why ?Small 𝑓𝑖 is always good !

Interval Scheduling

Problem



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Time

It must be the first one definitely !

Interval Scheduling

Problem



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Time

We can’t choose grey interval remove it

Interval Scheduling

Problem



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Time

Which one is better ?

Interval Scheduling

Problem



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Time


Interval Scheduling

Problem



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Time

Done ! You should know why greedy algorithm guarantees the global optimal

Interval Scheduling

Problem

Prove that this algorithm is optimal


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Interval Scheduling

Problem


Suppose that there is a solution 𝑠1 ≤ 𝑓1 ≤ 𝑠2 ≤ 𝑓2 ≤ … ≤ 𝑠𝑘 ≤ 𝑓𝑘,

and we found 𝑠1′ ≤ 𝑓1

′ ≤ 𝑠2′ ≤ 𝑓2

′ ≤ … ≤ 𝑠𝑜𝑝𝑡′ ≤ 𝑓𝑜𝑝𝑡′


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Interval Scheduling

Problem




′ ≤ 𝑠2′ ≤ 𝑓2

′ ≤ … ≤ 𝑠𝑜𝑝𝑡′ ≤ 𝑓𝑜𝑝𝑡′

We select the interval whose finish time is minimal.

∴ 𝑓1′ ≤ 𝑓1.


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Interval Scheduling

Problem




′ ≤ 𝑠2′ ≤ 𝑓2

′ ≤ … ≤ 𝑠𝑜𝑝𝑡′ ≤ 𝑓𝑜𝑝𝑡′

Also, we choose the second interval whose finish time is smallest

among intervals compatible with the first one.

∴ 𝑓2′ ≤ 𝑓2


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Interval Scheduling

Problem




′ ≤ 𝑠2′ ≤ 𝑓2

′ ≤ … ≤ 𝑠𝑜𝑝𝑡′ ≤ 𝑓𝑜𝑝𝑡′

Like this, we can always guarantee that 𝑓𝑘 ≤ 𝑓𝑘′. ∴ 𝑜𝑝𝑡 ≥ 𝑘

Also, 𝑜𝑝𝑡 ≤ 𝑘 because 𝑘 is solution


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Interval Scheduling

Problem




′ ≤ 𝑠2′ ≤ 𝑓2

′ ≤ … ≤ 𝑠𝑜𝑝𝑡′ ≤ 𝑓𝑜𝑝𝑡′

Like this, we can always guarantee that 𝑓𝑘 ≤ 𝑓𝑘′. ∴ 𝑜𝑝𝑡 ≥ 𝑘

Also, 𝑜𝑝𝑡 ≤ 𝑘 because 𝑘 is solution

Therefore, 𝑜𝑝𝑡 = 𝑘.


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Interval Scheduling

Problem



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Time

Time complexity : O(𝒏 𝒍𝒐𝒈 𝒏) for sorting

Interval Partitioning

Problem

− Lecture 𝑗 starts at 𝑠𝑗 and finishes at 𝑓𝑗

− Find minimum number of classrooms to schedule all lectures

so that no two occur at the same time in the same room


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Time


Problem





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Problem





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Time

We need 5 classrooms !


Problem

Idea : What is the minimum number of classrooms ?


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Problem



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TimeWe don’t know the solution, but we need at least 4 classrooms! Is it sufficient ?


Problem



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TimeWe don’t know the solution, but we need at least 5 classrooms!


Problem


∴ We need classrooms at least as many as the maximum number of

overlap


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TimeWe don’t know the solution, but we need at least 5 classrooms!


Problem

Is it sufficient ?


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Time


Problem

Is it sufficient ? Yes ! Consider this algorithm


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Time


Problem



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Time


Problem



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TimeI can reuse this color !


Problem



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Problem



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TimeDone !



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Prove that this algorithm needs classrooms no more than the maximum number of overlap



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Let 𝑀 be maximum number of overlap. Suppose that our algorithm needs 𝑀 + 1 classrooms



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Consider the situation that we need last classrooms when we consider lecture 𝑗.In other words, we need 1 more classroom although we already have 𝑀 classrooms



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To allocate another classroom, we have to go to “else” partIt means that lecture 𝑗 have no compatible classroom



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To allocate another classroom, we have to go to “else” partIt means that lecture 𝑗 have no compatible classroomTherefore, the maximum number of overlap have to be 𝑀 + 1. Contradiction.



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Implementation ?



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Implementation ?

By using priority queue, we can make O(𝒏 𝐥𝐨𝐠𝒏) algorithm

Fractional Knapsack


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Easy, just think about it

− I’ll provide this problem today

Other problems


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Problem

Given N points, find a maximum value of 𝑚 such that 𝑚 = |∆𝑦

∆𝑥|

Other problems


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Problem


∆𝑥|

Naïve approach O(𝑛2)

Other problems


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Problem


∆𝑥|

We can prove that two points have to be adjacent

It can’t be optimal !

Other problems


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Problem


∆𝑥|

We can prove that two points have to be adjacent O(𝑛 log 𝑛)

It can’t be optimal !

Other problems


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Problem

You have 2N cards, and each card have two numbers on both sides.

There are 2N locations to put your card on the table. For each

location, it has specific sign which is changed alternatively (+ or -).

Find the maximum value of the result of your calculation.

1 2 9 -1 -9 -8 7 4 4 -3 2 -1Cards (front/back)

Locations : + + +- - - = ?

Other problems


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Problem





1 2 9 -1 -9 -8 7 4 4 -3 2 -1Cards (front/back)

Locations : + + +- - - = ?

Other problems


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Problem





1

2 9

-1

-9

-8

7

4 4

-3

2

-1

Cards (front/back)

Locations : + + +- - - = 31

Other problems


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Solution

Let 𝑎𝑖 be the value of bigger side of card 𝑖.

𝑏𝑖 be the value of smaller side of card 𝑖.

If card 𝑖 which is located with (+) should have the value 𝑎𝑖

Otherwise, it should have the value 𝑏𝑖

Other problems


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Solution



If card 𝑖 which is located with (+) should have the value 𝑎𝑖

Otherwise, it should have the value 𝑏𝑖

Let 𝑆 = { 𝑖 | card 𝑖 is located in (+) side }

Let 𝑆′ = { 𝑖 | card 𝑖 is located in (-) side }

Other problems


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Solution



Then our result is represented by

𝑖∈𝑆

𝑎𝑖 −

𝑗∈𝑆′

𝑏𝑗

Other problems


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Solution




𝑖∈𝑆

𝑎𝑖 −

𝑗∈𝑆′

𝑏𝑗 =

𝑖∈𝑆

𝑎𝑖 −

𝑗∈𝑆′

𝑏𝑗 + (

𝑖∈𝑆

𝑏𝑖 −

𝑖∈𝑆

𝑏𝑖 )

Other problems


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Solution




𝑖∈𝑆

𝑎𝑖 +

𝑗∈𝑆′

𝑏𝑗 =

𝑖∈𝑆

𝑎𝑖 −

𝑗∈𝑆′

𝑏𝑗 + (

𝑖∈𝑆

𝑏𝑖 −

𝑖∈𝑆

𝑏𝑖 )

𝑖∈𝑆

𝑎𝑖 +

𝑖∈𝑆

𝑏𝑖 − (

𝑗∈𝑆′

𝑏𝑗 +

𝑖∈𝑆

𝑏𝑖 )

Other problems


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Solution




𝑖∈𝑆

𝑎𝑖 +

𝑗∈𝑆′

𝑏𝑗 =

𝑖∈𝑆

𝑎𝑖 −

𝑗∈𝑆′

𝑏𝑗 + (

𝑖∈𝑆

𝑏𝑖 −

𝑖∈𝑆

𝑏𝑖 )

𝑖∈𝑆

(𝑎𝑖 + 𝑏𝑖) − (𝑆𝑢𝑚 𝑜𝑓 𝑏)

Other problems


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Solution



𝑖∈𝑆


(Sum of 𝑏) is constant because 𝑏 is the smaller value of each card.

Other problems


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Solution



𝑖∈𝑆


(Sum of 𝑏) is constant because 𝑏 is the smaller value of each card.

Therefore, we have to choose 𝑖 by considering the value of 𝑎 + 𝑏

It means that a card whose 𝑎 + 𝑏 value is larger should be located in (+)

Other problems


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Solution

1. Sort cards with respect to the value of a + b

2. N cards whose value is larger should be located in (+)

3. Remaining N cards should be located in (-)

𝑶(𝒏 𝐥𝐨𝐠𝒏)

Poscat seminar 6

Education

Transcript of Poscat seminar 6