Greedy Algorithms

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Greedy Algorithms
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Announcements

• Exam 1
• See me for 2 extra points if you got 2(a) wrong.
• Lab Attendance
• 12 Labs, so if anyone needs to miss a lab you are
  covered since the lab attendance is out of 10.
• Assignment 3 is posted due 10/20/2010
• Exam 2 on 10/27/2010

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Algorithm Design Techniques

• We will cover in this class
• Greedy Algorithms
• Divide and Conquer Algorithms
• Dynamic Programming Algorithms
• And Backtracking Algorithms
• These are 4 common types of algorithms used to
  solve problems.
• For many problems, it is quite likely that at
  least one of these methods will work.

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Greedy Algorithms

• A greedy algorithm is one where you take the step
  that seems the best at the time while executing
  the algorithm.
• Previous Examples
• Huffman coding
• Minimum Spanning Tree Algorithms Kruskals and
  Prims
• Dijkstras Algorithm

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Coin Changing

• The goal is to give change with the minimal
  number of coins as possible for a certain number
  of cents using 1 cent, 5 cent, 10 cent, and 25
  cent coins.
• The Greedy Algorithm
• Keep giving as many coins of the largest
  denomination as possible until you have a
  remaining value less than the value of that
  denomination.
• Then you continue with the lower denomination and
  repeat until youve given out the correct change.
• If you think about it, this is the algorithm a
  cashier typically uses when giving out change.

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Coin Changing

• How do we know this algorithm always works?
• Consider all combinations of giving change,
  ordered from highest denomination to lowest using
  1-cent, 5-cent and 10-cent coins.
• 2 ways of making change for 25 cents are
• 10, 10, 1,1,1,1,1,
• 10, 5, 5, 5
• Since each larger denomination is divisible by
  each smaller one
• We can always make a mapping for each coin in one
  list to a coin or set of coins in the other list.
• For example
• 10 10 11111
• 10 5 5 5

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Coin Changing

• This argument doesnt work for any set of coins
  without the divisibility rule.
• Consider 1,6,10
• There is no way to match up these two ways of
  producing 30 cents
• 10 10 10
• 6 6 6 6 6
• Consider making change for 18 cents
• Largest denomination first gives you 10, 6, 1, 1
• However, 6, 6, 6 gives you the least amount of
  coins.

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Coin Changing Problem

• Extension to 25 cent coin
• Even though a 10 doesnt divide into 25, there
  are no values, multiples of 25, for which its
  advantageous to give a set of dimes over a set of
  quarters.

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Single Room Scheduling Problem

• Given a single room to schedule, and a list of
  requests, the goal of this problem is to maximize
  the total number of events scheduled.
• Each request simply consists of the group, a
  start time and an end time during the day.
• The Greedy Solution
• Sort the requests by finish time.
• Go through the requests in order of finish time,
  scheduling them in the room if the room is
  unoccupied at its start time.

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Single Room Scheduling Problem Example

• Say youve gone to a music festival
• There are so many shows and so little time!
• If each one of the shows schedule the first day
  are equally important for you to see,
• And you will only go to a show if you can stay
  the entire time,
• Maximize the number of shows you can go to.

• 5-7pm Green Day
• 7-9pm Weezer
• 730-930 Gorillaz
• 9-11pm 311
• 930pm-1030pm Coldplay
• 11pm-2am The Killers

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Single Room Scheduling Problem Example

• The Greedy Solution
• Sort the requests by finish time.
• Go through the requests in order of finish time,
  scheduling them in the room if the room is
  unoccupied at its start time.

• 5-7pm Green Day
• 7-9pm Weezer
• 730-930pm Gorillaz
• 930pm-1030pm Coldplay
• 9-11pm 311
• 11pm-2am The Killers

11pm 2am
The Killers
930pm-1030pm
Coldplay
7-9pm
Weezer
5-7pm
Green Day
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Single Room Scheduling Proof

• Prove that this algorithm does indeed maximize
  the number of events scheduled
• Shown on the board

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Multiple Room Scheduling

• Given a set of requests with start and end times,
  the goal here is to schedule all events using the
  minimal number of rooms
• Greedy Algorithm
• Sort all the requests by start time.
• Schedule each event in any available empty room.
  If no room is available, schedule the event in a
  new room.

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Multiple Room Scheduling Example

• Final Exam Schedule for Computer Science
• 7-10am CS1
• 7-10am Intro to C
• 9am-12pm OOP
• 10am-1pm OS
• 12-3pm CS 2
• 1-4pm Comp Arch.

• Greedy Algorithm
• Sort all the requests by start time.
• Schedule each event in any available empty room.
  If no room is available, schedule the event in a
  new room.

ROOM 1
7-10am CS 1
ROOM 2
7-10am Intro to C
ROOM 3
9am-12pm OOP
10am-1pm OS
12pm-3pm CS 2
1-4pm Comp Arch
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Multiple Room Scheduling Proof

• Let k be the number of rooms this algorithm uses
  for scheduling.
• When the kth room is scheduled, it MUST have been
  the case that all k-1 rooms before it were in
  use.
• At the exact point in time that the k room gets
  scheduled, we have k simultaneously running
  events.
• It's impossible for any schedule to handle this
  type of situation with less than k rooms.
• Thus, the given algorithm minimizes the total
  number of rooms used.

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Fractional Knapsack Problem

• Your goal is to maximize the value of a knapsack
  that can hold at most W units worth of goods from
  a list of items I1, I2, ... In. Each item has
  two attributes
• A value/unit let this be vi for item Ii.
• Weight available let this be wi for item Ii.
• The algorithm is as follows
• Sort the items by value/unit.
• Take as much as you can of the most expensive
  item left, moving down the sorted list. You may
  end up taking a fractional portion of the "last"
  item you take.

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Fractional Knapsack Example

• Say we have
• 4 lbs. of I1 available with a value of 50/lb.
• 25 lbs. of I3 available with a value of 40/lb.
• 40 lbs. of I2 available with a value of 30/lb.
• The algorithm is as follows
• Sort the items by value/unit.
• Take as much as you can of the most expensive
  item left, moving down the sorted list. You may
  end up taking a fractional portion of the "last"
  item you take.
• You will do the following
• Take 4 lbs of I1.
• Take 25 lbs. of I3.
• Tale 21 lbs. of I2.
• Value of knapsack 450 2540 2130 1830.

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Fractional Knapsack Proof

• Why is this maximal?
• Because if we were to exchange any good from the
  knapsack with what was left over, it is
  IMPOSSIBLE to make an exchange of equal weight
  such that the knapsack gains value.
• The reason for this is that all the items left
  have a value/lb. that is less than or equal to
  the value/lb. of ALL the material currently in
  the knapsack.
• At best, the trade would leave the value of the
  knapsack unchanged.
• Thus, this algorithm produces the maximal valued
  knapsack!

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References

• Slides adapted from Arup Guhas Computer Science
  II Lecture notes http//www.cs.ucf.edu/dmarino/
  ucf/cop3503/lectures/
• Additional material from the textbook
• Data Structures and Algorithm Analysis in Java
  (Second Edition) by Mark Allen Weiss
• Additional images
• www.wikipedia.com
• xkcd.com