Design and Analysis of Computer Algorithm Lecture 5-1

April 22, 2023

Design and Analysis of Computer AlgorithmLecture 5-1

Pradondet NilaguptaDepartment of Computer Engineering

This lecture note has been modified from lecture note for 23250 by Prof. Francis Chin

Design and Analysis of Computer Algorithm 2April 22, 2023

Greedy Method


Topics Cover

The General Method Activity -Selection Problem Optimal storage on Tapes Knapsack problem Minimal Spanning Tree Single Source shortest paths


Greedy Method: Definition

An algorithm which always takes the best immediate, or local, solution while finding an answer. Greedy algorithms will always find the overall, or globally, optimal solution for some optimization problems, but may find less-than-optimal solutions for some instances of other problems.


Example of Greedy Method (1/4)

Prim's algorithm and Kruskal's algorithm are greedy algorithms which find the globally optimal solution, a minimum spanning tree. In contrast, any known greedy algorithm to find an Euler cycle might not find the shortest path, that is, a solution to the traveling salesman problem.

Dijkstra's algorithm for finding shortest paths is another example of a greedy algorithm which finds an optimal solution.



If there is no greedy algorithm which always finds the optimal solution for a problem, one may have to search (exponentially) many possible solutions to find the optimum. Greedy algorithms are usually quicker, since they don't consider possible alternatives.



Consider the problem of making change: Coins of values 25c, 10c, 5c and 1c Return 63c in change

– Which coins? Use greedy strategy:

– Select largest coin whose value was no greater than 63c

– Subtract value (25c) from 63 getting 38– Find largest coin … until done



At any individual stage, select that option which is “locally optimal” in some particular sense

Greedy strategy for making change works because of special property of coins

If coins were 1c, 5c and 11c and we need to make change of 15c?– Greedy strategy would select 11c coin followed by

4 1c coins– Better: 3 5c coins


Greedy Algorithm

Start with a solution to a small subproblem Build up to a solution to the whole problem Make choices that look good in the short term

Disadvantage: Greedy algorithms don’t always work ( Short term solutions can be diastrous in the long term). Hard to prove correct

Advantage: Greedy algorithm work fast when they work. Simple algorithm, easy to implement


Greedy Algorithm

Procedure GREEDY(A,n)// A(1:n) contains the n inputs//

solution //initialize the solution to empty//for i 1 to n do

x SELECT(A)if FEASIBLE(solution,x)

then solution UNION(solution,x)endifrepeatreturn(solution)

end GREEDY


Activity-Selection Problem

The problem is to select a maximum-size set of mutally compatible activities.

Example We have a set S = { 1,2,…,n} of n proposed activities t

hat wish to use a resource, such as a lecture hall, which can be used by only one activities at a time.


Example

i si fi

1 0 62 3 53 1 44 2 135 3 86 12 147 8 118 8 129 6 1010 5 711 5 9


Brute Force

Try every all possible solution Choose the largest subset which is feasible Ineffcient (2n) choices


Greedy Approach

Sort by finish time


Activity-Selection Problem Pseudo code

Greedy_Activity_Selector(s,f)1 n <- length[s]2 A <- {1}3 j <- 14 for i <- 2 to n5 do if si > fj

6 then A <- A U {i}7 j <- i8 return A

It can schdule a set S of n activities in (n) time, assuming that the activities were already sorted


Proving the greedy algorithm correct

We assume that the input activities are in order by increasing finishing time

f1 < f2 < … < fn Activities #1 has the earliest finish time then it must b

e in an optimal solution.

k

1

1 possible solution

Activitiy 1


Proving (cont.)

k

1

Eliminate the activities which has a start time early than the finish time of activity 1


Proving (cont.)

1

Greedy algorithm produces an optimal solution


Element of the Greedy Strategy

Question? How can one tell if a greedy algorithm will solve

a particular optimization problem? No general way to tell!!! There are 2 ingredients that exhibited by most

problems that lend themselves to a greedy strategy– The Greedy Choice Property– Optimal Substructure


The Greedy Choice Property

A globally optimal solution can be arrived at by making a locally optimal (greedy) choice.

Make whatever choice seems best at the moment. May depend on choice so far, but not depend on any

future choices or on the solutions to subproblems


Optimal Substructure

An optimal solution to the problem contains within it optimal solutions to subproblems


Optimal Storage on Tapes

There are n programs that are to be stored on a computer tape of length L.

Each program i has a length li , 1 i n All programs are retrieved equally often, the expected

or mean retrieval time (MRT) is

nj

jtn1

)/1(


Optimal Storage on Tapes (cont.)

We are required to find a permutation for the n programs so that when they are stored on tape in the order the MRT is minimized.

Minimizing the MRT is equivalent to minimizing

jknjliID

k11

)(


Example

Let n = 3 and (l1,l2,l3) = (5,10,3)

Ordering I D(I)1,2,3 5 + 5 + 10 + 5 + 10 + 3 = 381,3,2 5 + 5 + 3 + 5 + 3 + 10 = 312,1,3 10 + 10 + 5 + 10 + 5 + 3 = 432,3,1 10 + 10 + 3 + 10 + 3 + 5 = 413,1,2 3 + 3 + 5 + 3 + 5 + 10 = 293,2,1 3 + 3 + 10 + 3 + 10 + 5 = 34


The Greedy Solution

Make tape emptyfor i := 1 to n do

grab the next shortest fileput it next on tape

The algorithm takes the best short term choice without checking to see weather it is the best long term decision.


Optimal Storage on Tapes (cont.)

Theorem 4.1 If l1 l2 … ln then the ordering ij = j, 1 j n

minimizes

Over all possible permutation of the ij See proof on text pp.154-155

n

k

k

jli j1 1


Knapsack Problem

We are given n objects and a knapsack. Object i has a weight wi and the knapsack has a capacity M.

If a fraction xi, 0 xi 1, of object I is placed into the knapsack the a profit of pixi is earned.

The objective is to obtain a filling of the knapsack that maximizes the total weight of all chosen objects to be at most M

maximize

subject to

and 0 xi 1, 1 I n

ni

iixp1

Mxwni

ii 1


Example

1020

30

50

$60 $100 $120

Item 1

Item 2

Item 3

knapsack


Knapsack 0/1

30

20

$120

$100

Total =$220

20

10

$100

$60

=$160

30

10

$120

$60

=$180


Fractional Knapsack

Taking the items in order of greatest value per pound yields an optimal solution

20

10

$100

$60

=$240Total

2030

$80


Optimal Substructure

Both fractional knapsack and 0/1 knapsack have an optimal substructure.


Example Fractional Knapsack (cont.)

There are 5 objects that have a price and weight list below, the knapsack can contain at most 100 Lbs.

Method 1 choose the least weight first– Total Weight = 10 + 20 + 30 + 40 = 100

– Total Price = 20 + 30 + 66 + 40 = 156

Price ($US) 20 30 66 40 60weight (Lbs.) 10 20 30 40 50



Method 2 choose the most expensive first– Total Weight = 30 + 50 + 20 = 100

– Total Price = 66 + 60 + 20 = 146

Price ($US) 20 30 66 40 60weight (Lbs.) 10 20 30 40 50

half



Method 3 choose the most price/ weight first– Total weight = 30 + 10 + 20 + 40 = 100

– Total Price = 66 + 20 + 30 + 48 = 164

Price ($US) 20 30 66 40 60weight (Lbs.) 10 20 30 40 50price/weight 2 1.5 2.2 1 1.2


More Example on fractional knapsac

Consider the following instance of the knapsack problem: n = 3, M = 20, (p1,p2,p3) = 25,24,15 and (w1,w2,w3) = (18,15,10)

(x1,x2,x3)1) (1/2,1/3,1/4) 16.5 24.252) (1,2/15,0) 20 28.23) ( 0,2/3,1) 20 314) ( 0,1,1/2) 20 31.5

iixw iixp


The Greedy Solution

Define the density of object Ai to be wi/si. Use as much of low density objects as possible. That is, process each in increasing order of density. If the whole thing ts, use all of it. If not, fill the remaining space with a fraction of the current object,and discard the rest.

First, sort the objects in nondecreasing density, so that wi/si w i+1/s i+1 for 1 i < n.

Then do the following


PseudoCode

Procedure GREEDY_KNAPSACK(P,W,M,X,n)//P(1:n) and W(1:n) contain the profits and weights respectively of the n obj

ects ordered so that P(I)/W(I) > P(I+1)/W(I+1). M is the knapsack size and X(1:n) is the solution vector//

real P(1:n), W(1:n), X(1:n), M, cu;integer I,n;

x ; //initialize solution to zero //cu M; // cu = remaining knapsack capacity //for i to n do

if W(i) > cu then exit endifX(I) 1;cu c - W(i);

repeatif I < n then X(I) cu/W(I) endif

End GREEDY_KNAPSACK


Proving Optimality

Let p1/w1 > p2/w2 > … > pn/wn

Let X = (x1,x2,…,xn) be the solution generated by GREEDY_KNAPSACK

Let Y = (y1,y2,…,yn) be any feasible solution We want to show that

0p)yx(n

1iiii


Proving Optimality (cont.)

If all the xi are 1, then the solution is clearly optimal (It is the only solution) otherwise, let k be the smallest number such that xk < 1.

1 1 1 1 1 1 1 1 1 1 1 1

1 2 n

1 1 .. .. 1 xk0 0 .. .. 0 0

1 2 nk



n

1iiii p)yx(

1 1 .. .. 1 xk0 0 .. .. 0 0

1 2 nk

1k

1i i

iiii wpw)yx(

k

kkkk w

pw)yx(

n

1ki i

iiii wpw)yx(



Consider each of these block

1k

1i i

iiii wpw)yx(

1k

1i k

kiii wpw)yx(

k

kkkk w

pw)yx( k

kkkk w

pw)yx(

n

1ki i

iiii wpw)yx(

n

1ki k

kiii wpw)yx(



n

1iiii p)yx(

n

1i k

kiii wpw)yx(

n

1iiii

k

k w)yx(wp

= W0p)yx(

n

1iiii

Since W always > 0, therefore

Design and Analysis of Computer Algorithm Lecture 5-1

Documents

Transcript of Design and Analysis of Computer Algorithm Lecture 5-1