03 Brute Force

8/4/2019 03 Brute Force

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Chapter 3Chapter 3

Brute ForceBrute Force

Copyright 2007 Pearson Addison-Wesley. All rights reserved.


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3-2Copyright 2007 Pearson Addison-Wesley. All rights reserved.A. Levitin Introduction to the Design & Analysis of Algorithms, 2nd ed., Ch. 3

Brute ForceBrute Force

A straightforward approach, usually based directly on theA straightforward approach, usually based directly on theproblems statement and definitions of the concepts involvedproblems statement and definitions of the concepts involved

Examples:Examples:

1.1. ComputingComputing aann ((aa > 0,> 0, nn a nonnegative integer)a nonnegative integer)

2.2. ComputingComputing nn!!

3.3. Multiplying two matricesMultiplying two matrices

4.4. Searching for a key of a given value in a listSearching for a key of a given value in a list


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Importance of Brute Force?

Applicable to a very wide variety of problems. For some important problems, it yields reasonable

algorithms.

The expense of designing a more efficient

algorithm may be unjustifiable.

Even if a brute-force algorithm is too inefficient in

general, it may still be useful for solving small-size

instances of a problem. For theoretical / educational purpose. A brute-

force algorithm can be used as a benchmark for

comparing other efficient alternatives.


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Brute-Force Sorting AlgorithmBrute-Force Sorting Algorithm

Selection SortSelection Sort Scan the array to find its smallest element andScan the array to find its smallest element andswap it with the first element. Then, starting with the secondswap it with the first element. Then, starting with the second

element, scan the elements to the right of it to find the smallestelement, scan the elements to the right of it to find the smallest

among them and swap it with the second elements. Generally,among them and swap it with the second elements. Generally,

on passon pass ii(0(0 ii n-n-2), find the smallest element in2), find the smallest element inAA[[i..n-i..n-1]1]and swap it withand swap it withAA[[ii]:]:

AA[0][0] . . .. . . AA[[ii-1] |-1] | AA[[ii], . . . ,], . . . ,AA[[minmin], . . .,], . . .,AA[[nn-1]-1]

in their final positionsin their final positions

Example: 7 3 2 5Example: 7 3 2 5


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Analysis of Selection SortAnalysis of Selection Sort

Time efficiency:Time efficiency:

Space efficiency:Space efficiency:

Stability:Stability:

SIMULATION


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)()(

2)1()1(

2)2()1(2

2

)1)(2()1)(1(

2

)1)(2(1)1(

)1()1(

]1)1()1[(1)(

2

2

0

2

0

2

0

2

0

2

0

2

0

1

1

nnC

nnnnn

nnnn

nnn

inin

innC

n

i

n

i

n

i

n

i

n

i

n

i

n

ij

==

=

=

==

++==

=

=

=

=

=

=

+=


7/333-7Copyright 2007 Pearson Addison-Wesley. All rights reserved.A. Levitin Introduction to the Design & Analysis of Algorithms, 2nd ed., Ch. 3

Bubble SortBubble Sort

a

Bubble sort is another example of applying brute-force toBubble sort is another example of applying brute-force tothe sorting algorithm.the sorting algorithm.

a Read section on Bubble sort in the textbook (page 100).Read section on Bubble sort in the textbook (page 100).


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Applying Brute-Force to the SearchingApplying Brute-Force to the Searching

ProblemProblem

a

Read section on Sequential Search in your textbook (pageRead section on Sequential Search in your textbook (page103).103).


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Brute-Force String MatchingBrute-Force String Matching

a patternpattern: a string of: a string ofmm characters to search forcharacters to search for

a texttext: a (longer) string of: a (longer) string ofnn characters to search incharacters to search in

a problem: find a substring in the text that matches the patternproblem: find a substring in the text that matches the pattern

Brute-force algorithmBrute-force algorithmStep 1 Align pattern at beginning of textStep 1 Align pattern at beginning of text

Step 2 Moving from left to right, compare each character ofStep 2 Moving from left to right, compare each character of

pattern to the corresponding character in text untilpattern to the corresponding character in text until

all characters are found to match (successful search); orall characters are found to match (successful search); or a mismatch is detecteda mismatch is detected

Step 3 While pattern is not found and the text is not yetStep 3 While pattern is not found and the text is not yet

exhausted, realign pattern one position to the right andexhausted, realign pattern one position to the right and

repeat Step 2repeat Step 2


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Examples of Brute-Force String MatchingExamples of Brute-Force String Matching

1.1. Pattern:Pattern: 001011001011Text:Text: 1001010110100110010111101010010101101001100101111010

2.2. Pattern:Pattern: happyhappyText:Text: It is never too late to have a happyIt is never too late to have a happy

childhood.childhood.

SIMULATION


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Pseudocode and EfficiencyPseudocode and Efficiency

Efficiency:Efficiency:


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a

In the worst case,In the worst case,

a It has been shown that the average case efficiencyIt has been shown that the average case efficiency

is linear (Consider typical word search in ais linear (Consider typical word search in a

natural language text).natural language text).

)()(

)1()( 2

mnnC

mmmnmmnnC

worst

worst

+=+=

)()()( nmnnCaverage =+=


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Brute-Force Polynomial EvaluationBrute-Force Polynomial Evaluation

Problem: Find the value of polynomialProblem: Find the value of polynomialpp((xx) =) = aannxx

nn++ aann-1-1xxnn-1-1 + ++ + aa11xx

11 ++ aa00

at a pointat a pointxx==xx00

Brute-force algorithmBrute-force algorithm


pp0.00.0

forforiinndowntodownto 00 dodo

powerpower11

forforjj11 totoiidodo //compute//computexxii

powerpowerpowerpowerxx

pppp ++ aa[[ii]] powerpowerreturnreturnpp


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)(2

)1(

)11(1)(

2

0

00 1

nnni

inC

n

i

n

i

n

i

i

j

+==

+==

=

== =


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Polynomial Evaluation: ImprovementPolynomial Evaluation: Improvement

We can do better by evaluatingWe can do better by evaluating

from right to left:from right to left:

Better brute-force algorithmBetter brute-force algorithm


ppaa[0][0]

powerpower11forforii11 totonndodo

powerpowerpowerpower xx

pppp ++ aa[[ii]] powerpower

returnreturnpp

)()( nnC


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Closest-Pair ProblemClosest-Pair Problem

Find the two closest points in a set ofFind the two closest points in a set ofnn points (in the two-points (in the two-dimensional Cartesian plane).dimensional Cartesian plane).

Brute-force algorithmBrute-force algorithm

Compute the distance between every pair of distinct pointsCompute the distance between every pair of distinct points

and return the indexes of the points for which the distanceand return the indexes of the points for which the distance

is the smallest.is the smallest.


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Closest-Pair Brute-Force Algorithm (cont.)Closest-Pair Brute-Force Algorithm (cont.)


How to make it faster?How to make it faster?


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a The algorithm can be made faster by avoiding computing squareThe algorithm can be made faster by avoiding computing square

roots. Is it possible for this problem?roots. Is it possible for this problem?

a ReplaceReplace

withwith

))()(( 22 jiji yyxxsqrtd +

22 )()( jiji yyxxd +


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)()1(

)2

)1()1((2

)2

)1)1)((1()11)1(((2

)(2)(2

)1)1((2

122)(

2

1

1

1

1

1

1

1

1

1

1 1

1

1 1

nnn

nnnn

nnnn

inin

in

nC

n

i

n

i

n

i

n

i

n

i

n

ij

n

i

n

ij

=

=

++=

==

++=

==

=

=

=

=

= +=

= +=


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Brute-Force Strengths and WeaknessesBrute-Force Strengths and Weaknesses

a StrengthsStrengths

wide applicabilitywide applicability

simplicitysimplicity

yields reasonable algorithms for some important problemsyields reasonable algorithms for some important problems(e.g., matrix multiplication, sorting, searching, string(e.g., matrix multiplication, sorting, searching, string

matching)matching)

a WeaknessesWeaknesses

rarely yields efficient algorithmsrarely yields efficient algorithms

some brute-force algorithms are unacceptably slowsome brute-force algorithms are unacceptably slow not as constructive as some other design techniquesnot as constructive as some other design techniques


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Exhaustive SearchExhaustive Search

A brute force solution to a problem involving search for anA brute force solution to a problem involving search for an

element with a special property, usually among combinatorialelement with a special property, usually among combinatorialobjects such as permutations, combinations, or subsets of aobjects such as permutations, combinations, or subsets of a

set.set.

Method:Method: generate a list of all potential solutions to the problem in agenerate a list of all potential solutions to the problem in a

systematic manner (see algorithms in Sec. 5.4)systematic manner (see algorithms in Sec. 5.4)

evaluate potential solutions one by one, disqualifyingevaluate potential solutions one by one, disqualifyinginfeasible ones and, for an optimization problem, keepinginfeasible ones and, for an optimization problem, keepingtrack of the best one found so fartrack of the best one found so far

when search ends, announce the solution(s) foundwhen search ends, announce the solution(s) found


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Example 1: Traveling Salesman ProblemExample 1: Traveling Salesman Problem

a

GivenGiven nn cities with known distances between each pair, findcities with known distances between each pair, findthe shortest tour that passes through all the cities exactlythe shortest tour that passes through all the cities exactly

once before returning to the starting cityonce before returning to the starting city

a Alternatively: Find shortestAlternatively: Find shortestHamiltonian circuitHamiltonian circuit in ain a

weighted connected graphweighted connected graph

a Example:Example:

a b

c d

8

2

7

5 3

4


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TSP by Exhaustive SearchTSP by Exhaustive Search

Tour CostTour Cost

aabbccdda 2+3+7+5 = 17a 2+3+7+5 = 17

aabbddcca 2+4+7+8 = 21a 2+4+7+8 = 21

aaccbbdda 8+3+4+5 = 20a 8+3+4+5 = 20

aaccddbba 8+7+4+2 = 21a 8+7+4+2 = 21aaddbbcca 5+4+3+8 = 20a 5+4+3+8 = 20

aaddccbba 5+7+3+2 = 17a 5+7+3+2 = 17

More tours?More tours?

Less tours?Less tours?

Only need to consider circuits starting from onevertex.

Only need to consider one tour direction.

E.g. No need to consideradcbaadcba


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a Efficiency:Efficiency:

Consider starting from the same vertex and only 4 verticesConsider starting from the same vertex and only 4 vertices

in graphin graph

vv00vv11vv22vv33vv00

3 X 2 X 1 = 6 tours3 X 2 X 1 = 6 tours

If we consider only one direction thenIf we consider only one direction then

(3 X 2 X 1) / 2 = 3 tours(3 X 2 X 1) / 2 = 3 tours


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So, for n vertices?So, for n vertices?

If we do not limit to starting from one vertex?If we do not limit to starting from one vertex?

and consider both directions?and consider both directions?

2

)!1(

2

1*2*...*)2(*)1( =

nnn

2

!

2

1*2*...*)2(*)1(* nnnn=

!2*2

!n

n=


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Example 2: Knapsack ProblemExample 2: Knapsack Problem

GivenGiven nn items:items:

weights:weights: ww11 ww22 w wnn

values:values: vv11 vv22 v vnn

a knapsack of capacitya knapsack of capacity WW

Find most valuable subset of the items that fit into the knapsackFind most valuable subset of the items that fit into the knapsack

Example: Knapsack capacity W=16Example: Knapsack capacity W=16

item weight valueitem weight value

1 2 $201 2 $20

2 5 $302 5 $30

3 10 $503 10 $50

4 5 $104 5 $10


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Knapsack Problem by Exhaustive SearchKnapsack Problem by Exhaustive Search

SubsetSubset Total weightTotal weight Total valueTotal value{1} 2 $20{1} 2 $20

{2} 5 $30{2} 5 $30

{3} 10 $50{3} 10 $50

{4} 5 $10{4} 5 $10

{1,2} 7 $50{1,2} 7 $50

{1,3} 12 $70{1,3} 12 $70{1,4} 7 $30{1,4} 7 $30

{2,3} 15 $80{2,3} 15 $80

{2,4} 10 $40{2,4} 10 $40

{3,4} 15 $60{3,4} 15 $60

{1,2,3} 17 not feasible{1,2,3} 17 not feasible{1,2,4} 12 $60{1,2,4} 12 $60

{1,3,4} 17 not feasible{1,3,4} 17 not feasible

{2,3,4} 20 not feasible{2,3,4} 20 not feasible

{1,2,3,4} 22 not feasible{1,2,3,4} 22 not feasible


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Each subset of a set ofn elements {e0, e1, , en-1} can be

represented as a binary number ofn-1 bits.

For example, consider the set { 3, 5, 2, 8, 9 }. The binary

number1 0 0 1 0

represents the subset {3, 8 }.

The number of subsets for a set ofn elements is 2n.

Hence,)2()( nnC =



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a

The Travelling Salesman and Knapsack problems areThe Travelling Salesman and Knapsack problems areexamples ofexamples ofNP-hard problemsNP-hard problems..

a Computer scientists believe that polynomial-time algorithmsComputer scientists believe that polynomial-time algorithms

do not exists for such problemsdo not exists for such problems

There are more sophisticated approaches which solveThere are more sophisticated approaches which solvesome but not all instances of those problems in less thansome but not all instances of those problems in less than

exponential time.exponential time.

Another approach is to use approximation algorithms.Another approach is to use approximation algorithms.


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Example 3: The Assignment ProblemExample 3: The Assignment Problem

There areThere are nn people who need to be assigned topeople who need to be assigned to nn jobs, one personjobs, one person

per job. The cost of assigning personper job. The cost of assigning person iito jobto jobjjis C[is C[ii,,jj]. Find an]. Find anassignment that minimizes the total cost.assignment that minimizes the total cost.

Job 0 Job 1 Job 2 Job 3Job 0 Job 1 Job 2 Job 3

Person 0 9Person 0 9 2 7 82 7 8

Person 1 6 4 3 7Person 1 6 4 3 7

Person 2 5 8 1 8Person 2 5 8 1 8

Person 3 7 6 9 4Person 3 7 6 9 4

Algorithmic Plan:Algorithmic Plan:Generate all legitimate assignments, computeGenerate all legitimate assignments, computetheir costs, and select the cheapest one.their costs, and select the cheapest one.

How many assignments are there?How many assignments are there?

Pose the problem as the one about a cost matrix:Pose the problem as the one about a cost matrix:


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9 2 7 89 2 7 8

6 4 3 76 4 3 7 5 8 1 85 8 1 8

7 6 9 47 6 9 4

AssignmentAssignment (col.#s)(col.#s) Total CostTotal Cost

1, 2, 3, 41, 2, 3, 4 9+4+1+4=189+4+1+4=18

1, 2, 4, 31, 2, 4, 3 9+4+8+9=309+4+8+9=30

1, 3, 2, 41, 3, 2, 4 9+3+8+4=249+3+8+4=24

1, 3, 4, 21, 3, 4, 2 9+3+8+6=269+3+8+6=26

1, 4, 2, 31, 4, 2, 3 9+7+8+9=339+7+8+9=331, 4, 3, 21, 4, 3, 2 9+7+1+6=239+7+1+6=23

etc.etc.

Assignment Problem by Exhaustive SearchAssignment Problem by Exhaustive Search

C =C =

Generate all

permutations


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a

Number of permutations to be considered isNumber of permutations to be considered is n!.n!. This problems domain grows exponentially; exhaustiveThis problems domain grows exponentially; exhaustive

search is impratical (except for very small instances of thesearch is impratical (except for very small instances of the

problem)problem)

a

Actually, there is a much more efficient algorithm for thisActually, there is a much more efficient algorithm for thisproblem called theproblem called theHungarian methodHungarian method..

a It is important to remember that, more often than not, thereIt is important to remember that, more often than not, there

are no known polynomial-time algorithms for problemsare no known polynomial-time algorithms for problems

whose domain grows exponentially with instance size.whose domain grows exponentially with instance size.


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3 33A L iti I t d ti t th D i & A l i f Al ith 2 d d Ch 3

Final Comments on Exhaustive SearchFinal Comments on Exhaustive Search

a

Exhaustive-search algorithms run in a realistic amount ofExhaustive-search algorithms run in a realistic amount oftimetime only on very small instancesonly on very small instances

a In some cases, there are much better alternatives!In some cases, there are much better alternatives!

Euler circuitsEuler circuits shortest pathsshortest paths

minimum spanning treeminimum spanning tree

assignment problemassignment problem

a In many cases, exhaustive search or its variation is the onlyIn many cases, exhaustive search or its variation is the only

known way to get exact solutionknown way to get exact solution

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03 Brute Force

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