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CPSC 320: Intermediate AlgorithmDesign and Analysis

July 21, 2014

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Course Outline

• Introduction and basic concepts

• Asymptotic notation

• Greedy algorithms

• Graph theory

• Amortized analysis

• Recursion

• Divide-and-conquer algorithms

• Randomized algorithms

• Dynamic programming algorithms

• NP-completeness

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Master Theorem

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Master Theorem

• Let be real constants, , and defined by:

• where could also be or . Then:

1. If for some , then .

2. If for some , then .

3. If for some , and for some and all large enough, then .

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In-class Exercises

Solve using recursion trees and the master theorem, and prove using substitution

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Problem

Suppose the following recurring relation:

The relation above is hard to evaluate, but let’s try (so )

If we assume that :

By the master theorem, this results in . So:

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Divide-and-Conquer Algorithms

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Binary to Decimal conversion

• Problem: given an n-bit binary integer, calculate its decimal representation

• Naïve Approach (given as an array of bits):

Algorithm BinaryToDecimal1()

If Then

Return “”

Else

Last Modulo(, 10)

Rest Divide(, 10)

Return BinaryToDecimal1(Rest) . Last

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Binary to Decimal Analysis

• Using recursion trees we can calculate that this is

• Can we do better?

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Faster Algorithm

• Approach: divide and conquer

• Divide the array in two parts, do each part separately

• Divide using an appropriate power of 10

• Choose power of 10 that uses at least half of the bits

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Faster Algorithm

Algorithm BinaryToDecimalFast(A)

While Do

Multiply()

Return BinaryToDecimalHelper()

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Faster Algorithm

Algorithm BinaryToDecimalHelper()

If Then

Return String()

FirstHalf Divide()

SecondHalf Modulo()

Return BinaryToDecimalHelper(FirstHalf, , ) .

BinaryToDecimalHelper(SecondHalf, , )

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Exponential

• Problem: given two numbers and , calculate

• Simple solution:

Function Exponential(, )

If Then

Return 1

Else

Return Exponential(,)

• This algorithm runs in

• Can we do better?

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Divide and Conquer

• Strategy: divide the problem in smaller problems, then solve smaller problems and merge the results

• Exponential: we know that , and

Function Exponential2(, )

If Then

Return 1

Else If Is Odd Then

Return * Exponential(,)

Else

Return Exponential(,)

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Time Complexity

• The new algorithm runs in

• From master theorem, we get

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Stock trading problem

• Assume a list of trading stock prices in the future:

• 100, 113, 110, 85, 105, 102, 86, 63, 81, 101, 94, 106, 101, 79, 94, 90, 97

• You want to find the maximum profit you can make by buying a stock on one day and selling on another

• Example: buy at 63, sell at 106, profit is 43.

• Greedy algorithms won’t always work

• Lowest or highest prices might not be optimal options

• Brute-force (check every possible pair) runs in

• Can we do better?

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Maximum Subarray

• Let’s consider the daily changes (difference from previous day):

• 13, -3, -25, 20, -3, -16, -23, 18, 20, -7, 12, -5, -22, 15, -4, 7

• Now the problem becomes: finding the range with largest sum

• In the example: 18, 20, -7, 12; sum is 43

• How does that help?

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Divide and Conquer

• If we divide the array in two, a maximum subarray may be in:

• The first half

• The second half

• In between (crossing division)

• We can calculate the best option for each, then select the largest

• First and second half can be solved by recursion

• Crossing subarray can be found in linear time

• Find largest sum that ends at midpoint

• Find largest sum that starts just after midpoint

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Algorithm

Algorithm FindMaxCrossingSubarray(, , , )

LeftSum

Sum

For DownTo Do

Sum Sum

If Sum > LeftSum Then

LeftSum Sum

MaxLeft

RightSum

Sum

For To Do

Sum Sum

If Sum > RightSum Then

RightSum Sum

MaxRight

Return (MaxLeft, MaxRight, LeftSum+RightSum)

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Algorithm

Algorithm FindMaxSubarray(, , )

If Then

Return (, , )

(LS, LE, LSum) FindMaxSubarray(, , )

(RS, RE, RSum) FindMaxSubarray(, , )

(CS, CE, CSum) FindMaxCrossingSubarray(, , , )

If LSum > RSum And LSum > CSum Then Return (LS, LE, LSum)

If RSum > LSum And RSum > CSum Then Return (RS, LE, RSum)

Return (CS, CE, CSum)

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Time Complexity

• Since finding the maximum crossing subarray is and finding the largest among three values is , we have

• From the master theorem, we get that this approach is