Divide and Conquer Summary Divide Identify one or more subproblems - - PowerPoint PPT Presentation

Divide and Conquer Summary

• Divide

– Identify one or more subproblems

• Conquer

– Solve some or all of those subproblems

• Combine

– Use the subproblems’ solutions to find large solution

Generic Divide and Conquer Solution

def myDCalgo(problem): if baseCase(problem): solution = solve(problem) #brute force if necessary return solution subproblems = Divide(problem) for sub in subproblems: subsolutions.append(myDCalgo(sub)) solution = Combine(subsolutions) return solution

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Karatsuba

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4 1 0 2 × 1 8 1 9

a b c d a b

+ = 10

𝑜 2

c d

+ = 10

𝑜 2

( )

a c

× 10𝑜

a d

10

𝑜 2

× ( ) ( )

b c

× +

b d

×

• 1. Break into smaller subproblems

+ +

Karatsuba

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10𝑜 𝑏𝑑 + 10

𝑜 2 𝑏𝑒 + 𝑐𝑑 + 𝑐𝑒 Can’t avoid these This can be simplified

𝑏 + 𝑐 𝑑 + 𝑒 = 𝑏𝑑 + 𝑏𝑒 + 𝑐𝑑 + 𝑐𝑒 𝑏𝑒 + 𝑐𝑑 = 𝑏 + 𝑐 𝑑 + 𝑒 − 𝑏𝑑 − 𝑐𝑒

One multiplication Two multiplications

a

×

b c d

Karatsuba Pseudocode

def dc_mult(x, y): n = length of larger of x,y if n == 1: return x*y a = first n/2 digits of x b = last n/2 digits of x c = first n/2 digits of y d = last n/2 digits of y ac = dc_mult(a, c) bd = dc_mult(b, d) adbc = dc_mult(a+b, c+d) – ac – bd return ac*10^n + (adbc)*10^(n/2) + bd

Divide Conquer Combine

Divide and Conquer Summary

• Divide

– Identify one or more subproblems:

• Conquer

– Solve some or all of those subproblems:

• Combine

– Use the subproblems’ solutions to find large solution:

Generic Divide and Conquer Solution

def myDCalgo(problem): if baseCase(problem): solution = solve(problem) #brute force if necessary return solution subproblems = Divide(problem) for sub in subproblems: subsolutions.append(myDCalgo(sub)) solution = Combine(subsolutions) return solution

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Karatsuba Pseudocode

def dc_mult(x, y): n = length of larger of x,y if n == 1: return x*y a = first n/2 digits of x b = last n/2 digits of x c = first n/2 digits of y d = last n/2 digits of y ac = dc_mult(a, c) bd = dc_mult(b, d) adbc = dc_mult(a+b, c+d) – ac – bd return ac*10^n + (adbc)*10^(n/2) + bd

Divide Conquer Combine

𝑈 𝑜 = 3𝑈 𝑜 2 + 𝑃(𝑜)

Maximum Sum Continuous Subarray Problem

The maximum-sum subarray of a given array of integers 𝐵 is the interval [𝑏, 𝑐] such that the sum of all values in the array between 𝑏 and 𝑐 inclusive is maximal. Given an array of 𝑜 integers (may include both positive and negative values), give a 𝑃(𝑜 log 𝑜) algorithm for finding the maximum-sum subarray.

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Naïve Solution

10

5 8

• 4

3 7

• 15

2 8

• 20 17

8

• 50 -5

22

2 1 3 4 5 6 7 8 9 10 11 12 13

What does a 𝑜 log 𝑜 recurrence look like?

• 𝑈 𝑜 =

Tree method

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𝑜 total / level log2 𝑜 levels

• f recursion

𝑜

𝑈 𝑜 = 2𝑈(𝑜 2 ) + 𝑜

𝑈 𝑜 = 𝑜

log2 𝑜 𝑗=1

= 𝑜 log2 𝑜

𝑜 2 𝑜 2 𝑜 4 𝑜 4 𝑜 4 𝑜 4

… … … …

1 1 1

…

1 1 1

𝑜 𝑜 2 𝑜 2 𝑜 4 𝑜 4 𝑜 4 𝑜 4 1 1 1 1 1 1

Divide and Conquer Θ(𝑜 log 𝑜)

13

5 8

• 4

3 7

• 15

2 8

• 20 17

8

• 50 -5

22

2 1 3 4 5 6 7 8 9 10 11 12 13

Divide in half Recursively Solve on Left Recursively Solve on Right

Divide and Conquer Θ(𝑜 log 𝑜)

14

5 8

• 4

3 7

• 15

2 8

• 20 17

8

• 50 -5

22

2 1 3 4 5 6 7 8 9 10 11 12 13

Divide in half Recursively Solve on Left 19 Recursively Solve on Right 25 Find Largest sum that spans the cut 2

• 13
• 6
• 3
• 7

1 6

• 20
• 42
• 37

13 5

• 12

8 Largest sum that ends here + Largest sum that starts here

Divide and Conquer Θ(𝑜 log 𝑜)

15

5 8

• 4

3 7

• 15

2 8

• 20 17

8

• 50 -5

22

2 1 3 4 5 6 7 8 9 10 11 12 13

Divide in half Recursively Solve on Left 19 Recursively Solve on Right 25 Find Largest sum that spans the cut 19 2

• 13
• 6
• 3
• 7

1 6

• 20
• 42
• 37

13 5

• 12

8 Return the Max of Left, Right, Center

𝑈 𝑜 = 2𝑈 𝑜 2 + 𝑜

Divide and Conquer Summary

• Divide

– Break the list in half

• Conquer

– Find the best subarrays on the left and right

• Combine

– Find the best subarray that “spans the divide” – I.e. the best subarray that ends at the divide concatenated with the best that starts at the divide

Typically multiple subproblems. Typically all roughly the same size.

Generic Divide and Conquer Solution

def myDCalgo(problem): if baseCase(problem): solution = solve(problem) #brute force if necessary return solution subproblems = Divide(problem) for sub in subproblems: subsolutions.append(myDCalgo(sub)) solution = Combine(subsolutions) return solution

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MSCS Divide and Conquer Θ(𝑜 log 𝑜)

def MSCS(list): if list.length < 2: return list[0] #list of size 1 the sum is maximal {listL, listR} = Divide (list) for list in {listL, listR}: subSolutions.append(MSCS(list)) solution = max(solnL, solnR, span(listL, listR)) return solution

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