Master Theorem - Introduction to Algorithms - Lecture Slides, Slides of Computer Science

Download Master Theorem - Introduction to Algorithms - Lecture Slides and more Slides Computer Science in PDF only on Docsity!

Solving Recurrences Continued

The Master Theorem

Introduction to heapsort

Review: Merge Sort

MergeSort(A, left, right) { if (left < right) { mid = floor((left + right) / 2); MergeSort(A, left, mid); MergeSort(A, mid+1, right); Merge(A, left, mid, right); } } // Merge() takes two sorted subarrays of A and // merges them into a single sorted subarray of A. // Code for this is in the book. It requires O(n) // time, and does require allocating O(n) space

Review: Solving Recurrences

• Substitution method
• Iteration method
• Master method

Review: Solving Recurrences

• The substitution method
  • A.k.a. the “making a good guess method”
  • Guess the form of the answer, then use induction to find the constants and show that solution works
  • Run an example : merge sort
    • T(n) = 2T(n/2) + cn
    • We guess that the answer is O(n lg n)
    • Prove it by induction
  • Can similarly show T(n) = Ω(n lg n), thus Θ(n lg n)

• T(n) = aT(n/b) + cn a(aT(n/b/b) + cn/b) + cn a 2 T(n/b 2 ) + cna/b + cn a 2 T(n/b 2 ) + cn(a/b + 1) a 2 (aT(n/b 2 /b) + cn/b 2 ) + cn(a/b + 1) a 3 T(n/b 3 ) + cn(a 2 /b 2 ) + cn(a/b + 1) a 3 T(n/b 3 ) + cn(a 2 /b 2 + a/b + 1) … a kT(n/b k) + cn(a k-1/b k-1^ + a k-2/b k-2^ + … + a 2 /b 2 + a/b + 1)

( ) cn n b

n aT T n

• So we have
  • T(n) = a kT(n/b k) + cn(a k-1/b k-1^ + ... + a 2 /b 2 + a/b + 1)
• For k = log (^) b n
  • n = b k
  • T(n) = a kT(1) + cn(a k-1/b k-1^ + ... + a 2 /b 2 + a/b + 1) = a kc + cn(a k-1/b k-1^ + ... + a 2 /b 2 + a/b + 1) = ca k^ + cn(a k-1/b k-1^ + ... + a 2 /b 2 + a/b + 1) = cna k^ /b k^ + cn(a k-1/b k-1^ + ... + a 2 /b 2 + a/b + 1) = cn(a k/b k^ + ... + a 2 /b 2 + a/b + 1)

( ) cn n b

n aT T n

• So with k = logb n
  • T(n) = cn(a k/b k^ + ... + a^2 /b 2 + a/b + 1)
• What if a < b?

( ) cn n b

n aT T n

• So with k = logb n
  • T(n) = cn(a k/b k^ + ... + a^2 /b 2 + a/b + 1)
• What if a < b?
  • Recall that Σ(xk^ + xk-1^ + … + x + 1) = (x k+1^ -1)/(x-1)

( ) cn n b

n aT T n

• So with k = logb n
  • T(n) = cn(a k/b k^ + ... + a^2 /b 2 + a/b + 1)
• What if a < b?
  • Recall that Σ(xk^ + xk-1^ + … + x + 1) = (x k+1^ -1)/(x-1)
  • So:
  • T(n) = cn ·Θ(1) = Θ(n)

( ) cn n b

n aT T n

( a b ) a b

a b a b

a b b

a b

a b

a k k k

k k

k − < − = − −

• • • • = + − + −

− 1

1 1

1 1 1 1 1 1 1

1 

• So with k = logb n
  • T(n) = cn(a k/b k^ + ... + a^2 /b 2 + a/b + 1)
• What if a > b?

( ) cn n b

n aT T n

• So with k = logb n
  • T(n) = cn(a k/b k^ + ... + a^2 /b 2 + a/b + 1)
• What if a > b?
  • T(n) = cn · Θ(a k^ / b k)

( ) cn n b

n aT T n

(( ) k )

k k

k k

k a b a b

a b b

a b

a b

a (^) = Θ −

• • • • = + − −

− 1

1 1 1 1

1 

• So with k = logb n
  • T(n) = cn(a k/b k^ + ... + a^2 /b 2 + a/b + 1)
• What if a > b?
  • T(n) = cn · Θ(a k^ / b k) = cn · Θ(a log n^ / b log n) = cn · Θ(a log n^ / n)

( ) cn n b

n aT T n

(( ) k )

k k

k k

k a b a b

a b b

a b

a b

a (^) = Θ −

• • • • = + − −

− 1

1 1 1 1

1 

• So with k = logb n
  • T(n) = cn(a k/b k^ + ... + a^2 /b 2 + a/b + 1)
• What if a > b?
  • T(n) = cn · Θ(a k^ / b k) = cn · Θ(a log n^ / b log n) = cn · Θ(a log n^ / n) recall logarithm fact: a log n^ = n log a = cn · Θ(n log a^ / n) = Θ(cn · nlog a^ / n)

( ) cn n b

n aT T n

(( ) k )

k k

k k

k a b a b

a b b

a b

a b

a (^) = Θ −

• • • • = + − −

− 1

1 1 1 1

1 

• So with k = logb n
  • T(n) = cn(a k/b k^ + ... + a^2 /b 2 + a/b + 1)
• What if a > b?
  • T(n) = cn · Θ(a k^ / b k) = cn · Θ(a log n^ / b log n) = cn · Θ(a log n^ / n) recall logarithm fact: a log n^ = n log a = cn · Θ(n log a^ / n) = Θ(cn · nlog a^ / n) = Θ(n log a^ )

( ) cn n b

n aT T n

(( ) k )

k k

k k

k a b a b

a b b

a b

a b

a (^) = Θ −

• • • • = + − −

− 1

1 1 1 1

1 