the recursion-tree method

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1 the recursion- method recurrence into a 1 recurrence into a 2 MCS 360 Lecture 39 Introduction to Data Structures Jan Verschelde, 22 November 2010

2 recurrence into a The for consists of two steps: 1 Guess the form of the solution. 2 Use mathematical induction to find constants in the form and show that the solution works. In the previous lecture, the focus was on step 2. Today we introduce the recursion- method to generate a guess for the form of the solution to the recurrence.

3 the recursion- method recurrence into a 1 recurrence into a 2

4 recurrence into a an Consider the recurrence relation T (n) =3T (n/4)+cn 2 for some constant c. We assume that n is an exact power of 4. In the recursion- method we expand T (n) into a : T (n) cn 2 T ( n 4 ) T ( n 4 ) T ( n 4 )

5 recurrence into a we expand T ( n 4 ) Applying T (n) =3T (n/4)+cn 2 to T (n/4) leads to T (n/4) =3T (n/16)+c(n/4) 2, leaves: c( n 4 )2 cn 2 c( n 4 )2 c( n 4 )2 T ( n 16 ) T ( n 16 ) T ( n 16 ) T ( n 16 ) T ( n 16 ) T ( n 16 ) T ( n 16 ) T ( n 16 ) T ( n 16 )

6 recurrence into a we expand T ( n 16 ) Applying T (n) =3T (n/4)+cn 2 to T (n/16) leads to T (n/16) =3T (n/64)+c(n/16) 2, leaves: cn 2 c( n 4 )2 c( n 4 )2 c( n 4 )2 c( n n c( n c( c( n c( n n c( c( n c( n n c(

7 the recursion- method recurrence into a 1 recurrence into a 2

8 recurrence into a the cost at We sum the cost at of the : cn 2 = cn 2 c( n 4 )2 + c( n 4 )2 + c( n 4 )2 = 3 16 cn2 c( n n + c( n + c( + c( n n + c( n + c( + c( n n + c( n + c( =( 3 cn 2

9 recurrence into a adding up the costs T (n) = cn ( ) cn2 + cn ( = cn ( ) ) + + The disappear if n = 16, or the has depth at least 2 if n 16 = 4 2. For n = 4 k, k = log 4 (n), we have: log 4 (n) T (n) =cn 2 i=0 ( ) 3 i. 16

10 recurrence into a geometric series Consider a finite sum first: n S n = 1 + r + r r n = r i. To find an explicit form of the solution we do i=0 rs n = r + r r n + r n+1 S n = 1 + r + r r n (r 1)S n = 1 + r n+1 So the explicit sum is S n = r n+1 1 r 1.

11 recurrence into a Applying to with r = 3 16 leads to geometric sum S n = n i=0 r i = r n+1 1 r 1 log 4 (n) T (n) =cn 2 i=0 ( ) 3 i 16 ( 3 ) log4 (n)+1 T (n) =cn

12 recurrence into a polishing the result Instead of T (n) dn 2 for some constant d, wehave Recall ( 3 ) log4 (n)+1 T (n) =cn log 4 (n) T (n) =cn 2 i=0 ( ) 3 i. 16 To remove the log 4 (n) factor, we consider T (n) cn 2 i=0 ( ) 3 i 16 = cn dn2, for some constant d.

13 the recursion- method recurrence into a 1 recurrence into a 2

14 recurrence into a verifying the guess Let us see if T (n) dn 2 is good for T (n) =3T (n/4)+cn 2. Applying the : T (n) = 3T (n/4)+cn 2 ( n ) 2 3d + cn 2 ( 4 ) 3 = 16 d + c n 2 = 3 (d ) 16 c n (2d) n2, if d 16 3 c dn 2

15 the recursion- method recurrence into a 1 recurrence into a 2

16 recurrence into a Consider T (n) =T (n/3)+t (2n/3)+cn. c n 3 log 3/2 (n) c n 9 2n c 9 cn = cn + = cn c 2n 9 c 2n = cn 4n c 9. + = O(n log 2 (n))

17 recurrence into a Summary + Assignments We covered 4.4 of Introduction to Algorithms, 3rd edition by Thomas H. Cormen, Clifford Stein, Ronald L. Rivest, and Charles E. Leiserson. Assignments: 1 Consider T (n) =3T (n/2)+n. Use a recursion to derive a guess for an asymptotic upper bound for T (n) and verify the guess with the. 2 Same question as before for T (n) =T (n/2)+n 2. 3 Same question as before for T (n) =2T (n 1)+1. Last homework collection on Monday 29 November: #1 of L-30, #1 of L-31, #3 of L-32, #2 of L-33, #1 of L-34. Final exam on Tuesday 7 December, 8-10AM in TH 216.

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