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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 L30, #1 of L31, #3 of L32, #2 of L33, #1 of L34. Final exam on Tuesday 7 December, 810AM in TH 216.
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