Data Structures and Algorithms
Solving Recurrence Relations
Chris Brooks
Department of Computer Science
University of San Francisco
Department of Computer Science โ University of San Francisco โ p.1/30
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Solving Recurrence Relations in Algorithm Analysis, Study notes of Data Structures and Algorithms
This document from the university of san francisco's department of computer science discusses methods for analyzing the running time of recursive programs using recurrence relations. Topics covered include the power function, the master theorem, and the method of iterated multiplication. The document also includes exercises for students to practice solving recurrence relations.
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Pre 2010
1 / 30
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Data Structures and Algorithms Solving Recurrence Relations
Chris Brooks Department of Computer ScienceUniversity of San Francisco
Department of Computer Science โ University of San Francisco โ p.1/
4-0:^
Algorithm Analysis
for^ (i=1;
i<=n*n;
i++)
for^ (j=0;
j<i;
j++) sum++;
Department of Computer Science โ University of San Francisco โ p.2/
4-2:^
Algorithm Analysis
for^ (i=1;
i<=n*n;
i++)
for^ (j=0;
j<i;
j++) sum++; Exact # of times
sum++
is executed: (^2) n โ^ i^ = i=
(^2) n (^2) (n+ 1)^2 n= 4 +^ n
โ^ ฮ(
(^4) n)
Department of Computer Science โ University of San Francisco โ p.4/
4-3:^
Recursive Functions
long^
power(long
x,^ long
n)
if^ (n
==^ 0)return^ 1; elsereturn
x *^
power(x,
n-1);
How many times is this executed?
Department of Computer Science โ University of San Francisco โ p.5/
4-5:^
Recurrence Relations
long^
power(long
x,^ long
n)
if^ (n
==^ 0)return^ 1; elsereturn
x *^
power(x,
n-1);
T^ (0) =
c^1
for some constant
c^1
T^ (n) =
c+^2
T^ (n
โ^ 1)
for some constant
c^2 Department of Computer Science โ University of San Francisco โ p.7/
4-6:^
Solving Recurrence Relations
T^ (0) =
c^1 T^ (n) =
T^ (n
โ^ 1) +
c^2
If we knew
T^ (n
โ^ 1)
, we could solve
T^ (n
T^ (n)
=^
T^ (n^
โ^ 1) +
c^2
Department of Computer Science โ University of San Francisco โ p.8/
4-8:^
Solving Recurrence Relations
T^ (0) =
c^1 T^ (n) =
T^ (n
โ^ 1) +
c^2
If we knew
T^ (n
โ^ 1)
, we could solve
T^ (n
T^ (n)
=^
T^ (n^
โ^ 1) +
c^2
T^ (n^
โ^ 1) =
T^ (n
โ^ 2) +
c^2
=^ T^ (
n^ โ^ 2) +
c+^2
c^2
=^ T^ (
n^ โ^ 2) + 2
c^2
T^ (n^
โ^ 2) =
T^ (n
โ^ 3) +
c^2
=^ T^ (
n^ โ^ 3) +
c+ 2^2
c^2
=^ T^ (
n^ โ^ 3) + 3
c^2
Department of Computer Science โ University of San Francisco โ p.10/
4-9:^
Solving Recurrence Relations
T^ (0) =
c^1 T^ (n) =
T^ (n
โ^ 1) +
c^2
If we knew
T^ (n
โ^ 1)
, we could solve
T^ (n
T^ (n)
=^
T^ (n^
โ^ 1) +
c^2
T^ (n^
โ^ 1) =
T^ (n
โ^ 2) +
c^2
=^ T^ (
n^ โ^ 2) +
c+^2
c^2
=^ T^ (
n^ โ^ 2) + 2
c^2
T^ (n^
โ^ 2) =
T^ (n
โ^ 3) +
c^2
=^ T^ (
n^ โ^ 3) +
c+ 2^2
c^2
=^ T^ (
n^ โ^ 3) + 3
c^2
T^ (n^
โ^ 3) =
T^ (n
โ^ 4) +
c^2
=^ T^ (
n^ โ^ 4) + 4
c^2
Department of Computer Science โ University of San Francisco โ p.11/
Solving Recurrence Relations
T^ (0) =
c^1 T^ (n) =
T^ (n
โ^ k) +
k^ โ^ c
for all 2
k
If we set
k^ =
n, we have: T^ (n)
=^
T^ (n^
โ^ n) +
nc^2
=^ T^ (0) +
nc^2 =^ c^1
+^ nc
2 โ^ ฮ(
n)
Department of Computer Science โ University of San Francisco โ p.13/
Building a Better
Power
Can we avoid making a linear number of function calls? long^
power(long
x,^ long
n)
if^ (n==0)
return
1;
if^ (n==1)
return
x;
if^ ((n
%^ 2)
==^ 0) return
power(x*x,
n/2);
elsereturn
power(x*x,
n/2)
*^ x;
Department of Computer Science โ University of San Francisco โ p.14/
Solving Recurrence Relations
T^ (n)
=^
T^ (n/
2) +^
c^3
Department of Computer Science โ University of San Francisco โ p.16/
Solving Recurrence Relations
T^ (n)
=^
T^ (n/
2) +^
c^3
T^ (n/
T^ (n/
c^3
=^ T^ (
n/4) +
c+^3
c^3
=^ T^ (
n/4) + 2
c^3
Department of Computer Science โ University of San Francisco โ p.17/
Solving Recurrence Relations
T^ (n)
=^
T^ (n/
2) +^
c^3
T^ (n/
T^ (n/
c^3
=^ T^ (
n/4) +
c+^3
c^3
=^ T^ (
n/4) + 2
c^3
T^ (n/
T^ (n/
c^3
=^ T^ (
n/8) +
c+ 2^3
c^3
=^ T^ (
n/8) + 3
c^3
T^ (n/
T^ (n/
c^3
=^ T^ (
n/16) +
c+ 3^3
c^3
=^ T^ (
n/16) + 4
c^3
Department of Computer Science โ University of San Francisco โ p.19/
Solving Recurrence Relations
T^ (n)
=^
T^ (n/
2) +^
c^3
T^ (n/
T^ (n/
c^3
=^ T^ (
n/4) +
c+^3
c^3
=^ T^ (
n/4)
c^3
T^ (n/
T^ (n/
c^3
=^ T^ (
n/8) +
c+ 2^3
c^3
=^ T^ (
n/8)
c^3
T^ (n/
T^ (n/
c^3
=^ T^ (
n/16) +
c+ 3^3
c^3
=^ T^ (
n/16) + 4
c^3
T^ (n/
T^ (n/
c^3
=^ T^ (
n/32) +
c+ 4^3
c^3
=^ T^ (
n/32) + 5
c^3
Department of Computer Science โ University of San Francisco โ p.20/