Recurrence relationclass 5
13
Solving Recurrences The Master Theorem
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Transcript of Recurrence relationclass 5
Solving RecurrencesThe Master Theorem
Recurrence Relations
Recurrences
• The expression:
is a recurrence.– Recurrence: an equation that describes a function in
terms of its value on smaller functions
1
22
1
)(ncn
nT
nc
nT
Recurrence Examples
0
0
)1(
0)(
n
n
nscns
0)1(
00)(
nnsn
nns
1
22
1
)(nc
nT
nc
nT
1
1
)(
ncnb
naT
nc
nT
Solving Recurrences
• Substitution method• Iteration method• Master method
Solving Recurrences
• The substitution method (CLR 4.1)– “Making a good guess” method– Guess the form of the answer, then use
induction to find the constants and show that solution works
– Examples:• T(n) = 2T(n/2) + (n) T(n) = (n lg n)• T(n) = 2T(n/2) + n ???
The Master Theorem
• Given: a divide and conquer algorithm– An algorithm that divides the problem of size
n into a subproblems, each of size n/b– Let the cost of each stage (i.e., the work to
divide the problem + combine solved subproblems) be described by the function f(n)
• Then, the Master Theorem gives us a cookbook for the algorithm’s running time:
The Master Theorem
• if T(n) = aT(n/b) + f(n) then
1
0
largefor )()/(
AND )(
)(
)(
)(
log)(
log
log
log
log
log
c
nncfbnaf
nnf
nnf
nOnf
nf
nn
n
nT
a
a
a
a
a
b
b
b
b
b
Using The Master Method
• T(n) = 9T(n/3) + n– a=9, b=3, f(n) = n– nlogb a = nlog3 9 = (n2)– Since f(n) = O(nlog3 9 - ), where =1, case 1
applies:
– Thus the solution is T(n) = (n2)
aa bb nOnfnnT loglog )( when )(
Simplified Masters Theorem
)(
)(
)(
log
log)(
log d
d
d
a
d
d
baif
baif
baif
nn
nn
n
nT
b
• In divide and conquer generally, a problem instance size of n is divided into two instances of n/2. More generally an instance of size n can be divided into b instances of size n/b, with 'a' of the needing to be solved. ( hence, 'a' and b are constants; a>=1 and b>1). Assuming that size n is a power of b, by simplifying the analysis, the recurrence for the running time
»T(n)=aT(n/b)+f(n)• Where f(n) is a function that accounts for the time
spent on dividing the problem into smaller ones and on combining their solutions.
More Examples of Master’s Theorem
• T(n) = 3T(n/5) + n• T(n) = 2T(n/2) + n• T(n) = 2T(n/2) + 1• T(n) = T(n/2) + n• T(n) = T(n/2) + 1
When Master’s Theorem cannot be applied
• T(n) = 2T(n/2) + n logn• T(n) = 2T(n/2) + n/ logn
