extended master method ( Method for solving Recurrence Relation)

1. UNIT-1
3RD YEAR (5TH SEM)
(2020-21)
Design and Analysis of
Algorithm (KCS503)
Submitted By:
Jeetesh Srivastava
Asst. Prof.(CSE)
UCEM, Prayagraj

2. 2
Extended Master
Method
Mr. Jeetesh Srivastava (DAA(KCS503))

3. Extended Master Method
Here, a >= 1, b > 1, k >= 0 and p is a real number.
 To solve recurrence relations using Master’s
theorem, we compare a with bk.
 Then, we follow the following cases-
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Mr. Jeetesh Srivastava (DAA(KCS503))

4. Extended Master Method (cont)
Case-01:
 If a > bk, then T(n) = θ (nlog
b
a)
Case-02:
 If a = bk and
If p < -1, then T(n) = θ (nlog
b
a)
If p = -1, then T(n) = θ (nlog
b
a.log2n)
If p > -1, then T(n) = θ (nlog
b
a.logp+1n)
Case-03:
If a < bk and
 If p < 0, then T(n) = O (nk)
 If p >= 0, then T(n) = θ (nklogpn)
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Mr. Jeetesh Srivastava (DAA(KCS503))

5. Example-1
Ex: Solve the following recurrence relation using
Master’s theorem- T(n) = 2T(n/2) + nlogn
 Solution: We compare the given recurrence relation
with T(n) = aT(n/b) + θ (nklogpn).
 Then, we have- a = 2, b = 2, k = 1, p = 1
 Now, a = 2 and bk = 21 = 2.
 Clearly, a = bk.
 So From Case 2, Since p = 1, so we have-
 T(n) = θ (nlog
b
a.logp+1n)
 T(n) = θ (nlog
2
2.log1+1n)
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Mr. Jeetesh Srivastava (DAA(KCS503))

6. Example-1
Mr. Jeetesh Srivastava (DAA(KCS503))
6
 Thus T(n) = θ (nlog2n)

7. Example-2
Ex: Solve the following recurrence relation using
Master’s theorem- T(n) = 2T(n/4) + n0.51
 Solution: We compare the given recurrence relation
with T(n) = aT(n/b) + θ (nklogpn).
 Then, we have- a = 2, b = 4, k = 0.51, p = 0
 Now, a = 2 and bk = 40.51 = 2.0279.
 Clearly, a < bk.
 So From Case 3, Since p = 0, so we have-
 T(n) = θ (nklogpn)
 T(n) = θ (n0.51log0n)
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Mr. Jeetesh Srivastava (DAA(KCS503))

8. Example-2
Mr. Jeetesh Srivastava (DAA(KCS503))
8
 Thus T(n) = θ (n0.51)

9. Example-3
Ex: Solve the following recurrence relation using
Master’s theorem- T(n) = √2T(n/2) + logn
 Solution: We compare the given recurrence relation
with T(n) = aT(n/b) + θ (nklogpn).
 Then, we have- a = √2, b = 2, k = 0, p = 1
 Now, a = √2 = 1.414 and bk = 20 = 1.
 Clearly, a > bk.
 So From Case 1,
 T(n) = θ (nlog
b
a) = θ (nlog
2
√2)
 T(n) = θ (n1/2)
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Mr. Jeetesh Srivastava (DAA(KCS503))

10. Example-3
Mr. Jeetesh Srivastava (DAA(KCS503))
10
 Thus T(n) = θ (√n)

11. Example-4
Ex: T (n) = 0.5T (n/2) + 1/n
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Mr. Jeetesh Srivastava (DAA(KCS503))

12. Example-4
 Solution:
T (n) = 0.5T (n/2) + 1/n ⇒ Does not apply (a < 1)
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Mr. Jeetesh Srivastava (DAA(KCS503))

13. Example-5
Ex: T (n) = 2nT (n/2) + nn
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Mr. Jeetesh Srivastava (DAA(KCS503))

14. Example-5
 Solution:
T (n) = 2nT (n/2) + nn ⇒ Does not apply (a is not
constant)
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Mr. Jeetesh Srivastava (DAA(KCS503))

15. Example-6
T (n) = 64T (n/8)− n2 log n
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Mr. Jeetesh Srivastava (DAA(KCS503))

16. Example-6
 Solution:
T (n) = 64T (n/8)− n2 log n ⇒ Does not apply (f(n) is
not positive)
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Mr. Jeetesh Srivastava (DAA(KCS503))

17. Example-7
T (n) = 3T (n/3) + √ n
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Mr. Jeetesh Srivastava (DAA(KCS503))

18. Example-7
 Solution:
T (n) = 3T (n/3) + √ n ⇒ T (n) = Θ(n) (Case 1)
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Mr. Jeetesh Srivastava (DAA(KCS503))

19. Example-8
T (n) = √ 2T (n/2) + log n
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Mr. Jeetesh Srivastava (DAA(KCS503))

20. Example-8
 Solution:
T (n) = √ 2T (n/2) + log n ⇒ T (n) = Θ(√ n) (Case 1)
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Mr. Jeetesh Srivastava (DAA(KCS503))

21. Example-9
T (n) = 4T (n/2) + n/ log n
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Mr. Jeetesh Srivastava (DAA(KCS503))

22. Example-9
 Solution:
T (n) = 4T (n/2) + n/ log n ⇒ T (n) = Θ(n2 ) (Case
1)
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Mr. Jeetesh Srivastava (DAA(KCS503))

23. Example-10
T (n) = 3T (n/3) + n/2
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Mr. Jeetesh Srivastava (DAA(KCS503))

24. Example-10
 Solution:
T (n) = 3T (n/3) + n/2 ⇒ T (n) = Θ(n log n) (Case 2)
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Mr. Jeetesh Srivastava (DAA(KCS503))

25. 25
Thank You
Mr. Jeetesh Srivastava (DAA(KCS503))