Master Theorem:
Practice Problems and Solutions

Master Theorem
The Master Theorem applies to recurrences of the following form:
T (n) = aT (n/b) + f (n)
where a ≥ 1 and b > 1 are constants and f (n) is an asymptotically positive function.
There are 3 cases:
1. If f (n) = O(nlogb a− ) for some constant

> 0, then T (n) = Θ(nlogb a ).

2. If f (n) = Θ(nlogb a logk n) with1 k ≥ 0, then T (n) = Θ(nlogb a logk+1 n).
3. If f (n) = Ω(nlogb a+ ) with > 0, and f (n) satisfies the regularity condition, then T (n) = Θ(f (n)).
Regularity condition: af (n/b) ≤ cf (n) for some constant c < 1 and all sufficiently large n.

Practice Problems
For each of the following recurrences, give an expression for the runtime T (n) if the recurrence can be
solved with the Master Theorem. Otherwise, indicate that the Master Theorem does not apply.
1. T (n) = 3T (n/2) + n2

2. T (n) = 4T (n/2) + n2

3. T (n) = T (n/2) + 2n

4. T (n) = 2n T (n/2) + nn

5. T (n) = 16T (n/4) + n

6. T (n) = 2T (n/2) + n log n

1 most

of the time, k = 0

1
7. T (n) = 2T (n/2) + n/ log n

8. T (n) = 2T (n/4) + n0.51

9. T (n) = 0.5T (n/2) + 1/n

10. T (n) = 16T (n/4) + n!

11. T (n) =

√
2T (n/2) + log n

12. T (n) = 3T (n/2) + n

13. T (n) = 3T (n/3) +

√
n

14. T (n) = 4T (n/2) + cn

15. T (n) = 3T (n/4) + n log n

16. T (n) = 3T (n/3) + n/2

17. T (n) = 6T (n/3) + n2 log n

18. T (n) = 4T (n/2) + n/ log n

19. T (n) = 64T (n/8) − n2 log n
20. T (n) = 7T (n/3) + n2

21. T (n) = 4T (n/2) + log n

22. T (n) = T (n/2) + n(2 − cos n)

2
Solutions
1. T (n) = 3T (n/2) + n2 =⇒ T (n) = Θ(n2 ) (Case 3)
2. T (n) = 4T (n/2) + n2 =⇒ T (n) = Θ(n2 log n) (Case 2)
3. T (n) = T (n/2) + 2n =⇒ Θ(2n ) (Case 3)
4. T (n) = 2n T (n/2) + nn =⇒ Does not apply (a is not constant)
5. T (n) = 16T (n/4) + n =⇒ T (n) = Θ(n2 ) (Case 1)
6. T (n) = 2T (n/2) + n log n =⇒ T (n) = n log2 n (Case 2)
7. T (n) = 2T (n/2) + n/ log n =⇒ Does not apply (non-polynomial difference between f (n) and nlogb a )
8. T (n) = 2T (n/4) + n0.51 =⇒ T (n) = Θ(n0.51 ) (Case 3)
9. T (n) = 0.5T (n/2) + 1/n =⇒ Does not apply (a < 1)
10. T (n) = 16T (n/4) + n! =⇒ T (n) = Θ(n!) (Case 3)
11. T (n) =

√
√
2T (n/2) + log n =⇒ T (n) = Θ( n) (Case 1)

12. T (n) = 3T (n/2) + n =⇒ T (n) = Θ(nlg 3 ) (Case 1)
13. T (n) = 3T (n/3) +

√
n =⇒ T (n) = Θ(n) (Case 1)

14. T (n) = 4T (n/2) + cn =⇒ T (n) = Θ(n2 ) (Case 1)
15. T (n) = 3T (n/4) + n log n =⇒ T (n) = Θ(n log n) (Case 3)
16. T (n) = 3T (n/3) + n/2 =⇒ T (n) = Θ(n log n) (Case 2)
17. T (n) = 6T (n/3) + n2 log n =⇒ T (n) = Θ(n2 log n) (Case 3)
18. T (n) = 4T (n/2) + n/ log n =⇒ T (n) = Θ(n2 ) (Case 1)
19. T (n) = 64T (n/8) − n2 log n =⇒ Does not apply (f (n) is not positive)
20. T (n) = 7T (n/3) + n2 =⇒ T (n) = Θ(n2 ) (Case 3)
21. T (n) = 4T (n/2) + log n =⇒ T (n) = Θ(n2 ) (Case 1)
22. T (n) = T (n/2) + n(2 − cos n) =⇒ Does not apply. We are in Case 3, but the regularity condition is
violated. (Consider n = 2πk, where k is odd and arbitrarily large. For any such choice of n, you can
show that c ≥ 3/2, thereby violating the regularity condition.)

3

Master theorm practive problems with solutions

  • 1.
    Master Theorem: Practice Problemsand Solutions Master Theorem The Master Theorem applies to recurrences of the following form: T (n) = aT (n/b) + f (n) where a ≥ 1 and b > 1 are constants and f (n) is an asymptotically positive function. There are 3 cases: 1. If f (n) = O(nlogb a− ) for some constant > 0, then T (n) = Θ(nlogb a ). 2. If f (n) = Θ(nlogb a logk n) with1 k ≥ 0, then T (n) = Θ(nlogb a logk+1 n). 3. If f (n) = Ω(nlogb a+ ) with > 0, and f (n) satisfies the regularity condition, then T (n) = Θ(f (n)). Regularity condition: af (n/b) ≤ cf (n) for some constant c < 1 and all sufficiently large n. Practice Problems For each of the following recurrences, give an expression for the runtime T (n) if the recurrence can be solved with the Master Theorem. Otherwise, indicate that the Master Theorem does not apply. 1. T (n) = 3T (n/2) + n2 2. T (n) = 4T (n/2) + n2 3. T (n) = T (n/2) + 2n 4. T (n) = 2n T (n/2) + nn 5. T (n) = 16T (n/4) + n 6. T (n) = 2T (n/2) + n log n 1 most of the time, k = 0 1
  • 2.
    7. T (n)= 2T (n/2) + n/ log n 8. T (n) = 2T (n/4) + n0.51 9. T (n) = 0.5T (n/2) + 1/n 10. T (n) = 16T (n/4) + n! 11. T (n) = √ 2T (n/2) + log n 12. T (n) = 3T (n/2) + n 13. T (n) = 3T (n/3) + √ n 14. T (n) = 4T (n/2) + cn 15. T (n) = 3T (n/4) + n log n 16. T (n) = 3T (n/3) + n/2 17. T (n) = 6T (n/3) + n2 log n 18. T (n) = 4T (n/2) + n/ log n 19. T (n) = 64T (n/8) − n2 log n 20. T (n) = 7T (n/3) + n2 21. T (n) = 4T (n/2) + log n 22. T (n) = T (n/2) + n(2 − cos n) 2
  • 3.
    Solutions 1. T (n)= 3T (n/2) + n2 =⇒ T (n) = Θ(n2 ) (Case 3) 2. T (n) = 4T (n/2) + n2 =⇒ T (n) = Θ(n2 log n) (Case 2) 3. T (n) = T (n/2) + 2n =⇒ Θ(2n ) (Case 3) 4. T (n) = 2n T (n/2) + nn =⇒ Does not apply (a is not constant) 5. T (n) = 16T (n/4) + n =⇒ T (n) = Θ(n2 ) (Case 1) 6. T (n) = 2T (n/2) + n log n =⇒ T (n) = n log2 n (Case 2) 7. T (n) = 2T (n/2) + n/ log n =⇒ Does not apply (non-polynomial difference between f (n) and nlogb a ) 8. T (n) = 2T (n/4) + n0.51 =⇒ T (n) = Θ(n0.51 ) (Case 3) 9. T (n) = 0.5T (n/2) + 1/n =⇒ Does not apply (a < 1) 10. T (n) = 16T (n/4) + n! =⇒ T (n) = Θ(n!) (Case 3) 11. T (n) = √ √ 2T (n/2) + log n =⇒ T (n) = Θ( n) (Case 1) 12. T (n) = 3T (n/2) + n =⇒ T (n) = Θ(nlg 3 ) (Case 1) 13. T (n) = 3T (n/3) + √ n =⇒ T (n) = Θ(n) (Case 1) 14. T (n) = 4T (n/2) + cn =⇒ T (n) = Θ(n2 ) (Case 1) 15. T (n) = 3T (n/4) + n log n =⇒ T (n) = Θ(n log n) (Case 3) 16. T (n) = 3T (n/3) + n/2 =⇒ T (n) = Θ(n log n) (Case 2) 17. T (n) = 6T (n/3) + n2 log n =⇒ T (n) = Θ(n2 log n) (Case 3) 18. T (n) = 4T (n/2) + n/ log n =⇒ T (n) = Θ(n2 ) (Case 1) 19. T (n) = 64T (n/8) − n2 log n =⇒ Does not apply (f (n) is not positive) 20. T (n) = 7T (n/3) + n2 =⇒ T (n) = Θ(n2 ) (Case 3) 21. T (n) = 4T (n/2) + log n =⇒ T (n) = Θ(n2 ) (Case 1) 22. T (n) = T (n/2) + n(2 − cos n) =⇒ Does not apply. We are in Case 3, but the regularity condition is violated. (Consider n = 2πk, where k is odd and arbitrarily large. For any such choice of n, you can show that c ≥ 3/2, thereby violating the regularity condition.) 3