Master’s Theorem | Master’s Theorem Examples

Master’s Theorem-

 

Master’s Theorem is a popular method for solving the recurrence relations.

 

Master’s theorem solves recurrence relations of the form-

 

 

where a >= 1, b > 1, k >= 0 and p is a real number

 

Solving Recurrence Relations-

 

To solve recurrence relations using Master’s theorem, we compare a with bk.

Then, we follow the following cases-

 

Case-01:

 

If a > bk, then T(n) = θ (nlogba)

 

Case-02:

 

If a = band

  • If p < -1, then T(n) = θ (nlogba)
  • If p = -1, then T(n) = θ (nlogba.log2n)
  • If p > -1, then T(n) = θ (nlogba.logp+1n)

 

Case-03:

 

If a < band

  • If p < 0,  then T(n) = O (nk)
  • If p >= 0, then T(n) = θ (nklogpn)

 

PRACTICE PROBLEMS BASED ON MASTER’S THEOREM-

 

Problem-01:

 

Solve the following recurrence relation using Master’s theorem-

T(n) = 3T(n/2) + n2

 

Solution-

 

On comparing the given recurrence relation with-

T(n) = aT(n/b) + θ (nklogpn)

we have-

a = 3

b = 2

k = 2

p = 0

 

Now, a = 3 and bk = 22 = 4.

Clearly, a < bk

So, we follow case-03.

Since p = 0, so we have-

T(n) = θ (nklogpn)

T(n) = θ (n2log0n)

 

Thus,

T(n) = θ (n2)

 

Problem-02:

 

Solve the following recurrence relation using Master’s theorem-

T(n) = 2T(n/2) + nlogn

 

Solution-

 

On comparing the given recurrence relation with-

T(n) = aT(n/b) + θ (nklogpn)

we have-

a = 2

b = 2

k = 1

p = 1

 

Now, a = 2 and bk = 21 = 2.

Clearly, a = bk

So, we follow case-02.

Since p = 1, so we have-

T(n) = θ (nlogba.logp+1n)

T(n) = θ (nlog22.log1+1n)

 

Thus,

T(n) = θ (nlog2n)

 

Problem-03:

 

Solve the following recurrence relation using Master’s theorem-

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

 

Solution-

 

On comparing the given recurrence relation with-

T(n) = aT(n/b) + θ (nklogpn)

we have-

a = 2

b = 4

k = 0.51

p = 0

 

Now, a = 2 and bk = 40.51 = 2.0279.

Clearly, a < bk

So, we follow case-03.

Since p = 0, so we have-

T(n) = θ (nklogpn)

T(n) = θ (n0.51log0n)

 

Thus,

T(n) = θ (n0.51)

 

Problem-04:

 

Solve the following recurrence relation using Master’s theorem-

T(n) = √2T(n/2) + logn

 

Solution-

 

On comparing the given recurrence relation with-

T(n) = aT(n/b) + θ (nklogpn)

we have-

a = √2

b = 2

k = 0

p = 1

 

Now, a = √2 = 1.414 and bk = 20 = 1.

Clearly, a > bk

So, we follow case-01.

So, we have-

T(n) = θ (nlogba)

T(n) = θ (nlog2√2)

T(n) = θ (n1/2)

 

Thus,

T(n) = θ (√n)

 

Problem-05:

 

Solve the following recurrence relation using Master’s theorem-

T(n) = 8T(n/4) – n2logn

 

Solution-

 

  • The given recurrence relation does not correspond to the general form of Master’s theorem.
  • So, it can not be solved using Master’s theorem.

 

Problem-06:

 

Solve the following recurrence relation using Master’s theorem-

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

 

Solution-

 

  • We can write the given recurrence relation as T(n) = 3T(n/3) + n.
  • This is because in the general form, we have θ for function f(n) which hides constants in it.
  • Now, we can easily apply Master’s theorem.

 

On comparing the given recurrence relation with-

T(n) = aT(n/b) + θ (nklogpn)

we have-

a = 3

b = 3

k = 1

p = 0

 

Now, a = 3 and bk = 31 = 3.

Clearly, a = bk

So, we follow case-02.

Since p = 0, so we have-

T(n) = θ (nlogba.logp+1n)

T(n) = θ (nlog33.log0+1n)

T(n) = θ (n1.log1n)

 

Thus,

T(n) = θ (nlogn)

 

Problem-07:

 

Form a recurrence relation for the following code and solve it using Master’s theorem-

 

A(n)
{
   if(n<=1)
     return 1;
   else
     return A(√n);
}

 

Solution-

 

  • We can write a recurrence relation for the given code as T(n) = T(n) + 1.
  • Here 1 = Constant time taken for comparing and returning the value.
  • We can not directly apply Master’s Theorem on this recurrence relation.
  • This is because it does not correspond to the general form of Master’s theorem.
  • However, we can modify and bring it in the general form to apply Master’s theorem.

 

Let-

n = 2m    ……(1)

then-

T(2m) = T(2m/2) + 1

 

Now, let T(2m) = S(m), then T(2m/2) = S(m/2)

So, we have-

S(m) = S(m/2) +1

Now, we can easily apply Master’s Theorem.

 

On comparing the given recurrence relation with-

S(m) = aS(m/b) + θ (mklogpm)

we have-

a = 1

b = 2

k = 0

p = 0

 

Now, a = 1 and bk = 20 = 1.

Clearly, a = bk

So, we follow case-02.

Since p = 0, so we have-

S(m) = θ (mlogba.logp+1m)

S(m) = θ (mlog21.log0+1m)

S(m) = θ (m0.log1m)

 

Thus,

S(m) = θ(logm)    ……(2)

 

Now,

  • From (1), we have n = 2m.
  • So, logn = mlog2 which implies m = log2n.

 

Substituting in (2), we get-

S(m) = θ(loglog2n)

Or

T(n) = θ (loglog2n)

 

To gain better understanding about Master’s Theorem,

Watch this Video Lecture

 

Next Article- Recursion Tree

 

Get more notes and other study material of Design and Analysis of Algorithms.

Watch video lectures by visiting our YouTube channel LearnVidFun.

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Master's Theorem | Master's Theorem Examples
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Master's Theorem | Master's Theorem Examples
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Master's Theorem is a popular method for solving the recurrence relations. Master's Theorem Examples. Master's Theorem Cases are used to solve the recurrence relations of the type T(n) = aT(n/b) + θ (nklogpn).
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