Example Problems - Master Method

 

Example 1

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

Solution

a=1,b=2a=1,\qquad b=2

Critical function

nlog⁡21=1n^{\log_21}=1

Since

f(n)=1=Θ(1)f(n)=1=\Theta(1)

Case 2

Answer

Θ(log⁡n)\boxed{\Theta(\log n)}

(Binary Search)

Example 2

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

Critical function

nlog⁡23=n1.585n^{\log_23} =n^{1.585}

Since

n=O(n1.585−ε)n = O(n^{1.585-\varepsilon})

Case 1

Answer

Θ(n1.585)\boxed{\Theta(n^{1.585})}

Example 3

T(n)=2T(n/2)+nlog⁡nT(n)=2T(n/2)+n\log n

Critical function

nn

Here,

f(n)=nlog⁡n=Θ(nlog⁡1n)f(n)=n\log n=\Theta(n\log^1n)

This matches Case 2 with k=1k=1.

Answer

Example 4 (Strassen's Matrix Multiplication)

T(n)=7T(n/2)+n2T(n)=7T(n/2)+n^2

Here,

a=7,b=2a=7,\qquad b=2

Critical function

nlog⁡27≈n2.807n^{\log_27} \approx n^{2.807}

Since

n2=O(n2.807−ε)n^2 = O(n^{2.807-\varepsilon})

Case 1

Answer

Θ(nlog⁡27)≈Θ(n2.81)\boxed{\Theta(n^{\log_27})\approx\Theta(n^{2.81})}

This is exactly why Strassen's algorithm is asymptotically faster than the classical O(n3)O(n^3)

Example 5

T(n)=16T(n/4)+nT(n)=16T(n/4)+n

Solution

a=16,b=4a=16,\quad b=4

Watershed

nlog⁡416=n2n^{\log_416} = n^2

Driving function

nn

Case 1

Answer

Example 6

T(n)=25T(n/5)+n2T(n)=25T(n/5)+n^2

Solution

a=25a=25
b=5b=5

Watershed

nlog⁡525=n2n^{\log_525} = n^2

Driving function

n2n^2

Exactly equal

Case 2

Answer

Example 7

T(n)=27T(n/3)+n2T(n)=27T(n/3)+n^2

Solution

Watershed

nlog⁡327=n3n^{\log_327} = n^3

Driving function

n2n^2

Case 1

Answer

Example 8

T(n)=8T(n/2)+n3T(n)=8T(n/2)+n^3

Watershed

nlog⁡28=n3n^{\log_28} = n^3

Equal

Case 2

Answer

Example 9

T(n)=4T(n/2)+n2log⁡nT(n)=4T(n/2)+n^2\log n

Watershed

n2n^2

Driving

n2log⁡nn^2\log n

Case 2

Answer

Example 10

T(n)=9T(n/3)+n2log⁡3nT(n)=9T(n/3)+n^2\log^3n

Watershed

n2n^2

Driving

n2log⁡3nn^2\log^3n

Case 2

Answer

Example 11

T(n)=3T(n/2)+n2T(n)=3T(n/2)+n^2

Watershed

nlog⁡23=n1.585n^{\log_23} = n^{1.585}

Driving

n2n^2

Case 3

Answer

Θ(n2)\boxed{\Theta(n^2)}

Example 12

T(n)=5T(n/2)+n4T(n)=5T(n/2)+n^4

Watershed

nlog⁡25=n2.322n^{\log_25} = n^{2.322}

Driving

n4n^4

Case 3

Answer

Θ(n4)\boxed{\Theta(n^4)}

Step 1: Identify
a,
b, and
f(n)

a=2,b=4,f(n)=1a=2,\qquad b=4,\qquad f(n)=1

Step 2: Find the watershed function

nlog⁡42n^{\log_4 2}

Since

log⁡42=12\log_4 2=\frac12

Therefore,

nlog⁡42=n1/2n^{\log_42}=n^{1/2}

Step 3: Compare f(n)f(n) with the watershed function

Driving function

f(n)=1f(n)=1

Watershed

n1/2n^{1/2}

Clearly,

1=O(n1/2−ε)1=O(n^{1/2-\varepsilon})

Choose

ε=14\varepsilon=\frac14

Step 4: Determine the case

Case 1


Final Answer

T(n)=Θ(n)\boxed{T(n)=\Theta(\sqrt n)}

Example:14

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

Step 1

a=2,b=4,f(n)=na=2,\qquad b=4,\qquad f(n)=\sqrt n

Step 2

Watershed

nlog⁡42=n1/2=nn^{\log_42} =n^{1/2} =\sqrt n

Step 3

Driving function

n\sqrt n

Exactly equals the watershed function.


Step 4

Case 2

Here

k=0k=0

Final Answer

T(n)=Θ(nlog⁡n)\boxed{T(n)=\Theta(\sqrt n\log n)}



Example - 15

T(n)=2T(n/4)+n log⁡2nT(n)=2T(n/4)+\sqrt n\,\log^2 n

Step 1

a=2,b=4a=2,\qquad b=4

Step 2

Watershed

nlog⁡42=nn^{\log_42} =\sqrt n

Step 3

Driving function

nlog⁡2n\sqrt n\log^2 n

This is

Θ(n1/2log⁡2n)\Theta \left( n^{1/2}\log^2 n \right)

Step 4

Case 2

because

k=2k=2

Final Answer

T(n)=Θ(nlog⁡3n)\boxed{T(n)=\Theta(\sqrt n\log^3 n)}

Example-16

T(n)=2T(n/4)+nT(n)=2T(n/4)+n

Step 1

a=2,b=4a=2,\qquad b=4

Step 2

Watershed

n1/2n^{1/2}

Step 3

Driving function

nn

Clearly,

n=Ω(n1/2+ε)n=\Omega(n^{1/2+\varepsilon})

Choose

ε=12\varepsilon=\frac12

Step 4

Check the regularity condition

We need

af(n/b)≤cf(n)af(n/b)\le cf(n)

Compute

2f(n/4)=2(n4)=n22f(n/4) = 2\left(\frac n4\right) =\frac n2

Since

n2≤12n\frac n2 \le \frac12n

Choose

c=12<1c=\frac12<1

Regularity condition holds.


Step 5

Case 3


Final Answer

T(n)=Θ(n)\boxed{T(n)=\Theta(n)}

Example - 17

(e) T(n)=2T(n/4)+n2T(n)=2T(n/4)+n^2

Step 1

a=2,b=4a=2,\qquad b=4

Step 2

Watershed

n1/2n^{1/2}

Step 3

Driving function

n2n^2

Clearly,

n2=Ω(n1/2+ε)n^2 = \Omega (n^{1/2+\varepsilon})

Choose

ε=1\varepsilon=1

Step 4

Check regularity

2f(n/4)=2(n4)2=n282f(n/4) = 2\left(\frac n4\right)^2 = \frac{n^2}{8}

Since

n28≤18n2\frac{n^2}{8} \le \frac18n^2

Choose

c=18<1c=\frac18<1

Regularity condition holds.


Step 5

Case 3


Final Answer

T(n)=Θ(n2)\boxed{T(n)=\Theta(n^2)}

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