lo02

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Technische Universitat MunchenFakultat f ur InformatikLehrstuhl f ur Effiziente AlgorithmenProf. Dr. Harald RackeChintan Shah

Wintersemester 2012Losungsblatt 2

16 November 2012

Effiziente Algorithmen und Datenstrukturen I

Aufgabe 1 (10 Punkte)Solve the following recurrence relations:

1. an = an−1 + 6n−1 with a0 = −15.

2. an = −an−1 + 9an−2 − 11an−3 + 4an−4 with a0 = −7, a1 = 4, a2 = 48 and a3 = 0.

Losungsvorschlag

1.

an = an−1 + 6n−1 (1)

an−1 = an−2 + 6n−2 (2)

Multiplying equation 2 by 6 and subtracting from equation 1 gives us

an = 7an−1 − 6an−2

The characteristic polynomial of this recurrence is λ2

−7λ + 6 = 0. Solving for λ

gives us λ = 6 or λ = 1. So the solution is of the form an = α1n + β 6n. Also,a0 = −15 gives us a1 = −14. Hence,

a0 = −15 = α + β

a1 = −14 = α + 6β

Solving for α and β gives us α = −151

5 and β = 1

5. So,

an = −151

5 +

6n

5

2. The characteristic polynomial of this recurrence relation is λ4

+λ3

−9λ2

+11λ−4 = 0or (λ− 1)3 · (λ + 4) = 0. So the solution is of the form an = α(−4)n + βn2 + γn + δ .Hence,

a0 = −7 = α + δ

a1 = 4 = −4α + β + γ + δ

a2 = 48 = 16α + 4β + 2γ + δ

a3 = 0 = −64α + 9β + 3γ + δ

Solving this system of linear equations gives us α = 1, β = 4, γ = 12 and δ = −8.So,

an = (−4)

n

+ 4n2

+ 12n− 8

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Aufgabe 2 (10 Punkte)Solve the following recurrence relations using generating functions:

1. an = an−1 + 2n−1 for n ≥ 1 with a0 = 2.

2. an = 3an−1 − 3an−2 + an−3 for n ≥ 3 with a0 = a1 = a2 = 1.

Losungsvorschlag

1. an = an−1 + 2n−1 for n ≥ 1 with a0 = 2.

A(z ) =n≥0

anz n

= a0 +

n≥1anz

n

= 2 +n≥1

(an−1 + 2n−1)z n

= 2 + z n≥1

an−1z n−1 + z

n≥1

2n−1z n−1

= 2 + z n≥0

anz n +

n≥0

2nz n

⇒ (1 − z )A(z ) = 2 + z n≥0

2nz n

= 2 + z

1 − 2z

= 2 − 3z

1 − 2z

⇒ A(z ) = 2 − 3z

(1 − 2z )(1 − z )

= 1

1 − 2z +

1

1 − z

=n≥0

2nz n +n≥0

z n

= n≥0

(2n + 1)z n

⇒ an = 2n + 1

2

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2. an = 3an−1 − 3an−2 + an−3 for n ≥ 3 with a0 = a1 = a2 = 1.

A(z ) =n≥0

anz n

= a0 + a1z + a2z 2

+n≥3

(3an−1 − 3an−2 + an−3)z n

= 1 + z + z 2 + 3z n≥3

an−1z n−1 − 3z 2

n≥3

an−2z n−2

+z 3n≥3

an−3z n−3

= 1 + z + z 2 + 3z (A(z ) − 1 − z ) − 3z 2(A(z ) − 1)

+z 3A(z )

⇒ (1 − 3z + 3z 2 − z 3)A(z ) = 1 + z + z 2 − 3z − 3z 2 + 3z 2

⇒ A(z ) = (1 − z )2

(1 − z )3

= 1

1 − z

=n≥0

z n

⇒ an = 1

Aufgabe 3 (10 Punkte)

Solve the following recurrence using a generating function:

an = an−1 + an−2 for n ≥ 2 with a0 = 0 and a1 = 1.

Losungsvorschlag

A(z ) =n≥0

anz n

= a0 + a1z + n≥2

anz n

= z + z n≥2

an−1z n−1 +

n≥2

an−2z n

= z + z (A(z ) − a0) + z 2A(z )

= z + zA(z ) + z 2A(z )

⇒ A(z ) = z

1 − z − z 2

Note that (1 − z − z 2) = (1 − αz )(1 − βz ) for

α = 1 + √ 52

and β = 1 −√ 52

3

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We want to find A, B satisfying

z

1 − z − z 2 =

A

1 − αz +

B

1 − βz

Solving, for A, B gives

A =1

α

1 − β

α

= 1

α− β =

1√ 5

B =1

β

1 − αβ

= 1

β − α = −1√

5

⇒ A(z ) = 1√

5

1

1 − αz − 1

1 − βz

=

1

√ 5 n≥0 αn

z

n

−n≥0 β

n

z

n=n≥0

(αn − β n)√ 5

z n

⇒ an = αn − β n√

5

= 1√

5

1 +

√ 5

2

n

−

1 −√

5

2

n

Aufgabe 4 (10 Punkte)Solve the following recurrence using a generating function:

an = 5an−1 − 8an−2 + 4an−3 for n ≥ 3 with a0 = 1, a1 = 3 and a2 = 11.

Losungsvorschlag

A(z ) =

n≥0anz

n

= a0 + a1z + a2z 2 +n≥3

(5an−1 − 8an−2 + 4an−3)z n

= 1 + 3z + 11z 2 + 5z (A(z ) − 3z − 1) − 8z 2(A(z ) − 1)

+4z 3A(z )

⇒ (1 − 5z + 8z 2 − 4z 3)A(z ) = 1 − 2z + 4z 2

⇒ A(z ) = 1 − 2z + 4z 2

(1 − z )(1 − 2z )2

= A

1−z

+ B

1−

2z +

C

(1−

2z )2 , (say)

4

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Solving for A, B,C gives

4 = 4A + 2B

−2 = −4A− 3B − C

1 = A + B + C

or A = 3, B = −4, C = 2.Therefore,

A(z ) = 3

1 − z − 4

1 − 2z +

2

(1 − 2z )2

= 3n≥0

z n − 4n≥0

2nz n + 2n≥0

(n + 1)2nz n

= n≥0

(3 + (n− 1)2n+1)z n

⇒ an = 3 + (n− 1)2n+1

5

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