Technische Universitat MunchenFakultat fur InformatikLehrstuhl fur Effiziente AlgorithmenProf. Dr. Harald RackeChintan Shah

Wintersemester 2012Losungsblatt 3

16 November 2012

Effiziente Algorithmen und Datenstrukturen I

Aufgabe 1 (10 Punkte)

1. Solve the following recurrence relation without using generating functions:

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

2. Give tight asymptotic upper and lower bounds for T (n):

T (n) = T (n− 1) + log n.

Losungsvorschlag

1.

an = an−1 + 2n−1 and

an−1 = an−2 + 2n−2 for n ≥ 2 gives us

an = 3an−1 − 2an−2 for n ≥ 2

The characteristic polynomial of this recurrence is

λ2 − 3λ+ 2 = 0

The roots are 2 and 1. So, the solution is of the form

an = α(2)n + β(1)n

⇒ a0 = 2 = α + β

⇒ a1 = 3 = 2α + β

Solving for α and β gives us α = β = 1. Hence,

an = 2n + 1

2.

T (n) =n∑

i=1

log(i)

= log(n∏

i=1

i)

= log(n!)

= Θ(n log n)

Aufgabe 2 (10 Punkte)Give tight asymptotic upper and lower bounds for T (n):

T (n) = T(n

2

)+ T

(n4

)+ T

(n8

)+ n.

Losungsvorschlag

We guess that T (n) = Θ(n).We first prove that T (n) = O(n) by induction. The inductive hypothesis is that T (n) ≤ cnfor some constant c. Then,

T (n) = T(n

2

)+ T

(n4

)+ T

(n8

)+ n

≤ cn

2+cn

4+cn

8+ n

=7

8cn+ n

= (1 +7c

8)n

≤ cn, if c ≥ 8

This shows that T (n) = O(n).Further, T (n) = T

(n2

)+ T

(n4

)+ T

(n8

)+ n = Ω(n).

Hence, T (n) = Θ(n).

Aufgabe 3 (10 Punkte)Consider the following procedure:RECURSIVE-SORT(A, i, j)if (A[i] > A[j]) then swap A[i]↔ A[j]if i+ 1 ≥ j then returnk ← b(j − i+ 1)/3cRECURSIVE-SORT(A, i, j − k)RECURSIVE-SORT(A, i+ k, j)RECURSIVE-SORT(A, i, j − k)

1. Argue that RECURSIVE-SORT(A, 1, n) correctly sorts a given array A[1 . . . n].

2. Analyze the running time of RECURSIVE-SORT using a recurrence relation.

Losungsvorschlag

1. We prove by induction on l = j − i + 1, the length of the array, that RECURIVE-SORT sorts the input. The basis of induction is l = 2. The basis is true, since weswap the first and last elements of the array if A[1] > A[n] = A[2]. For l ≥ 3, wehave k ≥ 1 and hence each of the 3 recursive calls is to a problem of smaller size.Hence by induction, we can assume that these 3 calls sort the respective subarrays.Further, note that after the first recursive call, the largest k elements can not becontained in A[1 . . . k]. Hence, after the second recursive call, the largest k elementsare contained in A[j− k+ 1 . . . j] and in sorted order. After the third recursive call,the elements other than the k largest elements are also sorted.

2

2. The running time T (n) is given by the recurrence relation

T (n) = 3T

(2n

3

)+ c, where c is a constant

By Case 1 of the Master Theorem, the running time is Θ(nlog 3

23)

or ω(n2.25).

Aufgabe 4 (10 Punkte)Given two n×n matrices A and B where n is a power of 2, we know how to find C = A ·Bby performing n3 multiplications. Now let us consider the following approach. We partitionA, B and C into equally sized block matrices as follows:

A =

[A11 A12

A21 A22

]B =

[B11 B12

B21 B22

]C =

[C11 C12

C21 C22

]Consider the following matrices:

M1 = (A11 + A22)(B11 +B22)

M2 = (A21 + A22)B11

M3 = A11(B12 −B22)

M4 = A22(B21 −B11)

M5 = (A11 + A12)B22

M6 = (A21 − A11)(B11 +B12)

M7 = (A12 − A22)(B21 +B22)

Then,

C11 = M1 +M4 −M5 +M7

C12 = M3 +M5

C21 = M2 +M4

C22 = M1 −M2 +M3 +M6

1. Convince yourself that the matrices Cij evaluated as above are indeed correct. Don’twrite anything to prove this.

2. Design an efficient algorithm for multiplying two n×n matrices based on these facts.Analyze its running time.

Losungsvorschlag

1. Refer Strassen’s Algorithm

2. The running time T (n) is given by the recurrence relation

T (n) = 7T(n

2

)+ cn2, where c is a constant

By Case 1 of the Master Theorem, the running time is Θ(nlog2 7

)or Θ (nx) where

x ≈ 2.807.

3