Discrete Mathematics 2. Exercise Sheet

Algorithmic

Discrete Mathematics 2. Exercise Sheet

Department of Mathematics SS 2012

PD Dr. Ulf Lorenz 2. and 3. May 2012

Dipl.-Math. David Meffert Version of June 21, 2012

Groupwork

Exercise G1 (Master-Theorem)

Determine, if possible, fixed bounds for the complexities of the recurrences (a) T(n) =4T(ⁿ₂) +n³,

(b) T(n) =4T(ⁿ₂) +n, (c) T(n) =4T(ⁿ₂) +n²logn, (d) T(n) =4T(ⁿ₂) +n². Hint:

Solution: Throughout the whole exercise we havelog_ba=log₂4=2.

(a) We have f(n) =n³. So f(n)∈Ω(n^2+"), because of0≤1·n³≤f(n). Furthermore4·f(₂ⁿ) =4ⁿ³

8 = ¹₂n³≤c·f(n), holds forc=¹₂. So by the third case of the Master-Theorem we concludeT(n)∈Θ(n³).

(b) We have f(n) =n. So f(n)∈O(n^2−")for"=1because of f(n)≤1·n. By the first case of the Master-Theorem we concludeT(n)∈Θ(n²).

(c) We havef(n) =n²logn. We immediately see f(n)6∈O(n^2−")and f(n)6∈Θ(n²). We want to show that third case of the Master-Theorem can’t be used either, because the second condition can’t be fulfilled. Lec∈(0, 1). We get

4·f(n

2)≤c·f(n)

⇔ 4·n²

4 log(n

2)≤c·n²log(n)

⇔ n²log(n

2)≤c·n²log(n)

⇔ log(n)−log(2)≤c·log(n)

⇔ log(n)−1≤c·log(n)

⇔ −1≤log(n)(c−1)

⇔ −1

≥log(n)

This can’t hold for all n ∈N because{log(n)| n∈N} is not bounded. So c∈ (0, 1)ist not a possible choice.

By this example we can see that although the Master-Theorem is quite powerful it can’t be used for alle types of recurrences.

(d) We havef(n) =n². Sof(n)∈Θ(n²)because of0≤1·n²≤f(n)≤1·n². By the second case of the Master-Theorem we concludeT(n)∈Θ(n²logn).

Exercise G2 (Complexity)

(a) Let f,t:N→Rbe functions with f ∈O(t). ProveO(f) +O(t)⊆O(t)andO(f) +O(f)⊆O(t).

(b) Does3³⁺ⁿ∈O(3ⁿ)hold?

(c) Does3³ⁿ∈O(3ⁿ)hold?

(d) Show thatO(f)·O(g) =O(f ·g)holds for f,g:N→R+.

Remark: For real valued functions f,g:N →R one just substitutes f(n), g(n)with|f(n)|,|g(n)|in the definition of O(g).

Solution:

(a) Letg,h:N→Rwithg∈O(f)andh∈O(t). By definition we getn_g,n_h∈N,c_g,c_h∈Rwith

|g(n)| ≤c_g|f(n)| and |h(n)| ≤c_h|t(n)|

for alln≥ n_g,n_h. Furthermore by assumption we getn_f ∈N and c_f ∈Rwith|f(n)| ≤c_f|t(n)|for all n≥n_f. Putting things together we conclude

|g(n) +h(n)| ≤c_g|f(n)|+c_h· |t(n)| ≤c_gc_f|t(n)|+c_h· |t(n)|

= (cgc_f +c_h)|t(n)|

for alln≥max{n_g,n_h,n_f}. So we getg+h∈O(t).

The second inclusion can be proved the same way or alternatively by showingO(f)⊆O(t).

(b) We have3³⁺ⁿ∈O(3ⁿ)because of3³⁺ⁿ=27·3ⁿ≤27·3ⁿfor alln∈N. (c) The term3³ⁿ=27ⁿis obviously not inO(3ⁿ).

(d) By definition we have

h∈O(f)⇔ ∃c_h,n_h h(n)≤c_hf(n) ∀n≥n_h and

k∈O(g)⇔ ∃c_k,n_k k(n)≤c_kg(n) ∀n≥n_k. So forh∈O(f)andk∈O(g)we have

(h·k)(n)≤c_kc_h(f ·g)(n) ∀n≥max{n_h,n_k}. and thereforeh·k∈O(f ·g). This proves the first inclusion.

For the second one letl∈O(f ·g). This means there existn_l∈Nandc_l∈Rwithl(n)≤c_l(f ·g)(n)for alln≥n_l. Now setl=f ·_f^l. Obviously f ∈O(f)and by dividing the last inequality by f(n)we get

l f

(n)≤c_lg(n)for all n≥n_l. So we have _f^l ∈O(g).

Exercise G3 (Algorithms)

(a) Given two algorithmsAandB:

• AlgorithmAhas complexityO(f).

• AlgorithmBhas complexityO(g).

We want to look at two new algorithms usingAandB.

Algorithm 1 INPUT :n∈N fori=1, ..., 100do

run algorithm A end for

fori=1, ...,ⁿ₂ do run algorithm B end for

Algorithm 2 if n≥30then

run algorithm A else

run algorithmus B end if

We already knowf ∈Ω(g). Determine the best possible estimates for the runtime of both algorithms.

(b) Take a look at algorithm3and determine the best possible estimate for its runtime. Justify you answer.

Algorithm 3 INPUT : n∈N m = n

whilem > 1do forj = 1,...,ⁿ

2do a=3·b c = a +b end for m =¹

2·m end while

Solution:

(a) For the runtime of algorithm1we get100·O(f) +ⁿ₂·O(g). Because we already knowf ∈Ω(g)we can summarize that toO(f ·h)withh(n) =n.

For algorithm2we notice that only the runtime forn→ ∞is important. So only the second case of theif-part is relevant. Hence algorithm2has runtimeO(f).

(b) We go through the outer looplogntimes. The inner loop we go throughⁿ

2 times. Ignoring any constant factors we getO(nlogn)for the runtime of the algorithm.

Exercise G4 (Sets) Order the functions

n²,p

n,n!,nⁿ,n

by their complexity. Start with lowest complexity and use theo-notation. Determinen₀dependend onc>0in every of those cases, too.

Remark:

f ∈o(g):⇐⇒ ∀c>0∃n₀∈N∀n≥n₀: 0≤f(n)<c g(n)

Solution:

pn∈o(n) n₀= ₁

c²

£

n∈o(n²) n₀= ₁

c

£

n²∈o(n!) n₀=max¦

6, ₁

c

£©

n!∈o(nⁿ) n₀=max¦

3, ₁

c

£©

For explanation: In last two cases we have chosenn₀≥6,n₀≥3becausen!>n³holds forn≥6andnⁿ>n·n!holds

Exercise H4 (Asymptotics) (14 points) (a) Prove that forr₁,r₂∈R₊we haven^r1∈O(n^r2)andr₁ⁿ∈O(r₂ⁿ)iffr₁≤r₂.

(b) Prove the following statements for functions f,t:N→R: i. O(f) +O(f)⊆O(f).

ii. O(f)·O(t)⊆O(f ·t).

iii. max{f,t} ∈Θ(f +t)for f,t≥0.

Solution:

(a) For alln∈N the statementn^r1 ≤ cn^r2 is equivalent ton^r1−r2 ≤c. The function n^x is bounded iff x ≤0, which meansr₁≤r₂.

The second statement can be proved the same way. For alln∈Nthe statementr₁ⁿ≤c r₂ⁿis equivalent to_r

1 r₂

n

≤c.

The functionxⁿis bounded iff x≤1, which meansr₁≤r₂.

(b) The proofs all work the same way in general by playing around with the definitions.

i. Letg,h:N→Rwithg,h∈O(f). By definition we havec_g,c_h∈Randn_g,n_h∈Nwith

|g(n)| ≤c_g|f(n)| and |h(n)| ≤c_h|f(n)|

for alln≥n_g,n_h. We conclude

|g(n) +h(n)| ≤ |g(n)|+|h(n)| ≤(c_g+c_h)|f(n) for alln≥max{n_g,n_h}, which meansg+h∈O(f).

ii. Letg,h:N→Rwithg∈O(f)andh∈O(t). By definition we havec_g,c_h∈Randn_g,n_h∈Nwith

|g(n)| ≤c_g|f(n)| and |h(n)| ≤c_h|t(n)|

for alln≥n_g,n_h. We conclude

|g(n)·h(n)|=|g(n)| · |h(n)| ≤c_g|f(n)| ·c_h|t(n)|= (cgc_h)|(f ·t)(n)|. for alln≥max{n_g,n_h}, which meansg·h∈O(f ·t).

iii. We want to proof the inequality

max{f,t}(n)≥ 1

2(f(n) +t(n)) (1)

pointwise for alln∈Nand therefore distinguish two cases. For everyn∈Nwith f(n)≥t(n)we get max{f,t}(n) =f(n) = 1

2(f(n) +f(n))≥1

2(f(n) +t(n)).

For all othern∈Nwitht(n)≥f(n)we get max{f,t}(n) =t(n) =1

2(t(n) +t(n))≥1

2(f(n) +t(n))

the same way. So by equation (1) we concludemax{f,t} ∈ Ω(f +t). Because of the obvious inequality max{f,g}(n)≤(f +g)(n)we getmax{f,g} ∈O(f +g). Hence we havemax{f,g} ∈Θ(f +g).

Exercise H5 (A sorting algorithm) (10 points)

The algorithmSortListsorts a sequence of numbers in ascending order.

Algorithm 4SortList(list)

INPUT: sequence of numbers,list=a₁, ...,a_n,a_i∈N ifn <=1then

returnlist else

leftlist=a₁, ...,a_dⁿ

2e

rightlist=a_dⁿ

2e+1, ...,a_n

return Sort(SortList(lelftlist),SortList(rightlist)) end if

Algorithm 5Sort(rightlist,leftlist) INPUT: two sequences of numbers:

rightlist=a₁, ...,a_l,leftlist=b₁, ...,b_k,a_i,b_i∈N newlist

whilerightlistandleftlistnot emptydo

iffirst element ofleftlist<= first element ofrightlistthen

append first element ofleftlisttonewlistand delete it fromleftlist else

append first element ofrightlisttonewlistand delete it fromrightlist end if

end while

whileleftlistnot emptydo

append first element ofleftlisttonewlistand delete it fromleftlist end while

whilerightlistnot emptydo

append first element ofrightlisttonewlistand delete it fromleftlist end while

returnnewlist

(a) Sort the sequence9, 10, 7, 3, 1, 2, 12, 9, 23in ascending order by using the algorithmSortList. Make sure to include detailed steps for the algorithm in your solution to indicate that you understand how it works.

(b) What is the runtime of the algorithmSortList?

Solution:

(a) We use the short termSfor the algorithmSortand writeS(rightlist;leftlist). For the algorithmSortListwe use the short termS L. So we have

S L(9, 10, 7, 3, 1, 2, 12, 9, 23) S(S L(9, 10, 7, 3, 1);S L(2, 12, 9, 23))

 S(S(S L(9, 10, 7);S L(3, 1));S(S L(2, 12);S L(9, 23)))

 S(S(S(S L(9, 10);S L(7));S(S L(3);S L(1)));S(S(S L(2);S L(12));S(S L(9),S L(23))))  S(S(S(S(S L(9),S L(10)); 7);S(3; 1));S(S(2; 12);S(9, 23)))

 S(S(S(S(9, 10); 7);S(3; 1));S(S(2; 12);S(9, 23)))  S(S(S(9, 10; 7);S(3; 1));S(S(2; 12);S(9, 23)))  S(S(7, 9, 10; 1, 3);S(2, 12; 9, 23))

 S(1, 3, 7, 9, 10; 2, 9, 12, 23)  1, 2, 3, 7, 9, 9, 10, 12, 23.

Now we want to show the the last step in detail and thereby demonstrate how theSortalgorithm works. We use S(rightlist;leftlist;newlist)to indicate all the steps and get

S(1, 3, 7, 9, 10; 2, 9, 12, 23;;) S(3, 7, 9, 10; 2, 9, 12, 23; 1) S(3, 7, 9, 10; 9, 12, 23; 1, 2)

 S(7, 9, 10; 9, 12, 23; 1, 2, 3) S(9, 10; 9, 12, 23; 1, 2, 3, 7)  S(10; 9, 12, 23; 1, 2, 3, 7, 9) S(10; 12, 23; 1, 2, 3, 7, 9, 9)  S(;; 12, 23; 1, 2, 3, 7, 9, 9, 10) S(;; 23; 1, 2, 3, 7, 9, 9, 10, 12)  S(;;;; 1, 2, 3, 7, 9, 9, 10, 12, 23) 1, 2, 3, 7, 9, 9, 10, 12, 23.

(b) The given algorithm is calledMergeSortand is a recursive algorithm. We haveT(1) =1and get the recurrence

T(n) = T(n 2)

| {z }

Sor t List(l e f t l ist)

+ T(n 2)

| {z }

Sor t List(r i ght l ist)

+ n

|{z}

Sor t

=2T(n 2) +n.

By the Master-Theorem(second case) we concludeT(n)∈Θ(nlogn).

Given algorithm 6. What does the algorithm? Determine its runtime.

Algorithm 6 INPUT :n∈N K1 = 2;

K2 = n;

whileK2 > K1do K2 = n/K1

ifdK2e==K2then return K1 else

K1=K1+1 end if end while return 0

Solution: The algorithm tests if a given numbern∈N is prime. This is done by checking all possible divisors from 2, . . . ,bp

nc. The checking part works by dividingnbyi∈ {2, . . . ,bp

nc}and looking if this fraction is a natural number.

If a divisor is found the algorithm returns this divisor and otherwise it returns0. In the case of output0the numbernis prime. The relevant part for the runtime (while-condition) is usedp

ntimes, so the runtime isΘ(p n)).