Algorithms Math Intro

8/18/2019 Algorithms Math Intro

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Assignment 1

1. Suppose that an algorithm A runs in time fA(n) = 2n2 + 7n and that an algorithm B runs intime fB(n) = 45n + 4, for prolems si!es of n. "or #hat $alues of n is algorithm A faster than

B. %hat is, for #hat n is fA(n) & fB(n)'

Sol

Algorithm runs in time gi$en e*pression f A(n)=2n2+7n

Algorithm runs in time gi$en e*pression f B(n)=45n+4

"or #hat $alues of n f A(n) & f B(n)

2n2+7n & 45n+4

2n2+7n 45n+4 &

2n2 -n / 4 &

n2 10n / 2 & (2 taen ommon)

a*2 + * + &

3t loos lie a uadrati ineuation, and to find the for #hat $alues of n its true #e ha$e to findthe roots of ineuation.

a = 1, =10, =2

− ±√ 2

=5−4

−(−19)±√361+8

−19±19.20

−19−19.20 = 10.1, −19+19.20 = .12 2 2 2 4

So, (n10.1) 6 (n+.1) &

n lies in the range of n & 10.1, n .18

i.e. for all n = .1,.2,.-9999.10.18 algorithm A is faster than algorithm B


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-. >o# man different arrangements for a 52 ard de are there'

>o# does this ompare to the 11

seonds that ha$e passed sine the Big Bang'

Sol ifferent arrangements for a 52 ard de is 52 = 52651659961 = .5617

Age of uni$erse sine ig ang is 1-.2 6 10 ears

1 ear = -15- seonds

age of uni$erse in seonds = 4-527526 10

== 3n times of 11

.4-5 6 11

Comparing different arrangements for a 52 ard de #ith the 11

seonds that ha$e passed sine the Big Bang

8.0658∗1067

1.5 6 1049

1.5 6 1050

18

0.4358 ∗ 10

%hus the arrangement of 52 is 1.5 6 1050 times the 1018 times seonds sine ig ang.

4. Sho# from definitions that :(S) = : (S D %) :(%) + : (S D not %) : (not %).

Sol Ei$en t#o e$ents S and % #hih are stohastiall independent or simpl independent

%hen onditional proailit in entire population ased on $alue %. :(S) = : (S D %)

:(S) = ∑ : (S= )

%his denotes the onditional proailit on entire population

S

T

Not T

%his denotes the onditional proailit of stohasti $ariale ased on $ariale % and the population is di$ided ased on %and not %


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%hus :(S) = :(: (S D %)) = ∑ ( = Ti) ∗ P (STi)

%hus

%hus :(S) = :(: (S D %)) = ∑ ( = Ti) ∗ P (STi) = : (%i = %) 6 : (SF%) + : ( %i = not%) 6 :(S F not %)

5. :ro$e euation 1. of our te*too (page 21).

Sol ∑2 = 2

3+3 2+

6 =1

%aing sum of ues for instane∑ =1

3

Griting in the form of sum of integers from to (i+1)- and sutrating the last term

== ∑ =1 3 = ∑(( + 1)3 − ( + 1)3)

∑ =1 3 =∑(( 3 + 3 2 +

3 + 1) − ( + 1)3∑ 3 = (∑ 3 + 3∑ 2 + 3∑ + ∑1) − ( + 1)3== ( + 1)3= 3∑ 2 + 3∑ + ∑1

==3 + 3

2 + 3 + 1 = 3∑

2 + 3∑ + ∑1 Ge no# ∑ = ( +1)

, ∑ 1 = + 1 82

%here fore 3 + 3 2 + 3 + 1

( +1)

/ ( + 1) = 3∑ 2

2

2 3+3 2+ = 3∑ 2 2

T! ∑ 2 = 2 3+3 2+ 6


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Biggest2 = ;istIi+1J@

8

8:rint Biggest 1, Biggest 2@

%he numer of omparisons this algorithm does in #orst ase is H1 omparisons for findinglargest integer, H omparisons for eliminating iggest1 from list and H11 for finding seondlargest numer

So in #orst ase m algorithm does H1+H+H2 = - H - omparisons.

7. :ro$e ;emma 1.1 of our te*too (page 51).

Sol ;emma 1.11. f " L(g) if and onl if g " M(f).

Suppose f " L(g) %here are t#o $ariales 1 and n1 suh that 1 and n1 suh that for n 1&= n,

f (n) & = 6g(n). %hen for n 1&= n,g(n) & = (1F)6 f (n).%he other diretion is pro$ed similarl.2. if f " N(g) then g " N(f).

f " N(g) means f " L(g) # M (g) B 1, g " L(f) # M (f) %hus g " N(f).

-. N defines an eui$alene relation on the funtions. (Setion 1.-.1 e*plains #hat is

neessar to demonstrate.) Kah set N (") 3s an eui$alene lass, #e all itomple* lass.

3f f " L(E) and g" L(>) then " " L(>)@ ie, L is transitive. So the are M, N, o and #. "or all f, f " N(g) then g " N(f) gi$es symmetry."or an f, f " N(f) and for an g " N(g) gi$es refle*i$it.

4. L(f + g) = L(mO*(f, g)).Similar euations hold for M and N. (%he are usefulto anal!e

omple* algorithms #here f and g ould desrie the #or done different parts of the algorithm.)


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"or L (f+g) = L(ma*(f,g)), %here are 1 and n1 suh that n n1 , h(n) &= 6( f +g)(n)for and n n1

%hen h(n) & = 2ma*( f ,g)(n).

. :ro$e or dispro$e ∑ =1 2

" $(2)

Sol "rom 5 ∑2 =

23+3

2+

=

3

+

2

+6 3 2 6 =1

means running time of the algorithm as n (input si!e) gets larger is proportional to 2 ut the sum of suares is ha$ing a 3 term #hih is not proportional to 2 so the sum of suares o$er an inter$al 1 to n does not elong to$( 2)

>ene dispro$ed ∑ =1 2 " $( 2).

0. ra# a deision tree for the inar searh algorithm (Algorithm 1.4 on pages555) for n = 17.

Algorithm

int inarSearh(intIJ K, int first, int last, int P)

1. if (last & first)

2. inde* = 1@

3. else if (last == first)

4. if (P == KIfirstJ)

5. inde* = first@

6. else

7. inde* = 1@

8. else

9. int mid = (first + last) F 2@10. if (P & = KImidJ)11. inde* = inarSearh(K, first, mid, P)@

12. else13. inde* = inarSearh(K, mid+1, last, P)@

14. return inde*@

$( 2)


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eision %ree for n = 17


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1. Add a ro# to %ale 1.1 (page 40) sho#ing the ma*imum input si!e that an e sol$ed in onehour.

Sol

ma*imum input si!e that an e sol$ed in one hour

1 hour 1 415-4 27272 1- -2

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Algorithms Math Intro