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Transformations(Sections

COMP157Oct^ 24,

)) (^1) ( log( 2 (^2) )( 2

2 (^10)

−=

n n ih h i

i

Stage^ 1:

Build^ heap

for^ a^ given

list^ of^ n

keys

worst‐case:^ C (

n )^ =^

C(n)^ є^ O(n)

Stage^ 2:

Repeat

operation

of^ root

removal

n ‐^1 times

(fix^ heap)

worst‐case:^ C (

n )^ =^

C(n)^ є^ O(n

log^ n)

Both^ worst

‐case^ and

average

‐case^ efficiency:

Θ( n log

n )

In‐place:

yes Stability:

no^ (e.g.,

nodes at level 1 1)

i

Analysis

of^ Heapsort −^1 ∑=^1 log 22 n i

i

Polynomial

Evaluation

Given^ a

polynomial

of^ degree

n p ( x )^ =^ a

n^ x+^ a^ nn

n ‐^1 x +^ ‐ 1 … +^ ax^1

+^ a^0

and^ a^ specific

value^ of

x ,^ find^

the^ value

of^ p^ at

that^ point.

Brute‐force

algorithm

p^ ←^ a^0

;^ power

←^1

Multiplications:

for^ i^ ←

1 to^ n^ do power^ ←^ power

*^ x p^ ←^ p^ +

a^ *^ poweri^

Additions:

return^

p

O( n )

n

n^22 =∑ i^1 = n^ n^ =^1 ∑ i =^0

Horner’s

Rule

Example:

p (x)^ =

(^4 3 2) x ‐^ x 2 +^3 x +

x^ – 5 (^3) = x (2 x (^2) ‐ x +^3 x

+^ 1)^ – 5

=^ x ( x (2 x

2 ‐^ x^ +^ 3)

+^ 1)^ – 5

=^ x ( x ( x (

x^ ‐^ 1)^ +^

3)^ +^ 1)^

Computing

by^ last

formula

(from^ inside

out)

leads^ to

a^ faster

algorithm x ( x ( x (2 x

‐^ 1)^ +

3)^ +^ 1)

‐^5

n multiplies

n additions

O( n )

Horner’s

Rule

(Python)

#^ Poly

coefficients

are^ stored

in^ P

#^ Example:

P^ =^ [1,

2,^ 3]

‐>^ 3x^

+^ 2x^

+^1

def^ Horner(P,

x): n^ =^ len(P)

‐^1

p^ =^ P[n]i^ =^ n‐

while^

(i>=0):p^ =^ x*p

+^ P[i] i^ =^ i‐

return

p print^

"Evaluate

2x^^

‐^ x^+

3x^^

+x^ ‐^5

at^ x=3"

print^

Horner([

‐1,2],

Computing

n a

-^ Obvious

brute

‐force

n‐^1 multiplies

-^ Divide

and^ conquer:

n^ a=^ (a

n/2^2 )^

log^ (n)^2

multiplies

Left

‐Right

Exponentiation

#^ Compute

a^n, #^ where

b^ is^

the^ binary

representation

of^ n

#^ Example:

b^ =^ [0,0,1,1]

‐>^ n^

=^12

def^ LeftRightBinaryExponentiation(a,b):

product

=^ a i^ =^ len(b)

‐^2

while^

(i>=0):product

=^ product*product if^ (b[i]==1):

product

=^ product*a

i^ =^ i‐

return

product

Computing

n a

Right‐to

‐left^ binary

exponentiation Scan^ n ’s

binary^

expansion

from^ right

to^ left^

and^ compute

n^ a as^ the

product

of^ terms

i^ (^2) a corresponding

to^ 1’s^ in

this^ expansion.

Example Compute

(^13) a by^ the

right‐to

‐left^ binary

exponentiation.

Here,^ n

=^13 =^

1101.^21

(^8) a

(^4) a

(^2) a

a^ :^

i^ (^2) a terms

(^8) a*^

(^4) a*

a^ :^

product

Efficiency:

same^ as

that^ of

left‐to

‐right^ binary

exponentiation

Problem

Reduction

-^ Prof

X^ noticed

that^

when

his^ wife

wanted

to

boil^ water,

she^ took

the^ kettle

from

the

cabinet,

filled

it^ with

water

and^ placed

it^ on

the^ stove. • One^ day,

while

his^ wife

was^ gone,

Prof^

X

needed

to^ boil

some

water.

He^ noticed

the

kettle

in^ the

sink.

-^ Solution:

Prof^

X^ put

the^ kettle

in^ the

cabinet

and^ proceeded

to^ follow

his^ wife’s

routine.

Problem

Reduction

This^ variation

of^ transform

‐and‐

conquer

solves

a

problem

by^ a^

transforming

it^ into

different

problem

for^ which

an^ algorithm

is^ already

available.To^ be^ of

practical

value,

the^ combined

time

of

the^ transformation

and^ solving

the^ other

problem

should

be^ smaller

than

solving

the

problem

as^ given

by^ another

method.

Analytic

Geometry

-^ Given

three

points:

p1^ =^ (x1,y1),

p2^ =^

(x2,y2), p

=^ (x3,

y3)

-^ Problem:

Is^ p

to^ left

of^ p1p2?

-^ Solution:

Only

if

(^01) (^11112233) det^

yx yx yx

Least

Common

Multiple

-^ lcm(m,n)

is^ smallest

integer

divisible

by^ both

m^ and

n

-^ Straightforward

solution:

multiply

prime

factorizations • Transform

problem:

lcm(m,n)

=^ mn

/^ gcd(m,n)

Now^

solve

gcd^ problem

using

Euclid’s

algorithm

Optimization

Problem

Reduction

-^ Suppose

we^ find

a^ minimum

of^ f(x)

and^ we

have^

an^ algorithm

for^ finding

maximum?

-^ Solution:

min^ [

f(x)^ ]

=^ ‐max

[^ ‐f(x)

]

Reduction

to^ Graph

-^ See

lecture

on^ state

‐space

search