Analysis of Heapsort

8/19/2019 Analysis of Heapsort

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Analysis of Algorithms

CMSC 142

Analysis of Heapsort


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3FEW UNIQUE RANDOM REVERSED ALMOST SORTED

MERGE SORT

INSERTION SORT

HEAP SORT


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4

Heap Types

• Max-heaps (largest element at root), have themax-heap property:

for all no!es i, e"#l$!ing the root%

A[PARENT(i)] ≥ A[i]

• Min-heaps (smallest element at root), have the

min-heap property: for all no!es i, e"#l$!ing the root%

A[PARENT(i)] ≤ A[i]


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&

A!!ing'eleting o!es

• e* no!es are al*ays inserte! at the +ottomlevel (left to right)

• o!es are remove! from the +ottom level (right

to left)


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-perations on Heaps

• Maintain'.estore the ma"/heap property MA0/HA5

• Create a ma"/heap from an $nor!ere! array

678/MA0/HA

• Sort an array in pla#e

HAS-.T

• riority 9$e$es


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:

Maintaining the Heap roperty

• S$ppose a no!e is smaller than a#hil!

8eft an! .ight s$+trees of i are ma"/heaps

• To eliminate the violation%

"#hange *ith larger #hil! Move !o*n the tree

Contin$e $ntil no!e is not smaller than

#hil!ren


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;

"ample

MA0/HA5(A, 2, 1 violates the heap property

A=2> ↔ A=4>

A=4> violates the heap property

A=4> ↔ A=?>

Heap property restore!


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?

Maintaining the Heap roperty

• Ass$mptions% 8eft an! .ight

s$+trees of i arema"/heaps

– A[i] may +esmaller than its

#hil!ren

Alg: MA0/HA5(A, i, n)1. l @ 8T(i)

2. r @ .HT(i)

3. if l ≤ n an! A[l] > A[i]

4B then largest l←

&B else largest i←

6. if r ≤ n an! A[r] > A[largest]

:B then largest r

←

8. if largest ≠ i

?B then e"#hange A[i] A[largest]↔

1


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1<

MA0/HA5 .$nning Time

• nt$itively%

• .$nning time of MA0/HA5 is O(lgn)

• Can +e *ritten in terms of the height of the heap,

as +eing O(h)

Sin#e the height of the heap is lgn

h

2h-(h)

////


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6$il!ing a Heap

Alg: 678/MA0/HA(A)

1. n length=A>

2B for i ← n/2 downto 1

3B do MA0/HA5(A, i, n)

•Convert an array A[1 n] into a ma"/heap (n ! length[A])

• The elements in the s$+array A[(n/2"1) .. n] are leaves

• Apply MA0/HA5 on elements +et*een 1 an! n/2

2

14 ;

1

1

:

4

3

? 1<

1

2 3

4 & :

; ? 1<

4 1 3 2 1 ? 1< 14 ; : A%


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.$nning Time of 678 MA0 HA

⇒ .$nning time% O(nlgn)

• This is not an asymptoti#ally tight $pper +o$n!

Alg: 678/MA0/HA(A)1. n length=A>

2B for i ← n/2 downto 1

3B do MA0/HA5(A, i, n) O(lgn)

O(n)


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• HA5 taDes O(h) ⇒ the #ost of HA5 on a no!e i isproportional to the height of the no!e i in the tree

Height 8evel

h


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1&


i

h

i

ihnnT

∑==

0

)( Cost of HA5 at level i ∗ n$m+er of no!es at that level

( )ihh

i

i−=∑

=0

2 .epla#e the val$es of ni an! hi #omp$te! +efore

h

h

i

ih

ih

220∑= −−

=

M$ltiply +y 2h +oth at the nominator an! !enominator an!

*rite 2i asi−2

1

∑=

=h

k

k

h k

0 22 Change varia+les% D h / i

∑∞

=

≤0 2k

k

k n The s$m a+ove is smaller than the s$m of all elements to ∞

an! h lgn

)(nO= The s$m a+ove is smaller than 2

.$nning time of 678/MA0/HA%T(n) ! O(n)


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1

Heapsort

• oal%

Sort an array $sing heap representations

• !ea%

6$il! a max-heap from the array

S*ap the root (the ma"im$m element) *ith the last

element in the array

Eis#ar!F this last no!e +y !e#reasing the heap siGe

Call MA0/HA5 on the ne* root

.epeat this pro#ess $ntil only one no!e remains


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1:

"ample% A=:, 4, 3, 1, 2>

MA0/HA5(A, 1, 4) MA0/HA5(A, 1, 3) MA0/HA5(A, 1, 2)

MA0/HA5(A, 1, 1)


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1;

Alg: HAS-.T(A)

1B 678/MA0/HA(A)

2B for i length[A]← downto 2

3B do e"#hange A[1] A[i]↔

4B MA0/HA5(A, 1, i # 1)

• .$nning time% O(nlgn) ### $an %e

sh&'n t& %e (nlgn)

O(n)

O(lgn)

n#1 times

- i


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1?

-perations

on riority $e$es

• Ma"/priority 9$e$es s$pport the follo*ingoperations%

S.T(, *)% inserts element * into set

0T.ACT/MA0()% removes an! ret$rns element of

*ith largest Dey

MA0M7M()% ret$rns element of *ith largest Dey

C.AS/I5(, *, +)% in#reases val$e of element

*Js Dey to + (Ass$me + ≥ *Js #$rrent Dey val$e)


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HA/MA0M7M

oal% .et$rn the largest element of the heap

Alg: HA/MA0M7M(A)1B return A[1]

.$nning time% O(1)

Heap A%

Heap/Ma"im$m(A) ret$rns :


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HA/0T.ACT/MA0oal%

"tra#t the largest element of the heap (iBeB, ret$rn the ma"

val$e an! also remove that element from the heap

!ea%

"#hange the root element *ith the last

e#rease the siGe of the heap +y 1 element

Call MA0/HA5 on the ne* root, on a heap of siGe n/1

Heap A% .oot is the largest element


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"ample% HA/0T.ACT/MA0

;

2 4

14

:

1

1

1<

? 3

a* ! 1-;

2 4

14

:

1

1<

? 3

Heap siGe !e#rease! *ith 1

4

2 1

;

:

14

1<

? 3

Call MA0/HA5(A, 1, n#1)


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HA/0T.ACT/MA0

Alg: HA/0T.ACT/MA0(A, n)1B if n 1

2B then error Eheap $n!erflo*F

3B a* A[1]←

4B A[1] A[n]←

&B MA0/HA5(A, 1, n#1) remaDes heapB return a*

.$nning time% O(lgn)


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HA/C.AS/I5

• oal%

n#reases the Dey of an element i in the heap

• !ea%

n#rement the Dey of A[i] to its ne* val$e

f the ma"/heap property !oes not hol! anymore%traverse a path to*ar! the root to fin! the proper

pla#e for the ne*ly in#rease! Dey

;

2 4

14

:

1

1

1<

? 3i

e0 [i] @ 1&


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2&

"ample% HA/C.AS/I5

14

2 ;

1&

:

1

1

1<

? 3

i

;

2 4

14

:

1

1

1<

? 3i

Key =i > @ 1&

;

2 1&

14

:

1

1

1<

? 3i

1&

2 ;

14

:

1

1

1<

? 3i


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2

HA/C.AS/I5

Alg: HA/C.AS/I5(A, i, +e0)

1B if +e0 A[i]

2B then error Ene* Dey is smaller than #$rrent DeyF

3B A[i] +e0←

4B while i > 1 an! A[PARENT(i)] A[i]

&B do e"#hange A[i] A[PARENT(i)]↔

B i PARENT(i)←

• .$nning time% O(lgn)

;

2 4

14

:

1

1

1<

? 3i

e0 [i] @ 1&


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2:

/∞

MA0/HA/S.T

• oal% nserts a ne* element into a ma"/

heap

• !ea%

"pan! the ma"/heap *ith a ne*

element *hose Dey is /∞

Calls HA/C.AS/I5 to

set the Dey of the ne* no!e to its

#orre#t val$e an! maintain the

ma"/heap property

;

2 4

14

:

1

1

1<

? 3

1&

;

2 4

14

:

1

1

1<

? 3


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2;

"ample% MA0/HA/S.T

/∞;

2 4

14

:1

1

1<

? 3

nsert val$e 1&%

/ Start +y inserting /∞

1&;

2 4

14

:1

1

1<

? 3

n#rease the Dey to 1&

Call HA/C.AS/I5 on A=11> 1&

:

;

2 4

14

1&

1

1

1<

? 3

:

;

2 4

1&

14

1

1

1<

? 3

The restore! heap #ontaining

the ne*ly a!!e! element


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2?

MA0/HA/S.T

Alg: MA0/HA/S.T(A, +e0, n)

1B hea#sie[A] @ n " 1

2B A[n " 1] #← ∞

3B HA/C.AS/I5(A, n " 1, +e0)

.$nning time% O(lgn)

/∞

;

2 4

14

:

1

1

1<

? 3


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3<

S$mmary

• Ke #an perform the follo*ing operations onheaps%

MA0/HA5 O(lgn)

678/MA0/HA O(n) HA/S-.T O(nlgn)

MA0/HA/S.T O(lgn)

HA/0T.ACT/MA0 O(lgn) HA/C.AS/I5 O(lgn)

HA/MA0M7M O(1)

A3erage O(lgn)


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HeapSort 8oop nvariant

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8-- LA.AT

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Alg: 678/MA0/HA(A)

1. n length=A>2B for i ← n/2 downto 1

3B do MA0/HA5(A, i, n)


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8esson from a HeapSort

3

Ang +$hay ay parang heapsort

Merong mas ma+a+a saJyo

Meron !in namang nasa itaas mo

Mas

magaling

aDo

Hi there


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3:

"ample% A=:, 4, 3, 1, 2>

MA0/HA5(A, 1, 4) MA0/HA5(A, 1, 3) MA0/HA5(A, 1, 2)

MA0/HA5(A, 1, 1)

ero tan!aan mo, !arating !in ang panahonN

Jm on the

top of the

*orl!

D t Dt D it

Analysis of Heapsort

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