AVERAGE CASE 66-70

8/8/2019 AVERAGE CASE 66-70

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AVERAGE CASE ANALYSIS OF

QUICK SORT

CA(n)= n+1+1/n CA(k-1)+CA(n-k)-----(1)

1eken

(n+1) is the number of element comparisons required byPARTITION on its first call

(@PARTITION (1, n+1) (n=5, say) with A(n+1)=infinitywith all elements in decreasing order stops with i6 and

p

5 and i>5 so total comparisons are 5+1 =6

for the listv = 75 65 55 45 35 + E

1 2 3 4 p5 i6


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QUICK SORT (Contd..) After the first call, one element is sorted and remaining n-1

elements are divided into two sets may be consisting of (0, n-1) or

(1, n-2) or or (n-1, 0) i.e. in general (k-1, n-k) elements 1e

ke

nclearly CA(0)=CA(1)=0

Multiplying both sides of (1) by n,

nCA (n)=n(n+1)+2(CA(0)+CA(1)+------+CA(n-1)--------(2)

(@ CA(k-1)+CA(n-k)=CA(0)+CA(n-1)+CA(1)+CA(n-2)+.+CA(n-1)+CA(0)1eken

=2(CA(0)+CA(1)+.+CA(n-1))


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QUICK SORT (Contd..)

Replacing n by n-1 in (2),

(n-1)CA(n-1)=n(n-1)+2(CA(0)+.+CA(n-2))..(3)

(2) - (3) gives

nCA(n)-(n-1)CA(n-1)=2n+2CA(n-1)

@nCA(n)=2n+(n+1)CA(n-1)(4)

CA(n)/(n+1)=CA(n-1)/n + 2/(n+1) {dividing (4) byn(n+1)


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CA(n)/ (n+1)=CA(n-1)/n +2/(n+1).(5)

Substituting n-1 instead of n in the above equation

CA(n-1)/n=CA(n-2)/ (n-1) + 2/n(6)

Substituting (6) in (5)

CA(n)/(n+1)=CA(n-2)/(n-1) +2/n +2/(n+1)

=CA(n-3)/(n-2) +2/(n-1) +2/n + 2/(n+1) .

=CA(1)/2+ 2 (1/k) =2 (1/k)

3eken+1 3eken+1

5 (CA(1) = 0)


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1/ke 2n+1 1/x dx = loge

n+1-loge2

1eken+1

@ CA(n)


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Iterative version of Quick Sort

Procedure : quick sort2(p,q)

Integer STACK (1:max), top

gobal A(1:n); local integer i

Integer p,qinteger p,q

top0

loop

while p


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Iterative version of Quick Sort(Contd..)if j-p < q-j then

stack (top+1)j+1

stack (top+2)q

qj-1

// store in stack the bands of larger file //

else

stack (top+1)p

stack (top+2)j-1pj+1

Endif

toptop+2

repeat


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Iterative version of Quick Sort(Contd..)

if top=0 then return

endif

qstack (top); pstack(top-1)toptop-2

repeat

end quicksort2


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Iterative version of Quick Sort Example

Example:

65,70,75,89,85,60,55,50,45, infinity

p = 1, q = 9, j10 partition(1,10)

j 5 j p not q - j

pj + 1 =

6< q =

941


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Iterative version of Quick Sort Example

so partition (6,10) gives value j 9

j p = 9-6 =3 not< q j = 9-9 = 0

stack(3)6

stack(4)8 8

p j+1=10 q9 6

and 6,8 are removed from stack 4

10 not < 9 so while is exited 1

(coming out of loop when top=0)

means there are no elements to

be sorted.


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The maximum stack space in iterative quick sort

The maximum stack space in iterative quick sort is

0(logn)

Proof : Let s(n) be the maximum stack space neededS(n) e 2 + s -(n-1/2) 5 2 for p,q and s -(n-1)/2is the stack space for processing the remaining

-(n-1)/2 elements

e 2+2 + S(-(n-1)/22

)e 2+2 +2+ S(- n-1/23)e 2+.......+2 (log2n times) +S (-n-1/2

log2n)


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The maximum stack space in iterative quick sort

(Contd..)

e 2log2n +S(-n-1/n , but S-n-1/n= 0

e 2log2n=2log

e

n/loge

2 5 loge

n=log2

n.loge

2

@ log2n=loge

n/loge2

@ S(n) = O(logn)

(log n=logen

and logn

2= logn

2)Quick sort is faster than merge sort for

average cases.


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SELECTION PROBLEM (Contd..)

integer n, k, m, r, j

m1; rn+1; A(n+1)infinity

loop

jr

call partition(m,j)

case

: k = j : return

:k


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SELECTION PROBLEM

(Contd..)

Procedure selection sort (A,n)

// sorts A(1)..A(n) using selection sort //

for k1 to n do

select (A,n,k)

B(k)A(k)

Repeat

end select sort


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ANALYSIS OF SELECTAssumptions

1.The n elements are distinct.

2. Each element in A(m:p) has an equal probability

of being the partitioning elements.

Partition requires 0(m-p) time

On each successive call to partitions m isincreased at least by 1 or p is decreased by at least

one.


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ANALYSIS OF SELECT (Contd..)

Hence at most n calls to Partitions may be made.

Thus the worst case complexity of SELECT is

1+2 +.+ n = n(n+1)/2 = O(n2

)(@First call to partition has n elements

2nd call to partitions has n-1 elements

:

nth call to partitions has 1 element.


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Average computing time of select (Contd..)

Average computing time of select is O(n)

Proof: Let TAk(n) be the average time to find the kth

smallest element in A (1:n), taken over all n!different permutations of n distinct elements.

Define TA(n) and R(n) as TA(n) = 1/n TAk(n)

and R(n) = max k{ TAk(n) } 1 e ken

TA(n) e R(n)

TAk(n) e cn + 1/n ( TAk-i(n-i)+ TAi(i-1))1ei


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cn is the time required by first call of partition, the partitionelement being the kth smallest with probability 1/n.

After partitioning you get kth smallest, k being any one 1..n.

Time for k-1th smallest of n-1

(k-2)n-1 smallest of (n-2)

1st smallest of (n-k+1) elements

Time for (k+1)1st smallest of k elements

(k+2)nd smallest of (k+1) elements

:

nth smallest of (n-1) elements


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Considering maximum on both sides

R(n) e cn + 1/n max { R(n-i) + R(i-1) ]---------(1)

k 1ei< k k < ie n

n-1 n-1

= cn + 1/n max { R(i) + R(i) } n u 2

k n-k+1 k

(By changing the index from n-i to i)


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We assume that c is chosen such that R(1) e

c and

we show by induction that R(n) e 4cn

Base : n=25n u2 onwards the relation (1) is valid.

R(n) e 2C + 1/2 max { R(1), R(1)}

e 2C+ 1/2 CR = (5/2) C = 2.5C < 4cn


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InductionHypothesis: Assume R(n) e 4cn for alln, 2 e n < m ------ (2)

Induction step: For n = m,

m-1 m-1

R(m) e cm + 1/m max { R(i) + R(i)} --- (3)k m-k+1 k

R(n) = max {TKA(n) } is a non-decreasing function ofn. k


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m-1 m-1

So R(i) + R(i) is maximized if k = m/2m-k+1 k

when m is even, and k = (m+1)/2 when m is odd.

Because

m-1 m-1 m-1 m-1max { R(i) + R(i)} = max { R(i) + R(i)},

k m-k+1 k (k=1) i=m i=1


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m-1 m-1 m-1 m-1 m-1 m-1

( R(i) + R(i)) ( R(i) + R(i))..., ( R(i) R(i)) }(K=2) i=m-1 i=2

(

k=m/2) i=m/2+1 i=m/2 (k=m-1) i=2 i=m-1

Thus if m is even, m-1 m-1 m-1R(m) e cm + 2/m R(i) 5 R(i) e R(i)

m/2 m/2+1 m/2

e cm + 2/m [R(m/2+R(m/2+1)++R(m-1)]

e cm + 2/m [4cm/2+4c(m/2+1)++4c(m-1)]

m-1

e cm + 8c/m i < 4cmm/2

AVERAGE CASE 66-70

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