Median, order statistics

Problem Find the i-th smallest of n elements.

i=1: minimum i=n: maximum i= or i= : median

Sol: sort and index the i-th element.

Ο(nlgn): worst case.

12

n

12

n

Randomized algorithm: (9.2) Divide & Conquer

≤A[q] ≥A[q]

k

p q r

Analysis: Lucky case:

T(n) ≤ T( ) + Θ(n) = Θ(n).

By case 3 of master method (chap 4)

Unlucky case:

T(n) = T(n-1) + Θ(n) = Θ(n2).

910n

109

log 10 1.n n

Analysis:

11

TMergesort(n) = 2 • TMergesort(n/2) + cte • n + cte

Forma Fechada de Recorrência

TMS (n) = 2 • TMS(n/2) + c • n + c= 2 •[ 2 • TMS(n/4) + c • n/2 + c ] + c • n + c = 4 • TMS(n/4) + 2c • n + 3c= 4 •[ 2 • TMS(n/8) + c • n/4 + c ] + 2c • n + 3c= 8 • TMS(n/8) + 3c • n + 7c= ...

TMS (n) = 2i • TMS(n/2i) + i • c • n + (2i-1) • c

Average case:

For upper bound, assume the i-th element always falls in larger side of partition.

1

1

1 1

2 2

1max 1, 1 max ,

1 2 1 2

n

k

n n

n nk k

T n T n T k n k nn

T n T k n T k nn n

if 2

max , if 2

1 2 12 2

1 2 12 2

nk kk n k

nn k k

n n n

n nn n n n

Solve By substitution method: assume T(n) ≤ cn.

We can pick c large enough so that c(n/4+1/2) dominates the Ο(n) term.

1 1

2 2

n 11 2

1 1

2 2

2

12 1 n n 12 22 2

1 12 2

3 1 4 2

n n

n nk k

n

k k

cT n ck n k n

n n

ck k n

n

n ncn

n

c n nc n nn

nc n cn

1

2

2 n

nk

T n T k nn

Worst case linear-time order statistics (9.3) Theoretical interest Idea: generate a good partitioning

element x.

Select (i){ Divide n elements into groups of 5 elements. Find the median of each group of 5. Use Select recursively to find the median x of the

medians. Partition the n elements around x

Let k = rank of x.5. If i=k then return x. If i<k then use Select recursively to find the i-th

smallest in the 1st part else use Select recursively to find (i-k)th smallest in the last part.

}

5n

5n

Analysis: At least ½ of 5-element medians ≤ x,

which is at least medians. At least elements ≤ x. For n ≥ 50, Similarly, at least n/4 elements ≥ x. Thus after partitioning around x, step 5 is

called on ≤ 3n/4 elements.

25 10n n

3 10n

3 .10 4n n

T(n) ≤ T(n/5) + T(3n/4) + Θ(n). By substitution: T(n) ≤ cn

3

5 419

20

20

,

cn cnT n n

cnn

cncn n

cn

if c is big enough!

Application of linear-time median algorithm: i-th order statistic:

{Find median xPartition input around x If i≤ then recursively find the i-th

element in the first half.

else recursively find (i- )th element in the 2nd half. }

T(n) = T(n/2) + Θ(n) = Θ(n).

12

n

12

n

Worst-case Θ(nlgn) quicksort:

{ Find median x and partition;

Recursively sort 2 halves;

}

T(n) = 2T(n/2) + Θ(n)

= Θ(nlgn).