randomized algorithms for selection and sorting prepared by john reif, ph.d. analysis of algorithms

Randomized Algorithms for Selection and Sorting

Prepared by

John Reif, Ph.D.

Analysis of Algorithms

Randomized Algorithms for Selection and Sorting

a) Randomized Samplingb) Selection by Randomized Samplingc) Sorting by Randomized Splitting:

Quicksort and Multisample Sorts

Readings

• Main Reading Selections:• CLR, Chapters 9

Comparison Problems

• Input set X of N distinct keys total ordering < over X

• Problems1) For each key x X

rank(x,X) = |{x’ X| x’ < x}| + 1

2) For each index i {1, …, N)select (i,X) = the key x X where i = rank(x,X)

3) Sort (X) = (x1, x2, …, xn)where xi = select (i,X)

Randomized Comparison Tree Model

Algorithm Samplerank s(x,X)

beginLet S be a random sample of X-{x}

of size soutput 1+ N/s [rank(x,S)-1]

end

Algorithm Samplerank s(x,X) (cont’d)

• Lemma 1– The expected value of

samplerank s(x,X) is rank (x,X)

Algorithm Samplerank s(x,X) (cont’d)

Let k = rank(x,X)

For a random y εX,

k-1 Prob (y< x) =

Nk-1

Hence E (rank(x,S)) = s 1N

Solving for k, we get

Nrank (x,X) = k = 1+ E[rank(x,S)-1] = E (samplerank(x,X))

s

proof

S is Random Sample of X of Size s

More Precise Bounds on Randomized Sampling (cont’d)

More Precise Bounds on Randomized Sampling

• Let S be a random sampling of X

• Let ri = rank(select(i,S),X)

-αi

2

N N Prob |r -i | > c log N N

s+1 s

Lemma

More Precise Bounds on Randomized Sampling (cont’d)

• ProofWe can bound ri by a Beta

distribution, implying

• Weak bounds follow from Chebychev inequality

• The Tighter bounds follow from Chernoff Bounds

i

2i 2

N (r ) = i

s+1i (s-i+1)

(r ) N(s+1) (s+2)

mean

Var

Subdivision by Random Sampling

• Let S be a random sample of X of size s

• Let k1, k2, …, ks be the elements of S in sorted order

• These elements subdivide X into

1 1

2 1 2

3 2 3

s+1

s+1 subsets x = {x ε X | x k }

X = {x ε X | k x k }

X = {x ε X | k x k }

X = {x ε X | x k}

Subdivision by Random Sampling (cont’d)

• How even are these subdivisions?

Subdivision by Random Sampling (cont’d)

• Lemma 3If random sample S in X is of size s and X is of size N,

then S divides X into subsets each of

-α(N-1) α (N) with prob 1 - N

ssize ln

Subdivision by Random Sampling (cont’d)

Proof• The number of (s+1) partitions of X is

• The number of partitions of X with one block of size v is

sN-1 (N-1)~

s s!

sN-v-1 (N-v-1)~

s s!

Subdivision by Random Sampling (cont’d)

• So the probability of a random (s+1) partition having a block size v is

-1

-1

N-v-1

s N-v-1 1 N-1~ 1 for Y=

N-1 N-1 v

s

1 (N-1) 1 ~ e = e N if v = N

s

1 since 1 e

s s

sY svsY

NY

Y

Y

lnY

Y

Randomized Algorithms for Selection

• “canonical selection algorithm”• Algorithm can select (i,X) input set X of N keys

index i {1, …, N}[0] if N=1 then output X[1] select a bracket B of X, so that

select (i,X) B with high prob. [2] Let i1 be the number of keys less

than any element of B[3] output can select (i-i1, B)

B found by random sampling

Hoar’s Selection Algorithm

• Algorithm Hselect (i,X) where 1 i N

beginif X = {x} then output x else choose a random splitter k X let B = {x X|x < k}if |B| i then output Hselect(i,B) else output Hselect(i-|B|, X-B)

end

Hoar’s Selection Algorithm (cont’d)

• Sequential time bound T(i,N) has mean

i N

j=1 j=i+1

1T (i, N) = N + T (i-j, N-j)+ T (i-j)

N

= 2 N + min (i, N-i) + o(N)

Hoar’s Selection Algorithm (cont’d)• Random splitter k X Hselect (i,X) has two

cases

Hoar’s Selection Algorithm (cont’d)

• Inefficient: each recursive call requires N comparisons, but only

reduces average problem size to ½ N

Improved Randomized Selection

• By Floyd and Rivest• Algorithm FRselect(i,X)begin

if X = {x} then output x else choose k1, k2 X such that k1< k2 let r1 = rank(k1, X), r2 = rank(k2, X)

if r1 > i then FRselect(i, {x X|x < k1})

else if r2 > i then FRselect(i-r1, {x X|k1x k2})

else FRselect(i-r2, {x X|x > k2})

end

Choosing k1, k2

• We must choose k1, k2 so that with high likelihood,

k1 select(i,X) k2

Choosing k1, k2 (cont’d)

• Choose random sample S X size s• Define

1

2

(s+1)k = select i ,

(N+1)

(s+1)k = select i ,

(N+1)

where = d s log N , d = constant

S

S

Choosing k1, k2 (cont’d)

• Lemma 2 implies

-α1

-α2

1 1

2 2

P rob (r > i) < N

and

P rob (r > i) < N

where r = rank (k , X)

r = rank (k , X)

Expected Time Bound

1 1

2 1 2

1 2 1 2 1

-

T(i,N) N + 2T(-,s)

+ Prob (r > i) T(i, r )

+ Prob (i > r ) T(i - r , N - r )

+ Prob (r i r ) T(i - r , r - r )

N+1N+2T(-,s)+2N N+T i, 2

s+1

N + min (i, N-i

2

3

) + o(N)

3 if we set and s = N log N = o(N)

Small Cost of Recursions

• Note• With prob 1 - 2N- each recursive

call costs only O(s) = o(N) rather than N in previous algorithm

Randomized Sorting Algorithms

• “canonical sorting algorithm”• Algorithm cansort(X)begin

if x={x} then output X elsechoose a random sample S of X of size sSort SS subdivides X into s+1 subsets

X1, X2, …, Xs+1

output cansort (X1) cansort (X2)… cansort(Xs+1)

end

Using Random Sampling to Aid Sorting

• Problem:– Must subdivide X into subsets of nearly

equal size to minimize number of comparisons

– Solution: random sampling!

Hoar’s Randomized Sorting Algorithm

• Uses sample size s=1• Algorithm quicksort(X)begin

if |X|=1 then output X elsechoose a random splitter k X

output quicksort ({x X|x<k})· (k)· quicksort ({x X|x>k})

end

Expected Time Cost of Hoar’s Sort

• Inefficient:• Better to divide problem size by ½

with high likelihood!

1

1T(N) N-1 + (T(i-1) T(N-i))

N

2 N log N

N

i

Improved Sort using

• Better choice of splitter is k = sample selects (N/2, N)

• Algorithm samplesorts (X)begin

if |X|=1 then output X choose a random subset S of X size

s = N/log N k select (S/2,S) cost time o(N)

output samplesorts ({x X|x<k})· (k) ·

samplesorts ({x X|x>k})end

Randomized Approximant of Mean

• By Lemma 2, rank(k,X) is very nearly the mean:

-αNProb(| rank(k,X) - | d α N log N) < N

2

Improved Expected Time Bounds

• Is optimal for comparison trees!

-1T(N) 2T(N ) + N T(N) N + o(N) +N - 1

log (N!)

Open Problems in Selection and Sorting

1) Improve Randomized Algorithms to exactly match lower bounds on number of comparisons

2) Can we de randomize these algorithms – i.e., give deterministic algorithms with the same bounds?

Randomized Algorithms for Selection and Sorting

Prepared by

John Reif, Ph.D.

Analysis of Algorithms