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Explore randomized algorithms for selection and sorting, including sampling, splitting, and efficient selection techniques. Understand the analysis of algorithms and precise bounds on randomized sampling. Discover improved sorting methods and approximations of mean values.
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Randomized Algorithms for Selection and Sorting Analysis of Algorithms Prepared by John Reif, Ph.D.
Randomized Algorithms for Selection and Sorting • Randomized Sampling • Selection by Randomized Sampling • 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 • Problems • For each key x Xrank(x,X) = |{x’ X| x’ < x}| + 1 • For each index i {1, …, N)select (i,X) = the key x X where i = rank(x,X) • Sort (X) = (x1, x2, …, xn) where xi = select (i,X)
Algorithm Samplerank s(x,X) begin Let 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)
More Precise Bounds on Randomized Sampling • Let S be a random sampling of X • Let ri = rank(select(i,S),X)
More Precise Bounds on Randomized Sampling (cont’d) • Proof We can bound ri by a Beta distribution, implying • Weak bounds follow from Chebychev inequality • The Tighter bounds follow from Chernoff Bounds
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
Subdivision by Random Sampling (cont’d) • How even are these subdivisions?
Subdivision by Random Sampling (cont’d) • Lemma 3 If random sample S in X is of size s and X is of size N, then S divides X into subsets each of
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
Subdivision by Random Sampling (cont’d) • So the probability of a random (s+1) partition having a block size v is
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 begin if 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
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 • AlgorithmFRselect(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
Choosing k1, k2 (cont’d) • Lemma 2 implies
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 else choose a random sample S of X of size s Sort S S subdividesX 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 else choose 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!
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:
Improved Expected Time Bounds • Is optimal for comparison trees!
Open Problems in Selection and Sorting • Improve Randomized Algorithms to exactly match lower bounds on number of comparisons • Can we de randomize these algorithms – i.e., give deterministic algorithms with the same bounds?
Randomized Algorithms for Selection and Sorting Analysis of Algorithms Prepared by John Reif, Ph.D.