Select

PROBLEM
The selection problem can be stated as follows: given an array A. of N elements and an integer K, 1<=K<=N, determine the Kth smallest element of A. and rearrange the array in such a way that this element is placed in A.K and all elements with subscripts lower than K have values not larger than A.K and all elements with subscripts greater than K have values not smaller than this.

ALGORITHM
Robert W. Floyd and Ronald L. Rivest have developed an improved average-time version. They say: "The results show that SELECT is at least near-optimal with respect to the number of comparisons used."

PRACTICE
Algorithms MODIFIND and SELECT are always faster than FIND; SELECT needs fewer comparisons than MODIFIND; MODIFIND needs fewer swaps than SELECT. The average time over 10 trials required by FIND, MODIFIND, and SELECT to determine the median of 10000 elements (strings) of length L<=6 (only numbers), L<=7, L<=500 was found experimentally

 

Selection problem - Comparisons of Algorithms
Algorithm L <= 6 L <= 7 L <= 500
FIND 0.957 1.475 4.104
MODIFIND 0.929 1.212 1.476
SELECT 0.602 0.828 0.985

 

IMPLEMENTATION
Unit: recursive internal function
 
Global variables: the array A. of arbitrary elements
 
Parameters: a positive integer N - number of elements in A., a positive integer K such that 1<=K<=N
 
Result: Reordering of input array such that A.K has the value it would have if A. were sorted, L<=I<=K will imply A.I<=A.K, and K<=I<=R will imply A.I>=A.K
 
Returns: A.K
 


SELECT: procedure expose A.
parse arg L, R, K
do while R > L
  if R - L > 600 then do
    N = R - L + 1; I = K - L + 1; Z = LN(N)
    S = TRUNC(0.5 * EXP(2 * Z / 3))
    SD = TRUNC(0.5 * SQRT(Z * S * (N - S)/N) *,
      SIGN(I - N/2))
    LL = MAX(L, K - TRUNC(I * S / N) + SD)
    RR = MIN(R, K + TRUNC((N - I) * S / N) + SD)
    call SELECT LL, RR, K
  end
  T = A.K; I = L; J = R
  W = A.L; A.L = A.K; A.K = W
  if A.R > T
    then do; W = A.R; A.R = A.L; A.L = W; end
  do while I < J
    W = A.I; A.I = A.J; A.J = W
    I = I + 1; J = J - 1
    do while A.I < T; I = I + 1; end
    do while A.J > T; J = J - 1; end
  end
  if A.L = T
    then do
      W = A.L; A.L = A.J; A.J = W
    end
    else do
      J = J + 1; W = A.J; A.J = A.R; A.R = W
    end
  if J <= K then L = J + 1
  if K <= J then R = J - 1
end
return A.K
 
EXP: procedure
parse arg Tr; numeric digits 3; Sr = 1; X = Tr
do R = 2 until Tr < 5E-3
  Sr = Sr + Tr; Tr = Tr * X / R
end
return Sr
 
SQRT: procedure
parse arg X; numeric digits 3
if X < 0 then return -1
if X=0 then return 0
Y = 1
do until ABS(Yp - Y) <= 5E-3
  Yp = Y; Y = (X / Yp + Yp) / 2
end
return Y
 
LN: procedure
parse arg X; numeric digits 3
M = (X + 1) / (X - 1); Ln = 1 / M
do J = 3 by 2
  T = 1 / (J * M ** J)
  if T < 5E-3 then leave
  Ln = Ln + T
end
return 2 * Ln

 

CONNECTIONS
Selection Problem
     Smallest and largest simultaneously
     Find
     Modifind
     Power - real exponent

Literature
Floyd R. W., Rivest R. L. Algorithm 489 The Algorithm SELECT - for Finding the ith Smallest of n Elements [M1]
CACM, March 1975, Vol. 18, No. 3, p. 173
Floyd R. W., Rivest R. L. Expected Time Bounds for Selection
CACM, March 1975, Vol. 18, No. 3, pp. 165-172
Zabrodsky V., FIND, SELECT, MODIFIND


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last modified 8th August 2001
Copyright 2000-2001 Vladimir Zabrodsky
Czech Republic.

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